Mathspace Year 9 WA Curriculum Flipbook by Mathspace

Year 9 Western Australia Curriculum

© 2023 Mathspace Pty Ltd Copyright Notice This Work is copyright. All rights are reserved. Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher. For permission to use material from this text or product, please email hello@mathspace.com.au For our full digital offering, visit mathspace.co Mathspace Pty Ltd Level 11, 477 Pitt St Sydney 2000 School Curriculum and Standards Authority. Mathematics Scope and Sequence of the mandated curriculum content: Pre-primary – Year 10 | Revised Curriculum for familiarisation in 2025. The School Curriculum and Standards Authority does not endorse this publication or product. Title: Mathspace WA Curriculum - Year 9: 2026 ISBN: 978-1-963022-79-7 Editors and lead authors: Erin Gallagher, Ronnel Jeppsen, Julia Reid, Sarah Butterworth, Anjelica Sieger, Jennifer George, Stephen George, Dot Yung Further reviewers: Melissa Archey, Warren Bardsley, Lianna Beeching, Kathryn Bennett, Benjamin Bui, Eric Chan, Cory Duker, Bahija Elmir, Wayne Gillie, Jeremy Hunter, Geoff Jamieson, Shahad Odeh, Natasha Pollard, Helen Soar, Patricia Toutzaris, Paul Unkles, Jie Yang Writing and development team: Adam Humphries, Neil Lopez, Ma. Lourdes Ona, Bernadette Barillo, Jelly Candido, Sheramie Castillo, Dhave Fernandez, Abegail Fontanilla, Regine Jara, Albert Jara, Shaira Llanita, Lesley Lopez, Kerby Mendoza, Mikaela Nicolas, Rolly Apdo, Valerie Baja, Marielle Belen, Michaela Belen, Christine Bibon, Allen Calabia, Adrian Doctolero, Dharent Fernandez, Janine Francisco, Rose Francisco, Rona Josue, Crystel Lontoc, Mae Lucid, Grace Manzano, Julie Manzano, Monisa Tacud, Paul Platero Images and Design team: Alex Duckmanton, Chastine Marquez, Keith Gimeno

Contents 1

Real numbers 1.01 1.02

Rational and irrational numbers Operations with integers Investigation: Construct irrational numbers Chapter 1 review

4 12 18

Indices 20 2.01 2.02 2.03 2.04 2.05

Multiplication law Division law Power of a power law Zero and negative indices Scientific notation Chapter 2 review

22 26 31 36 44 54

Algebra 56 3.01 3.02 3.03 3.04 3.05 3.06 3.07

Algebra review 58 Distributive law 67 Binomial expansion 71 Algebraic factors 76 Factorise by grouping 82 Optional: Add and subtract fractions with numeric denominators 88 Optional: Multiply and divide algebraic fractions with numeric denominators 93 Investigation: Calendar expressions Chapter 3 review 99

Contents mathspace.co

iii

Linear relationships 4.01 4.02 4.03 4.04 4.05 4.06 4.07 4.08 4.09 4.10 4.11 4.12

140 150 161 167 175 180 188 195 201 209

Quadratic relationships Graphs and features of quadratics Basic quadratics Solve algebraically Optional: Null factor law Chapter 5 review

216 225 236 244 249 255

Finance 258 6.01 6.02 6.03 6.04

104 112 128

Quadratics 214 5.01 5.02 5.03 5.04 5.05

Midpoint Gradient Distance between two points Investigation: Transformations of linear graphs Gradient-intercept form Linear graphs Equation of a straight line Direct proportion Optional: Indirect proportion Model linear equations Multi-step equations Optional: Equations with algebraic fractions Optional: Rearrange formulae Chapter 4 review

102

Investigation: Finances Cost, profit and revenue Income Simple interest Modelling investigation: Drone costs Re-investing with simple interest Investigation: Determining a suitable phone plan (Optional) Chapter 6 review

Mathspace WA Curriculum – Year 9 mathspace.co

260 270 282 291 293 306

Geometry 310 7.01 7.02 7.03 7.04

312 321 332 342

354

Trigonometry 360 8.01 8.02 8.03 8.04 8.05 8.06 8.07 8.08

Similar right-angled triangles Scale factors Transformations and similarity Investigation: Determine congruence or similarity Optional: Congruent triangles Investigation: Guessing distance Investigation: Scale diagrams Investigation: Scale models Chapter 7 review

Pythagoras’ theorem Investigation: Ratios in triangles Trigonometric ratios Side lengths Angles Applications of trigonometry Optional: Pythagoras in 3D Optional: Trigonometry in 3D Optional: The unit circle Chapter 8 review

362 372 385 396 406 413 421 429 448

Measurement 454 9.01 9.02 9.03 9.04 9.05 9.06

Area Composite shapes Surface area of prisms Volume of prisms Surface area of cylinders Volume of cylinders Investigation: Transform 2D shapes to make 3D objects Chapter 9 review

456 464 479 488 497 506 515

Contents mathspace.co

Probability 520 10.01 10.02 10.03 10.04 10.05 10.06

522 532 549 560 573 590 592 608

Statistics 612 11.01 11.02 11.03 11.04 11.05 11.06 11.07

Experimental probability Venn diagrams and two-way tables Arrays and simple tree diagrams With and without replacement Multi-stage probability Modelling investigation: Fish sampling Optional: Independent and dependent events Investigation: Simulation of compound events Chapter 10 review

Types of data Data displays Investigation: Opinion Polls Back-to-back stem and leaf plots Grouped histograms Optional: Frequency histograms and polygons Shape of data Investigation: Statistical reports in the media Compare data sets Investigation: Plan and conduct statistical investigations Chapter 11 review

614 621

Answers

704

Mathspace WA Curriculum – Year 9 mathspace.co

636 644 658 670 681 696

“Mathematics is the art of giving the same name to different things.” Henri Poincaré

Big ideas Using the correct order of operations on real numbers (rational and irrational numbers) ensures accurate calculations and meaningful estimates.

Curriculum Outcomes • Use flexible and efficient strategies for calculations involving the four operations with real numbers and express solutions in exact form or as an approximation • Investigate, define, compare and order real numbers, with equality and inequality symbols, including those expressed in scientific notation • (Optional) Explore to develop a sequence of steps to flexibly move between recurring decimals and fractions

1 Real numbers Chapter outline 1.01 1.02

Rational and irrational numbers Operations with integers Investigation: Construct irrational numbers Chapter 1 review

4 12 18

The world record for reciting π by a person was over 70 000 decimal places! How many can you memorise?

1.01 Rational and irrational numbers After this lesson, you will be able to… • identify and classify numbers as rational or irrational. • understand what surds are and how they are different from other irrational numbers. • estimate the value of irrational numbers using methods involving square numbers.

Rational and irrational numbers A rational number is a number that can be written as a fraction where both the numerator and denominator are integers. Rational numbers consist of: • Integers (positive and negative whole numbers, including zero) • Terminating decimals (decimals with a limited number of digits) • Recurring decimals (decimals with an inﬁnite number of digits, which have a repeating pattern) For example:

An irrational number cannot be expressed as a fraction where both the numerator and the denominator are integers. When expressed as a decimal, irrational numbers have inﬁnite digits, with no repeating pattern. For example:

π = 3.141 592 653 589 793… = 1.414 213 562 373 095… We often come across irrational numbers when working with square roots or other roots. Irrational numbers can take the form of: • The square root of a number that is not a perfect square, such as or • The cube root of a number that is not a perfect cube, such as or , and so on. Rational and irrational numbers together make up the set of numbers known as real numbers, as shown in the diagram. Real Numbers Rational numbers Integers 4

Whole numbers 1 0 1 2 3 4

Irrational numbers

Mathspace WA Curriculum – Year 9 mathspace.co

Example 1 Is each number rational or irrational? a

Create a strategy Can the number be expressed as a fraction where both the numerator and denominator are integers? If so, it is rational. Otherwise, it is irrational.

Apply the idea is a fraction that cannot be simpliﬁed. The denominator is an integer, but the numerator is not. This is an irrational number.

Create a strategy Is the number underneath the square root a perfect square? If so, it is rational. Otherwise, it is irrational.

Apply the idea 13 is not a perfect square, so

is irrational.

Reﬂect and check We can explore square root numbers using a calculator. Evaluate the and investigate the decimal representation, 3.605 551 27. We can be fairly certain that it isn’t rational because there appear to be no repeating patterns in the decimal digits.

Create a strategy This number is a decimal, so if it is a repeating or terminating decimal, then it is rational. If it is non-repeating and non-terminating, it is irrational.

Apply the idea is a recurring decimal, so it is a rational number.

Reﬂect and check If we evaluate

on the calculator, we get

. This shows that the decimal can be written as a

fraction, where both the numerator and denominator are integers, meaning it is rational.

1.01 Rational and irrational numbers mathspace.co

Create a strategy Evaluate the number and examine it to see if it is rational.

Apply the idea If we evaluate

on the calculator, we get 55.901 699 437 5 …. This is a non-terminating and

non-repeating decimal. is an irrational number.

Reflect and check If we used the definition of the power of 5, we can see if this can be simplified to a rational number or not.

Write in expanded form

Group pairs

Use that

Evaluate and simplify

Since 5 is not a perfect square, we know that

is irrational. So

is irrational.

Idea summary A rational number is a number that can be written as a fraction where both the numerator and denominator are integers. An irrational number is a number that cannot be written as a fraction of two integers. A surd is a type of irrational number. It is the square root of a number that is not a perfect square (or the cube root of a number that is not a perfect cube, etc.).

Estimate irrational numbers We can place irrational numbers on the number line if we know their (approximate) decimal equivalents. For example, π ≈ 3.14, so we can place π on the number line between 3.1 and 3.2. We can place surds approximately on a number line using perfect squares on either side of the number we are taking the square root of. For example, let’s look at

. The perfect squares on either side of 90 are 81 and 100. 81 < 90 < 100

Mathspace WA Curriculum – Year 9 mathspace.co

This means that and so So we can place

on the number line between 9 and 10.

In general, if a<b<c then

Example 2 Represent these values on the number line:

π2,

, −1.21,

Create a strategy Estimate the values of the numbers using decimals and plot the values on the number line.

Apply the idea Note that π is estimated as 3.14 π2 ≈ (3.14)2

Square 3.14

≈ 9.9

Evaluate .

Find the value of

Estimate the value of . than

The value of

by comparing to the closest square numbers that are bigger and smaller

is closer to

than

Plotting the values on the number line, we have: −2

−1

The numbers in order from smallest to largest are:

Reﬂect and check Comparing the numbers, we can see that:

1.01 Rational and irrational numbers mathspace.co

Idea summary We can estimate the values of surds by ﬁnding square numbers on either side of the number we are taking the square root of. If we want to estimate , we can ﬁnd square numbers a and c so that a<b<c then

1.01 Practice questions Understanding 1

Adventurer 1–4

Trailblazer 1–4

Match each number with its number type. a

Fraction

−7

Terminating decimal

4.65

Integer

iii

Surd

Which of these are integers?

Which of these are irrational?

Classify these as either rational or irrational: a

Explorer 1–4

−4

−0.7

Mathspace WA Curriculum – Year 9 mathspace.co

4.435

π2

5, 6, 7a–c, 8a–d, 9ab, 10ac, 11, 12, 14a–d, 15–17

Fluency 5

and

5, 6, 7a–d, 8a–d, 9cd, 10bc, 11, 12, 14c–f, 15–17

, we can say that

and

5, 7, 8e–h, 9cd, 10cd, 11, 13, 14e–h, 16–18

are numbers between 2 and 3.

Are these statements true or false? a

, therefore

and

, therefore

and

, therefore

and and

e 6

, therefore

Simplify each of these and classify them as either rational or irrational: b

, where n is odd

a e 9

, where n is even

−3

−2

−1

−2.2

0.5

Plot these sets of numbers on a number line: a

Plot these numbers on the number line: −4

Ex 2

Classify these as either rational or irrational:

g 8

are between 6 and 7.

are irrational numbers as they are not perfect squares.

a Ex 1

and

Which integer is each number closest to? a

1.01 Rational and irrational numbers mathspace.co

Select which surd is: a

Between 3 and 4 A

For a

: Complete the inequality with two consecutive perfect square numbers that 59 lies between: ⬚ < 59 < ⬚

Complete the inequality with two consecutive integers that

For a

Between 10 and 11 A

Between 8 and 9 A

Between 6 and 7 A

<⬚

⬚<

lies between:

Complete the inequality with two consecutive perfect square numbers that 22 lies between: ⬚ < 22 < ⬚ Complete the inequality with two consecutive integers that

lies between:

<⬚

⬚<

What two consecutive integers will complete these inequalities? a e

<⬚

⬚<

<⬚

⬚<

b f

⬚<

<⬚

c g

⬚<

<⬚

⬚<

<⬚

Which integer lies between

For each pair of numbers, select the number with the larger value: a

or 5

4π or 12 −1

6.2 × 10 or

b f

π2 or 144

or 11

c 0

or 2.5 × 10

or 4.8 × 10

Order the numbers in each part from smallest to largest. a

<⬚

⬚<

<⬚

or 3

or 8

1.3 × 102 or π3

For each pair of numbers, select the number with the smaller value: a

and

⬚<

, 3.7 × 101

Mathspace WA Curriculum – Year 9 mathspace.co

1.5 × 10−1,

Reasoning

21–23

19, 20, 22

Is it possible to ﬁnd two irrational numbers that multiply together to give a rational number? If so, give an example. If not, explain why not.

Is it possible to ﬁnd two irrational numbers that add together to give a rational number? If so, give an example. If not, explain why not.

A rectangle has length

and width

Is the perimeter of the rectangle rational or irrational?

Is the area of the rectangle rational or irrational?

When performing calculations involving π, there are a few common ways to do these, such as: • Using the π button on the calculator • Using 3.14 • Using Order the options from most to least accurate. Justify your answer.

Naomi was asked to estimate . She gave the answer 5.5, and her reason was that she knew the answer would be between 5 and 6. a

Explain how she knew the answer was between 5 and 6.

Is her estimate reasonable? Explain.

Problem-solving 24ab, 26 24

24bc, 25, 26

24–27

Pythagoras’ theorem says that c2 = a2 + b2, where c is the length of the hypotenuse and a and b are the lengths of the two shorter sides. Using Pythagoras’ theorem with the given triangle, answer these questions: a

Calculate the exact value of c if a = 12 and b = 5.

Calculate the exact value of b if c =

Calculate the exact value of c if a = 1 and b = 2.

Calculate the exact value of a if c = 7 and b = 6.

and a =

b cm

c cm

a cm

ISO paper (for example, A4 paper) always has a length that is times longer than the shorter side. How many times longer is the diagonal of a piece of ISO paper than its shorter side? Fiona has a sheet of card that is

long and

wide.

Fiona estimates the area of the card by using the closest integer value to each square root. What answer did she get?

Fiona then calculates the area using her calculator. What answer did she get, rounded to four decimal places?

What was the percentage error in her estimate, rounded to two decimal places?

1.01 Rational and irrational numbers mathspace.co

The spiral of Theodorus can be drawn to estimate the values of surds. The ﬁrst right triangle in the spiral has short sides of length 1. The length of the hypotenuse of the ﬁrst triangle is

The hypotenuse of the ﬁrst right triangle is then used as one of the short sides for the next right triangle (the other short side has a length of 1). The second right triangle will have a hypotenuse of length

, and so on. 1

Draw the ﬁrst six triangles of the spiral, starting with a triangle with short sides of 1 cm. Use a protractor to make sure your right angles are accurate. a

Measure the hypotenuse of the sixth right triangle to estimate the value of to one decimal place. Why might your answer to part (a) be different from the actual value of , rounded to one decimal place?

1 4

2 1

1 17

8 15

9 10

14 13

How might we draw the spiral differently to improve the accuracy of our estimate for ?

Try drawing the spiral as discussed in part (c) to improve your estimate of

1.02 Operations with integers After this lesson, you will be able to… • understand and apply the order of operations in evaluating expressions involving integers. • interpret and solve problems involving mixed operations with integers. • extend the understanding of the order of operations to fractions. • apply the rules of operations to solve problems involving fractions.

Mixed operations with integers We know how to follow the order of operations when evaluating expressions involving more than one operation. The order of operations is as follows: 1. Evaluate whatever is contained in brackets. 2. Evaluate any multiplication or division, reading from left to right. 3. Evaluate any addition or subtraction, reading from left to right. If we have negative integers in our expression, we still follow the exact same rules; we need to be careful when evaluating the terms involving negatives. 12

Mathspace WA Curriculum – Year 9 mathspace.co

Example 1 Evaluate: a 5 + 9 × (−6)

Create a strategy We ﬁrst need to evaluate the multiplication and then the addition.

Apply the idea 5 + 9 × (−6) = 5 + (−54) = −49

Evaluate the multiplication Evaluate

Reﬂect and check We can check with a calculator. Evaluating 5 + 9 × (−6) = −49 Since scientiﬁc calculators apply the order of operations, this is a good indication that we have applied the order of operations correctly.

b (7 − 8) × (−5)

Create a strategy We ﬁrst need to evaluate the brackets and then the multiplication.

Apply the idea (7 − 8) × (−5) = (−1) × (−5)

Evaluate 7 − 8

= 5 Evaluate

Example 2 Evaluate 21 ÷ (−7) × 2

Create a strategy Evaluate the multiplication and division from left to right.

Apply the idea 21 ÷ (−7) × 2 = −3 × 2 = −6

Evaluate 21 ÷ (−7) Evaluate

1.02 Operations with integers mathspace.co

Idea summary The order of operations is as follows: 1. Evaluate anything inside the brackets. 2. Evaluate any multiplication or division, reading from left to right. 3. Evaluate any addition or subtraction, reading from left to right.

Mixed operations with fractions We can apply the rules for mixed operations to fractions.

Example 3 Evaluate: a

Create a strategy The multiplication needs to be done before the addition.

Apply the idea

Simplify by removing common factors

Perform the multiplication

Use a common denominator for the addition

Perform the addition

Create a strategy The indices will need to be done ﬁrst. To do this, we need to express We can then do the addition and subtraction.

Mathspace WA Curriculum – Year 9 mathspace.co

as an improper fraction.

Apply the idea

Write

as an improper fraction

Evaluate the index:

Express 4 as

Evaluate the ﬁrst two terms

Use a common denominator for the remaining terms

Evaluate

We may write the answer as a mixed fraction.

so we can subtract the ﬁrst two terms

Reﬂect and check We can check with a calculator. Evaluating Since scientiﬁc calculators apply the order of operations, this is a good indication that we have applied the order of operations correctly.

Idea summary The order of operations applies when doing arithmetic with fractions.

1.02 Practice questions Understanding 1

Explorer 1–4

Adventurer 1–4

Trailblazer 1–4

Are these statements true or false? a

When evaluating an expression with mixed operations, you should always perform addition and subtraction before multiplication and division.

If an expression contains only multiplication and division operations, you should evaluate them from left to right.

Surds can be added in any order but not subtracted.

Two irrational numbers can be multiplied in any order and also divided in any order.

1.02 Operations with integers mathspace.co

Which of these expressions is equivalent to 7 − 4 + (−3)? A

−10 + 4

Ex 2

2×

−(5 − 2) + 4

−3

5a–d, 6a–d, 7a–d, 8a–d, 9ab

−7 + 8 − 2

− (−3)? C

2×9−3

5b–f, 6c–f, 7b–f, 8b–f, 9ac

5d–h, 6e–h, 7d–h, 8d–h, 9cd

−10 − 4 + 6

15 − (−11) − 5

(−24 + 18) − 13

−16 − (2 − (−7))

−9 − (−12) − (−3)

1 − (11 + (−14))

(34 − (−52)) − 27

Evaluate: a

−10 + 8 × (−3)

21 − 6 × (−7)

(−9 − 12) × (−2)

−11 × (−3) + 20

18 × (13 + (−15))

−30 × (−1 − 3)

(−4 × 14) − 4 × 3 − 17

−25 × (−3) + 2 × (−17 + 23)

Evaluate: a

−18 + 21 ÷ 7

72 ÷ 6 ÷ 2

(120 ÷ 5) × 3

54 ÷ 3 ÷ (−2)

6 − 25 ÷ 5 + 4

(−36) ÷ (−4) − 14

(−390) ÷ (−19 + 6)

(−100) ÷ (−4) ÷ (−5)

(−14) ÷ 2 − 10 ÷ (−2)

Evaluate: −7 + 8 × 6 ÷ 2

2 × (−25 − 20) ÷ (−7 + 2)

20 + (−14) − (−2) − 18

9 × (−6) + 28 ÷ (−7)

(28 × 5) ÷ (14 − 24) + 5

[7 × (6 − 3)] × [(8 + 2) ÷ 5]

[1 + (6 × 3 − 9)] ÷ [20 − (100 ÷ 10)]

Evaluate: b

Reasoning 10

4 − 10

Ex 3

7+4−3

Evaluate: a

2×3+3

Fluency

Ex 1

−7 − 4 − 3

Which of these expressions is equivalent to A

Which of these is not equivalent to −(5 × 2) + 4? A

7 + (−4) + (−3)

10, 11

c 11–13

d 12–14

Kaylah and Orion are evaluating the expression 10 ÷ 2 × 5. Here is their work: Orion: Kaylah: 10 ÷ 2 × 5 = 10 ÷ 10 =1 Who has the correct solution? Explain your answer.

Mathspace WA Curriculum – Year 9 mathspace.co

10 ÷ 2 × 5 = 5 × 5 = 25

An incorrect series of steps for solving an expression with mixed operations and integers is shown below. Identify the error and correct it. Expression: (−2) + 12 × 3 − (−4) (−2) + 12 × 3 − (−4) = 10 × 3 − (−4) = 30 − (−4) = 30 + 4 = 34

Determine the accuracy of this statement and provide the correct result if it is incorrect: 12 + (−63) ÷ (3) = −17

Do the expressions (−7 + 5) × (−3) and − 7 × (−3) + 5 × (−3) yield the same result? Explain why.

A square enclosure has an area of 20 m2. Explain why the side lengths cannot be rational.

Problem-solving 15–17 15

15–17

16–18

Tara is waiting for the next ﬂight to Los Angeles, which was scheduled to be in 36 minutes, but there is a 48-minute delay. She takes a nap and wakes up 24 minutes later. Calculate how much longer Tara has to wait before the plane departs.

A bus departs the depot with 8 passengers on board. At the ﬁrst stop, 4 people got off and 6 people got on. At the second stop, a further 3 people got off, and 7 more got on. Calculate the number of people on the bus when it leaves the second stop.

Luigi enters an elevator at the 7th floor (the ground floor being floor 0). The elevator goes down 3 floors, then up 9 floors and finally it goes down 2 floors, where Luigi gets out.

Which ﬂoor does Luigi get out on?

When Luigi gets off the elevator, how many ﬂoors from his starting point is he?

At the start of the week, Giovanni had $250 in his bank account. On Monday, he withdrew $39. On Wednesday, he deposited $15, and on Thursday, he deposited a further $45. Finally, he withdrew $115 on Friday. Find Giovanni’s ﬁnal bank balance at the end of the week.

Investigation: Construct irrational numbers

Investigate online

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1.02 Operations with integers mathspace.co

1 Chapter review 1

Classify these as either rational or irrational: a

−0.175

e 2

Which integer is each number closest to? a

What two consecutive integers will complete these inequalities? a

Between 7 and 8 A

Between 4 and 5 A

Select which surd is: a

Plot these sets of numbers on a number line: a

π+4

A square has side lengths of

c units.

Is the perimeter of the square rational or irrational?

Is the area of the square rational or irrational?

Abeny has a rectangular pool that is

long by

wide.

Abeny estimates the area of the pool using the closest integer value to each square root. What is her answer?

Abeny then calculates the area using her calculator. What answer did she get, rounded to four decimal places?

Mathspace WA Curriculum – Year 9 mathspace.co

Evaluate: a

−5 − 4 + 7

13 − (−3) − 7

−1 − (14 + (−3))

(10 − 20) + 30

Evaluate: a

−7 + 16 × (−1)

20 − 5 × (−9)

18 × (13 + (−15))

−20 × (−4) + 2 × (−6 + 6)

Evaluate: a

−8 + 30 ÷ 3

120 ÷ 4 ÷ (−5)

(−36) ÷ (−9) + 3

5 − 12 ÷ 6 + 14

Evaluate: a

−7 + 5 × 15 ÷ 3

11 × 9 ÷ (1 − 2) + 3

6 × (−4) + 35 ÷ (−7)

[5 + (10 × 15 − 15)] ÷ [10 − (30 ÷ 10)]

Evaluate: a

An incorrect series of steps for solving an expression with mixed operations and integers is shown. Identify the error and correct it. Expression: 15 − (−7) × 2 + (−9)

15 − (−7) × 2 + (−9) = 15 + 7 × 2 − 9

= 22 × 2 − 9

= 44 − 9

= 35

Braxton has a debt of $500. He repays $200, then borrows another $300. What is his new debt?

Doreen invested $15 000 in stocks. In the ﬁrst week, her investment tripled. In the second week, she gained $5 000. In the third week, she lost half of the total amount. What is the value of her investment after the third week?

Chapter 1 review mathspace.co

Big ideas Applying index laws simplifies expressions and makes it easier to work with very large or very small numbers.

Curriculum Outcomes • Extend and apply index laws with positive-integer indices and the zero index, to variable bases and simplify where appropriate • Extend and apply index laws with numerical expressions of base 10 to include negative-integer indices. Develop the relationship between these negative indices and equivalent fractions and decimals • Investigate very large and very small numbers and move flexibly between their exact and approximated scientific notation • Investigate, define, compare and order real numbers, with equality and inequality symbols, including those expressed in scientific notation

2 Indices Chapter outline 2.01 2.02 2.03 2.04 2.05

Multiplication law Division law Power of a power law Zero and negative indices Scientific notation Chapter 2 review

22 26 31 36 44 54

A grain of sand is about 10–3 metres across, proving that even the tiniest things can be understood through indices!

2.01 Multiplication law After this lesson, you will be able to… • use the multiplication law to simplify expressions with both numeric and algebraic bases.

Multiply terms with indices Interactive exploration

Discover this concept in action online

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Consider the product of a3 × a5. Notice that the terms share the same base, a. We can see that there are eight a’s being multiplied together. a 3 × a5 = a × a × a × a × a × a × a × a We can avoid having to write each expression in expanded form by using the multiplication law.

a m × an = a m + n where a, m and n are integers and a ≠ 0. In this example, the a is called the base, and the m and n are both called indices, powers or exponents. So, when we multiply terms with the same base: • The base remains the same. • We add the indices.

Example 1 Simplify: a 35 × 39

Create a strategy Use the multiplication law to simplify the expression.

Apply the idea 35 × 39 = 35 + 9 =3

Apply the multiplication law Simplify

b c7 × c6

Create a strategy Use the multiplication law to simplify the expression.

Mathspace WA Curriculum – Year 9 mathspace.co

Apply the idea c7 × c6 = c7 + 6 =c

Apply the multiplication law Simplify

c 5d5 × 3d3

Create a strategy Multiply the coeﬃcients, then use the multiplication law to simplify the parts of the terms with indices.

Apply the idea 5d5 × 3d3 = 5 × d5 × 3 × d3 5

=5×3×d ×d 5+3

= 15 × d

Expand Group the numeric values and algebraic components Multiply the numeric values and apply the multiplication law

= 15d

Simplify

d −3m2 n5 × 7m3 n

Create a strategy Multiply the coeﬃcients, then use the multiplication law to simplify the parts of the terms with indices. Work with the different bases in turn.

Apply the idea −3m2 n5 × 7m3 n = −3 × m2 × n5 × 7 × m3 × n 2

Expand

= −3 × 7 × m × m × n × n

roup the numeric values and algebraic G components

= −21 × m2 + 3 × n5 + 1

ultiply the numeric values and apply M the multiplication law

= −21m5 n6

Simplify

Create a strategy When a fraction is raised to a power, it means both the numerator and denominator are raised to the same power. So

Apply the idea

Apply the power of 3 to both numerator and denominator Evaluate 33 Group the numeric values and algebraic components Apply the multiplication law Simplify

2.01 Multiplication law mathspace.co

Idea summary When multiplying terms with the same base, add the indices: am × an = am + n If the terms also have coefficients, then multiply the coefficient first, keep the base the same and add the indices: bam × can = (b × c) am + n When a fraction is raised to a power, both the numerator and the denominator are raised to the same power:  a  b 

am bm

2.01 Practice questions Explorer

Understanding 1

a b c

Adventurer

1–3

Trailblazer 2, 3

1–3

Complete the expanded form: 35 = ⬚ × ⬚ × ⬚ × ⬚ × ⬚

Complete the expanded form: 34 = ⬚ × ⬚ × ⬚ × ⬚ Using (a) and (b), complete the expanded form:

35 × 34 = (⬚ × ⬚ × ⬚ × ⬚ × ⬚) × (⬚ × ⬚ × ⬚ × ⬚)

= 3⬚

Complete the statement: When you multiply terms that have the same bases you ⬚ the indices.

Write 53 in expanded form.

Write 2 × 2 × 2 × 2 in index form.

Write a5 in expanded form.

Write b × b × b × b × b × b in index form.

Trent stated that 22 + 32 is equal to 52. a

Use expanded form for each term to explain why Trent is incorrect.

Can 22 + 32 be simplified using the multiplication index law? 4a–i, 5a–d, 6a–c, 7ab, 8a–d

Fluency Ex 1a

4b–j, 5b–h, 6b–f, 7a–d, 8c–e

4c–k, 5e–l, 6c–h, 7c–h, 8d–f

Simplify: a

23 × 24

32 × 32

(−2) × (−2)

35 × 33

2 × 25

210 × 25

34 × 35

2 ×2 ×2

Mathspace WA Curriculum – Year 9 mathspace.co

3 ×3×3

54 × 5

73 × 77

Ex 1b

Simplify: a

Ex 1c

Ex 1d

x6 × x3 5

a 7 × a5 3

a4 × a

b ×b

y ×y

u ×u

b6 × b7

a6 × a − 3

x − 5 × x8

d7 × d −4

e −10 × e8

y3 × 3y

2x5 × 2x4

3a4 × 2a6

5y2 × 3y2

Simplify: a

5y2 × y3

−b × −b

7a × 3a

9u × 8u

5c5 d4 × 4c2 d3

Simplify: a

Ex 1e

y3 × y2

3x2 y2 × 2xy3 3 4

7a b × 2ab

−2u4 v2 × 3u6 v9

−3t6 u3 × −4t5 u3

5a3 b2 c4 × −5a2 b4 c6

5w5 x3 y9 × −3wx5 y

3abc × −3a2 b3 c4 × −3a2 bc3

Simplify: a

Reasoning

9, 10

10, 11

Explain why the expression a4 × b4 cannot be simpliﬁed using the multiplication law.

Marvin is asked to solve the expression (b2) × (b3) × (b7). He claims the result is b12. Is Marvin’s answer correct? Justify your answer.

Are (−2)3 and −23 equal? Explain.

Are (−2)4 and −24 equal? Explain.

Generalise a rule for when a negative is raised to an odd or even power.

Problem-solving 12, 13 12

12d–f, 13, 14

Fill in the missing number: a c e

12, 13

a2 × a⬚ = a11 5

⬚ × ⬚ = b

3mc⬚ × ⬚bm⬚ × 2b⬚c3 = −6b4 c5 m7

b d f

c⬚ × c7 = c8

2a2 b⬚ × ⬚a⬚b = 6a4 b3

a⬚b2 c4 × ⬚a⬚b⬚c⬚ =

1 2 3 6 a b c 4

A rectangle has a length a2 cm and width a3 cm. Write an expression for the area of the rectangle.

2.01 Multiplication law mathspace.co

If 2x × 2y = 210, x ≥ y. a

How many positive integer values of x and y are possible for this equation?

Why can we not list the possible solutions if we allow negative integers?

2.02 Division law After this lesson, you will be able to… • apply the division law to simplify expressions involving terms with indices.

Division law Interactive exploration

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Consider the division expression a3 ÷ a5. Notice that the terms share the same base. Writing out the terms in expanded form, it can be seen that we can remove three a‘s as a common factor.

We can avoid having to write each expression in expanded form by using the division law.

a m ÷ an = a m − n where a, m and n are integers and a ≠ 0. When dividing terms with a common base: • The base remains the same. • Subtract the indices. We can also write the division law as:

Example 1 Simplify: a 29 ÷ 24

Create a strategy Use the division law to simplify the expression.

Mathspace WA Curriculum – Year 9 mathspace.co

Apply the idea 29 ÷ 24 = 29 − 4 =2

Apply the division law Simplify

Reﬂect and check If asked for a basic numeral, we could evaluate 25 = 32.

Create a strategy Use the division law to simplify the expression.

Apply the idea Apply the division law Simplify

Create a strategy Handle the numeric and algebraic components separately. Simplify the numeric component, then use the division law to simplify the algebraic component.

Apply the idea Simplify the numeric component, 15 ÷ 5 = 3

Apply the division law

Simplify

Create a strategy Use the multiplication law to simplify the numerator. Then use the division law to simplify the expression.

2.02 Division law mathspace.co

Apply the idea Apply the multiplication law

Simplify

Apply the division law

Simplify

Reﬂect and check We can also use the division law ﬁrst and then the multiplication law. Use the division law

Evaluate the subtraction

Use the multiplication law

Evaluate the addition

Create a strategy Use the division law twice, once for each base.

Apply the idea Apply the division law for base x

Evaluate the subtraction

Apply the division law for base y

Evaluate the subtraction

Idea summary When dividing terms with the same base, subtract the indices:

If the terms also have coefficients, then divide the coefficient first, keep the base the same and subtract the indices:

Mathspace WA Curriculum – Year 9 mathspace.co

2.02 Practice questions Understanding 1

Complete the expression:

Is each statement true or false? a

a m ÷ a n = am ÷ n

310 ÷ 33

(−8)10 ÷ (−8)3

513 ÷ 512

e 5

am ÷ bn = am − n

am ÷ an = am − n

3a–g, 4a–e, 5a–f, 6a–d, 7a–d, 8ab, 9ab

Trailblazer 2

3b–h, 4b–f, 5c–h, 6d–h, 7c–f, 8a–d, 9a–d

3c–l, 4c–h, 5e–l, 6f–l, 7d–h, 8c–h, 9d–h

920 ÷ 99

(−5)8 ÷ (−5)2

77 ÷ 7

(−2)6 ÷ (−2)4

Simplify and evaluate as a basic numeral: a

Ex 1b

Adventurer 2

Simplify: a

Fluency Ex 1a

Explorer 1, 2abef

29 ÷ 25

48 ÷ 45

(−2)11 ÷ (−2)10

b6 ÷ b2

s13 ÷ s4

d20 ÷ d14

Simplify: a

a 9 ÷ a5

x7 ÷ x4

c7 ÷ c2

2.02 Division law mathspace.co

Ex 1c

Simplify: a

7 Ex 1d

Ex 1e

8b10 ÷ 2b6

b 30y11 ÷ 6y8

36c8 ÷ 9c3

32t13 ÷ 4t11

15l8 ÷ 3l2

6p13 ÷ 2p10

Simplify:

Simplify: a

f 4 g10 ÷ f 2 g2

m8 p13 ÷ mp3

40h11 j9 ÷ 8h7 j6

10–12

11–13

Simplify:

Reasoning

10, 12

and gets 5d2 as the result. Is Jane’s solution correct?

Jane simpliﬁes the expression Explain your reasoning.

Explain why m5 ÷ z3 is not equal to

Aidan says we can’t evaluate

If m6 = m × m × m × m × m × m then show using the division law why m0 = 1.

Mathspace WA Curriculum – Year 9 mathspace.co

. Becky says we can. Who is correct? Explain.

Problem-solving 14a–e, 16, 17 14

14b–f, 16, 17, 18a

14d–h, 15–18

Complete these statements: a

b9 ÷ b⬚ = b5

x⬚ ÷ x3 = x2

Using any of the digits from 1 to 9 at most once, complete the equation.

A rectangle has an area of 8a6 cm2. If one of the sides is of length 2a2 cm, ﬁnd an expression for the other side.

Simplify

Write 3 different expressions using the division law that are all equivalent to: a

4g4

16a2 b12 c3

2.03 Power of a power law After this lesson, you will be able to… • apply the power of a power index law to algebraic expressions. • solve problems that involve multiple index laws, including multiplication and division index laws.

Power of a power law Interactive exploration

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2.03 Power of a power law mathspace.co

Consider (a3)4.

(a3)4 =

= (a × a × a) × (a × a × a) × (a × a × a) × (a × a × a) = a3 × 4 (4 groups of 3 a’s) = a12 We can avoid having to write the expression in expanded form using the power of a power law.

(am)n = amn where a, m and n are integers and a ≠ 0: When simplifying a term with a power that is raised to another power: • The base does not change. • Find the product of the power. Since “powers of powers” involve expressions with brackets, it’s important to remember that every factor inside the brackets is raised to the outside power.

Example 1 Express in simpliﬁed index form: a (25)10

Create a strategy Use the power of a power law to simplify the expression.

Apply the idea (25)10 = 25 × 10 =2

Apply the power of a power law Simplify

b (w7)4

Create a strategy Use the power of a power law to simplify the expression.

Apply the idea (w7)4 = w7 × 4 =w

Apply the power of a power law Simplify

Mathspace WA Curriculum – Year 9 mathspace.co

c (2x2)10

Create a strategy Use the power of a power law to simplify the expression. Remember: the power outside the brackets applies to every factor inside the brackets.

Apply the idea (2x2)10 = 210 × (x2)10 2 × 10

= 1024 × x 20

= 1024x

Raise every factor inside the brackets to the power outside Apply the power of a power law Simplify

d (−3x3)3

Create a strategy Use the power of a power law to expand the brackets. Remember: the power outside the brackets applies to every factor inside the brackets. When dealing with negatives, remember that negatives raised to a positive power will be positive, and negatives raised to a negative power will be negative.

Apply the idea (−3x3)3 = (−3)3 × (x3)3 3×3

= −27 × x

Raise every factor inside the brackets to the power outside Apply the power of a power law

= −27x Simplify

( ) 2

2a  2a2  e  =   5  53 =

23 × a2 × 3

Create a strategy 53

6 a power law to simplify the expression. Use the power 8aof = 125 Remember: the power outside the brackets applies to every factor inside the brackets.

Apply the idea 3

( ) 2

2a  2a2  =  5    53

Raise every factor inside the brackets to the power outside

23 × a2 × 3 Apply the power of a power law = 53 8a6 = Simplify 125

2.03 Power of a power law mathspace.co

Create a strategy Use: • The power of a power law • The multiplication law • The division law

Apply the idea Apply the power of a power law

Simplify

Apply the multiplication law

Apply the division law

Idea summary When a term with a power is raised to another power, multiply the powers: (am)n = amn If the terms also have coeﬃcients, raise the coeﬃcient to the power, keep the base and multiply the indices: (bam)n = bnamn

2.03 Practice questions Understanding 1

( p × p × p) × ( p × p × p)

p 3 × p3

( p × p × p)2

p 3 × p2

( p × p × p) × ( p × p)

( p3)2 = p3 × 2

( pm)n = p⬚

True or false: ( p3)2 = p3 + 2

Complete the equation: ( p3)2 = p⬚

Mathspace WA Curriculum – Year 9 mathspace.co

Trailblazer 1–3

Adventurer 1–3

Are the expressions equivalent to ( p3)2?

a 3

Explorer 1–3

Fluency Ex 1a

Ex 1b

6 Ex 1c-e

Ex 1f

4a–d, 5a–d, 6a–d, 7abe

4d–h, 5e–h, 6e–l, 7b–e

4g–l, 5i–l, 6i–p, 7d–h

Simplify: a

(23)7

(34)4

(72)9

(53)2

(9 )

(11 )

10 3

(2 )

(512)5

(77)6

((−3)9)3

((−2)7)4

(136)5

Simplify: a

(j3)6

(w2)4

(t4)3

(a7)3

(h7)6

( p2)9

(a5)5

(b4)7

(c10)3

(e11)4

(gx)y

(wa)b

Simplify: a

(ab2)3

(wx3)4

(t2 u3)5

(a7 b2)4

(5c7)2

(ab4 c3)5

m (−2a3)5

(−3a2)3

(32)4 × (33)2

d (2h5)2

(3a2)3

Simplify: a

(25)3 × 27

Reasoning

8, 10

(a3)5 × (a7)4

8–10

Does (w2)3 = (w3)2? Explain.

Find a value of x so that ax × ax = (ax)x. Explain.

A student simpliﬁes the expression

(x2)10 ÷ (x4)2

and gets z4 as the result. Explain the student’s

mistake. What should the correct simpliﬁcation be?

Problem-solving 11a–d, 12 11

11b–e, 12, 13

11c–h, 13, 14

Complete each statement: a

(abc)⬚ = a3 b3 c3

c e g

(⬚m⬚n⬚)3 = 8m6 n9

d (0.1a⬚)2 × (10b4)⬚ = 0.1a6 b4

(4m2)⬚ × (⬚b⬚)2 = 256m6b4

2.03 Power of a power law mathspace.co

Write (16p)4 in the form ab, where a is a prime number.

Write

If the side length of a cube is 5a4, what is the volume?

as a product of prime numbers in index form.

2.04 Zero and negative indices After this lesson, you will be able to… • use index laws to simplify algebraic expressions involving zero and negative indices. • apply zero and negative index laws in real-world situations and problems.

The zero index Consider 25 ÷ 25. Using the division law, we can see that: 25 ÷ 25 = 25−5 = 20 However, we know that if we divide any number by itself, the result is 1. 25 ÷ 25 =

It follows that: 20 = 1 This is known as the zero index law.

a0 = 1 (provided a ≠ 0)

Example 1 Evaluate (6 × 19)0

Create a strategy Use the zero index law to evaluate the expression.

Apply the idea Everything is to the power of 0 so we can apply the rule to the whole expression. (6 × 19)0 = 1

Mathspace WA Curriculum – Year 9 mathspace.co

Example 2 Evaluate 5 × 20 + 3 × 70

Create a strategy Use the zero index law to evaluate the expression. Remember the order of operations for this question: 1. Indices 2. Multiplication 3. Addition In an expression like 5 × 20, the zero index applies to the base of 2, not to the 5.

Apply the idea 5 × 20 + 3 × 70 = 5 × 1 + 3 × 1

Apply the zero index law

=5+3

Evaluate the multiplication

Simplify

Idea summary For any non-zero base a, a0 = 1 That is, raising any non-zero number to the power of 0 will always result in 1.

Negative indices Interactive exploration

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We can use the division law to see that: 25 ÷ 23 = 25 − 3 = 22 When we divide, we subtract the indices. This is true even if subtracting the powers results in a negative. For example: 23 ÷ 25 = 23 − 5 = 2 − 2 Let’s generalise this to a rule. Consider a3 ÷ a5. We can write this in expanded form and remove the common factors.

a×a×a 1 = a × a × a × a× a a× a

2.04 Zero and negative indices mathspace.co

Writing this in index form:

This is the negative index law:

A reciprocal is a number that, when multiplied by another number, gives 1. For example the reciprocal of

× 3 = 1. Similarly, the reciprocal of

is 3 because

, and the reciprocal of

We can use this property of reciprocals to help with negative indices. When raising a base to a negative power: • Take the reciprocal of the expression. • Turn the power into a positive. Negative powers are used to represent fractional expressions. Using the equivalence we have just developed, we can also say that:

and in general,

Example 3 Express 6 − 10 with a positive index.

Create a strategy Use the negative index law.

Apply the idea Apply the negative index law

Example 4 Simplify

, giving your answer as a single power of 5.

Create a strategy Use index laws to simplify the expression.

Mathspace WA Curriculum – Year 9 mathspace.co

Apply the idea Use the power of a power law

Use the multiplication law

Use the division law

Simplify

Example 5 Simplify, giving your answers with positive indices: a 3q2 × 4q − 5

Create a strategy Apply algebraic conventions and index laws to simplify the expression.

Apply the idea Write in expanded form

Group the numeric and algebraic components together

Multiply the numbers and use the multiplication index law

Use the negative index law

Create a strategy Apply algebraic conventions and index laws to simplify the expression.

Apply the idea Write in expanded form

Divide the numbers by the common factor of 7

Apply the negative index law

Apply the multiplication law

Simplify

2.04 Zero and negative indices mathspace.co

Create a strategy Apply algebraic conventions and index laws to simplify the expression.

Apply the idea Apply the negative index law for fractions

Apply the power of a power law

Simplify

Idea summary For any base a,

That is, when raising a base to a negative power: • Take the reciprocal of the expression. • Turn the power into a positive. A negative power is used to represent a fractional expression.

2.04 Practice questions Understanding 1

Explorer 1, 2ad, 3, 4, 6

2 =2 20 = ⬚

2−1 = ⬚

Complete the sentence:

Each time the power of 2 decreases by 1, the number on the right is divided by ⬚.

2 =4

Complete the pattern.

13 − 10

Identify the base and the power. a

Trailblazer 1, 2cd, 3–6

Look at the pattern: 23 = 8

Adventurer 1, 2bd, 3–6

10 − 7

2−4

Mathspace WA Curriculum – Year 9 mathspace.co

(−5) − 8

, where a and b are non-zero numbers?

What is the value of A

It’s a mathematical convention without a reason.

When you divide any number by itself, you get 1, and x0 can be thought of as

Zero has no value, so it makes the number disappear.

Because 1 is the multiplicative identity.

The fraction gets inverted.

The fraction becomes a decimal.

The numerator and denominator both become negative.

The fraction becomes zero.

Is 2 − 3 less than or greater than 1? 7a–h, 8, 9ab, 10a–c, 11a–d, 12a–h, 13a–f, 14a–d, 15a–e, 16a–d, 17abij, 18abe, 19ae

7e–l, 8, 9bc, 10c–e, 11c–f, 12e–l, 13b–h, 14c–f, 15e–j, 16c–h, 17c–l, 18b–g, 19a–e, 20abe

7i–p, 8c, 9c, 10e–h, 11d–h, 12i–p, 13e–l, 14e–h, 15e–l, 16e–l, 17e–p, 18e–l, 19c–h, 20

Evaluate: a

33 × 30

(−3)0

−40

−1000

−9 (12 × 15)0

72 ÷ 72

25 × 110

(6 × 19)0

7 (14 × 18)0

Complete: a

When you have a fraction raised to a negative index, what happens?

m 8

Which statement explains why any non-zero number to the power of 0 is 1?

Fluency

Ex 1

7⬚ = 1

Complete the table: a

25 32

105 100 000

33 27

24 16

32 9

2−1

104 10 000

103

102

101

100

3−1

3−2

3−3

10 − 1

2.04 Zero and negative indices mathspace.co

Ex 2

Ex 3

Evaluate: a

22 + 2 + 20

7 × 30 + 32

(2 + 20) × 20

4 × 3 − 2 + 80

6 ÷ 2−1 + 4

(2 + 3 − 2) × 2 − 1

17 − 6

9−1

−8 − 11

Express with a positive index: a e

Ex 4

8 − 15 (−9)

−7

a−9

7 −6 (−12)

−8

−5

−45

p−2

3x − 4

7x − 9

k − 2 q3

2x − 8 y3 m − 5 r − 4 p4

8p − 3

m 3a4 b − 5 c − 7

4 − 2 k − 3 l7

Express with a negative index: a

Complete:

Simplify, leaving your answer in index form: a

511 ÷ 5 − 3

102 ÷ 103

2−3 × 2−4

3−4 ÷ 3−9

90 × 9 − 12

813 ÷ 8 − 9

7 − 7 ÷ 75

10025 ÷ 10026

i 16

Express with positive indices: a

5 × 50 − 3 × 20

Evaluate: a

67 × 6 − 7

48 × 4 − 6

3 − 8 × 311

55 × 5 − 7

4−4 × 2−4

1213 ÷ 127 ÷ 128

5−2 × 3−2

(−4)3 × (−4) − 7

712 ÷ 7 − 7 ÷ 717

Mathspace WA Curriculum – Year 9 mathspace.co

17 Ex 5a-b

Simplify, giving your answers with positive indices: a e

Ex 5c

× 4h

7a4 × 4a − 2 3y

−2

× 2y

5x4 × (−3x − 8)

−5

−4y × (−4y 12x5 ÷ 4x − 3

−5

)

3y − 2 × 4y − 3

(5mp)2 × mp − 2

9c6 ÷ 6c8

Simplify, giving your answers with positive indices: (2m) − 3

(4m − 6)4

(3p − 4) − 2

(3y2) − 2

Simplify: a

2y6 × 4y7 × 4y − 5

6y7 × 2y − 5 × 5y3

2x − 2 × 5x5 × 4x − 5

12x − 9 × 2x4 × 3x − 2

4y3 × 3y8 ÷ 2y − 1

−8y10 ÷ 2y − 4 ÷ y3

14x − 8 ÷ 2x × 3x6

(4x3) − 1 × ( y − 2) − 3

g 20

−4

5y9 × 4y − 3

40x

−2

÷ 5x ÷ 4x

−9

Simplify: a

(3x − 4 × 2x2)2

(10x3 y − 4)2

(10x3)3 × (2x4 y3) − 2

Reasoning

21, 22, 24

21–24

What is 04 equal to?

21, 23–25

Explain why 0 − 4 is undeﬁned.

Explain the difference between 2x0 and (2x)0.

When asked to ﬁnd an expression that is equivalent to −20, a student responded 1. Is this answer correct? Explain why or why not.

A student was writing 5a − 1 without negative indices and wrote working is incorrect, and write the correct answer.

Explain why the expression x − n is the same as

. Explain why their

. Does this hold true for negative values of x?

2.04 Zero and negative indices mathspace.co

Problem-solving 26, 27 26

26–28

26–30

Complete each statement: a c e

a⬚b3 × ⬚a2 b⬚ = 2ab − 1

b d

(gh2) ⬚g⬚ = g3 h − 8

Simplify, leaving your answers with positive indices:

(5–1 + 3–1)–1

If n = 2, determine the value of

If the surface area of a cube is 6s2, where s is the length of a side in centimetres, ﬁnd the surface area when s = 3 − 1 cm (without using a calculator).

Without using a calculator, what is the simpliﬁed form of (4 − 1 × 23)2?

2.05 Scientific notation After this lesson, you will be able to… • write numbers in scientific notation form. • perform mathematical operations using numbers in scientific notation. • understand the use of metric prefixes and convert measurements using scientific notation.

Scientific notation Scientific notation, sometimes referred to as standard form, is a compact way of writing very large or very small numbers. For example, • The sun has a mass of approximately 1.988 × 1030 kg, which is much easier to write than 1988 000 000 000 000 000 000 000 000 000 kg. • The mass of an atom of Uranium (one of the heaviest elements) is approximately 3.95 × 10 − 22 g. That is 0.000 000 000 000 000 000 000 395 g. It is called scientific notation because it is often used in scientific contexts. 44

Mathspace WA Curriculum – Year 9 mathspace.co

Interactive exploration

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In scientiﬁc notation, numbers are written in the form: a × 10n where a is a decimal number between 1 and 10 and n is an integer (positive, negative, or zero). Note that 1 ≤ a < 10, so a may be equal to 1, but cannot be 10. • A negative index indicates how many factors of ten smaller than a the value is. • A positive index indicates how many factors of ten larger than a the value is. • An index of zero indicates that the value is a because 100 = 1. There are lots of ways to express numbers using powers of ten. The decimal 0.032 can be written as: • • •

(scientiﬁc notation)

Example 1 Express each number in scientiﬁc notation: a 0.000 347

Create a strategy Use the scientiﬁc notation form a × 10n, where a is a decimal number between 1 and 10 and n is a negative integer.

Apply the idea To find the first part of our scientific notation we place the decimal point after the first non-zero digit, so a = 3.47 0.000 347 is 10 000 or 104 times smaller than 3.47. So: 0.000 347 = 3.47 × 10 − 4

b 84 245 000

Create a strategy Use the scientiﬁc notation form a × 10n, where a is a decimal number between 1 and 10 and n is a positive integer.

2.05 Scientific notation mathspace.co

Apply the idea To find the first part of our scientific notation we place the decimal point after the first non-zero digit, so: a = 8.4245 84 245 000 is 10 000 000 or 107 times bigger than 8.4245. So: 84 245 000 = 8.4245 × 107

Example 2 Convert the following from scientiﬁc notation to ordinary numbers: a 2 × 107

Create a strategy Use the fact that 107 = 10 000 000.

Apply the idea 2 × 107 = 2 × 10 000 000 = 20 000 000

Rewrite 107 Evaluate

b 3.62 × 10−4

Create a strategy Use the fact that 10−4 =

= 0.000 1.

Apply the idea 3.62 × 10−4 = 3.62 × 0.000 1 = 0.000 362

Evaluate 10−4 Evaluate

Idea summary In scientiﬁc notation, numbers are written in the form: a × 10n where a is a decimal number (1 ≤ a < 10) and n is an integer (positive, negative or zero). • A negative index indicates how many factors of ten smaller than a the value is. • A positive index indicates how many factors of ten larger than a the value is. • An index of zero indicates that the value is a because 100 = 1.

Mathspace WA Curriculum – Year 9 mathspace.co

Operations with scientific notation Just like ordinary numbers, we can add, subtract, multiply, and divide numbers written in scientiﬁc notation. To multiply a × 10m and b × 10n, we multiply a and b and add the powers of 10 (using the index laws): (a × 10m) × (b × 10n) = (a × b) × 10(m+n) To divide a × 10m by b × 10n, we divide a by b, and subtract the second power of 10 from the ﬁrst (following the index law on division):

When multiplying and dividing, it can happen that the answer is not in scientiﬁc notation, so we need to convert it. When adding and subtracting numbers in scientiﬁc notation, it is best to convert the numbers into decimal numbers.

Example 3 Simplify, giving your answers in scientiﬁc notation. a (6 × 107) × (4 × 10−2)

Create a strategy Multiply the number parts, multiply the power parts, then adjust the answer if necessary to ensure that it is in scientiﬁc notation.

Apply the idea (6 × 107) × (4 × 10−2) = 6 × 4 × 107 × 10−2 Remove the brackets and reorder the parts = 24 × 105 1

Multiply the numbers. Use the multiplication law 5

= 2.4 × 10 × 10 6

= 2.4 × 10

Express 24 in scientiﬁc notation onvert to scientiﬁc notation using the C multiplication law

b (2 × 102) ÷ (8 × 105)

Create a strategy When dividing numbers in scientific notation, divide the first number by the second number, divide the first power part by the second power part, then adjust the answer if necessary to ensure that it is in scientific notation.

2.05 Scientific notation mathspace.co

Apply the idea (2 × 102) ÷ (8 × 105) = (2 ÷ 8) × (102 ÷ 105) = 0.25 × 10−3 −1

ivide the number parts and divide the power D parts Use the division law

−3

= 2.5 × 10 × 10 −4

= 2.5 × 10

Express 0.25 in scientiﬁc notation onvert to scientiﬁc notation using the C multiplication law

c (3.2 × 10−2) + (2.3 × 10−3)

Create a strategy When adding or subtracting numbers given in scientiﬁc notation, it is often best to convert them to ordinary numbers.

Apply the idea (3.2 × 10−2) + (2.3 × 10−3) = 0.032 + 0.0023 = 0.0343

xpress both numbers as ordinary E decimals Add the decimals

−2

= 3.43 × 10

Express 0.0343 as scientiﬁc notation

Idea summary When multiplying and dividing numbers in scientiﬁc notation:

When adding and subtracting numbers in scientiﬁc notation, it is often best to convert the numbers into ordinary numbers.

Mathspace WA Curriculum – Year 9 mathspace.co

Metric prefixes Another way to write very large or very small measurements is using prefixes. We add prefixes to a unit to make the unit bigger or smaller by a power of ten. The metric prefixes kilo-, centi- and milli- are often used for lengths and weights. For example, one kilometre means 1 × 103 metres. Five milligrams means 5 × 10−3 grams. You may also be familiar with the prefixes tera-, giga- and mega- which are commonly used when discussing digital data. For example, one terabyte means 1 × 1012 bytes. Below are some commonly used metric prefixes: Prefix teragigamegakilohectodecadecicentimillimicronanopico-

Letter T G M k h da d c m μ n p

Size 1012 109 106 103 102 101 10−1 10−2 10−3 10−6 10−9 10−12

Example 4 Convert the measurements to the indicated unit, expressing your answer in scientiﬁc notation. a 52 nanoseconds to seconds.

Create a strategy Use the table showing the sizes of different metric preﬁxes.

Apply the idea 52 nanoseconds = 52 × 10−9 seconds Multiply by the size of a nanosecond to convert to seconds = 5.2 × 101 × 10−9 seconds −8

= 5.2 × 10

seconds

Write 52 in scientiﬁc notation Simplify

2.05 Scientific notation mathspace.co

b 0.07 Mb to bytes.

Create a strategy Use the table showing the sizes of different metric preﬁxes.

Apply the idea Mb means megabytes. 0.07 megabytes = 0.07 × 106 bytes Multiply by the size of a megabyte to convert to bytes = 7 × 10−2 × 106 bytes

Write 0.07 in scientiﬁc notation

= 7 × 10 bytes

Simplify

Idea summary We can use metric preﬁxes to write very large and very small measurements. Preﬁx teragigamegakilohectodeca-

Letter T G M k h da

Size 1012 109 106 103 102 101

Preﬁx decicentimillimicronanopico-

Letter d c m μ n p

Size 10−1 10−2 10 − 3 10 − 6 10 − 9 10 − 12

2.05 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

Choose the correct statement. In scientiﬁc notation, numbers are written in the form a × 10n where:

a is an integer between 1 and 10 (not equal to 10) and n is any decimal

a is a decimal number between 1 and 10 (not equal to 10) and n is a positive integer

a is a decimal number between 1 and 10 (not equal to 10) and n is any integer

a is any decimal number and n is any integer

Which is written correctly using scientiﬁc notation? A

8.56 × 1014

0.658 × 103

Mathspace WA Curriculum – Year 9 mathspace.co

608.56 × 1014

465.53 × 1021

Simplify using index laws: a

106 × 105

4ace, 5a–e, 6a–e, 7a–e, 8a–e, 9, 10ab, 12–14, 15a–c

Fluency

Ex 2

4bdf, 5c–h, 6c–h, 7c–h, 8c–h, 10c–f, 11, 15c–h

−4 d • 9.37 × 10 • 6 × 10 − 4 • 6 × 10 − 9

• 4.7 × 102 × 103 • 4.7 × 10 − 8 × 108 • 4.7 × 109 × 108

128

2000

884 000

84 626 000

23 456

6.14

9 700 000 000

0.000 74

0.000 059 32

Express in scientiﬁc notation: a

0.6

0.000 676

0.002

Express as ordinary numbers: a

3 × 109

4.13 × 104

9 × 103

8.97 × 105

5.03 × 105

3.014 × 103

8.0045 × 103

1.3008 × 107

4.7 × 10 − 6

2.13 × 10 − 7

6.4 × 10 − 5

Express as ordinary numbers: 5.28 × 10 − 2 8.29 × 10

7 × 10 − 3

3.47 × 10

−5

−2

9.72 × 10

⬚ = 7.3 × 10 − 4

Complete: a

0.002 = ⬚ × 10 − 3

⬚

4000 = ⬚ × 103

89 300 000 = 8.93 × 10

⬚ = 7.34 × 105

(3 × 104) × (2 × 105)

(4 × 10 − 3) × (2 × 10 − 2)

(6 × 108) × (3 × 10 − 3)

(8 × 107) ÷ (2 × 102)

(6 × 10 − 1) ÷ (2 × 10 − 4)

(3 × 10 − 1) ÷ (6 × 103)

c e 10

4bde, 5b–f, 6b–f, 7b–f, 8b–f, 10a–d, 11, 12, 14, 15a–f

Ex 3

Express in scientiﬁc notation:

For each set, indicate the largest and the smallest number. −7 −9 −2 a • 7.27 × 10 b • 7.31 × 10 c • 1.25 × 10 • 2.22 × 10 − 7 • 5.13 × 10 − 9 • 1.25 × 10 − 6 −7 −9 • 4.76 × 10 • 8.31 × 10 • 1.25 × 102 −2 e • 2.93 × 10 2 • 7.88 × 10 • 7.88 × 10 − 2

Ex 1

10 − 3 × 10 − 5

0.000 067 = 6.7 × 10⬚

Calculate, expressing your answers in scientiﬁc notation:

2.05 Scientific notation mathspace.co

Calculate, expressing your answers in scientiﬁc notation: a c

Ex 4

3 × 104 + 8 × 105 8

5.74 × 10 − 3.5 × 10

2.1 × 10 − 3 + 4 × 10 − 2

8.2 × 10 − 4 − 7.1 × 10 − 5

The thickness of coating on a camera lens is 0.000 002 metres. Express the thickness in scientiﬁc notation.

The distance between two stars is approximately 5 × 106 metres. Express this distance as a whole number.

The world’s oceans hold approximately 712 000 000 000 000 000 000 litres of water. Express this volume of water in scientiﬁc notation.

Use the table of preﬁxes to convert the measurements to the indicated unit, giving your answer in scientiﬁc notation. a

3.8 milliseconds to seconds

1.67 kL to L

340 pm to m

0.5 decibels to bels

Reasoning

16, 18

Preﬁx Letter Size teraT 1012 gigaG 109 0.022 µg to g megaM 106 63 ML to L kilok 103 0.8 terabytes to bytes h 102 1200 hectopascals to pascals hectodecada 101 decid 10 − 1 centic 10 − 2 millim 10 − 3 μ micro10 − 6 nanon 10 − 9 picop 10 − 12 16, 18

17, 18

Explain why a scientist would want to express a very large or small number in scientiﬁc notation.

b 18

Evaluate, expressing your answer in scientific notation to 2 decimal places: i

4.75 × 107 + 1.35 × 103

4.75 × 107 − 1.35 × 103

What do you notice about your answers to part (a)? Why did this happen?

Explain why each of these numbers is not correctly expressed in scientiﬁc notation. a

23.6 × 10 − 3

3.45 × 100.5

Problem-solving 19, 20 19

0.0006 × 102

20–22

9.73 × 82

19–22

Bone marrow in a person’s body produces 2.3 × 106 red blood cells each second. a

Express the amount of red blood cells as an ordinary number.

How many red blood cells are produced each minute? Express your answer in scientiﬁc notation.

Mathspace WA Curriculum – Year 9 mathspace.co

Sound travels at a speed of approximately 0.3 kilometres per second, while light travels at a speed of approximately 300 000 kilometres per second. a

Express the speed of sound in kilometres per second in scientiﬁc notation.

Express the speed of light in kilometres per second in scientiﬁc notation.

How many times faster does light travel than sound?

The mass of one helium atom is 6.7 × 10 − 24 g, and the mass of one iron atom is 9.3 × 10 − 23 g. a

Find the difference between the masses of the two atoms. Express your answer in scientiﬁc notation.

What percentage of the mass of an iron atom is a helium atom? Give your answer to one decimal place.

An experiment produces 3.14 × 10 − 1 g of a substance. Express this mass in kilograms.

The track left by a particle in an experiment is 2.76 × 10 − 5 m long. Express this distance in centimetres.

Did you know?

Bacteria can multiply incredibly fast due to the power of indices! In ideal conditions, some bacteria can double their population every 20 minutes. This means that if you start with just one bacterium, after 3 hours, you’d have over 500 bacteria! This exponential growth, modelled by indices, explains why food can spoil so quickly if left out. It’s the same principle scientists use to predict the spread of infections or to grow bacteria for research. Understanding indices helps us track these rapid changes and manage bacterial growth in everything from medicine to food safety!

2.05 Scientific notation mathspace.co

2 Chapter review 1

Simplify, leaving your answer in index form: a

73 × 75

(−2)2 × (−2)4

38 × 3h

b6 × b × b4

x12 × 5x3

e 2

c ×c

−5

Simplify and evaluate as a basic numeral: a

47 ÷ 44

(13)5

(97)2

52 × 50

a− 5

8a7 ÷ 4a6

((−2)2)2

( j13)2

(xy2 z3)5

−70

8 (13 × 17)0

4y−7

5−3 × 5−3

(3m)−2

2y3 × 4y−7

6x − 9 ÷ 2x × 7x7

(c3)5 ÷ (c8)7

83 ÷ 811

16x−7 ÷ 8x−3

Simplify:

Evaluate: a

Simplify, leaving your answers with positive indices:

(−5)20 ÷ (−5)t

Express with positive indices: a

Evaluate: a

Simplify, leaving your answer in index form: a

Simplify and evaluate as a basic numeral: a

(−6)11 ÷ (−6)9

Simplify, leaving your answer in index form: a

4 × 42 × 4 − 1

Mathspace WA Curriculum – Year 9 mathspace.co

3 − 2 + 8 × 70

Express in scientiﬁc notation: a

113 000

0.8

0.000 89

1.4 × 100

4 × 10 − 3

Express as a basic numeral: a

6 × 107

3.85 × 104

3.36 × 10 − 2

3.005 × 10 − 6

Calculate, expressing your answers in scientiﬁc notation: a

(2 × 104) × (4 × 105)

(3 × 10 − 3) × (2 × 10 − 4)

(6 × 107) ÷ (3 × 102)

(9 × 10 − 1) ÷ (3 × 107)

Calculate, expressing your answers in scientiﬁc notation: a

234

3.87 × 106 − 3.6 × 105

3.5 × 10 − 3 + 2 × 10 − 2

Use the table of preﬁxes to convert the measurements to the indicated unit, giving your answer in scientiﬁc notation. a

5.1 milliseconds to seconds

3.04 km to m

0.05 terabytes to bytes

0.089 µg to g

23 decibels to bels

Preﬁx teragigamegakilohectodecadecicentimillimicronanopico-

Letter T G M k h da d c m μ n p

Size 1012 109 106 103 102 101 10 − 1 10 − 2 10 − 3 10 − 6 10 − 9 10 − 12

A swimming pool has a length of a4 m, width of a3 m and depth of a m. What is the volume of the pool in cubic metres?

The population of Melbourne is P6 and the population of Margaret River is P2. How many times more populous, is Melbourne compared to Margaret River?

Write (27m)5 in the form ab, where a is a prime number.

Without using a calculator, what is the simpliﬁed form of: (165 ÷ 642) − 1 + 3 × 30

A right-angled triangle has a hypotenuse of 5 × 104 mm and a short side of 4 × 104 mm. Calculate the length of the third side, leaving your answer in scientiﬁc notation.

Chapter 2 review mathspace.co

Big ideas Simplifying, expanding, and factorising are interconnected processes that provide different ways to understand and manipulate algebraic expressions.

Curriculum Outcomes • Explore and apply the distributive law to expand and factorise algebraic expressions with a common algebraic factor, including collecting like terms where appropriate • Explore and apply the distributive law to expand binomial products, collecting like terms where appropriate • (Optional) Explore efficient strategies to simplify expressions that involve addition, subtraction, multiplication or division of algebraic fractions with an algebraic term in the numerator and a whole number denominator • Analyse, represent and interpret problems in real-world situations involving scientific notation, real numbers, linear equations with variables and/or brackets on either side of the equation, quadratic graphs and equations, direct proportion and/ or simple interest, earning income or income statements

3 Algebra Chapter outline 3.01 3.02 3.03 3.04 3.05 3.06 3.07

Algebra review Distributive law Binomial expansion Algebraic factors Factorise by grouping Optional: Add and subtract fractions with numeric denominators Optional: Multiply and divide algebraic fractions with numeric denominators Investigation: Calendar expressions Chapter 3 review

58 67 71 76 82 88 93 99

Algebra simplifications are just like the abbreviations we use in everyday life like BRB (Be Right Back), LOL (Laughing Out Loud) and MSR (Mathspace Rules).

3.01 Algebra review After this lesson, you will be able to… • review adding and subtracting algebraic terms, focusing on collecting like terms. • review multiplying and dividing algebraic terms, highlighting how indices may change. • apply the order of operations in the context of simplifying algebraic expressions.

Algebra review When working with algebraic terms, it can be helpful to remember some vocabulary. Index (power)

3 x Coefficient

Constant

+ 11

Variable

5x2 y + 3xy − 2xy − 8 Terms:

5x2 y, 3xy, −2xy, −8

Like terms: 3xy, and −2xy Add and subtract algebraic terms When adding and subtracting algebraic terms, we collect like terms by adding or subtracting their coeﬃcients, for example 2a + 3a = 5a Multiply and divide algebraic terms When multiplying and dividing algebraic terms, terms do not have to be like terms. We multiply or divide the coeﬃcients and may use index laws, for example 8a4 ÷ 2a = 4a3 When simplifying algebraic expressions we always follow the order of operations.

Example 1 Simplify: a 9a + 3b + 2a − 4b

Create a strategy There are two pairs of like terms, 9a and 2a, and 3b and −4b. Like terms can be collected together.

Mathspace WA Curriculum – Year 9 mathspace.co

Apply the idea 9a + 3b + 2a − 4b = 9a + 2a + 3b − 4b

Group like terms

= 11a – b

Combine like terms

Reﬂect and check We can identify two types of terms, terms in a and terms in b. Our solution has two terms, one in a and one in b.

b 5wy2 − 7w2 y + wy − 2w2 y + wy2

Create a strategy wy2 = w × y × y and w2 y = w × w × y, so terms in wy2 and terms in w2 y are unlike. Here: • 5wy2 and wy2 are like • −7w2 y and −2w2 y are like • wy does not have a like term, so it will not change

Apply the idea 5wy2 − 7w2 y + wy − 2w2 y + wy2 = 5wy2 + wy2 − 7w2 y − 2w2 y + wy 2

Group like terms

Combine like terms = 6wy − 9wy + wy

Reﬂect and check The original statement had three different types of terms: wy2, w2 y, and wy Our solution has three terms, one in wy2, one in w2 y, and one in wy.

Example 2 Simplify: a 5r × (−7)

Create a strategy Multiply the numeric components and then simplify the algebraic components.

Apply the idea 5r × (−7) = 5 × (−7) × r Group the numeric and algebraic components together = −35 × r

Multiply the numbers

= −35r

Simplify

3.01 Algebra review mathspace.co

b 3c2 × 5d3

Create a strategy Multiply the numeric components and then simplify the algebraic components.

Apply the idea 3c2 × 5d3 = 3 × 5 × c2 × d3 2

Group the numeric and algebraic components together

= 15 × c × d

Multiply the numbers

= 15c2 d3

Simplify

Reﬂect and check Since c2 and d3 have different bases, we cannot use the index laws to simplify them.

c 8ab2 × 3abc

Create a strategy Multiply the numeric components and then simplify the algebraic components.

Apply the idea 8ab2 × 3abc = 8 × 3 × a × a × b × b × b × c Group the numeric and algebraic components together = 24 × a2 × b3 × c Multiply the numbers, and then the variables in turn = 24a2 b3 c

Simplify

Reflect and check Looking at the original statement: • The coefficients are 8 and 3 and 8 × 3 = 24 • The first term has an a and the second term has an a, and a × a = a2 • The first term has a b2 and the second term has a b, and b2 × b = b3 • There is only one c

Example 3 Simplify: a 30m ÷ 5m

Create a strategy Remove common factors from the numeric components and then simplify the algebraic components. Often it is easier to work with algebraic division if it is written as a fraction.

Mathspace WA Curriculum – Year 9 mathspace.co

Apply the idea Write as a fraction

Remove the common factor of 5

Simplify the algebraic terms

Reﬂect and check • Considering the coeﬃcients, 30 ÷ 5 = 6 • Considering the algebraic part, m ÷ m = 1

Create a strategy Remove common factors from the numeric components and then simplify the algebraic components. Often it is easier to work with algebraic division if it is written as a fraction.

Apply the idea Introduce multiplication signs and bring out highest common factor (HCF) of numbers

Remove the common factor, 5

Remove the common factor, w

Remove the common factor, x

Remove the unnecessary multiplication signs

Reﬂect and check We can think about working on each factor separately: , then

, and then

3.01 Algebra review mathspace.co

Create a strategy Remove common factors from the numeric components and then simplify the algebraic components. Often it is easier to work with algebraic division if it is written as a fraction.

Apply the idea Reintroduce the multiplication signs Remove the common factors of 3, a and b

Group the numeric and algebraic components together

Simplify

Reﬂect and check We could use the approach we use with index laws:

Use multiplication law in numerator

Divide coeﬃcients and use division law

Simplify

Example 4 Simplify: a 3a2 + 5a × 7a

Create a strategy Use the order of operations and what we know about algebraic terms.

Apply the idea 3a2 + 5a × 7a = 3a2 + 35a2 = 38a

Perform the multiplication Add the like terms

Reﬂect and check The multiplication produces like terms that can be collected together, otherwise, we would not be able to simplify further.

Mathspace WA Curriculum – Year 9 mathspace.co

b (7a − 2a) × 3ab

Create a strategy Use the order of operations and what we know about algebraic terms.

Apply the idea (7a − 2a) × 3ab = 5a × 3ab 2

= 15a b

Collect the like terms in the brackets Perform the multiplication

Create a strategy Use the order of operations and what we know about algebraic terms.

Apply the idea Collect the like terms in the numerator and perform the multiplication in the denominator

Divide out the common factor of 15

Introduce multiplication signs

Divide out the common x and simplify.

Reﬂect and check We could also use the division and negative index laws: Collect the like terms in the numerator and perform the multiplication in the denominator

Divide out the common factor of 15

Division index law

Negative index law

Idea summary Simplifying algebraic expressions makes use of: • Adding and subtracting algebraic terms, which may use like terms • Multiplying and dividing algebraic terms, which may use index laws • Applying the order of operations

3.01 Algebra review mathspace.co

3.01 Practice questions Explorer

Understanding 1

Adventurer

1, 2

Trailblazer 1, 2

For the expression: 3x2 − xy + 5x − 3y − 9 + 5y a

How many terms are there?

What is the coeﬃcient of x?

What is the coeﬃcient of xy?

What is the constant term?

Identify any like terms.

Are these statements true or false? a

9xyz = 9zxy

−4 × 5 × a × b = −9ab

p5 = p × p × p × p × p

2x2 × 5xy cannot be simpliﬁed because they are not like terms.

c8 ÷ c2 = c4

x÷2=

2x + 3y = 5xy 3a–cij, 4a–j, 5a–j, 6a–cgh, 7a–g

5x + 8x

10y − 7y

7c + 3c − c

3x − 4x − x

3f–p, 4h–p, 5h–p, 6c–l, 7e–p

6c + 3d + 3c − d

7x + 4y − 2x + y

2x + 7y + 3x + z

3x + 7 − 2y + x

x2 + 6x + 5x + 10

5ab + 2bc − 2ab + 3cb

x2 − 4x + 4x − 16

5a2 b + 7ab2 + 2a2 b − 3ab2

m 5x − (−3x) + 7 + (−2)

3ab + 9cd − (−2ab) − (−cd)

x − (−3x) − 8x + x2

9 × 5u

7r × 4

xy + 2x y − 7 − 2xy + 3xy

Simplify: a c

2 × x × y × 3

4r × 6s

3u3 × 5v5

12p × 5q2

(−3c) × (−5d)

5e × (−5f )

(−3x) × (−10y)

6x2 × 2y2 × 3z2

9a × 3a × (−2a)

5a × 2b × a × 2b

3w3 x2 y × (−3)wx2 y3 × 2xy

−3e7 f 2 × 5e6 f 3 × e × f 2

3 2

4 3

5 3

m 5a b × 4a b o

3d–l, 4e–m, 5c–j, 6c–j, 7e–l

Simplify:

o Ex 2

What are the variables?

Fluency Ex 1

7c d × 3c × 2d

Mathspace WA Curriculum – Year 9 mathspace.co

5 Ex 3a, 3b

Ex 3c

Ex 4

Simplify: 6m ÷ 36

10ab ÷ 50ab

(9cd) ÷ 12de

(−12fg) ÷ (−3g2)

(−32p7) ÷ 8p7

Simplify:

(−3a) × (−5b) ÷ 10a2

5u × 8v ÷ 4w

11xy × 5z ÷ x2 y2

25g ÷ 5 × 7f

(8y ÷ 4x) × (12y ÷ 3x)

(9a × 4b) ÷ (5 × 2c)

Simplify: a

(7c − 3c) × 2c

(19t + 5t) ÷ 4u

(12a − 4a) ÷ (8 × 4a2)

5f − 3 × 4f + 7f

5 (3a × 2b) + 4a × (−3b)

2 (4x + x × 2x) ÷ 16x

3x5 × 7y4 − 2x3 y2 × 3x2 y2

Reasoning 8

15x6 ÷ 5x6

8, 9

p 9–11

Explain why we cannot combine 3a + 4b as 7ab.

Is it ever possible that 3a + 4b = 7ab? Explain your reasoning.

10–12

Explain why x2 and x are not like terms.

Is there any value of a for which 3a × 2a = 5a2? If so, give an example. If not, explain why.

Simplify the expression 7a + 5a × 2 in two different ways: • Perform the multiplication ﬁrst and simplify. • Repeat the problem and do the addition ﬁrst. Explain why an order of operations is needed when simplifying algebraic expressions. 3.01 Algebra review mathspace.co

Consider the problem

Identify and correct the error in each student’s work: Lawrence:

Marika:

Problem-solving 13–15, 17

14–16, 18

15–19

An isosceles triangle has 2 sides of length 3b cm. If the third side has a length of 2c cm, write a simpliﬁed expression for the perimeter of the triangle.

A rectangle has a length of (3x + y) cm and a width of (2x − 2y) cm. Find the perimeter of the rectangle in terms of x and y.

Addition pyramids work by the value in a box being equal to the sum of the two boxes below. The example shows how this works with numbers: 13 9 4

4 −1

Complete the addition pyramids. a

3a − b

5x − 2y 3x + 2y

4a + b a + 2b

4a + 3b

3x − y

A rectangle has a length 4 times its width. The width of the rectangle is 3a cm. Find an expression for the area of the rectangle.

A purse contains x $10 notes, y $5 notes, and z $1 coins. Write an expression for the number of dollars in the purse.

Emma bakes x batches of chocolate chip cookies, with y cookies in each batch.

How many chocolate chip cookies does Emma bake in total?

If there are an average of y chocolate chips in each cookie, and she bakes the same number of cookies on 3 consecutive days, how many chocolate chips does she use?

A piece of paper with dimensions 8.5 centimetres by 11 centimetres will have a square with length x cut out from each corner. Write and simplify an expression that can be used to represent the perimeter of the shape formed once the corners have been removed.

Mathspace WA Curriculum – Year 9 mathspace.co

3.02 Distributive law After this lesson, you will be able to… • review and apply the distributive law for both numerical and algebraic expressions. • use algebra tiles to illustrate the distributive law and create equivalent mathematical statements. • expand and simplify real-world algebraic expressions using the distributive law. • apply the distributive law to expand algebraic terms with brackets.

The distributive law Interactive exploration

Discover this concept in action online

mathspace.co

We use the distributive law to expand algebraic terms with brackets. The term on the outside of the brackets is multiplied by each term inside the brackets. A(B + C) = AB + AC When multiplying terms with indices, we may need to use the multiplication index law: am × an = am+ n

Example 1 Expand: a 6s (9t + 8)

Create a strategy Use the distributive law A (B + C) = AB + AC to expand the expression.

Apply the idea 6s (9t + 8) = 6s × 9t + 6s × 8 = 54st + 48s

Expand the expression Simplify

Reﬂect and check We can use arcs to help us remember to multiply each term inside the brackets by the factor outside of the brackets.

3.02 Distributive law mathspace.co

b 4u7 (4u7 + 2u6)

Create a strategy Use the distributive law A (B + C) = AB + AC to expand the expression. We will also use the multiplication index law am × an = am + n.

Apply the idea 4u7 (4u7 + 2u6) = 4u7 × 4u7 + 4u7 × 2u6 14

= 16u + 8u

Expand the expression Simplify

Reﬂect and check Remember that we multiply the coeﬃcients and add the indices for each product.

Example 2 Expand and simplify: x (x − 6) − 8 (x + 7).

Create a strategy Use the distributive law A (B + C) = AB + AC to expand the expression.

Apply the idea x (x − 6) − 8 (x + 7) = x2 − 6x − 8x – 56 2

= x − 14x – 56

Expand both brackets Simplify

Reflect and check When the factor outside of the brackets is negative, we need to multiply each term in the brackets by the negative.

Idea summary Expand means to write an algebraic expression without brackets. We can use the distributive law to expand an algebraic expression with brackets: A (B + C) = AB + AC and if the second term in the brackets is negative: A (B − C) = AB − AC

Mathspace WA Curriculum – Year 9 mathspace.co

3.02 Practice questions Understanding

Explain why 2(a − 3) is the same as (a − 3) × 2.

Insert (=) or (≠) in each statement.

3a(2a + b) ⬚ 3a × 2a + 3a × b

6m2 v(3 − 2m2 + v) ⬚ 6m2 v × 3 − 6m2 v × 2m2 + 6m2 v × v

4ab(3b + 2c − 6a) ⬚ 4ab × 3 + 4ab × c − 4ab × 6 4, 5a–i, 6a–ce, 7ac, 8a–df

Fluency 4

c 5

3 (2x + 4) = ⬚x + 12

4cd, 5h–p, 6d–h, 7cd, 8g–n

2 (−4s + 5) = ⬚s + ⬚

⬚ (3g − ⬚) = −6g + 14

Expand: a

a(b + 7)

m(m + 7)

y(1 + y)

4 (m2 + 10)

m(m − 3)

y(4y − 10)

m(8 − m)

5t(t + 14)

2b(2b + 16)

−5a(3a − 4) 3

−w(w + 1)

−m(11n + 19p)

7s (8t − 9)

6wy( y + w)

(5y − 6) (−3y2)

6a4 (3a6 + 6a4)

3u4 (8u + 10u3)

Expand: a

7y( y2 + 4)

−3x2 (5x − 2x4)

Expand: a

5 (⬚p + ⬚) = 15p − 30

m −4t(4t + 3) 6

4bc, 5d–l, 6abef, 7bc, 8d–jm

Complete the statements: a

Ex 2

Trailblazer 1–3

Ludivina states that (a3)(a3) = a9. Dortha states that (a3)(a3) = a6. Who is correct and explain why.

Ex 1b

Adventurer 1–3

Ex 1a

Explorer 1–3a

2f ( f − 8g − 7h)

6a(a − 5b + 3c)

(5x + 9y − 2) (4x)

2a2 (a2 + 4a − 7)

Expand and simplify: a

8 ( y + 6) + 3 ( y + 9)

h(h + 5) + 9 (h + 9)

x(x + 4) + 2 (x + 5)

3c(9c2 + 2c + 7)

4c(8c2 − 3c − 5)

8y2 − 7y + 9y( y − 5)

4y − 9y + 6y( y − 8)

5 (9x − 6y) − 3(8y + 2x)

9x(6x + 8y) − 7x(2y + 4x)

2 (4 (x − 3) + 5)

3(2(x – 4) + 5)

6u(5u + 8v) − 7u(4v + 2u)

3.02 Distributive law mathspace.co

Reasoning 9

9, 10

Explain the errors made by each student. i

Leon: −24x − 25

Jill: −2x − 25

iii

Claire: 24x − 150

Beryl: −24x − 150

What is the correct answer?

There are three consecutive integers. a

If x is the ﬁrst and smallest of the integers, write an expression for the: i

9, 11

Four students gave incorrect answers when expanding −6(4x − 25). a

9, 11

Second integer

Third integer

The sum of the ﬁrst, and two times the second is 15 more than two times the third. Write an equation for this.

What are the three integers?

Are there any other sets of three consecutive integers that satisfy the equation in part (b)? Explain how you know.

The perimeter of a circle is 3π + 2. Find and expand an expression that represents the perimeter of 5 of these circles.

Problem-solving 13–15 12

2 (x − 12) + 9 = 3x − ⬚

Write two or more expressions for the area:

By expanding each of the expressions given in part (a), show that they are equivalent and hence come up with a single expression for the area. Explain why x cannot be negative.

x+7

x+1

2 x

A rectangular garden has a length that is one metre less than twice the width (w). a

Write and expand an expression for the area of the rectangle.

When a border 2 metres wide is added around the garden, the total area (garden and border) is 2w2 + 7w + 12. What is the area of the border only?

The landscaper designing the garden has 100 square metres of pavers to use for the border. Determine the largest garden the landscaper can border before running out of pavers.

Expand and simplify −2a(a3 − 4a + 2) + 5a

The rule for the distributive law is that A(B + C) = AB + AC.

13–17

Complete the equation so that x = −7 is the only solution:

12–15

Write A(B + C) if A = x + 1, B = 2x and C = 3.

What is the expansion for A(B + C)?

Mathspace WA Curriculum – Year 9 mathspace.co

A rectangular prism has dimensions of width = x, length = 2x − 1 and height = 6x. a

Write an expanded expression for the surface area.

If materials needed to make the prism dictate that the width could be between 1 cm and 10 cm, ﬁnd the range of the possible surface area.

3.03 Binomial expansion After this lesson, you will be able to… • apply the distributive law for multiplying binomials, even when variables are involved. • recognise the pattern for squaring a binomial.

Binomial expansion A binomial is an expression with two non-like terms that are being added or subtracted. For example: • 3x + 2 • −x2 + 9 • 10xy2 − 140xyz A C

B BC

+ D

x x

x × x = x2

5 × x = 5x

+ 3

Consider the expression ( A + B) (C + D). We can represent it visually using this area model. This diagram shows that ( A + B) (C + D) = AC + AD + BC + BD. In other words, the area of the whole shape is the sum of the smaller areas.

Let’s look at using this model to expand (x + 3) (x + 5). Here we can see that: (x + 3) (x + 5) = x2 + 3x + 5x + 15 We can simplify this to:

3 × x = 3x

5 × 3 = 15

x2 + 8x + 15

3.03 Binomial expansion mathspace.co

Exploration Complete the area models for multiplication shown: 7

7x3

3x2

3x2(7x3 + 9)

3(7 + 9) Complete the new area models for multiplication:

x 20

−3

(10 − 3) (20 + 1)

(x − 3) (2x + 1)

What do the area models have in common?

What’s different about the area models?

How do you think multiplying binomials can be related to multiplying integers?

Example 1 Expand and simplify (x + 1) (x + 3)

Create a strategy Use the rule ( A + B) (C + D) = AC + AD + BC + BD.

Apply the idea (x + 1) (x + 3) = x × x + x × 3 + 1 × x + 1 × 3

Apply the rule

Simplify the products

Combine like terms

= x + 3x + x + 3 = x + 4x + 3

Idea summary We can expand the product of two binomial expressions using the rule: ( A + B) (C + D) = AC + AD + BC + BD

Mathspace WA Curriculum – Year 9 mathspace.co

Square of a binomial Sometimes binomials such as (a + b) (a + b) are written as (a + b)2. If we need to expand binomials like this, we can still follow the same structure. (a + b)2 = (a + b) (a + b) 2

= a + ab + ab + b 2

= a + 2ab + b

Write as binomial product 2

Binomial expansion

Simplify

Notice the middle term. It is 2ab, which is twice the product of a and b.

Example 2 Expand (x + 9)2

Create a strategy Use the pattern for squaring a binomial.

Apply the idea (x + 9)2 = (x + 9) (x + 9)

Write as a binomial product

Binomial expansion

Simplify

= x + 9x + 9x + 81 = x + 18x + 81

Example 3 Expand (5x − 2y)2

Create a strategy Use the pattern for squaring a binomial.

Apply the idea (5x − 2y)2 = (5x − 2y) (5x − 2y) 2

Write as a binomial product 2

= 25x − 10xy − 10xy + 4y 2

= 25x − 20xy + 4y

Binomial expansion Simplify

Idea summary Squaring a binomial means multiplying a binomial by itself. (a + b)2 = (a + b) (a + b) (a − b)2 = (a − b) (a − b) = a2 + ab + ab + b2 = a2 − ab − ab + b2 2 2 = a + 2ab + b = a2 − 2ab + b2

3.03 Binomial expansion mathspace.co

3.03 Practice questions Understanding 1

(8)2

(−5)2

(3x)2

(−12y)2

y( y − 5) − 8 ( y − 5)

5 ( y − 8) − ( y − 5)

y( y − 5) + y − 8

y( y − 8) − 5 ( y − 8)

The area of this ﬁgure is represented by the expression ( y + 3) ( y + 6).

Ex 2

Find the area of: i

Rectangle A

Rectangle B

iii

Rectangle C

Rectangle D

Write an equivalent, simpliﬁed expression for the area of the ﬁgure.

Is each statement true or false? a

(3x + 5)2 = 9x2 + 25

After completing a binomial expansion, adding products in different orders will always give the same answer.

The product (3x − 8) (3x + 8) will have two terms.

Fluency 5

Trailblazer 1cd, 2–4

Are the expressions equivalent to ( y − 5) ( y − 8)?

Ex 1

Adventurer 1cd, 2–4

Simplify:

Explorer 1–4

5a–cg, 6a–d, 7a–d, 8a–c, 9a–d

5c–g, 6c–h, 7b–e, 8c–e, 9b–e

5e–j, 6e–j, 7e–h, 8c–h, 9d–h

Expand and simplify: a

(x + 6) (x − 12)

(x + 2) (x + 6)

(v + 11) (v + 10)

(x − 4) (x − 7)

(m − 5) (m + 6)

(10m + 9) (m − 7)

(3w + 5) (7w + 6)

(3y − 3) (10y + 9)

(7y + 9) (−4y − 4)

Expand and simplify: a

(a + 3) (a − 3)

(x + 2) (x − 2)

( y + 5) ( y − 5)

(2x + 5) (2x − 5)

(4x + 6) (4x − 6)

(5x + 2) (5x − 2)

(3x + 7) (3x − 7)

(3y + 4) (3y − 4)

(9m + 8) (9m − 8)

(12x + 1) (12x − 1) (x − 3)2

(x − 5)2

Expand and simplify: a

(x + 7)2

(x + 4)2

(x + 6)2

(x − 2)2

Mathspace WA Curriculum – Year 9 mathspace.co

Ex 3

Expand and simplify: a

(3a + 4b)2

(2a − 3)2

(4m − 5n)2

(x − 2y)2

(2x − 7y)2

(6p + 3q)2

Expand: a c

(a + 6) (b − 12) 2

(q − 7) (t + 9) 2

(k − 2r) (k + 5)

(x + 5y) (x2 + y)

(3m − 4p) (m − p)

(4g2 + 3h) (6g + 5h3)

(3a3 − b2) (7a − 6b2)

(5x2 y + 4x) (7y3 − 12x5)

10–13

10–14

Reasoning

10–13

Explain why (x + 2)(x + 2) has three terms when expanded but (x − 2)(x + 2) has only two terms when expanded.

Marvin has a square bedroom. His sister Jane has a room that is 1 m shorter in length than Marvin’s room, but 1 m wider. a

Show that Marvin has the larger bedroom.

How much bigger is Marvin’s bedroom compared to Jane’s bedroom?

Are (x − 4)2 and (4 − x)2 equivalent expressions? Explain your answer.

Rahul attempts to expand the expression (x + 9)2 and gets a result of x2 + 81. Is his answer correct? Explain where you think Rahul has gone wrong and identify the correct expansion.

Anette decides to use the area model to expand and simplify (10x + 2) (10x + 3). To warm up, she decides to try a simple integer example of (10 + 2) (10 + 3): 10

100

100 + 20 + 30 + 6 = 156 She then tried this method on (10x + 2) (10x + 3) 10x

10x

100x

20x

30x

100x + 20x + 30x + 6 = 150x + 6 Find the error in Anette’s answer. Correct her work and solution. 3.03 Binomial expansion mathspace.co

Problem-solving 15a–d, 16a–d, 17ab, 19

15c–f, 16c–f, 17bc, 18, 20, 21

15e–h, 16e–h, 17bd, 18, 20, 21

Expand and simplify:

3 ( y + 3) ( y + 5)

8 (x − 2) (x − 6)

−(x + 5) (x + 6)

−9 ( y − 7) ( y + 6)

3 (−4x + 3) (4x − 3) g

4 (6x − 2) (x + 2)

−4 ( y + 1) ( y + 9)

Complete by filling in the blank:

(t + 8) (⬚) = t2 + 10t + 16 2

(t + 6) (⬚) = t2 + 17t + 66

(t − 2) (⬚) = t − 7t + 10

(t − 2) (⬚) = t2 + 4t − 12

(t − 5) (⬚) = t2 + 7t − 60

(v − 2) (v − 3) = v(v − 3) − ⬚ (⬚)

(t + 2) (t + 9) = t(t + 9) + ⬚ (⬚)

Expand:

(t − 2) (t + 8) = t(t + 8) − ⬚ (⬚)

(x + 3)3

(2x − 4)3

(2x + 3) (x + 5y − 4z)

(t + 3) (2t − 1) (3t + 5)

A square has a perimeter of 8x + 20. What is its area?

Find the area of a rectangle which has a length of (x + 4) cm and a width of (x + 3) cm.

What expression, when multiplied by n + 6, gives n2 + 14n + 48?

Consider the rectangular prism shown with its dimensions labelled. Write an expression for the volume, and then expand and fully simplify the result.

(x + 7) m (x + 5) m 4m

3.04 Algebraic factors After this lesson, you will be able to… • identify common factors in algebraic expressions. • rewrite algebraic expressions in their factorised form. • apply factorisation techniques to simplify algebraic expressions.

Factorise algebraic expressions Factorising can be thought of as being the reverse process to expanding. Expanding would be: 6(x + 3) = 6x + 18 Factorised

Mathspace WA Curriculum – Year 9 mathspace.co

Expanded

Factorising would be putting the expanded form into brackets by taking out the highest common factor (HCF): 6x + 18 = 6(x + 3) We can view this as the opposite of the distributive law: AB + AC = A(B + C) A is the highest common factor, which is taken out. This extends to three or more terms: AB + AC + AD = A(B + C + D) In any algebraic term, such as 12ab or 9ac, there can be numeric and algebraic components. When factorising we need to ﬁnd the highest common numeric factor and the highest common algebraic factor. In the case of 12ab and 9ac: • the highest common numeric factor is 3 • the highest common algebraic factor is a We can see this when we write the terms out in an expanded way: 12ab = 2 × 2 × 3 × a × b 9ac = 3 × 3 × a × c By identifying all the factors that are common to both terms we can see that the highest common factor of 12ab and 9ac is 3a.

Exploration The expanded form for 5x3 y4 z is 5 × x × x × x × y × y × y × y × z. The expanded form for 2y3 z5 is 2 × y × y × y × z × z × z × z × z. 1.

How many y’s are common in both expressions?

How many z’s are common in both expressions?

What is 5x3 y4 z + 2y3 z5 in factorised form?

What can we do to check our answer?

We can write 5x3 y4 z + 2y3 z5 = y3 z(5x3 y + 2z4) Expanded Factorised

To verify if you have factorised correctly, you can expand the result and check if it matches the original equation.

3.04 Algebraic factors mathspace.co

Example 1 Find the highest common factor for the following pairs of terms: a 10xy and 15yz

Create a strategy

Apply the idea

Break down the terms into factors.

10xy = 2 × 5 × x × y Rewrite 10xy 15yz = 3 × 5 × y × z Rewrite 15yz HCF = 5 × y = 5y

Select the common factors Simplify

b 6a2 b and 21ab2

Create a strategy Break down the terms into factors.

Apply the idea 6a2 b = 2 × 3 × a × a × b

Rewrite 6a2 b

21ab2 = 3 × 7 × a × b × b

Rewrite 21ab2

HCF = 3 × a × b

Select the common factors

= 3ab

Simplify

Example 2 Factorise −2s − 10

Create a strategy Check all the terms for common numeric and algebraic factors. Factorise using AB + AC = A (B + C).

Apply the idea There are no common algebraic factors in −2s and −10, but there is a common numeric factor, −2 −2s − 10 = −2 × s + (−2) × 5 = −2 (s + 5)

Write the terms as products of −2 Factorise

Reﬂect and check It is important to be careful about factorising a negative because the sign of each term will change. If the leading term in an expression is negative, then we should factorise the negative out.

Mathspace WA Curriculum – Year 9 mathspace.co

Example 3 Factorise 42x − x2

Create a strategy Check all the terms for common numeric and algebraic factors. Factorise using AB + AC = A (B + C).

Apply the idea There are no common numeric factors in 42x and − x2 but there is a common algebraic factor, x. 42x − x2 = x × 42 − x × x = x (42 − x)

Write the terms as products of x Factorise

Reﬂect and check Let’s check our factorisation of x(42 − x), by expanding it. x(42 − x) = x × 42 − x × x x(42 − x) = 42x − x2 This is the same as the original expression, so we know that we have factorised this correctly.

Example 4 Factorise 2ab2 + 16ab

Create a strategy Check all the terms for common numeric and algebraic factors. Factorise using AB + AC = A (B + C).

Apply the idea 2ab2 and 16ab have a common numeric factor of 2 and a common algebraic factor of ab. 2ab2 + 16ab = 2ab × b + 2ab × 8 = 2ab (b + 8)

Write the terms as products of 2ab Factorise

Idea summary We can factorise expressions by writing the highest common factor of the terms outside of the brackets: AB + AC = A(B + C) Factorising is the reverse of expanding.

3.04 Algebraic factors mathspace.co

3.04 Practice questions Explorer 1a–d, 2, 3a–c

Understanding 1

Trailblazer 1e–h, 2, 3

Find the highest common numeric factor of each pair of terms: a

8ab and 6cd

10x2 and 25yz

12j2 k and 16x2 y

18xyz2 and 6ab2 c

24a n y and 8b m h

36x7 y and 6a7 z

27ax2 z and 18pr2 k

100z10 a8 k5 and 20j10 b8 n5

2 3 6

2 6

Expand each expression to see which ones are a factorised form of −12x + 20x2: a

Adventurer 1d–f, 2, 3b–d

−4 × 3x + 5

−2x (6 − 10x)

4x (5 − 3x)

x(−12 + 20x)

All of these terms have a common factor of 5b. Write out the terms as a product of their factors, cross out the 5b that is common and determine what is remaining for each. a

10ab

15a2 b

5ab2

120a2 b2

8my and 12am

9bxm and 5bxa

2t2 + 2t = 2t (⬚ + ⬚)

Complete: 2

3y + 6y = ⬚ ( y + 2)

−y2 − 2y = ⬚ ( y + 2)

−m2 + 19m = ⬚ (m − 19)

8v − v2 = v (⬚ − ⬚)

4a + 12

5a − 10

15 + 20a

−18a + 16

4ab + 6a

11 km − mn

−12s + 10

8hj − 9gh

16jkt − 5kt

−35m + 49n

4 − 6a

−5c − 30

−8 − 12d

−w − 6w

−45mnq − 72mq

3x − x2

z2 + 6z4

−8w2 + 3w2 y

−63r2 + 28

−16a2 − 18a2 b

Factorise: −3a − 6 −4b − 5ab

−abc − 5a

k2 − 9k

5u2 − 15u

Factorise: a

y2 + 4y

z2 + 8z6 2

2 2

44uv − 8u v

5k t + 40k t

m −30w2 − 25w2 y

−10u2 v + 9uv2

y2 + 5y = y (⬚ + ⬚)

Factorise:

e 8

9xn and 4ax 13y2 an and 17my2 b

Ex 4

b d

Ex 3

9x and 8x 18b n and 7b m

5a2 b

4bdf, 5ef, 6g–j, 7c–h, 8f–n, 9f–h

Ex 2

Find the highest common algebraic factor of each term: a

5ab

4a–c, 5, 6a–e, 7a–f, 8a–f, 4ace, 5d–f, 6e–h, 7b–g, 9a–c 8d–j, 9d–f

Fluency Ex 1

Mathspace WA Curriculum – Year 9 mathspace.co

49p q − 28pq

Factorise: a

4x + 12 + 16yx 2

3x + 9 + 12yx

2x + 10x y + 8yx

4x2 y + 8xy2 + 12xz2

5ab2 c + 25bc3 + 100abc

30b2 c + 10abc + 20c2

5a2 b2 + 2ab − 3a2 b2 − 4ab

5x2 y2 + 15yz2 + 25y2 + 60y

Reasoning 10

10, 11

10–12

A student incorrectly factorised and wrote −5x2 − 15x = −5x (x − 3). Explain how to correct the error, and provide the correct answer.

Alex and Beth are both asked to factorise −5x + 10y. Alex wrote down −5(x − 2y) and Beth wrote down 5(2y − x). Who is correct? Explain.

Louis was asked to fully factorise 2x3 + 8x2 − 4x and gave as his answer 2(x3 + 4x2 − 2x). What has he done wrong?

Problem-solving 13–15 13

13, 14, 16, 18

14, 16–18

The rectangle shown has an area of 4x2 − 16x square units. What expression describes the length of the rectangle?

? 4x

A property management company has an area of land, 20n + 5n2 available for parking. When factorised, this represents the length and width of the block of land. a

Factorise the expression to determine the dimensions of the block.

Given that the length is longer than the width for all n > 1, which term is the length and which is the width?

The length is further divided into 5 equal sections, and one of these sections is fenced off for storage. What is the length of fencing needed?

You are to design a photo collage made of two large square photos with a side length of x and four smaller rectangular photos that have a height of x and a width of 4 units. a

Draw an example of what this arrangement would look like.

Find an algebraic expression for the area of the rectangle formed if the photos are all placed in a single row.

Fully factorise your answer from part (a) and then draw a different photo arrangement that would match these dimensions.

A rectangle has a width of 2xy cm and a length of 2(x − y) cm. a

Write an expression for the area of the rectangle, and expand it.

Write an expression for the length of a rectangle with the same area if its width is 4xy cm.

Write a factorised expression for the length of a rectangle with the same area if its width is: i

4 cm

xy cm

3.04 Algebraic factors mathspace.co

A container in the shape of a rectangular prism has a length of (2a − b) metres, a width of 6ab metres, and a height of 5ab metres. a

Write an expression for the volume of the prism, and expand it.

Write a factorised expression for the volume of a container that has twice the volume of the rectangular prism.

What percentage of the rectangular prism has been ﬁlled if water is poured in to a depth of 2ab m?

Write an expression for the percentage of the rectangular prism that has been ﬁlled if water is poured in to a depth of 3a metres.

Factorise 3t (t − 5) + 4 (t − 5)

3.05 Factorise by grouping After this lesson, you will be able to… • apply factorising by grouping in algebraic expressions. • identify the characteristics of a monic quadratic trinomial and the steps to factorise it. • apply factorising techniques to solve problems involving algebraic expressions with three or more terms.

Factorise by grouping Common factors can be more than 1 term. For example, 5 (a + 7) and b (a + 7) have a common factor of (a + 7). So, 5 (a + 7) + b(a + 7) = (5 + b) (a + 7) For expressions with 4 terms, sometimes separating them into 2 groups can help fully factorise the expression.

Mathspace WA Curriculum – Year 9 mathspace.co

Example 1 Factorise: 3x + 6 + 2x2 + 4x

Create a strategy Using inspection, determine the 2 groups that will likely end up with a common factor.

Apply the idea 3x + 6 + 2x2 + 4x = (3x + 6) + (2x2 + 4x) = 3 (x + 2) + 2x (x + 2)

Separate into two groups Factorise each group

= (x + 2) (3 + 2x) Identify (x + 2) as the common factor and ﬁnish the factorisation

Reﬂect and check There may be multiple ways to separate the original expression into two groups that share a common factor. For example, we could have factorised the expression this way: 3x + 6 + 2x2 + 4x = 2x2 + 3x + 4x + 6

Rearrange the terms

= (2x2 + 3x) + (4x + 6)

Separate into two groups

= x (2x + 3) + 2 (2x + 3)

Factorise each group

= (2x + 3) (x + 2) Identify (2x + 3) as the common factor and ﬁnish the factorisation

Notice that this is the same result we got before, the terms are simply rearranged.

Example 2 Fully factorise: 5(a + b) + v(a + b)

Create a strategy Factorise using the law, AB + AC = A(B + C).

Apply the idea Both terms have a common factor of (a + b) so we can factorise this out. 5 (a + b) + v (a + b) = (a + b) (5 + v)

Factorise

3.05 Factorise by grouping mathspace.co

Example 3 Factorise: 2x + xz − 40y − 20yz

Create a strategy Group the expression in pairs and factorise using the law, A (C + D) + B (C + D) = ( A + B) (C + D).

Apply the idea To factorise an expression with 4 terms, we factorise two pairs of terms, then take out the common factor binomial. 2x + xz − 40y − 20yz = (2x + xz) + (−40y − 20yz)

Group terms with common factors

= x (2 + z) − 20y (2 + z)

Factorise each pair

= (2 + z) (x − 20y)

Factorise

Idea summary When we factorise expressions with 4 terms, we may need to break the term into 2 groups and factorise each group individually. If there is a common factor between these two groups, we may need to factorise further.

Monic quadratic trinomials Interactive exploration

Discover this concept in action online

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A monic quadratic trinomial is in the form x2 + bx + c. It has three features: • A trinomial has 3 terms. • A quadratic has a highest power of 2. • A monic quadratic is one where the coefficient of x2 is 1. We factorise monic quadratic trinomials by ﬁnding the factors of the constant, c, that also add up to the coeﬃcient of x, b.

Mathspace WA Curriculum – Year 9 mathspace.co

Example 4 Factorise: x2 + 6x + 8

Create a strategy The constant is 8 and the coeﬃcient of x is 6. We need to ﬁnd two factors of 8 that add to 6.

Apply the idea The factors of 8 that add to 6 are 2 and 4. x2 + 6x + 8 = x2 + 4x + 2x + 8

Rewrite the 6x term as 4x + 2x

= (x + 4x) + (2x + 8) Separate into two groups = x (x + 4) + 2 (x + 4)

Factorise each group

= (x + 2) (x + 4)

Factorise using the common factor of (x + 4)

Idea summary When factorising a monic quadratic trinomial in the form ax2 + bx + c, we: 1. Find the factors of c ( p and q) that add up to b 2. Rewrite the middle term in terms of the factors: ax2 + px + qx + c, where px + qx = bx 3. Factorise by grouping.

3.05 Practice questions Understanding

Explorer 1–3

Adventurer

Trailblazer

1–3

For the expression 3(x + 7) + x(x + 7), what is the common factor?

Complete this statement: To factorise x2 + 9x + 18, we need to ﬁnd two numbers whose product is ⬚ and whose sum is ⬚.

Given that p < q, ﬁnd the values of p and q in each pair of equations: a

p + q = 19

p + q = −2

p + q = −10

pq = 90 pq = −63 pq = 24

3.05 Factorise by grouping mathspace.co

Fluency Ex 1

Ex 3

x2 + 5x + 6

x2 + 7x + 10

x2 + 6x + 9

x2 + 8x + 12

x2 − 3x + 2

x2 − 6x + 8

x2 − 5x + 6

x2 − 6x + 9

x + 11x + 24

x + 17x + 72

x − 4x + 3

x2 − 17x + 60

m x2 − x − 30

x2 + x − 20

x2 − 3x − 54

x2 − 3x − 70

−8 − 6x − x

−12 + 7x − x2

40 + 13x + x

35 − 12x + x

Factorise: a

y( y + 5) + 7 ( y + 5)

a (a − 4) − 3 (a − 4)

p( p − 3) + 6 ( p − 3)

5 (q + 4) − q (q + 4)

5 (r − 3) − r (r − 3)

6r (2r − s) − rs (2r − s)

7t (t + u) + 2u (t + u)

x ( y − z) − w ( y − z)

5y (4w + 3x) − z (4w + 3x)

8y ( y − 4) + 3 (4 − y)

Factorise: a

8x + xz − 16y − 2yz

24 + 3y + 8x + xy

7xy + wx + 7yz + wz

x2 + 2x + 5x + 10

x2 − 3x + 8x − 24

50 + 5y + 10x + xy

12xy + wx + 12yz + wz

x2 + 5x + 8x + 40

x2 − 5x + 10x − 50

2mp + 6 + 3p + 4m

5mp + 6 + 2p + 15m

2x + 18yz + 12xy + 3z

m 4x + 24yz + 32xy + 3z

a3 + 8a2 + a + 8

x2 − 4x + 4

Factorise: a e

x2 + 8x + 16 64 + 16x + x

Reasoning 8

4i–t, 5e–j, 6g–n, 7bdf

Factorise:

q Ex 2

4e–p, 5c–g, 6e–k, 7bdf

4a–h, 5a–d, 6a–f, 7ace

x2 + 10x + 25

x2 − 20x + 100

36 − 12x + x

8, 9

Identify and explain the error: 5xy − 5xz + 7z − 7y = 5x ( y − z) + 7 (z − y) = (5x + 7) ( y − z)

What is the largest possible integer value of k that will allow m2 + km + 24 to be factorised using whole numbers? Justify your answer.

Problem-solving 10a–d, 13 10

10e–h, 11, 12cd, 13, 14

Factorise each expression by ﬁrst factoring out a numeric factor: a

2x2 + 14x + 20

4x2 + 24x + 32

3x2 − 21x + 30

−2x2 − 18x + 44

−3x2 + 12x − 12

4x2 − 4x − 288

5x2 + 5x − 30

−4x2 + 12x + 40

2x3 + 16x2 + 30x

Factorise: a

10b–e, 11, 12ab, 13

x3 + 7x2 + 12x

Mathspace WA Curriculum – Year 9 mathspace.co

Factorise: a c

3f ( g + h) + (g + h)

3x2 (x + 4y) − 5y(x + 4y)

(2c − d) (c + 5d) − 3 (d − 2c)

Find the perimeter of the square.

A square has area t2 + 14t + 49. a

2y(2x2 + 3z) − (2x2 + 3z)

Find the side length of the square.

Find an expression for the shaded area in the ﬁgure in terms of x. Write your answer in factorised form. x

3 x+1 4

Did you know?

Algebra is the backbone of roller coaster design! Engineers use algebra to calculate the perfect curves, drops, and loops that create thrilling, yet safe, rides. By solving equations that involve speed, acceleration, and force, they ensure that the coaster delivers excitement without exceeding safety limits. Every twist and turn is carefully calculated to ensure that the G-forces feel exciting but not overwhelming for riders. Even the timing of the ride, from the initial climb to the final stop, is determined using algebraic models. The result is a smooth, exhilarating ride that feels like flying! This precise maths allows roller coasters to provide an adrenaline rush while keeping everyone safe. Without algebra, designing such complex, high-speed rides would be impossible. Algebra ensures the balance between excitement and safety on every thrilling roller coaster ride!

3.05 Factorise by grouping mathspace.co

3.06 Optional: Add and subtract fractions with numeric denominators After this lesson, you will be able to… • add and subtract fractions with algebraic numerators by finding a common denominator.

Add and subtract fractions with numeric denominators An algebraic fraction is a fraction with a pronumeral in the numerator or denominator. Adding and subtracting algebraic fractions with numeric denominators is very similar to adding and subtracting purely numeric fractions. We ﬁrst need to ﬁnd equivalent fractions with a common denominator, then we add or subtract the numerators. We may be able to simplify our answer by collecting like terms, and/or dividing the numerator and denominator by common factors.

Exploration Notice that

1. Write another equivalent fraction with a denominator of 12. 2. Write a fraction equivalent to

with a denominator of 15. Explain how you got this.

3. What would be a common denominator for

and

Example 1 Simplify:

Create a strategy The fractions already have a common denominator, so we can subtract the numerators. Simplify by collecting like terms and dividing the numerator and denominator by any common factors.

Mathspace WA Curriculum – Year 9 mathspace.co

Apply the idea

Subtract the numerators

Simplify the numerator

Simplify the fraction

Reﬂect and check To check our answer, substitute x = 1 into both the original expression and the simpliﬁed expression. We could use any value of x, but 1 is the simplest. Substitute x = 1 into the original expression:

Substitute x = 1

Simplify the numerators

Perform the subtraction

Evaluate Substitute x = 1 into the simplified expression: −x = −1

Substitute x = 1

Both the original expression and the simpliﬁed expression gave the same result, suggesting that the answer is correct.

Example 2 Simplify:

Create a strategy Rewrite the fractions with a common denominator, then add the numerators.

Apply the idea A common multiple of 3 and 5 is 15.

Rewrite the fractions to have a common denominator

Evaluate the products

Add the numerators

Collect like terms

3.06 Optional: Add and subtract fractions with numeric denominators mathspace.co

Example 3 Simplify: 8x −

Create a strategy Rewrite the ﬁrst term as a fraction with the same denominator as the second term, then subtract the numerators.

Apply the idea Write 8x as a fraction

Rewrite the ﬁrst fraction to have a denominator of 3

Evaluate the products

Subtract the numerators

Collect like terms

Reflect and check To check our answer, substitute x = 1 into both the original expression and the simplified expression. Substitute x = 1 into the simplified expression: Substitute x = 1

Evaluate the products

Rewrite the fractions with a common denominator

Subtract the numerators

Evaluate the difference

Substitute x = 1 into the simpliﬁed expression: Substitute x = 1

Evaluate the product

Both the original expression and the simpliﬁed expression gave the same result, suggesting that the answer is correct.

Mathspace WA Curriculum – Year 9 mathspace.co

Idea summary To add or subtract fractions where the numerators are algebraic terms: 1. Rewrite the fractions with a common denominator. 2. Add or subtract the numerators. 3. Simplify by collecting like terms and dividing the numerator and denominator by any common factors.

3.06 Practice questions Explorer

Understanding 1

and

For a

and

simpliﬁes to The expression

is equivalent to 2z.

4, 5a–e, 6a–e, 7a–c, 8ad

Fluency

Ex 1

and

Are these statements true or false? a

1–3

State a common denominator for these pairs of algebraic fractions: a

Trailblazer

1–3

Evaluate and simplify: a

Adventurer

1–3

and

simpliﬁes to .

b d

The expression

4, 5e–i, 6e–i, 7c–g, 8b–g

is equivalent to 3a. 4, 5g–l, 6h–l, 7e–h, 8c–h

Find a common denominator.

Simplify

Simplify: a

3.06 Optional: Add and subtract fractions with numeric denominators mathspace.co

Ex 2

Ex 3

Simplify: a

Simplify:

Simplify each expression, by ﬁrst simplifying the fractions (where possible). a

Reasoning

9, 12

9–12

10–12

Using mathematical vocabulary, explain how you can complete the equation.

Complete the equation:

Why is it important to simplify fractions before adding or subtracting them in algebraic equations?

Can we add or subtract fractions with different denominators directly? Explain your answer.

Problem-solving 13ac, 14a, 15, 17 13

Fully simplify these expressions: a

13acd, 14–17

Fully simplify these expressions: a

13a–c, 14ab, 15, 16a, 17

Mathspace WA Curriculum – Year 9 mathspace.co

Write an expression for the perimeter of the triangle shown, and simplify this expression. x 5

3x 10

Given y1 =

and y2 = − :

Find y3 if y3 = y1 + y2.

Sketch the graph of y3 on the Cartesian plane.

The fuel tank in Dario’s car holds p litres of petrol. After ﬁlling the tank, Dario uses tank driving to his friend’s house, then another

of the

of the tank driving to the beach. Write an

expression for number of litres of fuel left in the tank when Dario arrives at the beach, and simplify this expression.

3.07 Optional: Multiply and divide algebraic fractions with numeric denominators After this lesson, you will be able to… • apply the multiplication and division rules to algebraic fractions with numeric denominators. • use indices to simplify numerators and denominators in algebraic fractions. • identify and cancel common factors in the numerator and denominator.

Multiply fractions with numeric denominators We have multiplied numerical fractions in the past. When multiplying fractions, we multiply the numerators and multiply the denominators:

For example:

3.07 Optional: Multiply and divide algebraic fractions with numeric denominators mathspace.co

When it comes to working with algebraic fractions with numeric denominators and applying the four operations, the process is exactly the same as when we worked with numeric fractions. Let’s have a look at a simple example of multiplying two algebraic fractions with numerical denominators. We may need to use indices to simplify the numerators. Remember that: a × a = a2 and in general: am × an = am + n We should check if the ﬁnal answer can be simpliﬁed by removing a common factor in the numerator and denominator using that

= 1, a ≠ 0.

Example 1 Simplify

Create a strategy Multiply the numerators together and multiply the denominators together.

Apply the idea Multiply the numerators and denominators

Evaluate

Example 2 Simplify

Create a strategy Multiply the numerators together and multiply the denominators together.

Apply the idea Multiply the numerators and denominators

Evaluate

Simplify

Mathspace WA Curriculum – Year 9 mathspace.co

Reﬂect and check We could also have removed any common factors before we performed the multiplication:

Factorise the denominators

Remove the common factors

Simplify

Idea summary The product of two fractions is the product of their numerators divided by the product of their denominators. For B, D ≠ 0:

We may need to use indices to simplify the numerators. Remember that: a × a = a2 and in general am × an = am + n Before multiplying two fractions together, look for common factors that we can remove ﬁrst. This will make the resulting multiplication easier in most cases, as there will be less factors to deal with. We can use that = 1.

Divide fractions with numeric denominators When we divide fractions, we take the reciprocal of the second fraction and multiply by it. Sometimes we say ﬂip or invert when we take the reciprocal. Multiply by the reciprocal

Evaluate

The process for dividing is the same as when we divide numerical fractions, but we may need to use indices.

3.07 Optional: Multiply and divide algebraic fractions with numeric denominators mathspace.co

Example 3 Simplify

Create a strategy Multiply the ﬁrst fraction by the reciprocal of the second fraction.

Apply the idea

Multiply by the reciprocal Remove the common factor

Reﬂect and check Notice that if common factors are not removed, we would end up with very large numbers which would then need to simplify.

Multiply by the reciprocal Multiply the numerators and denominators

Simplify

Idea summary To divide by a fraction, multiply by the reciprocal. For B, C, D ≠ 0:

Simplify if possible by removing common numerical or algebraic factors.

3.07 Practice questions Understanding 1

Adventurer 1, 2, 3bc

Trailblazer 1, 2, 3bd

Evaluate: a

Explorer 1, 2, 3ab

Mathspace WA Curriculum – Year 9 mathspace.co

Are these statements true or false? a

To multiply two algebraic fractions, we multiply the numerators and multiply the denominators.

When multiplying or dividing algebraic fractions with numeric denominators, the denominators must be the same before performing the operation.

To multiply or divide algebraic fractions, you must ﬁrst ﬁnd a common denominator.

The order of operations still applies when multiplying or dividing algebraic fractions with numeric denominators.

Complete each simpliﬁcation. a

Fluency Ex 1

5 Ex 2

6 Ex 3

4a–d, 5a–d, 6a–d

4d–g, 5d–h, 6d–h

4d–h, 5g–l, 6g–l

Simplify: a

Simplify:

3.07 Optional: Multiply and divide algebraic fractions with numeric denominators mathspace.co

Reasoning 7

7, 9

8, 9

The following expression was simpliﬁed incorrectly:

a 8

Explain the mistake.

Amarion and Dean both simpliﬁed

Give the correct solution.

differently than the others in their group. Their

ﬁrst two steps are shown. Who could go on to be correct? Explain your answer. Amarion:

Dean:

Is each statement always true, sometimes true, or never true? Justify your answer. a

The quotient of two algebraic fractions is a numeric fraction.

The product of two numeric fractions is an algebraic fraction.

Problem-solving 10a, 11ab, 12a 10

. Find an expression

Simplify: a

10d, 11cd, 12cd, 13, 14

Simplify: a

10bc, 11bc, 12b, 13

Simplify: a

and

. Find an expression for

Investigation: Calendar expressions

Investigate online

Mathspace WA Curriculum – Year 9 mathspace.co

mathspace.co

3 Chapter review 1

Simplify: a

11y − 6y

7a + 2a − a

3x + 3y − (−x) + 4y

a2 + 5a − 4a + (−11a2)

Simplify: a

2r × 3s

6u2 × 7v4

5x × 8y2 × 2z2

−c8 d2 × 3c6 d × 2d2

Simplify: a

5b2 ÷ 25b

−(8xy) ÷ 12yz

4x2 × 7y ÷ 2xy2

Simplify: a

A construction company charges $2 000 for each job. They also charge labour costs of $50 per hour and $75 per hour for overtime. If x is the amount of regular hours worked and y is the amount of overtime hours worked, write an expression to represent the total cost of the job.

Expand: a

6f ( f 2 + 3)

7c3 (3d4 − 9)

5g (g − 4f + 8)

5 (7n − s) − 2 (6n − 3s)

4c2 − 7c + 3c(c − 11)

−a (11b + 9c)

−4a2(4a − 2a4)

−2g (2g + 4)

Expand and simplify:

a c 9

Expand: a

x (2x − 5)

5 (3 ( y − 2) + 7)

Complete the equation so that x = 11 is the only solution: 4(x − 8) + 13 = x +⬚

Expand and simplify: a

(b + 3) (b + 7)

(c − 11) (c − 8)

( f + 9) ( f − 9)

(12g + 10) (12g − 10)

(3d + 5) (d − 6)

(4c + 5d)2

(3e − 4) (5e + 6)

Expand and simplify: a

(h + 3)2

Chapter 3 review mathspace.co

Expand: a

(r − 8) (s + 7)

( p + 13w) ( p2 + w)

Find the area of a triangle with a base length of (2x − 5) mm and a height of (x − 1) mm.

Find the highest common algebraic factor of each term: a

7c2 − 28c

−8x − 4

−35xy − 21x

−3f 2g + f 2g2

5k + 35jk + 15

3st2 + 9s2t + 30t

Given that the length is longer than the width for all x > 1, which term is the length and which is the width?

If one of the longer sides of the playground is along the existing community hall, how much fencing is needed to fence the rest of the playground?

Factorise: x (x + 5) + 7 (x + 5)

7x (x − 3) + 2 (3 − x)

3a + ab − 6c − 2bc

x2 − 4x − 32

7a (a2 − 3b) − 13b (a2 − 3b)

Factorise: a

g2 + 2g + 6g + 12

3xy + 10 + 15x + 2y

Factorise: x2 + 11x + 28

x2 − 14x + 48

81 − 18x + x2

Factorise: 3x2 + 21x + 30

(Optional) Simplify: b

(Optional) Simplify each expression by first simplifying the fractions (where possible). a

a 23

16ab − ac

Factorise the expression to determine the dimensions of the playground.

a 22

a 21

3x + 9

The area of a rectangular playground is 18x2 + 3x. When factorised, this represents the length and width of the playground.

a 19

24xy2 and 10y2z

Factorise: a

13t3 and 7t

(Optional) The water tank in Maria’s house holds w litres of water. After filling the tank, Maria uses

of the tank for gardening, then another

of the tank for cleaning. Write an

expression for the number of litres of water left in the tank after cleaning, and simplify this expression.

100

Mathspace WA Curriculum – Year 9 mathspace.co

(Optional) Simplify: a

(Optional) Given A =

c ,B=

and C =

, find an expression for

d and simplify.

Did you know?

Algebraic equations play a crucial role in environmental science! For instance, scientists use linear and exponential equations to model pollution levels in rivers and oceans. By analysing these equations, they can predict how pollutants spread over time, assess the effectiveness of clean-up strategies, and help protect Australia’s unique marine ecosystems for future generations.

Chapter 3 review mathspace.co

101

Big ideas Equations can be solved more easily by manipulating algebraic expressions and applying a series of organised steps and techniques.

Curriculum Outcomes • Solve linear equations involving brackets and/or a variable on each side of the equation, and verify the solution by substitution • (Optional) Solve linear equations that involve simple algebraic fractions with numerical denominators and verify the solution by substitution • Use the Cartesian plane to explore finding the distance, gradient and midpoint between two points • Move flexibly between the equation of a line, represented by y = mx + c and its graph using the gradient and y-intercept. Graph the equation of a line represented in ax + by + c form • (Optional) Develop and use the algebraic formulas for finding the distance, midpoint and gradient between two points • (Optional) Rearrange formulae, including ax + by = c, to change the subject of the formula • Identify rates as direct proportion, represent algebraically and graphically and use both forms to predict unknown values and interpret in the context of the situation • (Optional) Investigate indirect proportion, represent algebraically and graphically, use both forms to predict unknown values and interpret in the context of the situation • Analyse, represent and interpret problems in real-world situations involving scientific notation, real numbers, linear equations with variables and/or brackets on either side of the equation, quadratic graphs and equations, direct proportion and/or simple interest, earning income or income statements • Use Pythagoras’ theorem to determine the perimeter and area of shapes involving right-angled triangles, in both exact and decimal approximation form. Investigate and apply the converse of Pythagoras’ theorem to establish whether a triangle is right-angled

4 Linear relationships Chapter outline 4.01 4.02 4.03

Midpoint Gradient Distance between two points Investigation: Transformations of linear graphs 4.04 Gradient-intercept form 4.05 Linear graphs 4.06 Equation of a straight line 4.07 Direct proportion 4.08 Optional: Indirect proportion 4.09 Model linear equations 4.10 Multi-step equations 4.11 Optional: Equations with algebraic fractions 4.12 Optional: Rearrange formulae Chapter 4 review

104 112 128 140 150 161 167 175 180 188 195 201 209

The longest distance between any two places on Earth is the distance between Chincha Alta in Peru and Siem Reap in Cambodia, which is over 20 000 kilometres!

4.01

Midpoint After this lesson, you will be able to… • use counting squares to find the midpoint of a line segment. • apply the midpoint formula to find the coordinates of a midpoint. • solve real-world problems involving the midpoint of a line segment.

Midpoint from horizontal and vertical segments The midpoint of a line segment is the point exactly halfway along the segment. A

That is, the distance from the midpoint to both endpoints is the same. AB = BC : means that the distance from A to B is the same as the distance from B to C.

Midpoint

Interactive exploration

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To find the midpoint of a line segment AB, we can count squares to find the lengths of the legs of the right-angled triangle with hypotenuse AB. Then we find the midpoint, the point halfway along, each leg. 12 11 10 9 8 7 6 5 4 3 2 1

length = 10

length = 6

1 2 3 4 5 6 7 8 9 10 11 12

Triangle with hypotenuse AB

12 11 10 9 8 7 6 5 4 3 2 1

length = 5

length = 3

1 2 3 4 5 6 7 8 9 10 11 12

The midpoint of each leg and the midpoint of AB

The coordinates of the midpoint of each leg helps us ﬁnd the midpoint of the hypotenuse. • The x-value of the horizontal leg’s midpoint is the x-value of the midpoint of AB. • The y-value of the vertical leg’s midpoint is the y-value of the midpoint of AB. The midpoint between two coordinates such as (0, 4) and (0, 9) can be found by finding the average of the two numbers 4 and 9. The midpoint of the y-coordinate is x-coordinate is 0. So, the midpoint is (0, 6.5).

104

Mathspace WA Curriculum – Year 9 mathspace.co

and the

Example 1 M is the midpoint of A (−8, −4) and B (2, 8). Find the coordinates of M.

8 7 6 5 4 3 2 1 −8 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 A

x 1 2

Create a strategy Draw a right-angled triangle with hypotenuse AB, then ﬁnd the midpoint of each leg.

Apply the idea 10

12 M

8 7 6 5 4 3 2 1

−9 −8 −7 −6 −5 −4 −3 −2 −1−1

The length of the horizontal leg is 10 units, so the midpoint will be 5 units to the left of B.

B (2, 8)

The length of the vertical leg is 12 units, so the midpoint will be 6 units above A. The coordinates of M are (−3, 2). x 1 2

−2 −3 −4

A (−8, –4)

Idea summary 12 11 10 9 8 7 6 5 4 3 2 1

To ﬁnd the midpoint of a line segment AB, draw a right-angled triangle with hypotenuse AB, and ﬁnd the midpoint of each leg.

• The x-value of the horizontal leg’s midpoint is the x-value of the midpoint of AB. • The y-value of the vertical leg’s midpoint is the y-value of the midpoint of AB. A

1 2 3 4 5 6 7 8 9 10 11 12

4.01 Midpoint mathspace.co

105

Midpoint formula The midpoint (M) of any two points has coordinates that are exactly halfway between the x-values and halfway between the y-values. This means we can find the average of the two x-coordinates to find the x-coordinate of the midpoint and the average of the two y-coordinates to find the y-coordinate of the midpoint.

(x1, y1)

ﬁrst endpoint of the line segment

(x2, y2)

second endpoint of the line segment

Example 2 M is the midpoint of A (4, −6) and B (8, −12). What are the coordinates of M ?

Create a strategy To ﬁnd the coordinates of M, we want to ﬁnd the average of the two x-coordinates and the average of the two y-coordinates. We can use the midpoint formula:

Apply the idea Write the midpoint formula

Substitute the coordinates (4, −6) as (x1, y1) and (8, −12) as (x2, y2)

Simplify each part

Simplify

The coordinates of the midpoint M are (6, −9).

Reﬂect and check We can check our answer by plotting the segment and the midpoint on a Cartesian plane.

106

Mathspace WA Curriculum – Year 9 mathspace.co

The length of the horizontal leg is 4 units, so the midpoint will be 2 units to the right of A.

x 2

The length of the vertical leg is 6 units, so the midpoint will be 3 units above B.

−2 −4 −6

This shows our answer is correct. The midpoint is at (6, −9).

−8 −10 −12

Idea summary For points A (x1, y1) and B (x2, y2), the midpoint (M) will be:

(x1, y1) (x2, y2)

ﬁrst endpoint of the line segment second endpoint of the line segment

4.01 Practice questions Understanding

Explorer

Adventurer

1–3

How many midpoints does a segment have?

Find the coordinates of the midpoint M. a

A (2, 8) and B (2, 2)

1–3

A (6, 5) and B (14, 7)

y 9 8 7 6 5 4 3 2 1

–4–3–2–1 –1 –2 –3 –4

Trailblazer

1–3

6 5

B x 1 2 3 4 5 6 7 8 9

M A

3 2 1

x 1 2 3 4 5 6 7 8 9 10 11 12 13 14

4.01 Midpoint mathspace.co

107

Segment AB and point M are plotted on the Cartesian plane. Is M the midpoint of AB? a

2 −4−3−2 −1

b 8 7

x 1 2 3 4 5 6 7 8 9

−2

−8

−6

Ex 1

−4

Fluency

4, 5a–e, 6ace, 7–9

x 1 2 3 4 5 6 7 8 9 10 11 12 13 14

4, 5c–g, 6bdf, 7–10

4, 5f–j, 6gh, 7–10

Find the coordinates of the midpoint M of segment AB. a

A (5, 4) and B (5, 10)

A (−9, −8) and B (−2, −8)

y −10−9 −8 −7 −6 −5 −4 −3 −2 −1 11 10 9 8 7 6 5 4 3 2 1

x 1

2 3 4 5 6 7 8 9

A (−7, −2) and B (1, 6)

A (−7, 3) and B (1, −7)

y 8 7 6 5 4 3 2 1

−9−8−7−6−5−4−3−2 −1 −1 −2 A −3 −4

108

−1 −2 −3 −4 −5 −6 −7 B −8 −9 −10

x 1 2 3

Mathspace WA Curriculum – Year 9 mathspace.co

3 2 1

−8 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8

x 1 2

A (−8, 5) and B (4, −3)

A (−6, 2) and B (10, −10)

y A

5 4 3

−8 −6 −4 −2 −2

2 1 −8−7−6−5−4−3−2 −1 −1

y A

−4

−6

1 2 3 4

−8

−2 −3

Ex 2

x 2 4 6 8 10

−10

Find the midpoint of segment AB. a

A (7, 3) and B (9, 9)

A (5, 7) and B (2, 3)

A (−9, 8) and B (−1, 10)

A (3, −2) and B (5, −11)

A (−11, −8) and B (−7, 0)

A (−14, −4) and B (4, 8)

A (−7, 2) and B (11, −14)

A (−13, −12) and B (−5, −2)

and

For the pair of points A and B: i

Find the coordinates of the midpoint, M.

Plot the segment AB and the point M on a Cartesian plane.

A (5, 5) and B (13, 9)

A (−12, −1) and B (6, 9)

A (−9, 1) and B (5, −7)

A (1, −11) and B (9, 7)

A (4, 5) and B (6, 9)

A (−10, −1) and B (4, 5)

A (−7, 5) and B (1, −9)

A (−12, −6) and B (4, −4)

The graph shows the annual net proﬁt (in millions) of a company over the last few years. It shows that its proﬁt has been growing approximately linearly from $19 million in 2008 to $39 million in 2014.

Profit (millions) (2014, 39)

What was the company’s net proﬁt in 2011? (2008, 19)

Year

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The graph shows a straight line that approximates the global life expectancy of a child over a period of 100 years. The graph shows the life expectancy to be 55 years in 1910, and 71 years in 2010.

Age

(2010, 71)

Use the midpoint formula to estimate the life expectancy of a child in 1960. (1910, 55)

Year

Lines of latitude and longitude measure position on the earth’s surface and work like coordinates. The first coordinate represents how far above or below the equator you are, and the second coordinate measures how far from Greenwich Mean Time you are. A plane starts its flight at (20° N, 62° E). It is bound for its destination at (49° N, 144° E). Find its position halfway through the flight.

The table shows the number of smartphone users in a particular city: a

Using the midpoint formula, estimate the number of smartphone users in 2009.

Using the midpoint formula, estimate the number of smartphone users in 2013.

Reasoning 11

12ab

Year 2007 2011 2015

Number of smartphone users 175 904 238 502 298 080

11, 12

11–13

What are the properties for the diagonals of a parallelogram?

ABCD is a quadrilateral with vertices A (−2, −1), B (−10, 3), C (−15, 7) and D (−7, 3). Find the midpoints of the diagonals. i

c 12

Is ABCD a parallelogram? Explain your answer.

For the points P (−3, −2), Q (19, 18) and R (10, 15).

Find the coordinates of T, the midpoint of PQ.

If T is the midpoint of RS, ﬁnd the coordinates of S. What type of quadrilateral is PRQS? Explain your answer.

Q R

5 −15 −10 −5 P−5 −10 −15

Explain how the midpoint formula is related to the average of two numbers.

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x 5

Problem-solving 14, 15a–d, 16 14

15e–j, 16, 18–23

For points A and M, where M is the midpoint of A and B: i

Find the coordinates of point B.

Plot the segment AB and the point M on a Cartesian plane.

A (1, −1) and M (4, 3)

A (−3, 1) and M (−4, −3)

If the midpoint of points A and B is M, what are the coordinates of A? a

15c–h, 16–19

B (9, 6) and M (7, 6)

B (10, −5) and M (8, −5)

B (9, 12) and M (7, 9)

B (11, 2) is M (8, −2)

B (−1, 0) and M (−3, 3)

B (12, 5) and M (8, 1)

B (2, −9) and M (−1, −7)

B (−4, −6) and M (1, 8)

B (−7, −9) and M (−5, −3)

B (−1, 0) and M (−4, −2)

A circle with centre at Point C (5, 3) has AB as its diameter. The Point A has coordinates (2, 7) as shown:

9 8 7 6 5 4 3 2 1

Find the coordinates of B and justify your answer.

−1 −1 −2 −3

y A C x 1 2 3 4 5 6 7 8 9 10 11

Find the coordinates of the midpoint of A (2x, 5y) and B (6x, y).

The points P, Q, R, S, T lie on a straight line such that PQ, QR, RS and ST have the same length. Find the coordinates of the points Q, R and S given P (−4, −2) and T (−12, 10).

M is the midpoint of AB. The coordinates of point A are (a, 12), the coordinates of point M are (c, 4), and the coordinates of B are (6, −4). If the length of AB = 20 units, ﬁnd the x-coordinates of A and M.

For the points A (−10, 10) and B (16, 4): a

Find the coordinates of the midpoint M of AB.

For what value of k will the line y = 6x + k bisect (cut in half) the line segment AB?

Find the coordinates of a point that is three-quarters of the way from point A (−3, 1) to point B (1, 7).

A segment AB has endpoints midpoint of the segment.

M (4a + 2, 5b − 3) is the midpoint of S (20, −12) and T (−18, 6). a

Find the value of a.

and B (−7r, 5s). Find the coordinates of the

Find the value of b.

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4.02

Gradient

After this lesson, you will be able to… • identify increasing and decreasing lines. • calculate the gradient using the “rise over run” method. • apply the gradient formula using the coordinates of two points on a line. • examine the special case of horizontal and vertical lines and their gradients.

Increasing and decreasing lines The gradient of a line is the measure of its slope. The gradient value describes: • If a line is increasing left to right (positive gradient) • If a line is decreasing left to right (negative gradient) • How much the line moves vertically (change in the y-value) compared to how much it moves horizontally (change in the x-value)

Interactive exploration

Discover this concept in action online

Positive gradient:

Negative gradient: 4

1 −4 −3 −2 −1

−1

−4 −3 −2 −1

−1

−3

−3 −4

Mathspace WA Curriculum – Year 9 mathspace.co

x 1

−2

−4

Line with increasing gradient, going up from left to right

−2

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mathspace.co

Line with decreasing gradient, going down from left to right

Example 1 Describe the gradient of the line as increasing or decreasing. 4

3 2 1 −4 −3 −2 −1

−1

x 1

−2 −3 −4

Create a strategy

Apply the idea

Identify if the line is increasing or decreasing from left to right.

The line is decreasing from left to right, which means that the line has a negative gradient.

Idea summary If the gradient is positive, the line is increasing from left to right. If the gradient is negative, the line is decreasing from left to right.

Gradient as rise over run The gradient can be calculated by using the formula: m= Note: we use the variable m to denote the value of the gradient. The rise and run are found by forming a right-angled triangle using any two points on the line. y

This line has a gradient of

2 1 −3

−2

−1 −1

3 run

4 rise

x 1

−2

In this case, the gradient is positive because, over the 3-unit increase in the x-values, the y-value has increased. Remember that increasing lines have positive gradients, and decreasing lines have negative gradients.

Interactive exploration

Discover this concept in action online

mathspace.co

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7 6 5

y y = 4x + 1 m=4 y = 2x + 1 m=2

4 3 2 1 −1

−1

x 1

Graphs of y = 2x + 1 and y = 4x + 1 The steeper the graph, the larger the value of m.

Example 2 Find the gradient of the line shown using points A and B, by creating a right-angled triangle and identifying the rise and run.

8 7 6 5

4 3

2 1

x 1

2 3 4

5 6 7 8 9

Create a strategy We can draw a triangle to ﬁnd the rise and run, and then use the gradient formula.

Apply the idea 9 8 7 6 5 4 3 2 1

We can now see that this line has a run of 3 and a rise of 2.

B 2 rise A

x 1

114

3 run

2 3 4 5 6 7 8 9

Mathspace WA Curriculum – Year 9 mathspace.co

Reﬂect and check The line is increasing so it should have a positive gradient.

Idea summary We calculate the gradient by using the formula:

Gradient

rise is the vertical measurement run is the horizontal measurement

Gradient formula 6

We have just been using

A (3, 6)

Gradient =

5 4 3 2 1 −1

x 1

−2

B (7, −2)

to ﬁnd the gradient of a line. This works when it’s easy to see the graph, and construct a rightangled triangle with easy to measure lengths of rise and run. We can extend this thinking to use the coordinates of points A and B and construct a general formula.

Let’s look at an example.

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The line is decreasing so should have a negative gradient, and it does. 7

For any two points (x1, y1) and (x2, y2), we can ﬁnd the gradient using the formula:

Gradient =

5 4

(x2, y2)

Gradient =

2 1 −1

−1

(x1, y1) 1

x 3

Example 3 What is the gradient of a line that passes through the points A(3, 5) and B(−2, 10)?

Create a strategy Use the gradient formula.

Apply the idea Use the gradient formula

Substitute the coordinates

Evaluate

Reﬂect and check By doing a quick sketch of the line we can see that the line is decreasing, so it should have a negative gradient.

y 10 8 6 4 2 −2 −2

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x 2

Idea summary For any two points (x1, y1) and (x2, y2), we can ﬁnd the gradient using the formula:

Gradient = (x1, y1) (x2, y2)

are the coordinates of the ﬁrst point are the coordinates of the second point

Gradient of horizontal and vertical lines On horizontal lines, the y-value is the same for every point on the line. 4 A (−4, 4) B (−1, 4) 3

y C (3, 4)

The equation of this line is y = 4.

1 −4 −3 −2 −1

Coordinates A, B, and C all have the same y-coordinate. Every point on the line has a y-value equal to 4.

x 1

−1

Since gradient is calculated by

and there is

no rise, then:

−2 −3 −4

The gradient of a horizontal line is always 0. On vertical lines, the x-value is always the same for every point on the line.

A (−3, 8)

Coordinates for A, B, and C all have the same x-coordinate, x = −3.

The equation of this line is x = −3.

Since gradient is calculated by

no run, then:

4 B (−3, 3)

and there is

3 2 1

−4 −3 −2 −1 −1 −2 −3 C (−3, −3)

x 1

Division by 0 is undefined. The gradient of a vertical line is always undefined.

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Example 4 What is the gradient of the line passing through A(−3, −1) and B(−3, 5)?

Create a strategy Observe that the x-coordinates have the same value, so the points lie on the same vertical line.

Apply the idea The gradient of any vertical line is undeﬁned, thus the gradient of the line passing through A(−3, −1) and (−3, 5) is undeﬁned.

Reﬂect and check Use the gradient formula in an attempt to calculate the gradient. Gradient = Write the gradient formula

Simplify

Since there is a 0 on the denominator, the gradient is undeﬁned.

Idea summary Horizontal lines have a gradient 0. Vertical lines have a gradient that is undeﬁned.

4.02 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

A line is drawn on the Cartesian plane.

Which of these describe the relationship between the gradients of the segments?

Gradient AB is equal to gradient AC.

Gradient AB is twice gradient AC.

Gradient AC is twice gradient AB.

1 −4 −3 −2 −1

Mathspace WA Curriculum – Year 9 mathspace.co

B 1

−1

118

2 A

x 3

Identify whether each graph has a positive gradient or negative gradient. y

1 −4 −3 −2 −1

x 1

−1

−4 −3 −2 −1

−2

−3

−4

Identify whether each graph has a zero gradient or undefined. b

2 1 −4 −3 −2 −1

−1

1 −2 −1

−3 −4

Fluency

−2

−1

−2

Ex 1

x 1

−1

x 1

−2

−5

−3

−6

−4

4–9, 10ab, 11a–el, 12, 15a–d

4–9, 10cd, 11e–l, 12–14, 15b–e

6, 7, 8ef, 9, 10d, 11e–n, 12–14, 15c–f, 16

Identify if the lines are increasing or decreasing, from left to right. a

b 4

4 3

1 −4 −3 −2 −1

−1

−2

x 1

−4 −3 −2 −1

−1

x 1

−2

−3

−4

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Identify if the line segments have a positive or negative gradient. a

1 −4 −3 −2 −1

−1

x 1

−4 −3 −2 −1

−1

−2

−3

−4

9 8 7 6 5 4 3 2 1

−4 −3 −2 −1−1

3 2 1 −4 −3 −2 −1 x 1

−1

−2

−2 −3 −4

−3 −4

Order the gradients of these graphs from smallest to largest. A

B 4

4 3

1 −4 −3 −2 −1

120

x 1

Describe the gradient of these lines: a

−1

x 1

−4 −3 −2 −1

−1

−2

−3

−4

Mathspace WA Curriculum – Year 9 mathspace.co

x 1

1 −4 −3 −2 −1

−1

x 1

−4 −3 −2 −1

−2

Ex 2

x 1

−1

−2

−3

−4

For each of these intervals between points A and B: i

Find the rise.

Find the run.

6 4

2 1

x 1

2 3 4

f 4

x 3

2 1

−6

−4 −3 −2 −1

−8

−1

−10

−2

−14

1 2 3 4

−4

−12

−6 −5 −4 −3 −2 −1 −2 A

y 10

4 2

8 6

−4 −3 −2 −1 A −2

−9 −8 −7 −6 −5 −4 −3 −2 −1

14 12

−4 −3 −2 −1

5 6 7 8 9

Find the gradient.

iii

x 1

−3 −4

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Find the gradient of each line passing through points A and B: a

4 B

1 −3 −2

−1

x 1

−1 −1

−8 −7 −6 −5 −4 −3 −2 −1 −1

−4 −3 −2 −1

−1

−3 −4

Sketch the line that passes through the given points.

Find the gradient.

(1, 3) and (0, 6)

(2, 1) and (8, −2)

(−1, 0) and (0, −5)

(−6, −6) and (6, 2)

Find the gradient of the line that passes through the given points: a

(6, 4) and (3, 4)

(1, −4) and (0, 0)

(2, −6) and (0, 0)

(−1, 0) and (0, 3)

(−4, 7) and (1, 10)

(−6, 5) and (4, 5)

(10, 5) and (15, −5)

(−8, 20) and (−4, 10)

(−2, −5) and (−9, −12)

(−3, −1) and (−5, 1)

(−15, −35) and (−30, −25)

and

−2

1 2 3

−2

and

Ex 4

Ex 3

−1

−3

−2

y B

What is the gradient of the line passing through A(5, −2) and (5, −3)?

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A line is graphed where A (2, 0) and B (0, 4) both lie on the line. y 4 B 3 2 1 x

A −1

−1

As x increases, what happens to the value of y? 14

A line is graphed where A (−5, 0) and B (0, 2) both lie on the line. y 4 3 2

1 A −5 −4 −3 −2

x −1

1 −1

As x increases, what happens to the value of y?

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123

Find the gradient of each line. a

b 6

1 −4 −3 −2 −1

−1

x 1

−2

−4 −3 −2 −1

−3

−1

x 1

1 −7 −6 −5 −4 −3 −2 −1

−3

f 7

x 1

−2 −3 −4

124

−1

x 1

−2

−4

−4 −3 −2 −1

y 6

−2

1 −4 −3 −2 −1

−1

−2

−4

x 1

Mathspace WA Curriculum – Year 9 mathspace.co

3 2

1 −4 −3 −2 −1

−1

−2

x 1

The line y = 5x + 2 is graphed: 4

3 2 1 −4 −3 −2 −1

x 1

−1

−2 −3 −4

Find the y-value of the point on the line where x = 5.

Find the gradient of the line.

Reasoning 17

17, 19, 20

17–20

The points A (−11, −9), B (−5, 1) and C (−2, 6) are plotted on the Cartesian plane. a

Find the gradient of AB.

Find the gradient of BC.

Do the points A, B and C lie in a straight line? Explain your answer.

Explain why a vertical line has an undeﬁned gradient.

A line is drawn on the Cartesian plane. Tom uses points A and C to ﬁnd the gradient of the line, and Johannes uses points B and C to ﬁnd the gradient of the line. Will their results be the same or different? Explain your answer. 4

3 2 B 1

−4 −3 −2 −1

−1

−2 −3

x 1

−4

Two gradients are calculated. • Line 1 gradient = 3 • Line 2 gradient = 5 Which line is steeper, (increases faster)? Explain why.

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Problem-solving 21ab, 23–26

21cd, 22–24, 26, 27

Complete the coordinates using the points and gradients provided. a c

21bc, 23–26

(4, −3) and (1, ⬚), gradient = −2

(5, 3) and (⬚, 63), gradient = 4

b d

(5, 3) and (2, ⬚), gradient = −4

(11, ⬚) and (−20, 16), gradient =

The points A (26, m − 24), B (−1, m) and C (−10, 9) are plotted on the Cartesian plane. Find m, given that A, B and C all lie on the same line.

Pair up these points to create 3 lines, one with a positive gradient, one with a negative gradient and one with a gradient of 0. A (3, 5), B (4, 6), C (−1, 8), D (7, 5), E (−3, 10), F (4, 12)

Two lines L1 and L2 have equations y = x − 3 and y = −x + 5 respectively. The lines and their point of intersection have been graphed: a

When x = 4, ﬁnd the y-coordinate of the corresponding point on L1.

When x = 4, ﬁnd the y-coordinate of the corresponding point on L2.

Find the gradient of the two lines: i

Find the product of their gradients.

The graph illustrates Jay’s walk to school along a straight track: a b

The line has a positive gradient, what does this mean in the context of the problem? What does the gradient context?

represent in this

6 5 4 3 2 1 −3−2 −1 −1 −2 −3 −4 −5 −6

y L2

x 1 2 3 4 5 6 7 8 9 L1

Distance (m) 250 200 150 100 50 Time (min) 0

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The lines plotted here show the distance of an object in metres over time measured in seconds. i

Distance (metres) 10 9 8 7 6 5 4 3 2 1

Time (seconds) 0

iii

Time (seconds) 0

Distance (metres)

Time (seconds) 0

Distance (metres) 10 9 8 7 6 5 4 3 2 1

10 9 8 7 6 5 4 3 2 1

Distance (metres)

Time (seconds) 0

For each graph: a Is the gradient positive, negative or neither? b What is the value of the gradient? c What does this value mean in the context of the graph? 27

A car begins at the starting line of a drag race and travels 100 metres down the track in 5 seconds. a Create a graph that represents this context. b Is the gradient of the line positive or negative? c Find the gradient of the line. d What does the gradient of the line represent?

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4.03 Distance between two points After this lesson, you will be able to… • recognise the relationship between Pythagoras’ theorem and the distance formula. • calculate the distance between two points using Pythagoras’ theorem. • solve problems requiring the calculation of distances between points on the Cartesian plane. • apply the distance formula in special cases, such as horizontal or vertical lines.

Distances on the Cartesian plane Exploration B (−1, 8) 9

Triangle ABC is shown.

C (3, 8)

6 3 −4 −3 −2 −1 −3

x 1 2 3 4 5 6

−6 −9

2. Explain how to ﬁnd the distance from B to C. 3. How can we use the distance from A to C and the distance from B to C to ﬁnd the distance from A to B?

−12 −15 −18

1. Explain how to ﬁnd the distance from A to C.

A (3, −18)

We can use Pythagoras’ theorem to help us ﬁnd distances and lengths on the Cartesian plane. We already know that Pythagoras’ theorem can help us calculate the side lengths in a right-angled triangle. a2 + b2 =

shorter side lengths hypotenuse The value c is used to represent the hypotenuse which is the longest side of the triangle. The other two lengths are represented by a and b.

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We can also use Pythagoras’ theorem to ﬁnd the distance between two points on the Cartesian plane.

3 2 1 −3 −2

−1

x 1

−1 A

−2

Draw the right-angled triangle where the interval is the hypotenuse. Then, we can use the horizontal and vertical distances as the lengths of the sides.

AC2 = AB2 + BC2

−3

Example 1 The points A (−3, −2), B (−3, −4), and C (1, −4) are the vertices of a right-angled triangle, as shown on the Cartesian plane.

y 2 1 −3

−2

−1

x 1

−1 A

−2 −3

−4

a What are the lengths of AB and BC?

Apply the idea Using the graph, AB is 2 units and BC is 4 units.

b Use Pythagoras’ theorem to ﬁnd the length of the interval AC to three decimal places.

Create a strategy Use Pythagoras’ theorem AC2 = AB2 + BC2 with the values we identified in part (a).

Apply the idea Let the length of AC be denoted by c. Use Pythagoras’ theorem

Substitute a = 2 and b = 4

Evaluate the indices

Evaluate the addition

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129

Take the square root of both sides

Evaluate

Since AC is denoted by c, the length of AC is 4.472 units.

Reﬂect and check When we want to know the exact distance, we leave the answer in square root form, so the exact distance between points A and C is units. Recall that there are two answers to c2 = 20: a positive result and a negative result.

Since lengths and distances are always positive, we only state the positive result.

Idea summary If the line is horizontal, the difference of the x-coordinates of the points is its distance. If the line is vertical, the difference of the y-coordinates of the points is its distance. c2 = a2 + b2

y 5

We can use Pythagoras’ theorem to ﬁnd the distance between any two points, by drawing a right-angled triangle where the interval is the hypotenuse. Then we ﬁnd the shorter side lengths and use Pythagoras’ theorem to solve for the hypotenuse. The result will be the distance between the two points.

4 b

3 c 2 1

a 1

x 4

The distance formula We can generalise the method for ﬁnding the distance between any two points, (x1, y1) and (x2, y2), using Pythagoras’ theorem. Create a right-angled triangle.

y (x2, y2)

We can label the hypotenuse as d, the horizontal side as a, and the vertical side as b. By Pythagoras’ theorem: d2 = a2 + b2

d (x1, y1)

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y (x2, y2)

The horizontal distance (a) is the size of the difference between the two x-coordinates. b = y 2 – y1

a = x2 − x1 The vertical distance (b) is the size of the difference between the two y-coordinates.

(x1, y1) a = x2 – x1

b = y2 − y1

Use Pythagoras’ theorem to ﬁnd the length of d:

This equation is known as the distance formula.

horizontal vertical distance distance (width) (height) d distance between the two points (x1, y1) coordinates of the first point (x2, y2) coordinates of the second point

Example 2 Find the distance between Point A (−2, −9) and Point B (2, −14), rounded to two decimal places.

Create a strategy Determine the coordinates, then use the distance formula.

Apply the idea The coordinates are A (−2, −9) and B (2, −14). Use the distance formula

Substitute (x1, y1) = (−2, −9), (x2, y2) = (2, −14)

Evaluate the brackets

Evaluate the indices

Evaluate the addition

Evaluate and round

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131

Reﬂect and check y −3 −2

−1

−2

We could have found the answer by plotting the points on a Cartesian plane. Since the distance formula is derived from Pythagoras’ theorem, we will get the same answer.

x 2

−4 −6 −8

−10

5 units

−12 −14 4 units

Idea summary The distance between two points (x1, y1) and (x2, y2) is given by:

d (x1, y1) (x2, y2)

is the distance between two points are the coordinates of the ﬁrst point are the coordinates of the second point

4.03 Practice questions Understanding

Explorer 1, 2ab, 3ab, 4

Adventurer 1, 2ac, 3ac, 4

Trailblazer 1, 2cd, 3cd, 4

What is the formula for ﬁnding the distance between two points on a Cartesian plane?

Find the distance between the two points, A and B. a

−4 −3 −2 −1 −1 −2 −3 −4 −5

132

5 4 3 2 1

x 1

A (2, 5) and B (2, 8)

Mathspace WA Curriculum – Year 9 mathspace.co

5 4 3 A 2 1 −5 −4 −3 −2 −1 −1 −2 −3 B −4 −5

x 1 2 3 4 5

A (−7, −1) and B (−7, −6)

Find the distance between the two points, A and B. a

y −6 −4 −2

8 10

−1

−2

−6 −5 −4 −3 −2 −1 −1 B −2

−3 A

−4

A (3, 2) and B (9, 2)

x 1 2 3 4 A

−3

−5

y 2

−4

A (−4, 5) and B (−7, 5)

Is each statement true or false? a

If two points have the same y-coordinate, subtract the smaller x-coordinate from the larger x-coordinate to ﬁnd the distance between them.

To ﬁnd the distance between two points on the same vertical line, calculate the positive difference between their x-coordinates.

If the distance between two points is zero, the points lie on the same horizontal line.

The formula for the distance between two points is derived from Pythagoras’ theorem.

Did you know?

Drone deliveries rely on calculating the distance between two points to plan efficient flight paths. This ensures packages arrive faster while using less battery power.

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5ab, 6ab, 7a, 8a, 9ab, 10, 11a–c, 12ab

Fluency 5 Ex 1

5ac, 6bc, 7a, 8b, 9bc, 10cd, 11cd, 12, 14

5cd, 6d, 7b, 8b, 9, 10, 11ef, 13–15

For each triangle: i

Find the length of interval PQ.

iii

Find the length of PR, rounded to two decimal places.

Find the length of interval QR.

P (−6, 5), Q (−6, 2) and R (−2, 2)

P (−1, 9), Q (−1, 2) and R (−5, 2)

6 P

5 3 Q

1 x −7 −6 −5 −4 −3 −2 −1

−6

P (−4, 5), Q (−4, 2) and R (4, 2)

−5

−4

−3

4 3 Q −4 −3 −2 −1

134

R x 1

Mathspace WA Curriculum – Year 9 mathspace.co

−1

−4 P

x 1

−3

1 −1

−1 −2

1 x

y −4 −3 −2

−2

P (−3, −5), Q (−3, −1) and R (1, −1)

y P

−5 −6

2 R

Use Pythagoras’ theorem to ﬁnd the length of side AC, rounded to two decimal places. a

b 9

3 2

x 1

2 3 4 5 6 7 8 9

y A

1 x

2 x −3 −2 −1 −2

−1 B −2

A 0 1

−14 −12 −10 −8 −6 −4 −2

2 3 4 5 6

−4

Use Pythagoras’ theorem to ﬁnd the length of side AC, as an exact value. a

−7 −6 −5 −4 −3 −2 −1 −1 A B −2 −3 −4 −5 −6 −7 −8 C −9

7 6 5

4 3 2 1

0 1

2 3 4 5 6 7 8 9

y x 1 2 3

The vertices of a triangle are listed. i

Sketch the triangle on the Cartesian plane.

Find the length of AC as an exact value using Pythagoras’ theorem.

A(−4, 3), B(−4, −4), C(2, −4)

A(6, 4), B(6, −2), C(−7, −2)

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135

Find the distance of the given point P from the origin. a

P (12, 16)

P (−7, 24)

y P

24 21

14 12

x 4

P (−5, −4)

x y

2 3 4 5 6 7 8 9

y −6 −5 −4 −3 −2

−1 −1 −2 −3 −4

−5

Use the distance formula to ﬁnd the length of the interval AC. Write your answer in exact form. a

y 1

−5 −4 −3 −2

A −1 −1 −2

−3

−4

C 0

136

−7 −6 −5 −4 −3 −2 −1

P (7, −9) −1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −11

Mathspace WA Curriculum – Year 9 mathspace.co

−5

x 1

y A

6 5 4 3 2 1

9 8 7 6 5 4

−4−3−2−1 −1 −2 −3 A −4

2 1

−2 −1 Ex 2

y C

x 1 2 3 4 5 6 7 8 9

Find the distance between Point A and Point B. Write your answer in exact form. a

A (1, 4) and B (7, 12)

A (4, 2) and B (−8, −7)

A (−5, 5) and B (3, −3)

A (−1, 9) and B (−4, 1)

A (1.4, −3.8) and B (6.2, −10.2)

and

Find the length of the segment indicated, rounded to two decimal places. A (−1, 5), B (3, 7), C (0, 6), D (−4, −5), E (−3, 0), F (2, −4)

Find the length of the segment indicated, rounded to two decimal places. A (0, 5), B (−5, 9), C (0, −2.3), D (−1.5, −5),

Given P (4, 3), M (−3, −4) and N (−7, 1): a

Find the distance from P to M, rounded to two decimal places.

Find the distance from P to N, rounded to two decimal places.

Which point is further from P ?

For the points A (12, 3) and B (14, 0): a

Find the length of AB rounded to two decimal places.

If M is the midpoint of AB, ﬁnd the length of AM rounded to two decimal places.

Reasoning 16

16, 17

For the points M (−9, −1) and N (1, 5): a

Find the exact distance from M to the origin.

Find the exact distance from N to the origin.

Which point is closer to the origin? Explain your answer.

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For the points A (−1, −7), B (4, −2) and C (0, −3), ﬁnd: a

The exact length of AC.

The exact length of BC.

What do you notice about the distance between the points?

Is C the midpoint of AB? Justify your answer.

Problem-solving 18, 19

20, 22–23, 25

20, 21, 24, 26–28

K is the midpoint of A (3, 1) and C (15, −7). Find the distance from A to K rounded to one decimal place.

A triangle has vertices at A (1, −1), B (−3, −4) and C (5, −4). a

Find the length of sides: i

b 20

iii

Is this triangle equilateral, isosceles or scalene?

Find the length in exact form: i

A triangle has vertices at A (9, −2), B (2, −8) and C (3, 5). a

iii

How can we show that the triangle has a 90° angle and two 45° angles?

Does the triangle have a 90° angle and two 45° angles?

Given the points P (10x, −6x) and Q (4x, 2x), where x > 0: a

Find the distance between P and Q in terms of x.

Find the coordinates of the midpoint of the segment PQ in terms of x.

The points M (−3, −5) and R (4, −12) are the endpoints of an interval. N (−1, −7) is a point on this interval. a

Find the exact distance between the points: i

M and N

N and R

What is the ratio MN : NR?

Find the perimeter of the parallelogram whose vertices are A (3, 2), B (−2, 7), C (2, 9) and D (7, 4). Round your answer to one decimal place.

△PQR has vertices P (2, −6), Q (−9, −17) and R (−5, 1). a

Find the exact length of the sides: i

138

Is the triangle right-angled?

Mathspace WA Curriculum – Year 9 mathspace.co

iii

P (−4, −4), Q (5, −7), R (9, −11) and S (0, −8) are the vertices of a quadrilateral. a

Find the exact length of the sides: i

Classify the quadrilateral.

ABCD is a rhombus whose vertices are A (1, 2), B (3, 10), C (11, 12) and D (9, 4). Find: a

The exact length of diagonals:

i AC ii BD The exact area of the rhombus.

△PQR is isosceles. y 11 P 10 9 8 7 6 5 4 3 2 1 Q

iii

Find the exact length of the diagonals: i

d R x 1 2 3 4 5 6 7 8 9 10 11

Find the area of the triangle.

Find the exact length of PR.

Find the exact value of d, the perpendicular from Q to the side PR.

A circle with centre at point C (3, 4) has point A (−12, 12) lying on its circumference. Find: a

The radius of the circle.

The exact circumference of the circle.

The distance between point (20, 4) and the centre.

Does the circle also pass through the point (20, 4)? Explain your answer.

Investigation: Transformations of linear graphs

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4.04 Gradient-intercept form After this lesson, you will be able to… • understand and apply the gradient-intercept form of a line. • explore the effects of changing the gradient and y-intercept on the equation and graph of a line. • interpret the significance of positive, negative and zero gradients. • determine the y-intercept of a line and understand its role in the equation. • derive the equation of a line from a graph.

The gradient-intercept form of a line Interactive exploration

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All linear equations can be written in the form:

y = mx + c m the gradient c the y-value of the y-intercept For the gradient: • If m < 0, the gradient is negative and the line is decreasing. • If m > 0, the gradient is positive and the line is increasing. • If m = 0, the gradient is 0 and the line is horizontal. • The larger the magnitude of m, the steeper the line. For the y-intercept: • If c is positive, the line crosses the y-axis above the origin. • If c is negative, the line crosses the y-axis below the origin. An equation of the form y = mx + c is known as the gradient-intercept form of a line because we can easily identify both the gradient, m, and the value of the y-intercept, c.

Example 1 Find the equation of a line whose gradient is −8 and crosses the y-axis at −9. Express your answer in the form y = mx + c.

Create a strategy

Apply the idea

Substitute the gradient for m and the value of the y-intercept for c.

From the information, we have m = −8 and c = −9, so the equation is: y = −8x − 9

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Idea summary For all equations in gradient-intercept form, y = mx + c: • The value of m is the gradient. The larger the value of m, the steeper the line is. An increasing line has a positive gradient and a decreasing line has a negative gradient. • The value of c is the y-value of the y-intercept. The line passes through the point (0, c).

The equation of a line from a graph Exploration During a snowstorm, the initial depth of snow is 2 centimetres. The snow falls at a rate of 1.5 centimetres per hour. The graph and equation both represent this situation. Equation:

y, Depth (cm) 22 20 18 16 14 12 10 8 6 4 2

y = 1.5x + 2

x, Time (hours) 1 2 3 4 5 6 7 8 9 10 11 12131415

Choose any point on the line, substitute its coordinates into the equation, and evaluate.

Repeat with two more points on the line. What do you notice?

Choose any point not on the line, substitute its coordinates into the equation, and evaluate.

Repeat with two more points not on the line. What do you notice?

Substitute a value between 1 and 15 into the equation for x and evaluate.

Explain the relationship between the equation and the points on the graph.

To find the equation of a line from a graph we need the gradient and y-intercept. Find the gradient (m) of the line. Using the coordinates of any two points. Identify the y-intercept by finding where the line crosses the y-axis. The coordinates will be in the form (0, c), but we only need the y-value (c) to create the equation. Once we have identified these two features, we can write the equation of the line in the form y = mx + c. 4.04 Gradient-intercept form mathspace.co

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Example 2 Find the equation of the line graphed on the Cartesian plane.

y 7 6 5 4 3 2 1 −4 −3 −2 −1

−1

x 1

a What is the gradient of the line?

Create a strategy

Apply the idea

Find two points that lie on the line and use the gradient formula.

We can easily identify A (0, 4), and B (1, 1). Use the gradient formula Substitute the coordinates Simplify numerator and denominator

Evaluate

b What is the value of the y-intercept?

Create a strategy

Apply the idea

We can identify the y-intercept by ﬁnding where We can see that the line crosses the y-axis at 4. the line crosses the y-axis. The coordinates will be The value of y is 4 when x = 0. So c = 4. in the form (0, c) but we only need the y-value (c). c Write the equation in gradient-intercept form.

Create a strategy

Apply the idea

Substitute the values found in parts (a) and (b) into the equation y = mx + c.

The equation is y = −3x + 4.

Idea summary We can ﬁnd the equation of a line of the form y = mx + c by ﬁnding the gradient for the value of m, and the y-value of the y-intercept for the value of c.

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4.04 Practice questions Understanding

Explorer 1–4

Adventurer

Trailblazer

1–4

In the equation of a line, y = mx + c, what do m and c represent?

Describe the effect on the graph of the line, if m is: a

Negative

Zero

In each of the equations, identify the value of m and c. a

Positive

y = 15x − 10

y = 7x

For the graph shown: a

What is the gradient of the line?

By how much does the y-value change as the x-value increases by 1?

10 9 8 7 6 5 4 3 2 1 −1 −1 −2

y=6−x

x 1

Did you know?

The gradient-intercept form of a linear equation, written as y = mx + c, is a powerful tool in everyday problem-solving! Urban planners use it to model straight paths like roads or railways— m represents the slope or steepness of the path, while c shows where it crosses the vertical axis. By adjusting these values, planners can design efficient routes that account for elevation changes and ensure accessibility across different terrains. 4.04 Gradient-intercept form mathspace.co

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5, 6a–e, 7, 9, 10, 12–14, 17, 19ab, 20a

Fluency 5

5bd, 6d–g, 7, 8ab, 9, 12c–f, 15–17, 19bc, 20b

For each graph: i

What is the y-intercept?

−4 −3 −2 −1

−2

−3

−4

6 5 4 3 2 1

4 3 2 1

x 1 2 3 4 5 6

−2 −3 −4 −5 −6

144

x 1

−1

−2

−1

−6−5−4 −3−2 −1−1

What is the gradient?

Ex 1

5cd, 6d–g, 8, 11, 12c–f, 16, 18, 19cd, 20b

–3 –2

–1

x 1

–2 –3 –4

Find the equation in the form y = mx + c, for: a

A line with a gradient 2 and y-intercept 5

A line with a gradient −6 and y-intercept 9

A line with gradient

and y-intercept 3

A line with gradient

and passes through the point (0, 3)

A line with gradient 8 and passes through the point (0, −4)

A line with gradient 0 and passes through the point

A line that has the same gradient as y = 9 − 8x and the same y-intercept as y = −5x − 3

Mathspace WA Curriculum – Year 9 mathspace.co

Ex 2

For each equation: i

What is the gradient, m?

What is the value of the y-intercept, c?

y = −2x + 9

y = −4x − 8

y = 6 + 8x

y = 3 (4x − 3)

By rewriting the equation in the form y = mx + c: i

What is the gradient, m?

What is the value of the y-intercept, c?

9x − y − 8 = 0

3y = 12x − 15

hat are the values of the gradient, m, and the W y-intercept, c?

Write the equation of the line in gradient-intercept form.

By substituting into your equation from part (b), find the value of y when x = 27

y 5 4 3 2 1 −2

−1

x 1

−1

−2

This graph shows the relationship between the time (x) in minutes and distance ( y) in kilometres of a high-speed train ride at a constant speed.

y (kilometres) 5

Find the equation of the line in gradient- intercept form.

Find how far the high-speed train travelled, y, after 29 minutes.

3 2 1 x (minutes) 0

This graph shows the relationship between the time (x) in seconds and distance ( y) in metres for a sprint at a constant speed that Jocel did. a

Find an equation of the line in gradient- intercept form.

By substituting into your equation from part (a), find how far Jocel ran, y, after 50 seconds.

10 9 8 7 6 5 4 3 2 1

y (metres)

x (seconds) 0 1 2 3 4 5 6 7 8 9 10 4.04 Gradient-intercept form mathspace.co

145

Find the equation of the line in gradient-intercept form. a

−1

−2

−1

2 4 6 8 10 12 14 16 18

−1

−2

−3

−4

−5

y 2

1 x −1

−1

−2

−1

−2

−1

−3

−2

f 2

4 3

−2

1 −1

−3 −4

Find the equation in gradient-intercept form of a line that has a gradient of 5 and has each of these y-intercepts: a

y=3

y = −6

Find the equation in gradient-intercept form of a line that has a y-intercept of 3 and has each of these gradients: a

146

1 −2

−2

−1

m=4

Mathspace WA Curriculum – Year 9 mathspace.co

m = 0.6

A line has a gradient of −3 and crosses the y-axis at y = −2. a

Find the equation of the line in the form y = mx + c.

Does the point (−2, 4) lie on this line?

A line has a gradient of −3 and a y-intercept at (0, 8). a

Find the equation of the line in the form y = mx + c.

Does the point (8, −31) lie on this line?

For the two points given: i

Find the gradient of the line passing through the points.

Find the equation of the line in the form y = mx + c.

(0, 2) and (2, 6)

(0, −9) and (5, 1)

(0, 2) and (−7, 44)

(0, −3) and (1, 4)

Find the equations of these lines in the form of y = mx + c : a

Passes through the points A (0, 4) and B (3, 0)

Passes through the points A (0, −3) and B (9, 0)

Passes through the points A (0, −1) and B (−1, 0)

Passes through the points A (0, −9) and B (−45, 0)

Sketch on a Cartesian plane: a

The line with a y-intercept of 3 and gradient of

The line with equation y = 2x + 5

The line with a y-intercept of −2 and gradient of −3

The line with equation

Write the equation of the line in gradient-intercept form.

Find the y-coordinate of the y-intercept.

iii

Find the x-coordinate of the x-intercept of the line.

Sketch the line on a Cartesian plane.

A line that has gradient of −2 and passes through the point (3, −8).

A line that has gradient of

and passes through the point (−10, 4).

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Reasoning 21

21, 22, 26

22, 23, 25, 26

24–27

Look at the three lines:

• Equation 1: y = x + 1 • Equation 2: y = 2x + 1 • Equation 3: y = 4x + 1 a

What do all the equations have in common?

What do all the graphs have in common?

Describe all lines that have the form y = mx + 1.

y=x+1

2 1

x −2

−1

−1 4x + 1 −2 y = 2x + 1

Look at the three lines: 8

• Equation 1: y = 2x + 4 • Equation 2: y = 2x + 8 • Equation 3: y = 2x − 4

What do all the equations have in common?

What do all the graphs have in common?

Describe all lines that have the form y = 2x + c.

4 y = 2x + 4

−8 −6 −4 −2 −2

x 2

−4 y = 2x − 4 −6 y = 2x + 8 −8

Given: • Line A: 5x + 3y + 5 = 0 • Line B: 7x + 6y − 3 = 0

Express both lines in the form y = mx + c.

Which line is steeper? Justify your answer.

Which of these lines is steeper? Justify your answer. Line A: 2x + 5y − 5 = 0 Line B: 4x + 4y + 1 = 0

The equation y = −3x + 5 represents a straight line. Is the graph of this line increasing, decreasing, vertical or horizontal? Justify your answer.

Explain the difference between the graphs of the equations y = 2x and y = −2x.

Is it possible for a vertical line to be represented by an equation in the gradient-intercept form y = mx + b? Why or why not?

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Problem-solving 28 28

28–30

28–32

A farmer noticed that the growth of his corn plants follows a linear relationship with the number of days since planting. He observed that on day 4, the plants were 8 cm tall, and on day 16, they were 36 cm tall. Find the equation of the line representing the relationship between the height of the corn plants ( y) and the number of days since planting (x).

For each equation: What is the gradient, m?

What is the value of the y-intercept, c?

x − y = 13

3x − 9y − 27 = 0

3x − 4y − 28 = 0

3x − 10y = −2

16x + 12y = 10

−9x + 9y = 27

A straight line has gradient −1 and goes through the points (0, 2) and (a, −6). a

Write the equation of the line in the form y = mx + b.

Find the value of a.

A car rental company charges a ﬁxed fee per day for renting a car, plus an additional fee for each kilometre driven. The company found that a customer who drove 30 kilometres in a day was charged $45, while another customer who drove 60 kilometres in a day was charged $75. Find the equation representing the cost ( y) as a function of the number of kilometres driven (x).

A gardener charges a callout fee of $85 plus $28 per hour. Write an equation to represent the total amount charged, y, by the gardener as a function of the number of hours worked, x.

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4.05 Linear graphs After this lesson, you will be able to… • use a table of values to sketch a linear graph. • identify x and y-intercepts and how to use them to sketch a linear graph. • use the gradient and a single point to sketch a linear graph. • apply knowledge of sketching linear graphs to solve problems.

Sketch a graph using a table of values We can use a table of values to sketch a graph. x y

This is a table of values. The ﬁrst row contains the independent variable (in this case, x). The second row contains the dependent variable, which is the variable that depends on the value of x.

To create a table of values, we will need to ﬁnd the y-value for each x-value in the table. We can do this by substituting each x-value into the equation.

Example 1 For the equation y = 2x − 4: a Complete the table of values. x y

Create a strategy Substitute each x-value into the given equation.

Apply the idea For x = 0, y=2×0–4 = −4

Substitute x = 0 Evaluate

y=2×1–4 = −2

Substitute x = 1 Evaluate

y=2×2–4 =0

Substitute x = 2 Evaluate

y=2×3–4 =2

Substitute x = 3 Evaluate

For x = 1,

For x = 2,

For x = 3,

x y

150

0 −4

1 −2

2 0

3 2

Mathspace WA Curriculum – Year 9 mathspace.co

b Using the table of values, plot the points onto a Cartesian plane.

Apply the idea 4

3 2 1 −4 −3 −2 −1

−1

x 1

−2 −3 −4

c Using the points plotted above, sketch the line that passes through two points.

Create a strategy Connect the plotted points.

Apply the idea 4

Reﬂect and check Notice that the line passes through all of the plotted points.

y = 2x − 4

2 1 −2 −1 −1

x 1

−2 −3 −4

Idea summary We can sketch a line by using a table of values. We do this by: • Completing a table of values by substituting the x-values, to ﬁnd the corresponding y-value. • Plotting the points onto a Cartesian plane. • Drawing a line through these points.

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Sketch a graph using its intercepts x and y-intercepts are often some of the easiest points to find from an equation to use to plot the line. • The x-intercept is where the line crosses the x-axis. It occurs when the y-value is 0. To find the x-intercept, set y = 0 in the equation and solve for x. • The y-intercept is where the line crosses the y-axis. It occurs when the x-value is 0. To find the y-intercept, set x = 0 in the equation and solve for y.

Interactive exploration

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Example 2 For the linear equation y = 2x − 2: a What are the coordinates of the y-intercept?

Create a strategy

Apply the idea

Substitute x = 0 into the given equation.

y=2×0–2 = −2 The coordinates are (0, −2).

b What are the coordinates of the x-intercept?

Create a strategy Substitute y = 0 into the given equation.

Apply the idea Write the equation

Substitute y = 0

Add 2 to both sides

Evaluate

Divide both sides by 2

Evaluate

The coordinates are (1, 0).

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Substitute x = 0 Evaluate

c Sketch the line y = 2x − 2.

Create a strategy Plot the points found and draw a line through them.

Apply the idea 4

3 2

y = 2x − 2

1 −3 −2 −1

−1

x 1

−2 −3 −4

Example 3 For the equation 2x − 2y = −4: a Complete the table of values. x y

Create a strategy Because we need to substitute multiple values of x into the equation, start by rearranging the equation and then substituting each value.

Apply the idea Start by rearranging the equation: 2x − 2y = −4 2x − 2y + 2y = −4 + 2y 2x = −4 + 2y

Write the equation Add 2y to both sides of the equation Simplify

2x + 4 = −4 + 2y + 4

Add 4 to both sides of the equation

2x + 4 = 2y

Simplify

x+2=y

Divide both sides by 2 Simplify

Now we can use this equation to evaluate y for the x-values in the table.

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For x = 0, y=x+2 =0+2 =2 For x = 1, y=x+2 =1+2 =3 For x = 2, y=x+2 =2+2 =4 For x = 3, y=x+2 =3+2 =5 x y

0 2

1 3

2 4

3 5

b Using the table of values, plot the points onto a Cartesian plane.

Apply the idea 7

6 5 4 3 2 1 −4 −3 −2 −1 −1 −2

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x 1 2 3 4 5 6 7

c Using the points plotted above, sketch the line that passes through two points.

Create a strategy Connect the plotted points.

Apply the idea

Reﬂect and check 7

Notice that the line passes through all of the plotted points.

6 5 4 3 2 1 −4 −3 −2 −1 −1

2x − 2y = −4 x 1 2 3 4 5 6 7

−2

Idea summary To sketch a line using the x and y-intercepts: • Set y = 0 to ﬁnd the x-intercept. • Set x = 0 to ﬁnd the y-intercept. If the equation is in gradient-intercept form ( y = mx + c), we don’t need to let x = 0, because we know that the y-intercept is c.

Sketch a graph using its gradient and one point We can sketch a graph by identifying one point and its gradient. For example, if a line passes through the point (−2, −1) and has a gradient of 3: 4

First, we plot the point (−2, −1)

3 2 1 −3 −2

−1

x 1

−2 −3 −4

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155

We then use the gradient to ﬁnd the second point.

y 4 3

We know that the gradient is 3 or

2 run 1

in fraction

form. We also know that the gradient is x

–3 –2 –1 rise –1 –2

So the rise = 3 and the run = 1.

–3 –4

We then connect the two points with a line.

4 3 2 1 −3 −2

−1

x 1

−1

−2 −3 −4

We can also graph a line by identifying the gradient and the y-intercept, from the equation, when it is in the form y = mx + c. We know that the y-intercept occurs at c, and the gradient is m. y 6

For example, if we have the equation , we know the y-intercept is at (0, 3) and as the

gradient is , another point will be at

(0 + 5, 3 + 1) = (5, 4).

3 2 1

x 1

Note: if the gradient is negative, we will move down instead of up, when ﬁnding the second point.

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Example 4 Sketch the line that passes through the point (−1, 2) and has a gradient of .

Create a strategy We can draw a line by plotting any two points on the line.

Apply the idea 5

We have been given only one point, so we want to use the gradient to ﬁnd a second point.

A gradient of run is 2.

3 2 1 −4 −3 −2 −1

−1

. The rise is 1 and the

Starting with the point (−1, 2), we move two units to the right (run of 2) and then one unit up (rise of 1).

x 1

is already written as

When we do that, we have the second point (1, 3). Using the two points, we can plot the line.

−2 −3

Example 5 Sketch the line that has a gradient of −3 and an x-intercept of −5.

Create a strategy Use the gradient to ﬁnd another point from the x-intercept.

Apply the idea 4

The x-intercept has coordinates of (−5, 0).

The gradient is

the run is 1. We can use this to ﬁnd another point on the line.

2 1 −5 −4 −3 −2 −1 −3

1 −1

so the rise is −3 and

x 1

By moving right 1 unit and down 3 units, we get to the point (−4, −3).

−2 −3 −4

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Now, we can draw our line through these two points.

3 2 1 −5 −4 −3 −2 −1

−1

x 1

−2 −3 −4

Idea summary To sketch a graph using its gradient and one point: • We plot the given point. • Use the gradient to ﬁnd the second point. • If the gradient is positive, we go up for the rise. • If the gradient is negative, we go down for the rise. • Connect the two points with a line. If we are given a line in gradient-intercept form y = mx + c remember that: • The y-intercept is (0, c). • The gradient is m.

4.05 Practice questions Understanding 1

Explorer 1, 2

Adventurer 1–3

Trailblazer 1–3

When sketching the graph of a straight line using its intercepts, what are you speciﬁcally finding? A

The point(s) where the line is steepest.

The points where the line crosses the x-axis and the y-axis.

The point(s) where the line is farthest from the origin.

The point where the line crosses the x-axis.

Fill in each blank to make a true statement. To graph

, plot the coordinate (0, ⬚) on the y-axis, then rise ⬚ units up and run ⬚

units to the right which gives the point ⬚, then draw a line passing through the points ⬚ and (3, 6). 3

158

Explain how to find the x-intercept of the line −x − 8y = 8?

Mathspace WA Curriculum – Year 9 mathspace.co

4, 6, 7a–d, 10–12

Fluency Ex 1

5, 6, 7e–h, 10, 11cd, 12, 13

For each equation: i

Complete the table of values.

Sketch the graph of the line on a Cartesian plane.

y=x+9

y = 3x + 3

y = 3x − 4

y = −2x + 8

y = 4x

Complete the given table of values.

Sketch the graph of the line on a Cartesian plane. y = −3x − 1

Ex 3

y = −x + 1

−2

x x −3 0 3 6 y y

y = 2x − 9

x x −3 0 3 6 y y

Ex 2

−1

x y

For each equation:

5, 8–10, 11cd, 13

For the equation y = −x − 2: a

Complete the table of values.

Sketch the line on a Cartesian plane.

Find the coordinates of the y-intercept.

Find the coordinates of the x-intercept.

Find the y-value when x = 3.

Find the x-value when y = −4.

x y

−2

−1

iii

Sketch the line.

For each equation: i

Find the x-intercept.

y=x+1

y = 3x − 1

y = 3x − 3

y = −2x + 4

Find the y-intercept.

For the equation 2x − 3y = 18: a

Complete the table of values.

Sketch the line on a Cartesian plane.

Find the coordinates of the y-intercept.

Find the coordinates of the x-intercept.

Find the y-value when x = 6.

y = −x + 4

y = −5x − 4

y = −3x + 6 x y

−3

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For each equation: i

Ex 4

Ex 5

Find the x-intercept.

Sketch the line.

iii

3x + y − 1 = 0

9x + 2y + 18 = 0

−6x + 2y + 18 = 0

−20x + 5y − 40 = 0

Sketch these lines: a

A line that passes through the point (−1, 2) and has a gradient of 4

A line that passes through the point (−2, 4) and has a gradient of −3

Sketch: a

A line that has a y-intercept of −2 and whose gradient is 4

A line that has an x-intercept of −2 and whose gradient is 3

A line that has a y-intercept of 2 and whose gradient is

A line that has a y-intercept of 7 and whose gradient is

For each equation: i

Find the gradient of the line.

Find the coordinates of the point on the line where x = 0.

iii

Sketch the line.

y = 3x − 3

y = −2x − 1

y=x−1

Sketch using the gradient and y-intercept: y = 4x − 1

Reasoning

14, 15

b 14, 15

14–16

Elisabeth and Arthur’s teacher asks them to sketch a graph that passes through the point (−1, 3) and has a gradient of

. Both students sketch different graphs.

Elisabeth’s graph 7

Arthur’s graph y

1 −4 −3 −2 −1 −1

x 1

Who was correct? Why?

160

Find the y-intercept.

Mathspace WA Curriculum – Year 9 mathspace.co

1 −4 −3 −2 −1 −1

x 1

Does the graph of y = −4x + 1 rise from left to right? Why?

The equations of y = 7x − 1 and y = 7x + 3 have the same gradient but with different y-intercepts. What can you conclude about their graphs? Why?

Problem-solving 17, 18

17, 18

17–19

Mary is saving money to buy a new bike. She currently has $50 saved and plans to save $10 per week until she reaches her goal. Write an equation representing the relationship between the number of weeks (x) and the amount of money saved ( y). Graph the corresponding line and find how many weeks it will take for Mary to save $150.

For each equation:

Find the gradient of the line.

Find the coordinates of the point on the line where x = 0.

iii

Sketch the line.

6x − 3y + 18 = 0

−6x + 3y + 24 = 0

The perimeter of a rectangle is 10 cm. The length (L) is L cm. a

Write an expression for the width (W ) of the rectangle in terms of L.

Use a table of values to show the relationship between the length (L) and width (W ).

Sketch a graph to visually show the relationship.

Usually, when sketching graphs, there shouldn’t be any endpoints. Should this graph have any endpoints? Why?

4.06 Equation of a straight line After this lesson, you will be able to… • apply the point-gradient formula to find the equation of a line.

The point-gradient formula We know that to ﬁnd the gradient given two points, we use:

We can use one of the points, say (x1, y1), with the equation y = mx + c to ﬁnd c. y1 = mx1 + c where c = y1 − mx1 4.06 Equation of a straight line mathspace.co

161

If we put this in the equation y = mx + c, we get y = mx + y1 − mx1

Substitute c = y1 − mx1

y − y1 = mx − mx1

Subtract y1 from both sides

y − y1 = m (x − x1)

Factor out m

This gives us a new generalisation that we can use to ﬁnd the equation of any line with gradient m, and point (x1, y1). We call this the point-gradient formula.

Example 1 A line passes through point A (8, 2) and has a gradient of 2. Write the equation of the line in gradient-intercept form.

Create a strategy Use the point-gradient formula and rearrange the equation into gradient-intercept form.

Apply the idea y − y1 = m (x − x1)

Start with the point-gradient formula

y − 2 = 2 (x − 8)

Substitute the values for m = 2 and (x1, y1) = (8, 2)

y − 2 = 2x – 16

Expand the brackets

y = 2x – 14

Add 2 to both sides

Reﬂect and check It’s also possible to ﬁnd the equation another way, using just the gradient-intercept form. Let’s take a look: y = mx + c

Start with the gradient-intercept form

2 = 2 (8) + c

Substitute the values for m = 2 and (x, y) = (2, 8)

2 = 16 + c

Evaluate the multiplication

−14 = c

Subtract 16 from both sides

Since we now know the gradient and y-intercept, we can write the equation as y = 2x − 14.

Example 2 A line passes through the points (2, −7) and (−5, 7). a Find the gradient of the line.

Create a strategy Use the gradient formula,

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Apply the idea Write the gradient formula Substitute (x1, y1) = (2, −7) and (x2, y2) = (−5, 7) Evaluate the subtraction = −2

Evaluate

The gradient of the line is m = −2.

b Using the point-gradient formula, express the equation of the line in gradient-intercept form.

Create a strategy Use the point-gradient formula, y − y1 = m (x − x1), substitute the gradient and the coordinates of one point, then rearrange into the form y = mx + c.

Apply the idea y − y1 = m (x − x1)

Use the point-gradient formula

y − (−7) = −2 (x −2)

Substitute (2, −7) and m = −2

y + 7 = −2 (x − 2)

Evaluate the adjacent signs

y + 7 = −2x + 4

Expand the brackets

y + 7−7 = −2x + 4 −7 y = −2x −3

Subtract 7 from both sides Evaluate

Idea summary To ﬁnd the equation of a line given the gradient and a point, use the point-gradient formula:

y − y1 = m (x − x1) m (x1, y1)

is the gradient are the coordinates of the given point

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4.06 Practice questions Understanding 1

Explorer

Adventurer

1–4

Trailblazer

1–4

For each linear relationship, complete the statement: As x increases by 1, y changes by ⬚. a

1 −4 −3 −2 −1

−1

x 1

−4 −3 −2 −1

−2

−3 −4

1 −8

(12, 0)

x y

x+y=9

0 5

2 7

4 9

6 11

8 13

? c

y=x+6

y = −x − 6

y = −x + 6

Identify the gradient of the line.

Identify a point on the line.

y − 6 = −3 (x − 1)

y=x−6

For each equation:

164

−3

(2, 7)

Which is the gradient-intercept form of the equation: x − y + 6 = 0? A

−1

−4

Does the point lie on the line a

x 1

−2

x −3 −2 −1 0 y 4 1 −2 −5

Mathspace WA Curriculum – Year 9 mathspace.co

y+3=x+8

Fluency Ex 1

Ex 2

5a–e, 6, 7ac, 9ab

5b–f, 6, 7bc,9a–c

5e–j,6,8,9bd–f, 10, 11

Find the equation of each line. a

A line that passes through the point (−5, −4) and has a gradient of −4.

A line that passes through the point (−4, 3) and has a gradient of 4.

A line that passes through the point (7, −6) and has a gradient of −3.

A line that passes through the point (−4, 3) and has a gradient of −9.

A line that passes through the point (−6, −3) and has a gradient of −2.

A line that passes through the point

A line that passes through the point (4, 3) and has a gradient of

A line that passes through the point

A line that passes through the point (8, 1), and has a gradient of .

A line that passes through the point (−4, 5) and has a gradient of

and has a gradient of 2. .

and has a gradient of 7.

For each line described: i

Find the equation.

A line has gradient 2 and passes through the point (−5, −3).

A line has gradient

Sketch the graph.

and passes through the point (−2, 2).

For each line described: i

Find the gradient.

A line passes through the points (3, −3) and (5, −11).

A line passes through the points (−7, 1) and (5, −2).

A line passes through the points (−6, 7) and (−8, −4).

Find the equation of a line which has a gradient of answer in the form ax + by + c = 0.

For each table of values:

Find the equation.

and crosses the x-axis at −10. Write your

Find the equation of the line in the form y = mx + c.

Complete the table.

x 1 2 3 4 y −3 −6 −9 −12

x y

−28

−4 −9

−2 −8

0 −7

2 −6

x −15 1 2 3 y 82 74 66

4 58

x y

−8 21

−4 13

3 −1

6 −7

x −3 −1 1 3 y 18 2 −14 −30

x y

−9 −45

−6 −30

3 15

11 55

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165

Find the equation of the line that: a

Passes through the points (9, −5) and (0, 40).

Passes through the points (−2, −4) and (10, 20).

Passes through the points (10, 6) and (15, −24).

Passes through the points (−3, −13) and (9, −21).

Does the equation have a gradient of 7 and pass through the point (1, 4)? a

Reasoning

12, 13

d 12–15

Is it possible to ﬁnd the equation of a line if you are only given the coordinates of intercepts? Explain.

In the point-gradient form of the equation of a line, y − y1 = m (x − x1), what would the formula become if the line is horizontal?

If an equation is given in point-gradient form, y − y1 = m (x − x1), how could you sketch the graph of the line?

Write down the equations of three lines that pass through the point (1, 3). Explain how your lines are different.

Problem-solving 17, 18

16–18

16–20

Find the equation of the line that passes through (−4, 6) and is parallel to the line 5x − 4y + 2 = 0.

The equation of a line is 2x + y − 8 = 0.

Find the x-intercept of the line.

Find the equation of a line with a gradient of −4 that passes through the x-intercept of the given line.

A line has a gradient of

and passes through the midpoint of A (−6, −6) and B (8, 8).

Find the coordinates of M, the midpoint of AB.

Find the equation of the line.

A circle with centre C (11, 13) has a diameter with endpoints A (5, 14) and B ( p, q). a

Find the value of p.

Find the value of q.

Find the equation of the line passing through B with gradient .

The line 4x + y = 12 intersects the x-axis at P and the y-axis at Q. Let R be the origin. Find the area of △PRQ.

166

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4.07 Direct proportion After this lesson, you will be able to… • define what is directly proportional relationship. • determine the constant of proportionality in a direct proportion. • identify a proportional relationship in a graph or table of values. • solve problems involving directly proportional quantities.

Direct proportion Two quantities are said to be directly proportional if an increase in one leads to an increase in the other, or if a decrease in one leads to a decrease in the other. This means that they change at the same rate. For example, if you earn $18 per hour, your total income depends on the number of hours you work. The more hours you work, the more money you make. We can write this as: Income = $18 × hours worked or y = 18x where y is your income, and x is the number of hours worked. The constant of proportionality is the number that links the two quantities together. In this case, it’s 18, because for every hour you work, you earn $18. Notice that this constant is the same as the slope or gradient of the line when you graph the relationship. For directly proportional relationships, the starting point ( y-intercept) is always zero, meaning if you work 0 hours, you earn 0 dollars. The general equation for any directly proportional relationship is:

y = kx k

is the constant of proportionality

The constant of proportionality is also called the unit rate. While it can be any number except zero, we usually focus on positive values for k.

Interactive exploration

Discover this concept in action online

mathspace.co

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167

Example 1 Consider the values in each table. Which of them could represent a directly proportional relationship between x and y? Table A: x y

1 2

Table B: 2 8

3 18

x y

4 32

1 2

2 4

3 6

4 8

Create a strategy Determine which table shows that the y-values are a constant multiple of the x-values.

Apply the idea Table A x

2 3 4 ×2 ×4 8 18 32

Multiplier has changed not proportional

Table B x y

2 3 4 ×2 ×2 ×2 ×2 2 4 6 8

Multiplier has stayed the same proportional

Example 2 Consider the equation P = 90t. a State the constant of proportionality.

Create a strategy

Apply the idea

For equations in the form y = kx, k is the constant of proportionality.

The constant of proportionality in P = 90t is 90.

b Find the value of P when t = 2.

Create a strategy Substitute the given value into the equation.

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Mathspace WA Curriculum – Year 9 mathspace.co

Apply the idea P = 90t

Write the equation

= 90 × 2

Substitute t = 2

= 180

Evaluate

Example 3 Oliver is making cups of fruit smoothie. The amount of bananas and strawberries he uses is shown in the proportion table. Strawberries Bananas

8 5.5

16 11

24 16.5

32 22

40 27.5

a Sketch the graph of this proportional relationship.

Create a strategy Use two pairs of values from the table as coordinates of points in the graph.

Apply the idea 11 10 9 8 7 6 5 4 3 2 1

The points (8, 5.5) and (16, 11) have been plotted, and a line drawn through them.

Bananas

Strawberries 0

10 12 14

Reﬂect and check Conﬁrm that the other points (24, 16.5), (32, 22), and (40, 27.5) lie on the same line you have created. b What is the unit rate of this relationship?

Create a strategy The unit rate is the gradient or constant of proportionality.

Apply the idea 11 10 9 8 7 6 5 4 3 2 1

We can see in the graph that rise = 5.5 and run = 8.

Bananas

5.5 units

8 units

Strawberries 0 2

10 12 14

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c Select ALL the statements that describe the proportional relationship. A

For every 5.5 bananas Oliver uses, he adds 8 strawberries.

The unit rate of bananas in respect to strawberries is

For every 8 bananas, William uses 5.5 strawberries.

The unit rate of bananas in respect to strawberries is

Create a strategy Use the table given in the question, and the unit rate from part (b).

Apply the idea From the table, when William uses 5.5 bananas, he adds 8 strawberries. So option A is correct. The unit rate is

bananas per strawberry. So option B is correct.

Reﬂect and check We can also see in the graph that for every 5.5 bananas Oliver uses, he increases or adds 8 strawberries.

Bananas

11 10 9 8 7 6 5 4 3 2 1

5.5 units

This also shows why option A is correct.

8 units

Strawberries 0

10 12 14

Idea summary A directly proportional relationship is a linear relationship with a y-intercept of zero:

y = kx k

is the constant of proportionality

4.07 Practice questions Understanding 1

Adventurer

Trailblazer

1–3

Find the gradient of each linear relationship represented in the table. a

170

Explorer 1–3

x 1 2 4 8 y 3 6 12 24

Mathspace WA Curriculum – Year 9 mathspace.co

x y

0 14

2 9

4 4

6 −1

Which graphs indicate that y is directly proportional to x? a

1 −4 −3 −2 −1

−1

x 1

−4 −3 −2 −1

−2

−3 −4

−4 −3 −2 −1

−1

x 1

−2

−3

−4

Is each statement true or false? a

In a directly proportional relationship, the constant of proportionality represents the gradient of the line.

The graph of any directly proportional relationship will go through the point (0, 0).

The graph of a directly proportional relationship can be curved.

A directly proportional relationship can be increasing or decreasing.

The equation

Fluency 4

−2

−4

−1

x 1

−1

−3

−4 −3 −2 −1

represents a direct proportionality between two quantities y and x.

4, 6, 7ab, 8, 10, 12, 14ac

5, 6, 7bc, 9–12, 14bc

Given that y = 2x, ﬁnd the missing values in the table.

Given that y = 0.4x, ﬁnd the missing values in the table.

5, 6, 7cd, 9–11, 13, 14bd, 15

x y

4.07 Direct proportion mathspace.co

171

Ex 1

Which could represent a directly proportional relationship between x and y? a

x 1 3 5 7 y 20 16 12 8

x y

1 16

5 12

6 8

20 4

x 1 2 3 4 y 5 20 45 80

x y

1 5

2 10

3 15

4 20

Find the equation relating the given variables in each table. a

a 0 1 2 3 b 0 9 18 27

x y

0 −10

4 −2

9 8

16 22

s 0 5 10 15

t 8

Ex 2

Ex 3

q is directly proportional to r and q = 10 when r = 20. a

Graph the relationship between q and r.

Find the gradient of the line.

Express q in terms of r.

Find the value of q when r = 30.

Find the value of r when q = 22.

a is directly proportional to b and a = 54 when b = 9. a

Graph the relationship between a and b.

Find the gradient of the line.

Express a in terms of b.

Find the value of a when b = 7.

Find the value of b when a = 60.

Given A = 70w: a

Find the value of A when w = 4.

What is the constant of proportionality?

Given

What is the constant of proportionality?

Find the value of M when n = 50.

William is making rice. The amount of rice and water he uses is in the table. Rice (g) Water (mL)

172

100 210

110 231

120 252

130 273

140 294

Sketch the graph of the proportional relationship.

Find the unit rate of the relationship between rice and water.

Write two statements that describe the proportional relationship.

Mathspace WA Curriculum – Year 9 mathspace.co

The table shows the proportional relationship between hours spent ﬁshing and the number of ﬁsh caught:

Hours Fish caught

4 9

8 18

12 27

Sketch the graph of the proportional relationship.

Find the unit rate of the relationship.

By considering parts (a) and (b), make a conclusion about the number of ﬁsh caught.

For each table: i

Sketch the graph.

x 5 10 15 y 9.5 19 28.5

x 15 25 35 45 y 36 60 84 108

ii 20 38

25 47.5

x y

10 7.5

20 15

30 22.5

40 30

55 132

x y

1 1.25

3 3.75

5 6.25

7 8.75

Find the constant of proportionality, k.

Reasoning

16, 17

50 37.5 9 11.25

when x = 3:

If y and x are directly proportional, and a

Find the constant of proportionality.

Write the equation for y in terms of x.

16, 17

16, 18

Two quantities are directly proportional. If one of the quantities increases by a factor of 7 (is multiplied by 7), what happens to the other quantity?

Jimmy and Ben are both saving money for a vacation. Jimmy can represent his savings over , where x represents the number of days and y represents the

time using the equation

savings. Ben saves $7 every 4 days. a

Complete the table showing Ben’s savings over time. Days Amount saved

8 14

12 21

Graph the amount each of them have saved over time.

Explain how to determine who is saving more per day.

A real estate agent advertised a house for sale online and in the newspaper. The agent received 4 online enquiries each day and 2 enquiries each day through the newspaper advertisement. a

Let y represent the number of online enquiries received x days after initially advertising the house. Write the equation for y based on x. The line representing the number of online enquiries has been graphed. On the same set of axes, graph the line representing the number of newspaper enquiries after x days.

y 40 35 30 25

Online

20 15 10 5

x 0

Why is the line representing the number of online enquiries steeper than the line representing the number of newspaper enquiries?

Why do both lines pass through the origin?

4.07 Direct proportion mathspace.co

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Problem-solving 19–22

20–24

20, 23–26

Write an equation involving a constant k such that m varies directly with p.

The revenue of a company, r, varies directly with the number of sales, n. Write an equation involving a constant k.

Paul paints 15 plates every 6 hours. a

Complete the proportion table: Hours worked Plates painted

b 22

6 15

12 20

174

24 40

36 60

48 80

Complete this table for Harry: Number of weeks Number of books read

Harry and Carl love reading. Harry reads 16 books every 12 weeks. Carl has kept a table of his reading habits:

Sketch the graph of this proportional relationship.

Number of weeks Number of books read

12 30

12 16

48 32

60 80

Who reads the fastest?

Derek and Yuri are building robot kits. Derek can make 2 robots every 3 hours. Yuri can make 3 robots every 6 hours. a

Sketch the graphs of both proportional relationships on the same Cartesian plane.

Who is the faster builder?

The original of a printed image measures 8.5 cm in width and 51 cm in length. When a customer wants to print a copy of the original, they are offered prints in various sizes, but the width and length are in the same ratio as the original so that the photo does not appear distorted. a

Find the constant of proportionality in the equation length = k × width.

Find the length of a copy if the width of the copy is 14 cm.

The height, h, of a regular tetrahedron varies directly with its side length, s. A particular tetrahedron with a side length of 3 cm has a height of 2.45 cm. a

Find the constant of proportionality, k, to two decimal places.

Express h in terms of s.

Find the height of a different tetrahedron with a side length of 2 cm.

The amount A (in grams) of a substance dissolved in water is directly proportional to the capacity C (in litres) of water used. When 5 litres of water is used, 15 grams of the substance is dissolved. a

Find an equation relating the amount and capacity.

How much of the substance will be dissolved in 7 litres of water?

Mathspace WA Curriculum – Year 9 mathspace.co

4.08 Optional: Indirect proportion After this lesson, you will be able to… • express an inverse relationship mathematically and interpret its graph. • apply the knowledge indirect proportions to solve related problems.

Indirect proportion Some quantities have what we call an inverse relationship. As one of the quantities gets closer to zero, the other one gets further from zero, and vice versa. The relationship between speed and time is inversely proportional. For example, if a truck driver needs to drive 1000 km to deliver a load, the faster the driver travels, the less time it will take to cover this distance. Speed (km/h) Time (hours)

60 16.7

70 14.3

80 12.5

90 11.1

100 10

110 9.1

120 8.3

Plotting these values, the graph of the curve is given by: Each point on the curve corresponds to a different combination of speed and time, and all the points on the curve satisfy the equation

Time (hours) 35 30 T=

, or TS = 1000, where T is the time

taken and S is the speed travelled.

(60, 16.7)

(100, 10)

10 5 0

1000 S

Speed (km/h) 20 40 60 80 100 120 140 160

We can see from both equations that as average speed increases, the time taken will decrease. This is what we call inverse proportion, also known as inverse variation. Mathematically, we write this as y ∝

to say “y is inversely proportional to x”.

Two quantities x and y that are inversely proportional have an equation of the form: y = or xy = k or x = where k ≠ 0. When we graph this relationship, it will be non-linear.

4.08 Optional: Indirect proportion mathspace.co

175

Example 1 The equation y =

represents an inverse relationship between x and y.

a Complete the table of values: −5

x y

−4

−3

−2

−1

Create a strategy Substitute the values of x into y =

Apply the idea

Reﬂect and check

For x = −1, we get y = 6.

We could check our values by sketching a graph of the relationship using technology and see

For x = −2, we get y = 3. Similarly, substituting the other values of x, the complete table of values is given by: −5

x y

−4

−3

−2

−1

that y =

looks like this: 6 5 4 3 2 1 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4

b Is y = A

x 1

2 3 4

increasing or decreasing when x < 0?

Increasing

Decreasing

Create a strategy

Apply the idea

Consider the table of values from part (a). Look what happens to y as the values of x increase.

Based on the table of values, y values increase as the values of x increase. The correct answer is Option A.

c Describe the rate of change of the function when x < 0. A

As x increases, y increases at a faster and faster rate.

As x increases, y increases at a slower and slower rate.

As x increases, y increases at a constant rate.

Create a strategy Look at how much the value of y changes by each time.

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Apply the idea Let’s consider some values from the table: −4

−3

−2

Based on the table, y increases by

from x = −4 to x = −3. Then y increases by 1 from x = −3 to x = −2.

This means that as x increases, y increases at a faster and faster rate, so the correct answer is Option A.

Idea summary Two quantities, x and y, that are inversely proportional have an equation of the form: y=

or xy = k or x = ,

where k ≠ 0.

4.08 Practice questions Understanding

Explorer 1, 2, 3a–d

Adventurer

Trailblazer

1–3

Is each statement true or false? a

The equation y =

For an inversely proportional relationship, as x gets closer to zero, y gets further from zero.

In an indirectly proportional relationship, as one quantity increases, the other decreases.

If y is inversely proportional to x and x doubles, then y also doubles.

represents a direct proportionality between two quantities, y and x.

Determine whether each of the following is an example of a direct variation or an inverse variation: a

The variation relating the distance between two locations on a map and the actual distance between the two locations.

The variation relating the number of workers hired to build a house and the time required to build the house.

Does each relationship between x and y show direct proportion, inverse proportion, or neither? a e

x2 = y

d xy = −7

y = 2x2 − 7x − 4

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Fluency 4

4, 5ab, 6, 7ab, 9, 12

4, 5cd, 6–8, 10–12

If y is inversely proportional to x, and y = 20 when x = 10: a

Find the constant of variation, k.

Express y in terms of x.

Find the value of y when x = 5.

The equation xy = 7 can be described as a function where y varies inversely with x and has a constant of proportionality, k = 7. Find the constant of proportionality, k, for the following:

Ex 1

178

y = 4 when x = 3

when x = −2

y = −2 when x = 3

when x = −3

Does each table represent an inversely proportional relationship between x and y? a

x 1 2 3 4 y 3 1.5 1 0.75

x y

1 36

2 18

3 12

4 9

x 1 5 6 10 y 3 75 108 300

x y

1 4

2 5

3 6

4 7

The equation

represents an inverse relationship between x and y.

Complete the table of values:

Describe the rate of change of the function when x < 0.

x y

increasing or decreasing when x < 0?

If y is inversely proportional to x, and y = 24 when x = −6: a

Find the constant of variation, k.

Express y in terms of x.

Find the value of y when x = 8.

Find the value of x when y = 16.

q is inversely proportional to r, and q = 10 when r = 20. a

Graph the relationship between q and r.

Find the constant of proportionality.

Express q in terms of r.

Find the value of q when r = 30.

Given that M varies inversely with n, and M = a

What is the constant of proportionality?

Find the value of M when n = 50.

Mathspace WA Curriculum – Year 9 mathspace.co

when n = 10:

−5

−4

−3

−2

−1

If s varies inversely with t, and s =

when t = 6:

Find the constant of proportionality, k.

Write the equation for s in terms of t.

William is using a machine to cook rice. The machine adjusts the cooking time based on the amount of rice, and the relationship is inversely proportional. The data is shown in the table: Rice (g) Time (min)

100 30

110 27.3

120 25

130 23.1

140 21.4

Sketch the graph of the inverse proportional relationship.

Find the constant of proportionality.

Two statements that describe the inverse proportional relationship.

Reasoning

14, 15

Two quantities are inversely proportional. If one of the quantities increases by a factor of 7, what happens to the other quantity?

A real estate agent advertised a house for sale online and in the newspaper. The agent received 4 online enquiries on the first day, and the number of online enquiries decreased over time. The agent also received 2 newspaper enquiries on the first day, which also decreased over time.

Let y represent the number of online enquiries received x days after initially advertising the house. Assume y is inversely proportional to x. Write the equation for y in terms of x.

Why does the line representing the number of online enquiries decrease at a faster rate than the line representing the number of newspaper enquiries?

Boyle’s law describes the relationship between pressure and volume of a gas of ﬁxed mass under constant temperature. The pressure for a particular gas can be found using

where P has units kg/cm2 and V has units cm3. a

What is the pressure if the volume is 1 cm3?

What happens to the pressure as the volume increases? Explain.

Can the pressure ever get to 0 kg/cm2? Explain.

Problem-solving 16 16

17, 18, 20

17–20

The height, h, of a structure varies inversely with its base length, s. A particular structure with a base length of 3 cm has a height of 2.45 cm. a Find the constant of proportionality, k, rounded to two decimal places. b Express h in terms of s. c Find the height of a different structure with a base length of 2 cm.

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The values in the table represent the relationship between m and p, where m is inversely proportional to p. Find the values of a and b.

p m

4 63

6 b

7 36

a 28

The time, t, taken by a typist to type up a document is inversely proportional to his typing speed, s. That is, the quicker the typing speed, the less time it will take. It takes a typist 20 minutes to type a particular document, typing at a speed of 61 words per minute. a Find the constant of variation, k. b How long, in minutes, will it take a typist with a typing speed of 30.5 words per minute to type up the document?

A particular shelter has cylindrical pillars that can hold a maximum load, L, that varies directly as the fourth power of the diameter, d, and inversely as the square of the height, h. If 6 m tall pillars with diameters of 1 m can support a maximum load of 18 metric tons, ﬁnd the maximum load that can be supported by a pillar 2 m tall and

m in diameter.

Write an equation involving the constant k for the revenue of a company, r, that varies inversely with the number of sales, n.

4.09 Model linear equations After this lesson, you will be able to… • identify and differentiate between independent and dependent variables in linear equations. • model real-world situations using linear equations. • apply the concept of slope and y-intercept in the context of linear equations. • solve problems using linear equations.

Application of linear equations We can combine all the skills we have seen related to linear functions and apply them to real-world situations. These often involve the collection of data, which is commonly displayed in a table of values. The independent variable is the variable you can control or change, and is displayed in the ﬁrst row of the table, followed by the dependent variable, which is affected by this change, in the second row. Once we know that the relationship between the variables is linear, we can work out the gradient and y-intercept and express the relationship as a linear equation (or function).

Interactive exploration

Discover this concept in action online

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mathspace.co

We can identify relevant information through various features of the graph: • The slope of the line, expressed as a rate, will be the change in the dependent variable, for each unit change in the independent variable. • The y-intercept will be the initial value of the dependent variable.

Example 1 Petrol costs a certain amount per litre. The table shows the cost of various amounts of petrol in dollars: Number of litres (x) Cost of petrol ( y)

0 0

10 16.40

20 32.80

30 49.20

40 65.60

a How much does petrol cost per litre?

Create a strategy The cost per litre is a rate of change; this is the same as calculating the gradient. Divide the cost of the petrol ( y) by the number of litres (x).

Apply the idea Divide 16.40 by 10

Evaluate

b By using your answer from part a, write an equation linking the number of litres of petrol pumped (x) and the cost of the petrol ( y).

Create a strategy

Apply the idea

The cost of petrol is equal to the cost per litre times the number of litres pumped.

We know the cost of petrol is $1.64 We can write the equation as y = 1.64x.

c How much would 47 litres of petrol cost at this unit price?

Create a strategy

Apply the idea

Using the equation from part (a), subtitute x = 47.

y = 1.64x

Write the equation

= 1.64 × 47

Substitute x = 47

= $77.08

Evaluate

d In the equation, y = 1.64x, what does 1.64 represent? A

The unit rate of cost of petrol per litre.

The number of litres of petrol pumped.

The total cost of petrol pumped.

Create a strategy

Apply the idea

By looking at part (a), 1.64 represents the rate at which y increases.

The correct answer is option A: The unit rate of cost of petrol per litre.

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Example 2 Kerry currently pays $50 a month for her internet service. She is planning to switch to a ﬁbre optic cable service. a Write an equation for the total cost T of Kerry’s current internet service over a period of n months.

Create a strategy To ﬁnd the total cost, we need to multiply the cost of internet a month by the number of months.

Apply the idea Total cost = T, Cost of internet per month = $50, Number of months = n T = 50n

b For the ﬁbre optic cable service, Kerry pays a one-off amount of $1200 for the installation costs and then a monthly fee of $25. Write an equation of the total cost T of Kerry’s new internet service over n months.

Create a strategy The $1200 represents an initial value, which can be represented with a y-intercept (c). To form the full equation, we multiply the monthly fee by the number of months, and then add it to the installation fee.

Apply the idea Total cost = T, Monthly fee = $25, Number of months = n, Installation fee = $1200 T = 25n + 1200

c Fill in the table of values for the total cost of the current internet service, given by T = 50n. n

T (dollars)

Create a strategy Substitute the value of n in T = 50n.

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Apply the idea For n = 18,

For n = 1, T = 50 × 1 = 50

T = 50 × 18 = 900

Substitute n = 1 Evaluate

Substitute n = 18 Evaluate

For n = 24,

For n = 6, T = 50 × 6 = 300

T = 50 × 24 = 1200

Substitute n = 6 Evaluate

Substitute n = 24 Evaluate

For n = 12, T = 50 × 12 = 600

Substitute n = 12 Evaluate

The complete table of values for the total cost of the current internet service: n T (dollars)

1 50

6 300

12 600

18 900

24 1200

d Fill in the table of values for the total cost of the ﬁbre optic cable service, given by T = 25n + 1200. n

T (dollars)

Create a strategy Substitute the value of n in T = 25n + 1200.

Apply the idea For n = 18,

For n = 1, T = 25 × 1 + 1200 Substitute n = 1 = 1225 Evaluate

T = 25 × 18 + 1200 Substitute n = 18 = 1650 Evaluate For n = 24,

For n = 6,

T = 25 × 24 + 1200 Substitute n = 24 = 1800 Evaluate

T = 25 × 6 + 1200 Substitute n=6 = 1350 Evaluate For n = 12, T = 25 × 12 + 1200 Substitute n = 12 = 1500 Evaluate

The complete table of values for the total cost of the ﬁbre optic internet service: n T (dollars)

1 1225

6 1350

12 1500

18 1650

24 1800

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Idea summary • The independent variable is displayed in the ﬁrst row of the table, followed by the dependent variable, in the second row. • The slope is the constant rate of change. If the x-value is increasing by 1, then it will be the increase in the y-values. • The y-intercept is the value of y when x = 0. This may not actually be given in the table, so you may have to count backwards using the rate of change.

4.09 Practice questions Understanding 1

1–2

Trailblazer 1–2

When 4 is added to a certain number, the answer is 6.

Subtracting 10 from a certain number gives a result of 9.

Eleven times a certain number is 88.

A certain number divided by 5 gives a result of 25.

When 13 is subtracted from four times a certain number, the result is 21.

When 11 is added to eight times a certain number, the answer is −5.

When a certain number is subtracted from 20 and this result is then multiplied by 6, the result is −10.

The sum of −1 times a certain number and 4 is divided by 7, which gives a result of 9.

Which is shown ﬁrst in a table of values, the independent or the dependent variable? 3, 4, 7–10

3–5, 7–9, 11

3, 5–8, 11, 12

The table shows the linear relationship between the length of a cellphone call and the cost of the call: Length of call in minutes (x) Cost in dollars ( y)

184

Adventurer

For each statement, write a linear equation using x to represent the unknown.

Fluency Ex 1

Explorer

1 7.6

2 14.4

3 21.2

Write a linear equation to represent the relationship between the cost of a call, y, and the length of the call, x.

Find the cost of a 6-minute call.

Mathspace WA Curriculum – Year 9 mathspace.co

In a study, scientists found that the more someone sleeps, the quicker their reaction time. This table displays the ﬁndings: Number of hours of sleep (x) Reaction time in seconds ( y)

0 6

1 5.8

5 5

How much does the reaction time decrease for each extra hour of sleep? Write a linear equation relating the number of hours of sleep, x, and the reaction time, y.

Calculate the reaction time for someone who has slept 4.5 hours.

Calculate the number of hours someone sleeps if they have a reaction time of 5.5 seconds.

A racing car starts the race with 150 L of fuel. From there, it uses fuel at a rate of 5 L per minute. Complete this table of values: 0

Write a linear equation linking the number of minutes passed, x, and the amount of fuel left in the tank, y.

How many minutes will it take for the car to run out of fuel?

A diver starts at the surface of the water and begins to descend below the surface at a constant rate. This table shows the depth of the diver, in metres, over 4 minutes: Number of minutes passed (x) Depth of diver in metres ( y)

Ex 2

4 5.2

Number of minutes passed (x) Amount of fuel left in the tank ( y)

3 5.4

2 5.6

0 0

1 1.4

2 2.8

3 4.2

4 5.6

Calculate the increase in depth each minute.

Write a linear equation for the relationship between the number of minutes passed, x, and the depth, y, of the diver.

Calculate the depth of the diver after 6 minutes.

Calculate how long the diver takes to reach 12.6 m beneath the surface.

Some friends spent a total of $5.00 on stickers and badges at a store. A sheet of stickers costs $0.50 and a badge costs $1.50. a

Write a linear equation that represents the relationship between the number of sheets of stickers, x, and the number of badges, y, the friends bought for $5.00.

If the friends bought 2 badges, how many sheets of stickers did they buy?

The amount of medication M (in milligrams) in a patient’s body gradually decreases over time t (in hours) according to the equation M = 1050 − 15t. a

After 61 hours, how many milligrams of medication are left in the body?

Calculate the number of hours it will take for the medication to be completely removed from the body.

A carpenter charges a callout fee of $150 plus $45 per hour. a

Write a linear equation to represent the total amount charged, y, by the carpenter as a function of the number of hours worked, x.

Find the total amount charged by the carpenter for 6 hours of work. 4.09 Model linear equations mathspace.co

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A particular restaurant has a ﬁxed weekly cost of $1300 and receives an average of $16 from each customer. a

Write an equation to represent the net proﬁt, y, of the restaurant for the week as a function of the number of customers, x.

Find the restaurant’s net proﬁt if it has 310 customers for the week.

Mario is running a 100 km ultramarathon at an average speed of 9 km/h. a

Write a linear equation to represent the distance Mario has left to run, y, as a function of the number of hours since the start, x.

Find the distance Mario will have left to run after 4.5 hours.

Mohamad is taking his new car out for a drive. He had only driven 50 kilometres in it before and is now driving it down the highway at 75 km/h. a

Write a linear equation to represent the total distance, y, that Mohamad had driven in his car as a function of the number of hours, x.

Find the total distance Mohamad will have driven in his car if his current drive begins at 5:10 p.m. and ﬁnishes at 7:25 p.m.

Reasoning 13

13, 14, 16

Paul has just purchased a prepaid phone, which he intends to use exclusively for sending text messages, and has purchased some credit along with it. After sending 11 text messages, he has $34.39 of credit remaining and after sending 19 text messages, he has $30.31 of credit remaining. The relationship between the amount of credit remaining, y, as a function of the number of text messages, x, is y = −0.51x + 40.

Explain what the slope of the line represents in this context.

Explain what the y-intercept represents in this context.

Find the amount of credit Paul will have left after sending 36 text messages.

A baseball is thrown vertically upward by a baseball player when they are standing on the ground, and the velocity of the baseball, V (in metres per second), after T seconds is given by V = 120 − 32T. When T = 4, there is a negative value of V. Explain what this means in context.

The table shows the water level of a well that is being emptied at a constant rate with a pump: Time in minutes (x) Water level in metres ( y)

2 26.8

5 25

8 23.2

The linear equation which represents the water level, y, as a function of the minutes passed, x, is y = −0.6x + 28

186

Explain what the slope of the function represents in this context.

State the y-intercept.

Explain what the y-intercept represents in this context.

Mathspace WA Curriculum – Year 9 mathspace.co

After Anri starts running, her heartbeat increases at a constant rate. The equation which describes this relationship can be written as y = 3x + 49, where x represents the number of minutes passed, and y represents the heartbeat. a

Explain what the value of 49 represents in this context.

Explain what the slope represents in this context.

Problem-solving 17–19 17

18, 19, 21

17, 19–21

A mobile phone salesman earned $600 in a particular week during which he sold 26 phones, and $540 in another week during which he sold 20 phones. a

Write a linear equation to represent the relationship between the weekly earnings of the salesman, y, and the number of phones sold, x.

Find how much the salesman will earn in a week during which he sells 36 phones.

The cost of a taxi ride is given by C = 5.5t + 3, where t is the duration of the trip in minutes. a

Calculate the cost of an 11-minute trip.

For every extra minute the trip takes, how much more will the trip cost?

What could the constant value of 3 represent in this context?

Lucas is creating digital pixel art. On the ﬁrst day of his project, he created an artwork with a size of 8 megabytes. Inspired by minimalist art, he decided that each subsequent pixel art will be 0.5 megabytes smaller than the previous one. a

What will be the size of the pixel art he creates on the 10th day?

Write a formula to represent the size, S, of the pixel art Lucas creates on the Dth day.

On which day will his digital pixel art’s size be just 1 megabyte?

A house is running a smart speaker in their household for eight hours every day of the year. The price of a smart speaker is $50 and they cost R dollars per hour to run.

Write an equation for the total cost (C dollars) of running this smart speaker for one year in terms of R.

Find the value of R if the total cost of running the smart speaker for one year is $650. Round your answer to two decimal places.

If the smart speaker is permanantly left on (i.e. 24 hours a day), show that the cost does not triple.

Claude is 32 years old and keeps active by running at their local park. a

Use this formula to ﬁnd their maximum heart rate in beats per minute (bpm). Maximum Heart Rate = 220 − Age in Years

Write a linear equation to represent the maximum heart rate, m, as a function of age, a.

Claude will get the most beneﬁt from this exercise if their heart rate is between 50% and 85% of their maximum heart rate. Between what two heart rates should Claude be aiming for to get the most beneﬁt from this exercise?

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4.10 Multi-step equations After this lesson, you will be able to… • solve multi-step equations by using appropriate steps like expanding brackets and combining like terms. • apply the method of clearing fractions by finding the common denominator in equations with fractional terms. • demonstrate the process of isolating variables using inverse operations in equations. • identify errors in solving equations and correct them.

Equations with more than three steps We’ve solved linear equations with up to three steps. For three or more steps, we will have more terms that we need to move or combine. For the equations with more than three steps, we may follow these steps: 1. If necessary, expand any brackets and/or multiply both sides by the common denominator of any fractions. 2. Combine like terms on both sides of the equation. 3. Get all the terms with variables on one side and all the constants to the other side using inverse operations. 4. Solve for the variable. We can still check our answer by substituting the solution into the original equation.

Example 1 Solve 4 (2x + 5) − 3 = 37

Create a strategy Follow the steps for solving equations with more than three steps and solve for x.

Apply the idea Write the equation

Expand the brackets

Combine like terms

Subtract 17 from both sides

Divide both sides by 8

Simplify the fraction

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Reﬂect and check Substitute

into the left-hand side of the equation and check if the resulting value will be the

same value as the right-hand side of the equation. Write the left-hand side of the equation Substitute x =

Evaluate the multiplication inside the brackets

Evaluate the addition inside the brackets

Evaluate the product

Evaluate This is the same value as the right-hand side of the equation. This confirms that x = solution to the equation.

is the

Example 2 Solve 9 (x − 108) − 3 (x + 96) = 96 − 6 (x − 48)

Create a strategy Expand the brackets and solve for x.

Apply the idea Write the equation

Expand the brackets

Simplify

Add 6x to both sides, add 1260 to both sides

Evaluate

Divide both sides by 12

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Reﬂect and check As an alternative strategy, we can divide each term by 3 before expanding brackets to work with smaller numbers: Write the equation

Divide each term by 3

Evaluate

Expand the brackets

Simplify

Add 2x to both sides, add 420 to both sides

Evaluate

Divide both sides by 4

Both strategies yield the same solution, x = 137.

Example 3 Solve

Create a strategy We can start by multiplying both sides by 2 to clear the fractions. Then, we can expand the brackets and solve for a.

Apply the idea Write the equation

Multiply both sides by 2

Expand the outer brackets

Expand the inner brackets

Combine like terms

Add 63 to both sides

Subtract 6a from both sides

Divide both side by 7

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Reﬂect and check Let’s check our solution by substituting a = 6 back into the original equation. Write the equation Substitute a = 6 Evaluate the expressions inside the brackets Evaluate the expressions inside the outer brackets Evaluate the product Combine like terms Rewrite the fractions with common denominator Evaluate the left-hand side Both sides are equal, so a = 6 is the correct solution.

Idea summary To solve equations with more than three steps, start by expanding any brackets and clearing any fractions if necessary. Then combine like terms, get all the terms with variables to one side, and solve for the variable.

4.10 Practice questions Understanding

Explorer

Adventurer

1, 2

Trailblazer 1–3

Complete the solution for the given equation. 5 − 2(3x − 4) = 3(2x + 1) − 7x 5 − 6x⬚8 = ⬚x + ⬚ − 7x 13 − 6x = −x + 3

13 = ⬚x + 3

⬚ = 5x

⬚=x

Write the equation ⬚ the brackets ⬚

Add ⬚ to both sides

Subtract ⬚ from both sides

Divde both sides by ⬚

4.10 Multi-step equations mathspace.co

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For each equation, is the given solution correct or incorrect? a

x = 2, for −2x + 13 = −4 (2x + 1) + 3 (5x − 2)

y = 3, for 3 (3y − 8) − 5 (3y − 8) = 4 (y − 2) − 6 (y − 2)

p = 4, for 2 (3p + 7) + 3 (p − 3) = 8 (p + 10)

d = −9, for

Write the equation without any fractions. Do not solve. b

a c

Fluency

4 Ex 1

5 Ex 2

7 Ex 3

4, 5

4–6, 7a-e

Solve: a

−4a − 11 = −5(a − 2a)

15 − (3b + 1) = 6 (b + 1) + 2b

−6y − 4 (4y − 8) = 3 (−12y − 1)

4 (3h − 9) − 12 = 12 − 8 (h − 9)

− (−8 − 3d) = −2 (1 − d) + 2d

12h + 4 = 5 (6h + 5) + 60

3q + 4 = 6 (5q − 5) + 88

8c − 6 = −8 (6c − 7) + 42

4 (8y + 5) = 3 (6y + 8) + 108

−5 (−5d + 5) = 8 (4d − 8) + 37

9 (2y − 5) − 3y = 10 − 5 (y + 9)

4 (3a + 4) + 5 (a − 3) = 6 (a + 10)

3 (2b + 1) − 3 (3b − 1) = 2 (b + 1)

8 − 6 (2h + 1) = 4 (3h − 1) + 6 (h + 10)

2 (3c + 5) − (3c − 9) = 12 − 2 (3c + 12)

Solve: a

16 (5z + 8) + 2 (3z − 5) = 7 (z + 11) − (2z − 8)

2 (3t − 8) − 4 (3t − 8) = 12 (t − 2) − 6 (t − 2)

−8 (9d + 8) − 6 (8d + 2) = −12 (d + 2) − 7 (d − 3)

9 (c + 5) − 6 (2c + 8) = −5 (c − 2) − 6 (c + 3)

12 − (z − 12) − 6 (2z + 8) = −5 (−z − 2) + 8 (z + 2)

Solve: a

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Reasoning 8

8–10, 12

9–12

Two students, Clara and David, attempted to solve the equation 2z + 6 = 3z − 5. Their workings are shown: Clara’s working:

2z + 6 = 3z − 5 −z + 6 = −5

z−6=5

z = 11

David’s working: 2z + 6 = 3z − 5 −z + 6 = −5

−z = −11

−11 = z

Which student provided the correct solution and why?

Explain the mistake made by the other student.

Alex attempted to solve the equation 4 (2x + 5) − 3 = 37. His working is shown:

4(2x + 5) − 3 = 37

Start with the equation

4(2x + 5) = 40

Add 3 to both sides

2x + 5 = 10

Divide both sides by 4

2x = 15

Add 5 to both sides

x = 7.5

Divide both sides by 2

a 10

What was his mistake?

Solve the equation correctly.

Two students, Alice and Bob, attempted to solve the equation 2 (x + 4) − 3 (x − 3) = x + 7. Their workings are shown: Alice’s working:

2 (x + 4) − 3 (x − 3) = x + 7

2x + 8 − 3x + 9 = x + 7 −x + 17 = x + 7

2x = 10

x=5

Bob’s working:

2 (x + 4) − 3 (x − 3) = x + 7

2x + 8 − 3x + 9 = x + 7 −x + 17 = x + 7

−2x = 10

x = −5

Which student provided the correct solution and why?

Explain the mistake made by the other student.

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193

Solve the equation:

(14p − 28) = −8p + 16 in two different ways.

Method 1: Expand the brackets ﬁrst

Method 2: Clear the fraction ﬁrst

Was one method easier than the other? Explain your answer.

Write an equation that could take at least four steps to solve that meets the criteria. Justify your answer. A solution of a = 3

Problem-solving 13, 14, 16

13–17

A solution of b =

14–18

If the perimeter of this triangle is 263 cm, ﬁnd the value of x.

7x − 21

5x + 28

3x − 14

The cost of a cricket ball is 136 cents more than the cost of a rugby ball. a

Write an expression for the price of a cricket ball in dollars.

If 8 cricket balls and 9 rugby balls cost $119, write a linear equation representing the situation.

Find the cost of a rugby ball.

Find the value of a cricket ball.

One side of a parallelogram is 4 cm shorter than an adjacent side ( y). The perimeter of this parallelogram is 12 cm. a

Write an expression for the length of the shorter side.

Write an equation for the perimeter.

Find the dimensions of the parallelogram

A rectangle with a height of 4x + 8 cm and a width of 4x has the same perimeter as a square with side length 3x + 9 cm. Find the value of x.

This rectangle has a perimeter of 126 + 3y centimetres.

4y + 8

Find the value of y. 4y + 3

Right now, Bianca’s father is 42 years older than Bianca. 4 years ago, her father was 4 times as old as her. Solve for y, Bianca’s current age.

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4.11 Optional: Equations with algebraic fractions After this lesson, you will be able to… • apply the multi-step process in solving equations with fractions. • use the lowest common multiple to clear fractions from equations. • apply inverse operations to isolate the variable and solve equations.

Equations with algebraic fractions To solve equations involving fractions with numerical denominators, we can solve as we did with equations of three or more steps, but at some point we will clear the fractions. We can clear fractions by ﬁnding the lowest common multiple of the denominators and multiplying both sides by it. Using the lowest common multiple will eliminate the fractions and make it easier to solve the equation.

Exploration Let’s look at part of the process for solving this equation. 1

Write the equation

Multiply by the lowest common multiple

Simplify the fraction

1. What do you notice about the coeﬃcients outside the brackets in the last step and the denominators in the ﬁrst step? 2. Will this pattern always hold? For example, does it work for 3. What will happen to the constant, 1, when solving

? ?

Here are some things to remember: • We often start with the inside brackets and work our way out. • The distributive law can still be used if there are more than two terms in the grouping symbol. • We must multiply the whole side of an equation by the lowest common multiple, not just the fractional terms. It is essential to simplify the equation and check your solution to ensure accuracy in your ﬁnal answer.

4.11 Optional: Equations with algebraic fractions mathspace.co

195

Example 1 Solve

Create a strategy We can start by expanding the brackets, and then clear the fraction. Once we have done that we can solve by using inverse operations.

Apply the idea Write the equation

Expand the brackets

Multiply both sides by 8

Combine like terms on the left-hand side

Subtract 15x from both sides

Subtract 21 from both sides

Divide both sides by −14 and simplify

Reﬂect and check An alternative method for solving this problem involves ﬁrst multiplying through by 8, and then expanding the brackets.

Keep in mind that when using this strategy, we multiply each group, not each individual term.

Example 2 Solve

Create a strategy First, we need to ﬁnd the lowest common multiple to clear the fraction. Then we can solve using inverse operations to isolate the variable.

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Apply the idea Write the equation

Multiply both sides by the lowest common multiple, 10

Distribute the 10

Simplify the fractions

Expand the brackets

Add 2x to both sides

Subtract 50 from both sides

Divide both sides by 27

Reﬂect and check We can substitute the value of x to check the answer. Write the equation

Substitute

Multiply both sides by the lowest common multiple, 10

Evaluate the brackets then expand

Multiply both sides by 27

Evaluate both sides

Since both sides are equal, we can conﬁrm that our answer is correct.

Example 3 Solve

Create a strategy First, we need to ﬁnd the lowest common multiple to clear the fractions. Then we can solve using inverse operations to isolate the variable.

4.11 Optional: Equations with algebraic fractions mathspace.co

197

Apply the idea Write the equation

Multiply both sides by the lowest common multiple, 12

Distribute the 12

Simplify the fractions and perform the multiplications

Expand the brackets

Combine like terms

Subtract 73 from both sides and add x to both sides

Divide both sides by −7 and simplify

Reﬂect and check We can substitute the value of x to check the answer. Write the equation

Substitute

Write each term as a fraction with a denominator of 12

Multiply both sides by 12 to clear them

Evaluate the brackets

Expand the brackets and simplify

Multiply both sides by 7

Evaluate both sides

Since both sides are equal, we can conﬁrm that our answer is correct.

Idea summary Solving equations with fractions involves simplifying the equation, clearing fractions by multiplying both sides by the lowest common multiple, and solving for the variable using inverse operations.

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4.11 Practice questions Understanding 1

x = 1 is a solution for the equation

is a solution for the equation

is a solution for the equation 4x − 3 = 3

is a solution for the equation

Ex 2

Fluency

5a-dg, 6, 7a–d

Fill in the blanks: b

4a–d, 5, 6, 7a–f

5gh, 6e–h, 7, 8a–f

Solve: a

Ex 1

Trailblazer 1–3

Is each statement true or false?

Adventurer 1–3

What is the lowest common denominator of each set of fractions? a

Explorer 1, 2

Solve:

4.11 Optional: Equations with algebraic fractions mathspace.co

199

Ex 3

Solve:

Solve: a

Reasoning 9

9, 10

Dylan attempted to solve the equation

9–11

. Here is his work:

8 × (3x + 6) = 4 − 2 × x

Cross multiply

24x + 48 = 4 − 2x

Expand the brackets and simplify

26x = −44

Add 2x to both sides and subtract 48 from both sides Divide both sides by 26 and simplify

Identify where Dylan has made an error and explain what it is.

Solve the equation correctly, showing the correct steps of work.

Complete the equation so it is true for all values of x.

Give an example of a linear equation with fractions that has no solutions when simplified. Justify your answer.

Problem-solving 12, 14 12

12–14

A store sells gadgets and tools. On a given day, it sold a total of 36 items. The number of gadgets sold is x, and the number of tools sold is 36 − x. Profits are

200

14–16

and

dollars from tools. The total profit was $76:

Write an equation using fractions to represent the total profit.

Find x, the number of gadgets sold.

Mathspace WA Curriculum – Year 9 mathspace.co

dollars from gadgets

A worker completes a task by assembling parts in two stages. In the first stage, she assembles x parts, taking hours. In the second stage, she assembles the remaining parts, hours, where the total number of parts is 24. The total time for the task is 5

taking hours.

Find x, the number of parts assembled in the first stage. 14

If x = 2 and

At an annual charity event, the money raised was distributed in several different ways. One-third of the funds came from the silent auction, art sales comprised one-fourth of the total, and direct in-person donations made up one-ﬁfth of the revenue. The food booth raised $250. How much money was raised at the charity event? Round your answer to two decimal places.

, ﬁnd the value of k.

, ﬁnd the value of r.

4.12 Optional: Rearrange formulae After this lesson, you will be able to… • apply correct algebraic operations such as addition, subtraction, multiplication, and division to rearrange equations effectively. • solve real-world problems that require making different variables the subject.

Rearrange formulae In mathematics and science, there are many formulas that are written in a specific way.

Exploration Let’s look at the formula for speed:

s d t

speed distance time

4.12 Optional: Rearrange formulae mathspace.co

201

1. Use the formula to find each of the following: • The speed at which a person travels if they drive 100 km in 75 min • The distance travelled when walking 5 kph for 30 min • The distance travelled by a car driving 85 kph for 5 hr • The time it will take to travel 1 000 km if you walk 80 m/min • How fast a plane is travelling if it can fly 2789 km in 6 hr • How long it will take to walk to the store 2 km away if you walk 100 m/min 2. Which ones took the most effort to solve? 3. If we always wanted to find the distance, given time and speed, how could we write the formula to make these calculations more efficient? A variable is another term for a letter, but it is specifically used in the context of formulas and functions to represent a value that can change or vary. By rearranging a formula for a variable of interest, we can make these calculations more efficient. To rearrange an equation and make a specific variable the subject, we need to perform various algebraic operations such as addition, subtraction, multiplication, and division. The key is to isolate the desired variable on one side of the equation while maintaining the balance of the equation. Here are some general steps to follow: • Identify the variable that you want to make the subject. • Perform algebraic operations on both sides of the equation to isolate the desired variable, remembering to use the correct order of operations. • Simplify the equation as much as possible, and always double-check your steps to ensure accuracy.

Example 1 Rearrange kx = m + nx to make x the subject.

Create a strategy Notice the like terms. Rearrange so that all the terms in x are on one side of the equation.

Apply the idea Write the equation

Subtract nx from both sides

Take out the common factor of x

Divide both sides by (k − n) to make x the subject

Reﬂect and check When the variable we want to make the subject appears more than once in the equation, we should get them all on the same side and factorise it out.

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Example 2 Make M the subject of

Create a strategy We need to have one instance of M in the equation, so we ﬁrst need to clear the fraction and then factorise.

Apply the idea Write the equation

Multiply both sides by (M + A)

Simplify

Expand the brackets

Subtract M from both sides

Subtract PA from both sides

Factorise each side of the equation

Divide both sides by (P − 1)

Simplify

Reﬂect and check We can substitute values for A and P, to see if we get the same value solving for M in the original equation and using the rearranged equation. However, this does not guarantee we haven’t made any errors. It is worth checking every step for errors.

Example 3 Make R the subject of V = IR − E.

Create a strategy I is multiplied by R, then E is taken away to give V. Reverse these operations.

Apply the idea Write the equation

Add E to both sides

Divide both sides by I

Write R as the subject

4.12 Optional: Rearrange formulae mathspace.co

203

Example 4 Find the gradient and use the gradient and intercept method to sketch the graph of 6x + 2y − 12 = 0

Create a strategy Make y the subject of 6x + 2y − 12 = 0, which will put the equation in the form y = mx + c. From this form, we can identify the gradient (m) and the y-intercept (c).

Apply the idea Make y the subject of 6x + 2y − 12 = 0. 6x + 2y − 12 = 0 2y − 12 = −6x

Write the equation Subtract 6x from both sides

2y = −6x + 12

Add 12 to both sides

y = −3x + 6

Divide both sides by 2

From the equation y = −3x + 6, the gradient is −3 and the y-intercept is 6. Sketch the graph. Plot the y-intercept (0, 6) and then apply the gradient of −3. Step across 1 unit and down 3 units. y 7 6 5 4 3 2 1

−1

Draw the line. y 7 6 5 4 3 2 1 −1

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x 1

Idea summary We can rearrange equations and formulas to make different variables the subject. This can make it easier to do repeated calculations and for graphing equations that are not in the form y = mx + c. Carefully using inverse operations, we can rearrange equations in a similar way to solving equations. It is very important to keep the equation balanced by doing operations on both sides. However, instead of operating on numbers, we are operating on variables.

4.12 Practice questions Understanding

Explorer 1–4

Adventurer

Trailblazer

1–4

Why do we rearrange formulae?

For the formula, Q = m × c × ΔT:

1–4

• Q is the amount of heat absorbed in Joules, (J) • m is the mass in grams (g) • c is the heat capacity (J/g °C) • ΔT is the temperature change, (°C) We are given the amount of heat absorbed, the mass, and temperature change for ten different substances. Which variable should we make the subject to ﬁnd the unknown values more eﬃciently? 3

To make p the subject of R = pq − 4pt2, which option could be a ﬁrst step? A

Divide pq − 4pt2 by p

Multiply both sides of the equation by 4pt2

Factorise p from pq − 4pt2

Add p to one side of the equation

Is each statement true or false? a

When we rearrange a formula, we will get a numerical value.

When rearranging a formula, the formula may involve numerical and algebraic terms.

We use inverse operations to make a variable the subject.

When we rearrange a formula, we will eliminate all of the other variables.

Fluency Ex 1

5a–d, 6a–e, 7a–c, 8ab, 10, 11a–f, 12ab, 14–17

5–7, 8a–c, 9, 10, 11d–h, 12bc, 14–16, 18

5–9, 11e–j, 12cd, 13–15, 18, 19

Make x the subject of these equations: a

y=x+z

y = 5 (8 + x)

3y = 2x + z

y = 5 (3 − x)

4.12 Optional: Rearrange formulae mathspace.co

205

Make m the subject of these equations: a e

b = −2am + 10

Ex 4

x = 8k (n + m)

x = −7m (p + q)

y = −5mx − 14

x = −7k (l − m)

8x3 = 4y − 12

9x2 + 3y = 12

Make k the subject of these equations: a

Ex 3

y = 3mx − 5

Make y the subject of these equations: a

Ex 2

Make p the subject of the equation M = pq − 7pr2.

Make v2 the subject of the formula for kinetic energy,

Make y the subject of these equation:

y − x + 1 = 0

−4y −8x + 12 = 0

7y + 14x − 21 = 0

−6y + 24x − 30 = 0

5y − 15x + 25 = 0

3y + 6x − 9 = 0

−2y + 4x + 8 = 0

10y − 20x − 30 = 0

6x + 4y − 18 = 0

4x − 3y − 12 = 0

Identify the gradient and y-intercept and graph each equation. a

5x + 2y − 10 = 0

−4x + y + 16 = 0

6x + 3y − 12 = 0

−12x + 4y − 24 = 0

The bend allowance for sheet metal is given by the formula

Make T, the thickness of the sheet, the subject of the formula. 14

Make r2 the subject of these formulas: The volume of a cone is given by the formula

The area of a circle is given by the formula A = π r2.

The area of a sector of a circle is given by the formula

Make the mentioned variables the subject in the following equations: a

206

E = mc2, make m the subject 2

A = P (1 + r)t, make P the subject

P = I R, make R the subject

V = π r2 h, make h the subject

, make h the subject

D = b2 − 4ac, make c the subject

, make z the subject

g = hx2 k, make h the subject

, make c the subject

d = uv4 − t, make u the subject

Mathspace WA Curriculum – Year 9 mathspace.co

The formula for the force of an object is F = ma, where F is the force, m is the mass of the object, and a is the object’s acceleration. Write an equation that could be used to solve for the mass of the object.

The formula for the perimeter of a rectangle is P = 2l + 2w, where P is the perimeter, l is the length, and w is the width. Write an equation that could be used to ﬁnd the length of the rectangle.

Betsy knows the formula to convert Celsius into Fahrenheit is

. She wants to

convert a temperature from Fahrenheit to Celsius. Find an equation that she could use to do this conversion. 19

The displacement of an object is given by used to ﬁnd v0.

Reasoning

20, 22

. Write an equation that could be

20–23

Describe the similarities and differences of solving 3x + y = z and 3x + 8 = 2. Explain your thinking.

Two students, Alice and Bob, are given the Pythagora’s theorem equation to rearrange: c2 = a2 + b2 Alice rearranges the equation to make a2 the subject, while Bob rearranges the equation to make b2 the subject. Their results are as follows: Alice: a2 = c2 − b2 Bob: b2 = c2 + a2 Determine which student, if any, is correct in their rearrangement of the equation and explain your reasoning.

Pamela attempted to make r the subject of the equation

and followed these steps:

Multiplied both sides by R + r

Divided both sides by L

Subtract R from both sides of the equation

What was Pamela’s mistake? Explain and correct her error.

4.12 Optional: Rearrange formulae mathspace.co

207

Ruﬁno has submitted the following work to write the equation for the mass of an object, m, given the kinetic energy, KE, and velocity, v:

Was Ruﬁno correct? If so, explain each step of his working. If not, ﬁx his error(s).

Problem-solving 24, 25

The area, A, of a triangle is given by the formula and b is the length of its base.

208

24–27

, where h is the height of the triangle

Write an equation that could be used to solve for h.

Find the height of a triangle with a base of 12 cm and an area of 18 cm2.

The volume of a sphere is given by the formula

, where r is the radius of the sphere.

Rearrange the formula to make, r the subject of the formula.

Calculate the radius of a sphere with a volume of 91 cm3 to two decimal places.

When a certain amount of money, PV, is invested, its future value, FV, is given by the formula FV = PV periods.

24, 25, 27

, where r is the rate per time period and n is the number of time

Make PV, the present value, the subject of the equation.

How much must be invested now, if it is to grow to $3000 in 10 years with interest of 4% compounded annually? Round to the nearest whole dollar.

The volume of the cylinder that ﬁts exactly into a sphere is given by the formula:

Rearrange the formula so that r is the subject.

What is the radius of the sphere if the volume of the cylinder is 13.75π mm3 and the height of the cylinder is 5 mm? Round your answer to the nearest millimetre.

Mathspace WA Curriculum – Year 9 mathspace.co

4 Chapter review 1

Find the coordinates of the midpoint M of segment AB. a

A (−5, −1) and B (3, 5)

A (−5, 3) and B (1, −7)

y 7 6 5 4 3 2 1

−8 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8

x 1 2 3 4

1 2

A (−2, 2) and B (−2, −2)

A (−4, 9) and B (5, 9)

If the midpoint of points A and B is M, what are the coordinates of A? a

Find the midpoint of segment AB. a

3 2 1

−6 −5 −4 −3 −2 −1 −1 A −2 −3

B (4, 8) and M (−1, 7)

B (−4, −9) and M (2, −3)

Describe the gradient of the lines as negative, zero, positive, or undefined. a

2 1

−2 −1

−1

−7 −6 −5 −4 −3 −2 −1

x 1

−2 −3

−3

−4

−5

−6

−7

d 8

4 3

1 −1

x 1

5 4 3

−2

−3

−4

x 1

−1

−2

−4 −3 −2 −1

x 1

Chapter 4 review mathspace.co

209

Find the gradient of the line that passes through the given points: a

(5, 1) and (2, 7) and

(−9, −5) and (−9, −3)

(−1.6, 8) and (0.2, 8)

Complete the coordinates using the points and gradient provided.

(3, −3) and (5, ⬚), gradient = −3

Use Pythagoras’ theorem to find the length of side AC, rounded to two decimal places. a

y 4

b C

2 1 1

1 1

−2

−3

−5 −4 −3 −2 −1 −1

−2

−3 −2 −1 −1

−3

−4

Use the distance formula to find the length of the interval AB. Write your answer in exact form. a

y A

−4 −3 −2 −1 −1

2 1 −4 −3 −2 −1 −1

x 1

−3

−4

−2

x 1

−5

Find the perimeter of the parallelogram whose vertices are A (−5, 1), B (−14, 3), C (−17, −1) and D (−8, −3). Round your answer to one decimal place.

For each equation:

210

What is the gradient, m?

What is the value of the y-intercept, c?

y = −4x + 12

Mathspace WA Curriculum – Year 9 mathspace.co

5x − 2y = 17

hat are the values of the gradient, m, W and the y-intercept, c?

y 5

Write the equation of the line in gradient-intercept form.

Find the value of y when x = 20

4 3 2 1 −2

−1

x 1

−1

−2

For the equation y = −2x + 5: a

Complete the table of values.

Sketch the graph of the line on a Cartesian plane. x y

−3

−1

Sketch the line that has a y-intercept of 3 and whose gradient is .

For each equation:

Find the x-intercept.

iii

Sketch the line.

y = 4x − 6

Find the y-intercept.

3x − 5y = 15

Find the equation of each line. a

A line that passes through the point (−7, −13) and has a gradient of 4.

A line that passes through the points (−9, 5) and (−12, 14).

A line has a gradient of

and passes through the midpoint of A (5, 4) and B (−1, −10).

Find the coordinates of M, the midpoint of AB.

Find the equation of the line.

Could the table represent a directly proportional relationship between x and y? If yes, write the equation relating x and y. a

x 1 3 5 7 y 2 6 10 14

If y varies directly with x, and

x y

1 3

2 12

3 27

4 48

when x = 4:

Find the constant of proportionality, k.

Write the equation for y in terms of x.

Chapter 4 review mathspace.co

211

(Optional) If y is inversely proportional to x, and y = 30 when x = 5: a

Explain how we can find the constant of proportionality, k.

Express y in terms of x.

Find the value of y when x = 10. represents an inverse relationship between x and y.

(Optional) The equation a

Complete the table of values.

Describe the rate of change of the function when x < 0.

x y

-5

-4

-3

-2

-1

increasing or decreasing when x < 0?

(Optional) At ProPainters, painters must paint at a minimum rate of 60m2/hr. The equation represents the time, t, it takes p painters to paint the outside of a house with a surface area of 1200 m2.

Find the constant of proportionality, k.

Find the time it would take 4 painters to finish painting 1200 m2.

(Optional) In an art competition, the total prize money is distributed evenly among the winners. The graph shows the relationship between the number of winners and the prize money each winner receives. Determine whether or not there is an inverse variation between the number of winners and the prize money each receives. If so, find the constant of variation.

Prize money

No. of winners 0 5 10 15 20 25 30 35 40 45 50 55

(Optional) The total costs of a seminar, including venue hire, materials, and refreshments, amount to $560. These costs are shared equally among the participants. a

What is the cost per participant if the seminar is attended by two people?

Let the number of participants be x, and the cost per participant be y. Formulate an equation to represent this relationship.

Complete the table with values based on the equation, rounding to two decimal places when needed. x y

212

55 50 45 40 35 30 25 20 15 10 5

Sketch the graph based on the equation.

Describe the relationship between x and y.

Mathspace WA Curriculum – Year 9 mathspace.co

A self-serve, frozen yoghurt company charges customers by the amount of frozen yoghurt in grams. The table of values shows the costs of various weights. Amount in grams (x) Total cost ( y)

0 0

25 $1

50 $2

75 $3

How much does frozen yoghurt cost per gram?

How much would a cup of frozen yoghurt that weighs 315 g cost?

If a person paid $11.24 for their cup of frozen yoghurt, how much frozen yoghurt did they have in their cup?

A driver starts driving 150 km from home. They are driving at a rate of 80 km per hour. a

Complete this table of values: Number of hours passed (x) Number of kilometres left to drive ( y)

0.5

1.5

Write a linear equation linking the number of hours passed, x, and the number of kilometres left to drive, y.

How many minutes will it take for the car to reach home?

Solve: a

− 4 (−3r − 4) −5 (−r + 3) = −6 (−r −10)

4 (3a − 2) −2 (5a + 1) = 3 (a − 2)

−9 (−2x + 5) + 3x = −10 + 5 (−x − 9)

5 − 3 (3n − 2) = 6 (4n + 1) + 4 (n − 5)

Solve: a

(8b − 12) = −3b + 5

(4d − 2) =

(6d − 3)

(6x − 18) = x − 4 = − (y − 1) + 2

(Optional) Solve:

(Optional) Make m the subject of these equations: a

e 29

100 $4

x = 5k (n − m)

6x2 − 2m = 15

(Optional) Make the mentioned variables the subject in the following equations: a

W = Fd, make F the subject

S = I (1 + rt), make I the subject

Q = n2 L, make n the subject

, make H the subject

(Optional) The volume of a cylinder is given by the formula V = π r2h, where r is the radius of the base and h is the height of the cylinder. a

Rearrange the formula to make r the subject of the formula.

Calculate the radius of a cylinder with a volume of 200 cm3 and a height of 10 cm. Round your answer to two decimal places. Chapter 4 review mathspace.co

213

Big ideas Understanding quadratic relationships requires identifying their key features which highlight various characteristics of the function.

Curriculum Outcomes • Determine and explain why there are up to two solutions to a quadratic equation of the form ax2 = k and verify the possible solution/s by substitution • (Optional) Solve quadratic equations in factorised form using the null factor theorem and verify the solution/s by substitution • Use a table of values to plot points and graph quadratic functions of the form y = ax2, describe key features and make connections to the algebraic solution/s of ax2 = k

5 Quadratics Chapter outline 5.01 5.02 5.03 5.04 5.05

Quadratic relationships Graphs and features of quadratics Basic quadratics Solve algebraically Optional: Null factor law Chapter 5 review

216 225 236 244 249 255

Quadratics can model real-life situations, like the projectile motion of a basketball flying through the air!

5.01 Quadratic relationships After this lesson, you will be able to… • define and recognize a quadratic equation in the form y = ax2, where a ≠ 0. • distinguish quadratic equations in the form y = ax2 from non-examples. • identify quadratic relationships from their equations, graphs, and tables of values, focusing on the form y = ax2. • define a parabola and explain its connection to quadratic equations of the form y = ax2. • determine whether a relationship is linear, quadratic (in the form y = ax2), or neither, based on a graph or a table of values.

Quadratic equations Equations of the form y = ax2, where a is a coefficient and not equal to zero, are called quadratic equations. The highest power of x is 2, and there are no other terms like x or constants. Examples y = 5x2 y = −x2

Non-examples y=

Reason Not a quadratic as

and has a constant

Not a quadratic as there is a square root

Has a degree of 3, not 2

y=x +x

Example 1 Which two of these equations are quadratic equations of the form y = ax2? A y = −2x2

y = x2 + 3

y = x3

y = 3x2

E y = x2 − x

Create a strategy

Apply the idea

Look for equations where the only term is ax2, with The correct answers are option A: y = −2x2 and no additional terms like x or constants, and the option D: y = 3x2. Both have only an x2 term. highest power is 2.

216

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Idea summary A quadratic equation of the form y = ax2 has a single term where the power of x is 2, and a ≠ 0.

Quadratic relationships A quadratic relationship of the form y = ax2 can be identified from its equation, its table of values, and its graph’s shape. The most basic quadratic equation is y = x2 (where a = 1). We can create a table of values for this equation and then plot the points to see the general shape. The curve that is formed is known as a parabola. x y = x2

−3 (−3)2 = 9

−2 (−2)2 = 4 y

(−3, 9)

(3, 9)

8 7 6 5 (−2, 4)

(2, 4)

3 2 (−1, 1)

−5 −4 −3 −2 −1 −1

(1, 1) x (0, 0) 1 2 3 4 5

−1 (−1)2 = 1

0 02 = 0

1 12 = 1

2 22 = 4

3 32 = 9

This graph shows a quadratic relationship. • The graph is U-shaped. In this case, it is open upwards. We call this concave up. • The graph is symmetrical about a vertical line. In this case, the graph is symmetrical about the line x = 0, the y-axis. We call this line the axis of symmetry. • It has a turning point (a vertex) which could be a minimum or maximum value for y. In this example, the turning point is (0, 0) and we can see that there are no values of y below this. This is a minimum turning point.

We can also identify a quadratic relationship by looking at the table of values. A reminder that with linear relationships there is a constant ﬁrst difference for evenly spaced x-values. The ﬁrst difference is the difference between consecutive y-values.

Linear relationship

−3

−2

−1

−2

−5

First Difference 7 − 10 = −3 4 − 7 = −3 1 − 4 = −3 −2 − 1 = −3 −5 − (−2) = −3 −8 − (−5) = −3

−8 In linear relationships, the first difference is constant. 3

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The second difference is the difference between consecutive ﬁrst differences. If the second difference is a non-zero constant, we have a quadratic relationship. Let’s look at a table of values for y = x2.

Quadratic relationship

x −3 −2 −1 0 1

y 9 4 1 0 1

2 3

4 9

First Difference Second Difference 4 − 9 = −5 1 − 4 = −3 0 − 1 = −1 1−0=1

−3 − (−5) = 2 −1 − (−3) = 2 1 − (−1) = 2

4−1=3 9−4=5

5−3=2

3−1=2

In quadratic relationships, the second difference is constant. We can see the ﬁrst differences are: −5, −3, −1, +1, +3, +5. The ﬁrst differences have a difference of 2 each time, this is the second difference.

Example 2 Are the graphs of the relationships shown linear, quadratic or neither? a 4

3 2 1 −4 −3 −2 −1 −1

x 1

−2 −3 −4

Create a strategy Linear forms a line, quadratic forms a parabola (U-shape) and other shapes are neither.

Apply the idea

Reﬂect and check

The graph is an upside down U-shape, has a maximum value of y = 0, and is symmetrical, so is quadratic.

We could make a table of values, and examine how the y-values are changing for a consistent change in x.

x y

−3 4

−2 1.5

−1 −1

0 −3.5

1 −6

2 −8.5

3 −11

Create a strategy

Apply the idea

Determine if the table of values has constant ﬁrst differences (linear) or non-zero constant second differences (quadratic) or neither.

The y-values are decreasing by a constant value of 2.5 for each consecutive increase in x by 1. Therefore, the table shows a linear relationship.

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Example 3 Complete this table that represents a quadratic relationship: x y

−3

−2 2

−1 0.5

0 0

1 0.5

3 4.5

Create a strategy Determine if the minimum or maximum value is shown in the table. If it is, we can use the property of symmetry to help complete any missing values. To find any remaining values, we want to find the first and second differences and use them to complete the table.

Apply the idea In this case since the y-values of x = 1 and x = −1 are equal, the minimum value occurs at the x-value directly between −1 and 1. The minimum value occurs at (0, 0). Since the quadratic is symmetrical about this point, we can use this fact to ﬁnd other points. We are given the point (−2, 2). Using symmetry, we know (2, 2) must also be a point. For the same reason, we know when x = −3, the output is 4.5. x y

−3 4.5

−2 2

−1 0.5

0 0

1 0.5

2 2

3 4.5

To find the last two values, we want to find the first and second differences. x

−3 −2

−1

4.5

0.5

4.5

−2.5 −1.5 −0.5 0.5 1.5 2.5 +1

This means the second difference is +1, so we can now ﬁnd the remaining values. When x = 4, the output will be 2.5 + 1 = 3.5 more than the output for when x = 3: 4.5 + 3.5 = 8 The output for x = 5 will be 3.5 + 1 = 4.5 more than the previous output. 8 + 4.5 = 12.5 x y

−3 4.5

−2 2

−1 0.5

0 0

1 0.5

2 2

3 4.5

4 8

5 12.5

Reﬂect and check In this case the quadratic equation describing this relationship is y = 0.5x2. We will learn techniques to derive this a little later, but for now you can test against the table of values.

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Idea summary • Quadratic relationships have non-zero, constant second differences. • The graph of a quadratic relationship is a parabola. It is U-shaped, symmetrical, and has a minimum or maximum turning point (vertex).

1 −4 −3 −2 −1 −1

x 1

−4 −3 −2 −1 −1

−2

−3

−4

Minimum turning point

Maximum turning point

5.01 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

Copy and complete this table representing a quadratic relationship. x y First Difference Second Difference 0 0 −2 − 0 = −2 1 −2 2 −8 3 −18 −14 − (−10) = −4 −32 − (−18) = −14 −32

220

Do these statements describe a linear or quadratic relationship? a

Has a constant rate of change.

The graph is symmetrical about a vertical line.

First differences between the y-values is a constant.

Second differences between the y-values is a non-zero constant.

The graph is a straight line.

Second differences between the y-values is zero.

The graph is U-shaped.

Has a changing rate of change.

Has a turning point on the graph.

The equation has a squared term.

Mathspace WA Curriculum – Year 9 mathspace.co

Create a table of values for step number versus number of squares.

Describe the pattern in words.

Find the number of squares for Step 5.

Step 1

Fluency Ex 1

Step 2

Step 3

4, 5, 6ac, 7ab, 8ab

Step 4

4, 5, 6ab, 7ab, 8bc

4, 5, 6cd, 7cd, 8bc

Identify whether each equation is a quadratic equation. a e

2x − 3y − 2 = 0 2

y = (x − 3)

b f

m 8x2 = y

y = 2x2 2

y = −5x

y = (x2)2

y = 5x3 − x2

y = 6x

y = 2x − 3

y = –30x2

Did you know?

When water is shot from a fountain at an angle, the stream doesn’t travel in a straight line — it forms a parabolic arc. That’s because the only force acting on the water after it leaves the nozzle is gravity, which pulls it downward at a constant rate. The water’s horizontal motion remains steady, but the vertical motion slows down, stops at the highest point (the turning point), then speeds up again as it falls — exactly like the path of a parabola described by a quadratic equation. Fountain designers use this type of modelling to calculate the perfect angle, speed, and height so the water lands where it should — whether it’s for safety in public parks or artistic displays in places like gardens, hotels, and theme parks. 5.01 Quadratic relationships mathspace.co

221

Ex 2a

Do these graphs show relationships that are linear, quadratic or neither? a

8 6

4 2

x −5

−5

−2

y 5

x −5

5 −5

−5

−10 −15

5 x

x −5

−5

Ex 2b

Do these tables of values show relationships that are linear, quadratic, or neither? a

222

x −3 −2 −1 0 1

y 27 12 3 0 3

x −3 −2 −1 0 1

Mathspace WA Curriculum – Year 9 mathspace.co

y −9 −6 −3 0 3

x −2 −1 0 1 2

y −8 −2 0 −2 −8

x 0 1 3 6 10

y 9 4 1 4 9

Ex 3

Complete these tables that represent quadratic relationships. a

x y

−3 ⬚

−2 1

−1 0.25

0 0

1 0.25

−1 0.75

3 ⬚

4 64

−2 ⬚

2 ⬚

0 0

−2 ⬚

−1 −0.4

1 ⬚

−3 ⬚

For these patterns:

3 2.25

5 100

2 ⬚

6 144

7 196

1 0.75

2 3

0 0

1 −0.4

3 ⬚

4 ⬚

2 −1.6

3 ⬚

4 ⬚

5 ⬚

6 ⬚

8 ⬚

4 ⬚

9 ⬚

10 ⬚

5 ⬚

Describe the pattern in words.

How many squares or hexagons will be in the next ﬁgure?

iii

Is the relationship between ﬁgure number and number of shapes linear or quadratic?

5.01 Quadratic relationships mathspace.co

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Reasoning 9

9–11

10–12

Explain why this table of values does not show a quadratic relationship. x y

9, 10

−2 2

−1 0

0 −2

1 0

2 2

Compare these equations, tables and graphs. Do they have a quadratic or linear relationship? Justify your answer. a

y1 = 3x + 1 and y2 = 9x2

x x y1 −3 −8 −3 −2 −5 −2 −1 −2 −1 0 1 0 1 4 1 2 7 2 3 10 3

y2 13.5 6 1.5 0 1.5 6 13.5 y2

x −5

−5

5 −5

−5

Create a quadratic relationship using an equation in the form of y = ax2, table of values, or a graph. Justify that it is quadratic in two different ways.

Identify if these statements about this incomplete table of values are always, sometimes or never true and explain why. x y

−2

−1

If the y-values are all positive, then it is not quadratic.

If the y-values go from positive, to negative, and back to positive, then it cannot be a linear relationship.

If the y-values are all the same number, then it is a quadratic relationship.

Problem-solving 13, 14 13

13–15

A quadratic pattern has an equation y = 2x2. It starts with x = 1 and y = 4. Draw ﬁgures to show the ﬁrst four steps geometrically using shapes.

224

13–16

Mathspace WA Curriculum – Year 9 mathspace.co

A community garden is designed as a large square with a perimeter of 40 m. Inside, it is divided into smaller square sections, each with side length s metres. The total area of the garden is given by A = 4s2, and this must equal the area of the large square. Explore possible integer side lengths for the smaller squares and calculate the corresponding total area and perimeter of the large square. Complete the table to find which side length fits the garden’s design.

Side length of small square (m) 2 3 4 5 6

⬚ ⬚ ⬚ ⬚ ⬚

Perimeter of large square (m) ⬚ ⬚ ⬚ ⬚ ⬚

Determine the integer side length of each smaller square so that the total area A = 4s2 matches the area of the large square with a perimeter of 40 m. What is the resulting total area?

Verify that smaller squares, each with side length s = 5, can contribute to a large square with a perimeter of 40 m, and confirm the total area is 4s2.

Consider the table shown: x y

Total area (sq m)

−2 16

−1 4

0 0

1 4

2 16

Describe the pattern of the output in words.

Determine if the rule would be a linear or quadratic relationship.

Write the rule that relates x and y.

A quadratic relationship has a second difference of 5. Create a table of values that could represent the relationship.

5.02 Graphs and features of quadratics After this lesson, you will be able to… • identify the key features of a parabola: intercepts, turning points, and the axis of symmetry. • graph quadratic relationships using a table of values.

Identify key features Intercepts are where the parabola crosses the horizontal (x) and vertical ( y) axes. • The x-intercepts are where the parabola crosses the x-axis. This occurs when y = 0. • The y-intercept is where the parabola crosses the y-axis. This occurs when x = 0. 5.02 Graphs and features of quadratics mathspace.co

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Note, that there can be 0, 1, or 2 x-intercepts.

x-intercepts 4 2

−6 −4 −2 −2

−4

y-intercept

For now, we will just identify the intercepts from a graph, later we will look at doing it algebraically. Parabolas can have a maximum or minimum value. We call this a turning point, or vertex. If a parabola opens upwards, we call this concave up and it will have a minimum value. If a parabola opens downwards, we call this concave down, and it will have a maximum value. 3

1 −1

1 −3 −2 −1

−3 −4

−2

−5

Max value (Maximum turning point)

−3 −2 −1

−2

Min value (Minimum turning point)

Concave up, with minimum value

−1

x 1

−3

Concave down, with maximum value

Parabolas are symmetrical. The axis of symmetry is the dotted line pictured here. It has an equation of the form x = ⬚, and you can see that this x-value will be the same as the x-value for the maximum or minimum turning point.

3 2 1 −4 −3 −2 −1 −1 −2 −3 −4

Example 1 For each graph, identify: 1. the turning point 2. whether the turning point is a maximum or minimum 3. the equation of the axis of symmetry 4. any intercepts that can be identiﬁed

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x 1

y 16 14 12 10 8 6 4 2 −4

−2

Apply the idea 1. The graph has a turning point at (0, 0). 2. This graph has a minimum turning point because it is concave up. 3. The equation of the axis of symmetry is the vertical line through the turning point. The equation is x = 0. 4. The graph has x-intercepts at (0) and (0, 0) and a y-intercept at (0). b

y 50 25 x −10−8 −6 −4 −2 −25

2 4 6 8 10

−50 −75 −100

Apply the idea 1. The graph has a turning point at (0, 0). 2. This graph has a maximum turning point because it is concave down. 3. The equation of the axis of symmetry is the vertical line through the turning point. The equation is x = 0 4. The graph has an x-intercept at (0, 0) and a y-intercept at (0, 0).

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Idea summary The key features of a quadratic graph are: • Turning point: point where the graph has a maximum or minimum and it changes between increasing and decreasing. • The minimum or maximum value of the graph is the y-value of the turning point. • y-intercept: point where the graph crosses the y-axis. • x-intercept(s): point(s) where the graph touches or crosses the x-axis, there may not be any. • Axis of symmetry: vertical line that cuts the graph in half; it passes through the turning point.

Plot from table of values If we have an equation for a quadratic function we can create a table of values to create a series of points to plot.

Example 2 Consider the quadratic function y = 2x2. a Complete the table of values from x = −2 to x = 2 and graph the parabola. −2

x y

−1

Create a strategy Substitute each x-value into y = 2x2 to complete the table. Plot the points and draw a smooth curve.

Apply the idea y = 2x2

Write the equation 2

= 2(−2)

Substitute x = −2

=2×4

Evaluate the square

Evaluate the product

Similarly, compute the y-values for x = −1, 0, 1, 2: x y

228

−2 8

−1 2

0 0

1 2

2 8

Mathspace WA Curriculum – Year 9 mathspace.co

Plot these points and draw the parabola: y 8 7 6 5 4 3 2 1 −2

−1

x 1

Reﬂect and check The vertex at (0, 0) is the minimum point, typical for quadratics of the form y = ax2 with a > 0.

b Solve 2x2 = 18 and verify the solution by graphing y = 2x2 and y = 18.

Create a strategy Solve 2x2 = 18 algebraically to find the solution. Then, plot y = 2x2 and y = 18 to identify the intersection points and confirm the solution.

Apply the idea Solve the equation:

Write the equation

Divide both sides by 2

Take the square root of both sides

Evaluate the square root

The solutions are x = −3 and x = 3. Plot the two graphs to confirm their points of intersection:

y 24 21

y = 18 (−3, 18)

(3, 18)

15 12 9

From the graph, we can see that y = 18 and y = 2x2 intersect at the points (−3, 18) and (3, 18). This confirms that x = −3 and x = 3 are the solutions to 2x2 = 18.

y = 2x2

6 3 −4 −3 −2 −1

x 1

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Idea summary The graph of a quadratic relationship is called a parabola. To graph it, we can complete a table of values and plot the points on the Cartesian plane.

5.02 Practice questions Understanding 1

Explorer

Adventurer

1, 2

Trailblazer

1–3

Describe these curves as concave up or concave down. a

1 −1

−1

x 1

−1

−2

230

−1

x 1

−2

−3

−4

Determine if the statements are true or false. a

The axis of symmetry is a vertical line passing through the turning point.

A minimum turning point is the highest point on the graph.

The x-intercepts are the points where the graph cuts the y-axis.

The y-intercept occurs where x = 0.

Describe the relationship between the turning point and the axis of symmetry.

Mathspace WA Curriculum – Year 9 mathspace.co

Fluency 4

4–6, 7abeg, 8

4–6, 7acfg, 8, 9

4, 6, 7c-eh, 8, 9

For each graph: i

Describe the graph as concave up or concave down.

Is the turning point a maximum or minimum point?

x −4 −3 −2 −1

4 20

−5 15 −10

−15

5 x −2

−1

x −3 −2

−1

1 y

−2

12 10

−4

−6

−8

4 x

−1

Ex 1

For each graph: i

Identify the turning point.

State whether the turning point is a maximum or minimum.

iii

Identify the equation of the axis of symmetry.

Identify any intercepts that can be identified.

−1

x 1

−5

−10

−15 −20

5 x −1

−25

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Ex 2

For equation y = −7x2 : a

Complete the table of values for x from −2 to 2. −2

x y

−1

Plot the points and join with a parabolic shaped line.

Solve −7x2 = −7 and verify the solution by graphing y = −7x2 and y = −7.

For each equation: i

Create a table of values from x = −2 to x = 2.

Plot the points and join with a parabolic shaped line.

y = −11x2

y = 9x2

y = −16x2

y = 10x2

y = 2x

y = −13x2

y = −x2

y = 12x2

This diagram shows only part of the graph of a quadratic relationship. 6 4 2

y x 1 2 3 4 5 6 7 8 9 10

−2 −4 −6 −8 −10 −12 −14 −16 −18

Complete the table of values for the graph. x y

232

−3

−2

−1

Does the vertex represent a minimum or a maximum point?

If we knew the equation of the curve, what would be the sign of the coefficient of x2?

For the equation y = 18x2: i

State whether the graph is concave up or down.

Find the turning point.

iii

Sketch the graph using at least 5 points.

Mathspace WA Curriculum – Year 9 mathspace.co

Reasoning

10, 11, 13, 14

10–12, 14

Explain why the graph of y = x2 has a minimum value of 0.

Determine whether this statement is always true, sometimes true, or never true. Justify your answer with examples or counter-examples. “A parabola has either a maximum or minimum value, but not both.”

Justify the signiﬁcance of the y-value of the turning point in determining the range of values of the y-values of a quadratic function.

Match each equation to its graph. A

y = 6x2

y −2

B −1

y = 19x2 1

C x

y = –11x2

y = −7x2

−10

−20

−30

−40

10 x

iii

−2

−1

−2

−1

x y

−10

−20

−30

−40 x

−2

−1

Explain how the value of a affects the solutions to ax2 = 18 and their positions on the graph.

5.02 Graphs and features of quadratics mathspace.co

233

Match the given quadratic equations with their corresponding graphs and features. Equations: A

y = 13x2

y = −20x2

y = 21x2

Parabola concave down, turning point at (0, 0), symmetric about the y-axis.

Parabola concave up, turning point at (0, 0), symmetric about the y-axis.

y = −12x2

Graphs and features: a

x −2

−1

−10 40

−20

−30

−40

−50

x −2

Parabola concave down, turning point at (0, 0), symmetric about the y-axis.

−1

Parabola concave up, turning point at (0, 0), symmetric about the y-axis.

x −2

−1

−10 40 −20 30 −30 20 −40 10 −50

x −2

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−1

Problem-solving 17 15

15–17

The profit of a company is modeled by the quadratic equation P = 22x2, where P is the profit in thousands of dollars and x is the number of items produced (in hundreds). a

Is the parabola concave up or down? Does it have a minimum or maximum value?

Find the coordinates of the turning point (vertex) of the parabola.

What is the minimum value of the parabola, and what does it represent?

Solve 22x2 = 88 and interpret the solutions.

Painting a multi-storey building requires extra effort to set up each additional floor. So a painter decides to charge proportionally to the square of the number of floors. He charges $1000 for a 1-floor building, $4000 for a 2-floor building and so on. a

Complete the table for the amount charged for 2, 3, … 6 floors. Number of floors Amount charged

15–17

1 $1000

2 $4000

Plot the graph for the amount charged against the number of floors.

Determine the equation for the amount charged ( y) as a function of the number of floors (x).

What would be the cost to paint an 8-floor building?

The painter charged $ 400 000 for a building. How tall was the building?

A company’s cost to produce items is given by C = 23x2, where C is the cost in dollars and x is the number of items produced. a

Create a table of values for x from 0 to 4 and sketch the graph.

Solve 23x2 = 368 and explain what the solutions mean in this context. Verify your solutions using the graph below. C 350 300 250 200 150 100 50 −5 −4 −3 −2 −1

x 1 2 3 4 5

What is the significance of the vertex in this context?

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5.03 Basic quadratics After this lesson, you will be able to… • understand and explain the concept of vertical reflections in quadratic graphs. • interpret and analyse the effect of a negative coefficient on the graph of a quadratic equation. • understand and explain the concept of dilation in quadratic graphs. • interpret and analyse the effect of changing the coefficient of the quadratic term on the graph of a quadratic equation. • apply understanding of these concepts in solving problems and interpreting real-world scenarios.

Basic form Interactive exploration

Discover this concept in action online

The basic parabola is represented by the equation y = x2. It forms a U-shaped curve that opens upwards.

y 4 3 2 1 −4 −3 −2 −1 −1

mathspace.co

x 1

−2

If the coefficient of x2 is negative (meaning the equation looks like y = −x2 ), the parabola flips upside down. This is called a vertical reflection over the x-axis. In the basic parabola, y = x2, all the y-values are positive or zero. When the parabola is reflected (flipped), like in y = −x2, the y-values become negative or zero.

−3 −4

The features of y = x2 and y = −x2 are: Turning point Axis of symmetry x-intercept(s), also called the roots or solutions to the quadratic Description

236

y = x2 Minimum turning point at (0, 0) x=0

y = −x2 Maximum turning point at (0, 0) x=0

(0, 0)

The graph of y = x2 represents a basic parabola.

The graph of y = x2 that is vertically reflected across the x-axis (flipped upside down)

Mathspace WA Curriculum – Year 9 mathspace.co

Dilations Interactive exploration

Discover this concept in action online

mathspace.co

A dilation stretches or compresses a graph based on a scale factor. The equation y = ax2 represents a vertical dilation of y = x2 by a factor of a. The terms vertical enlargement and vertical compression describe how a graph changes in relation to the x-axis: • Vertical enlargement occurs when the graph moves away from the x-axis, making it appear narrower. This happens for a > 1. • Vertical compression happens when the graph moves towards the x-axis, making it appear wider. This happens when 0 < a < 1. y 8

1 −3 −2

−1

x 1

A vertical enlargement makes the graph narrower

−3 −2

−1

x 1

A vertical compression makes the graph wider

The features of y = ax2 and y = −ax2 are: Turning point Axis of symmetry x-intercept(s), also called the roots or solutions to the quadratic

Description

y = ax2 Minimum turning point at (0, 0) x=0

y = −ax2 Maximum turning point at (0, 0) x=0

(0, 0)

The graph of y = x2 that has been dilated by a factor of a.

The graph of y = −x2 that has been dilated by a factor of a.

Vertically compressed for 0 < a < 1

Vertically compressed for 0 < a < 1 Vertically enlarged for a > 1

Vertically enlarged for a > 1

5.03 Basic quadratics mathspace.co

237

Example 1 Graph y = −x2 and y = −4x2 on the same Cartesian plane. Mark the turning point and intercepts on the graph. Which graph is vertically enlarged?

Create a strategy To graph the parabolas, create a table of values and plot the points on the same Cartesian plane. To decide which is vertically enlarged, look for the graph that is narrower.

Apply the idea Create a table of values by substituting x-values into each equation. x y = −x2 y = −4x2

−2 −4 −16

−1 −1 −4

0 0 0

1 −1 −4

2 −4 −16

Both graphs have a turning point at (0, 0). This is also the x and y-intercept for both quadratics. y −4 −3 −2 −1 −2

x 1

−4 −6 −8 −10 −12 −14 −16

The graph of y = −4x2 is narrower so it is vertically enlarged.

Reﬂect and check When looking at the table of values, notice that the y-values for y = −4x2 are multiplied by 4, which moves them further away from the x-axis. This is what causes the parabola to become steeper, or narrower.

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Example 2 Order the quadratic equations from most narrow (vertically enlarged), to least steep (vertically compressed).

Create a strategy Compare the values of a in each equation. When a = 1, the graph is the basic parabola. When a > 1, the graph is narrower (vertically enlarged) than the basic parabola. When 0 < a < 1, the graph is less steep (vertically compressed) than the basic parabola.

Apply the idea For the equation y = x2, a = 1. This is the graph of the basic parabola. For the equation

. Since

, the graph will be less steep (vertically compressed)

than the basic parabola. For the equation y = 5x2, a = 5. Since 5 > 1, the graph will be narrower (vertically enlarged) than the basic parabola. In order from most narrow to least steep:

Reﬂect and check We can check our answer by graphing all three quadratics on the same Cartesian plane. y 6

y = 5x2

y = x2

1 2 x 2

4 3 2 1 −5

−4

−3

−2

−1

x 1

5.03 Basic quadratics mathspace.co

239

Idea summary The equation y = ax2 represents a vertical dilation of y = x2 by a factor of a. • a > 1 represents a vertical enlargement • 0 < a < 1 represents a vertical compression 8

1 −3 −2

−1

x 1

−3 −2

A vertical enlargement makes the graph narrower

−1

x 1

A vertical compression makes the graph wider

5.03 Practice questions Understanding

Explorer 1–4

Adventurer 1–4

Trailblazer 1–4

List two equations for quadratics that are narrower (vertically enlarged) than y = x2.

List two equations for quadratics that are wider (vertically compressed) than y = x2.

What is different and what is the same when comparing the graphs y = x2 and y = −x2?

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The graph of y = x2 is shown with a dashed line. Does the parabola with a solid line show a vertical enlargement or a vertical compression? a

1 −4 −3 −2 −1

x 1

−4 −3 −2 −1

Fluency Ex 1

−4 −3 −2 −1

5, 6a, 7a, 8–10

5, 6b, 7b, 8–10

y = x2 and y = 3x2

y = −2x2 and y = −x2

Graph both quadratics on the same Cartesian plane. Mark the turning point and intercepts on the graph. Which graph is vertically compressed? a

Ex 2

5, 6a, 7a, 8–10

Graph both quadratics on the same Cartesian plane. Mark the turning point and intercepts on the graph. Which graph is vertically enlarged? a

x 1

−4 −3 −2 −1

y = x2 and

and y = −x2

Order the quadratic equations from most narrow (vertically enlarged), to least steep (vertically compressed). a

y = 0.2x2, y = 2x2, y = x2

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241

Match each graph to the equation: i

y = 2x2

y = −x2

b 4

−4 −3 −2 −1 −1

−2

−3

−4

d 4

−4 −3 −2 −1 −1

−2

242

y = x2

x 1

−4 −3 −2 −1 −1

y = −3x2

x 1

iv y

iii

−2

−3

−4

Match each description to the equation: i

y = −2x2

Narrower than y = x2, a reﬂection of y = x2 on the x-axis.

Wider than y = x2, shares the same turning point as y = x2.

Narrower than

Same vertical enlargement as part (a) and is a reﬂection of part (a) on the x-axis.

iii

y = 2x2

but wider than y = 2x2. Has the same x-intercept as y = 2x2.

Mathspace WA Curriculum – Year 9 mathspace.co

Each of the graphs shown have an equation of the form y = ax2. By using a point on the graph, ﬁnd the value of a, and the equation of each graph. a

y 5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

4 3 2 1

Reasoning

4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6

1 2 3 4 5

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

1 2 3 4 5

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6

5 4 3 2 1

11, 12

1 2 3 4 5

x 1 2 3 4 5

12, 13

Using a table of values, explain why y = x2.

Compare the similarities and differences in the graphs of y = 2x2 and y = (2x)2 and y = −2x2.

Without plotting the graph, explain why y = −3x2 is non-positive for all values of x.

Problem-solving 14, 16a–c 14

is wider (vertically compressed) compared to

14, 16c–e, 17

14, 15, 16d–f, 17

A square paving tile has side length s cm and a height of 3 cm. a

Write an expression for the volume of 10 square paving tiles stacked on top of each other.

If the volume of the 10 square paving tiles is 1920 cm3, ﬁnd the side length, s, of each paving tile.

Graph y = x2 and the line y = 2x + 2 on the same Cartesian plane. Shade the area between the two graphs. Find an estimate for the area of the shaded region.

5.03 Basic quadratics mathspace.co

243

Find the equation of the quadratic function of the form y = ax2 for quadratics that have the given coordinates on the curve. Leave a as a fraction if needed.

(−9, 9)

a e

(4, 8)

(10, −20)

(1, −4)

(−2, 40)

A social media company’s total advertisement sales (S, in thousands of dollars per week) is proportional to the square of the number of countries (C) they have a presence in.

If S = 16 thousand when C = 3, then: a

What is the quadratic relationship between S and C?

How many countries does the company need to be in to triple the sales from 16 thousand?

5.04 Solve algebraically After this lesson, you will be able to… • understand that a quadratic equation can have one, two, or no real solutions. • apply algebraic manipulation to solve quadratic equations. • solve a variety of quadratic equations using algebraic methods.

Solve algebraically Recall that a quadratic can have one, two, or no real solutions. y

One real solution

Two real solutions

No real solutions

Some forms of quadratics can be solved using a few steps of algebraic manipulation. The key thing to remember for solving algebraically is that when you take the square root of a number

244

, the answers are

Mathspace WA Curriculum – Year 9 mathspace.co

and

Example 1 Solve 5x2 = 180 for x.

Create a strategy Use inverse operations to isolate x on one side of the equation.

Apply the idea

Write the equation

Divide both sides by 5

Square root both sides

Evaluate the square root

Reflect and check Our two solutions are x = 6 and x = −6. To check these, substitute each into the left-hand side of the equation and check if the resulting value will be the same value as the right-hand side of equation. For example, substitute x = 6: Left-hand side = 5x2

Write the left-hand side of the equation 2

=5×6

Substitute x = 6

= 5 × 36

Evaluate the power

= 180

Evaluate

This is the same as the right-hand side of the equation, so the value of x = 6 is correct.

Example 2 Solve

for x.

Create a strategy Use inverse operations to isolate x on one side of the equation.

Apply the idea Write the equation

Multiply both sides by

Square root both sides

Evaluate the square root

5.04 Solve algebraically mathspace.co

245

Reﬂect and check Our two solutions are

and

. To check these, substitute each into the left-hand side of the

equation and check if the resulting value will be the same value as the right-hand side of equation. For example, substitute

: Write the left-hand side of the equation

Substitute x =

Evaluate the power

Evaluate the multiplication

Simplify This is the same as the right-hand side of the equation, so the value of x =

is correct.

Example 3 Solve −4x2 + 196 = 0 for x.

Create a strategy Use inverse operations to isolate x on one side of the equation.

Apply the idea Write the equation

Subtract 196 from both sides

Divide both sides by −4

Square root both sides

Evaluate the square root

Reﬂect and check Our two solutions are x = 7 and x = −7. To check these, substitute each into the left-hand side of the equation and check if the resulting value will be the same value as the right-hand side of equation. For example, substitute x = −7: Left-hand side = −4x2 + 196 Write the left-hand side of the equation = −4 × (−7)2 + 196

Substitute x = −7

= −4 × 49 + 196

Evaluate the power

= −196 + 196 Evaluate the product = 0 Evaluate This is the same as the right-hand side of the equation, so the value of x = −7 is correct.

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Idea summary Quadratic equations can have up to two solutions. Quadratic equations in the form x2 = k (where k > 0) have two solutions.

5.04 Practice questions Understanding

Are the following statements true or false?

The graph of a quadratic equation is a parabola.

Quadratic equations can potentially have three solutions.

The symbol ± means “plus or minus.”

To solve for x we must do the inverse operation.

The equation 2x2 − 288 = 0 has two solutions. One of those solutions is x = 12. What is the other solution?

4–7, 9–13

4–8, 11–13

x2 = 2

x2 = 25

x2 = 121

x2 = 294

Solve: a

4y2 = 100

25y2 = 36

−3k2 = −12

8d2 = 288

−5c2 = −320

5j2 = 405

12b2 = 108

−6n2 = −600

2x2 − 242 = 0

4x2 − 100 = 0

−12k2 + 108 = 0

−25v2 + 64 = 0

9x2 − 4 = 0

e g 8

4–7, 9–13

Solve: a

Ex 3

Trailblazer 1–3

Identify the similarities and differences between a linear equation ax + k = 0 and a quadratic equation ax2 + k = 0.

Fluency

Ex 2

Adventurer 1–3

Ex 1

Explorer 1–3

81x − 16 = 0

Solve the equation for x, in terms of a and k. −ax2 + k = 0

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A baker makes cookies using multiple trays. Each tray produces a number of cookies such that twice the square of the number of trays equals 72 cookies in total. Let x represent the number of trays. What is the possible number of trays?

The pressure in a water tank, p in atmospheres (atm), has 125 less than five times its square resulting to zero. What is the pressure of the water tank?

The temperature change in a freezer, t (in degrees Celsius), has 16 less than four times its square resulting to zero. What is the temperature change, given that it represents a decrease below zero?

The number of beats per second in a song, b (in beats per second), has 50 less than two times its square resulting to zero. What is the number of beats per second?

The debt in a bank account, d (in dollars), has 192 less than three times its square resulting to zero. What is the debt written on the account?

Reasoning

14–16

14–17

Explain why the equation 3x2 − 27 = 0 has two solutions and verify the solution(s) by substitution.

Explain why the equation 2x2 + 4 = 0 has no real solutions.

Identify when Ted made an error and explain what he should have done to correct the error: Write the equation Divide both sides by 25 Take the square root of both sides

Identify when Samantha made an error and explain what she should have done to correct the error: Write the equation Add 8 to both sides Divide both sides by 36 Take the square root of both sides

Problem-solving 20, 21 18

19–21

A rocket’s altitude during a test flight is modelled by this quadratic relationship, where h = 0 initially represents the height in metres and t is the time in seconds. h(t) = 4t2 When will the rocket be exactly 16 metres above the ground?

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18–22

Mathspace WA Curriculum – Year 9 mathspace.co

A small business tracks its weekly profit during a promotion, where P = 0 initially represents profit in hundreds of dollars, and t is time in weeks. The profit at any time is modelled by:

P(t) = 8t2 When will the profit be exactly 72 hundred dollars (that is $7200)? 20

Gian thinks of a square number. When he squares it and doubles the result, he gets 338. a

Write an equation representing this situation, where Gian’s number is x.

What could Gian’s number be?

A square has a side length of x cm. Its area, when tripled is equal to 75cm2.

Write an equation for the tripled area of the square, A, in terms of x.

Substitute A = 75 and solve for x, stating both possible solutions.

Determine the realistic side length of the square.

A stone is dropped from a bridge 0.8 metres above a river, with a horizontal push that keeps it moving sideways. Its height (in metres) above the water is modelled by the equation h = −3.2t2 + 0.8, where t represents the time (in seconds) after the stone is dropped. a

What is the height of the stone when it is dropped?

At what time will the stone hits the water?

If the stone travels horizontally at a constant speed, and the ratio of the horizontal distance to the time it takes to reach the water is 5 : 3, what is the horizontal distance the stone travels before hitting the water?

5.05 Optional: Null factor law After this lesson, you will be able to… • apply the null factor law to solve quadratic equations.

The null factor law The null factor law states that if a product of two or more factors is equal to 0, then at least one of those factors must be equal to 0. For example, if ab = 0 then either a = 0 or b = 0 We can use this to solve quadratic equations that are not in the form x2 = k. If we can rearrange a quadratic equation into the form (x − p) (x − q) = 0, then we know that either x − p = 0 or x − q = 0. We can solve the quadratic equation by solving each of these linear equations.

5.05 Optional: Null factor law mathspace.co

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Example 1 Solve: a (x − 6) (x + 7) = 0

Create a strategy

Apply the idea

Using the null factor law, set both factors to 0 and solve the resulting linear equations.

x−6=0 x=6

Set the ﬁrst factor to 0 Add 6 to both sides

x+7=0

Set the second factor to 0

x = −7

Subtract 7 from both sides

So the solutions are x = 6 and x = −7.

Reﬂect and check To verify, substitute each solution back into the original equation (x − 6) (x + 7) = 0. Check x = 6: LHS = (x − 6) (x + 7)

Write the formula

= (6 − 6) (6 + 7)

Substitute x = 6

= (0) (13)

Evaluate the equations inside brackets

Evaluate

Check x = −7: LHS = (x − 6) (x + 7)

Write the formula

= (−7 − 6) (−7 + 7) Substitute x = −7 = (−13) (0)

Evaluate the equations inside brackets

Evaluate

Both solutions satisfy the equation, conﬁrming they are correct.

b 4m (m + 5) = 0

Create a strategy Using the null factor law, set both factors to 0 and solve the resulting linear equations.

Apply the idea 4m = 0 Set the ﬁrst factor to 0 m = 0 Divide both sides by 4 m + 5 = 0 Set the second factor to 0 m = −5 Subtract 5 from both sides So the solutions are m = 0 and m = −5.

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Reﬂect and check To verify, substitute each solution back into the original equation 4m(m + 5) = 0. Check m = 0: LHS = 4m(m + 5) Write the formula = 4 × 0 × (0 + 5)

Substitute m = 0

= 0

Evaluate

Check x = −5: LHS = 4m(m + 5)

Write the formula

= 4 × (−5) × (−5 + 5)

Substitute m = −5

= 4 × (−5) × (0)

Evaluate the equation inside bracket

Evaluate

Both solutions satisfy the equation, conﬁrming they are correct.

Idea summary The null factor law states that if a product of two or more factors is equal to 0, then at least one of those factors must be equal to 0. If we can rearrange a quadratic equation into the form (x − p) (x − q) = 0, then we know that either x − p = 0 or x − q = 0. We can solve the quadratic equation by solving each of these linear equations.

5.05 Practice questions Understanding 1

Trailblazer 2, 3cf

0 = x(x − 5)

2 = x2 + 3x − 10

12 = (x − 7) (x + 2)

0 = x(x + 4) + 5

A quadratic equation can be written in its factorised form (x − a) (x − b) = 0. Write the values for a and b for these quadratic equations. a

Adventurer 1, 2, 3be

Which of these can you solve directly using the Null Factor Law? A

Explorer 1, 2, 3ad

(x − 3) (x − 5)

(x − 6) (x + 3)

x(x − 4)

Construct a monic quadratic equation in factorised form based on the given solutions. a

x = 5 and x = 1

x = −3 and x = 8

x = −2 and x = 9

x = 0 and x = −7

and x =

x=4

5.05 Optional: Null factor law mathspace.co

251

4a-i, 5, 6a-e, 7a-d, 8ab

Fluency Ex 1

j ( j + 7) = 0

x (2x − 9) = 0

−5m (m + 5) = 0

(m + 5) = 0

(10x − 9)2 = 0

(4k − 9)2 = 0

(−3 + 7x) = 0

(e − 4) (e − 2) = 0

(s + 6) (s + 7) = 0

(8x − 5) (3x − 7) = 0

(3w + 8) (5w − 7) = 0

(3x − 17) (2x − d) = 0

−2( y + 1) ( y − 5) = 0

Solve: −(x − 3) (x + 8) = 0

−3(m + 4) (m − 2) = 0

−(p − 7) (p + 3) = 0

−5(t + 6) (t − 1) = 0

−4(q + 9) (q − 3) = 0

Solve:

a c

252

Find the value(s) of x when y = 0 for: a

y = x(x + 7)

y = 5x(x + 3)

y = 3x(x + 3)

y = −2x(x − 7)

e 8

b d

e 7

4c-l, 9, 10cd, 13

Solve:

4a-i, 5, 6f, 7ef, 8c-j, 9, 10ab, 11, 12

Solve: a

k(k + 10) = 0

p(p − 7) = 0

(t + 7) (t + 8) = 0

(w − 6) (w + 5) = 0

(r − 8) = 0

(n − 9) (n + 2) = 0

(s − 6) (s − 17) = 0

(m − 10) (m + 2) = 0

(v − 4) (v + 3) = 0

3(u + 4) (u − 2) = 0

Solve for x in terms of a, using the null factor law: a

(x − a) (x + 2a) = 0

(3x + a) (2x − 5a) = 0

(4x − 3a) (x + a) = 0

(5x + 2a) (3x − 7a) = 0

Find the value of b in terms of t for each equation: a

(3t − b) (t + 2) = 0

(2t + 5) (−t − b) = 0

−4(2t + b) (−t + 3) = 0

t(4t − b) = 0

Mathspace WA Curriculum – Year 9 mathspace.co

Find the value of k that makes each equation true when x = 3 is a solution: a

(x − 3) (x − k) = 0

(2x + 1) (kx − 6) = 0

−(x + 4) (3x − k) = 0

(5x − k) (x − 1) = 0

Solve each equation to ﬁnd the possible values of L, where L represents a length in metres. a

(L − 5) (L + 2) = 0

(L + 3) (L − 7) = 0

−(L − 4) (L + 1) = 0

(L + 6) (L − 2) = 0

A machine’s output depends on a setting k. The product of two adjustments is zero when the machine stops. Find the settings k at which the machine stops if the adjustments are given by:

Reasoning

14, 15

14–17, 18ab

14, 16, 17, 18c

Jade used the null factor law to solve the equation (x − 4) (x + 6) = 0 and found the solutions x = −4 and x = 6. Is Jade’s answer correct? Explain why.

Milo claims that for any equation of the form (x − a) (x − b) = 0, changing the order of the factors to (x − b) (x − a) = 0 will give different solutions. Is Milo correct? Explain your reasoning using the Null Factor Law, and provide an example to support your answer.

Explain why the quadratic equation will always have two solutions and state the two solutions. cx2 + dx = 0

Consider the equation (2x − 5) (x + 3) = 0. Sasha says that because there are two factors, there must always be exactly two distinct solutions. Is Sasha correct? Explain your reasoning using the Null Factor Law, and discuss what happens if the factors were identical instead.

Solve for x in each of these equations: i

(x − 4) (x − 5) = 0

iii

(x + 1)2 = 0

−(x + 2) (x − 7) = 0

Use technology to sketch these graphs and determine the coordinates of the x-intercepts of each graph: i

y = (x − 4) (x − 5)

iii

y = (x + 1)2

y = −(x + 2) (x − 7)

What do you notice about your answers in parts (a) and (b)?

5.05 Optional: Null factor law mathspace.co

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Problem-solving 19, 20 19

20–23

21, 23, 24

Find the value of a if: a

The non-zero solution of the equation x (x + a) = 0 is x = 12.

The non-zero solution of the equation x (x + a) = 0 is x = −3.

A small bakery tracks its daily profit based on the number of specialty cakes sold. The profit P (in dollars) depends on the number of cakes c beyond a baseline, given by the equation P = (2c + 5) (c − 3). Determine when the bakery breaks even (i.e., when profit is zero).

Neville needs a sheet of paper x cm by 13 cm for an origami giraffe. The local origami supply store only sells square sheets of paper. The lower portion of the image below shows the excess area, A, of paper that will remain after Neville cuts out the x cm by 13 cm piece. The excess area, in square centimetres, is given by the equation: x A = x (x − 13) a

At what length of x will the excess area be zero?

Does the graph of A have a minimum or maximum turning point?

For what value of x will Neville be able to make an origami giraffe with the least amount of excess paper?

13 x A

Engineers are optimising a ticketing system for a theme park. The visitor ﬂow during the day can be modelled by the equation: V = −5(t − 2) (t − 8) where V represents the deviation in visitor numbers from a baseline of 50 (with V = 0 indicating 50 visitors), and t is the time in hours since the park opened. To handle peak crowds, an additional ticket booth activates when the number of visitors ﬁrst reaches 50 and deactivates when it next returns to 50. How many hours will the extra booth be operational?

A farmer uses a water pump system with two settings that affect flow rate. The total flow deviation F (in liters per minute) from a standard rate is modelled by: F = −2(3w − 4) (w + 5) where w is the pump’s adjustment in units. The farmer needs the flow to match the standard rate at F = 0. Find the adjustment settings that achieve this, and recommend which setting is more practical if w must be positive.

A drone delivery service tests a ﬂight path where the height deviation H (in metres) from a target altitude is given by H = (5h − 2) (h − 1), with h as the time in minutes since takeoff. The drone must reach the target altitude two times during the ﬂight for successful delivery. Calculate the times when this occurs, and determine the earliest time if h must be non-negative.

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5 Chapter review 1

Does the equation represent a quadratic relationship between x and y? a

4x − 6y − 3 = 0

y = 2x3 − 3x2 + x

x y −2 12 −1 3 0 0 1 3 2 12

−1 0 1 2 3

1 2 4 8

For each graph, identify: i

If the graph is concave up or down.

The turning point.

iii

The equation for the axis of symmetry.

The x and y-intercepts.

−8 −6 −4 −2 −2

x 6

−4

−6 −8

−10 −12

−14 −16

y = (3x2 − 1)2

Does the table of values show a relationship that is linear, quadratic, or neither? a

y = 6x2

In an environmental study, the temperature of the water in a lake is monitored throughout the day. The graph shows the temperature (c) in Celsius change over time (d) in days. Use the graph to approximate answers to each question.

x −8

−4

d 2 4 6 8 10 12 14 16 18 20

−50 −100

What was the initial temperature?

−150

When did the temperature reach 0 °C? What was the maximum temperature reached?

−200

c d

When is the temperature decreasing?

−250 −300

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Match each description to the equation. i

y = 3x2

Narrower than y = x2 and opens upwards.

Wider than y = x2 and has a minimum point.

Wider than y = −x2 and opens downwards.

Narrower than y = −0.5x2 and has a maximum point.

y = −0.5x2

iii

y = −x2

The graph shown has an equation of the form y = ax2. By using a point on the graph, ﬁnd the value of a, and hence the equation of the graph.

y = 0.5x2

4 3

2 1 −5 −4 −3 −2 −1 −1 −2

x 1 2 3 4 5

−3 −4 −5

For each quadratic equation: i

Sketch the graph.

Describe the graph as concave up or concave down.

iii

Is the turning point a maximum or minimum point?

y = −2.5x2

y = −1.25x2

y = 0.8x2

x2 − 49 = 0

Solve: a

3m2 = 0

49x − 121 = 0

16y2 − 64 = 0 2

x2 − 225 = 0 2

4x − 96 = 0

5x − 25 = 0

24x2 − 48 = 0

(5r + 6)2 = 0

(2s − 7)2 = 0

(Optional) Solve: a

p(p − 4) = 0

3q(q + 2) = 0

(t + 3) (t −5) = 0

(4u − 1) (2u + 3) = 0

(Optional) The Sunshine Bakery is projecting its daily profit over the next month. The profit P (in dollars) at any time d (in days) is modeled by the equation: P = −2(d − 8)2 When will the profit be zero?

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(Optional) Maria is designing a rectangular banner with dimensions w m by 9 m for a school event. The supplier provides square sheets of fabric. The excess fabric area, E, after cutting out the banner from a square sheet is shown below. The excess area (in square meters) is given by: E = w(w − 9) a

At what widths w will the excess area be zero?

For what value of w will Maria minimize the excess fabric?

9 w E

Did you know?

Whenever you kick a footy, shoot a basketball, serve a tennis ball, or throw a frisbee, the object follows a curved path through the air. This path is a parabola—the classic U-shape graph of a quadratic function. By adjusting the initial speed and angle of launch, players can control the arc. Coaches and athletes use video analysis and motion tracking software that applies quadratic modelling to improve accuracy and performance. Chapter 5 review mathspace.co

257

Big ideas Accurate calculation and interpretation of interest, earnings, and profit are crucial for effective money management in earnings, investments, and loans.

Curriculum Outcomes • Explore, explain and perform calculations that relate to earning income. Identify the elements of an income statement/payslip, including employer superannuation contributions and income tax as a deduction from gross income • Develop and use the simple interest formula to solve problems relating to saving and borrowing • Analyse, represent and interpret problems in real-world situations involving scientific notation, real numbers, linear equations with variables and/or brackets on either side of the equation, quadratic graphs and equations, direct proportion and/ or simple interest, earning income or income statements • (Optional) Use authenticated websites to explore and compare different savings account options based on their characteristics (interest rates, fees, withdrawal policy) or compare price, quality, terms and conditions of goods and services, such as phone plans and digital subscriptions

6 Finance Chapter outline 6.01 6.02 6.03 6.04

260 270 282 291 293 306

Long-term saving is a great way to set yourself up in the future. If you saved and invested $1 each week from your 5th birthday until you turned 65, how much would you have?

Investigation: Finances

Investigate online

mathspace.co

6.01 Cost, profit and revenue After this lesson, you will be able to… • model revenue and cost using linear equations. • define what is a break-even point. • apply the formulas to calculate profit and loss. • solve problems involving profit, revenue and costs.

Revenue and cost Businesses keep track of two important things: • Revenue: This is the money the business earns from selling products or services. • Cost: This is the money the business spends to operate, such as buying materials or paying workers. There are two types of costs: • Fixed costs: Costs that do not change, no matter how many items are sold (like rent). • Variable costs: Costs that change depending on how many items are sold (like the cost of materials). We can model both revenue and cost using linear equations. In these equations: • The number of items sold (or services provided) is the independent variable (x). • The revenue or cost in dollars is the dependent variable ( y). Key points: • The slope of each line represents how much the cost or revenue increases for each item sold (cost per item or revenue per item). • The y-intercept (where the line crosses the y-axis) represents the fixed costs (costs that exist even when no items are sold). y 10 9 8 7 6 5 Break-even point 4 Cost 3 2 Revenue 1 x 0 1 2 3 4 5 6 7 8 9 10

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A business breaks even when the revenue and cost are the same - meaning the business isn’t losing or making money yet, but it’s covering its costs. In a graph, the break-even point is where the revenue line and cost line meet. To find this point, we set the revenue equal to the cost: Revenue = cost By solving this equation, we can determine when a business will break even.

Example 1 Luana sells handmade scarves from a stall at her local market for $25. Each scarf costs Luana $15 to make, and the cost to have a stall at the market for a day is $100. a Determine the revenue from selling 20 scarves.

Create a strategy Multiply the revenue of $25 per scarf by 20 scarves.

Apply the idea R = Selling price per scarf × Number of scarves

Find revenue

= 25 × 20 Substitute known values = $500 Evaluate

b Write an equation for the daily revenue, R, as a function of x, the number of scarves sold in a day.

Create a strategy Multiply the revenue of $25 per scarf by x.

Apply the idea R = 25x

Write the equation

c Write an equation for the daily cost, C, as a function of x.

Create a strategy The cost is $15 per scarf plus a ﬁxed $100 for the stall, so the equation needs to multiply 15 by x and add 100.

Apply the idea C = 15x + 100

Write the equation

Example 2 Sven decides to start teaching a ﬁtness class. The predicted cost (C) and revenue (R) from running the class are shown on the graph.

72 66 60 54 48 42 36 30 24 18 12 6

Dollars

C = 6x + 24

R = 12x

Participants 0

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261

a Find the revenue and cost if Sven has 2 participants.

Create a strategy Use the graph to ﬁnd the revenue and cost when x = 2.

Apply the idea On the revenue line, when x = 2, y = 24, so the revenue when there are 2 participants is $24.

Dollars

72 66 60 54 48 42 36 30 24 18 12 6

C = 6x + 24

On the cost line, when x = 2, y = 36, so the cost when there are 2 participants is $36.

R = 12x

Participants 0

Reﬂect and check Alternatively, we could have used the equations: R = 12x

Write the equation for revenue

= 12 × 2

Substitute x = 2

= $24

Evaluate

C = 6x + 24

Write the equation for the cost

= 6 × 2 + 24 Substitute x = 2 = $36

Evaluate

b How many participants does Sven need to break-even?

Create a strategy Use the graph to ﬁnd the point of intersection of the revenue and cost lines.

Apply the idea The point of intersection is (4, 48). So if Sven has 4 students, he will break-even.

Reﬂect and check Alternatively, we could have solved this algebraically:

262

R=C

Equate revenue and cost

12x = 6x + 24

Substitute the expressions

6x = 24

Subtract 6x from both sides

x=4

Divide both sides by 6 to solve for x

Mathspace WA Curriculum – Year 9 mathspace.co

Idea summary Revenue and cost can be modelled using linear equations. The break-even point, where the two lines intersect, is when revenue and cost are the same.

Profit and loss If revenue is greater than cost, a business is making money (making a profit (P)). If revenue (R) is less than cost (C), a business is losing money (making a loss). Profit can be calculated using the formula: Profit = Revenue − Cost If the profit is zero, then the business is breaking even. If the profit is negative, then the business is making a loss. A profit of −$1000 is a loss of $1000. If the profit is positive, then the business is making a profit.

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Example 3 Thao umpires hockey games on the weekend and is paid $50 per game. His costs work out to around $10 per game in travel, plus a ﬁxed $210 for yearly registration to the hockey association. Thao umpires 52 games in a year. a Calculate the annual revenue.

Create a strategy Multiply the revenue per game by the number of games.

Apply the idea Revenue = Pay per game × Number of games = 50 × 52

Substitute the values

= $2600

Evaluate

b Calculate the annual cost.

Create a strategy Multiply the cost per game by the number of games and add the ﬁxed cost.

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Apply the idea Cost = Cost per game × Number of games + Flat registration fee = 10 × 52 + 210 Substitute the values = $730 Evaluate c Calculate the proﬁt that Thao will make in a year.

Create a strategy Subtract the cost from the revenue.

Apply the idea Profit = Revenue – Cost = 2600 – 730 = $1870

Proﬁt formula Substitute the values Evaluate

In a year of umpiring, Thao will earn a proﬁt of $1870.

Example 4 Tayna has a small business selling bunches of cut ﬂowers. The weekly cost (C) and revenue (R) for the business, based on the number of bunches sold (x), are shown in the graph.

Dollars 600 500 400 C = 20x + 200

300 200

a Determine the proﬁt (or loss) that Tayna makes if she sells 10 bunches in a week.

R = 60x

100

Bunches 0 1 2 3 4 5 6 7 8 9 10

Create a strategy Find the difference between the revenue and the cost when x = 10.

Apply the idea On the cost line, when x = 10, y = 400. So the cost when selling 10 bunches is $400.

Dollars 600

On the revenue line, when x = 10, y = 600. So the revenue when selling 10 bunches is $600.

500 400 300

Cost = 20x + 200

The difference between 600 and 400 is 200, so the proﬁt is $200.

200 100

Revenue = 60x Bunches 0 1 2 3 4 5 6 7 8 9 10

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The revenue is greater than the cost, so Tayna will make a proﬁt.

Mathspace WA Curriculum – Year 9 mathspace.co

Reﬂect and check Alternatively, we could have solved this algebraically: Profit = revenue – cost

Write the equation for proﬁt

= 60x − (20x + 200)

Substitute the expressions

= 40x – 200

Simplify

= 40 × 10 – 200

Substitute x = 10

= $200

Evaluate

b Determine the proﬁt (or loss) that Tayna makes if she sells 2 bunches in a week.

Create a strategy Find the difference between the revenue and the cost when x = 2.

Apply the idea On the cost line, when x = 2, y = 240. So the cost when selling 2 bunches is $240.

Dollars 600

On the revenue line, when x = 2, y = 120. So the revenue when selling 2 bunches is $120.

500 400 300

Cost = 20x + 200

The difference between 240 and 120 is 120, so the loss is $120.

200 100

The revenue is less than the cost, so Tayna will make a loss.

Revenue = 60x Bunches 0 1 2 3 4 5 6 7 8 9 10

Reﬂect and check Alternatively, we could have solved this algebraically: Profit = 4x – 200 = 40 × 2 – 200

Write the simpliﬁed equation for proﬁt from (a) Substitute x = 2

= −$120 Evaluate A proﬁt of −$120 is a loss of $120.

Idea summary Profit can be calculated using the formula: Profit = Revenue − Cost A profit of zero dollars is breaking even. A negative profit is a loss.

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6.01 Practice questions Understanding 1

5 Ex 1a

Ex 1b

Ex 1c

266

Adventurer 1–3

Trailblazer 1–3

Identify the formula for calculating proﬁt: A

Profit = Cost – Revenue

Profit = Revenue − Cost

Profit = Revenue – Loss

Profit = Loss − Revenue

Money spent is called loss.

Is each statement true or false? a

Money earned is called revenue.

A proﬁt of −$500 is a loss of $500.

The break-even point is when revenue and cost are equal.

When given the graph for revenue and cost, how can we ﬁnd the break-even point?

Fluency 4

Explorer 1–3

4–11, 13, 15

4–11, 13, 15, 17

4–10, 12, 14, 16, 17

Determine the proﬁt (or loss) made if: a

The revenue is $6590 and the cost is $1250.

The cost is $250 and the revenue is $980.

The revenue is $12 500 and the cost is $13 000.

The cost is $720 and the revenue is $240.

Calculate the revenue if: a

68 drinks are sold for $4.80 each.

A personal trainer runs a group session with 12 people, earning $9 per person.

15 pairs of jeans are sold for $84 per pair.

Write an equation for revenue (R) if: a

A restaurant sells x meals for $26 each.

A computer store sells x computers for $1890 each.

Calculate the cost if: a

A pop-up clothing store spends $12 per shirt for 150 shirts, plus a ﬁxed cost of $450 to hire space in a shopping centre.

A bakery spends $1800 on rent for the month and purchases 540 kilograms of ﬂour at $1.50 per kilogram.

Write an equation for the cost (C) in each scenario: a

An artist creates x paintings. Supplies like canvas and paint cost $170 for each painting, and they also spend a ﬁxed $220 on advertising.

A furniture maker rents a workshop for $440 and purchases materials to make x items. The cost of materials is $520 per item.

Mathspace WA Curriculum – Year 9 mathspace.co

Ex 2

Fatu makes cakes and sells them at a market stall. The graph shows her cost (C) and revenue (R) where x is the number of cakes sold. a

Find the revenue if Fatu sells 20 cakes.

Find the cost if Fatu sells 20 cakes.

What is the proﬁt or loss from selling 20 cakes?

How many cakes does Fatu need to sell to break-even?

What are the ﬁxed costs for Fatu?

Dollars 400 300 C = 10x + 100

200 100

R = 20x

0 2 4 6 8 10 12 14 16 18 20

Each week Vince sells x motors. The weekly revenue and cost for his business are shown in the graph. a b c

Find the cost and revenue if Vince doesn’t sell any motors. Find the cost and revenue if Vince sells 8 motors. How many motors must Vince sell to cover his costs?

Dollars 60 000 50 000 40 000 30 000

C = 2500x + 10 000

20 000 10 000

R = 5000x 0

x 8 10 12 14 16

The revenue and cost for a business are given by R = 120x and C = 40x + 380 respectively, where x is the number of items sold. a

Graph both the revenue (R) and cost (C) lines on the same Cartesian plane.

Find the revenue per item.

Find the fixed part of the cost.

Find the revenue if 5 items are sold.

Find the cost if 12 items are sold.

For each of the businesses with revenue (R) and cost (C) descriptions, ﬁnd: i

The revenue per item

The ﬁxed costs

iii

The proﬁt or loss value for 3 items

The proﬁt or loss for 10 items

Business A selling ﬂoor lamps: R = 150x, C = 40x + 400

Business B selling USB drives: R = 25.5x, C = 1250 + 10x

Business C selling digital keyboards: R = 3500x, C = 10 000 + 1250x

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13 Ex 3

Ex 4

Aser has a website design business with a revenue of $450 per website. His costs are $80 per website, plus a ﬁxed $240 for the week. Next week, Aser will design 4 websites. a

Determine next week’s revenue.

Determine the cost for the week.

Calculate the proﬁt that Aser will make next week.

Wasim sells gourmet mushrooms to local restaurants for $40 per box. Each box of mushrooms costs them $20 to produce, and the cost to market and distribute the mushrooms to restaurants each week is $120. In a slow week, Wasim sells just 5 boxes. a

Determine the revenue for the week.

Determine the cost for the week.

Does Wasim make a proﬁt or a loss?

Calculate the proﬁt (or loss) that Wasim makes that week.

Sasha collects washing machines which are no longer Dollars working. After ﬁxing the machines she sells them on. 2500 Her monthly cost (C) and revenue (R) are shown in the graph, where x is the number of washing 2000 machines sold. a

Determine the proﬁt (or loss) that Sasha makes if she sells 4 washing machines in a month.

Determine the proﬁt (or loss) that Sasha makes if she sells 1 washing machine in a month.

C = 250x + 500

1500 1000 500 R = 500x 0

Each week, Kathie makes x reusable bags and sells them online. Her weekly revenue (R) and cost (C) are shown in the graph. a

Determine the proﬁt (or loss) that Kathie makes if she sells 20 bags in a week.

Determine the proﬁt (or loss) she makes if she sells 0 bags in a week.

Solve the equation 35x = 10x + 100 to ﬁnd the number of bags Kathie must sell in a week to break-even.

268

x 4

700 Dollars 650 600 550 500 450 400 350 300 250 200 150 100 50 R = 35x 0

C = 10x + 100

x 10

Ismene runs a catering business that supplies platters of food for $50 each. Ingredients cost her $25 for each platter and other ﬁxed costs like advertising come to $200 for a month. a

Write equations for the monthly cost C and revenue R, if Ismene sells x platters in a month.

How many platters does Ismene need to sell each month to break-even?

How much proﬁt (or loss) will Ismene make each month if she sells 4 platters?

Mathspace WA Curriculum – Year 9 mathspace.co

Reasoning 18

19, 20

For this graph: a

Using this example of cost and revenue, explain why 800 it can be diﬃcult for some businesses when the proﬁt 720 640 margin per item is so small and costs start to rise.

Consider what might happen to the break-even point for example if the cost per item went up from 10 to 11 and they were unable to raise the revenue per item. b

18–20

Is it harder for a business to break even if the ﬁxed price rises from $120 to $200, or if the cost per item rises from $10 to $10.50?

y = 10x + 120

560 480 400 320 240 160 80

y = 12x 10

20 30 40 50 60

The cost of most businesses has a variable (per item) part and a ﬁxed part. a

Consider the following example, where the cost is purely variable. R = 20x C = 5x

Will the business ever make a loss? Justify your answer.

Give an equation for the proﬁt (P) in terms of x.

Consider the following example, where the cost is purely ﬁxed. R = 20x C = 500

Describe the graph of the cost.

Give an equation for the proﬁt (P) in terms of x.

Naia is thinking about opening a smoothie stand. Her predicted weekly revenue (R) and cost (C), where x is the number of smoothies sold, are: R = 8x C = 9x + 600 a

Graph the revenue (R) and cost (C) lines on the same Cartesian plane.

Explain why Naia will never break-even with this model.

What does Naia need to change in order for the smoothie stand to make a proﬁt?

Problem-solving 21, 22 21

21, 22

21–23

Salvio is considering opening a restaurant. He has calculated his ﬁxed costs at $8 000 each month. He expects the cost to produce each meal will be $15. a

Write an equation for Salvio’s monthly cost, C, if he sells x meals.

Determine the number of meals that will result in a monthly cost of $20 000.

What revenue per meal will cause Salvio’s restaurant to break-even when the monthly cost is $20 000?

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Helene makes wooden bowls. Each bowl costs her $20 to make and she sells them for $70 each. She also spends money on advertising and rent for the workshop each month (her ﬁxed costs). a

Calculate the revenue when Helene sells 6 bowls in a month.

If Helene breaks even when she sells 6 bowls in a month, determine the ﬁxed part of the cost.

Find the proﬁt made if Helene sells 20 bowls.

The revenue and cost for a small business are given by R = 110x and C = 50x + 390

respectively, where x is the number of items sold. a

Graph the revenue (R) and cost (C) lines on the same Cartesian plane.

Find the proﬁt if 10 items are sold.

Find the proﬁt if 10 items are sold, as a percentage of the revenue. Round your answer to one decimal place.

Give an equation for the proﬁt, in terms of x.

Give an expression for the proﬁt as a percentage of the revenue, in terms of x.

6.02 Income After this lesson, you will be able to… • understand the concepts of salary, wage, and commission as different ways of earning income. • calculate and compare salaries using time conversions. • model the income from wages using linear equations. • model the income from commission using linear equations. • apply the understanding of salary, wage, and commission to real-world scenarios.

Salary Income is the amount of money a person earns. One common way to earn income is through a salary. A salary is a set amount of money you get over a certain time period, usually a year. When earning a salary, you get paid the same amount no matter how many hours you work (unless there are special arrangements like overtime or bonuses). For example, a teacher is paid the same annual (yearly) salary, regardless of how much time they spend preparing lessons and marking assessments. We can calculate and compare income using these time conversions:

1 year = 12 months 1 year = 26 fortnights 1 year = 52 weeks

We avoid using 1 month ≈ 4 weeks as it’s not accurate. If converting between months and weeks, convert to an annual value ﬁrst. 270

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Example 1 Derek is considering three different job offers and wants to choose the one that will pay him the greatest salary: • Offer 1: $2917 per month • Offer 2: $955.73 per week • Offer 3: $52 170 per year a Find the annual income for Offer 1.

Create a strategy Convert the monthly salary to its annual equivalent.

Apply the idea Annual income = 2917 × 12

Multiply the monthly salary by 12

= $35 004

Evaluate

Reﬂect and check Verify the calculation by dividing the annual income by 12. $35 004 ÷ 12 = $2917 The value matches the monthly salary, so it is correct. b Find the annual income for Offer 2.

Create a strategy Convert the weekly salary to its annual equivalent.

Apply the idea Annual income = 955.73 × 52 = $49 697.96

Multiply the weekly salary by 52 Evaluate

Reﬂect and check Verify the calculation by dividing the annual income by 52. $49 697.96 ÷ 52 = $955.73 The value matches the weekly salary, so it is correct. c Which job offer should Derek accept?

Create a strategy

Apply the idea

Compare the annual incomes and choose the greatest.

• Offer 1: $35 004 per year • Offer 2: $49 697.96 per year • Offer 3: $52 170 per year Offer 3 has the highest annual income, so Derek should choose Offer 3.

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Idea summary A salary means earning a set amount of income (regardless of the hours worked). To calculate and compare income, use the conversions: 1 year = 12 months = 26 fortnights = 52 weeks

Wage Another way to earn income is with a wage. A wage is based on how much money you earn per hour (or another time period). Since your income depends on how long you work, we can model this using a linear equation: Income = Rate of pay × Time worked

y = kx y is the income k is the rate of pay x is the time worked

Sometimes, the rate of pay can increase under certain conditions. For example, if you work on a public holiday, you might be paid 1.5 times your normal rate (this is called time-and-a-half) or even double your normal rate (double-time). Wages and salaries are paid differently because they account for different work conditions. For example: • A job that is paid by salary usually comes with extra benefits like sick leave or holiday leave. • Someone with a wage might have a higher hourly rate but doesn’t get paid leave benefits.

Example 2 Naziha earns $385 per hour as a lawyer. a Calculate Naziha’s weekly wage if she works 30 hours in a week.

Create a strategy Multiply the hourly rate of pay by the number of hours.

Apply the idea Income = 385 × 30 = $11 550

Write the multiplication Evaluate

Reﬂect and check Verify the calculation by dividing the weekly wage by 30. $11 550 ÷ 30 = $385 The value matches the weekly wage, so it is correct.

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b Write an equation for Naziha’s weekly wage ( y) if she works x hours per week.

Create a strategy Multiply the hourly rate by x.

Apply the idea y = 385x

Write the equation

Reﬂect and check Check by substituting known values, x = 30 and y = 11 550 to see if the result is true. 11 550 = 385 × 30 11 550 = 11 550 This equation is true, so our equation is correct.

c Determine the number of hours that Naziha worked in a week if she earned $12 801.25

Create a strategy

Apply the idea y = 385x

Use the equation from part (b).

12 801.25 = 385x

Write the equation Substitute the income

x = 33.25 hours Divide both sides by 385 to solve for x Naziha worked 33.25 hours, or 33 hours and 15 minutes.

Example 3 Waqas worked a total of 18 hours last fortnight at his normal hourly rate of pay. He was paid a wage of $390.96 a What is his normal hourly rate of pay?

Create a strategy Divide the income by the number of hours worked.

Apply the idea Divide income by hours worked

Evaluate

Reﬂect and check The rate of pay is approximately $20 per hour. Multiplying by the hours worked gives the income: 20 × 18 = $360, which is close to $390.96

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b O ver the next fortnight, Waqas works for 14 hours at the normal rate and 4 hours at double time. Calculate his total pay for the fortnight.

Create a strategy Multiply each pay rate by the hours worked, then add these.

Apply the idea Normal income = 21.72 × 14 = $304.08

Multiply the pay rate by hours worked Evaluate

Double time pay rate = 2 × 21.72 Double the normal pay rate = $43.44 per hour Double time income = 43.44 × 4 = $173.76 Income = 304.08 + 173.76 = $477.84

Evaluate Multiply by hours worked at double time Evaluate Add each part of the income Evaluate

Reﬂect and check Working for 4 hours at double time is equivalent to working for 8 hours at the normal rate. Multiplying the normal rate by the total equivalent hours: 21.72 × (14 + 8) = $477.84

Idea summary A wage means that income is based on the number of hours worked.

y = kx y is the income k is the rate of pay x is the time worked

Commission Another way to earn income is through a commission. Being paid on commission means that you earn money based on the total value of items that you sell. Usually, commission is a percentage of the total value of what you sell. For example, a real estate agent might earn 2% of the selling price of each house that they sell. Employees paid by commission may also be paid a retainer. This is a set amount of money paid over a given time period. A retainer ensures that an employee still earns some money, even if they don’t make any sales.

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Interactive exploration

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Income from commission can be modelled using a linear equation: Income = Commission × Value of items sold + Retainer

y = mx + c y is the income m is the commission rate (as a decimal) x is the total value of items sold c is the retainer

Example 4 Wentao is an art dealer. The gallery where he works pays him a retainer of $1100 per fortnight. He also earns a commission of 20% of the total value of art that he sells. a Determine Wentao’s income for a fortnight in which he sells art worth $8500.

Create a strategy Multiply the commission rate by the total value of art sold, and add the retainer.

Apply the idea Income = 20% × 8500 + 1100

Write the equation

= 0.2 × 8500 + 1100

Convert the percentage to a decimal

= $2800

Evaluate

b Write an equation for Wentao’s fortnightly income ( y) based on the total value of art sold (x).

Create a strategy Multiply the commission rate by x, and add the retainer.

Apply the idea y = 0.2x + 1100

Write the equation

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c Calculate the total value of art that Wentao must sell to make $4000 in a fortnight.

Create a strategy Use the equation from part (b).

Apply the idea y = 0.2x + 1100

Write the equation

4000 = 0.2x + 1100

Substitute the income

2900 = 0.2x

Subtract 1100 from both sides

x = $14 500

Divide both sides by 0.2 to solve for x

Reﬂect and check Substitute the answer back into the equation to check: y = 0.2 × 14 500 + 1100

Substitute x = 14 500

= $4000

Evaluate

Example 5 Jenina works at a car dealership. The monthly income that she earns ( y) is shown in the graph, where x is the total value of her car sales for the month. y 7000 6000 5000 4000 3000 y = 0.05x + 2000

2000 1000

x 0

20 000 40 000 60 000 80 000

a Find Jenina’s monthly retainer, and her commission as a percentage of the total value of car sales.

Create a strategy

Apply the idea

The coeﬃcient of x (the slope of the line) is the commission rate and the y-intercept is the retainer.

The coeﬃcient of x is 0.05, so the commission is 5% of the value of car sales.

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The y-intercept is 2000, so the retainer is $2000 per month.

b Find Jenina’s monthly income if she sells cars which have a total value of $80 000.

Create a strategy Use the graph to ﬁnd the y value when x = 80 000.

Apply the idea When x = 80 000, y = 6000.

y 7000

So Jenina’s income is $6 000.

6000 5000 4000 3000 y = 0.05x + 2000

2000 1000

x 0

20 000 40 000 60 000 80 000

Reflect and check Alternatively, we could have used the equation: y = 0.05x + 2000

Write the equation

= 0.05 × 80 000 + 2000

Substitute x = 80 000

= $6000

Evaluate

Idea summary Commission means that income is based on the total value of items sold.

y = mx + c y m x c

is the income is the commission rate (as a decimal) is the total value of items sold is the retainer

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6.02 Practice questions Understanding

Explorer 1–3

Adventurer 1–3

What is the difference between a wage and a salary?

True or false:

Commission means that income is based on the number of items sold.

Double time means that your normal hourly rate of pay is doubled.

A retainer is a set amount of money earned, even if an employee doesn’t make any sales.

Income from commission can be modelled using a linear equation: y = mx + c. What is the correct description for the variables? A

y is the income, m is the retainer, x is the total value of items sold and c is the commission rate

y is the income, m is the commission rate, x is the retainer and c is the total value of items sold

y is the income, m is the commission rate, x is the total value of items sold and c is the retainer

Fluency 4 Ex 1

5 Ex 2

Trailblazer 1–3

4–7, 9, 10, 12, 14, 15, 17, 18, 21

4, 5, 8–11, 13, 14, 16, 19, 20

Find the annual salary of a worker whose weekly pay is $332. Assume there are 52 weeks in a year. Christa earns $21 per hour working as a call centre operator. If she works 14 hours per week, how much is her weekly wage?

Roald’s annual salary is $100 482. Calculate his monthly income.

Bianca’s monthly salary is $5841.50. Calculate her fortnightly pay rate.

Yousaf is considering three different job offers and wants to choose the one that will pay him the most. • Offer 1: $4570 per month • Offer 2: $1020 per week • Offer 3: $53 450 per year

278

Find the annual income for Offer 1.

Find the annual income for Offer 2.

Which job offer should Yousaf accept?

James has a job that pays him $24 per hour and he works 38 hour weeks. Calculate to the nearest dollar his: a

Weekly income

Fortnightly income

Annual income

Monthly income

Katarina has a job that pays a salary of $76 000 per year. Calculate each to the nearest dollar: a

Monthly income

Fortnightly income

Weekly income

Hourly rate for a 38 hour week

Mathspace WA Curriculum – Year 9 mathspace.co

Pauline works at a tutoring company for 34 hours a week at a rate of $35 per hour. How much more will she earn per week if her wage is increased to $43 per hour?

Gurdip earns $52 per hour as a chef.

14 Ex 3

17 Ex 4

Write an equation for Gurdip’s weekly income, y, if he works x hours in a week.

How many hours does Gurdip need to work in a week to earn $2158?

Jmiaa works as a gardener and earns $36 per hour. a

Write an equation for Jmiaa’s monthly income, y, if she works x hours in a month.

Calculate Jmiaa’s annual income if she works 150 hours each month.

Ouafaa earned a total of $627 after working a total of 22 hours as a waitress. a

Calculate her hourly rate of pay.

Calculate Ouafaa’s weekly income if she works for 18 hours at the normal rate and 4 hours at double time.

Norik works as a trade assistant and earns $25 per hour. a

Calculate Norik’s annual income if he works for 38 hours each week at his normal rate of pay.

Calculate Norik’s annual income if each week he works for 32 hours at the normal rate, and 6 hours at time-and-a-half.

Neil normally works from 6:00 a.m. to 5:00 p.m. Monday to Friday, and from 10:00 a.m. to 3:00 p.m. on weekends. His normal weekly income is $1602. a

How many hours does he work per week?

Calculate his normal hourly rate of pay.

One week he works 60 hours at his normal rate, and 5 hours at double time. Calculate his income for the week.

Xiaoyi is a real estate agent. She earns a retainer of $502 per week and a commission of 1% of the total value of her property sales. Determine Xiaoyi’s income for a week in which she sells a property worth $540 000.

This year, Irene has made sales with a total value of $571 130. Her annual retainer is $39 500. If Irene earns a commission of 3.5% of the value of her sales, calculate her annual income.

Sahil earns a retainer of $1700 per month plus a commission of 2% of the total value of his sales. a

Write an equation for Sahil’s monthly income, y, based on the total value of his sales, x.

Calculate Sahil’s monthly income if he makes sales worth $248 900.

Calculate the total value of sales that Sahil must make in a month to earn $5200 in income.

Cythia works as a ﬁnancial adviser and earns a commission of 0.5% of the value of her clients’ proﬁts. Her retainer is $800 per week. a

Write an equation for Cythia’s weekly income, y, based on the total value of her clients’ proﬁts, x.

Calculate Cythia’s weekly income if her clients were to make a total proﬁt of $85 000 each week.

What would her clients’ total weekly proﬁts need to be for Cythia to earn an annual income of $70 200? 6.02 Income mathspace.co

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21 Ex 5

Jarvis is a travel agent. His monthly income ( y) is shown in the graph, where x is the total value of bookings he takes. a

Find the value of the monthly retainer that Jarvis earns.

3500 3000

Find the commission as a percentage of the total 2500 2000 value of bookings he takes.

Find Jarvis’ monthly income if he takes bookings with a total value of $5000.

What is the total value of bookings that Jarvis needs to take to earn a monthly income of $4000? 22, 23

4000

Reasoning 22

4500

y = 0.1x + 2500

1500

1000

23–25

500

x 0

5000

10 000

15 000

23–25

Mbacke has two part-time jobs and is comparing the income from each. His job as a software developer pays $2850 per month, while his position as a manager at a computing store pays $680 per week. He multiplies his weekly income as a manager by 4 to get a monthly income of $2720 and concludes that he earns more as a software developer. Is Mbacke correct? Justify your answer.

Elna has two job offers, both of which pay a retainer and commission. • Offer A pays a weekly retainer of $500 and a commission of 10% of the total value of sales. • Offer B pays a weekly retainer of $1000 and a commission of 5% of the total value of sales. Elna thinks that she can make about $4000 per week in sales in either role. Explain which offer she should choose.

An employee works for a hours at their normal hourly rate of pay and b hours at double time. If the rate of pay is $20 per hour, give an expression for:

Weekly income in terms of a and b

Weekly income in terms of a, if

Weekly income in terms of b, if a = 3b

Compare these two jobs and discuss situations where one job may be more beneﬁcial than the other. • Job A: Wage of $28.50 per hour, for 30 hours per week, 40 weeks of the year. • Job B: A salary of $40 000 a year, which includes 4 weeks paid holidays.

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Problem-solving 26, 27 26

26, 28, 29, 31

26–31

Nico earned $43 100 in one year and worked an average of 30 hours per week. a

What was his hourly rate of pay?

If his hourly rate of pay increased by $5, how much would his annual income increase by?

Calculate the percentage increase in his income, rounded to one decimal place.

Asmaa earns $24 per hour as a sales assistant. If she works on a Sunday, she is paid timeand-a-half. For Asmaa to earn $540 per week, a

How many hours would she need to work at her normal rate?

How many hours would she need to work on a Sunday only?

How many hours would she need to work on a Sunday, if she also works for 12 hours at her normal rate?

Lavern is considering two job offers. • Offer A pays $35 per hour. • Offer B pays an annual salary of $65 520. For how many hours each week would Lavern need to work under Offer A to to earn the same or higher income than Offer B?

Arona earns a commission of 3.5% of the total value of her sales, plus a weekly retainer. In a week where she sold items with a total value of $25 200, her weekly income was $1122. Determine the value of Arona’s weekly retainer.

Luigi is paid a retainer of $2100 per month plus a commission on his mobile phone sales. Last month Luigi sold mobile phones with a total value of $13 680 and earned a monthly income of $2373.60. Determine his commission rate as a percentage of his total sales value.

Missy bakes cookies and sells them in their canteen. The graph shows the cost, revenue and the profit where x is the number of cookies sold. Each line has been labelled. 400

y (dollars)

300 200

100 -100

B 2

C x (cookies) 8 10 12 14 16 18 20

-200

Match lines A, B and C to the profit, cost or revenue.

How many cookies must be sold to reach the break-even point?

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6.03 Simple interest After this lesson, you will be able to… • define what simple interest is and how it is calculated. • convert different time periods for calculating interest. • understand loan repayments and the formula used to calculate them. • solve problems involving calculating simple interest and determining loan repayments.

Simple interest Simple interest is a way to calculate how much extra money you will earn (if you invest) or pay (if you borrow) over time. It is a percentage of the starting amount, called the principal. Simple interest = Principal amount × Interest rate × Time

I = PRT

I is the interest (the extra money you earn or owe) P is the principal (the amount you start with) R is the rate of interest per time period, as a decimal T is the number of time periods (the duration of the loan/investment)

The units of time for R and T must be the same. For example, if an annual interest rate is used, then the time must be in years. Note that “p.a.” is an abbreviation of “per annum” and means per year. We can use the following conversions:

1 year = 4 quarters = 12 months = 26 fortnights = 52 weeks = 365 days

When you borrow money, you must pay back both the principal and the interest. When you invest money, you own both the principal and the interest. In both cases, the total amount is: A=P+I

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Example 1 $906 is invested at 5% p.a. simple interest. Dave wants to know the number of years it will take the investment to grow to $1132.50 a Calculate the interest that will be earned on the investment.

Create a strategy Use the formula for the total amount.

Apply the idea A=P+I

Write the formula

1132.50 = 906 + I

Substitute the values of A and P

I = 1132.50 – 906

Subtract 906 from both sides to solve for I

= $226.50

Evaluate

b Calculate the number of years it will take the investment to grow to $1132.50

Create a strategy Use the simple interest formula.

Apply the idea

Write the formula

Substitute I, P, and R

Convert the percentage to a decimal

Evaluate the multiplication

Divide both sides by 45.3 to solve for T

Evaluate

Reﬂect and check Each year the investment of approximately $900 is earning 5% interest. 10% of $900 is $90, so 5% is $45. After 5 years the investment has earned 5 × 45 = $225 interest, bringing the total amount to about $1125, which is close to $1132.50

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Example 2 Calculate the simple interest earned on an investment of $6050 at 0.7% per quarter for 3 years.

Create a strategy To use the simple interest formula, the interest rate and time period both need to be in the same unit. In this case, quarters. Three years has 3 × 4 = 12 quarters. So now the interest rate of per quarter matches the time period.

Apply the idea I = PRT Write the simple interest formula = 6050 × 0.7% × 12

Substitute the values

= 6050 × 0.007 × 12 Convert the percentage to a decimal = $508.20

Evaluate

Reﬂect and check We could have left the time in years and converted the quarterly rate to an annual rate. To ﬁnd the annual rate, we multiply the quarterly rate of 0.7% by 4 to get 2.8%. Then I = 6050 × 2.8% × 3 which gives $508.20 again.

Example 3 The simple interest on a loan of $6600 over 33 months is $1252.35 Find the annual interest rate as a percentage, rounded to one decimal place.

Create a strategy Use the simple interest formula.

Apply the idea Since we are looking for the annual interest rate, we should convert the time to years (by dividing by 12 months).

Write the formula

Substitute I, P, and T (in years)

Evaluate the multiplication

Divide both sides by 18 150

Evaluate the division

Convert to a percentage

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Reﬂect and check We could have left the time in months. Then the rate found would be a monthly value which we could convert to an annual rate by multiplying by 12: 1252.35 = 6600 × R × 33 R = 0.575% per month = 6.9% p.a.

Idea summary We can calculate the simple interest for a loan or investment using the formula:

I = PRT I P R T

is the interest is the principal (initial amount) is the rate of interest per time period (as a decimal) is the number of time periods

The time periods for R and T must be consistent. The total amount owed or in the account is: A=P+I

Loan repayments Loans are usually repaid in a number of instalments (repayments), rather than all at once. We can use the following formula to calculate loan repayments:

Repayment amount =

A is the total amount (principal plus interest) n is the number of repayments

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Example 4 Katrina takes out a loan to purchase a surround sound system. The total amount to be repaid is $95 000. a If Katrina makes 19 equal loan repayments, what is the value of each repayment?

Create a strategy Use the loan repayments formula.

Apply the idea Write the formula

Substitute the values

Evaluate

b If Katrina pays the loan back in 2 years with repayments of $11 875, how often did she make repayments?

Create a strategy Use the loan repayments formula.

Apply the idea Write the formula Substitute the values Multiply both sides by n Divide both sides by 11 875 Evaluate Divide by 2 years to ﬁnd repayments per year Katrina made quarterly repayments (1 repayment every 3 months).

Idea summary

Repayment amount A n

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is the total amount (principal plus interest) is the number of repayments

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6.03 Practice questions Understanding 1

Adventurer

Trailblazer

1–3

Calculate the simple interest on each scenario: a

An investment of $45 000 at 6% p.a. for 2 years

A 3-year loan of $3000 at 12%. p.a.

How can we convert: a

A time period from months to years?

A monthly interest rate to an annual rate?

True or false: a

A higher interest rate will result in more interest being paid on a loan.

A higher principal will result in less interest being paid on a loan.

A larger number of repayments over the same time period will result in a lower repayment amount.

Fluency 4

Explorer 1–3

4, 6, 7, 9–11, 12ace, 13, 15, 5–7, 9–11, 12bdf, 14, 15, 16a, 17, 18 16b, 17, 18

5, 6, 8–11, 12bdf, 14, 15, 16c, 17, 19

Homer borrowed $7600 from a bank at a simple interest rate of 4.5% for one year. a

How much interest will Homer need to pay after 1 year?

Calculate the total amount Homer will have to repay the bank at the end of 1 year.

Find the total amount to be paid on a 4-year, $7000 loan at 11% p.a. ﬂat interest.

Toshko invests $12 500 at 4% p.a. simple interest and wants to know the number of years it will take for the investment to grow to $15 000.

Ex 1

Calculate the interest that will be earned on the investment.

Calculate the number of years it will take the investment to grow to $15 000.

For how many years does $7050 need to be invested for it to grow to $7950, if simple interest is paid at 7% p.a.? Round your answer correct to two decimal places.

Ajaz invests $5680 at a simple interest rate of 5% p.a. and earns $426 in interest.

Calculate the number of years taken to earn the interest.

Find the number of quarters taken to earn the interest.

Find the equivalent annual simple interest rate. a

0.63% per month

0.03% per day

1.96% per quarter

The simple interest rate for an investment is 6% p.a. Find the percentage rate, rounded to two decimal places where necessary, used when calculating interest: a

Quarterly

Monthly

Weekly

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11 Ex 2

Calculate the simple interest earned on these investments, rounded to two decimal places. a

$5720 is invested at 10% p.a. for 63 weeks.

$9780 is invested at a rate of 5% p.a. for 17 months.

$9020 is invested at 1% per month for 2 years.

$5440 is invested at 0.3% per fortnight for 566 days.

Calculate the ﬁnal value of these investments over the given time period rounded to two decimal places: a

$4570 at 4% p.a. for 28 months

$9980 at 7% p.a. for 26 weeks

$1430 at 8% p.a. for 130 days

$6170 at 1.2% per month for 3 years

$9060 at 1% per quarter for 4 years

$4620 at 0.1% per fortnight for 3 years

Maria is given three investment account options by her bank: • Option 1: Interest is earned weekly at a simple interest rate of 0.14% per week. • Option 2: Interest is earned monthly at a simple interest rate of 0.63% per month. • Option 3: Interest is earned yearly at a simple interest rate of 7.3% p.a. Which option should Maria choose if she wants to earn as much interest as possible?

Sean is offered three different loan plans when borrowing money from the bank: • Option 1: Interest is earned daily at a simple interest rate of 0.02% per day. • Option 2: Interest is earned fortnightly at a simple interest rate of 0.34% per fortnight. • Option 3: Interest is earned quarterly at a simple interest rate of 1.98% per quarter. Which option should Sean go for if he wants to pay as little interest as possible?

For his investment into government bonds, Buzz was paid simple interest of 9% p.a. Calculate the size of Buzz’s initial investment if he earned $294.12 interest after 24 months.

Find the annual interest rate as a percentage. Round your answer to the nearest whole percent.

Ex 3

17 Ex 4

The simple interest on an investment of $5670 over 36 months is $1360.80

The simple interest on a loan of $25 100 over 39 weeks is $1694.25

The simple interest on a loan of $7500 over 50 months is $1250

Gitte takes out a loan to purchase a new fridge. She makes 16 equal loan repayments. The total loan amount paid is $1280. Find the value of each repayment.

Iain takes out a loan to purchase a jet ski. He makes 15 equal loan repayments of $5239. Calculate the total amount paid back on the loan.

Summer borrows money to pay for a vet bill and wants to pay it back in a year. If the total loan amount is $2470 and her repayments are $95, how often must Summer make repayments?

Reasoning

20, 21

20–22

Yarel borrows $1200 at a simple interest rate of 8% p.a. To calculate the interest he will owe over 22 months, Yarel uses the calculation I = 1200 × 0.08 × 22. Explain why Yarel’s calculation is not correct, and how he can ﬁnd the correct value.

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Ariela wants to invest her savings. She is considering two options: • Option 1: Simple interest rate of 0.5% per month for two years • Option 2: Annual rate of 2.5% simple interest, for 42 months Which option will result in Ariela earning the most interest? Justify your choice.

Explain why a shorter loan term (T value) will result in less interest paid, even though the monthly repayments may be higher.

Problem-solving 23–26 23

24–27

24, 26–30

Tara takes out a car loan of $6000 at a simple interest rate of 2% per quarter. She plans to repay the loan over 2 years through regular monthly repayments. a

Calculate the total interest that Tara will accrue over the duration of the repayment.

Calculate the value of each repayment of the loan.

The graph shows the amount of simple interest charged each year by a particular bank, on a 8-year loan.

Interest ($) 350 300

Find the total amount of simple interest charged on a loan of $5000.

Calculate the simple interest rate per annum 200 charged by the bank on these 8-year loans. Give your answer as a percentage. 150

250

100 50 Principal ($) 0

1000 2000 3000 4000 5000

Mohcin and Emily are looking to invest $11 600 and $18 300 respectively for 6 years. The simple interest rate applied is determined by the following table: Principal Between $4800 and $14 800 More than $14 800 to $24 800 Greater than $24 800

Quarterly rate 1% 1.25% 1.5%

How much interest will Mohcin earn on his investment?

How much interest will Emily earn on her investment?

How much will they earn if they invest their combined principal amounts?

How much more interest will they earn by combining their principal amounts?

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The graph shows the amount of money in a savings account earning simple interest. a

Find the principal amount.

Calculate the interest earned after 10 months.

Calculate the monthly rate of simple interest.

Amount ($) 1400 1200 1000 800 600 400 200 0

Time (months) 5

An amount is invested at 0.5% per quarter, simple interest, for 8 years. What loan term would earn the same amount of interest if the interest rate was 0.8% per quarter?

A borrower received a loan of $5500 which was repaid in monthly instalments of $269 over 36 months. Calculate the flat rate of interest per annum as a percentage, rounded to two decimal places.

At the start of a month, Lana had $7000. She invested some of it in a friend’s startup and the rest in a savings account with a simple interest rate of 3% per month. By the end of the month, her friend’s startup gave her a 10% return on her investment and her total worth was $7410. Use equations to calculate how much she invested in her friend’s startup.

Shawn is going to purchase a car. He has two options: • Option 1: Using dealer finance which uses simple interest of 5.6% over 3 years for the full purchase amount of $5599. • Option 2: By paying an initial deposit of $1000, and then financing the rest at a higher rate of 7.2% for 3 years. Which option should Shawn take if he wants to pay the least amount in total for the car?

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MODELLING INVESTIGATION

Drone costs Model costs for a school science fair drone project. Analyse material and battery choices, represent costs and savings mathematically, and interpret findings to propose a budget within constraints.

Objectives • Analyse the drone project cost scenario and identify key variables and constraints • Represent costs and savings using real numbers, linear equations, direct proportion and simple interest • Interpret findings to propose a budget and communicate implications

Analyse the drone cost scenario You are planning a school science fair drone project. Analyse preferences for frame materials and battery types, considering costs and savings to meet budget constraints. Choose a frame material cost: • Plastic: $50-$80 • Carbon fibre: $100-$150 • Other-research your own Determine other variables: • Battery cost − Standard: $30-$50 − High-capacity: $60-$80 • Savings principal: $100-$200 • Simple interest rate: 3.5%-5% per annum The total cost per drone must be less than $300 and savings must cover at least 50% of the cost within 6 months.

Investigate 1.

Record your chosen costs, principal and interest rate. Why did you select these values?

List one assumption and one constraint.

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291

Determine your dataset and analysis Use real numbers, linear equations, direct proportion and simple interest to model the drone’s total cost and savings plan.

Interactive exploration

Discover this concept in action online

mathspace.co

Calculate the total cost, savings with simple interest and verify that the cost is less than $300 per drone and that the savings is greater than 50% of the total cost after 6 months.

Investigate 1.

Calculate the total cost and savings for your values.

Use linear inequalities to verify the cost and savings constraints. Are they satisfied?

Interpret your plan Interactive exploration

Discover this concept in action online

mathspace.co

If costs increase by 10%, what happens to the cost of the drone and savings goals? Also explore your model to determine what happens if one of the savings variables changes. Propose a budget for 50 drones.

Investigate 1.

Does your budget meet cost and savings goals? Is it realistic for the science fair? Why?

How does adjusting the principal affect savings?

How did the 10% cost increase impact your model, what can you adjust to accommodate the increase and keep the cost of the drone down?

Present your budget plan to the class. How do assumptions (fixed costs) affect it?

Discussion

292

What was the most challenging part of modelling the drone costs? Why?

How could you improve your budget with more funds or different materials?

Compare your budget with a classmate’s. How do your costs and savings differ?

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6.04 Re-investing with simple interest After this lesson, you will be able to… • use tables to calculate simple interest over different time periods. • compare the effects of charging interest on the total amount versus only on the principal amount. • apply these concepts to real-world scenarios involving loans and investments.

Using tables for simple interest We have used the formula I = PRT to calculate the total interest accrued at the end of the term of a loan or investment. An alternative approach is to use a table which shows how interest is calculated each time period and how repayments affect the balance. The opening balance is the amount owed, (or in the account), at the start of each time period. The closing balance is the amount left at the end of the time period after repayments. The table shows the monthly balances for a loan of $20 000 at a simple interest rate of 1% per month, with monthly repayments of $1500. Month 1 2 3 4

Opening balance 20 000 18 700 17 400 16 100

Interest 200 200 200 200

Repayment 1500 1500 1500 1500

Closing balance 18 700 17 400 16 100 14 800

The interest is calculated based on the opening balance, so even though the balance decreases each month, the interest stays the same at $200. Each month, the interest is added to the opening balance, and the repayment is subtracted to give the closing balance. The closing balance then becomes the opening balance for the next month. If we continue the table until month 16, we get: Month 13 14 15 16

Opening Balance 4400 3100 1800 500

Interest 200 200 200 200

Repayment 1500 1500 1500 700

Closing Balance 3100 1800 500 0

In month 16, the repayment is only $700, because the opening balance plus interest is less than the normal repayment amount. This brings the closing balance to zero. When using a table to calculate the interest on an investment, there will be no repayments and the ﬁnal balance will not be zero (as the balance is always increasing).

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Exploration Using a spreadsheet tool, see if you can design a spreadsheet of the table mentioned. As an extra challenge, can you design it so that you can explore different interest rates, different opening balances, or different loan terms?

Example 1 Tirsa invests $425 000 in an account that pays 1.5% simple interest per quarter. Construct a table showing the opening balance, interest and closing balance each quarter for the ﬁrst year.

Create a strategy Each quarter, add the interest to the opening balance to get the closing balance, then copy this into the next row as the opening balance of the next quarter.

Apply the idea One year is four quarters. Quarter 1 will have an opening balance of $425 000. Each quarter the interest will be: I = 1.5% × 425 000

Write an expression for quarterly interest

= 0.015 × 425 000

Convert the percentage to a decimal

= $6 375

Evaluate

Quarter 1 2 3 4

Opening Balance 425 000 431 375 437 750 444 125

Interest 6 375 6 375 6 375 6 375

Closing Balance 431 375 437 750 444 125 450 500

Reﬂect and check We can use the simple interest formula (with T = 4 quarters) to check our table: I = PRT = 425 000 × 1.5% × 4 = $25 500 A=P+I = 425 000 + 25 500 = $450 500

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Example 2 Inas takes out a short-term loan of $1000 at a simple interest rate of 2% per month and makes monthly repayments of $220. a Create a table showing the opening balance, interest, repayment and closing balance for each month. Continue the table until the closing balance is zero.

Create a strategy Each month, add the interest and subtract the repayment to get the closing balance, then copy this into the next row as the opening balance.

Apply the idea Month 1 will have an opening balance of $1000. Each month the interest will be: I = 2% × 1000

Write an expression for monthly interest

= 0.02 × 1000

Convert the percentage to a decimal

= $20

Evaluate

Month 1 2 3 4 5

Opening Balance 1000 800 600 400 200

Interest 20 20 20 20 20

Repayment 220 220 220 220 220

Closing Balance 800 600 400 200 0

b How long did it take Inas to pay off the loan and what was the total amount repaid?

Create a strategy Use the table from (a).

Apply the idea The last row of the table (with a closing balance of zero) was month 5, so it took Inas 5 months to pay off the loan. Inas made 5 monthly repayments of $220 so the total amount repaid was 5 × 220 = $1100.

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c What was the total interest paid?

Create a strategy Use the table from (a).

Apply the idea It took Inas 5 months to pay off the loan and the interest was $20 each month. The interest paid was 5 × 20 = $100.

Reﬂect and check From (b) we know that the total amount repaid was $1100. The principal was $1000 so the interest paid was 1100 − 1000 = $100. We could also use the simple interest formula to check our table: I = PRT = 1000 × 2% × 5 = $100

Example 3 Danita takes out a $30 000 car loan where simple interest is charged quarterly. The last quarters of the loan balances are shown in the table. Quarter 15 16 17 18

Opening Balance 5500 3750 2000

Interest 750 750 750

Repayment 2500 2500 2500

Closing Balance 3750 2000 250

a Complete the last row of the table for quarter 18.

Create a strategy The opening balance for quarter 18 will be the closing balance for quarter 17. The interest will be the same as previous rows. Add the interest to the opening balance to calculate the ﬁnal repayment.

Apply the idea The opening balance plus the interest is 250 + 750 = $1000. Since this is less than the normal repayment amount, the last repayment will be reduced to $1000, to bring the closing balance to zero. Quarter 15 16 17 18

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Interest 750 750 750 750

Repayment 2500 2500 2500 1000

Closing Balance 3750 2000 250 0

b How many years did it take Danita to pay off the loan?

Create a strategy Use the table from (b).

Apply the idea The last row of the table (with a closing balance of zero) was quarter 18, so it took Danita 18 quarters to pay off the loan. Divide the number of quarters by 4 to ﬁnd the time in years

c Calculate the quarterly simple interest rate charged on this loan.

Create a strategy Find the interest as a percentage of the principal

Apply the idea Write the expression for $750 as a percentage of $30 000

Evaluate

Idea summary We can use tables to ﬁnd the opening and closing balances for a loan or investment each time period. • The opening balance is the amount owed or in the account at the start of each time period. This is either the principal (for the ﬁrst row) or the closing balance of the previous row. • The closing balance is the amount owed or in the account at the end of the time period. The closing balance is calculated by adding the interest accrued and subtracting any repayments.

Interest on the total amount We are familiar with simple interest, where interest is calculated as a percentage of the principal. Under simple interest, the interest amount is always the same. A more common (but more complex) situation is where interest was calculated as a percentage of the total amount in the loan or investment each time period. This means that the interest can change each time period.

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Exploration Complete the table for a loan of $20 000, with monthly repayments of $1500. For each row, calculate the interest by ﬁnding 1% of the opening balance. Round values to two decimal places. Month 1 2 3 4 … 12 13 14 15

Opening Balance 20 000 18 700

Interest 200 187

Repayment 1500 1500

Closing Balance 18 700 17 387

… 4963.11

… 49.63

… 1500

… 3512.74

What happened to the interest when it was calculated as a percentage of the opening balance, instead of a percentage of the principal (simple interest)?

Was the total interest more or less than it would have been under simple interest?

If there were no repayments (if this was an investment rather than a loan), how would your answers to the above change?

Example 4 Saﬁa invests $15 000 in an account that pays interest of 0.5% of the total balance per month. a Construct a table showing the opening balance, interest and closing balance for the ﬁrst four months. Round values to two decimal places.

Create a strategy Month 1 will have an opening balance of $15 000. Each month the interest will be 0.5% of the opening balance. Add the interest to get the closing balance, then copy this into the next row as the opening balance.

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Apply the idea Month 1 Interest = 0.5% × 15 000

Multiply the opening balance by the interest rate

= 0.005 × 15 000

Convert the percentage to a decimal

= $75

Evaluate

Month 1 Closing Balance = 15 000 + 75 = $15 075 Month 2 Interest = 0.005 × 15 075 = $75.38 Month 2 Closing Balance = 15 075 + 75.38 = $15 150.38

Add the interest to the opening balance Evaluate Multiply the opening balance by the interest rate Evaluate Add the interest to the opening balance Evaluate

Continuing this process for the next two months gives: Month 1 2 3 4

Opening Balance 15 000 15 075 15 150.38 15 226.13

Interest 75 75.38 75.75 76.13

Closing Balance 15 075 15 150.38 15 226.13 15 302.26

b How much interest has Saﬁa earned on her investment in four months?

Create a strategy The closing balance is the total amount (principal plus interest from each month), so the total interest earned is the closing balance minus the principal.

Apply the idea I = 15 302.26 − 15 000 Write the subtraction = $302.26

Evaluate

c Compare your answer to part (b) with the interest that Saﬁa would have earned with 0.5% simple interest per month.

Create a strategy Use the simple interest formula.

Apply the idea I = PRT Write the formula = 15 000 × 0.5% × 4

Substitute the values

= 15 000 × 0.005 × 4 Convert the percentage to a decimal = $300

Evaluate

When the interest was calculated as a percentage of the total amount, Saﬁa earned $302.26 interest. This is $2.26 more than the simple interest where interest was calculated as a percentage of the principal only.

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Example 5 Emeric borrowed $340 000 to purchase a house. The table shows the opening balance, interest, repayment and closing balance for each month at the end of the loan term. Month 189 190 191 192

Opening Balance 9288.53 6349.26 3392.36 417.71

Interest 55.73 38.10 20.35 2.51

Repayment 2995 2995 2995 420.22

Closing Balance 6349.26 3392.36 417.71 0

a Determine the monthly interest rate charged on this loan. Write your answer as a percentage rounded to one decimal place.

Create a strategy Find the interest as a percentage of the opening balance

Apply the idea Write the expression for interest as a percentage of the opening balance for month 189 Evaluate and round to one decimal place

Reﬂect and check We could have used values from a different row of the table. Below is the calculation using the interest and opening balance of month 190:

b How many years did it take Emeric to pay off the loan?

Create a strategy Use the last row of the table.

Apply the idea The last row of the table (with a closing balance of zero) was month 192, so Emeric took 192 months to pay off the loan. = 16 years

300

Divide the number of months by 12 to ﬁnd the time in years

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c What was the total value of Emeric’s repayments?

Create a strategy

Apply the idea

Add all of the repayments from the table.

Emeric made 191 monthly repayments of $2995 then a ﬁnal repayment of $420.22, so the total amount repaid was 191 × 2995 + 420.22 = $572 465.22.

d Calculate the interest Emeric paid.

Create a strategy Subtract the principal from the total repayments.

Apply the idea I = 572 465.22 − 340 000

Subtract the principal from the total repayments

= $232 465.22

Evaluate

Idea summary Simple interest is calculated as a percentage of the principal, so the interest amount is always the same. Interest can also be calculated as a percentage of the total amount (opening balance) each time period, which means that the interest amount changes.

6.04 Practice questions Understanding

Explorer 1–2

Adventurer 1–2

Trailblazer 1–2

How is the closing balance of a loan calculated each time period?

True or false: a

The amount of simple interest each time period is always the same.

The repayment amount each time period is always the same.

The opening balance of an account is the closing balance of the previous time period.

Simple interest is when interest is calculated as a percentage of the total amount.

For an investment, calculating interest as a percentage of the total amount results in more interest accruing.

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Fluency 3 Ex 1

Ex 2

7 Ex 3

4, 6, 8, 10, 12

Vidas has $44 000 in a bank account that earns 2% simple interest per quarter. Construct a table showing the opening balance, interest and closing balance each quarter for the ﬁrst year.

How long does it take for the investment to grow to $46 640?

How much interest is earned in total over the year?

Lexie takes out a loan of $2500 at a simple interest rate of 3% per month and makes monthly repayments of $700. a

Create a table showing the opening balance, interest, repayment and closing balance for each month. Continue the table until the closing balance is zero.

How many months does it take Lexie to pay back the loan?

What was the total interest paid?

Abilio borrows $27 800 at a simple interest rate of 1.5% per quarter and makes quarterly repayments of $3200. a

Create a table showing the opening balance, interest, repayment and closing balance for each quarter of the ﬁrst year.

What was Abilio’s total repayments for the ﬁrst year?

Yousra takes out a home loan for $315 000 where simple interest is charged monthly. The last months of the loan balances are shown in the table. Month 177 178 179 180

302

3, 5, 7, 9–11

Lifeng invests $73 000 in an account that pays 0.1% simple interest per month. Construct a table showing the opening balance, interest and closing balance for the ﬁrst six months.

3, 5, 7, 9–11

Opening Balance 7000 5250 3500 1750

Interest 1575 1575 1575 1575

Repayment 3325 3325 3325 3325

How many years did it take Yousra to pay off the loan?

Calculate the simple interest rate charged on this loan.

What was the total amount repaid?

Find the total amount of interest Yousra paid.

Mathspace WA Curriculum – Year 9 mathspace.co

Closing Balance 5250 3500 1750 0

Elijah borrows $8000 where simple interest is charged fortnightly. The table shows the loan balances for the duration of the loan. Fortnight 1 2 3 4 5

9 Ex 4

11 Ex 5

Interest 80 80 80 80 80

Repayment 1700 1700 1700 1700

Calculate the ﬁnal repayment.

How many weeks did it take Elijah to pay off the loan?

What simple interest rate was charged on the loan?

Calculate the total amount that Elijah repaid.

How much interest did Elijah pay over the loan term?

Closing Balance 6380 4760 3140 1520 0

Page invests $9400 in an account that pays interest of 0.5% of the total balance per month. a

Complete the table showing the opening balance, interest and closing balance for the ﬁrst four months. Round values to two decimal places. Month 1 2 3 4

Opening Balance 8000 6380 4760 3140 1520

Opening Balance 9400 9447

Interest 47 47.24

Closing Balance 9447 9494.24

How much interest did Page earn on her investment in four months?

Compare your answer to part (b) with the interest that Saﬁa would have earned with 0.5% simple interest per month.

The opening balance of an investment is $8720 and the interest accrued that month is $34.88 a

What is the closing balance for the month?

What is the monthly rate of interest, as a percentage of the opening balance?

Susana takes out a car loan where interest is charged on the total amount each month. Three of the last four months of Susana’s repayments are shown in the table. Month 69 70 71

Opening Balance 868.48 622.82 375.93

Interest 4.34 3.11 1.88

Repayment 250 250 250

Closing Balance 622.82 375.93 127.81

Calculate the monthly interest rate charged on this loan rounded to one decimal place.

Complete the next row of the table for month 72, rounding your answers to two decimal places.

How many years did it take for her to pay off the loan?

Calculate her total repayments.

If her loan was for $15 000, calculate the total interest paid on the loan.

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Mr and Mrs Maximilian have a mortgage. Some of the ﬁnal months of their repayments are shown in the table. Month 146 147 148 149 150

Opening Balance 26 452.84 21 651.24 16 813.62 11 939.72 7029.27

Interest 198.40 162.38 126.10 89.55 52.72

Repayment 5000 5000 5000 5000 5000

Closing Balance 21 651.24 16 813.62 11 939.72 7029.27 2081.99

Calculate the monthly interest rate charged on this loan. Write your answer as a percentage rounded to two decimal places.

Complete the next row of the table for month 151, rounding your answers to two decimal places.

How long, in years and months, did it take for them to pay off the loan?

Calculate the total repayments.

If they paid $302 097.60 in interest, how much did they initially borrow?

Reasoning

13, 14

13–15

Explain why the ﬁnal repayment on a loan is sometimes less than the normal repayment amount.

Ibryam is considering two different options for an investment. • Option A pays simple interest at a rate of 9% p.a. • Option B pays interest on the total amount at a rate of 0.75% per month Which option would earn Ibryam the most interest? Justify your answer.

When paying off a loan, you can usually choose to make higher monthly repayments. Discuss the beneﬁts of making higher monthly repayments when being charged simple interest, and when interest is calculated as a percentage of the total amount.

Problem-solving 16, 17 16

304

16–18

Angela has $100 to invest. a

Calculate the interest she would earn each year for the ﬁrst four years at a simple interest rate of 10% p.a.

Find the interest she would earn each year for the ﬁrst four years if annual interest were calculated as 10% of the total amount in the account.

Graph your answers from (a) and (b) as points connected by lines on the same Cartesian plane where x is time in years and y is interest earned each year.

Comment on the differences between the graphs of the two types of interest.

Mathspace WA Curriculum – Year 9 mathspace.co

Teodor is taking out a loan of $3000 at a monthly interest rate of 0.5% of the total amount. a

Use tables to calculate the time it takes Teodor to repay the loan, and the total amount of interest paid, if he makes monthly repayments of: i

$600

$800

How much money does Teodor save by making monthly repayments of $800 instead of $600?

Ashly is considering two different options for a loan of $5600. She can afford to make monthly repayments of $1200. • Option A charges simple interest at a rate of 0.4% per month. • Option B charges interest on the total amount at a rate of 0.6% per month. a

Calculate the total interest paid with: i

Option A

Option B

On the same Cartesian plane, graph both Option A and Option B, using x as the time in months and y as the amount owing on the loan.

Which option should Ashly choose and why?

Investigation: Determining a suitable phone plan (Optional) Investigate online

mathspace.co

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6 Chapter review 1

Calculate the proﬁt (or loss) for each scenario: a

The revenue from selling handmade crafts is $7590 and the cost of materials and labor is $1500.

The cost of producing organic juices is $320 and the revenue from selling them is $1100.

The revenue from a tech workshop is $14 000 but the expenses for the venue, speakers, and materials amounted to $14 500.

The cost of setting up a pop-up store is $820 but the revenue from the sales was only $340.

Maria sells artisanal cheeses to local cafes for $50 per wheel. Each wheel of cheese costs her $30 to produce, and the cost to advertise and deliver the cheeses to cafes each week is $150. In a particular week, Maria sells 4 wheels. a

Determine the revenue for the week.

Determine the total cost for the week.

Does Maria make a proﬁt or a loss?

Calculate the proﬁt (or loss) that Maria makes that week.

Lila earned $38 500 in one year, and worked an average of 25 hours per week. a

What was her hourly rate of pay?

If her hourly rate of pay increased by $4, how much would her annual income increase by?

What is this increase as a percentage of her original income? Round your answer to two decimal places.

Mayca is a livestock sales agent and her weekly income ( y) is shown in the graph, where x is the total value of sales that she makes.

1600

Find Mayca’s weekly retainer (in dollars) and commission (as a percentage).

1200

a b c

Calculate Mayca’s annual income if she makes sales worth $10 000 each week. Mayca wants to make a fortnightly income of $3200. What is the total value of sales she must make each week?

1800

1400 1000 800

y = 0.05x + 600

600 400 200

x 0

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5000 10 000 15 000 20 000

Determine the simple interest earned on the following investments, rounded to two decimal places. a

$6200 is invested at 8% p.a. for 50 weeks.

$10 500 is invested at a rate of 6% p.a. for 15 months.

$8300 is invested at 1.5% per month for 3 years.

$5800 is invested at 0.4% per fortnight for 480 days.

Shin took out a loan of $8500 from a credit union at a simple interest rate of 5% p.a. for two years. a

How much interest will Shin need to pay after 2 years?

Calculate the total amount Shin will have to repay the credit union at the end of 2 years.

Juana invests $85 000 in a savings account which pays 0.15% simple interest per month. Construct a table showing the opening balance, interest, and closing balance for the ﬁrst six months.

Elise decides to invest her inheritance of $30 000 in a premium account which offers an interest of 2% of the total balance per month. a

Construct a table showing the opening balance, interest, and closing balance for the ﬁrst four months. Round values to two decimal places.

How much interest did Elise earn on her investment in the four months?

What monthly rate of simple interest would give the same interest as your answer to (b)?

Give your answer as a percentage rounded to two decimal places. 9

Carlos takes out a personal loan where interest is charged on the total amount each month. Three of the last four months of Carlos’s repayments are shown in the table. Month 57

Opening Balance 945.60

Interest 4.73

Repayment 280

Closing Balance 670.33

670.33

3.35

280

393.68

1.97

280

115.65

Calculate the monthly interest rate charged on this loan to one decimal place.

Complete the next row of the table for month 60, rounding your answers to two decimal places.

How many years did it take for him to pay off the loan?

Calculate his total repayments.

If his loan was for $14 000, calculate the total interest paid on the loan.

This graph shows the cost, the revenue and the profit from making and selling x units of a certain good. Each line has been labelled. a

What letter corresponds to the graph of: i

the profit?

iii

the revenue?

the cost?

How many units must be sold to reach the break-even point?

y (thousand $)

50 40 30 20 10 −10 −20

A B 20

x units 100

−30

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Marco is planning to start a coffee shop. He estimates his fixed costs to be $10 000 each month. The cost to produce each cup of coffee is expected to be $5. a

Write an equation for Marco’s monthly cost, C, if he sells x cups of coffee.

Determine the number of cups of coffee that will result in a monthly cost of $15 000.

How much revenue per cup is needed for Marco’s coffee shop to break even if the monthly cost is $15 000?

Aria is a freelance graphic designer and a part-time tutor. They are comparing their income from both roles. As a graphic designer, they earn $3150 per month, while tutoring pays her $750 per week. Aria multiplies their weekly income from tutoring by 4 to get a monthly income of $3000 and concludes that they earn more as a graphic designer. Is Aria correct? Justify your answer.

Leo has two consulting offers, both of which pay a base salary and a bonus for each project completed: • The Strategist package provides a monthly base salary of $1200 and a bonus of 15% for each project’s value. • The Executive package provides a monthly base salary of $1800 and a bonus of 8% for each project’s value. Leo believes he can complete projects worth $5000 each month in either role. Explain which offer he should choose.

Jaden borrows $1500 at a simple interest rate of 6% p.a. To calculate the interest he will owe over 18 months, Jaden uses the calculation I = 1500 × 0.06 × 18. Explain why Jaden’s calculation is not correct, and how he can ﬁnd the correct value.

Choi has $150 to invest.

Calculate the interest he would earn each year for the ﬁrst four years at a simple interest rate of 8% p.a.

Use a table to ﬁnd the interest he would earn each year for the ﬁrst four years, if annual interest were calculated as 8% of the total amount in the account.

Graph your answers from (a) and (b) as points connected by lines on the same Cartesian plane where x is time in years and y is interest earned each year.

Which of the two scenarios returned the larger investment and by how much?

Nutu is evaluating two different options for a loan of $6500. He plans to make monthly repayments of $1400. • Option 1 charges simple interest at a rate of 0.5% per month. • Option 2 charges interest on the total amount at a rate of 0.7% per month. a

Calculate the total interest paid with: i

308

Option 1

Option 2

On the same Cartesian plane, graph both Option A and Option B, using x as the time in months and y as the amount owing on the loan.

Which option should Nutu choose to minimise the amount of interest paid?

Mathspace WA Curriculum – Year 9 mathspace.co

“The individual investor should act consistently as an investor and not as a speculator. This means… he must be mathematically competent to recognize value.” Benjamin Graham

Big ideas Understanding similarity in right-angled triangles, scale factors, and transformations enables solving for missing measurements, making real-world comparisons, and estimating distances or dimensions using proportional relationships and scale models.

Curriculum Outcomes • Construct similar figures by enlargement and reduction and use this to establish, explain and apply properties of similar figures • Apply the properties of similarity to determine scales, lengths and angles of reallife figures from scale drawings, maps, plans and photographs • (Optional) Apply deductive reasoning and use a sequence of logically connected statements to produce proofs of congruent triangles • Explore to identify and describe conditions for triangles to be congruent. Use this to determine unknown sides or angles in pairs of congruent triangles and explain reasoning • Analyse, represent and interpret problems in real-world situations involving the perimeter and area of composite shapes and right-angled triangles, enlargement and reduction of similar figures, finding unknown side lengths and angles using trigonometric ratios, scale in similar figures and/or volume, capacity and surface area of right prisms and cylinders

7 Geometry Chapter outline 7.01 7.02 7.03 7.04

312 321 332 342

354

Bonsai trees are miniaturised versions of full-sized trees, just like scale models in mathematics!

7.01 Similar right-angled triangles After this lesson, you will be able to… • understand the conditions that make right-angled triangles similar. • identify similar right-angled triangles by comparing angles and side ratios. • identify corresponding sides and angles in similar right-angled triangles. • use ratios to confirm that two right-angled triangles are similar. • solve for missing side lengths or angles in similar right-angled triangles using ratios or Pythagoras’ theorem. • apply knowledge of similar right-angled triangles to real-world problems.

Similar right-angled triangles In similar triangles, all corresponding angles are equal and all corresponding sides are proportional, or in the same ratio. Here are two similar, right-angled triangles, △UST and △CAB. We can see that: • ∠U corresponds to ∠C • ∠S corresponds to ∠A • ∠T corresponds to ∠B • TU corresponds to BC • SU corresponds to AC • ST corresponds to AB

S 8 52°

38° 10

T A

52°

38° 25

Each pair of angles are equal: ∠U = ∠C = 90°

∠S = ∠A = 52°

∠T = ∠B = 38°

The ratios of the corresponding sides are equal:

The symbol for similarity is ‘∼’, so we would write △UST ∼ △CAB.

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Exploration These four triangles are similar:

45° 5

50 45°

Triangle A

Triangle B

Triangle C

Triangle D

How could we show that Triangle A and Triangle B are similar?

How could we show that Triangle A and Triangle D are similar?

How could we show that Triangle B and Triangle C are similar?

What information do we need to show that two right-angled triangles are similar?

For two right-angled triangles, they are similar if: 1. They have an acute angle in common, let’s call it θ - AA or AAA similarity. • This means that all angles are the same since they both have a right angle, have one acute angle in common, the other angle will be (180 − 90 − θ ) ° A S

θ U

UST

CAB

2. They have two or three pairs of corresponding sides that are in the same ratio - RHS (right angle- hypotenuse-side) or SAS (side-angle-side) similarity • If we know two pairs of sides are corresponding, then we can use Pythagoras’ theorem to show that the third pair is also in the same ratio.

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A S c ck

UST

CAB

If we can show two triangles are similar, then we can solve for missing sides and angles.

Example 1 For these two triangles:

4 2

60°

30°

a Are the triangles similar?

Create a strategy Determine whether all the corresponding angles are equal or whether all the corresponding sides are proportional.

Apply the idea

Reﬂect and check

From the diagram, we can see that the Since the hypotenuse of the larger triangle is 4 and corresponding angles in both triangles are equal. the smaller triangle is 2, the larger triangle will have By AAA similarity, these triangles are similar. all lengths double those of the smaller triangle. b Find x.

Create a strategy Since we know they are similar, we can set up a proportional ratio and solve for the unknown.

Apply the idea

Ratio of hypotenuses to sides opposite 30° angle

Simplify

Reﬂect and check The scale factor is 2, so we knew that this side length would be double the corresponding length on the smaller triangle.

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Example 2 A

These two triangles are similar:

1 Q

a°

45°

a Find the value of a.

P B

Create a strategy Since these two triangles are similar, this means that corresponding angles are equal. We may also use properties of right-angled triangles.

Apply the idea Since △ABC is a right-angled triangle, this means: ∠A + ∠B + 90° = 180° Sum of interior angles in a triangle ∠A + ∠B = 90°

Subtract 90° from both sides

∠A + 45° = 90°

Substitute ∠B = 45°

∠A = 45°

Subtract 45° from both sides

Since the triangles are similar, we know that corresponding angles are equal. a can correspond to ∠A or ∠B since they are equal. a = 45°

Reﬂect and check Another strategy is to use that △PQR is isosceles since it has two sides that have length 1. This means that ∠Q = ∠P. Since △PQR is a right-angled triangle, this means ∠Q + ∠P = 90° So ∠P = ∠Q = 45° a = 45° b Find the value of b.

Create a strategy Use the ratios of corresponding sides to set up an equation and solve for b.

Apply the idea The corresponding sides are: • AB and PQ • CA and RP • BC and QR

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Side BC corresponds to side QR.

Corresponding sides are proportional

Substitute known values

Take the reciprocal of both sides

Reﬂect and check Alternatively, we could use that they are both isosceles triangles to say that b = 4.

Idea summary For two right-angled triangles, they are similar if: • They have an acute angle in common - AA or AAA similarity as we already know they have a right angle in common. • They have two or three pairs of corresponding sides that are in the same ratio - RHS (right angle-hypotenuse-side) or SAS (side-angle-side) similarity If we can show two triangles are similar, then we can solve for missing sides and angles.

7.01 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

The two triangles shown are similar: a

Trailblazer

1–3 B

Identify the angle corresponding to: i

∠D

∠B

DC corresponds to which side in △RPQ?

CB corresponds to which side in △RPQ?

60°

iii

∠C

R 30° Q

In the diagram, △ECD ∼ △LMN. a

What must be true about each pair of corresponding angles?

What must be true about each pair of corresponding sides?

N C

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For two right-angled triangles △ABC and △XYZ, is each statement true or false? a

Since they are both right-angled triangles, they are similar.

If they have one acute angle in common, they are similar.

If they have the same hypotenuse length, they are similar.

If △ABC ∼ △XYZ, then

Fluency 4

4, 5, 6abcf, 7

If △ABC ∼ △XYZ, then

4, 5, 6acef, 7, 8

4, 5, 6c–f, 7, 8

For each set, identify which two triangles are similar. a

20 11

C 9

D 7 10

5 4

4 Ex 1a

Determine whether each pair of triangles is similar.

30°

45° 41.4°

48.6°

45°

30°

60°

2 2 30°

2 60°

4 3

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Ex 1b

Find the missing length y of each pair of similar right-angled triangles. a b 17.5 mm 7 cm

2 cm

32° 42 mm 12 mm

y mm

y cm

5 cm

58°

3 cm

14 m 7m

3 2 cm

30°

45°

9 2 cm

y cm

60° 3m

D 3m

11.4 m C 2m

ym A

E ym

3.7 m

7 Ex 2

Find the value of the variables in each pair of similar, right-angled triangles: a

s°

8 55°

r°

1.6

3 n

2 y° 10

y cm

16 cm

5 cm x

x cm 8 cm

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Find the value of the variables in each triangle. Express your answer in exact form. a

x cm

30° 30° 3 cm

y cm

45°

x cm

45° y cm

60°

60° 3 cm

Reasoning 9

9, 10

9–11

Is the information given enough to show

△ABD ∼ △DEF ? Justify your answer. a

∠B = ∠E

Explain how you know this pair of triangles is similar: 5

34° 9

22.45

27 56°

Explain why all the triangles are similar. J

H D x° A

Problem-solving 12, 13 12

12–14

13–15

Jenny wants to find the height of her school’s flag pole. During recess, she measures the length of the flag pole’s shadow to be 335 cm. Her friend then measures her own shadow which turns out to be 95 cm. The triangles formed by casting these shadows are similar. If Jenny is 160 cm tall, what is the height of the flag pole? Round your answer to the nearest whole centimetre.

160 95

335

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A ﬂagpole that stands 4.9 metres high casts a shadow of 4.5 metres. At the same time, the shadow of a nearby building falls at the same point S. The shadow cast by the building measures 13.5 metres. Find h, the height of the building.

4.9 m 4.5 m

13.5 m

The design for a slide at a water park is shown. Assume the two triangles created are similar. How tall is the slide? Round your answer to two decimal places.

2.0

1.2

Solve for the height of the roof. Round your answer to two decimal places.

8.2 m

5.1 m

9.66 m

320

.9 25

Mathspace WA Curriculum – Year 9 mathspace.co

7.02 Scale factors After this lesson, you will be able to… • understand and explain the concept of a scale factor. • calculate the scale factor between two similar figures. • use the scale factor to find unknown lengths. • solve problems using scale factors in practical situations.

Scale factor The scale factor is the ratio of corresponding sides in similar ﬁgures. An enlargement is an increase in size without changing the shape. This means a scale factor greater than 1.

A reduction is a decrease in size without changing the shape. This means a scale factor between 0 and 1.

Interactive exploration

Discover this concept in action online

mathspace.co

Corresponding angles of similar figures are equal. Corresponding sides of similar figures are in the same ratio. If we know the scale factor between two similar figures, then any length in one figure can be used to calculate the corresponding length in the other figure.

Example 1 These two shapes are similar.

61°

6 16

68°

7 104°

x° 8

127° 14

7.02 Scale factors mathspace.co

321

a What is the scale factor from the larger to the smaller quadrilateral?

Create a strategy Since this is a reduction, it should be between 0 and 1. We can divide the corresponding side lengths of the smaller shape by the larger shape.

Apply the idea We are told the shapes are similar, so only need to check one pair, but you can pick any pair. We’ll show all pairs from shortest to longest. Shortest sides =

= Second shortest sides =

= Second longest sides =

= Longest sides =

Smaller shape divided by larger shape Simplify Smaller shape divided by larger shape Simplify Smaller shape divided by larger shape Simplify Smaller shape divided by larger shape Simplify

The scale factor for the reduction is .

Reﬂect and check If any of these had been different, then the shapes would not be similar.

b What is the size of the angle marked x?

Create a strategy Identify the sides that correspond to the sides marked 6 and 7, and ﬁnd the measure of the angle between those sides.

Apply the idea The sides in the large shape that correspond to the sides of length 6 and 7 in the small shape are 12 and 14 in the large shape. The angle between them is 127°. x = 127

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Example 2 The triangles in this diagram are similar: A′

6 5 4 3 2 1

−6−5−4−3−2 −1 −1 −2 −3 −4 −5 −6

B′ B x 1 2 3 4 5 6 C

C′

What is the scale factor from △ABC to △A′B′C′?

Create a strategy Determine the ratio between two corresponding points’ distances from the origin.

Apply the idea Points A and A′ correspond to each other. Point A is 3 units away from the origin and point A′ is 6 units away from the origin. Divide 6 by 3

Evaluate

Reﬂect and check This is an enlargement, so the scale factor should be greater than 1.

Example 3 Consider the following triangles: z

Y C

30° 12

30°

85° A

65°

B Z 7.02 Scale factors mathspace.co

323

a Is △ABC similar to △XYZ?

Create a strategy To check if two triangles are similar, we can use the angle-angle-angle or side-angle-side similarity tests. We can’t use RHS since this is not a right-angled triangle.

Apply the idea The interior angles of ABC are C = 30°, A = 85° and B = 180 − 30 − 85 = 65°. The interior angles of XYZ are Z = 30°, Y = 65° and X = 180 − 30 − 65 = 85°. Since all corresponding angles are equal, we have that △ABC ∼ △XYZ.

b What is the scale factor from ABC to XYZ?

Create a strategy Set up a ratio using corresponding sides with the side of XYZ in the numerator and the side of ABC in the denominator.

Apply the idea

Ratio of known corresponding sides

Substitute known values

Simplify the scale factor

Reﬂect and check The scale factor should be greater than 1 since it is an enlargement.

c What is the value of z?

Create a strategy Set up a proportional equation using corresponding sides.

Apply the idea

Ratio of corresponding sides are equal

Substitute using labels on triangles

Simplify the scale factor

Multiply both sides by 6

Evaluate

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Reflect and check Since z is on the larger triangle, we should expect that it is longer than the corresponding side on the smaller triangle. Another way to do this would be to first find the scale factor, and then multiply the length of the corresponding side by it. z = AB × scale factor = 6 × 1.5 =9

Example 4 Adonis has a poster on A4 paper that he wants to enlarge to be on A0 paper. 297 mm 210 mm

126 mm

1189 mm h 841 mm

a What is the enlargement factor? Round your answer to the nearest whole number.

Create a strategy Divide one side of the large poster by the corresponding side in the small poster.

Apply the idea

Reﬂect and check

Dividing the horizontal length of the large poster by We would have achieved the same answer if we the horizontal length of the small poster, we get: had used the vertical lengths instead:

The enlargement factor is 4.

which rounds to 4.

b What is the height of the tallest mountain in the A0 size poster?

Create a strategy Multiply the height of the tallest mountain in the small poster by the enlargement factor from part (a).

Apply the idea 126 × 4 = 504 mm

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Idea summary The scale factor between two similar ﬁgures is the ratio of lengths between the larger and smaller ﬁgure. An enlargement will have a scale factor greater than 1. A reduction will have a scale factor between 0 and 1.

7.02 Practice questions Understanding 1

Explorer 1–4

Adventurer

Trailblazer

1–4

For the two similar triangles: a

Which side in △LMN corresponds to the side DE?

Which side in △LMN corresponds to the side DC?

C D

These two quadrilaterals are similar: a

Which side in the larger shape corresponds to side AB?

Which angle in the larger shape corresponds to ∠ADC?

T B

D C

326

Is each statement true or false? a

All squares are similar.

All rectangles are similar.

All isosceles triangles are similar.

All right-angled isosceles triangles are similar.

The sum of the interior angles of a triangle is 180°.

The sum of the interior angles of a quadrilateral is 360°.

For two similar shapes, the longest side in one shape must correspond to the longest side in the other shape.

If a shape is dilated by a scale factor of , will the new shape be larger or smaller than the original?

Mathspace WA Curriculum – Year 9 mathspace.co

5, 6aceg, 7a–c, 8ac, 9–11

Fluency Ex 1

5, 6bdfg, 7a–c, 8ac, 10, 11

5c–f, 6bdfh, 7b–d, 8bd, 10, 11

Find the value of the unknown angles for each of the similar ﬁgures: a

I G

b G

63°

28°

a° 10

b° H

H N

x° 9

31°

y°

64°

A 71°

112°

D B F

R Q 28

q°

p°

G k°

F j°

80°

N N

108° 150° 90°

6 E

P 6

117° 75°

44°

s° r°

M O

x°

E y°

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Find the value of the unknown sides for each of the similar ﬁgures: a

S 18

15 f

c d 12

T X

C 5

C A

j 33

S a V

f e P

U 22

E 10

R 24

S X

8 Z

f W

S R

328

18 B x

15 Q G 10

21 35

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10.5

C 12

Q 11

y S

Ex 2

Find the scale factor for each of the similar shapes: a

y 10 8 6 4 2

E B

−4 −3 −2 −1−1 A −2 −3 W −4

9 8 7 6 5 E 4 3 2 1

−8 −6 −4 −2 −2 −4 −6 −8 −10

x 2

C x 1

Z 4

For each of the similar triangles: i

Find the scale factor.

x cm

6 cm

Find the value of the variable.

a cm

4 cm

24 cm 28 cm

12 cm

21 cm

P 24

16 B

y 10 B 8 6 C 4 Y 2

A D

F −8 −6 −4 −2 −1

Ex 3

2 4 6 8 10

10 y 9 8 7 6 5 4 B 3 2 1

D A −10−8 −6 −4 −2 −2 −4 −6 −8 −10

8 C

8 R

m Q

Determine the scale factor between the two circles: 12

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A 4 in × 6 in (10 cm × 15 cm) picture of a sun is enlarged. a

What is the enlargement factor?

If the diameter of the sun in the 4 × 6 photo is 6.35 cm, what is the diameter of the sun in the 8 in × 12 in photo?

2.5 in

d 6 in

12 in

4 in 8 in

Alistair and Abigail are rearranging their room. They measured their room and are trying to make a scale model to see what might ﬁt in different places. 4m 1m 12 cm

closet

Ex 4

0.6 m

What is the scale factor from the room to the drawing?

Draw the scale model with all sides labelled.

Reasoning 12

13, 14

12–14

A square has side length 3 cm and a rectangle has dimensions 18 × 6 cm. Nila said that because all of the angles are equal, the AAA similarity tests means they are similar with a scale factor

= 3.

Is she correct? Justify your answer. 13

Brady took a picture of a lighthouse in the ratio 1 : 1, then posted it online. Who is right, Brady or Dorothy? Explain. 100 x 300 px

Brady Look how tall this lighthouse is Dorothy What did you do to the picture? It is not that tall in real life. Brady I just enlarge it by 3 times, I didn’t distort it or anything

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A ﬁgure was enlarged by a scale factor of 3. What scale factor could be used to get it back its original size? Explain.

Problem-solving 15–17 15

16–18

17–19

An equilateral triangle of side length 6 cm is to be enlarged by a factor of 5. a

What will be the side length of the resulting triangle?

What will be the size of each angle in the resulting triangle?

Mae wants to insert a picture into a document. She enlarges it by a factor of 12 but it becomes too blurry, so she reduces the resulting picture by a scale factor of 4. Find the overall scale factor from the original to the ﬁnal size.

A 1.1 m high fence casts a shadow of length 1.5 m. At the same time, a lamp post of height 3.9 metres casts a shadow of length L metres. Find the value of L, rounding to one decimal place.

A swimming pool of length 40 m and width 20 m is to be represented on a scale drawing with a scale of 1 : 2000. Find the scale dimensions of the pool in centimetres. a

Find the length in centimetres.

Find the width in centimetres.

Giuseppe took a picture of a tall waterfall, which is in the ratio 2 : 3 then he visits a photo centre that does prints from digital photos. This is their pricing guide: $0.10

$0.35

$2.50

4" × 6" (15 cm × 10 cm) Digital Photo Print

5" × 7" (12 cm × 17 cm) Digital Photo Print

8" × 10" (20 cm × 25 cm) Digital Photo Print

$3.50

$12

$13

8" × 12" (20 cm × 30 cm) 12" × 16" (30.5 cm × 40.6 cm) 12" × 18" (30.5 cm × 45.7 cm) Digital Photo Print Digital Photo Print Digital Photo Print

For each option, would the picture be distorted or cropped, or would it display correctly? i

4" × 6" (10 cm × 15 cm)

5" × 7" (13 cm × 18 cm)

iii

8" × 10" (20 cm × 25 cm)

8" × 12" (20 cm × 30 cm)

12" × 16" (30 cm × 41 cm)

12" × 18" (30 cm × 46 cm)

He bought a frame which holds 8" × 10" (20 cm × 25 cm) pictures. i

If he scales the picture down using the shorter side, what area of the picture will get cropped off?

If he doesn’t maintain the original ratio, and distorts the photo, will the waterfall look taller or wider than in reality?

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7.03 Transformations and similarity After this lesson, you will be able to… • define and explain the concept of dilations and how it results in similar shapes. • calculate the scale factor given a pair of similar shapes. • determine whether a given pair of shapes are similar by checking the ratio of corresponding sides and equality of corresponding angles.

Dilations A dilation will result in a shape that is enlarged or reduced, so all the side lengths will increase or decrease in the same ratio. The scale factor is the value by which we increase or decrease the side lengths of the shape. A scale factor of 3 means the new shape will have side lengths 3 times as big, whereas a scale factor of

means the new shape will be

as big as the original.

Here, the smaller triangle is dilated by a scale factor of 2 to produce the larger triangle.

If any of these ratios were not equal to 2, then this would not be a dilation.

In general, • If the scale factor is greater than 1, the dilated shape will be larger than the original. • If the scale factor is greater than 0 but less than 1, the dilated shape will be smaller than the original.

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Example 1 Determine whether the given diagrams show a dilation. If yes, ﬁnd the scale factor.

4 5

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Create a strategy To check if the diagrams represent a dilation, we need to check if the ratios of the corresponding sides of the two triangles are equal. If the ratios are equal, then the diagram shows a dilation.

Apply the idea We can then calculate the scale factor by ﬁnding the ratio of a side of the larger triangle to the corresponding side of the smaller triangle. The ratio of the sides of the larger triangle to the corresponding sides of the smaller triangle is:

So, the scale factor is 3.

Idea summary A dilation will result in a shape that is enlarged or reduced, so all the side lengths will increase or decrease in the same ratio. A scale factor is the value by which we increase or decrease the side lengths of the shape. • A scale factor greater than 1 is an enlargement. • A scale factor between 0 and 1 is a reduction.

Similarity While reﬂections, rotations and translations result in a shape congruent to the original shape, dilations will result in a shape that is similar to the original shape. Compare these triangles. What do you notice about the sides? What about the angles?

60°

60° 60°

60°

60° 2

You might have noticed that both triangles have the same angle measures, but they have different side lengths. Speciﬁcally, the smaller triangle has been dilated by a scale factor of 3. All of the corresponding sides between the two shapes are in the same ratio, or proportional, and all corresponding angles between the two shapes are equal. This means the two triangles are similar.

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Can we say that these two quadrilaterals are similar? 7

7 4

4 7

All corresponding angles between the two shapes are equal, but the corresponding sides are not proportional. If we look at the proportions for two pairs of sides, we have 7 and 8 . This is not equal. 4 7 The same scale factor has not been applied to each side of the smaller rectangle to create the larger one. Therefore, the two quadrilaterals are not similar. To prove that two figures are similar, one must be able to be transformed to the other using only rotations, translations, reflections, and dilations.

Example 2 Are these triangles similar?

B 24 26

A E 10

12 F 5

Triangle DEF is the image of triangle ABC.

Create a strategy Determine if the same scale factor has been applied to corresponding sides.

Apply the idea The hypotenuse is easy to identify in both triangles. Find the scale factor here: Test the scale factor on the other sides. AB = 24, so we need a side that has length of 12. AC = 10, so we need a corresponding side that has length of 5.

This confirms △ ABC is similar to △ DEF.

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Example 3 Are these triangles similar? 42

3 8

Create a strategy Identify whether or not all the corresponding sides are in the same ratio.

Apply the idea The first triangle has side lengths 48, 42 and 18. We’ve written the lengths in descending order. Let’s do the same for the other triangle: 8, 7, 3 If these triangles have a constant scale factor, then we can compare the scale factors for corresponding side lengths.

So, there is a common scale factor of 6, and the triangles are similar.

Idea summary Two shapes are similar when: • All corresponding sides are in the same ratio. • All corresponding angles are equal.

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7.03 Practice questions Understanding

Explorer 1–4

Adventurer

Trailblazer 2–5

1–4

What characteristics are required for two shapes to be similar?

What is the difference between congruent shapes and similar shapes?

Determine whether each statement is true or false.

Any dilation with a scale factor less than or equal to 1 is a reduction.

Any dilation with a scale factor greater than 1 is an enlargement.

After dilating a ﬁgure, all corresponding angles will be congruent.

All rectangles are similar.

All squares are similar.

All equilateral triangles are similar.

All isosceles triangles are similar.

Identify each transformation as a congruence transformation or a similarity transformation. a

A translation of 3 units left and 8 units down.

A reﬂection across the y-axis.

A dilation by a factor of 0.6.

A rotation of 90° anticlockwise.

Provide an example of a transformation that satisﬁes each statement: a

After the transformation, the side lengths of the transformed image are equal to the side lengths of the original object.

After the transformation, the angle measures of the transformed image are equal to the angle measures of the original object.

After the transformation, the side lengths and angle measures of the transformed image are equal to the side lengths and angle measures of the original object.

Fluency 6

6ac, 7, 8, 9bc, 10, 11, 12ab, 13, 14, 16, 17, 19

6ac, 7, 8, 9bc, 11, 12ac, 15, 16, 18, 20

Draw a labelled diagram of the figure after it has been dilated by the given factor: a

Scale factor of 3

Scale factor of 4

2 2

336

6bd, 7, 8, 9ad, 11, 12cd, 15, 16, 18, 20

Mathspace WA Curriculum – Year 9 mathspace.co

Scale factor of

8 14

8 6

Ex 1

Determine whether the diagram shows a dilation. If yes, ﬁnd the scale factor of the smaller shape to the larger one. a

5 15

6 18

3 10

For each diagram: i

Are the shapes congruent or similar?

Name the transformation that takes ﬁgure A to ﬁgure B.

A B

B A

A B

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Ex 2

For each pair of ﬁgures: i

Identify the scale factor.

List the transformations that would take the original ﬁgure to the image.

A 4 B B′ 2

A′

B′

3 6

C′

A D

D′

C′

B 3 C

B 24

3 9

D′

A′

A B′ 10

12 C′

C′

B′

A′

In the diagram, △ECD is similar to △LMN. a

Match the corresponding angles by ﬁlling in the blanks: ∠D corresponds to ∠⬚ ∠E corresponds to ∠⬚

What must be true about these pairs of angles?

Which side in △LMN corresponds to CD?

Which side in △LMN corresponds to CE?

C N

D Ex 3

Determine whether each pair of triangles is similar. If they are, find the scale factor of the larger shape to the smaller one. a

3 2 4

2 4 9

338

∠C corresponds to ∠⬚

Mathspace WA Curriculum – Year 9 mathspace.co

Find the scale factor for each pair of similar ﬁgures: a

A B

b B

A D

10 A′

B′

4 D′

C′

A′ B′ 25 B

D′

B′

C′ C

A′

C′

A′

DE corresponds to which side in △LMN ?

DC corresponds to which side in △LMN ?

C 5

Consider the two similar triangles: a

C′

M C

N L

Consider the two similar triangles: a

What is the corresponding angle in △RPQ for: i

∠D

∠B

CD corresponds to which side in △RPQ?

CB corresponds to which side in △RPQ?

iii

∠C Q

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The smaller quadrilateral has been reﬂected, then enlarged and ﬁnally rotated to form the larger quadrilateral: a

Are the two shapes congruent, similar, or neither?

Which side in the larger shape is corresponding to AB?

Which angle in the larger shape is corresponding to ∠ADC?

E F

The two given triangles are similar:

10 cm

Find the missing length y.

4 cm

y cm 7 cm

For each pair of similar triangles determine the value of x: a

b 26°

8 cm

30° 6 cm

x°

12 cm x° 14

9 cm

The two triangles are similar: a

Comparing the side lengths of the two triangles, ﬁnd the ratio of corresponding sides.

Find the value of each variable: i

iii

14° 5

m° n° 4

122°

44°

s° 14

The two given trapeziums are similar: a

What is the enlargement factor?

Calculate the size of x.

5 11

127°

x 127° 44

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In each diagram the two quadrilaterals are similar. a

If a = 6 cm, d = 11 cm and g = 5 cm, ﬁnd the value of k.

If r = 9 m, s = 17 m and c = 5 m, ﬁnd the value of d.

Reasoning

21, 22

21, 23

p d

Determine whether the two given shapes are similar. Explain your answer.

Explain how you know these pairs of ﬁgures are similar:

81 20 79°

s q

22–24

a b

b 1.15

60°

78°

2.3 30° 2

60° 2.30

4.6 30° 4

If two shapes are congruent, are they also similar? Explain your answer.

If two shapes are similar, are they also congruent? Explain your answer.

Problem-solving 25, 26

25, 26

25–27

If a square with a perimeter of 20 cm is dilated by a factor of 0.6, ﬁnd the side length of the dilated square.

If a square with an area of 25 m2 is dilated by a factor of 0.4, ﬁnd the side length of the dilated square.

A video phone call includes a full-screen image of the person being called and a smaller image of the caller in the corner of the screen (see image for reference). Consider making a video call on the smartphone which is 1782 pixels by 828 pixels in dimension. Suppose the smaller image is

th the size of

the entire screen. What would be the dimensions of the smaller image in pixels?

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Investigation: Determine congruence or similarity

Investigate online

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7.04 Optional: Congruent triangles After this lesson, you will be able to… • use the four tests to determine congruence and non-congruence between triangles.

Test for triangle congruence Recall that two triangles are congruent if one can be moved by translation, rotation, and/or reflection to lie exactly on top of the other. For example, if we reflected one of these triangles, rotated it, and translated it, we could place it directly on top of the other. We can see that three pairs of corresponding sides are equal, and three pairs of corresponding angles are equal. When deciding whether or not two triangles are congruent, we do not need to know the lengths of all the sides and the measures of all the angles. We need only three pairs of sides or angles in one of the following combinations. Side-side-side congruence (SSS) If two triangles have three equal side lengths, then the triangles must be congruent. Try this yourself with three straight objects — once you put them together, you can rotate, translate, and reflect the triangle to make every other possible combination:

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Since each triangle is made from the same three sides, they pass the side-side-side or SSS congruence test. Side-angle-side congruence (SAS) The side-angle-side (SAS) rule is a reliable method to determine if two triangles are congruent. This rule requires two triangles to have one pair of corresponding sides that are equal in length, and the angles enclosed by these sides must also be identical. When these conditions are met, the second pair of corresponding sides is guaranteed to be equal, confirming triangle congruence.

In the diagrams, you can see two triangles with these congruence conditions. Notice how each triangle has: • one side marked with a single hash, indicating these are equal in length • another side marked with a double hash, indicating these are equal in length • an angle between them marked with an arc, indicating these are equal in length The congruence of these elements according to the SAS rule ensures that the entire triangles are congruent, meaning every corresponding side and angle in one triangle exactly matches the corresponding side and angle in the other triangle.

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There is no side-side-angle (SSA) congruence. Right angle-hypotenuse-side congruence (RHS) If two right-angled triangles have hypotenuses with equal lengths and another pair of equal sides, then the triangles must be congruent.

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Notice that the right angle is not included between the sides — this is an exception to the side-side-angle rule, but we do not call it SSA. Instead, this congruence test is called right angle-hypotenuse-side, or RHS. Angle-angle-side (AAS) and Angle-side-angle (ASA) congruence If you know two angles, then by the fact that the sum of angles in a triangle is 180°, then we know all three angles. The AAS and ASA congruence use the fact that the angles are equal, and that we have one pair of matching sides. ASA - is two angles and the included side between them. AAS - is two angles and a non-included side.

In summary, if two triangles do not satisfy these tests, one of two things could be true: • The two triangles are deﬁnitely not congruent, or • We don’t have enough information to know whether or not they’re congruent Two triangles are deﬁnitely not congruent if there is a pair of sides or angles that are in the same relative position but are not equal to each other.

Example 1 Consider the triangles: a Are these triangles congruent? 34° 9

56°

Create a strategy We can use the fact that the sum of the measure of the interior angles of a triangle is 180° to find the other angle of Triangle B. Then, use the appropriate congruence test to check whether they are congruent or not.

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Apply the idea We can find the unlabelled angle of Triangle B by using 180° − (56° + 34°). This means that the unlabelled angled is 90°. This is the same as the angle from Triangle A. Because both triangles are right angle, we can try the RHS congruence for checking. We see that both the hypotenuses measures 9 and both legs measures 5, and so the triangles must be congruent.

b What congruence test does this pair satisfy? A

SSS

SAS

AAS

RHS

Create a strategy Choose the congruence test used from part (a).

Apply the idea We found from part (a) that two triangles are congruent by right angle-hypotenuse-side or RHS. The answer is option D.

Example 2 D

Consider the diagram:

a Are △ABD and △CDB deﬁnitely congruent?

Create a strategy Take note of the markings used for congruence and determine if any of the congruence tests are satisﬁed.

Apply the idea Looking at △ABD and △CDB in the diagram, we have one pair of common sides, which tells that those two sides are equal, BD = DB. Based on the given markings in the diagram, we also have sides CD = AB. The parallel line marks on sides CD and AB also give us a pair of equal angles which are angles ∠ABD and ∠CDB, since they are alternate interior angles on parallel lines. So, we have enough information to deﬁnitely say that △ABD and △CDB are equal. So, △ABD and △CDB are definitely congruent.

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b What congruence test does this pair satisfy? A

SSS

SAS

AAS

RHS

Create a strategy Take note of what the triangles share in common.

Apply the idea △ ABD and △CDB have a common side DB = BD, another pair of equal sides CD = AB, and a pair of equal angles ∠CDB = ∠ABD which are between the equal sides. This gives us side-angle-side congruence or SAS. Option B is the correct answer. c Select the three statements that, when put together, establish congruence for this test. Make sure each reason is correct as well. A

∠ADB = ∠CBD (Alternate angles on parallel lines)

AD = CB (Corresponding sides on parallel lines)

∠ABD = ∠CDB (Corresponding angles on parallel lines)

AB = CD Given

BD is common

∠ABD = ∠CDB (Alternate angles on parallel lines)

Apply the idea We have previously established that: • BD is a common side, • AB = CD from the same markings on the diagram, and • ∠ABD = ∠CDB since they are alternate interior angles on parallel lines. So, the three correct statements are options D, E, and F.

Idea summary These are the four congruence tests that can be used to prove two triangles are congruent: • Side-side-side, or SSS: The triangles have three pairs of equal sides. • Side-angle-side, or SAS: The triangles have two pairs of equal sides with an equal included angle. • Right angle-hypotenuse-side, or RHS: Both triangles have a right angle, equal hypotenuses, and equal sides. • Angle-angle-side, or AAS: The triangles have two pairs of equal angles and one pair of equal sides.

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7.04 Practice questions Explorer

Understanding

Adventurer

1–3

Trailblazer

1–3

If two triangles are congruent, what do we know about their corresponding sides and angles?

Decide whether these conditions can be used to show two triangles are congruent:

Three congruent sides

Three congruent angles

Two congruent sides and a non-included angle

Two congruent legs of a right triangle

Name the triangle congruence test described by each statement. a

Two consecutive angles and a side length are equal.

Corresponding side lengths are equal.

Two pairs of corresponding angles and an included side are equal.

Have right angles and equal length of hypotenuse and another side.

Two corresponding sides and included angles are equal.

Fluency 4

4–10, 13, 14, 18

4, 6, 8, 10, 14, 16, 18

4, 6, 10–12, 14–19

Are the two triangles congruent? If yes, name the congruence test they satisfy. a

R S

4 R

63°

28°

63°

28°

84°

4 R

U B

84°

U 4 T

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I G

63°

28°

63°

28°

8 U

5 4

B F

Triangle B Triangle A

Triangle C

7 31° 73°

Ex 1

31°

73°

Which two triangles are congruent?

Which congruence test do the triangles satisfy?

Triangle B Triangle A Triangle C 9

9 6 6

Which two triangles are congruent?

Which congruence test is satisﬁed?

Triangle A Triangle B 4

8 9

Which two triangles are congruent?

Which congruence test is satisﬁed?

Mathspace WA Curriculum – Year 9 mathspace.co

Triangle C 4

348

73°

31°

Triangle B Triangle A 85°

Triangle C 7

31° 7

85° 31°

Which two triangles are congruent?

Which congruence test is satisﬁed?

85° 31°

Triangle A Triangle B

Triangle C

5 4

5 27° 4

27°

Which two triangles are congruent?

Which congruence test is satisﬁed?

Triangle A

Triangle B

Triangle C

5 5

Which two triangles are congruent?

Which congruence test is satisﬁed?

Triangle A

41°

T riangle B

41°

Triangle C Triangle D

Which two triangles are congruent?

Which congruence test is satisﬁed?

Triangle A Triangle B

4 41°

7 41°

81° 42° 7

42° 81°

Which two triangles are congruent?

Which congruence tests are satisﬁed?

Triangle C

81°

Triangle D

7 57°

57° 81° 7

7.04 Optional: Congruent triangles mathspace.co

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Which congruence test do △ABD and △CDB satisfy?

Which two triangles are congruent?

Which congruence test does this pair satisfy?

8 6 3

E 5

6 D

KM and LN are straight line segments: a

Which two triangles are congruent?

Which congruence test does this pair satisfy?

Which two triangles are congruent?

Which congruence test does this pair satisfy?

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Which congruence test do △EFG and △EHD satisfy?

Name the sides and angles of the triangles that satisfy this congruence test.

Ex 2

hich congruence test do △PQS and △RSQ W satisfy?

Name the sides and angles of the triangles that satisfy this congruence test.

hich congruence test do △ABX and △DCX W satisfy?

Name the sides and angles of the triangles that satisfy this congruence test.

R C A

Reasoning

20, 22

21, 22

22, 23

If we can use angle-side-angle (ASA) to show two triangles are congruent, explain why the triangles will also pass the angle-angle-side (AAS) congruence test.

Explain why angle-angle-angle (AAA) is not a valid triangle congruence test. Draw a diagram to support your answer.

Draw two line segments that bisect each other. Draw two new lines that connect the endpoints of the original segments. Will a pair of congruent triangles always be formed? Explain.

If we have △ABC and △PQR, where: • ∠A = ∠P • AB = PQ • BC = QR Is there enough information to know for sure that the two triangles are congruent? Draw a diagram to support your answer. 7.04 Optional: Congruent triangles mathspace.co

351

Problem-solving 25, 26 24

24, 25, 27, 28

24, 26, 28

△PQR and △PSR are shown in the diagram. a

What additional information is needed so that the triangles are congruent by RHS?

What additional information is needed so that the triangles are congruent by SSS?

What additional information is needed so that the triangles are congruent by SAS?

△JKN and △MLN are shown in the diagram.

R J

What additional information is needed so that the triangles are congruent by SSS?

What additional information is needed so that the triangles are congruent by SAS?

What additional information is needed so that the triangles are congruent by AAS?

Consider △PQX and △RSX where X is the centre of the circle. a

Explain how we know sides QX, XP, SX, and XR all have the same length.

What additional information is needed so that the triangles are congruent by SSS?

What additional information is needed so that the triangles are congruent by SAS?

S P

X is the centre of both circles. a

Which congruence test do triangles △ABX and △DCX satisfy?

Name the sides and angles of the triangles that satisfy this congruence test. Justify your reasoning.

X C

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For each diagram: i

Name the congruence test that the two triangles satisfy.

Solve for x.

35 − 6x

6x°

4x + 5

(5x − 7)°

(3x + 15)°

(8x − 18)°

d (3x + 5)°

(7x − 23)°

Investigation: Guessing distance

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Investigation: Scale diagrams

Investigate online

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Investigation: Scale models

Investigate online

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7 Chapter review 1

Determine whether each pair of triangles is similar. a

60°

30° 60°

30°

7.5

4 3

Find the missing length y of each pair of similar, right-angled triangles: a

13 cm

6 mm

y mm

12 cm

21 mm

18 mm y cm

18 cm

Find the value of the variables in each pair of similar, right-angled triangles: a

s° 45 7 62°

r°

b a°

10.5 cm

b cm

3 cm 65°

7 cm

A person that is 1.6 m tall casts a shadow of 1.8 metres. At the same time, the shadow of a nearby tree falls at the same point S. The shadow cast by the building measures 5.4 metres. a

Identify two similar triangles. Explain how you know they are similar.

Find the height of the tree.

1.6 m S 1.8 m 5.4 m

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The design for a ramp at a skateboard park is shown. The two triangles created are similar. How tall is the ramp?

pm 2.25 m 1.5 m 18 m

For each of the similar triangles: i

Find the scale factor.

30 cm

Find the value of z.

z cm

35 mm 21 mm

30 cm

25 cm

10 mm

z mm

Find the value of the variables in each of the similar figures: a

b 11

g°

a° 117°

83°

b° f° 29°

Find the value of the variables in each of the similar figures: a

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A rectangle has the dimensions 4 × 7. Jay dilates the rectangle vertically by a scale factor of 2 to get a rectangle with dimensions 8 × 7. Jay said that because he dilated the first rectangle and all angles are equal, the rectangles are similar with scale factor

4 7

Is Jay correct? Justify your answer. 10

Lyor wants to print a photograph for a frame. He magnifies it by a factor of 8 but it doesn’t fit the frame, so he shrinks the enlarged photo by a scale factor of 2. Determine the overall scale factor from the original to the final size.

A football ﬁeld of length 105 m and width 68 m is to be depicted on a scale diagram with a scale of 1 : 500.

Find the length on the scale diagram in centimetres.

Find the width on the scale diagram in centimetres.

The dimensions of the triangle shown are to be enlarged using a scale factor of 2.

Find the enlarged length of: a

20 m

26 m B 23 m C

Consider the quadrilaterals: a b

′ ′ ′

′

Is quadrilateral A B C D an enlargement of quadrilateral ABCD?

Find the scale factor.

5 D

C’ 12

B’

12 A’

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D’

Triangle ABC has been reduced to form a smaller triangle A′B′C′. Find the scale factor.

B 6 cm B’

A C

6 cm

A’

7 cm

C’ 7 cm

The two quadrilaterals in the diagram are similar. Find the value of b.

b 8

(Optional) Consider the given diagram:

What information is needed so that △PQR and △PSR are congruent?

(Optional) Consider the given diagram:

R J

What information is needed so that △JKN and △MLN are congruent?

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(Optional) Consider the given diagram: a

Which congruence test do triangles △ABD and △CDB satisfy?

What are the three ways in which the triangles satisfy this congruence test?

Triangle B

Triangle C

Triangle D 4 9

7 5

(Optional) Consider the triangles: Triangle A

Identify the two congruent triangles.

State the congruence test that this pair satisﬁes.

8 9

Did you know?

Fashion designers use transformations to flip, rotate and scale clothing patterns—ensuring perfect fits on different body types. They often begin with a basic pattern and use geometric transformations to adapt it into different sizes or styles. Reflection, or flipping, allows them to create symmetrical designs by mirroring one side across a fold line—this is common for elements like sleeves, pant legs, or front and back panels. Rotation helps them correctly position curved seams and darts, or reorient pattern pieces on the fabric for cutting. Dilation, or scaling, is especially important when adjusting patterns for different body sizes while maintaining the same proportions. This real-world use of transformations shows how geometry plays a key role not only in engineering, but also in creative industries like fashion, costume design, and tailoring.

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“Geometry is a language that gives us a way to describe the shape of space and the nature of the universe.” Marcus du Sautoy

Big ideas Pythagoras’ theorem and trigonometric ratios provide the tools for calculating unknown sides and angles with practical applications in the real world.

Curriculum Outcomes • Use Pythagoras’ theorem to determine the perimeter and area of shapes involving rightangled triangles, in both exact and decimal approximation form. Investigate and apply the converse of Pythagoras’ theorem to establish whether a triangle is right-angled • (Optional) Explore and apply Pythagoras’ theorem and trigonometry to simple situations involving right-angled triangles in three-dimensional contexts projected to twodimensions • Use similarity to investigate and explain the constancy of the sine, cosine and tangent ratios for a given angle in right-angled triangles. Choose and use a trigonometric ratio to determine the length of an unknown side or the size of an unknown angle • (Optional) Explore the relationship between sine and cosine ratios and the unit circle, determine their approximate values for angles from 0° to 360°, and identify pairs of angles that share the same ratio value • Analyse, represent and interpret problems in real-world situations involving the perimeter and area of composite shapes and right-angled triangles, enlargement and reduction of similar figures, finding unknown side lengths and angles using trigonometric ratios, scale in similar figures and/or volume, capacity and surface area of right prisms and cylinders

8 Trigonometry Chapter outline 8.01 8.02 8.03 8.04 8.05 8.06 8.07 8.08

362 372 385 396 406 413 421 429 448

The Eiffel Tower in Paris relies on trigonometric principles in its design to make it strong!

8.01

Pythagoras’ theorem After this lesson, you will be able to… • apply Pythagoras’ theorem to find the length of the hypotenuse in a right-angled triangle. • use Pythagoras’ theorem to find the lengths of the other sides in a right-angled triangle. • explore the derivation of Pythagoras’ theorem using squares and triangles. • determine whether a triangle is right-angled using the converse of Pythagoras’ theorem. • use the theorem to solve real-world problems involving right-angled triangles.

Pythagoras’ theorem Exploration For the squares shown in the diagram: b

What is the area of the inside square, where length is c?

What is the area of the four triangles, where height is a and base is b?

What is the area of the outside square, where length is (a + b) ?

Equate the area of the inside square plus the four triangles to the area of the outside square. Can you derive Pythagoras’ theorem?

The hypotenuse is labelled c and the two smaller sides are a and b. a2 + b2 = c2

Pythagoras’ theorem states that in a right-angled triangle, the sum of the squares of the lengths of the two shorter sides is equal to the square of the length of the hypotenuse.

We can use this theorem to find unknown side lengths in right-angled triangles.

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Example 1 Find the length of the unknown side c in the triangle.

Create a strategy

Apply the idea 2

Use Pythagoras’ theorem: c = a + b .

Write the formula

Substitute a = 40 and b = 9

Evaluate the squares

Add the terms

Take the square root of both sides

Evaluate the square root

Example 2 Find the length of the unknown side x in the triangle. 15

Create a strategy Use Pythagoras’ theorem: a2 + b2 = c2.

Apply the idea We can substitute the following a = x, b = 12, and c = 15 into the formula: Write the formula

Substitute the values

Evaluate the indices

Evaluate the difference

Take the square root of both sides

Evaluate the square root

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Idea summary

a2 + b2 = c2

Pythagoras’ theorem relates the three sides of a right-angled triangle. a and b are the two shorter sides, and c is the longest side, called the hypotenuse.

Converse of Pythagoras’ theorem The converse of Pythagoras’ theorem states that if the square of the longest side of a triangle equals the sum of the squares of the other two sides, then the triangle is right-angled.

Exploration Consider a triangle with side lengths 5, 12, and 13. Test whether it is right-angled: 1. Identify the longest side and label it c. 2. Compute c2 and a2 + b2. 3. If they are equal, what can you conclude about the triangle?

Example 3 A triangle has legs of length 8 cm and 15 cm. a Determine if the triangle is right-angled, given a third side of length 19 cm.

Create a strategy Use the converse of Pythagoras’ theorem: check if the square of the longest side equals the sum of the squares of the other two sides. If true, the triangle is right-angled.

Apply the idea Let c = 19, a = 8, and b = 15: c2 = a2 + b2

Write the formula

19 = 8 + 15

Substitute the values

361 = 64 + 225

Evaluate the squares

361 ≠ 289

Evaluate the addition

Since 361 ≠ 289, the triangle is not right-angled.

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b Find the perimeter of the right-angled triangle.

Create a strategy Use Pythagoras’ theorem to conﬁrm the hypotenuse, then add all three sides to calculate the perimeter.

Apply the idea Let a = 8 cm and b = 15 cm. Using Pythagoras’ theorem: c2 = a2 + b2

Write the formula

c = 8 + 15

Substitute a = 8 and b = 15

c = 64 + 225

Evaluate the squares

c = 289

Evaluate the addition

Take the square root of both sides

c = 17

Simplify

Therefore, the perimeter of the right-angled triangle is 8 + 15 + 17 = 40 cm. c Find the area of the right-angled triangle.

Create a strategy Use the formula for the area of a triangle: A =

× base × height

Apply the idea Let the base = 8 cm and the height = 15 cm: A right triangle image with measurements base = 8 cm, height = 15 cm and a hypotenuse of 17 cm Area =

× base × height

Write the formula

× 8 × 15

Substitute the values

Evaluate = 60 cm2

Idea summary The converse of Pythagoras’ theorem states that if c2 = a2 + b2 for side lengths a, b, and c (where c is the longest), the triangle is right-angled.

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8.01 Practice questions Understanding 1

Explorer 1–4

Adventurer

Trailblazer

1–4

Are these statements true or false? a

Pythagoras’ theorem can only be applied to a right-angled (90°) triangle.

The side directly across from the right angle is known as the opposite.

To ﬁnd the unknown length, we must know the length of the other two sides.

The hypotenuse is the longest side.

For this right-angled triangle, write the equation that can be used to ﬁnd x. 19

Using the diagram: a

Three different right-angled triangles

The hypotenuse of △ABC

iii

The shortest side of △ABC

The perpendicular height of △ABC from B

The hypotenuse of △ABD

Can we ﬁnd DC given: i

AD and BC?

AD and BD?

iii

BD and BC?

AB, AD, and BC?

Determine whether each set of side lengths could belong to a right-angled triangle: a

366

Identify:

3, 4, 5

5, 12, 13

Mathspace WA Curriculum – Year 9 mathspace.co

7, 8, 10

9, 12, 16

5ac, 6aceg, 7ac, 8ace, 9, 10, 12

Fluency 5 Ex 1

5ac, 6aceg, 7ac, 8ace, 10–12

5bd, 6bdfh, 7bd, 8bdf, 10–13

For each triangle, ﬁnd the length of the unknown side c. a

21 m

c cm

20 m

48 cm

14 cm

For each triangle, ﬁnd the length of the hypotenuse. Give your answer as an exact value. a

17 cm

c cm

9 cm

3 cm

c cm

d 4 c 21

The two smallest side lengths of a right-angled triangle are 6 and 9.

The two smallest side lengths of a right-angled triangle are 14 and 17.

The two smallest side lengths of a right-angled triangle are 6 and 7.

The two smallest side lengths of a right-angled triangle are 17 and 28.

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7 Ex 2

For each triangle, ﬁnd the length of the unknown side: a

y 12

29 m

d 82 m

80 m

20 m

For each triangle, ﬁnd the length of the unknown side as an exact value. a

14 m

b cm

17 m

15 cm

23 cm

d 42

31 m

A right-angled triangle q, has a hypotenuse of length 27 and another side of length 9.

A right-angled triangle r, has a hypotenuse of length 39 and another side of length 13.

Violet parks her car in a parking lot and walks 6 km north to get to a river. She walks 8 km east along the riverbank. If she is now h km directly from her car, ﬁnd the value of h.

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8 km 6 km

h km

TVs are measured in inches along their diagonal. A 65 in TV has dimensions of 144 cm by 81 cm. Use this to ﬁnd the value for 65 inches to the nearest cm.

144 cm 81 cm

65 in

In designing a diamond-shaped picture frame, the vertical length is 48 cm and the horizontal length is 14 cm. How long will each diagonal side of the diamond be? Round your answer to two decimal places.

48 cm

14 cm

A triangle has legs of length 6 cm and 12 cm: a

Determine if the triangle is right-angled, given a third side of length 14 cm.

Find the perimeter of the right-angled triangle, rounded to one decimal place.

Find the area of the right-angled triangle.

Yasmina is standing in the northwest corner of a rectangular field. A walking path runs along the diagonal of the field. If she walks 250 m along the diagonal path, then walks due west 240 m, she will reach the southwest corner of the field. If the distance between the northwest and southwest corners of the field is x m, determine the value of x.

Reasoning

14–17

14–18

A triangle is formed by segments of lengths 10 cm, 24 cm, and 26 cm. Demonstrate using the converse of the Pythagoras’ theorem whether or not this is a right triangle.

Ella tried to solved for a on the Pythagoras’ formula c2 = a2 + b2. Her working is shown: c2 = a2 + b2 c 2 – b 2 = a2

c2 – b2 = a2 c–b=a

What was her mistake? Justify your answer.

Explain why she did not include the negative square root in step 3.

Solve the equation correctly.

Using Pythagoras’ theorem: a

Can a triangle with sides of length 5 cm, 7 cm and 9 cm be a right-angled triangle? Justify your answer.

Can a triangle with sides of length 7.5 cm, 10 cm and 12.5 cm be a right-angled triangle? Justify your answer. 8.01 Pythagoras’ theorem mathspace.co

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Given a rectangle with side lengths of 16 cm and 32 cm, what is the length of the diagonal rounded to two decimal places? Explain why the two diagonals in a rectangle are equal.

A triangle with sides 3, 4, and 5 units is a right-angled triangle. If the triangle is enlarged by a factor of 2, does Pythagoras’ theorem still hold? Justify your answer.

Problem-solving 19, 20, 24, 25 19

20, 21, 23–25

21–27

A surveyor measures three sides of a piece of land in the shape of a quadrilateral, as shown in the diagram. a

Find the value of d.

Find the area of the land.

If one acre is 4046.86 m2, how many acres is this plot of land? Round your answer to one decimal place.

111 m 167 m 119 m

Using the given diagram, ﬁnd the value of x rounded to one decimal place. 6.4 m

xm 3m 3.9 m

AB is perpendicular to the radius of the circle with centre O.

21.2 cm

Using the diagram, ﬁnd:

The radius of the circle, rounded to one decimal place.

The length of BC, rounded to 1 decimal place.

A farmer wants to build a fence around the entire perimeter of his land, as shown in the diagram. The fencing costs $37 per metre. a

Find the length x, rounded to two decimal places.

Find the length y, rounded to two decimal places.

How many metres of fencing does the farmer require, if fencing is sold by the metre?

370

At $37 per metre of fencing, how much will it cost him to build the fence along the entire perimeter of the land?

Mathspace WA Curriculum – Year 9 mathspace.co

24 cm

A 1m F x

5m 7m B

E2 m

G y

In the diagram, △ABE ∼ △ACD, with BE = 3.2, DE = 9 and CD = 8. a

Find the length of EA.

Find x, the length of AB.

A x E

3.2

9 D

A chocolate box in the shape of a triangular prism has a base of width 6 cm, a height of 8 cm and a length of 16 cm. a

Find the length x.

Every edge of the chocolate box is to be sealed with special gold strips. What length of gold strip is required?

A yard has a rectangular patio that measures 4 m by 7 m. A path runs from one corner diagonally to the opposite corner, with a measured length of 8 m. Is this path truly the direct corner-to-corner diagonal? Explain using the converse of the Pythagoras’ theorem.

8 cm

x 16 cm 6 cm

4m 8m

A ramp is being built in a shopping centre to make it more accessible. The building is 68 cm off the ground and the ramp is to start 1.2 m away from the doorway. Calculate the length of the ramp, rounded to the nearest mm.

Jesmond and Bilal are meeting each other for a mountain top picnic. Bilal took a wrong turn and is walking on a trail that is parallel to the trail that Jesmond is on. Bilal is at a height of 1870 m and Jesmond is at a height of 2340 m. The horizontal distance between them is 1080 m. a

What is the direct distance between the two friends? Round your answer to the nearest metre.

Going off trail, how long will it take for Bilal to walk directly to Jesmond at a speed of 1.42 m/s? Round your answer to the nearest minute.

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Investigation: Ratios in triangles

Investigate online

8.02

mathspace.co

Trigonometric ratios

After this lesson, you will be able to… • identify the hypotenuse, opposite side, and adjacent side in a right-angled triangle. • define and use the trigonometric ratios sine, cosine, and tangent. • express the trigonometric ratios as fractions of side lengths. • solve problems involving trigonometric ratios.

Ratios in right-angled triangles Hypotenuse

Opposite

θ Adjacent

AC is the hypotenuse in this right-angled triangle as it is opposite the 90° angle. BC is opposite the angle θ, so we can refer to it as the opposite side. AB is next to the angle θ, so we can refer to it as the adjacent side.

Interactive exploration

Discover this concept in action online

mathspace.co

For similar right-angled triangles (triangles which have the same angles), the ratio of any pair of sides will always be the same. Trigonometry has special ratios of side lengths in right-angled triangles. These are the ratios we will explore:

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Example 1 For this right-angled triangle:

θ B

a Which side is the hypotenuse?

Create a strategy The hypotenuse is the side opposite the right angle.

Apply the idea

Reﬂect and check

The hypotenuse is side AB.

We will often label sides that are opposite a vertex with the same letter as the label of the vertex but lower case, so we can also call this side c.

b For the angle θ, which side is the opposite?

Create a strategy The opposite side across from the angle.

Apply the idea

Reﬂect and check A

Notice that the opposite side is the one side that does not have an endpoint at the vertex where the angle is located.

α C

θ B

Side AC is opposite the angle θ.

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c For angle α, which side is the opposite?

Create a strategy Starting from angle α we look across the triangle.

Apply the idea

Reﬂect and check

Side BC is opposite angle α.

Notice that in the same triangle, we can have different sides labelled as the opposite side depending on which angle we are starting from.

d For angle α, which side is the adjacent side?

Create a strategy The adjacent side is one of the sides next to the angle, but is not the hypotenuse.

Apply the idea

Reﬂect and check

Side AC is adjacent to angle α.

We can check that this side is not the hypotenuse and not the opposite side.

Example 2 For the angle θ, what is the value of the ratio

A 15 17

Create a strategy First we can identify the opposite and adjacent sides from angle θ, then we can ﬁnd their ratio.

Apply the idea Starting from angle θ, side AC is the opposite and side BC is the adjacent side.

A Opposite 15 17 Hypotenuse

Now we can ﬁnd the ratio: C

8 Adjacent

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We cannot simplify this any further.

Reﬂect and check It can help to label the hypotenuse as well to check that we have labelled the other sides correctly.

Idea summary

Opposite

C Hypotenuse

θ A

Adjacent

AC is opposite the right angle, so we can call it the hypotenuse. AB is next to the angle θ, and is not the hypotenuse, so we call it the adjacent side. BC is opposite the angle θ, so we can call it the opposite side.

For any triangle, we can ﬁnd these ratios:

If we change the side lengths, but keep the angles the same, to make similar triangles, then the ratios do not change.

Trigonometric ratios We can define the ratios we have seen as the trigonometric ratios.

If we are given one of the angles in a right-angled triangle and all of the side lengths, we can write the trigonometric ratios as fractions of the side lengths.

Interactive exploration

Discover this concept in action online

mathspace.co

The ratios are often abbreviated to H

θ A

Where H is the hypotenuse, O is the opposite side and A is the adjacent side. Note that we can also abbreviate sine, cosine and tangent to sin, cos and tan.

A helpful technique to remember these ratios is to use the mnemonic SOH-CAH-TOA.

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Example 3 Which trigonometric ratio relates the given sides and the angle θ ? 8

θ 15

Create a strategy Remember the mnemonic SOH-CAH-TOA. Consider the relationship between the given sides and the angle θ.

Apply the idea Based on the given triangle, the triangle has side lengths that are opposite and adjacent to the angle θ. This means that the trigonometric ratio for this triangle is tangent

Example 4 Write the expressions for sine, cosine and tangent for the angle θ.

A 25

θ 7 C

Create a strategy Label the triangle for the hypotenuse, opposite and adjacent sides. Identify the ratios.

Apply the idea With respect to the given angle θ, the opposite side is AC, the adjacent side is BC, and the hypotenuse is AB, that is O = 24, A = 7 and H = 25.

A H

Use the sine ratio

Substitute the values

θ A C

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Use the cosine ratio

Substitute the values Use the tangent ratio

Substitute the values

Reﬂect and check Note that it is customary in trigonometry to leave fractions as improper fractions.

Example 5 Find the value of tan θ in △ABC.

A 10

θ B

Create a strategy Label the sides of the triangle with reference to angle θ. Label the missing side length x. Find the missing side length (x) using Pythagoras’ theorem. Use the trigonometric ratio for tangent.

Apply the idea From the diagram, the opposite side of the given angle θ, is the missing side length x, so O = x, the adjacent side is A = 8, and the hypotenuse is H = 10.

A x (O) B

(H)

θ 8 ( A)

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. Use Pythagoras’ theorem to ﬁnd the value of x. Use Pythagoras’ theorem

Substitute the values and x

Subtract 82 from both sides

Evaluate

Take the square root of both sides

Evaluate

The length of the opposite side x is 6 and we can substitute this length into the trigonometric ratio for tangent. Use the tangent ratio

Substitute the values

Simplify

Idea summary

Hypotenuse

θ A

For a right-angled triangle, the trigonometric ratios are given by:

Opposite

Adjacent

A helpful technique to remember these ratios is to use the mnemonic SOH-CAH-TOA.

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8.02 Practice questions Explorer

Understanding 1

1–4

Adventurer

Trailblazer

1–4

Which side is the hypotenuse? a

Z Y

Z X

Which angle is opposite the hypotenuse? a

A B

Fill in the blanks. a

Determine if the statement is true or false based on the given triangles. a

These triangles are similar triangles.

The angles in each triangle have the same values.

One triangle is twice as large as the other.

The value for sin(45) in the ﬁrst triangle will have the same value in the second triangle.

2 45°

45°

45° 2

4 45° 4

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5–7, 8aceg, 9ab, 10, 11

Fluency

Ex 1

5, 6, 7ace, 8bdfh, 9bc, 12a

5, 6, 7bdf, 8bdfh, 9cd, 12

Identify the adjacent and opposite sides to the angle θ. a

C C

θ B

Z Z

θ Y

Ex 2

For the angle θ :

Find the length of the hypotenuse.

Find the length of the opposite side.

Find the length of the adjacent side.

d Ex 3

48 cm

θ A

What is the value of the ratio

Which trigonometric ratio relates to the given sides and the angle θ? a

7 5

θ θ B

Mathspace WA Curriculum – Year 9 mathspace.co

C A

9 31

380

52 cm

θ B

Ex 4

Write the ratios for each triangle: i

sin θ

cos θ

A 21

iii

tan θ

C 61 20

A 34

5 4

θ B

13 B

53 C A

B 7

B 17

A 15 8

α A

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9 Ex 5

For each of the right-angled triangles: i

Find the value of x.

Write the ratio for tan θ.

B 8

B 29

21 C

In terms of the side lengths AB, AC and BC, write the ratios for: a

sin θ

sin α

cos θ

cos α

tan θ

tan α

For each right-angled triangle: i

Identify the ratio that relates the given angle and sides.

Write the trigonometric ratio.

b 6

θ 13

d w

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25°

7.5

For each triangle: i

Find the value of sin θ.

Find the value of cos θ. A

Reasoning

13–15

A right-angled triangle is adjusted so that the length of the side opposite an angle is doubled while the adjacent side length is unchanged. By doing this the hypotenuse and the angle will change. How do the sine, cosine and tangent ratios change?

Using the diagram:

What is sin θ?

What is cos α?

Is the value of the ratio for sin θ and cos α the same?

Is this true for all right-angled triangles? Explain your answer.

Using the diagram shown, show that

4 3 C

= tan θ.

A a° 13

Problem-solving 16–18 16

16–19, 21

θ°

18–22

Find the value of tan θ.

19 15

θ° 28

If sin θ = : a

Which angle is represented by θ?

Find cos θ.

Find tan θ.

5 4 C

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A 383

If tan θ =

Which angle is represented by θ?

Find cos θ.

Find sin θ.

If cos θ = a

Find sin θ.

Find tan θ.

△PQR is a right-angled triangle and S is the midpoint of QR as shown.

If ∠SPQ = θ, show that QR = 2a tan θ.

Using the diagram: T 56.64 m 57 m

28.32 m

19 m B

42 m

A 21 m

Using the triangle created by the building, ﬁnd the ratio of the opposite side to the adjacent side of ∠TAB.

The tree has a height of 19 metres. For the smaller triangle created by the tree, ﬁnd the ratio of the opposite side to the adjacent side of ∠HAC.

Using the triangle created by the tree, ﬁnd the ratio of the adjacent side to the hypotenuse for ∠HAC.

Using the triangle created by the building, ﬁnd the ratio of the opposite side to the hypotenuse for ∠TAB.

Beth is building a house. She wants the roof height to be between 3.5 m and 5 m. She must decide the angle measure to use for the slant of the roof when the slant height is d m. Write an inequality that Beth can use in order to ensure that her roof height will be within the desired range. Also find an appropriate value for the slant height d.

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Roof height

d x°

50 metres

8.03

Side lengths

After this lesson, you will be able to… • evaluate trigonometric ratio expressions using technology. • apply trigonometric ratios to find unknown angles and side lengths in rightangled triangles. • solve problems that require the application of trigonometric ratios to find a missing side length.

Evaluate trigonometric ratios We can evaluate trigonometric expressions like sin 42° or cos 71° using a calculator. When evaluating trigonometric expressions, make sure that the calculator is in degrees mode.

Exploration Use sin 30° = 0.5 and sin 60° = 0.866 (rounded to 3 decimal places) to answer these questions. 1.

What is the value of (sin 30°) × 2?

What is the value of sin(30° × 2)?

Are they the same?

What is the value of (sin 60°) ÷ 2?

What is the value of sin(60° ÷ 2)?

Are they the same?

Why do you think this is?

Example 1 Evaluate these using a calculator, round your answer to two decimal places. a sin (45)

Create a strategy Use a calculator and make sure it is in degrees mode.

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Apply the idea Using a calculator we get: sin 45 = 0.707 106 781 2 = 0.71

Round the answer

Reﬂect and check If your calculator is in radian mode, your answer would have been sin 45 = 0.8509. Change it to degree mode, then evaluate sin 45 again.

b cos (45)

Create a strategy Use a calculator and make sure it is in degrees mode.

Apply the idea Using a calculator we get: cos 45 = 0.707 106 781 2 = 0.71

Round the answer

c tan (45)

Create a strategy Use a calculator and make sure it is in degrees mode.

Apply the idea tan 45 = 1

Reﬂect and check We can use a right isosceles triangle with side lengths of 1 to check our answers to the previous parts. Using Pythagoras’ theorem to ﬁnd the length of the hypotenuse:

45° 1 45° 1

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Use the sine ratio

Substitute known values

Evaluate with a calculator Use the cosine ratio

Substitute known values

Evaluate with a calculator Use the tangent ratio

Substitute known values

Simplify

Example 2 Evaluate 7 cos 77° to two decimal places.

Create a strategy Use a calculator and make sure it is in degrees mode.

Apply the idea Enter this into the calculator: 7 cos 77° = 1.574 657 38 = 1.57

Round the answer

Reﬂect and check A way to check to make sure your calculator is in degrees is to evaluate sin (30°) and check that you get 0.5.

Idea summary Evaluate trigonometric expressions like sin 42° and cos 71° using the trigonometric functions on a calculator and entering the desired angle. When evaluating trigonometric expressions, make sure that the calculator is in degrees mode.

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Find the side Hypotenuse

θ Adjacent

Trigonometric ratios for a right-angled triangle:

Opposite

Steps to ﬁnd a missing side in a right-angled triangle: 1. Identify the reference angle. 2. Label the side lengths as hypotenuse, opposite, and adjacent. 3. Choose the ratio which has one of the required sides given and the missing side: • Sine: opposite and hypotenuse • Cosine: adjacent and hypotenuse • Tangent: opposite and adjacent 4. Substitute the information you know into the chosen trigonometric ratio. 5. Solve for the unknown.

Example 3 f

Find the value of f rounded to two decimal places.

53°

Create a strategy Use the trigonometric ratio that relates the two sides and substitute the given values. In this case: • Label the sides for opposite, adjacent and hypotenuse. • Write an equation using the cosine ratio as we have the angle, the adjacent side and the hypotenuse. • Solve for f.

Apply the idea Use the cosine ratio

Substitute the values and f

Multiply both sides by 12

Evaluate using a calculator

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Reﬂect and check To check our answer, we can evaluate both sides of the initial equation with the cosine ratio. Use the cosine ratio

Substitute the values

Evaluate both sides on a calculator

They aren’t the exact same because we rounded to only two decimal places, but they are close enough that we can tell our answer is correct.

Example 4 Find the value of h rounded to two decimal places.

21° hm

16 m

Create a strategy Use the trigonometric ratio that relates the two sides and substitute the given values. In this case: • Label the sides for opposite, adjacent and hypotenuse. • Write an equation using the sine ratio as we have the angle, the opposite side and the hypotenuse. • Solve for h.

Apply the idea Use the sine ratio

Substitute the values and h

Multiply both sides by h

Divide both sides by sin 21°

Evaluate using a calculator

Reﬂect and check The hypotenuse should always be longer than any other sides. It is helpful to check this when ﬁnding or using the hypotenuse.

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Example 5 A sailor looks up at a 13° angle to the top of 50 m tall lighthouse. Find the distance d metres from the sailor’s boat to the base of the lighthouse.

50 m 13° A

Create a strategy Use the trigonometric ratio that relates the two sides and substitute the given values. In this case: • Label the sides for opposite, adjacent and hypotenuse. • Write an equation using the tangent ratio as we have the angle, the opposite side, and the adjacent side. • Let d represent the length AB, the distance that we need to ﬁnd. Solve for d.

Apply the idea Use the tangent ratio

Substitute the values and d

Multiply both sides by d

Divide both sides by tan 13°

Evaluate using a calculator

Idea summary Using the trigonometric ratios:

we can solve for any of the side lengths in a right-angled triangle. Maintain equality by performing the same operation to both sides of the equation, and use a calculator to evaluate the trigonometric expressions.

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8.03 Practice questions Explorer

Understanding

1–3

Adventurer

Trailblazer 2, 3

1–3

In a right-angled triangle, the side opposite the right angle is called the hypotenuse. What are the other two sides called?

For each diagram, is there enough information to ﬁnd the missing length? a

b bm

27°

θ°

12 m

12 cm

44°

25° g cm

16.68 cm a cm

What is the ratio that will help ﬁnd the missing side? Do not ﬁnd the value of the missing side: a

If the angle is 45 degrees and the adjacent side is 7 cm, what is the length of the opposite side?

If the angle is 30 degrees and the hypotenuse is 10 cm, what is the length of the opposite side?

If the angle is 60 degrees and the adjacent side is 5 cm, what is the length of the hypotenuse?

Fluency 4 Ex 1

5 Ex 2

4a–c, 5a–c, 6a–c, 7, 8, 12

4c–f, 5c–f, 6cdf, 7a–c, 8a–c, 9–12

4f–h, 5dgh, 6egh, 7cdf, 8d–f, 9–12

Evaluate and round to three decimal places: a

sin 82°

cos 28°

tan 84°

cos 70°

tan 47°

sin 39°

cos 54°

tan 5°

Evaluate and round to three decimal places: a

9 sin 63°

12 cos 7°

8 tan 28°

5 cos 35°

1.8 tan 56°

4.2 sin 23°

2.1 sin 56.5°

0.75 tan 1.22°

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7 Ex 3

Evaluate and round to three decimal places: a

Find the value of the variable, rounded to three decimal places: a

f mm

33°

56°

6 cm

y cm

11 m 25°

11 mm

44°

12 mm

15 m 15.75 cm

58°

x cm

42°

y mm

8 Ex 4

Find the value of the variable, rounded to three decimal places: a

4 cm

41°

b mm

25° g cm

hm 63° 3.75 mm

10 m

43 m

xm 32°

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11.5 m

42°

f mm 23.2 mm 26°

Find the length of AC, rounded to one decimal place.

C 63°

Find the value of a, rounded to the nearest centimetre. 12 cm

44°

16.68 cm a cm

Find the value of the variable, rounded to two decimal places: a

54°

16 cm 42°

26 mm

f cm h mm

12 Ex 5

Ambrose’s home is located at point A, and Ambrose is standing at point B.

A 2313 m

If d is the distance between Ambrose and their home, ﬁnd the value of d rounded to two decimal places.

55°

Reasoning 13

13, 14

13–15

Given the value of tan θ = , ﬁnd the value of b, rounded to two decimal places. Explain your thinking. bm

q° 10 m

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Jin asks his friend Ethan if he can construct a right-angled triangle with an angle of 30° and side lengths of 10 cm and 20 cm. Ethan says it is not a right-angled triangle. Look at Ethan’s work and describe the error.

Pythagoras’ theorem

Evaluate: i

sin 80° and cos 10°

sin 67° and cos 23°

iii

sin 45° and cos 45°

sin 32° and cos 58°

What is the pattern developed in part (a)?

If sin (θ ) = x and cos (α ) = x, what is the relationship between θ and α?

Problem-solving 16–18 16

17–19

Find a rounded to two decimal places.

Find b rounded to one decimal place.

Find d rounded to the nearest whole number.

18–21

b cm 37°

37° 5 cm

d cm

Find the length of AD, rounded to two decimal places.

Find the length of BD, rounded to two decimal places.

Find the length of AB, rounded to two decimal places.

C 9 B

Two ﬂag posts of height 12 m and 17 m are erected 20 m apart. Find the length, l, of the string needed to join the tops of the two posts. Round your answer to one decimal place.

60°

35°

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lm 17 m

12 m 20 m

394

a cm

A is a point on the circumference of the circle with centre O. OB is 24 cm long and cuts the circle at C. A

B 62°

24 cm

Find the radius of the circle, r, rounded to one decimal place.

Find the length of BC rounded to one decimal place.

For the following diagrams: i

Find y, rounded to two decimal places.

Find w, rounded to two decimal places.

iii

Find the value of x, rounded to one decimal place.

y w 39°

67°

68 63 37°

46° w

x y

A safety fence is constructed to protect tourists from the danger of a 9 metre tall eroding castle toppling down. A surveyor measures the angle from a distance to the top of the castle as 10°. She then walks towards the fence and takes a second measurement to the top of the tower of 22°. How far did she walk between taking the two measurements? Round your answer to the nearest metre.

9m 10°

22°

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8.04

Angles

After this lesson, you will be able to… • differentiate between trigonometric ratios and their inverses. • apply inverse trigonometric ratios to find missing angles in right-angled triangles. • understand the relationship between the tangent ratio and the gradient of a line. • solve problems involving the calculation of missing angles in right-angled triangles and the determination of the gradient of a line.

Inverse trigonometric ratio To ﬁnd a missing angle in a right-angled triangle, we can calculate the inverse of the trigonometric ratio. The inverse effectively undoes the trigonometric ratio. If sin θ = x, then we undo the sine by using the inverse operation sin−1. sin θ = x sin−1(sin θ ) = sin−1 x

θ = sin−1 x We use the inverse function on the calculator to evaluate the expressions. When evaluating inverse trigonometric functions, make sure that your calculator is in degrees mode.

Exploration Evaluate the following, where θ = 0.5: 1.

sin−1 θ and

. Are they the same?

cos−1 θ and

. Are they the same?

tan−1 θ and

. Are they the same?

While we may represent the inverse trigonometric functions using an index of −1, they are not reciprocals of the original functions,

To ﬁnd missing angles in right-angled triangles, we require at least two known sides in order to use trigonometric ratios.

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Hypotenuse

Trigonometric ratios for a right-angled triangle:

Opposite

Adjacent

To determine which trigonometric ratio we can use to ﬁnd the unknown angle, we must label each of the two known side lengths of the triangle as either opposite, adjacent or the hypotenuse. We then solve for the angle by applying the inverse of that ratio.

Example 1 If cos θ = 0.256, ﬁnd θ. Round your answer to the nearest degree.

Create a strategy Use the inverse cosine to ﬁnd the angle θ.

Apply the idea cos θ = 0.256 −1

θ = cos (0.256)

Write the cosine Use the inverse cosine to solve for θ

= 75.167 152…°

Evaluate using a calculator

= 75°

Round to the nearest degree

Example 2 Find the value of θ to the nearest degree.

Create a strategy Use the trigonometric ratio that relates the two sides then apply the inverse ratio.

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Apply the idea With respect to the given angle of θ, the side AC = 8 is the opposite side and the side AB = 10 is the hypotenuse so we can use the sine ratio. Write the sine ratio

Substitute the values

Use the inverse sine to solve for θ

Evaluate and round to the nearest degree

Example 3 The person in the picture sights a pigeon above him. If the angle the person is looking at is α, ﬁnd α in degrees. 19 m a 13 m

Round your answer to two decimal places.

Create a strategy Use the trigonometric ratio that relates the two sides then apply the inverse ratio.

Apply the idea With respect to the given angle of α, the side length of 13 m is adjacent and the side length of 19 m is opposite, so we can use the tangent ratio. Write the tangent ratio

Substitute the values

Use the inverse tangent to solve for α

Evaluate and round to two decimal places

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Idea summary To ﬁnd an unknown angle in a right angled triangle: • Label the sides opposite, adjacent and hypotenuse. • Write an equation using the trigonometric ratio that relates to the angle and the two given sides. • Solve for the angle using the inverse ratio.

Tangent and gradient A special relationship exists between the tangent ratio and the gradient of a line. If we wanted to ﬁnd the gradient (m) of this straight line, we

would use m = A

In this case, the rise AB is opposite to the angle θ and the run CB is adjacent to θ. x

θ C

This means that

The gradient of a line is equal to tan θ, where θ is the angle between the line and the positive direction of the x-axis: m = tan θ

Example 4 A line has a gradient of 6. Find the angle between the line and the positive direction of the x-axis. Write your answer to the nearest degree.

Create a strategy Use the relationship between the gradient and tan θ, then apply the inverse tangent.

Apply the idea m = tan θ

Write the equation relating gradient and tan θ

6 = tan θ

Substitute m = 6

−1

θ = tan (6)

Use the inverse tangent to solve for θ

= 80.537 677…°

Evaluate using a calculator

= 81°

Round to the nearest degree

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Example 5 Point T (25, 6) lies on the given line passing through the origin:

y 7 6

5 4 3 2 1

x 5

Find θ, the angle between the line and the positive x-axis. Round your answer to two decimal places.

Create a strategy Use the relationship between the gradient and tan θ, then apply the inverse tangent.

Apply the idea Write the equation relating gradient and tan θ

Substitute m =

Substitute the values

Use the inverse tangent to solve for θ

Evaluate and round to two decimal places

Idea summary The gradient of a line is equal to tan θ, where θ is the angle between the line and the positive direction of the x-axis: m = tan θ

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8.04 Practice questions Explorer 1, 2a–c, 3

Understanding 1

Adventurer 1, 2b–d, 3

Name the opposite side to angle θ.

Name the adjacent side to angle θ.

Name the opposite side to angle α.

Name the adjacent side to angle α.

Name the angle that is opposite the hypotenuse.

Trailblazer 2d–f, 3 A α

θ B

With reference to the angle θ, ﬁnd the value of these ratios: i

iii b

34 5

A 21

θ B

θ 5

θ θ

Fill in the missing information. ⬚ trigonometric functions (sin−1, cos−1, ⬚) help you ﬁnd the unknown ⬚ in a right-angled triangle when you know the side lengths. On the other hand, regular trigonometric functions (sin, ⬚, tan) are used to work out the ratios of the side lengths of a ⬚ triangle when one of the angles is already known. 4a–c, 5a–dm, 6, 7a–e, 8, 10a–c, 11ab, 12, 13

Fluency 4

4c–e, 5e–hn, 6, 7c–h, 8, 10a–c, 11bc, 13, 14

4f–h, 5i–lo, 6, 7d–j, 8, 9, 10d–f, 11bd, 13, 14

Evaluate, rounded to the nearest degree: a e

sin−1(0.7035) −1

cos (0.6484)

b f

cos−1(0.4447) −1

tan (8.1519)

c g

tan−1(0.5085) −1

sin (0.8176)

sin−1(0.2734)

cos−1(0.7035)

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5 Ex 1

Find θ, rounded to the nearest degree: a

sin θ = 0

cos θ = 1

sin θ = 0.210

tan θ = 1.739

cos θ = 0

tan θ = 1

cos θ = 0.296

tan θ = 0.869

tan θ = 0

cos θ = 0.5

cos θ = 0.899

tan θ = 2.2

m 6

Find θ, rounded to the nearest hundredth of a degree: a

7 Ex 2a

sin θ = 0.8613

cos θ = 0.4825

tan θ = 8.958

Find the unknown angles, marked by a variable, rounded to the nearest degree: a b c B 7.9 cm

27 cm

39 cm

24 C

x A x

e A

i 63.6 cm

x A

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68.5 cm

15 cm

θ 10

8 Ex 3

The person in the ﬁgure sights a pigeon above him. If the person is looking at an angle of θ above horizontal, ﬁnd θ, rounded to the nearest minute.

17 m

θ 12 m

Write an expression for the gradient of the line that is inclined at an angle of 33° to the positive x-axis.

Find the gradient of a line, rounded to one decimal place, that is inclined to the positive x-axis, at the angle:

11 Ex 4

10°

30°

50°

20°30′

40°15′

60°45′

For each given gradient, ﬁnd the angle between the line and the positive x-axis. Round your answer to the nearest degree. a

Consider the given line which has an angle of θ with the positive x-axis: a

Find the gradient of the line.

Find θ. Round your answer to two decimal places.

x 18

13 Ex 5

Point P (5.2, 5.72) lies on the given line passing through the origin: Find θ, the angle between the line and the positive x-axis. Round your answer to two decimal places.

8 7 6

5 4 3 2 1

x 1

2 3 4 5 6 7 8 9

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Point P (15, 7) lies on the given line passing through the origin: Find θ, the angle between the line and the positive x-axis. Round your answer to two decimal places.

y 7

6 5 4 3 2 1

x 5

Reasoning 15

15, 18

15, 16, 18

15–18

An isosceles triangle has equal side lengths of 10 cm and a base of 8 cm as shown. a

Calculate the size of angle A, rounded to one decimal place.

Explain why the cosine function is the most appropriate choice in this case.

10 cm

A°

A° 8 cm

When sin θ = cos θ, what angle will θ be? Explain your answer.

In a right-angled triangle, if the sine ratio of an angle is greater than the cosine ratio of the same angle, what can you conclude about the angle? Explain your reasoning.

If a gradient is doubled then the angle between the line and the positive x-axis is also doubled. Is this statement true or false? Explain your reasoning. (Hint: Questions 10 or 11 above might help.)

Problem-solving 19–21 19

20–23, 25

20, 22–25

Find x, y and z, rounded to two decimal places. 10

z x 30

Find x and y rounded to one decimal place.

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Consider the graph of the line

and the triangle

whose vertices are the origin, the x-intercept and the y-intercept.

3 2

The angle θ is the acute angle in the triangle that the line makes with the x-axis. a b

1 x

State the gradient of the line.

−3 −2

Find the value of tan θ.

−1

−1 −2

a b

ind the size of the angle x in degrees, rounded to two F decimal places. Find the size of the angle y in degrees, rounded to two decimal places.

During rare parts of Mercury and Venus’ orbit, the angle from the Sun to Mercury to Venus is a right angle.

19 cm

x 33 cm Venus

115 000 000 km

The distance from Mercury to the Sun is 60 000 000 km. The distance from Venus to the Sun is 115 000 000 km.

8 cm

Mercury

60 000 000 km

What is the angle θ, from Venus to the Sun to Mercury? Round your answer to the nearest degree. 24

During a large-scale renovation of a historical town, city planners decided to connect the bell tower of an old church to the base of the new city library with a glass walkway. If the top of the bell tower stands at a height of 80 metres from the ground and the base of the library is 200 metres away from the tower’s base, ﬁnd the angle θ, at which the walkway should be built to connect them. Round your answer to two decimal places.

△PQR and △QRS are both right-angled triangles as shown. a

Find the length of QR. Round your answer to two decimal places.

Find the value of θ. Round your answer to two decimal places.

Q P 7 cm 12 cm

θ 15 cm

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8.05 Applications of trigonometry After this lesson, you will be able to… • translate real-world scenarios into right-angled triangle problems. • solve real-world problems involving trigonometry. • interpret the results of trigonometric calculations in the context of real-world scenarios.

Applications of trigonometry Remember that the trigonometric ratios for a right-angled triangle are:

To ﬁnd unknown sides in right-angled triangles, we require at least one angle (other than the right angle) and one other side length. To ﬁnd missing angles in right-angled triangles, we require at least two known sides and we use the inverse trigonometric ratios. To solve trigonometry problems, in a real-world scenario, we can follow these steps: 1. Draw the diagram as a right-angled triangle and label the hypotenuse, opposite and adjacent. 2. Label the given side lengths and angle measures. 3. Choose the trigonometric ratio that applies to the given information. 4. Substitute the known values into the trigonometric ratio and solve for the unknown. 5. Give the answer in the context of the situation. This section will need you to choose the correct trigonometric relationship and method to solve to problems.

Example 1 Ilhana is sitting 3.2 metres from the wall of her room. She spots a spider in the lower corner of the wall. If Ilhana is looking down at an angle of 19°, what is the height, x m, of Ilhana’s line of sight above the ground? Round your answer to one decimal place. 3.2 m 19° xm

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Create a strategy Use the trigonometric ratio that relates the two sides and substitute the given values.

Apply the idea With respect to the given angle of 19°, the side of length x is opposite and the side of length 3.2 m is adjacent, so we can use the tangent ratio. Write the tangent ratio

Substitute the values and x

Multiply both sides by 3.2

Evaluate using a calculator

Ilhana’s sightline is 1.1 m above the spider.

Example 2 A tree of lights is formed by 9 strings of lights that are each 350 cm long. The strings are all joined at one point to form the top of a tree, and the other end of each string is attached to a point on the ground.

θ° 350 cm

305 cm

A 305 cm pole stands in the centre, from the top of the tree to the ground. Find the angle, θ, one string of lights makes with the centre pole, rounded to two decimal places.

Create a strategy Identify the relevant trigonometric ratio, then solve for the angle θ.

q 350 cm 305 cm

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Apply the idea With respect to the given angle θ, 305 is the length of the adjacent side and 350 is the length of the hypotenuse, so we can use the cosine ratio. Write the cosine ratio

Substitute the values

Apply the inverse to solve for θ

Evaluate using a calculator

Idea summary The trigonometric ratios for a right-angled triangle are:

To ﬁnd missing sides in right-angled triangles, we require at least one acute angle and one other side length. To ﬁnd missing angles in right-angled triangles, we require at least two known sides and we use the inverse trigonometric ratios.

8.05 Practice questions Understanding 1

Explorer 2–4

Adventurer 1–4

A man standing at point C, is looking at the top of a tree at point A.

Trailblazer 1–4 D

If you were to use θ σ

to solve for an angle, which angle would you be solving for? 2

408

A lifeguard standing at the top of the observation deck at point B, is looking down at the shore at point C.

Which angle is equivalent to 90 − α?

Which two labelled angles are equal?

Mathspace WA Curriculum – Year 9 mathspace.co

B B α

θ A

Rayna is standing 20 m from a tree and measures the angle from the ground to the top of the tree to be 43°. If she wants to ﬁnd the height of the tree, which trigonometric ratio does she need to use?

h 43° 20 m

A ladder is leaning against a wall. The angle between the ladder and the ground is 44° and the ladder reaches to the top of the wall, which is 1.36 m high. a

Sketch a diagram of the situation.

If you want to solve for the length of the ladder, which trigonometric ratio should you use?

Fluency 5

5–10, 12, 14

5–7, 9–11, 13, 14

6, 7, 10, 11, 13–15

Find the height of the tree, h, rounded to two decimal places. 23 m

29°

6 Ex 1

A helicopter is ﬂying at an altitude of 198 m. The angle from the helicopter to the landing pad is 44°.

44°

What is the distance, d, between the helicopter and the landing pad? Round your answer to the nearest whole number.

198 m

Landing Pad

7 Ex 2

A boy ﬂying his kite releases the entire length of his string which measures 27 m, so that the kite is 18 m above him as shown in the diagram. Find θ, rounded to two decimal places.

27 m

18 m

A slide casts a shadow 5.66 m along the ground. The distance between the end of the shadow and the top of the slide is 7.84 m. Find θ, rounded to two decimal places.

7.84 m

θ 5.66 m

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A ladder measuring 2.36 m in length is leaning against a wall. If the angle the ladder makes with the ground is y°, ﬁnd y, rounded to two decimal places. 2.36 m

1.25 m

A wooden support which is 2.79 m long is preventing a fence from falling over. The angle that the wooden support makes with the ground is θ. a

Sketch a diagram of the situation.

Find θ, rounded to two decimal places.

A suspension bridge is being built. The top of the concrete tower is 35.5 m above the bridge and the connection point for the cable is 65.9 m from the tower. Assume that the concrete tower and the bridge are perpendicular to each other.

cable

Find the length of the cable to two decimal places.

Find the angle the cable makes with the bridge. Round your answer to two decimal places.

35.5 m 65.9 m

During a particular time of the day, a tree casts a shadow of length 24 m. The height of the tree is estimated to be 7 m. a

Sketch a diagram of the situation.

Find the angle θ formed by the length of the shadow and the line from the end of the shadow to the top of the tree. Round your answer to two decimal places.

The air traffic controller is communicating with a plane in flight approaching an airport for landing. The plane is 10 369 m above the ground and is still 23 444 m from the runway. If θ ° is the angle at which the plane should approach, find θ, rounded to one decimal place.

10 369 m

θ° 23 444 m

410

Georgia is riding her bike up a hill that has an incline of 5°. If she rides her bike 817 m up the hill, ﬁnd d, the horizontal distance from where she started, rounded to the nearest metre.

Mathspace WA Curriculum – Year 9 mathspace.co

817 m 5°

A rocky ledge is 261 m high. From the top of the ledge, a boat is at an angle of 8° below horizontal. If the boat is d m from the foot of the cliff, ﬁnd d, rounded to two decimal places.

8° 261 m d

Reasoning 16

16, 17

16–18

18, 19

Luca says that he can use the tangent ratio to ﬁnd angle θ. Is Luca correct? Explain your answer. 30 m

θ 23 m

A diagonal path through a rectangular garden is 10 m long. The path makes an angle of 26°48′ with the longer side of the garden. a

If the length of the shorter side of the garden is y m, which trigonometric ratio should we use to solve for y? Explain your answer.

Find y, rounded to two decimal places.

A ladder of length 3.46 m needs to make an angle with the wall of between 10° and 20° for it to be safe to stand on. At smaller angles, the ladder is at risk of toppling backwards when the climber leans away from it. At larger angles, the ladder may lose its grip on the ground. The base of the ladder is placed at 190 cm, 42 cm or 86 cm from the wall. At which of these base lengths is the ladder safest to climb? Explain how you found your answer.

At a seaside resort, two sailors, Vern and Raz, are having a debate. Vern argues that as their ship drifts away from the anchor, the angle the anchor line makes with the water surface decreases. However, Raz believes that the angle increases. The ship dropped anchor off the coast and the anchor fell 83 m to the sea bed. During the next 4 hours, the ship drifted 105 m. Who is correct and what is the angle x between the anchor line and the surface of the water, rounded to the nearest degree?

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Problem-solving 20–22 20

20, 22–24

A jet takes off and leaves the runway at an angle of 34°. It continues to ﬂy in this direction for 7 min at a speed of 630 km/h before levelling out. a

20–22

T = 7 min V = 630 km/h T=0

Find the distance, in metres, covered by the jet just before it levels out.

h 34°

If the jet reaches a height of h m just before levelling out, calculate h, to the nearest metre.

The angle from an observer’s horizontal line of sight to the top of a building is 65°. The observer is 20 metres from the base of the building, and the distance from the ground to her eyes is 1.6 m. a

Sketch a diagram of the situation.

Find the height of the building rounded to two decimal places.

Milad is on a beach, standing 46 metres from the base of a lighthouse. The angle from the ground to the top of the lighthouse is 54°. θ°

54° 85 m

46 m

Find the height of the lighthouse, rounded to one decimal place.

Milad’s picnic spot is 85 metres from the base of the lighthouse. What is the angle below horizontal from the top of the lighthouse to the picnic spot, θ°? Round your answer to two decimal places.

In the diagram shown, OQ is parallel to ST. Angle TPQ is 36°. Angle OQR is 25°. If QT = 15 metres and RS = 18 metres, ﬁnd the size of angle SPR. Round your answer to two decimal places.

Liam is standing at a distance from a tall monument. The angle below horizontal from the top of the monument to the ground at Liam’s feet is 48°. Liam then walks a further 40 metres away from the monument, and from there, the angle from the ground to the top of the monument is 28°.

Mathspace WA Curriculum – Year 9 mathspace.co

O 18 m

15 m

48°

If the height of the monument is 40.8 metres, ﬁnd the original distance between Liam and the base of the monument, rounded to one decimal place. 412

28° 40 m

8.06 Optional: Pythagoras in 3D After this lesson, you will be able to… • recognise and identify right-angled triangles in different 3D shapes like cubes, cones and prisms, and relate them to 2D slices of those shapes. • calculate the lengths of sides and diagonals in 3D objects using right-angled triangles and Pythagoras’ theorem. • apply Pythagoras’ theorem to real-world problems.

Pythagoras in 3D We can identify right-angled triangles in 3D space the same way that we do in 2D space, by looking for a right angle. In 3D space, right angles occur between lines that are perpendicular in the same 2D plane. In other words, if two lines meet at a right-angle in some 2D slice, the triangle formed with these two lines is a right-angled triangle. F H

D A

We can look at 2D slices of this cube.

If we take a slice through the vertices A, B, C and D we can see that any triangle determined by three of these vertices will be right-angled, since this slice gives us a square.

In fact, any of the 2D slices that corresponds to a face of the cube will give us multiple possible right-angled triangles. F

Taking a 2D slice through the vertices F, H, B and D gives us a rectangle, which means that we still form a right-angled triangle from any three of these vertices.

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F G

E H C

D A

If we return to 3D space, we can see how these right-angled triangles look in 3D.

We can perform a similar exercise with other 3D solids like cones and prisms to ﬁnd other rightangled triangles in 3D space:

In most straight-edged solids, the diagonals formed by joining two non-adjacent vertices will often be the hypotenuse of some right-angled triangle in that solid.

Interactive exploration

Discover this concept in action online

mathspace.co

We can ﬁnd lengths in 3D using these steps as guidance: 1. For the desired length, draw a 2D right-angled triangle involving that length using as many vertices as possible. 2. Find the other two lengths of the 2D right-angled triangle. This may involve drawing another triangle and Pythagoras’ theorem. 3. Use Pythagoras’ theorem to ﬁnd the desired length.

Example 1 A square prism has sides of length 9 cm, 9 cm and 16 cm as shown.

A D

y cm

C z cm

a Calculate the length of HF, leaving your answer in exact form.

Create a strategy We can use Pythagoras’ theorem to ﬁnd the length of HF in the triangle △FGH.

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9 cm B 9 cm F

16 cm

Apply the idea E

We can start by drawing the 2D slice and labelling the rightangled triangle that can be used to ﬁnd the length of HF.

z cm

16 cm

9 cm

HF2 = HG2 + GF2 2

Use Pythagoras’ theorem

z = 9 + 16

Substitute the lengths

z = 337

Evaluate the powers and addition

Take the square root of both sides

So the value of HF is

b Find the length of the diagonal DF, rounded to two decimal places.

Create a strategy We can use Pythagoras’ theorem to ﬁnd the length of DF in the triangle △DHF.

Apply the idea D 9 cm H

B y cm

337 cm

We can now draw the 2D slice which includes DF and label the right-angled triangle.

Use Pythagoras’ theorem

Substitute the lengths

Evaluate the squares

Evaluate the addition

Take the square root of both sides

Evaluate and round

So the value of DF is 20.45 cm.

Reflect and check If we were just asked to find DF without first being asked to find HF, would you have used this same strategy or something different? There are many ways to find DF, but in all cases you will need to solve for one other length that isn’t given on the diagram.

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Example 2 A soft drink can has a height of 13 cm and a radius of 5 cm. Find L, the length of the longest straw that can ﬁt into the can (so that the straw is not bent and ﬁts entirely inside the can).

Round your answer down to the nearest centimetre, to ensure it ﬁts inside the can.

Create a strategy Use the Pythagoras’ theorem to ﬁnd the length L using the height and radius of the can.

Apply the idea Making a 2D slice, L is the hypotenuse of the right-angled triangle. The other sides are the height, which is 13 cm, and the diameter of the can, which is 2 × 5 = 10 cm.

10 cm

13 cm

c2 = a2 + b2 2

Use Pythagoras’ theorem 2

L = 13 + 10 2

L = 269 L=

Substitute the height, diameter and L Evaluate the powers and addition Take the square root of both sides

= 16

Evaluate and round

The length of the longest straw to the nearest centimetre is 16 cm.

Idea summary We can ﬁnd right-angled triangles in a 3D solid by taking 2D slices. Once we identify right- angled triangles, we can use Pythagoras’ theorem to ﬁnd unknown lengths in the solids.

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8.06 Practice questions Understanding

Explorer 1

Adventurer

Trailblazer

1, 2

Can Pythagoras’ theorem be applied to any triangle? If not, specify the type of triangle for which it can be applied.

The diagram shows a rectangular prism with side lengths a, b and c. We want to ﬁnd the length of the diagonal from B to H:

What triangle could we use to work out the length x? Once we ﬁnd x, can we then work out length y?

c d

Write an expression for y in terms of c and x.

Fluency

3−5, 7, 9, 11, 13

C D

Do we know all the lengths in △BFH? If not, which length can we work out from the given information?

3−5, 7, 9, 12, 13

x H

3, 4, 6−8, 10, 12, 13

All edges of the cube are 5 cm long. x is the diagonal of the base and y is the diagonal through the centre of the cube. Find x and y, rounded to two decimal places.

y cm

x cm

4 Ex 1

A square-based prism has dimensions of 12 cm by 12 cm by 15 cm. Calculate, with answers rounded to two decimal places: a

The length of HF.

The length of DF.

D B

C H G

15 cm

A rectangular prism has dimensions as labelled on the diagram. Find each length in exact form:

E 12 cm

A D H

B C

E G

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Consider the triangular prism:

Find, rounded to two decimal places: a

The value of EC.

The value of XC.

The value of BF.

13 cm D

C 6 cm

Consider the rectangular prism. Find, rounded to two decimal places: a

The value of EG.

The value of AG.

Find the length of DG to two decimal places.

Calculate the area of the right-angled triangle △ADG.

A D

Write an expression for the length of BF in terms of BD and DF.

Find the length of AF in terms of AB, BD and DF.

If AB = 19, BD = 30 and DF = 43, ﬁnd the length of AF, rounded to two decimal places.

Find the length of AF if AB, BD and DF increase by 10. Round your answer to two decimal places.

E F

C A

D B 10

3 C 15 F

The square pyramid has its apex, V, aligned directly above the centre of its base, W. a

Calculate the length of AW, rounded to two decimal places.

Find the length of VW, rounded to two decimal places.

F 3 cm

11 cm

A triangular prism has dimensions as shown. Find the exact length of: a

4 cm

This triangular prism shaped box, labelled ABCDEF, needs a diagonal support inserted as shown: a

F 3 cm

11 cm

4 cm

A new diagonal DG has been added.

8 cm

26 cm D A

C B

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16 cm

Consider the cone with slant height of 13 m and perpendicular height of 12 m: a

Find the length of the radius, r.

Find the diameter of the cone’s base.

13 m 12 m

13 Ex 2

A right-angled triangular divider has been placed inside a box, as shown: a

Calculate the area of the triangular divider, rounded to two decimal places.

Calculate the perimeter of the triangular divider, rounded to two decimal places.

D 12 cm C 12 cm 17 cm

A soft drink can has a height of 11 cm and a radius of 4 cm. Find L, the length of the longest straw that can ﬁt into the can.

4 cm

Round your answer down to the nearest cm to ensure it ﬁts inside the can.

Reasoning 14

14, 16

14−16

Using the diagram, explain how to show that the length of the diagonal AG is

. Name all triangles used.

D b

G a

The diagram shows a rectangular-based pyramid ABCDE, where F is the centre of the base. Suppose that M is the midpoint of AD and that N is the midpoint of CD.

In the diagram, the length of AG is

H E

If we were going to calculate the surface area of the pyramid, state the additional lengths we would need to calculate, giving them in terms of a, b and c. Name the triangles that would be needed for the calculations. 16

A c

11 cm

L cm

B b

F a

units.

Given that a ≥ b ≥ c, ﬁnd a set of integer values of a, b and c that will give the correct length for AG. Is this the only set that works? Explain.

A c E

b F

G a

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419

Problem-solving 17−19 17

17−20

19−23

A rectangular prism shaped garden shed has a height of 2.4 m, length of 0.94 m and width of 4.2 m. a

Sketch a diagram of a wooden plank that is placed at the greatest possible length inside the garden shed.

Find the length of the plank of wood, rounded to one decimal place.

A cylindrical can of drink has a circumference of 7π cm and a height of 12 cm. Find the length of the longest straw that can ﬁt into the can. Round your answer down to the nearest cm to ensure it ﬁts inside the can.

A juice container has the shape of a rectangular prism. It needs a straw that must extend 20 mm beyond the container while touching the furthest corner of the base. 110 mm

Calculate the length of the straw needed. Round your answer to the nearest millimetre. x

30 mm

65 mm

A builder needs to carry lengths of timber along a corridor in order to get them to where he is working. There is a rightangled bend in the corridor along the way. The corridor is 2 m wide and the ceiling is 2.5 m above the ﬂoor:

Calculate the maximum length of timber, in exact form, that 2.5 m would ﬁt in the corridor when held parallel to the ground.

Since the corridor has a height of 2.5 m, he can ﬁt a longer piece of timber around the corner by angling it so that it reaches from the ﬂoor to the ceiling.

Find the maximum length of timber that is able to ﬁt through the corridor, rounded to two decimal places. 21

Consider the net of a pyramid with a rectangular base: • APQRC is a straight line • PA = 12 cm, RC = 9 cm, HG = 15 cm, and EH = 8 cm • EA = HA and FC = GC • When the net is folded, points A, B, C, and D all meet at S, which lies vertically above the point Q along PR.

B E A

F C

G D

Show that △PSR is a right-angled triangle.

Find the area of △PSR.

Find the height of the pyramid SQ.

Using the formula pyramid.

, ﬁnd the volume of the

S E

R Q

P H

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Consider the rectangular prism aquarium with dimensions of 25 cm by 30 cm by 48 cm. The top has points JKLM while the base has points PQRS, respectively. A tree branch is resting at points J and T, where T is 40 cm away from Q.

K M

Find the length of PT to the nearest tenth.

Find the length of the tree branch JT to the nearest tenth.

Find the length ST to the nearest tenth.

A ﬁsh at point L swims to point P. Find the distance, LP, that the ﬁsh travelled, to the nearest tenth.

25 cm

40 cm 48 cm

Find the smallest integer value of h, so that a straw of length 12 cm can ﬁt into the box shown:

h cm 6 cm 8 cm

8.07 Optional: Trigonometry in 3D After this lesson, you will be able to… • apply trigonometric ratios and Pythagoras’ theorem in 3D.

Trigonometry in 3D The key difference between 3D and 2D space when using trigonometry is that there are a lot more triangles we can ﬁnd. Using this abundance of triangles, we can use multiple applications of trigonometry to ﬁnd previously unknown side lengths and angles in a given problem.

Example 1 Consider the given rectangular prism: a Find the length x.

y cm 8 cm x cm θ 12 cm

9 cm

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Create a strategy

Apply the idea 2

Use Pythagoras’ theorem c = a + b .

The lengths x, 9, and 12 create a right-angled triangle on the base of the prism where x is the hypotenuse. x2 = 92 + 122 Use Pythagoras’ theorem x2 = 225

Evaluate

x = 15

Take the square root of both sides

b Find the length of the prism’s diagonal y.

Create a strategy

Apply the idea 2

Use Pythagoras’ theorem c = a + b .

The lengths x = 15, 8, and y create a right-angled triangle where y is the hypotenuse. y2 = 82 + 152 Use Pythagoras’ theorem y2 = 289

Evaluate

y = 17

Take the square root of both sides

c Find the angle θ to the nearest degree.

Create a strategy Use a trigonometric ratio.

Apply the idea In the triangle with angle θ, x = 15 is the adjacent side, and y = 17 is the hypotenuse. So we can use the cosine ratio to ﬁnd θ. Use the cosine ratio

Find the inverse cosine of both sides

Evaluate and round

Example 2 Asad and Billie are trying to find the height of a pole, but can’t reach the top of the pole. Asad is standing at point A and is 100 m from Billie at point B. On the diagram, ∠BAC = 30° and ∠DBC = 48°.

D Pole B

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a Find BC, rounded to two decimal places.

Create a strategy

Apply the idea

We can use △ABC to ﬁnd BC. Since we know one side and one angle, and it is a right-angled triangle, we can use a trigonometric ratio.

Since we have the adjacent side and are looking for the opposite side, we will use the tangent ratio.

B 100 m A

30° C

The length of BC is approximately 57.74 m.

Reflect and check If we use 57.74 m for BC in other calculations, there will be some error introduced as it is a rounded value. b Find the height of the pole.

Create a strategy

Apply the idea

We can use △BCD as it is a right-angled triangle and we know one side and one angle.

Since we have the adjacent side and are looking for the opposite side, we will use the tangent ratio.

48°

The height of the pole is approximately 64.13 m.

57.74 m C

Idea summary To solve trigonometric problems in 3D, we can use any of the tools that we have previously learnt: • Pythagoras’ theorem • Trigonometric ratios

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8.07 Practice questions Understanding 1

Explorer 1−3

Adventurer

Trailblazer

1−3

A straw with length L has slipped inside a cylindrical can.

4 cm

Use Pythagoras’ theorem to write a rule which could be used to ﬁnd L.

11 cm

L cm

This box, shaped like a triangular prism, needs a diagonal support inserted between A and F as shown.

E F

Write an expression for the length of BF in terms of BD and DF. C A

D B

All edges of the cube are 5 cm long. a

Draw a labelled diagram of the triangle on the base of the cube which has x as its hypotenuse.

Find the value of x, rounded to two decimal places.

Draw a labelled diagram of the triangle through the middle of the cube which has y as its hypotenuse.

Find the value of y, rounded to two decimal places.

y cm

x cm

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Fluency 4

4−8, 10, 11, 14

4−6, 8, 9, 11−14

All edges of the cube are 7 cm long: a

Find the length of these sides, rounded to two decimal places: i

4, 5, 8, 10, 11, 14

B C

Find these angles to the nearest degree: i

∠EGH

∠EGA

iii

∠AHG

∠AGH

E H

A square prism has sides of length 3 cm, 3 cm and 14 cm. a

Find z, the length of diagonal HF rounded to two decimal places.

Find θ, rounded to two decimal places.

3 cm B

D 3 cm

θ°

z cm

14 cm G

A rectangular box has a triangular divider placed inside it. a b

If z = AC, ﬁnd the value of z, rounded to two decimal places.

7 cm

Find the area of the divider, rounded to two decimal places. A

8 Ex 1

Consider the triangular prism. Round all answers to two decimal places. a

Find the size of ∠AED.

Find the length of CE.

Find the length of BE.

Find the size of ∠BEC.

Find the length of CX.

Find the size of ∠BXC.

C 10 cm

z cm 19 cm

B 3

Consider the rectangular prism: a

Find the length x.

Find the length of the prism’s diagonal y, rounded to two decimal places.

Find angle θ to the nearest degree.

y cm 10 cm x cm 16 cm

θ 12 cm

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Consider the cube: a

Find the size of: i

iii

Calculate the length of x, rounded to two decimal places.

Calculate the length of y, rounded to two decimal places.

y cm

β x cm

Find the angle θ to the nearest degree.

α 1 cm

A 25 cm × 11 cm × 8 cm cardboard box contains an insert (the shaded area) made of foam: a

8 cm

Find the length of b, the base of the foam insert, rounded to two decimal places.

Find the area of the triangular-shaped insert, rounded to the nearest square centimetre.

Find the value of θ rounded to the nearest degree.

b cm

11 cm

25 cm

A, C, and X are three points in a horizontal plane and B is a point vertically above X as shown:

Find these lengths to the nearest metre: a

36° 45°

If z = AW, calculate the value of z.

If θ = ∠VAW, ﬁnd the value of θ.

82 m

Consider the right pyramid with a square base with side length 12 cm. The edge length VA is 23 cm. a

23 cm

z cm

θ° A

426

The pyramid has a square base with side length 8 cm and all slanted edges are 12 cm in length. Round all answers to two decimal places. a

Find the length of MD.

Find the size of ∠PDM.

Find the height of the pyramid, MP.

Find the length of MN.

Find the length of PN.

Find the size of ∠PNM.

Find the size of ∠PDC.

Mathspace WA Curriculum – Year 9 mathspace.co

12 cm

C N

M A 8

14 Ex 2

Antonio and Blake are trying to ﬁnd the height of a pole, but can’t reach the top of the pole. Antonio is standing at point A and is 80 m from Blake at point B. On the diagram ∠BAC = 27° and ∠DBC = 55°. D

Pole

How far is Blake from the bottom of the pole, rounded to two decimal places?

How tall is the pole, rounded to two decimal places?

Reasoning 15

15, 16

A cone has radius 7 cm and a slant height of 13 cm. a

Using a diagram, explain how it is possible to ﬁnd angle θ, at the top of the cone, even though it is not in a right-angled triangle.

θ 13 cm

Find the size of θ, rounded to the nearest degree.

7 cm

A pyramid has a square base with side length 14 m and a vertical height of 24 m.

Find the length of the edge AE using two different methods. 24 m D A C 14 m B

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Problem-solving 17 17

17, 18

18, 19

A tree stands at the corner of a square playing ﬁeld. Each side of the square is 130 m long. At the centre of the field, the top of the tree is at an angle of 22° as shown:

130 m 22° 130 m

Find the distance from the tree to the centre of the ﬁeld. Round your answer to two decimal places.

Find the height of the tree. Round your answer to two decimal places.

Find the size of the angle from the corner of the ﬁeld opposite the tree to the top of the tree. Round your answer to the nearest degree.

Find the size of the angle from an adjacent corner of the ﬁeld to the top of the tree. Round your answer to the nearest degree.

In the tetrahedron shown, the angles ∠VBC, ∠VBA, and ∠ABC are all right angles. V

6 cm B 6 cm 6 cm A

Find these distances, rounded to two decimal places: i

iii

Find the size of angle ∠VCA. Round your answer to the nearest degree. Explain how you found your answer.

A cone has a diameter of 16x and a slant height of 17x. Find the perpendicular height of the cone.

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8.08 Optional: The unit circle After this lesson, you will be able to… • divide the unit circle into four quadrants and describe the unique trigonometric properties of each. • identify and use reference angles to find trigonometric ratios in any quadrant. • recognise and apply the signs of trigonometric ratios in different quadrants. • apply the knowledge of the unit circle to solve problems.

The unit circle In right-angled triangles, we can use trigonometric ratios to describe the relationship between a given angle and two sides. The trigonometric ratios are:

Opposite

Hypotenuse

θ A

Adjacent

Exploration Two special triangles are each drawn in a circle with a radius of 1. 1

θ = 45°

θ = 30° −1

1 0

−1

1 45°

60° 1

−1

1. Calculate the lengths of the legs of each triangle, keeping your answer in fraction form. 2. Write the coordinates of the point where each hypotenuse meets the circle. 3. Write the coordinates of the point where each hypotenuse meets the circle in terms of sine and cosine.

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If we place a right-angled triangle on the Cartesian plane at the origin and make the hypotenuse have a length of 1, we can write trigonometric ratios in terms of x and y. The trigonometric ratios become:

P (x, y) 1

θ −1

−1

If we look at any right-angled triangle with a hypotenuse of 1, the endpoint of the hypotenuse will lie on a circle with radius 1, known as the unit circle. The radius of the unit circle, which creates the angle with the positive x-axis, is called the terminal arm. unit circle

terminal arm −1

P (cos θ ,sin θ )

θ is the angle that the terminal arm makes with the positive x-axis when rotating anticlockwise.

Since cos (θ ) = x and sin (θ ) = y, the coordinates of the endpoint can be represented by:

θ x

P(cos θ, sin θ )

The ratio , which is the gradient of the hypotenuse, can be represented by: −1

This will be true in all quadrants.

Example 1 The graph shows an angle a with its terminal side intersecting the circle at

. 1 a −1 −1

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x 1

a Find the value of sin a.

Create a strategy The value of sin a is the y-coordinate of point P.

Apply the idea

Reﬂect and check If we draw a vertical line from point P to the x-axis, we get the triangle shown.

b Find the value of cos a.

Create a strategy The value of cos a is the x-coordinate of point P.

Apply the idea

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Reﬂect and check We can check this answer by using the diagram from the previous part.

c Find the value of tan a.

Create a strategy , the gradient of the line from the origin to the point.

Apply the idea Write the gradient formula

Substitute values of x and y

Evaluate the division

Reﬂect and check Checking the answer with the diagram once more:

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Example 2 Find the coordinates of P(x, y), rounded to two decimal places.

y 1

P 72°

−1

x 1

−1

Create a strategy When the angle in the first quadrant is not a known angle (0°, 30°, 45°, 60°, 90°), use a calculator to evaluate. Recall cos θ = x and sin θ = y.

Apply the idea

Reflect and check

Using a calculator, we find:

When using a scientific or graphing calculator, be sure it is set to degree mode. The other mode (radian mode) will give you different values since it is used for different units.

cos 72° = 0.31 sin 72° = 0.95 The coordinates of the point are (0.31, 0.95).

Idea summary The unit circle is a circle with a radius of 1. A point on the unit circle, after having rotated by an angle of measure θ in the anticlockwise direction, can be represented by P(cos θ, sin θ ). 1

P (x, y) 1

−1

θ x

• sin θ = y, or the height of the triangle. • cos θ = x, or the length of the base of the triangle. •

, or the gradient of the hypotenuse.

−1

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Trigonometric ratios in all four quadrants The unit circle shows us that sin θ, cos θ, and tan θ exist for angles larger than what can be contained in a right-angled triangle. (cosθ , sinθ )

Previously, we defined:

y 1

θ −1

1 x

−1

We can still use these definitions in any quadrants, but the signs of x and y will change based on the quadrant they are in.

Interactive exploration

Discover this concept in action online

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We know that sin θ is equal to the y-coordinate (height), cos θ is equal to the x-coordinate (length), and tan θ is the gradient of the line from the origin to the point. y (cosθ , sinθ )

cosθ tanθ

sinθ

−1

Second quadrant: • y is positive: sin θ is positive. • x is negative: cos θ is negative. • Gradient is negative: tan θ is negative.

−1 y

θ −1

1 x

sinθ tanθ cosθ

(cosθ , sinθ )

434

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Third quadrant: • y is negative: sin θ is negative. • x is negative: cos θ is negative. • Gradient is positive: tan θ is positive.

−1

tanθ cosθ −1 S sin positive

(cosθ , sinθ )

A all positive (cosθ , sinθ )

θ −1

T tan positive

sinθ

Fourth quadrant: • y is negative: sin θ is negative. • x is positive: cos θ is positive. • Gradient is negative: tan θ is negative.

−1

C cos positive

A helpful mnemonic for positive trigonometric function values is “All Stations To Central” (ASTC). This can help you remember the signs of the trigonometric ratios in each quadrant. • In quadrant 1, all ratios are positive. • In quadrant 2, only sin θ is positive. • In quadrant 3, only tan θ is positive. • In quadrant 4, only cos θ is positive.

Example 3 Will tan 195° have a positive or a negative answer?

Create a strategy Think about which quadrant the angle is in and what the sign of tangent will be in that quadrant.

Apply the idea The angle 195° is between 180° and 270°, so it lies in quadrant 3. In quadrant 3, both x and y are negative, so

tan 195° will have a positive answer.

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Reflect and check If we graph the angle and terminal arm, we will see that the terminal arm is increasing from left to right, so it has a positive gradient.

y 1 195°

−1

1 −1

Idea summary y

(−, +)

(+, +)

−x

(−, −) III

S sin positive

(+, −)

−1

T tan positive

tan θ is the gradient of the line,

−y

A helpful mnemonic for positive trigonometric function values is “All Stations To Central” (ASTC).

y 1

436

We can determine the sign of trigonometric functions by relating the point P (x, y) on the unit circle to the trigonometric functions P (cos θ, sin θ ). In the quadrants where x is positive or negative, cos θ will be positive or negative. In the quadrants where y is positive or negative, sin θ is positive or negative.

−1

Mathspace WA Curriculum – Year 9 mathspace.co

1 x

C cos positive

• • • •

In quadrant 1, all ratios are positive. In quadrant 2, only sin θ is positive. In quadrant 3, only tan θ is positive. In quadrant 4, only cos θ is positive.

Angles beyond the ﬁrst quadrant Angles between 0° and 90° are in quadrant 1. Second

First

Angles between 90° and 180° are in quadrant 2. Angles between 180° and 270° are in quadrant 3. Angles between 270° and 360° are in quadrant 4. A boundary angle marks the edge of a quadrant, so 0°, 90°, 180°, 270°, and 360° are boundary angles.

Third

Fourth

Interactive exploration

Discover this concept in action online

mathspace.co

For any angle beyond quadrant 1, we can draw an angle in the ﬁrst quadrant that creates a congruent triangle to the one formed by the original angle. In the congruent triangles, the angles at the origin will be congruent. y

60°

1 3 , 2 2

60° angle

1 3 P − , 2 2 120°

60°

120° angle

210° 60°

−

300° 60°

−

210° angle

−

300° angle 8.08 Optional: The unit circle mathspace.co

437

The relative acute angle, or reference angle, is the smallest angle from the x-axis to the terminal arm. It is shown in each quadrant in the diagram.

90° y 1 B

180° −1

180° − θ

180° + θ

360° − θ

0°

In each quadrant, the reference angle is equivalent to: • Quadrant 1: θ • Quadrant 2: 180 − θ • Quadrant 3: 180 + θ • Quadrant 4: 360 − θ

D −1 270°

We can use the relative acute angle to ﬁnd trigonometric ratios in any quadrant. We can use these steps to evaluate any trigometric ratio: 1. Sketch the angle on the unit circle. 2. Identify the sign of the ratio based on the quadrant the terminal arm is in. 3. Find the reference (relative acute) angle. 4. Rewrite the ratio using the sign and reference angle. 5. Evaluate using special angles or a calculator.

Example 4 Point P on the unit circle shows a rotation of 330°. What acute angle in the ﬁrst quadrant can 330° be related to?

x P

Create a strategy We want to draw an acute angle with the closest x-axis.

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Mathspace WA Curriculum – Year 9 mathspace.co

Apply the idea 330° is between 270° and 360°, so it is in the fourth quadrant.

So the positive x-axis is closest. x P

This diagram shows that an acute angle 360° − θ is related to θ in the fourth quadrant.

330°

360° − 330° = 30° which is the reference angle. x

30° P

Example 5 For sin 300°: a Write an equivalent ratio using an acute reference angle.

Create a strategy Find the quadrant the angle lies in and determine the sign of the ratio. Then we can ﬁnd the closest x-axis to ﬁnd the reference angle.

Apply the idea Starting at the positive x-axis and rotating 300° anticlockwise, we move into the fourth quadrant.

90° y sin positive

180°

all positive

θ θ

0° 360°

This diagram shows that an angle of 360° − θ in the fourth quadrant is equivalent to an acute angle of θ in the ﬁrst quadrant for the trigonometric ratios. Sine is negative in the fourth quadrant.

tan positive

cos positive 270°

sin 300° = −sin(360° − 300°) = −sin 60°

Find the acute angle and use the sign Evaluate the angle

8.08 Optional: The unit circle mathspace.co

439

b Evaluate sin 300°, without a calculator.

Create a strategy We can now use a special triangle to evaluate.

Apply the idea Using our special triangle, we can see that: 30° 2

60° 1

So we have that: From above

Using special triangles

Reﬂect and check We can check our answer using a calculator or the GeoGebra applet in this section by looking at the y-coordinate.

Idea summary y

90° 1

180° −1

180° − θ

180° + θ

360° − θ

D −1 270°

First quadrant Second quadrant Third quadrant Fourth quadrant

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Mathspace WA Curriculum – Year 9 mathspace.co

Reference angle θ 180° − θ 180° + θ 360° − θ

0°

The relative acute angle, or reference angle, is the angle made with the closest x-axis. It is shown, in our diagram, in each quadrant

8.08 Practice questions Understanding 1

Explorer

Adventurer

1−4

Trailblazer

1−4

1−5

In which quadrant is the angle located? a

299°

5°

160°

229°

40°

310°

138°

344°

Use the special triangles to complete the table. sin

cos

30°

tan

30° 45° 60°

45°

60° 1

45°

Evaluate on your calculator to three decimal places. a

tan 300°

sin 25°

cos 108°

tan 171°

cos 344°

sin 226°

sin 270°

cos 360°

Is each statement true or false? a

All points on the unit circle have positive coordinates.

We cannot evaluate trigonometric ratios for angles greater than 90°.

All trigonometric ratios are positive in the first quadrant.

The reference angle is always acute.

The reference angle is made with the closest x-axis.

A point P (x, y) on the unit circle makes an angle of θ with the positive x-axis. y 1

θ −1

x 1

−1

What is the radius of the unit circle?

What is cos θ ?

What is sin θ ?

What is tan θ ?

Write the coordinates of P (x, y) in terms of θ. 8.08 Optional: The unit circle mathspace.co

441

6ab, 7, 8a-d, 9a-d, 10, 11a-f, 12, 13, 16a-g

Fluency Ex 1

6bc, 7, 8c-g, 9c-g, 10, 11c-h, 14, 15, 16e-l

For each diagram, find the value of: i

sin a

cos a

iii

tan a

P −

3 4 P  ,  5 5 θ

21 20 , 29 29

a −1

−1

1 a

−1

Find the coordinates of P (x, y), rounded to two decimal places. a

y 1

−1

40°

442

160°

−1

1 −1

Ex 3

x 1

−1

x 1

−1

Ex 2

6cd, 7, 8c-h, 9c-h, 10, 11e-l, 14−16

x 1

−1

Will the ratio have a positive or a negative answer? a

sin 143°

tan 284°

cos 143°

cos 284°

sin 284°

cos 269°

sin 170°

cos 320°

Mathspace WA Curriculum – Year 9 mathspace.co

Ex 4

Write the relative acute angle on the unit circle that is equivalent to: a

310°

100°

160°

245°

192°

138°

256°

289°

For each statement, fill in the blank with = or ≠. a d

Ex 5a

cos 300° ⬚ cos 60°

sin 150° ⬚ sin 30°

−cos 110° ⬚ cos 70°

tan 275° ⬚ tan 85°

tan 145° ⬚ −tan 35°

Write an equivalent ratio using an acute reference angle. a

sin 215° ⬚ sin 35°

sin 106°

cos 149°

tan 124°

sin 240°

cos 315°

tan 140°

sin 174°

cos 210°

tan 295°

sin 320°

cos 111°

tan 264°

Find the value of y.

Consider the point (x, y): i

Find the value of x.

(x, y) 1

135°

225°

x 1 (x, y)

The diagram shows a unit circle with point

marked on the circle. Point P represents a rotation of 45° anticlockwise around the origin from the positive x-axis: a

Find the exact value of sin 45°.

The 45° reference angle has been marked in all quadrants. For each quadrant, find the relative angle. i

Quadrant 2

Quadrant 3

iii

Quadrant 4

Find the exact coordinates of point S.

Write sin 315° in terms of an equivalent ratio of 45°.

Find the exact value of sin 135°.

P 45°

−1

1 x

−1

y 1 Q −1

P 45° 45°

45° 45°

1 x S

−1

8.08 Optional: The unit circle mathspace.co

443

The diagram shows a unit circle with point

y 1

marked on the circle. Point P represents a rotation of 60° anticlockwise around the origin from the positive x-axis: a

Find the exact value of the ratio tan 60°.

The 60° reference angle has been marked in all quadrants. For each quadrant, find the relative angle. i

Quadrant 2

Quadrant 3

iii

Quadrant 4

Find the exact coordinates of point R.

Write tan 240° in terms of an equivalent ratio of 60°.

Find the exact value of the ratio tan 240°.

1 −1

60°

1 x

−1

y 1

−1

60°

The diagram shows a unit circle with point

−1

1 x

y 1

marked on the circle. Point P represents a rotation of

30° anticlockwise around the origin from the positive x-axis: a

Find the exact value of cos 30°

The 30° reference angle has been marked in all quadrants. For each quadrant, find the relative angle: i

Quadrant 2

Quadrant 3

iii

Quadrant 4

Find the exact coordinates of point Q.

Write cos 150° in terms of an equivalent ratio of 30°.

Find the exact value of cos 150°.

−1

y 1 Q −1

P 30° 30°

x S

−1

Mathspace WA Curriculum – Year 9 mathspace.co

30° 30°

444

30°

Ex 5b

Evaluate without using a calculator. a

sin 120°

tan 135°

cos 240°

cos 300°

sin 135°

cos 225°

tan 315°

sin 240°

cos 120°

tan 300°

sin 210°

tan 330°

Reasoning 17

17, 18, 20

18−20

In which quadrant is angle θ located? Explain. a

sin θ > 0 and cos θ < 0.

tan θ < 0 and sin θ > 0.

tan θ < 0 and cos θ < 0.

tan θ > 0 and sin θ > 0.

The points (1, 0), (0, 1), (−1, 0) and (0, −1) are signiﬁcant points on the unit circle. a

Explain how the points can be used to ﬁnd the value of sin 0°, sin 90°, sin 180°, and sin 270°.

Use the points to ﬁnd the value of cos 0°, cos 90°, cos 180°, and cos 270°.

Using the Pythagoras’ theorem and the unit circle, show that sin2 θ + cos2 θ = 1.

y 1

(cosθ , sinθ )

θ −1

x 1

−1

The sine, cosine and tangent ratios are all positive for angles in the ﬁrst quadrant.

y 1

This unit circle is labelled with the letters A, S, T, C in the quadrants. Explain how these letters help identify the sign of the ratios for angles in each quadrant.

1 −1

−1

8.08 Optional: The unit circle mathspace.co

445

Problem-solving 21, 23, 25a 21

21−24, 25a

22−26

Find, rounded to two decimal places:

The coordinates of P (x, y)

The new coordinates of P if it was reﬂected across the y-axis.

sin 114°

66°

cos 114°

tan 114°

Suppose that a

Find sin θ.

, where 270° < θ < 360°. b

Find tan θ. θ 5

Given: w = 65°, x = 10°, y = 44°, z = 380°, a = 460°, b = 550° Is the statement true or false?

tan a, cos(90° + x) and sin(180° + y) are negative.

sin x, cos z, and tan b are all positive.

Angles w, x, and y are in Quadrant 1.

sin x, cos y, and tan w are all positive.

Angles z, a, and b are in Quadrant 1.

Given: w = 253°, x = 265°, y = 193°, z = 462°, a = 538°, b = 407° Is the statement true or false?

446

Angles z, a, and b are in Quadrant 3.

sin z and sin b are both negative.

tan(z − 90°) and tan(z + 90°) are both positive.

The angles w, x, and y all have negative cosine values.

Angles w, x, and y are in Quadrant 3.

Mathspace WA Curriculum – Year 9 mathspace.co

The diagram shows points P, Q, R and S, which represent rotations of 62°, 118°, 242° and 298° respectively around the unit circle. a

If point P has coordinates (a, b), write the coordinates of the following points in terms of a and b: i

iii

Q 180°

tan 118°

iii

tan 298°

P (a, b) 0°

−1

62° R −1

Is the ratio equivalent to −tan 62° or tan 62°?

tan 242° 270°

Express in terms of tan x, where x < 90°: i

90°

tan(180° − x)

tan(180° + x)

iii

tan(360° − x)

For the trigonometric expressions: i

Find the value of the relative acute angle.

Find the exact value of the trigonometric expression.

sin 480°

tan 540°

cos 780°

sin 720°

Did you know?

Ceiling fans provide a real-life example of uniform circular motion, where each blade moves around a central axis at a constant speed. This idea is straight from the unit circle, where angles are measured from the positive x-axis in a circular direction. Engineers use this model to design fans with blades that are evenly spaced, balanced, and spin safely at the intended speed. This is important because it helps prevent wobble or imbalance in the fan’s rotation, which can lead to noise, wear, or even mechanical failure. It also plays a key role in maintaining energy efficiency, as a well-balanced fan requires less power to operate smoothly. In addition, accurate angle calculations ensure that the airflow is consistent and symmetrical, providing even cooling across the room. 8.08 Optional: The unit circle mathspace.co

447

8 Chapter review 1

For each triangle, ﬁnd the length of the hypotenuse. Give your answer as an exact value. a

11 7

9 5 c

For each triangle, ﬁnd the length of the unknown side as an exact value. a

12 m

g cm

14 cm

23 cm 18 m

A right-angled triangle has one leg measuring 9 cm and the other leg measuring 12 cm. a

Find the length of the hypotenuse.

Find the area.

Find the perimeter.

Explain why Pythagoras’ theorem cannot be used to ﬁnd the value of c in this triangle.

63°

28°

Rosalind and Natalia leave their hotel room and walk 400 m east to get to the main street in the city. Then, they walk due south for 1.8 km. If d is the shortest distance between the pair’s current location and their hotel, ﬁnd the value of d in metres. Round your answer to the nearest metre.

Suella is building vegetable garden according to the given diagram. Suella wants the fence to go around the entire edge of the garden. Find the amount of fence that she needs to purchase.

400 m

270 cm

190 cm 300 cm

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Mathspace WA Curriculum – Year 9 mathspace.co

1.8 km

On one particular night, the light from a lighthouse can be seen up to 19.3 kilometres away. A boat travelling due north sails past the lighthouse. The boat is closest to the lighthouse when it is due east of it, as shown in the diagram. At that point, the boat is 16.1 kilometres from the lighthouse.

19.3 km

Find the distance in which the light will be visible from the boat, d. Round your answer to one decimal place.

19.3 km

16.1 km d km

Which trigonometric ratio relates the given sides and the angle θ ? a

θ 15

Write the ratios for each triangle: i

sin θ

cos θ

iii b

tan θ

For each of the right-angled triangles: i

Find the value of y.

Write the ratio for tan θ.

θ B

If sin θ =

Describe the type of triangle this is and explain why.

Give examples of three different triangles that would result in the ratio sin θ =

Chapter 8 review mathspace.co

449

Use technology to evaluate and round to two decimal places: a

sin 40°

tan 79°

2.5 tan 51°

0.85 cos 3.15°

cos 14°

8 sin 58°

Find the value of the variable, rounded to two decimal places: a

25 m

s cm

61°

vm 16 cm 27°

40°

ym g mm

48 m

30°

f 45 cm

48°

26°

1567 m

h cm

Find θ rounded to the nearest degree: a

sin θ = 0.5

e 15

28.5 mm

cos θ = 0.7071

tan θ = 0.577

tan θ = 0.8391

cos θ = 0.2588

sin θ = 0.8192

Find the unknown angles, marked by a variable, rounded to the nearest degree: a

z°

j° 29

w°

35 60

m° 14

A line has a gradient of 3. Find the angle between the line and the positive direction of the x-axis. Round your answer to one decimal place.

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Mathspace WA Curriculum – Year 9 mathspace.co

Point Q (6, 7) lies on the given line passing through the origin:

8 7

Find θ, the angle the line makes with the positive direction of the x-axis. Round your answer to two decimal places.

6 5 4 3 2 1

x 1

Trusses are used for bridges and roofs, and they can have various designs. One design, shown in the diagram, is called a Howe truss. a

Find θ.

Find x rounded to two decimal places.

2 3 4 5 6 7

3m 3m

8 9

3m 3m

Gianna is relaxing on the beach outside of her hotel. She looks upward at an angle of 25° and sees the balcony of her hotel room. If she is 92 m from the hotel, ﬁnd H, the vertical height of her hotel room, rounded to the nearest metre. Hm 25° 92 m

Alena starts her hike 3 km away from a mountain’s base. The mountain stands 800 m tall. After hiking towards the mountain, she is only 2.5 km from the base. Ezekiel believes that as Alena moves closer, the angle between the ground and the peak increases, while Celia thinks the angle decreases. Who is correct? Justify your answer.

The top 6 stories of a building are 25.8 m in height. From a point 70 m from the base of the building, the angle from the ground to the point where the top 6 stories begin is 22°. What is the angle from the ground to the top of the building, θ ? Round your answer to one decimal place.

25.8 m

θ 22°

70 m

Chapter 8 review mathspace.co

451

(Optional) A rectangular prism has dimensions as labelled on the diagram. AC

Find each length in exact form: a

F 6

(Optional) A triangular prism has dimensions as shown. Explain the process for ﬁnding: a

4 D

C 15

(Optional) A can has a height of 14 cm and a radius of 5 cm. Find L, the length of the longest stick that can ﬁt into the can. Round your answer down to the nearest cm, to ensure it ﬁts inside the can.

(Optional) Consider the rectangular prism: a

Find the length x.

Find the length of the prism’s diagonal y.

Find angle θ to the nearest degree.

y cm 8 cm x cm θ 12 cm

9 cm

(Optional) A, C, and X are three points in a horizontal plane and B is a point vertically above X as shown:

Find these lengths to the nearest metre: a

XB 37° A

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Mathspace WA Curriculum – Year 9 mathspace.co

45° 86 m

(Optional) A pyramid has a square base with side length 8 m and a vertical height of 25 m. a

Find the length of the edge AE using two different methods.

If θ is the measure of the surface ∠BEA, ﬁnd θ rounded to the nearest degree.

25 m D A C 8m B

(Optional) Use the diagram to find the value of: a

sin b

cos b

tan b

y 1 b −1

x 1

−1

(Optional) Will the ratio have a positive or a negative answer? a

cos 309°

tan 225°

cos 92°

sin 318°

(Optional) Write an equivalent ratio using an acute reference angle: a

sin 200°

cos 230°

sin 115°

tan 171°

(Optional) For the point (x, y): a

Find the value of x.

Find the value of y.

y (x, y) 1 120° x

Chapter 8 review mathspace.co

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Big ideas Calculating the area, surface area, and volume of various shapes like polygons, prisms, and cylinders involves applying the right formulas, breaking down complex shapes, and ensuring accurate measurements.

Curriculum Outcomes • Explore, explain and use efficient strategies to determine the perimeter and area of composite shapes involving triangles, quadrilaterals and/or circles, (including sectors), using appropriate units • Establish, explain and apply formulas to determine the volume, capacity and surface area of cylinders, using appropriate units • Explore and explain efficient strategies to determine the surface area of right prisms using appropriate units • Use dynamic geometry software to explore and construct familiar objects in threedimensions using transformations of two-dimensional figures • Analyse, represent and interpret problems in real-world situations involving the perimeter and area of composite shapes and right-angled triangles, enlargement and reduction of similar figures, finding unknown side lengths and angles using trigonometric ratios, scale in similar figures and/or volume, capacity and surface area of right prisms and cylinders

9 Measurement Chapter outline 9.01 9.02 9.03 9.04 9.05 9.06

Area Composite shapes Surface area of prisms Volume of prisms Surface area of cylinders Volume of cylinders Investigation: Transform 2D shapes to make 3D objects Chapter 9 review

456 464 479 488 497 506 515

Graffiti artists can create 50 m of 10 cm-wide lines with just one 400 ml spray can! Can you figure out how thick the paint will be?

9.01 Area After this lesson, you will be able to… • identify and use appropriate units for measuring area. • calculate the area of common polygons using given formulas. • understand and apply the formula for the area of a circle, including special cases like semicircles and quarter circles. • interpret and solve real-world problems related to area calculation.

Area of polygons Area is the amount of space enclosed inside a two-dimensional shape. From smallest to largest, the common units used for measuring area are: • square millimetres (mm2) • square centimetres (cm2) • square metres (m2) • square kilometres (km2) A hectare is a special unit of area often used to measure the area of a piece of land. 1 ha = 10 000 m2. A summary of the area formulas for common polygons: Rectangle

Square

Parallelogram

Rhombus

Kite

Triangle

x y

h b

Trapezium

a h b

Example 1 Find the area of the triangle. 10 m

456

Mathspace WA Curriculum – Year 9 mathspace.co

Create a strategy Use the formula for the area of a triangle:

Apply the idea For this triangle the base is 8 m and the height is 10 m. Write the formula

Substitute the values

Evaluate

Idea summary A summary of the area formulas for common polygons: Rectangle Square Parallelogram Rhombus

w l

Kite

x y

Triangle

x y

Trapezium

Area of circles We can calculate the area of a circle using the formula:

A = p r2

A is the area r is the radius of the circle

When you are required to give the exact value of the area of a circle, you will leave your answer in terms of p. To find the area of a semicircle (half a circle) we find the total area, then find a half of that by dividing by 2. To find the area of a quarter of a circle we find the total area, then find a quarter of that by dividing by 4.

9.01 Area mathspace.co

457

Example 2 Find the area of the circle as an exact value.

3 cm

Create a strategy Use the formula for the area of a circle: Area = π × r2

Apply the idea Area = π × r2

Write the formula

=π×3

Substitute r = 3 into the formula 2

= 9π cm

Evaluate, leaving your answer as an exact value

Reﬂect and check It can be helpful when using exact values to ﬁnd an approximate answer. A rough estimate for π is 3, so a rough estimate for this area would be 9 × 3 = 27 cm2.

Example 3 If the diameter of a circle is 21 cm, ﬁnd its area rounded to one decimal place.

Create a strategy Find the radius by halving the diameter. Use the area of a circle formula to ﬁnd the area, and round the answer.

Apply the idea r = 21 ÷ 2

Halve the diameter

= 10.5 cm

Evaluate

Area = π r

Write the formula for area of a circle 2

= π × 10.5

= 346.4 cm

Substitute the radius 2

Evaluate and round to one decimal place

Reﬂect and check Since the diameter is 21 cm, a square drawn around the circle would have an area of just over 20 × 20 = 400 cm2. The area of the circle should be less than this, and it is.

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Mathspace WA Curriculum – Year 9 mathspace.co

Idea summary We can calculate the area of a circle using the formula:

A = π r2 A is the area r is the radius of the circle

9.01 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

For the given diagram, is each statement true or false?

a b c d e 2

We must ﬁnd the lengths of all the sides of the triangle before we can ﬁnd its area. The area of the triangle is half of the area of the rectangle. An appropriate unit would be m2. The height of the triangle is 9 m. For the rectangle, it must be l = 9 m and w = 6 m, not l = 6 m and w = 9 m.

A circle has a diameter of 10 km. i

Is each area value correct or incorrect?

Is each area value exact or approximate?

10π km2

78.54 km2

314.16 km2

25π km2

Convert to the unit indicated: a

3 m to centimetres

7 mm to centimetres

0.27 km2 to square metres

50 000 m2 to hectares

9.01 Area mathspace.co

459

4a–h, 5, 8ab, 10ac, 11ac, 12

Fluency 4 Ex 1

4e–l, 5, 6, 8cd, 10b–d, 11c, 13ab

4g–n, 5–7, 8bd, 9, 10bd, 11bd, 13cd

Find the area of these shapes: a

c 29 mm

8 cm

12 cm

9 cm

10 mm

6m 27 cm 9m 11 cm 57 cm

i 14 cm

5m 30 cm 6m

10 m

18 m

9 cm 18 cm

4 cm 23 mm

25 mm

6 mm

4 mm 16 mm 13 mm

460

12 mm

Mathspace WA Curriculum – Year 9 mathspace.co

9 mm

17 cm

Complete the table for bases, heights and areas of parallelograms. Base 5 mm 15 cm

Height 2 mm 5 cm

7.5 m

Ex 2

Area 105 cm2 50 cm2 37.5 m2

A harbour has a pier in the shape of a trapezium with a perpendicular height of 8 m. One base of the pier has a length of 9 m and the other has a length of 5 m. Using a labelled diagram, find the area of the pier.

Buzz used some scrap paper to make a birthday card in the shape of a parallelogram. The base of the card is 18 cm long, and the perpendicular height is 16 cm. Using a labelled diagram, find the area of the card.

Find the area of these triangles with its given side lengths to one decimal place: a

Base length of 5 cm and a perpendicular height of 8 cm

Base length of 3.9 cm and a perpendicular height of 1.7 cm

Base length of 6 cm and a perpendicular height of 9 cm

Base length of 17.4 cm and a perpendicular height of 9.2 cm

The faces on a 4-sided die are all triangular. Each face has a base length of 13 mm and a perpendicular height of 20 mm. Find the area of one face of the die.

Find the exact area of these circles. a

b 22 mm

13 cm

d 8.5 m

11 cm

9.01 Area mathspace.co

461

11 Ex 3

Find the area of these circles to one decimal place. a

19 cm

24 mm

32 cm 14 m

The radius of a circular baking tray is 10 cm. Find its area, rounding to two decimal places.

Find the radius, rounded to two decimal places, of a circle that has an area of: a

16π cm2

Reasoning 14

289π mm2

14, 15

812.25π cm2

14–16

10π m2

16–18

Explain how the areas of the following shapes are related to the area of a rectangle: a

Triangle

Parallelogram

Kite

Rhombus

Will doubling each side of the rectangle also double the area of the rectangle? Justify your answer.

6m 9m

Antonio states that if the radius of a circle is halved then the area will also be halved. Nadine believes that the area will be a quarter of the original area. Who is correct? Explain your answer using an example.

A circle has a circumference of 16π. Explain how to ﬁnd the area of the circle.

A trapezium is put together with an identical trapezium to make a parallelogram: a

462

Find an expression for the base of the entire parallelogram in terms of a and b.

Find an expression for the area of the entire parallelogram in terms of a, b and h.

Find an expression for the area of one trapezium in terms of a, b and h.

Mathspace WA Curriculum – Year 9 mathspace.co

Problem-solving 19–21 19

20–25

20, 22–26

Consider a rectangular park that is 500 m long and 300 m wide: a

Calculate the area of the park in square metres.

Calculate the area of the park in hectares. (Note: 10 000 m2 = 1 ha)

300 m

500 m

Five farms around Australia were measured and their dimensions, in metres, were recorded in the table: a

Find the area of each farm in square metres.

Which farms have an area of more than 1 ha?

Which farms have an area of less than 1 ha?

Which farm has an area of exactly 1 ha?

What is the average area of the farms?

Farm Length (m) Width (m) Area (m2) 1 200 100 2 350 21 3 100 24 4 400 40 5 100 100

Katrina is digging a rectangular garden bed to plant some new hedging. a

If the garden bed is 0.5 m wide and 8 m long, calculate the total area of the garden bed.

Each hedge plant ﬁlls an area that is the equivalent of 0.8 square metres. How many hedge plants are needed to ﬁll Katrina’s garden bed?

Sharon has purchased a rectangular piece of fabric that is 12 m long and 7 m wide. Find the area of the largest triangular piece she can cut out from it.

An area measuring 2016 cm2 is to be paved with identical tiles in the shape of parallelograms. Each tile measures 14 cm along the base, and has a perpendicular height of 6 cm. a

Determine the area that each tile covers.

Calculate the number of tiles needed to cover the whole area.

6 cm 14 cm

Some car parks require the cars to park at an angle as shown.

4.4 m

The dimensions of the car park are as given, where each individual parking space has a perpendicular length of 4.9 m and a width of 4.4 m.

4.9 m

Determine the area needed to create an angled carpark suitable for 7 cars. 25

A circular lake has a diameter of 50 metres. There’s a small circular island in the lake with a diameter of 10 metres. Calculate the area of the lake water, not including the island.

The design shown is made using a large circle and two smaller circles of diameter 3 cm. Find the area of the shaded region correct to one decimal place.

3 cm

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9.02 Composite shapes After this lesson, you will be able to… • identify the basic shapes within a composite shape. • calculate the area of basic shapes within a composite shape. • apply the addition and subtraction methods to calculate the area of composite shapes. • calculate the perimeter of composite shapes by summing the lengths of straight line segments and arc lengths.

Perimeter of composite shapes In practical problems, shapes often consist of combinations of sectors and polygons. These shapes are referred to as composite shapes. For composite shapes, such as those shown in the diagram, the perimeter can be calculated by adding the lengths of all straight line segments and arc lengths. It is useful to add up the lengths in a consistent order. This ensures that no side is missed, and all the lengths are accounted for. By following the outline of the shape in a systematic way, such as moving clockwise or anticlockwise, you can avoid skipping any sides and guarantee an accurate calculation of the total perimeter.

There are instances where not all side lengths of a shape are known. In many applications, such as surveying the dimensions of a block of land, taking direct measurements can be time-consuming. Therefore, calculating certain side lengths using logical reasoning or formulas like Pythagoras’ theorem can be a more eﬃcient approach.

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Example 1 For this ﬁgure:

5m 8m 3m

18 m

a Determine the length x.

Create a strategy The side length x will be equal to the difference between the other two horizontal side lengths. Use the fact that the horizontal side on top has the same length as the two horizontal sides at the bottom added together.

Apply the idea 5=x+3

Write the equation

x=2m

Subtract 3 from both sides

b Determine the length y.

Create a strategy The side length y will be equal to the difference between the other two vertical side lengths. Use the fact that the vertical side on the left has the same length as the two vertical sides on the right added together.

Apply the idea 18 = y + 8

Write the equation

y = 10 m

Subtract 8 from both sides

c Calculate the perimeter of the ﬁgure.

Create a strategy The perimeter of a ﬁgure is the sum of all of its sides. Use the results found in parts (a) and (b).

Apply the idea Perimeter = 8 + 3 + y + x + 18 + 5

Add the sides

= 8 + 3 + 10 + 2 + 18 + 5

Substitute the values of x and y

= 46 m

Evaluate

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Reﬂect and check The perimeter of this shape can also be found without the calculation of x and y. The vertical side on the left is 18 m which will be equal to the total length of the vertical edges on the right. The horizontal length on the top is 5 m, which will be equal to the total length of the horizontal edges on the bottom. Therefore, the perimeter is calculated by: Perimeter = (2 × 5) + (2 × 18) = 46 m

Write the equation Evaluate

Example 2 Determine the perimeter of the shape, rounded to two decimal places 14 cm 5 cm

Create a strategy The perimeter of a ﬁgure is the sum of the sections making up its boundary. This shape has two straight sections, a semicircle of radius radius 5 cm.

cm = 2.5 cm, and a quarter circle of

The circumference of a circle is given by C = 2π r.

Apply the idea The top straight edge is 14 cm, the bottom straight edge is 14 + 5 cm and we have the semicircle and the quarter circle. Write the equation

Evaluate each part

Evaluate

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Example 3 Consider this composite shape: xm 100 m

120 m

40 m

a Determine the length of the side labelled, x, in metres, rounded to two decimal places.

Create a strategy Form a right-angled triangle, with the side x as a hypotenuse, identify the lengths of the sides and use the Pythagoras’ theorem to solve for x.

Apply the idea xm 120 m

100 m

The perpendicular height of the right-angled triangle will be the length minus 40 m. So h = 100 − 40 = 60 m. The base of the right-angled triangle will be half of 120 m. So b =

= 60 m.

40 m

Use the Pythagoras’ theorem and solve for x: Write the formula

Substitute b = 60 and h = 60

Evaluate

Take the square root of both sides

Evaluate and round to two decimal place

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b Determine the perimeter of the shape, in metres, rounded to two decimal places.

Create a strategy Add the lengths of all straight line segments that form the boundary of the shape.

Apply the idea Perimeter = 2x + (2 × 40) + 120

Write the equation

= (2 × 84.85) + (2 × 40) + 120

Substitute the value of x

= 369.70 m

Evaluate

Idea summary Composite shapes consist of combinations of sectors and polygons. The perimeter of composite shapes can be calculated by adding the lengths of all straight line segments and arc lengths.

Area of composite shapes There are two main methods for ﬁnding the area of a composite shape: the addition method and the subtraction method. The addition method involves identifying the basic shapes, calculating the area of the basic shapes and adding them all together to ﬁnd the total area of the composite shape. This composite shape is made up of two semicircles and a trapezium.

The subtraction method involves identifying the basic shapes, calculating the area of the basic shapes and subtracting the area of the smaller shapes (that were cut out) from the area of the larger shape to ﬁnd the total area of the composite shape. This composite shape is made up of a trapezium and a semicircle. We could use either of these two methods, and sometimes we may need to use both.

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Example 4 24 mm

Consider the composite shape. 32 mm

16 mm

18 mm 56 mm

a List the basic shapes that make up the composite shape.

Apply the idea The composite shape is made up of a parallelogram, a triangle, and a rectangle.

b Find the total area of the figure.

Create a strategy

24 mm

Add the area of the simple shapes that make up the composite shape.

32 mm 16 mm

18 mm 56 mm

Apply the idea The composite shape is made up of a parallelogram, a triangle, and a rectangle so we need to find the area of each and add them. Area of parallelogram = bh

Start with the area formula of a parallelogram

= 24 × 32

Substitute the base and height 2

= 768 mm

Evaluate Start with the area formula of a triangle

Substitute the base and height

Evaluate

Area of rectangle = lw

Start with the area formula of a rectangle

= 56 × 18

Substitute the height and width 2

= 1008 mm

Total area = 768 + 256 + 1008 = 2032 mm

Evaluate Add the areas Evaluate

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Reﬂect and check Find the area of the nearest rectangle that would encompass the whole shape. Area of rectangle = lw

Start with the area formula of a rectangle

= 56 × 50

Substitute the length and total width 2

= 2800 mm

Evaluate

Given the area of the nearest rectangle is 2800 mm2, it is expected that the actual shape is less than this as we can see a part cut out. The actual area of the shape is 2032 mm2 which is less, so it’s a reasonable answer.

Example 5 Consider the composite shape.

7 cm 6 cm

4 cm

a Which basic shapes make up this composite shape? A

One trapezium and one rectangle minus two semicircles.

One trapezium and one rectangle minus two quarter-circles.

One trapezium and one parallelogram minus two quarter-circles.

One trapezium and one rectangle.

12 cm

Create a strategy We can draw some lines on the composite shape to break it up into its components.

Apply the idea By drawing lines we can see a trapezium at the top, and a rectangle at the bottom with two quarter-circles cut out from each side.

The correct answer is option B: One trapezium and one rectangle minus two quarter-circles.

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b Find the area of the composite shape. Round your answer to two decimal places.

Create a strategy We need to add the areas of the trapezium and the rectangle together, then take away the areas of the two quarter-circles.

Apply the idea

Start with the area formula of a trapezium

Substitute a, b and h Evaluate Rectangle area = lw Start with the area formula = 12 × 4 Substitute l, w = 48 cm2 Evaluate The area of the 2 quarter-circles is the same as the area of 1 semicircle.

Start with the area formula

Substitute r

Evaluate and round

Composite area = 57 + 48 − 25.13 Combine the areas Evaluate = 79.87 cm2

Reﬂect and check Find the area of the nearest rectangle that would encompass the whole shape. Area of rectangle = lw Start with the area formula of a rectangle = 12 × 10 2

= 120 cm

Substitute the length and total width Evaluate

Given the area of the nearest rectangle is 120 cm , it is expected that the actual shape is less than this as we can see parts cut out. The actual area of the shape is 79.87 cm2 which is less, so it’s a reasonable answer.

Idea summary There are two main methods for ﬁnding the area of a composite shape: • The addition method involves breaking the composite shape up into simple shapes and adding the areas together. • The subtraction method involves ﬁnding the area of the larger shape and subtracting the smaller areas that were cut out.

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9.02 Practice questions Understanding 1

Explorer 1–3

Adventurer

Trailblazer

1–3

Are these statements true or false? a

The area of a composite shape can be found by dividing it into smaller, basic shapes, ﬁnding the area of each of these shapes, and then adding these areas together.

The area of a composite shape can be found by subtracting the larger shape from the smaller shape.

The area of a composite shape is always expressed in square units.

If a composite shape is made up of two identical rectangles, the total area can be found by calculating the area of one rectangle and then doubling it.

ame the two geometric shapes that make up this N composite shape.

Explain how to ﬁnd the total area.

6 cm

8 cm

10 cm

ame the two geometric shapes that make up this N composite shape.

Explain how to ﬁnd the total area. 12 cm 4 cm

Fluency Ex 1

5, 8a-e, 9, 10, 12bc, 13ab, 14a

4, 6b, 8a-ce, 9, 10, 12, 13a-c, 14

Consider this shape: a

Determine the value of x.

Determine the value of y.

Calculate the perimeter of the shape.

4, 6acd, 8c-f, 9, 11, 12b-d, 13–15 4m x

y 2m 12 m

Find the area of rectangle A.

Find the area of rectangle B.

Find the total area of the given shape.

6m 13 m

B 3m

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6 Ex 2

Calculate the perimeter of each shape, rounded to one decimal place: a

35°

34 cm

12 cm 17 cm

2.9 cm

52° 32°

32°

3.2 cm 1.2 cm

26 cm

7 Ex 3

8 Ex 4

Consider this trapezium:

Calculate the value of a.

Calculate the value of b.

Calculate the value of x. Round your answer to two decimal places.

Calculate the perimeter of the trapezium, rounded to two decimal places.

8 cm

6 cm

E x cm

a cm

C 14 cm b cm

Find the total area: a

2 cm

c 4 mm

8 cm 14 cm

6 mm 19 mm

9 mm

21 mm

13 mm

5 cm

8 cm

8 mm

13 mm

7.7

3 cm

6 cm 19 cm

6 cm 11 cm

20 cm

15 cm

15 8

6 cm 14.6

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Find the shaded area: a

7 cm

b 9 cm

6 cm 6 cm 6 cm

16 cm

17 cm

d 6 cm 3 cm

12 cm

14 cm

4 cm

Calculate the perimeter of the shape. 2 mm

1 mm 6 mm

2 mm

For each of these composite shapes: i

What basic shapes make up the composite shape?

Find the area, rounded to two decimal places.

8 cm 5 cm 13 cm

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11 cm

Find the area of the composite shapes. Round your answer to one decimal place. a

27 cm

18 cm 22 cm

5 cm

6 cm

13 Ex 5

Find the shaded area in the diagrams. Round your answer to one decimal place. a

6 cm 7 cm 10 cm

d 36 mm

20 cm

Find the shaded area, rounded to two decimal places: a

18 m

28 cm

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Eddie designs a plot of land which contains his house and garden. a

Find the total area of the plot of land.

Find the area of the garden.

18 m 11 m

Garden 23 m House

Reasoning 16

17, 18

16–18

16, 18, 19

A rectangle with dimensions width, w, and length, 2w, has a square with side length s cut out of it. a

Write an expression for the area, A, of the composite shape.

Does the area change if the position of the square changes?

A piece of origami paper, originally in the shape of a parallelogram, is folded along its shortest diagonal. Explain how to ﬁnd the total area covered by the folded paper.

18 cm

8 cm

Joshua and Lila are calculating the area of a composite shape that consists of a rectangle and a semicircle. The rectangle is 10 cm long and 6 cm wide. The diameter of the semicircle is the same as the width of the rectangle. Joshua calculates the area of the rectangle and the semicircle separately and then adds them together. Here’s his solution:

Lila argues that they can calculate the area of the entire shape as a whole, without separating the rectangle and the semicircle. She calculates the area as: (10 cm × 6 cm) + π × (3 cm)2 Which student’s approach is correct and why?

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Ava and Max are calculating the area of a composite shape that consists of a square of side lengths 5 m and a quarter-circle.

Ava’s approach is to calculate the area of the square and the quarter-circle separately. Here’s her calculation:

However, Max argues that the radius of the quarter-circle should be the same as the side of the square, and calculates the area of the quarter-circle as:

Who is correct in this situation and why?

Problem-solving 20, 23, 24

20–24

22, 23, 25–27

A square has a quarter of a circle attached to one of its sides. The square has side length 9 cm and the quarter of a circle has radius equal to the sides of the square. Find the total area, rounded to two decimal places.

In this diagram, each square of the grid has a side length of 30 m. A jogger runs the perimeter of the park four times each morning: a

Calculate the distance of one lap around the park.

Determine the total distance the jogger will run each week, expressing your answer in kilometres.

You are tiling a kitchen ﬂoor, which has a rectangular shape measuring 4.5 m in length and 3 m in width. Along the edges of the room, there is a border of square tiles, each measuring 15 cm on a side. The remaining interior of the ﬂoor will be covered with square tiles measuring 30 cm on a side. How many of each type of tile are needed?

A rectangular park is 30 m wide and 50 m long, inside the park there is a playground area. The playground consists of a square sandbox with sides measuring 3.2 m and a circular swing set section with a radius of 3 m. a

Draw a possible map of the playground.

What is the area of park excluding the playground areas? Round your answer to two decimal places.

The council is going to fertilise the lawn in the park (the area excluding the playground), for each square metre of lawn they need 250 g of fertiliser. How many kilograms of fertiliser is needed? Round your answer to two decimal places.

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The hotel lobby room is to be tiled with slate tiles as shown. The cost of tiling the room is $48/m2.

5m 2m

Find the area to be tiled.

How much will it cost to tile the room?

HOTEL LOBBY

A landscape architect designs a pond for his backyard, surrounded by a brick wall composed of a straight section and a circular arc with a radius of 5 m. The wall costs $210 per metre to build.

Calculate the total cost of the wall.

The landscaping plan includes a garden positioned at the top, bordered by a circular arc and the property boundary. 5m

Find the area of the lawn in square metres.

Find the area of the garden in square metres. Round your answer to two decimal places.

c d

If roll-on lawn costs $12 per square metre, how much will it cost to cover the lawn and the garden?

Garden Paved Area Lawn

Find the paved area in square metres. Round your answer to two decimal places.

Paving costs $100 per square metre. Find the cost to pave the paved area.

12 m

House 10 m

The pattern for a simple t-shirt features a semicircular neck hole, found in the front piece only. 20 cm 25 cm

15 cm

70 cm

10 cm 50 cm Front

478

Back

Find the area of the back of the t-shirt.

Find the area of the front of the t-shirt, rounded to one decimal place.

If the front and back pieces are both cut from two separate rectangular pieces of fabric with dimensions 90 cm by 80 cm, how much fabric is wasted? Round your answer to the nearest square centimetre.

The fabric has to be purchased from a roll that is 80 cm wide and costs $8 per metre, how much will the fabric cost for 100 t-shirts?

Mathspace WA Curriculum – Year 9 mathspace.co

9.03 Surface area of prisms After this lesson, you will be able to… • calculate the surface area of common prisms. • solve problems involving the surface area of prisms.

Surface area of prisms The surface area of a prism is the sum of the areas of all the faces. To ﬁnd the surface area of a prism, we can determine the basic shapes we need to add together. The net of a prism will show all of the faces as 2D shapes.

Another way to calculate the surface area for rectangular prisms and cubes is to use the formula. For any faces that are the same, we can ﬁnd the area of one face and multiply it by how many times that face appears. A rectangular prism has surface area: SA = 2 × top + 2 × front + 2 × side = 2lw + 2wh + 2lh = 2 (lw + lh + wh) For length, l, height, h, and width, w. A cube has surface area: SA = 6 × Area of front = 6s2 where s is the side length of the cube. The surface area of a triangular prism can be found by ﬁnding the sum of the areas of the two triangular faces and the three rectangular faces. The units for surface area will be in square units like mm2, cm2, and m2.

Interactive exploration

Discover this concept in action online

mathspace.co

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Example 1 4 cm

Find the surface area of the triangular prism shown.

3 cm 5 cm

19 cm

Create a strategy Add the areas of these ﬁve faces, where the top and bottom face are identical: +

Apply the idea Start with the formula for the area of a triangle, multiplied by 2

Substitute the values of b and h

Evaluate

Area of left rectangle = 19 × 3 2

= 57 cm

Area of front rectangle = 19 × 5 2

= 95 cm

Area of back rectangle = 19 × 4 2

= 76 cm

Multiply the length and width Evaluate Multiply the length and width Evaluate Multiply the length and width Evaluate

Surface area = 12 + 57 + 95 + 76 Add the areas = 240 cm2 Evaluate

Example 2 Find the surface area of the trapezoidal prism shown. Give your answer to the nearest two decimal places.

10 cm 5 cm 3 cm 13 cm

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Create a strategy Add the areas of the 6 faces together where the front and back face are identical trapeziums.

Use Pythagoras’ theorem to ﬁnd the missing dimension of the sloped side of the trapezoidal prism.

10 cm 5 cm

3 cm

a 13 cm

Apply the idea We can ﬁnd the area of each face using the following formulas: Start with the formula for the area of a trapezium formula, multiplied by 2

Substitute the dimensions

Evaluate

Left rectangle = 5 × 3 Area of a rectangle = length × width = 15 cm2 Evaluate Top rectangle = 10 × 3 Area of a rectangle = length × width = 30 cm2

Evaluate

Bottom rectangle = 13 × 3 Area of a rectangle = length × width = 39 cm2

Evaluate

We can ﬁnd the missing dimension of the sloped side using Pythagoras’ theorem:

Pythagoras’ theorem

Substitute a and b Evaluate

Take the square root of both sides

Evaluate Right rectangle = 5.830 952 × 3 Area of a rectangle = length × width = 17.49 cm2 Evaluate and round to two decimal places Surface area = 115 + 15 + 30 + 39 + 17.49 2

= 216.49 cm

Add the areas Evaluate

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Reﬂect and check If we were to unfold the prism into its net, we would get two trapezium faces and a large rectangle through the middle. 10 cm c cm

5 cm 13 cm

3 cm

c cm

5 cm

10 cm

The large rectangle will have a width equal to the height of the prism, and a length that is the perimeter of the trapezium (5 + 13 + 10 + 5.83 = 38.83 cm). 10 cm 5.83 cm

5 cm 3 cm

13 cm

5 cm 38.83 cm

Then the surface area is: SA = 2 × Atrapezium + Arectangle = 2 × 57.5 + 38.83 × 3 = 216.49 cm2

Example 3 A birthday gift is placed inside the box shown. What is the minimum amount of wrapping paper needed to wrap this gift? (Assume the box is in the shape of a rectangular prism).

12 cm 14 cm

5 cm

Create a strategy In a rectangular prism there are 6 rectangular faces. We have two faces that are 5 cm × 14 cm, two faces that are 14 cm × 12 cm and two faces that are 5 cm × 12 cm.

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Apply the idea SA = 2 (lw + lh + wh)

Sum the areas of the faces

= 2(14 × 5 + 14 × 12 + 5 × 12) Substitute the values = 596 cm2

Evaluate

Reflect and check If we wanted to do it without the formula, we could find the area of the top, front, and side, multiply each by 2 to get the bottom, other side, and back. Then finally add them all up. Atop = 12 × 5

Use the area of a rectangle to solve the top area 2

= 60 cm

Evaluate

Afront = 14 × 5

Use the area of a rectangle to solve the front area 2

= 70 cm

Evaluate

Aside = 12 × 14

Use the area of a rectangle to solve the side area 2

= 168 cm

Evaluate

SA = 2 × Atop + 2 × Afront + 2 × Aside

Adding up all of the areas

= 2 × 60 + 2 × 70 + 2 × 168

Substitute

= 120 + 140 + 336

Perform the multiplication

= 596 cm

Evaluate

This matches what we found using the formula.

Idea summary The surface area of a prism is the sum of the areas of each face.

9.03 Practice questions Understanding 1

Explorer

Adventurer

1–3

Trailblazer 1–3

How many faces does each prism have? a

Rectangular prism

Triangular prism

Cube

Hexagonal prism

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Are these statements true or false? a

Surface area can be found by ﬁnding the area of the base and multiplying it by the height.

The net of a prism can be used to help calculate its surface area.

A prism is made up of two congruent, parallel, polygonal bases that are connected by rectangular faces.

No other face of a prism has the same area as the base.

Which net matches this rectangular prism? 6 9 10

A 10

6 6 6 6

9 6 9 6

9 6 9 9 10 9

9 10 9

10 10 10

4, 5ace, 6ac, 7–10, 12a, 13a

Fluency 4

9 6

4, 5ace, 6bc, 7–10, 12, 13

5bdf, 6cd, 7–13

The cube has a side length of 6 cm: a

Draw the net.

Find the surface area of the cube. 6 cm 6 cm

Find the surface area: a

3 cm

12 cm

d 8 cm

12 m

3m 5m

3 cm 6 cm

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f 9 mm 8 mm 10 mm

18 m

6 Ex 1

Find the surface area: a

15 cm

b 7 cm

3 cm

20 cm 9 cm

18 cm 9 cm

12 cm

4 cm 3 cm 5 cm

12 cm

2 cm

19 cm 21 cm

8 cm

Find the surface area rounded to two decimal places when necessary. a

12 cm

Ex 2

9 cm 5 cm

11 cm 7 cm

4 cm

5 cm

3 cm 12 cm

17 cm

The ﬁgure is a trapezoidal prism. a

Find the value of d rounded to one decimal place.

Find its surface area.

d cm 2 cm

12 cm

6 cm 15 cm

9 Ex 3

A birthday gift is placed inside the box shown: Find the minimum amount of wrapping paper needed to wrap this gift.

6 cm 9 cm

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Sally is building a storage chest in the shape of a rectangular prism. The chest will be 93 cm long, 80 cm deep, and 18 cm high. Find the surface area of the chest.

The swimming pool has the dimensions shown. The sides and base of the pool are to be tiled. Calculate the area to be tiled, rounding your answer to two decimal places.

2.7 m

Find the value of x and the surface area of each prism. a

8 cm

13 cm

x 16 cm

6 cm

13 m 1.2 m

22 cm 24 cm

Find the surface area rounded to two decimal places when necessary. a

9 cm

5 cm 6 cm

10 m

12 cm

Reasoning

14, 15

14–16

14–17

If a cube and a rectangular prism have the same volume, will they also have the same surface area? Justify your answer with an example.

John and Ryan are asked to determine the surface area of a rectangular prism with dimensions 8 units by 5 units by 3 units. Ryan argues that the surface area is 120 square units, while John believes it’s 158 square units. Who is correct and why?

A triangular prism and a rectangular prism have the same base area and the same height. Vlad suggests that they will have the same surface area. Do you agree? Why or why not?

Write an expression for the surface area of this trapezoidal prism. Justify your formula.

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d a

c l

Problem-solving 18, 20, 21 18

18, 20–23

19, 20, 22–25

A cube has surface area of 1032 cm2. a

Find the area of one of the square faces.

Find the length of one side of the cube, rounded to two decimal places.

Find the side length of a cube that has a surface area of 726 cm2.

A rectangular box has a surface area of 312 m2. If width and length of the box are 8 m and 2 m respectively, ﬁnd the height of the box.

Find the surface area of the object shown.

5 cm 9 cm 2 cm

12 cm 18 cm

A box that has a rectangular base 2b centimetres in length and 3b centimetres in width. The height of the box is b centimetres. Calculate the surface area of the box in terms of b.

A rectangular prism has a length of 4x metres, width of 3x metres and height of 2x metres.

Determine the surface area in terms of x.

If the dimensions of this prism are increased by a factor of 2, calculate the increase in the surface area in terms of x.

The diagram shows a high rise building.

19 m

Find the surface area of the external walls of the building.

If 2% of the surface area is taken up by metal frames and the rest is glass, how many square metres of glass are used on the walls? Round your answer to one decimal place.

18 m

45 m

15 m

Dramona is creating a rectangular prism-shaped box with a volume of 360 cm3. The box must have a square base, and the height must be twice the length of one side of the base. Determine the surface area of the box in square centimetres, rounding your answer to one decimal place.

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9.04 Volume of prisms After this lesson, you will be able to… • apply the volume formula to find the volume of various prisms. • find the unknown dimensions of prisms using the volume formula. • distinguish between units of volume and units of capacity. • apply formulas to convert between different units of volume and capacity. • solve real-world problems involving the volume of prisms.

Volume of prisms Volume is the amount of space a three-dimensional object takes up. A prism is a three-dimensional object with two identically shaped ends joined by rectangular faces. For all prisms: Volume = area of the base × height of the prism V = Abase × H The ﬁrst step in the calculation is to ﬁnd the area of the base, which may be a basic shape or a composite shape. Here are some helpful formulas for area that we have used previously:

Square

Rectangle

Triangle

Parallelogram

Trapezium

w l

h b

a h b

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Here are some volumes that we have seen before: Vrectangular prism = length × width × height V=l×w×h Height Width

Length

Vcube = side3 V = s3

s s

Vtriangular prism = Area of base × prism height V = Area of triangle × height

Height

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Example 1 Find the volume of:

A = 35 cm2

8 cm

Create a strategy Use the volume formula: V = Abase x H

Apply the idea V = Abase x H = 35 × 8

Write the volume formula Substitute the base area and height

= 280 cm

Evaluate

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Example 2 Find the volume of this hexagonal prism 10 cm 12 cm 6 cm 5 cm

Create a strategy Find the area of the hexagonal face, then multiply by the height of the prism.

10 cm

6 cm

5 cm

Apply the idea To ﬁnd the area of the base, we need to cut the hexagonal base into two triangles and a rectangle, then ﬁnd the sum of the areas of each part.

6 cm 2.5 cm 5 cm

The height of the triangle will be For the triangular parts: Write the formula for the area of a triangle

Substitute b = 6 and h = 2.5

Evaluate

For the rectangle: A=l×w

Write the formula for the area of a rectangle

=6×5 = 30 cm

Substitute l = 6 and w = 5 2

Evaluate

The area of the base is the sum of the parts: 7.5 cm2 + 7.5 cm2 + 30 cm2 = 45 cm2 V = Abase × H = 45 × 12

Substitute the area and height 3

= 540 cm

490

Write the volume formula Evaluate

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Reﬂect and check We could also have found the volume by breaking the entire prism into a rectangular prism and two triangular prisms and then adding up the volumes.

Example 3 A prism has a volume of 630 cm3. If it has a base area of 70 cm2, ﬁnd the height of the prism.

Create a strategy Use the formula for the volume of a prism and solve for the unknown height, H.

Apply the idea V = Abase × H

Write the formula

Substitute the values of the volume and the base area

Divide both sides by 70

Evaluate

Idea summary The volume of any prism is given by:

V = Abase × H Abase is the area of the base of the prism H is the height of the prism We may need to divide the base of a prism into basic shapes to ﬁnd its area.

Units of volume and capacity Volume is the amount of space a solid takes up. Capacity is the amount that something can hold. Volume is measured using units such as: • cubic millimetres (mm3) • cubic centimetres (cm3)

• cubic metres (m3 ) • cubic kilometres (km3)

The units of capacity include megalitres (ML), kilolitres (kL), litres (L) and millilitres (mL). When we consider liquids, the relationship between capacity and volume can be expressed as: 1 ML = 1000 kL 1 kL = 1000 L = 1 m3 1 L = 1000 mL = 1000 cm3 1 mL = 1 cm3

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Example 4 This water bottle has a volume of 700 cm3.

a What is its capacity in millilitres?

Create a strategy Use the conversion for 1 cm3 = 1 mL.

Apply the idea Volume = 700 cm3

Given

Capacity = 700 mL

Change the units from cm3 to mL

b What is its capacity in litres?

Create a strategy Remember that 1 L = 1000 mL.

Apply the idea Capacity = 700 ÷ 1000 = 0.7 L

Divide by 1000 to convert from mL to L Evaluate

Idea summary Volume is measured using units such as cubic millimetres (mm3), cubic centimetres (cm3), cubic metres (m3), and cubic kilometres (km3), The units of capacity include megalitres (ML), kilolitres (kL), litres (L) and millilitres (mL). 1 ML = 1000 kL 1 kL = 1000 L = 1 m3 1 L = 1000 mL = 1000 cm3 1 mL = 1 cm3

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9.04 Practice questions Explorer

Understanding 1

1–3

Adventurer

Trailblazer

1–3

Identify the base shape and the heights for these prisms: a

2 cm 8 cm 4 cm

21 cm

d 5 cm

11 cm

Capacity ⬚ mL ⬚ mL 750 mL ⬚ kL ⬚L ⬚ mL

Are these statements true or false? a

The volume of a prism is calculated by multiplying the area of the base by the height of the prism.

A rectangular prism with the same height and base area as a triangular prism will have a different volume.

Calculating the volume of a prism requires you to draw the net of the prism.

The units of volume are always cubed units, regardless of the type of prism. 4, 5, 6a, 7, 9–12

Fluency 4 Ex 1

16 cm

Complete the table: Volume 1 cm3 50 cm3 ⬚ cm3 1 m3 1 m3 1 m3

3 cm

10 cm

13 cm

4, 5c–f, 6a–c, 7–10, 12–14

4, 5c–f, 6b–d, 8–10, 12, 14, 15

Find the volume: a

Area = 21 cm2 6 cm

13 m

Area = 12 m2

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Find the volume: a

50 cm

14 cm

4 cm 6 cm

9 cm 5 cm

7 mm 6 cm

4 mm 20 mm

8 cm 5 cm

f 10 cm

7m 5m

6 Ex 2

6 cm

40 mm

Find the volume: a

b 12 mm

8 mm

5m 4m

8 mm

16 mm

26 mm

12 mm

8 cm

6 cm 8 cm 5 cm

3 cm 1 cm

494

3 cm

4 cm 5 cm

A garden bed is 10 m in length, 4 m in width and 40 cm in height. Find the volume of soil in m3 that will be needed to ﬁll up the garden bed.

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A box is 1 m long, 30 cm high and 40 cm wide. What is the volume of the box in cubic centimetres?

A prism has a volume of 360 cm3. If it has a base area of 60 cm3, ﬁnd the height of the prism.

Ex 3

The volume of the triangular prism is 117 m3.

Find the value of y.

7.6 m

4m 6.5 m

A storage box needs to have a volume of 153 090 cm3, a height of 63 cm and a width of 81 cm. Find the length of the box.

A bottle of orange juice has a volume of 450 cm3.

Ex 4

What is its capacity in millilitres?

What is its capacity in litres?

A container has the shape of a rectangular prism, with dimensions 50 cm, 80 cm, and 20 cm. Find its volume in litres.

A medical refrigerator used to store containers has dimensions 65 cm × 52 cm × 26 cm. The containers have dimensions 50 mm × 40 mm × 20 mm.

Find the capacity of one container in millilitres.

Find the volume of the fridge in cubic millimetres.

How many containers can fit in the refrigerator?

The rectangular prism has a capacity of 1224 kL.

Find the length, b, of the rectangular prism. bm 17 m

Reasoning

16, 17

16, 18, 20

16, 18–20

A triangular prism and an octagonal prism have the same base area but different heights. Is it possible for them to have the same volume?

Stuart wants a locker which will give him the most space for his books at school. Locker A is 40 cm long, 30 cm wide and 160 cm tall. Locker B is 45 cm long, 30 cm wide and 145 cm tall. Which locker should Stuart choose? Justify your answer.

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Explain what will happen to the volume of a rectangular prism if the length, width, and height are doubled.

If two prisms have the same volume, does it mean they have the same base area or height? Justify your answer with an example.

A certain prism has a larger base area than another but a shorter height. Can it have a smaller volume? Justify your answer with an example.

Problem-solving 22–24

21–24, 27

23–29

A length of gutter pipe has the shape of a rectangular prism and has a capacity of 1245 mL. The base is 10 cm wide, and the length is 1 m. What is the depth of the pipe?

An aquarium is in the shape of a rectangular prism with dimensions 120 cm in length, 60 cm in width, and 75 cm in height. If the aquarium needs to be ﬁlled to 90% of its capacity, determine the amount of water needed in litres.

A rectangular prism has a base area of 28 cm2 and a height of 7 cm. If the dimensions of the base are both doubled, ﬁnd the volume of the rectangular prism.

Calculate the volume of a single sheet of paper, given that a pile of 500 sheets of paper measures 230 mm wide, 298 mm long and 30 mm tall.

A blue postage box has a base area of 65 cm2 and a height of 13 cm. A red postage box has the same height but triple the base dimensions of the blue box. a

Find the volume of the red postage box.

How many of the blue postage boxes can ﬁt into the red postage box?

Find the dimensions of two rectangular prisms, A and B. Use each digit from 1 to 9 exactly once across both prisms to form the numbers for their dimensions. The volume of Prism A must be exactly double the volume of Prism B.

The roof of a house is ﬂat and measures 16 m by 15 m in the shape of a rectangle.

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Find the area of the roof.

If a water tank collects all the water from the roof, how many litres of water would it collect if there had been 20 mm of rain?

How many millimetres of rain need to fall on the roof in total, to ﬁll a water tank with a capacity of 24 000 L?

A rectangular box has the dimensions x cm by x cm by y cm. The sum of the dimensions is equal to 80 cm. a

Write down the formula for y in terms of x.

What is the volume of the box in terms of x, in expanded form?

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A garage with dimensions shown has a cement ﬂoor. 2.5 m 5m

2.2 m

The cement ﬂoor is 0.4 m thick. What volume of cement is needed for the ﬂoor?

If a ventilation fan circulates 4 m3 of air per minute, how long would it take to circulate all the air in the garage?

9.05 Surface area of cylinders After this lesson, you will be able to… • calculate the surface area of cylinders. • solve problems involving the surface area of cylinders.

Surface area of a cylinder A cylinder has three faces: two identical circular bases and a curved surface that joins the two bases together. The surface area of a cylinder is the sum of the areas of these three faces.

The area of a circle is A = π r2. Now we need to ﬁnd the area of the curved face that connects the two circles.

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When unrolled the curved surface is a rectangle. One side length is equal to the height of the cylinder and the other side length is the circumference of the base circle. The area of the curved part of a cylinder is 2π rh, where r is the radius and h is the height. 2π r

2π r h

h r

To ﬁnd the surface area of the whole cylinder, we need to add the area of the top and bottom circles to the area of the curved part. Surface area of a cylinder = 2 × (Area of circular base) + area of curved face = 2 × (π × r2) + 2π r × h

SAcylinder = 2π r2 + 2π rh r the radius of the cylinder h the height of the cylinder

Example 1 Consider the following cylinder. 7m 4m 4m 7m

Find the area of the curved part of the cylinder, rounded to two decimal places.

Create a strategy Use the curved part area formula of the cylinder: Curved part area = 2π rh

Apply the idea Curved part area = 2π rh

Write the formula

= 2π × 4 × 7 2

≈ 175.93 m

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Substitute the values of r and h Evaluate and round

Example 2 Consider the cylinder shown in the diagram. 5 cm

6 cm

a Find the surface area of the cylinder in square centimetres. Round your answer to one decimal place.

Create a strategy Find the radius then use the formula for the surface area of a cylinder.

Apply the idea Write the formula of the radius in terms of diameter (d) Substitute the value of d Evaluate Write the surface area formula of the cylinder Substitute the values of r and h Evaluate and round

Reﬂect and check If we had used an approximation for π, like 3.14, then our answer would be slightly different, SA = 150.7 cm2.

b Use your answer from part (a) to ﬁnd the surface area of the cylinder in square millimetres.

Create a strategy Use the fact that 1 cm2 = 100 mm2.

Apply the idea Surface area = 150.8 × 100 mm2 = 15 080 mm2

Convert cm2 to m2 Evaluate

Reﬂect and check If we used the exact values and had not rounded in part (a), then our answer would be 4800π ≈ 15 079.6 mm2. In this case, they would be the same when rounded to the nearest square millimetre.

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Example 3 The area of the circular face on a cylinder is 8281π m2. The total surface area of the cylinder is 25 662π m2. a Find the radius of the cylinder, r.

Create a strategy Use the formula for the area of the circle: A = π r2, and equate the area in the formula with the given area.

Apply the idea π r2 = A

Write the formula Substitute the value of A

Divide both sides by π

Take the square root of both sides

Evaluate

Reﬂect and check We only take the positive square root as r is a length, so must be positive.

b Find the height h of the cylinder.

Create a strategy Use the formula for the surface area of the cylinder: SAcylinder = 2π r2 + 2π rh

Apply the idea Surface area = 2π r2 + 2π rh

Write the formula Substitute the values of r and the surface area

Simplify each term

Subtract 16 562π from both sides

Evaluate the subtraction

Divide both sides by 182π

Evaluate

The cylinder has a radius of 91 m and a height of 50 m.

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Idea summary

Surface area of a cylinder = 2π r2 + 2π rh r h

is the radius is the height of the cylinder

9.05 Practice questions Understanding

Explorer

Adventurer

1–3

Trailblazer 1–3

In terms of π and x, ﬁnd the:

Circumference of the circle

Area of the circle

Is each statement true or false? a

The surface area of a cylinder includes the area of both bases and the curved surface.

If you double the radius of a cylinder, the surface area will also double.

A cylinder that is taller always has a larger surface area than a shorter one.

For this cylinder: 4 cm 8 cm

4 cm

8 cm

Does this net match the cylinder?

To ﬁnd the surface area of the cylinder, is the radius 8 cm and the height 4 cm? 9.05 Surface area of cylinders mathspace.co

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Fluency 4 Ex 1

4, 5, 6ab, 7, 8a–c, 9

4, 5, 6bc, 7, 8c–e, 9

4–7, 8d–f, 9–11

For each pair of cylinders and their nets, ﬁnd the curved surface area of the cylinder. Round your answers to two decimal places. a

b 3m

9m 5m

7m 3m

Find the area of the curved part of a cylinder with a radius of 9.2 m and a height of 15.1 m. Round your answer to one decimal place.

For each cylinder: i

Find the curved surface area of the cylinder, rounded to two decimal places.

Find the total surface area of the cylinder, rounded to two decimal places.

2 cm

3m 6 cm

4m 8m 9m

502

Find the exact surface area of a cylinder with diameter 6 cm and height 21 cm by leaving your answer in terms of π.

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8 Ex 2

Find the surface area of each cylinder, rounded to two decimal places. a

5 cm 7m

13 cm

23 cm

5 cm 87 cm 10 cm

800 mm

200 mm

53 cm

55 cm

9 Ex 3

The area of the circular face on a cylinder is 225π m2. The total surface area of the cylinder is 3210π m2. a

Find the radius of the cylinder.

Find the height of the cylinder.

This cylinder has a surface area of 16 173 m2: 26 m

Find the height of the cylinder. Round your answer to the nearest whole number.

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This cylinder has a surface area of 54 425 cm2:

71 cm

Find the height of the cylinder. Round your answer to the nearest whole number. h cm

Reasoning

12, 13

12–14

Matea suggests that if the height of a cylinder is doubled then the surface area of the cylinder will be doubled. Is he correct? Justify your response with an example.

Consider a cylinder with a ﬁxed surface area. If you increase the radius of this cylinder, what will happen to its height? Explain your reasoning.

If you have a toilet paper roll (a cylinder) and you want to increase the total amount of paper on the roll, will this affect the surface area of the roll and paper combined? Justify your answer.

Problem-solving 15–20

15–22

15, 18, 20–25

A cylindrical can of radius 7 cm and height 10 cm is open at one end. Find the external surface area of the can. Round your answer to two decimal places.

Paul is using a toilet paper roll for crafts. He has measured the toilet paper roll to have a diameter 4 cm and a length 10 cm.

4 cm

Find the external surface area of the toilet paper roll. Round your answer to two decimal places. 17

10 cm

A paint roller is cylindrical in shape. It has a diameter of 6.2 cm and a width of 38.1 cm. Find the area painted by the roller when it makes one revolution. Round your answer to two decimal places.

The diagram shows a water trough in the shape of a half cylinder. Find the exact surface area of the outside of this water trough, leaving your answer in terms of π.

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3.15 m

1.21 m

Amy and Vincent each have a cylinder. Amy’s cylinder has a diameter of 8 cm and a height of 9 cm. Vincent’s cylinder has a diameter of 9 cm and a height of 8 cm. a

Find the surface area of Amy’s cylinder, rounded to two decimal places.

Find the surface area of Vincent’s cylinder, rounded to two decimal places.

Which cylinder has a larger surface area and by how much?

If a spherical ball with a radius of 4.9 m ﬁts exactly inside a cylinder, ﬁnd the surface area of the cylinder rounded to one decimal place.

The two identical spherical balls with radii of 1.8 m ﬁt exactly inside a cylinder. Find the surface area of the closed cylinder rounded to one decimal place.

Find the surface area of the ﬁgure, rounded to two decimal places.

3 cm 5 cm 5 cm

12 cm

Find the surface area of the brickwork for this silo rounded to two decimal places. Assume that there is a brick roof and no ﬂoor.

3.16 m

5.42 m

A beekeeper has a unique beehive made up of two complete cylinders of beeswax. a

Write down an expression for the total surface area of the beehive. Use h to represent the height of the smaller cylinder. Leave your answer in terms of π.

If the total surface area is 336π cm2, determine the radius of the base of the beehive, rounded to two decimal places, given that h = 8 cm.

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At a luxury hotel’s lobby, there’s an ornate chocolate fountain made from a cylindrical block of solid chocolate. This block has been cut in half vertically to allow for a cascade of ﬂowing chocolate. If the original chocolate cylinder had a height of 2.3 m and a diameter of 150 mm, how much chocolate surface is exposed after the cut? Answer in m2 and round your answer to two decimal places.

9.06

Volume of cylinders

After this lesson, you will be able to… • calculate the volume of a cylinder using its base area and height. • calculate the volume of fractions of cylinders. • calculate the volume of a cylinder with a hollow section, such as a pipe. • solve problems involving the volume of cylinders.

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A cylinder is very similar to a prism with a circular base. The volume of a cylinder can be found in a similar way: Volume = Area of base × Height Height

Remember the area of a circle is equal to π r2. So, the volume of the cylinder can be written as: Volume = Area of base × Height of prism

Base

= π r2 × h

Radius

= π r2 h where r is the radius.

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Example 1 Find the volume of a cylinder rounded to one decimal place if its radius is 5 cm and its height is 13 cm.

Create a strategy Use the formula of the volume of a cylinder.

Apply the idea We have been given values for r = 5 and h = 13 into the formula. V = π r2 h

Use the formula 2

= (π × 5 ) × 13

Substitute the values

= π × 25 × 13

Evaluate the squares

= 1021.0 cm3

Evaluate

Reﬂect and check As an exact value, this would be 325π cm3.

Idea summary The volume of the cylinder is:

V = π r2 h r is the radius of the cylinder h is the height of the cylinder

Volumes formed from part of a cylinder For prisms and cylinders, we have seen that:

V = Abase × h V Abase h

volume of the solid area of the end face height of the solid

When we are asked to find the volume of half or a quarter of a cylinder, we multiply the full volume by the fraction we are trying to find. To find the volume of a tube, we can find the area of the end face (base) by subtracting the smaller circle from the larger one. We can then use the area of the base to find the volume of the solid.

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Example 2 Calculate the volume of the solid. Assume that the solid is a quarter of a cylinder. Round your answer to one decimal place.

14 cm 5 cm

Create a strategy Multiply the volume of a cylinder formula by a quarter.

Apply the idea We are given r = 5 and h = 14. Multiply the formula by

Substitute the values

Evaluate

Example 3 A 13 cm concrete cylindrical pipe has an outer radius of 6 cm and an inner radius of 4 cm as shown. Find the volume of concrete required to make the pipe, rounded to two decimal places.

13 cm 4 cm 6 cm

Create a strategy Use the formula of the volume of a cylinder.

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Apply the idea To ﬁnd the base area, subtract the area of the inner circle from the outer circle. Abase = Aouter circle − Ainner circle 2

Find the area of the pipe face

=π×6 −π×4

Area of circle is A = π r2

= 36π − 16π

Evaluate the squares

= 20π cm

Evaluate as an exact value

V = Abase × h

Use the formula

= 20π × 13 = 816.81 cm

Substitute the values 3

Evaluate 3

The volume of concrete required is 816.81 cm .

Reﬂect and check We can identify the volume in one set of workings: Use the formula V = Abase × h = (π × (6)2 − π × (4)2) × 13 cm3 3

= (π × 36 − π × 16) × 13 cm

Substitute the values Evaluate the squares

Evaluate = 816.81 cm This can help with accuracy, because if we had rounded the area of the base to 20π = 62.83 cm2, then our answer would be V = 62.83 × 13 = 816.79 cm3. Being off by 0.02 sometimes does not make a difference, but if we make that rounding error for 1 000 000 pipes, then it can add up to be signiﬁcant.

Idea summary When asked to calculate the volume of part of a cylinder, multiply the total volume by the portion of the fraction required. To find the volume of a tube, we can find the area of the end face (base) by subtracting the smaller circle from the larger one. We can then use the area of the base to find the volume of the solid.

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9.06 Practice questions Understanding 1

Explorer 1–3, 5

Adventurer 1–3, 4a, 5

Trailblazer 1–5

What is the shape of the base of the solid?

6 cm

10 cm

What fraction of the cylinder is shown? a

19 cm 14 cm 8 cm 5 cm

Calculate the area of each circle, rounded to two decimal places. a

14 cm

8 cm

Find the shaded area as an exact value. a

32 mm

12 mm 20 mm

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10 mm

Is each statement true or false? a

The volume of a cylinder is found by multiplying the area of the base by the height.

If two cylinders have the same height but different radii, they will have the same volume.

If two cylinders have the same volume, they must have the same height and radius.

If the radius of a cylinder is halved, the volume of the cylinder will be reduced by half. 6ab, 7ab, 8a–c, 9ab, 10ab, 11ab

Fluency 6 Ex 1

6bc, 7bc, 8c–e, 9bc, 10bc, 11bc

6cd, 7cd, 8d–f, 9cd, 10bd, 11bd, 12

Find the volume of each of the cylinders. Round your answers to one decimal place. a

A cylinder with a radius of 3 cm and height of 16 cm.

A cylinder with a radius of 9 m and height of 22 m.

A cylinder with a diameter of 17 cm and height of 25 cm.

A cylinder with a diameter of 2.8 cm and height of 31 mm.

Find the volume of each of the cylinders as an exact value. a

A cylinder with a radius of 5 mm and height of 14 mm.

A cylinder with a radius of 8 cm and height of 14 cm.

A cylinder with a diameter of 38 km and height of 10 km.

A cylinder with a diameter of 32 m and height of 120 cm.

Find the volume of each of the cylinders. Round your answers to one decimal place. a

5 cm

10 cm

15 cm 35 cm

d 20 cm 4m 55 cm

34 cm

f 9.8 cm 80 mm 26 cm

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Find the volume of each of the cylinders as an exact value. a

8 mm

17 cm

22 mm 20 cm

37 km

26 m 21 km 17 m

10 Ex 2

Find the volume of each solid, rounding your answers to one decimal place. a

8 mm 21 mm 14 cm

5 cm

3 cm

12 m

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18 mm

11 Ex 3

Find the volume of each of the concrete cylindrical pipes. Round your answers to two decimal places. a

b 7 cm 5 cm 11 cm

15 m 5m 8m

d 5 cm 3 cm 1.4 m

13 m 3m 7m

Find the volume of the ﬁgure shown, as an exact value.

3 mm

Reasoning

13, 15

13–15

12 mm

13–15

There are two types of cylindrical juice cans available for Lyn to purchase at her local store. One has a diameter of 16 cm and a height of 18 cm, and the other has a twice the diameter and height of the smaller can. Lyn believes the volume will double, is she correct? Justify your answer.

A cylindrical cup has the dimensions as shown: a

Calculate the volume of the cup in cm3, rounded to two decimal places.

Convert your answer from part (a) to the nearest mm3.

Calculate the exact volume of the cup in cm3.

Convert your answer from part (c) to mm3, then round to the nearest mm3.

Explain why part (d) is more accurate than part (b)? Does this matter?

6 cm

10 cm

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Bob has a choice between two types of cylindrical soup cans at his local store. The ﬁrst type has a diameter of 12 cm and a height of 14 cm, and the second type has a diameter of 14 cm and a height of 12 cm. Assuming the price is the same for both cans, which one should Bob buy to get the maximum amount of soup? Justify your answer using the formula for the volume of a cylinder.

Problem-solving 16–18

16–19

18, 20, 21

A cylindrical lip gloss container has a volume of 35 cm3 and the radius of 2 cm. Using the rule for volume of a cylinder, ﬁnd the height of the cylindrical lip gloss container, rounding your answer to one decimal place.

A cylindrical spa has a radius of 1.8 m and the water is 160 cm deep. Find the volume of water in the spa in litres. Round your answer to two decimal places.

A cylindrical candle has a radius of 46 mm and a height of 0.13 m. Find the exact volume of the candle in cubic centimetres.

A dumbbell is made of three cylindrical sections. The outer sections both have a diameter of 12 cm, and the inside section has a diameter of 4 cm.

12 cm

If each section is 10 cm long, calculate the total volume of the dumbbell. Round your answer to two decimal places. 20

4 cm

10 cm 10 cm 10 cm

A solid is formed by extruding a sector as shown in the diagram. The sector has a radius of 6 cm and is extruded to a height of 12 cm. Find the volume of the solid, leaving your answer in terms of π. A

60° 6 cm

12 cm

A hollow cylindrical pipe has the dimensions shown: a

Calculate the volume of the pipe, rounded to two decimal places.

Calculate the cost of the pipe if 1 cm3 of metal weighs 5.7 g, and costs $0.50/ g.

15 cm

2.4 cm

1.8 cm

Round your answer to the nearest cent.

Investigation: Transform 2D shapes to make 3D objects

Investigate online

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9 Chapter review 1

Find the area of these shapes: a

b 16 cm

8m 35 cm 12 m

28 mm 5 mm

25 mm

15 mm

12 mm

Find the exact area of the circle.

Find the area to two decimal places. 7m

Find the radius of a circle that has an area of 361π mm2

Jake is setting up a rectangular playground for children.

If the playground is 22.5 m wide and 8 m long, calculate the total area of the playground.

Each piece of play equipment requires an area that is the equivalent of 7.5 square metres. How many pieces of playground equipment can Jake ﬁt in the playground?

A garden path is shaped like a rectangle with a semicircle at one end. The rectangular part is 10 m long and 4 m wide. The semicircle is attached to one of the shorter sides. a

Calculate the perimeter of the entire path. Round your answer to two decimal places.

If the curved part of the semicircle is made of stone, and the rest of the perimeter is bordered with timber edging, how much timber edging is needed?

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Find the total area: a

7 mm

5 mm 21 mm 10 mm 15 mm

9 mm

14 mm

23 mm

Find the shaded area: a

9 cm

b 4 cm

16 cm

8 cm

18 cm 6 cm

Find the shaded area, rounded to two decimal places: a

9 cm 24 cm 18 cm

516

A pool has a rectangular shape measuring 9 m in length and 6 m in width. Around the outside of the pool, there is a walkway made of square stones with side lengths of 60 cm. How many square stones are needed? Justify your answer.

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Find the surface area rounded to two decimal places when necessary. a

10 cm

21 cm

3 cm 9 cm 4 cm

8 cm

14 cm

12 cm

8 cm

5 cm 6m

For the triangular prism shown: a

Find the value of x.

Find the surface area.

17 m

28 m 30 m

A storage container is shaped as a rectangular prism and has a length of 3x metres, width of x metres and height of 4x metres. If the dimensions of this container are doubled, determine the increase in the surface area in terms of x. Justify your answer.

Find the volume: a

5 cm

b 4m

12 cm

8 cm

8 mm

4 mm 11 mm

3 mm

14 mm 10 mm 10 mm

20 mm

30 mm

14 mm

A pentagonal prism has a volume of 175 cm3. If it has a base area of 25 cm2, ﬁnd the height of the prism. Area = 25 cm2 h cm Chapter 9 review mathspace.co

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A rectangular box has dimensions 45 mm × 35 mm × 25 mm. Find the capacity of one box in millilitres.

A water storage tank is in the shape of a rectangular prism with dimensions 130 cm in length, 65 cm in width, and 80 cm in height. If the tank needs to be ﬁlled to 85% of its capacity, determine the amount of water needed in litres.

Calculate the surface area of each cylinder, rounded to two decimal places. a

b 15 cm

8 mm

40 cm

12 mm

The area of one circular face of a cylinder is 144π m2. The total surface area of the cylinder is 1296π m2. a

Find the radius of the cylinder.

Find the height of the cylinder.

A cake has two tiers. The top layer has a radius of 7 cm, and the bottom layer has a radius of 10 cm. Each layer is 6 cm high. 7 cm

6 cm 6 cm

10 cm

Find the amount of light green icing needed to cover the top tier of the cake, rounded to two decimal places.

Find the amount of dark green icing needed to cover the bottom layer of the cake, rounded to two decimal places.

Calculate the volume of each cylinder. Round your answers to one decimal place. a

7 mm

15 mm

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20 m

18 m

Calculate the volume of each solid rounded to one decimal place. a

25 mm

10 mm 14 cm 4 cm

Find the volume of the concrete cylindrical pipe. Round your answer to two decimal places.

10 m

A cylindrical can has a volume of 120.1 cm3 and a height of 8 cm. Find the radius of the cylindrical can rounded to two decimal places.

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Big ideas • Probability helps us make predictions about future events by analysing past events or modelled data, and the rules can change depending on the type of experiment. • Tools like Venn diagrams, arrays, and tree diagrams help organise and calculate probabilities, making it easier to understand and analyse data.

Curriculum Outcomes • Construct sample spaces to show outcomes for two-stage chance experiments both with and without replacement. Assign probabilities to outcomes and make informal connections to independent and dependent events • (Optional) Identify independent and dependent two-stage chance events using P (A and B) = P (A) × P (B) and sample spaces, such as tree diagrams to determine the probability of independent events • Analyse data with multiple variables represented in tables, describe using statistical measures and relative frequencies to make inferences • Explore, choose and create graphical or visual representations and justify choice with regards to context, purpose, data type and intended audience • Conduct repeated two-stage chance experiments and simulations, both with and without replacement, to produce datasets, including through the use of digital tools. Discuss, compare and interpret variation and estimated probabilities for compound events

10 Probability Chapter outline 10.01 10.02 10.03 10.04 10.05

Experimental probability Venn diagrams and two-way tables Arrays and simple tree diagrams With and without replacement Multi-stage probability Modelling investigation: Fish sampling 10.06 Optional: Independent and dependent events Investigation: Simulation of compound events Chapter 10 review

522 532 549 560 573 590 592 608

Astronomers use statistics to calculate the probability of finding other planets like Earth.

10.01

Experimental probability

After this lesson, you will be able to… • differentiate theoretical and experimental probability. • calculate the experimental probability of an event. • calculate the expected number of occurrences for specific outcomes based on experimental probabilities. • interpret data from a frequency table to calculate experimental probability. • solve problems involving experimental probability.

Relative frequency Experimental probability, also called relative frequency, is based on experiments, observations and trials. Imagine throwing 200 thumbtacks and trying to ﬁnd the probability of a thumbtack falling on its back or its pointed end.

Pointed end

Back

How it falls Back Pointed end

Frequency 73 127

Here are the results recorded in a frequency table.

We can use this experimental data to predict a future event. The probability of the event occurring is:

We can add the relative frequency in a column to the frequency table.

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How it falls

Frequency

Back

Pointed end

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Relative frequency

Sometimes “experiments” involve looking at historical data. The following table shows the winner of the Eurovision Song Contest from 1999 to 2018: Year 1999 2000 2001 2002 2003 2004 2005 2006 2007 2008

Winning country Sweden Denmark Estonia Latvia Turkey Ukraine Greece Finland Serbia Russia

Year 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018

Winning country Norway Germany Azerbaijan Sweden Denmark Austria Sweden Ukraine Portugal Israel

Using the data of past winners, we could calculate the probability that Sweden will win the next Eurovision Song Contest. Based off the data, the probability that Sweden wins the Eurovision Song Contest is:

We can use this information to predict how many of the next 60 contests Sweden could win. This is calculated by multiplying the relative frequency by the number of trials. In this case, Sweden could expect to win:

We might expect Sweden to win 9 contests in the next 50 years. Note: There are often many other factors that might affect Sweden winning in the next 50 years, and sometimes past performance cannot be used as a predictor for future performance. In these cases, we need to consider other factors in the interpretation of results.

Interactive exploration

Discover this concept in action online

mathspace.co

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Example 1 Ryan ﬂips a coin 20 times. a How many times would he expect a head to appear?

Create a strategy Find the probability of tossing a head, then multiply it by the number of trials.

Apply the idea The theoretical probability of tossing a head is . Write the formula

Multiply by the number of trials

Evaluate Ryan can expect that a head will appear 10 times.

b Ryan then tossed the coin 20 times and recorded that heads came up 11 times. Write the relative frequency of getting a head as a fraction.

Apply the idea Write the formula

Substitute the values

c Comment on the expected result compared to the actual outcome.

Apply the idea Ryan expected 10 heads, and resulted in 11. With such a small number of trials, the variance is negligible.

Example 2 To prepare for the week ahead, a restaurant keeps a record of the number of each main meal ordered throughout the previous week. Find the probability of randomly selecting chicken. Meal Chicken Beef Lamb Vegetarian

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Frequency 21 55 44 31

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Create a strategy Find the total amount of meals ordered the previous week. Calculate the probability. Convert to a percentage.

Apply the idea Total number of meals = 21 + 55 + 44 + 31 Add the frequencies from the frequency table = 151

Evaluate Write the formula

Substitute the given values

Multiply by 100%

Evaluate and round

Example 3 This graph shows the outcomes from a coin toss experiment.

Frequency

Coin Toss Experiment

Heads Tails Outcome

a How many times was the coin ﬂipped?

Create a strategy The total frequency will be the number of heads + the number of tails.

Apply the idea Based on the graph, the frequency of heads tossed is 26 and the frequency of tails tossed is 14. Total number of flips = 26 + 14

Sum the outcomes for all the possible events

= 40 Evaluate

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b What was the relative frequency of heads in this experiment?

Create a strategy Use the formula for finding the relative frequency of an event =

Apply the idea Based on the graph, the frequency of heads tossed is 26 and the total number of trials based on part (a) is 40. Divide 26 by 40

Simplify the fraction

Idea summary • To calculate the relative frequency (experimental probability):

• To calculate the expected number of trials with a speciﬁc outcome: Expected no. of favourable trials = Relative frequency × Total number of trials • Experimental probability does not need to equal theoretical probability. Experimental probability should get closer to the theoretical probability value as the number of trials increases.

10.01 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

If the relative frequency of an event is 1, the event: A

Happened for most of the trials

Didn’t happen

Happened on each trial

In what situations should we consider using experimental probability rather than theoretical probability?

True or False: As the number of trials in an experiment increases, the experimental probability gets closer to the theoretical probability.

Fluency 4

4–7, 9, 10, 12–15, 17, 19, 20, 22, 23, 26

4–7, 9–14, 16, 18–20, 22, 24, 26, 27

A bag contains 24 red marbles and 26 blue marbles. What is the theoretical probability of drawing a red marble?

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4, 5, 7–11, 13, 14, 17–19, 21, 24, 25, 27, 28

Ex 1

A bag contains 28 red marbles, 29 blue marbles, and 27 black marbles. a

What is the theoretical probability of drawing a blue marble?

A single trial is drawing a marble from the bag, writing down the colour, and putting it back. If this trial is repeated 400 times, how many blue marbles should you expect?

Mohamad watched cars drive past his house for ﬁve minutes. In this time, 12 cars went by, and 10 of those cars were white. What was the relative frequency of a white car driving past?

Kenji rolled a 6-sided die 30 times.

How many times would he expect a 3 to appear?

After rolling the die 30 times, he recorded that 3 had appeared 17 times. Write the relative frequency of rolling a 3.

Rosey has a bag with 2 red balls, 2 blue balls, and 2 green balls in it. She took a ball out of the bag and returned it 36 times, recording the colour each time. a

How many times would she expect to get a red ball?

After she ﬁnished, she noticed that she had drawn a red ball 15 times. Write the relative frequency of getting a red ball as a fraction.

A retail store served 989 customers in October, and there were 21 complaints during that month. What is the relative frequency that a customer complains? Round your answer to the nearest whole percent.

In the past year, an insurance company found that out of 2523 claims made, 1492 of them were from drivers under the age of 25.

Find the relative frequency that a claim is ﬁled by someone under the age of 25. Round your answer to the nearest whole percent.

Find the relative frequency that a claim is ﬁled by someone aged 25 or older. Round your answer to the nearest whole percent.

Five schools compete in a basketball competition. The results from the last season are given in the table: Match Lakehurst v St Trinian’s Summer Heights v Marquess Lakehurst v Ackley Bridge St Trinian’s v Marquess Summer Heights v Ackley Bridge

Winner

Match St Trinian’s v Ackley Bridge Marquess v Lakehurst Lakehurst v Summer Heights Marquess v Ackley Bridge St Trinian’s v Summer Heights

Lakehurst Marquess Ackley Bridge Marquess Ackley Bridge

Winner St Trinian’s Lakehurst Summer Heights Ackley Bridge St Trinian’s

What is the relative frequency that Ackley Bridge wins a match? 12

At the main traﬃc light in town, when traveling from north to south, the green light is illuminated for 27 seconds, followed by the yellow light for 3 seconds, and then the red light for 10 seconds. This cycle then repeats. If a car approaches the traﬃc light, what is the probability that the light will be: a

Green?

Yellow?

Red? 10.01 Experimental probability mathspace.co

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13 Ex 2

A food festival keeps track of the number of each type of dessert sold during the weekend: a

How many desserts were sold in total?

Calculate the relative frequency that a customer will purchase a chocolate dessert. Round your answer to the nearest whole percent.

Find the relative frequency of rolling a number greater than or equal to 3.

Find the relative frequency of not rolling a prime number.

Find the relative frequency of rolling an even number.

Find the relative frequency of rolling a 1, 2, 3, 4, 5 or 6.

Boxes of matchsticks are examined and the number of matchsticks in each box is recorded in the table:

Have 89 matchsticks?

Have more than 90 matchsticks?

Have less than 90 matchsticks?

This frequency table shows the number of people that came to donate blood and their respective weights: a

If this group is considered to be representative of the population, what is the probability that someone in the population weighs 50 to 54 kg?

What is the probability that someone weighed between 60 and 69 kg inclusive?

A die is rolled 60 times and the results are recorded in the following table: Using this frequency table, ﬁnd the probability of rolling a:

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Frequency 108 72 60 0

Outcome 1 2 3 4 5 6

Frequency 17 14 12 13 12 16

Number of matchsticks 87 88 89 90 91 92 93

Number of Boxes 0 0 7 5 6 5 3

Weight in kg (x) 40–44 45–49 50–54 55–59 60–64 65–69 70–74 Total

Frequency ( f ) 4 1 1 3 10 8 2 29

Number 1 2 3 4 5 6

Frequency 11 12 12 12 11 2

The table shows the results of rolling a die multiple times:

If the number of matchsticks in another box were counted, what is the relative frequency that it will:

Dessert Chocolate Vanilla Strawberry Mango

3 or higher

3 or lower

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How many times was the experiment repeated?

Find the probability of tossing: i

3 tails

at least 2 heads

iii

at least 1 tail

only 1 head

Outcome HHH HHT HTH HTT THH THT TTH TTT

This bar graph shows the outcomes of a six-sided die rolling experiment:

Frequency 13 13 14 11 13 14 11 13

Rolling a 6-sided die experiment 14

How many times was the die rolled?

What was the relative frequency of rolling a 4 in this experiment?

10 8 6 4 2 0

The size of several earthquakes was measured over a period of time, and the results are presented in the graph: a b

Find the probability that an earthquake measures 5, 6, or 7. Find the probability that an earthquake measures less than 5.

This histogram shows the number of people who waited in line for a rollercoaster at a theme park, and the duration of their wait time: a

Find the probability that someone waited in line for 40 minutes or less.

Find the probability that someone had to wait for at least 50 minutes to be served.

3 4 5 Outcome

6 5 4 3 2 1 0

7 Number of Earthquakes

Number of people

Ex 3

The table shows the outcomes of tossing three coins multiple times:

Frequency

100 90 80 70 60 50 40 30 20 10 0

1 2 3 4 5 6 7 8 9 10 Size of Earthquakes

35 40 45 50 55 60 65 Time (mins)

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This histogram shows the heights of people that were surveyed at a particular rollercoaster ride in a theme park: a

Estimate the probability that someone randomly chosen was between 130 and 150 cm tall.

Find the probability that someone randomly chosen was at most 130 cm tall.

30 25 Frequency

20 15 10 5 0

90 110 130 150 170 190 Height (cm)

The relative frequency that a commuter uses public transport is 30%. Out of 200 commuters, how many would you expect to use public transport?

If the probability of an event is , how many times would you expect it to occur in 76 trials?

A medical student is predicting how many people in their town will have certain genes. The population of their town is 1 600 000.

The probability of a person having the gene for red hair is 3%. How many people in the town are expected to have this gene?

The probability of a person having the gene for tetrachromacy is 0.03%. How many people in the town are expected to have this gene?

A factory produces tablets. In March, it produced 7000 tablets, of which 140 were found to be faulty. a

What is the probability that a tablet produced by the factory is faulty?

The factory plans to produce 8000 tablets in April. How many should they expect to be faulty?

A transportation company recorded the results of the types of vehicles used for deliveries during the past month. The table shows the number of deliveries made using different types of vehicles: Vehicle Type Number of Deliveries a How many deliveries were made in total during the month? Bicycle 120 Car 80 b Calculate the probability of a delivery being made using a bicycle. Motorcycle 100 c Which type of vehicle was used for the Truck 100 majority of deliveries? d

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If an additional 600 deliveries are expected next month, how many deliveries do you expect will be handled using motorcycles?

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This table shows the number of books borrowed from a local library during a month: a

What is the probability of a book being borrowed on the 5th of the month? Round your answer to the nearest whole percent.

What was the highest probability of a book being borrowed on any day during the month? Round your answer to the nearest whole percent.

What is the probability of a book being borrowed at least once during the entire month? Round your answer to the nearest whole percent. 29

Borrowed 36 32 35 30 33 32

29, 30

Beth is testing a coin to see if it is fair. She has ﬂipped the coin 100 times and recorded 52 tails. a

What is the probability of ﬂipping tails with this coin? Write your answer in decimal form.

If Beth wants to communicate the fairness of the coin to the manufacturer based on her experiment, what should she tell them?

Beth tested the coin again. This time she ﬂipped the coin 1000 times, and recorded 504 tails. What is the probability of ﬂipping tails based on this second experiment? Write your answer in decimal form.

Considering the results of the two experiments, can Beth conclude whether the coin is fair? Explain your answer.

Maria is tossing a coin. She continues tossing the coin until she gets a Tail. Her ﬁrst set of tosses resulted in Heads, Heads, Tails, so she stopped after three tosses. She then repeated the experiment 19 more times and recorded her results in the following table:

Number of Tosses before a Tail appears 1 2 3 4 5

Frequency 7 3 4 3 3

Based on Maria’s experiment, what is the probability that it takes 5 tosses of the coin before a Tail appears?

Theoretically, what is the probability that it takes 5 tosses of a coin before a Tail appears?

Was the experimental probability Maria found the same as the theoretical probability? If not, why might there be a difference?

Problem-solving 31, 32 31

1st 5th 10th 15th 20th 25th

Number of Books 50 45 48 46 49 47

Reasoning 29

Day

Homer used a spinner to choose “YES” or “NO”. The table shows three different days, how many times he spun the spinner, and how many times it landed on “YES”:

31, 32

Day 1 2 3

31–33

Number of spins Number of “Yes” 6 3 30 9 84 20

Create a possible spinner that Homer used in his experiment.

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The column graph shows the four countries that university students applied to for exchange in the last month:

Worker Applications by Country

Frequency

30 20 10 0

Germany

Philippines Country

USA

What is the relative frequency of the country with the fewest applications?

If the monthly applications are the same throughout the year, how many people in total would apply for the UK over the next six months?

If the monthly applications remain constant, how many months would it take for the Philippines to receive more than 500 applications?

Consider the table showing the status of a domino game being played by four players:

Each player draws as many dominos as they like from a bag. A domino has two numbers from 1 to 6. A high value domino occurs if the sum of its two numbers is greater than or equal to 6 while a low value domino occurs otherwise.

Player A B C D

Number of Number of high dominoes drawn valued dominoes 10 9 7 1 8 7 6 4

A player wins by randomly selecting a low value domino from their drawn sample. At this point in the game, which player has the lowest chance of winning?

10.02 Venn diagrams and two-way tables After this lesson, you will be able to… • translate data between a Venn diagram and a two-way table. • organise data using Venn diagrams and two-way tables. • interpret data presented in Venn diagrams and two-way tables.

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Venn diagrams A Venn diagram is a helpful tool when organising data into categories, especially when categories overlap. A Venn diagram, used in probability calculations, tells us how many favourable outcomes there are in each category. The rectangle is called the sample space. The sum of everything inside the rectangle represents all the events, and the sum of the probabilities of each section within the rectangle will be 1. In this case, 5 + 6 + 3 + 15 = 29 is the total. Football

The sum of all the numbers in the left circle tells us how many students play football. We can see that 5 + 6 = 11 students play football.

Baseball

P (Football) =

15 Football

Everything in the outside box does not ﬁt into either category.

Baseball

There are 15 students who do not play football or baseball.

P (Not Football nor Baseball) = 15

Football

The very middle section highlighted here represents students that are in both categories. It represents the 6 students who play football and baseball.

Baseball

P (Football and Baseball) = 15

Football

The part of the circle highlighted here are all the students who play football but do not play baseball. We could refer to these as “the students who only play football”.

Baseball

Using notation: P (F ∩ B) =

P (Football and not Baseball) = 15

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In Venn diagrams, special notations help represent various situations. The entire rectangle represents the sample space, denoted by the letter ‘S’.

The entire area inside both circles is shaded, representing the union of events A and B, which is written as A ∪ B.

In two overlapping circles, the area where the overlap is shaded indicates the intersection of events A and B, which is written as A ∩ B.

S A

The area inside the rectangle but outside the circle that is shaded within the sample space represents the complement of event A. It can be represented by A′, A or Ac.

Example 1 A group of students were asked about their siblings. The two categories show if they have at least one brother, and if they have at least one sister.

Have a brother

Have a sister

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a What is the probability that a student selected from the group will have at least one sibling?

Create a strategy Use the probability formula:

Apply the idea Number of students with at least 1 sibling = 6 + 5 + 9 Sum the numbers inside the circles = 20 Evaluate Total number of students = 20 + 8 Sum the numbers inside and outside the circles = 28 Evaluate

Write the formula

Substitute the values

Simplify The probability that a selected student will have at least one sibling is .

b What is the probability that a student chosen from the group will have at least one brother?

Create a strategy Use the probability formula:

Apply the idea Number of students with at least 1 brother = 5 + 9 Sum the numbers inside the ‘Have a brother’ circle = 14

Evaluate

Based on part (a), the total number of students is 28. Write the formula

Substitute the values

Simplify The probability that a selected student will have at least one brother is .

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c What is the probability that a student chosen from the group will not have a sister?

Create a strategy Use the probability formula:

The section from the Venn diagram that relates to “Does not have a sister” is everything outside of the “Have a sister” circle.

Apply the idea Number of students who have no sister = 9 + 8 Sum the numbers outside the ‘Have a sister’ circle = 17

Evaluate

Based on part (a), the total number of students is 28. Write the formula

Substitute the values

The probability that a selected student will not have a sister is

Idea summary Venn diagrams are a type of diagram that uses circles to group and organise things. The overlap or intersection, ∩, will usually use the word “and”. Combined areas or unions, ∪, will usually use the word “or”. The whole area contained in the rectangle is called the sample space.

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Two-way tables Two-way tables represent data that is classiﬁed by two criteria.

Enjoys drawing Doesn’t enjoy drawing

Right 4 2

The table has two criteria: a student’s main writing hand, and whether or not a student enjoys drawing.

Left 9 15

To read a two-way table, look at the column and row that a cell is in. For example, there are 9 students who are left-handed and enjoy drawing, and 2 students who are right-handed that don’t enjoy drawing. Tables will often include totals of each column, row, and the total sum. This is found by adding up every cell in that row or column. Enjoys drawing Doesn’t enjoy drawing Total

Right 4 2 6

Left 9 15 24

Total 13 17 30

The categories of the rows and columns should be chosen such that each person or object can only go in one of the cells.

Example 2 This table describes train departures from a train station during the months of May and June. May June

Departed on time 113 108

Departed late 31 33

a Complete the table by adding totals for the columns and rows.

Apply the idea May June Total

Departed on time 113 108 221

Departed late 31 33 64

Total 144 141 285

b How many trains departed during May and June?

Create a strategy Add all the values in the original table, or use the total from part (a).

Apply the idea Number of trains = 113 + 108 + 31 + 33 = 285

Add Evaluate

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c What percentage of the trains in June were delayed? Write your answer as a percentage to one decimal place.

Create a strategy Find the number of delayed trains in June as a percentage of the total trains departing in June.

Apply the idea According to the table in part (a), the total number of June trains is 141. Find the 33 delayed June trains as a percentage of the total number of June trains: Percentage delayed =

Write the fraction

= 0.234 04 …

Evaluate the division

≈ 23.4%

Convert to a percentage

Reﬂect and check This percentage also tells us that if a train in June was randomly selected, the probability that it would have been delayed is 23.4%.

d What fraction of the total number of trains across the two months departed on time in May?

Create a strategy Find the number of on time trains in May as a fraction of the total trains.

Apply the idea From part (a), we found that there are 285 total trains. We can see from the table that 113 departed on time in May. Write as a fraction

Reﬂect and check This fraction also tells us that the probability that a train selected at random departed on time in May is

Idea summary Two-way tables represent data that is classiﬁed by two criteria.

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Two-way tables and Venn diagrams A two-way table presents similar information to a Venn diagram. We can convert between a two-way table and a Venn diagram by matching up their parts. Lefthanded

Righthanded

Entered

Didn’t enter

The column marked “Left-handed” is a category used in the Venn diagram and the row marked “Entered” is the other category used.

Entered

Left-handed

This image shows how a typical two-way table and Venn diagram are related.

The number that is in both categories, 4, will go in the overlap of the two circles. The remaining value, 9, in the selected row, represents the “Entered and right-handed” students and will go into the “Entered” circle but not in the overlap because they are not “Left-handed”. Any cells that are in neither the highlighted row nor the highlighted column will go into the surrounding box.

Example 3 A student makes a Venn diagram of students who are late to school, and students who catch the bus to school.

Late to school

Caught bus

Using the Venn diagram, complete this table: Late

Not late

Total

Caught bus Didn’t catch bus Total

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Create a strategy We can use the Venn diagram to construct the table by looking at the corresponding regions.

Apply the idea The categories are Late/Not late and Caught bus/Didn’t catch bus.

Caught bus Didn’t catch bus Total

Late 13

Not late

Total

The top left cell is for the number of people who are late to school and caught the bus, that is, 13 people. The 9 that is written outside of the circles refers to 9 students who did not do either of these things, so it goes in the ‘not late’ and ‘didn’t catch a bus’ cell.

Total

3 is the number of students that caught the bus and are not late, so it goes in the top right cell. 7 is the number of students that did not catch the bus and are late, so it goes in the bottom left cell.

Late 13

Not late 3

Late 13

Not late 3

Total 16

The total for the ‘Caught bus’ row is 13 + 3 = 16, which matches the total inside the ‘Caught bus’ circle.

The total for the ‘Late’ to school column is 20, and this matches the total inside the ‘Late’ to school circle in the Venn diagram. We can complete the other totals by summing rows and columns.

Reﬂect and check We can check if the table is correct using the totals columns. The totals of the horizontal row (20 + 12) should be the same as the total for the vertical totals column (16 + 16).

Idea summary We can convert between a two-way table and a Venn diagram by matching up corresponding parts.

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10.02 Practice questions Understanding 1

Explorer 1–4

Adventurer

Trailblazer

1–4

Match each Venn diagram with what the shaded area represents. i

Sample space

A′, the complement of event A

iii

A ∩ B, the intersection of events A and B

A ∪ B, the union of events A and B

A B

b S

A gardener creates this diagram of her favourite plants:

Flowering plants

How many of the plants are ﬂowering plants?

How many of the plants are both ﬂowering and succulent?

Sunflower

How many of the plants are either ﬂowering or succulent, but not both?

Daisy

A B

Succulent plants

Orchid

Tulip

For the given two-way table, is each statement true or false?

A B

Christmas cactus

Aloe Vera

Crown of Thorns

Bamboo

Ivy

21 11 32

23 45 68

Total 44 56 100

Events A and B are complementary.

100 is the total number of outcomes.

We add all nine numbers in the table to get the total number of outcomes.

The number of favourable outcomes for event A is 21.

If 23 was a typo, and meant to be 32, that is the only number that we would need to change to correct the error.

Total

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38 students were asked whether they prefer online or in-person classes, and whether they like group projects. Data is recorded in this two-way table: Prefer online classes Prefer in-person classes

Ex 1

How many students prefer online classes?

How many students prefer online classes or like group projects, or both?

How many students prefer at most one of the two options (either online classes or liking group projects, but not both)?

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5–11, 13, 15, 16

5–10, 12, 13, 15, 16

A small magazine asked people from different states to send in a vote on whether they supported Daylight Saving Time. The results are shown in this ﬁgure: a

What proportion of people voted “YES”?

What proportion of “NO” votes are from NSW?

Considering just the voters from NSW and ACT, what proportion of the votes are NSW “YES” votes?

Clara creates a Venn diagram about the musical instruments she could assign to a student in the music class. She is told that the instrument will be chosen at random. a

What is the probability that the chosen instrument will be a stringed instrument that isn’t traditional?

What is the probability that the chosen instrument won’t be traditional?

Dislike group projects 5 13

Fluency 5

Like group projects 8 12

5–7, 9, 10, 14, 15, 17

NSW YES NO

ACT YES NO

QLD YES NO

NSW YES NO

ACT YES NO

QLD YES NO

NSW YES NO

ACT YES NO

QLD YES NO

NSW YES NO

QLD YES NO

Traditional

Stringed

What is the probability that the chosen instrument will be traditional?

Coach Aaron creates a Venn diagram about potential team captains based on their leadership skills and goal shooting ability:

Leadership skills

Good shooter

How many players are there on Coach Aaron’s team, excluding Aaron?

What is the probability that a randomly selected team captain has leadership skills or is a good shooter?

Coach Aaron thinks a player is suitable to be a captain if they have both leadership skills and are a good shooter. How many suitable players are there on the team?

Aaron is told that the team captain will be chosen randomly. What is the probability that the chosen captain will be suitable?

Mathspace WA Curriculum – Year 9 mathspace.co

A group of students were asked about the dogs or cats they have as pets and if they walk them regularly or not. The results are in the two-way table. Fill in the missing values.

10 Ex 2

Dogs Cats Total

240 cyclists were surveyed on whether they would support the use of smart glasses during competitive races. The results are detailed below: a

Fill in the missing value in the table.

What is the probability that a cyclist chosen at random is an amateur that does not support using smart glasses?

45 2

Does not walk regularly 20

Support Professional Amateur

Total

Do not support 48 40

The table below illustrates the take-offs of ﬂights from an airport for the months of May and June: Month May June Total

Walks regularly

Took off on time 135 140

Delayed 45 60

Total

Complete the table.

How many ﬂights took off during May and June?

Find the percentage of ﬂights in June that were delayed. Round your answer to one decimal place.

Find the percentage of the total number of ﬂights over the two months that took off on time in May. Round your answer to one decimal place.

Find the exact probability that a ﬂight selected at random in June would have taken off on time.

Find the exact probability that a ﬂight selected randomly over the 2 months was delayed.

A beverage preference survey asked individuals to select their preferred drink: tea or coffee. Alongside this, participants were also asked about their peanut butter preference. The table below summarises the responses:

Tea Coffee Total

Likes smooth peanut butter 45 40

Likes crunchy peanut butter 60 35

Total

Complete the table.

How many individuals participated in the survey?

If one person is chosen at random, what is the probability that they prefer tea and like smooth peanut butter?

If one person is chosen at random, what is the probability that they prefer coffee?

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15 Ex 3

In a study, some individuals were asked whether they enjoy listening to classical music. A partially completed two-way table of the results is shown: a

Complete the table.

Of those in the study, one is chosen at random. Find the probability that they said they don’t enjoy classical music.

Teenagers Seniors Total

Total

18 12

40 60

A group of individuals were asked whether they liked sushi or pizza. The result is summarised in the table: a

Complete the table.

How many people like both foods?

If one person is chosen at random, what is the probability that they like neither food?

If one person is chosen at random, what is the probability that they like only one of the foods?

Likes sushi Doesn’t like sushi

Likes pizza

Doesn’t like pizza

Total

In a study, some people were asked whether they were vegetarian or not. 27 responders said they were vegetarian, of which 8 were children. 18 children said they were not vegetarian, and 21 adults said they are not vegetarians. a

Complete the table.

Calculate the proportion of responders who are vegetarian.

Calculate the proportion of adults who are vegetarian.

What is the probability that a person chosen at random is an adult?

What is the probability that a person chosen at random is a child who is not a vegetarian?

Not Vegetarian vegetarian

Total

Children Adults Total

A librarian notes that 30 books are checked out in a day. The books are categorised based on whether they are ﬁction and whether they have been borrowed. a

Using the Venn diagram, complete the table: Borrowed

Available

Total

Fiction Nonﬁction Total

544

Don’t enjoy

Enjoy

What is the probability that a book is ﬁction and available?

What is the percentage of books that are ﬁction and have been borrowed?

Mathspace WA Curriculum – Year 9 mathspace.co

Fiction

Borrowed

Students in Luke’s class were asked if they owned a laptop and if they owned a tablet. The following two-way table shows the results of the survey: Owns a tablet Doesn’t own a tablet Total

Owns a laptop Doesn’t own a laptop 5 4 10 9 15 13

Total 9 19 28

Construct a Venn diagram that represents the information provided in the two-way table.

What is the probability that a student owns both a laptop and a tablet?

150 random individuals were surveyed regarding their preference for reading material format and their employment type. The individuals were categorised as either freelancers or fulltime employees, and as preferring physical or digital books. a

Construct a Venn diagram that represents the information provided in the two-way table. What is the probability that an individual prefers digital books and is a full-time employee?

Prefers physical books Prefers digital books Total

Freelancer

Full-time employee

Total

150

What is the proportion of individuals who are freelancers?

Is it true to say that individuals who are freelancers generally prefer physical books?

Reasoning

18, 19

18–20

The results from a recent survey showed that 56 people speak Spanish or French. Of these, 36 speak Spanish, and 18 speak both Spanish and French. The number of people who speak French is asked. Trent’s answer is 20 and Joanna’s answer is 38. Who is right? Why?

In a class, 5 people play both football and tennis, 13 people in total play tennis, and 11 in total play football.

How many people only play football?

If a random student is chosen from the group, what is the probability that the student only plays tennis?

Goran believes that 14 people only play 1 sport. Yujuan believes that 19 people only play 1 sport. Who is correct? Justify your answer.

Four more people enter the class and are asked to contribute their answers to the survey. They all said that they don’t play either. How does this affect your results from part (b)?

At a school, 142 students are asked to join either the soccer club or the basketball club (or both). 75 students join the soccer club and 78 students join the basketball club. a

Samuel calculates that 11 students joined both clubs. Explain how he got his answer.

Can we determine what is the most popular club from this survey? Why or why not?

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Problem-solving 21, 23, 25, 26, 30 21

22, 24, 26, 27, 30

How many of the movies are horror ﬁlms?

How many of the movies ﬁt into only one genre?

How many of the movies are an action films combined with at least one other genre?

If she randomly chooses a movie, what is the probability that it is a movie that is classiﬁed as:

546

Action

Comedy

Rosey is deciding which movie to watch from a streaming website. A Venn diagram of her options sorts movies into three categories based on genre:

5 3

6 13 Horror

action and horror?

comedy only?

iii

comedy?

comedy or action?

action and comedy?

not horror?

Linda is planning to redesign her garden. She catalogues her plants based on certain criteria. A Venn diagram of her garden sorts the plants into three categories: Flowering plants, plants that require direct sunlight, and plants taller than 1 metre.

22, 24, 26–30

How many plants are there in Linda’s garden?

Linda wants to choose a ﬂowering plant. What is the probability she will select a ﬂowering plant?

Linda is looking for a plant that ﬂowers but doesn’t grow taller than 1 metre. What is the probability that she will select a suitable plant?

Linda wants a ﬂowering plant that grows taller than 1 metre. What is the probability that she will select a suitable plant?

Flowering

Direct sunlight

4 7

2 13

Taller than 1 metre

A student is analysing the favourite art activities among her classmates. 22 people enjoy sketching, 19 enjoy pottery, and 7 enjoy both sketching and pottery. a

Construct a Venn diagram that represents the given information.

How many people only enjoy pottery?

What is the probability that a student randomly selected from the group: i

only enjoys sketching?

enjoys sketching or pottery but not both?

iii

enjoys sketching and pottery?

Mathspace WA Curriculum – Year 9 mathspace.co

In a group of 190 primary and senior students, 88 are primary students. The students fell into three categories of travel to school: by bus, car, or walking. 109 students get to school by bus, of which 52 are primary students. 50 students get to school by car, of which 20 are primary students. a

b 25

a senior student?

a senior student that catches the bus to school?

iii

either a senior student or a student that catches a bus but not both?

What is another way to describe the groups of students discussed in part (a)?

In a study, a number of people were asked whether they were musicians or not. 29 responders said they were a musician, of which 8 were children. 19 children said they were not musicians, and 12 adults said they were not musicians. a

What is the probability that a student chosen at random is:

What is the probability that a student chosen at random is: i

a musician?

an adult musician?

iii

a musician, but not an adult?

What is another way to describe the same groups from part (a)?

A manager is assessing the participation of employees in two different training programs: leadership development and project management. Out of a team of 40 members, 32 attended the leadership development program and 15 attended the project management training. a

Given that everyone in the team attended at least one program, ﬁnd out how many employees attended both programs.

Construct a Venn diagram that represents the given information.

What is the probability that an employee attended only the leadership development program?

A jeweller has sorted 25 gems based on two categories: • Based on the colour – emerald • Based on the cut - rectangular a

Put the information into a two-way table.

If a gem is selected at random, what is the probability that it is: i

An emerald or rectangular?

Not emerald or rectangular?

iii

Both emerald and rectangular or neither?

Ruby Emerald Amethyst

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The employees of Koala Airlines were discussing where they should hold their end of year party. Of all of the employees: • 53 would go to a restaurant. • 67 would go to a bowling alley. • 76 would go to a theme park. • 16 selected both the restaurant and bowling alley. • 29 selected both the bowling alley and theme park. • 17 selected both the restaurant and theme park. • 4 would not go to a restaurant or a bowling alley or to the theme park. • 5 would go to all three places.

Construct a Venn diagram that represents the given information.

How many people in total work at Koala Airlines?

If the restaurant isn’t available, what proportion of the employees still have another option?

What is the probability that an employee chosen at random had chosen: i

the bowling alley or the theme park?

the bowling alley?

iii

not the bowling alley?

A gardener observed that some plants were affected by pests. 8 plants showed signs of aphid infestation, 14 plants showed evidence of mite damage, and 12 plants had slug damage. Only one plant had signs of all three pests. There were two plants with aphid and mite damage but no slug damage; there were three plants with aphid and slug damage but no mite damage. There were 25 affected plants in total in the garden. a

Draw a Venn diagram to illustrate the information.

What is the probability that if they chose one affected plant at random, it would show mite and slug damage but not aphid damage?

It costs $1.20 per plant to treat for aphids and $2.50 per plant to treat for mites. The gardener cannot use the individual treatments together; instead, if a plant has both aphids and mites, they have to use the all purpose spray that costs $4.10 per plant. What is it going to cost to treat the aphids or mites?

The treatments were 60% successful. Create a new Venn diagram for the plants that are still affected, and work out the probability that a plant at random has aphids but not mites.

During a food quality check, a new device is introduced to determine if fruits have any pesticides. To validate its precision, a batch of 200 fruits was taken. Some are known to have pesticides and others are organic (have no pesticides). Given this two-way table:

Fruit has pesticides Fruit is organic Total

548

Detected Pesticides 40

Detected as Organic

Total 60

200

Complete the table.

For what percentage of fruits did the device detect pesticides?

Based on these results, what is the probability that the device accurately detects if a fruit has pesticides?

Mathspace WA Curriculum – Year 9 mathspace.co

10.03 Arrays and simple tree diagrams After this lesson, you will be able to… • interpret arrays to identify all possible outcomes and calculate probabilities. • use tree diagrams to track multi-stage experiments. • construct and interpret tree diagrams to identify outcomes and calculate probabilities. • solve problems involving arrays and tree diagrams.

Arrays A table, also known as an array, is useful for showing all possible outcomes of two-stage events in the rows and columns.

H T

1 H1 T1

2 H2 T2

3 H3 T3

4 H4 T4

5 H5 T5

6 H6 T6

For example, if we tossed 1 coin and 1 die, we can show the outcomes for the coin along the ﬁrst column on the left (H, T) and the outcomes for the die across the top row (1, 2, 3, 4, 5, 6).

Each cell in the array is an outcome of rolling a die and tossing a coin. There are 12 possible outcomes in the sample space. We can use the probability formula to determine probabilities from the array.

Example 1 The table shows the possible results for the event of rolling 2 six-sided dice and summing the total. 1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

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What is the probability that a player rolls two dice that sum to 8?

Create a strategy

Apply the idea From the table, we can see that there are 6 × 6 = 36 possible outcomes in total, 5 of which are an 8. 6

Substitute the values

Reﬂect and check Let’s check, the ﬁve ways that we can get an 8 are (5, 3), (3, 5), (6, 2), (2, 6) and (4, 4).

Example 2 This spinner is spun and a six-sided die is rolled. The product of their respective results is recorded.

a Create an array that shows the sample space for the experiment.

Create a strategy Place the die numbers on the left and spinner numbers on top, then ﬁll each cell with their product.

Apply the idea The array for the product of the spun spinner and rolled six-sided die is given by:

Die

550

1 2 3 4 5 6

2 2 4 6 8 10 12

Spinner 3 3 6 9 12 15 18

Mathspace WA Curriculum – Year 9 mathspace.co

5 5 10 15 20 25 30

8 8 16 24 32 40 48

b What is the probability of an odd product?

Create a strategy Use the array from part (a) then the probability formula.

Apply the idea Using the array, we can see that there are 24 total outcomes and 6 are odd products. This means that the probability of an odd product is: Write the formula

Substitute the odd and total outcomes

Simplify

Reﬂect and check The only way to get an odd product is to multiply two odd numbers. There are three odd numbers on the die, 1, 3, 5, and two odd numbers on the spinner, 3, 5, so there are 3 × 2 = 6 possible odd products. This matches what we found with the array.

c What is the probability of rolling a 5 on the die and scoring an even product?

Create a strategy Use the array from part (a) to ﬁnd the number of times a 5 is rolled and an even product is scored. Then use the probability formula.

Apply the idea Using the array in part (a), we can see that there are 24 total outcomes and 2 outcomes resulted from rolling a 5 and scoring an even product. This means that the probability of rolling a 5 on the die and scoring an even product is given by: Write the formula Substitute the 2 favourable outcomes and 24 total outcomes

Simplify

10.03 Arrays and simple tree diagrams mathspace.co

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d What is the probability of spinning a 3 on the spinner or scoring a product which is a multiple of 4?

Create a strategy Use the array from part (a) to find the number of times a 3 is spun or a product which is a multiple of 4 is scored. Then use the probability formula.

Apply the idea Based on the constructed array from part (a), there are 24 total outcomes and 16 outcomes that are either a spinner value of 3, or a product that is a multiple of 4. 5

3 24 6

12 20

32 8

5 40 10 15 25 6 48 12 18 30

This means the probability of spinning a 3 on the spinner or scoring a product which is a multiple of 4 is given by:

Write the formula

Substitute the 16 favourable outcomes and 24 total outcomes

Simplify

Idea summary An array is useful for showing all possible outcomes of two events. The rows display one event and the columns display the second event.

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Tree diagrams A tree diagram is useful in multi-stage experiments. The important components of the tree diagram are: • Branch • Probability (only shown when the events have unequal probabilities) • Outcome

Probability Branch

0.1

Outcome

0.9

T T

T H T

T H

Here is a three-stage example where all the outcomes are equally likely. There are 8 outcomes and each has a probability of : • TTT • TTH • THT • THH • HTT • HTH • HHT • HHH

In tree diagrams where all branches are equally likely, we can find the probability in two ways: 1. By counting the number of favourable branches and dividing by the total number of branches. 2. Using the complementary events and doing P ( A) = 1 − P ( A′) For some scenarios, the options for the second split of branches will be different from the first step because the option from the first step is no longer available.

Example 3 Three fair coins are tossed and the outcomes are displayed in the tree diagram shown.

T T

T H T

T H

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a List the sample space of the event.

Create a strategy List all of the possible outcomes from the tree diagram.

Apply the idea The array for all the possible outcomes of three fair coins tossed is given by: H H H H T T T T

1st Toss H H H H T T T T

2nd Toss 3rd Toss H H H T T H T T H H H T T H T T

b Find the probability of obtaining at least one head.

Create a strategy Express the number of times one or more heads occurs as a fraction of the total possible outcomes.

Apply the idea Based on the given tree diagram, one or more heads occurs on 7 occasions. This means that the probability of obtaining at least one head is

Reﬂect and check Another way to do this would be to reword this as “not all tails”. This would mean:

Use complementary events

Only one branch has all tails, TTT

Simplify

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c What is the probability of obtaining TTH in this sequence?

Create a strategy

Apply the idea

Look for how many times this outcome (TTH) occurs in the sample space.

Based on the given tree diagram, there is one occurrence of TTH out of 8 possible outcomes. This means that the probability of obtaining TTH in the sequence is

d What is the probability of obtaining two heads and a tail in any order?

Create a strategy

Apply the idea

Look for how many times we can get 2 heads and 1 tail.

Based on the given tree diagram, there are three possible ways to get 2 heads and one tail out of 8 possible outcomes. This means that the probability of obtaining THH in any order is:

We could have the tail first THH, the tail second HTH or the tail third HHT.

Idea summary A tree diagram is useful in tracking multi-stage experiments. Each branch represents one event of the multi-stage events.

10.03 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

Are these statements true or false? a

In a two-dimensional array, every row must contain the same number of elements.

We can use a two-dimensional array for a three-stage experiment.

We can use a tree diagram for a three-stage experiment.

Each cell in an array is equally likely.

Lukas uses a tree diagram to represent an experiment where a fair six-sided die is rolled three times. Does each pair of events have an equal probability? a

Rolling an even and an odd.

Rolling a ﬁve and not rolling a ﬁve.

Rolling a prime and not rolling a prime.

10.03 Arrays and simple tree diagrams mathspace.co

555

A player is rolling two dice and calculating their sum. They draw an array of all the possible dice rolls with the sum of two dice. List the sample space for the sum of two dice.

Fluency Ex 1

4, 6–8, 10–12

4–9, 11, 12, 13a

A player rolls 2 six-sided dice and the sum is recorded in an array.

less than 5

more than 6

7 or higher

an even number

What is the probability that the sum is 6 or less?

What is the probability that the sum is more than 6?

What is the sum of the probabilities from (a) and (b)?

Explain the sum in part (c).

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

George has 4 different coloured shirts: blue (B), white (W ), red (R) and yellow (Y ). He also has 2 different patterns of ties: dots (D) and stripes (S). Blue (B) B, D B, S

White (W ) W, D W, S

Red (R) R, D R, S

How many different combinations are possible?

Find the probability that he is wearing: A blue shirt and a striped tie

Yellow (Y ) Y, D Y, S

A yellow shirt and a dotted tie

iii

A red shirt

A red or white shirt

A striped tie

A white shirt or a striped tie

A coin is tossed and a die is rolled at the same time. a

Complete the array to show the possible outcomes: Heads Tails

556

3 4 5 6 7 8 9

Using the table from question 4, showing the results of the sum of 2 six-sided dice:

Dots (D) Stripes (S)

2 3 4 5 6 7 8

4, 5, 8, 9, 11, 13

1 2 3 4 5 6

What is the probability that sum is

1 2 3 4 5 6

1 2 3 4 5 6 7

1 H, 1

3 H, 3 T, 2

4 H, 4 T, 3

6 H, 6 T, 5

How many possible outcomes are there?

Determine the probability of rolling a 1.

Determine the probability of rolling an even number.

Determine the probability of tossing heads and rolling an even number.

Determine the probability of tossing tails and rolling an odd number.

Mathspace WA Curriculum – Year 9 mathspace.co

This spinner is spun and a six-sided die is rolled. The sum of their results is recorded. a

Construct an array to represent all possible outcomes.

Find the probability of an even sum.

Find the probability of rolling a 3 on the die and scoring an odd sum.

Find the probability of spinning a 4 on the spinner or scoring a sum of 7.

Find the probability of the sum being greater than 8.

These two spinners are spun and the result of each spin is recorded:

Complete the given array. How many possible outcomes are there?

b c

Find the probability that the spinner lands on a consonant and an even number.

Find the probability that the spinner lands on a vowel or a prime number.

1 2 3

1, A 2, C

Xavier is choosing an outﬁt for the day and has 3 shirts (cyan, pink and white) and 4 ties (black, grey, red and yellow) to select from. a

Complete the given array.

Ties

Ex 2

Black (B) Grey (G) Red (R) Yellow (Y )

Cyan (C) C, B C, G ⬚, ⬚ C, Y

Shirts Pink (P ) ⬚, B P, G P, R ⬚, ⬚

White (W ) W, B W, ⬚ W, R W, Y

How many different outﬁts are possible?

What is the probability that he wears a black tie?

What is the probability that he wears a white shirt?

What is the probability that he wears a pink shirt or a yellow tie, but not both?

10.03 Arrays and simple tree diagrams mathspace.co

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Ex 3

Each day Chris chooses one accessory, a hat or a tie. There is an even chance that Chris wears a tie or hat on any given day. The tree diagram represents the possibilities of Chris’ choice on three consecutive days.

H H

H T H

What is the total number of outcomes?

List the outcomes.

Find the probability that Chris:

Wears a hat on all three days.

Wears a hat on the ﬁrst day, a hat on the second day and a tie on the third day.

iii

Wears a hat once and a tie twice in any order over the three days.

Doesn’t wear a hat on the ﬁrst day

Wears a tie at least once

In a town in Australia, only ravens (R) and sparrows (S) are found. Samuel waits to see two birds. He is equally likely to see each type of bird. a

Find the probability that at least one of the birds he sees is a sparrow.

Find the probability that the last bird he sees is a raven.

Find the probability that he only sees 1 sparrow.

R S

All possible outcomes of red or black cards that a person could have if they draw three cards from a deck.

All the ways the names of three candidates; Alvin, Sally and Peter, can be listed on a ballot paper. 14, 15

14–16

Construct a tree diagram showing:

Reasoning

14–16

Students in a Mathematics class were asked to use either a tree diagram or an array to list the sample space, when rolling 2 dice. Which method would be more eﬃcient? Justify your answer.

A fair coin is tossed 8 times and the result was heads every time. What is the probability that the next toss will also be heads?

Frankie has a ten-sided die and Myah has a six-sided die. Do they both have the same chance of rolling a 6 and then a 2? Explain.

Problem-solving 17, 18 17

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17–19, 21

20–22

A fair die is rolled twice. Find the probability of rolling: a

A 3 and a 6 in any order

A 3 and then a 6

A double

An odd and an even number

Mathspace WA Curriculum – Year 9 mathspace.co

Two spinners labelled 1 to 4 are spun at the same time and the results are added: Spinner 1

Spinner 2

Find the probability of:

An even result

A result greater than 6

A result less than 5 and even

A result less than 5 or even

A two-digit number is formed by spinning the two spinners, with Spinner 1 being the ﬁrst digit and Spinner 2 being the second digit in the number: Spinner 1

Spinner 2

Find the probability of the number formed being:

Equal to 25

Even

Prime

Divisible by 5

Divisible by 5 and odd

Containing at least one 2

Two dice are rolled, and the combination of numbers rolled on the dice is recorded. a

find the probabilities for the two numbers rolled: i

P(1 and 4)

P(1 then 4)

iii

P(difference = 4)

P(product = 12)

P(difference ≤ 2)

P(difference = 3)

vii P(product = 20) b

viii P(difference ≤ 1)

If the numbers appearing on the uppermost faces are added, determine whether these statements are true or false: i

A sum greater than 7 and a sum less than 7 are equally likely.

A sum greater than 7 is more likely than a sum less than 7.

iii

A sum of 5 or 9 is more likely than a sum of 4 or 10.

An even sum is more likely than an odd sum. 10.03 Arrays and simple tree diagrams mathspace.co

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These two spinners are spun and the product of their spins is recorded. Find the probability that:

The product is less than 15.

The product is odd.

A 7 was spun, given that the product is less than 20.

The product is less than 40.

The product is odd, given that a 4 was spun.

5 2

Melinda loves board games and has 4-sided, 6-sided, 10-sided, 12-sided, and 20-sided dice. Two different dice are rolled and their sum is recorded. a

If the probability of a sum greater than 30 is

If the probability of a sum of 3 is

If the probability of an even sum is , which two dice were rolled?

, which two dice were rolled?

10.04 With and without replacement After this lesson, you will be able to… • differentiate between selecting items with and without replacement. • construct tree diagrams to represent multi-stage experiments. • calculate probabilities for scenarios where items are either replaced or not replaced. • evaluate real-world examples to apply the concept of replacement in probability.

With and without replacement Selecting items with replacement means putting each selected item back before you select another item. When we select with replacement, the probabilities at each stage stay the same. Sample space B

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Mathspace WA Curriculum – Year 9 mathspace.co

For example, when randomly selecting a pen from a pencil case with two blue pens and one red pen, the probability of getting a blue pen on the ﬁrst selection is . If we get a blue pen ﬁrst and replace it before we make the second selection, the probability of getting a blue pen on the second selection is still , as shown in the tree diagram. The probability of selecting two blue pens is .

Selecting items without replacement means that we don’t put each selected item back before selecting another. When we select without replacement, the probabilities at each stage will change. Sample space B

In the same pencil case, if we get a blue pen ﬁrst and don’t replace it, then for our next selection there is still one red pen, but only one blue pen remaining. This means that the probability of getting a blue pen on the next selection is . The probability of selecting two blue pens is now .

When constructing tree diagrams for experiments without replacement, we must think carefully about the number of each item left after each selection, and change our probabilities accordingly.

Exploration To decide which chores Jacques needs to do, he pulls out pieces of paper from a hat. The options are sweeping (S), mopping (M), and vacuuming (V). No chore is done twice on the same day. Day 1 chore

Day 2 chore

Day 1 chore S

V M

V V

Day 2 chore S M V S M V S M V

Tree diagram 1: two chores on the same day Tree diagram 2: two chores on different days 1.

Which tree diagram shows the experiment with replacement?

What is the probability he will need to sweep and mop if they are done on the same day?

What is the probability he will need to sweep and mop if they are done on different days?

Is there a difference in the probabilities when selecting with or without replacement?

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Example 1 Esther was given four animal crackers as a snack. She has a donkey (D), an elephant (E), a goat (G), and a hippo (H). She eats one cracker and then another. a Is this an example of with or without replacement?

Create a strategy With replacement means the ﬁrst choice is still available for the second choice. Without replacement means the ﬁrst choice is eliminated in the second selection.

Apply the idea Since the ﬁrst cracker is eaten, it is not available in the second round. This means it is without replacement.

b List the sample space for the two crackers she eats if the order she eats them matters to her.

Create a strategy Construct a tree diagram to easily list all the outcomes, making sure that none are missed.

Apply the idea First cracker

Second cracker Sample space E G H

DE DG DH

D G H

ED EG EH

D E H

GD GE GH

D E G

HD HE HG

So the sample space is {DE, DG, DH, ED, EG, EH, GD, GE, GH, HD, HE, HG}.

Reﬂect and check Since for the ﬁrst animal cracker she has 4 choices and then after eating one she has 3 choices for the next one, it makes sense that there are 4 × 3 = 12 outcomes in the sample space.

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Example 2 Construct a tree diagram showing all the ways a captain and a vice-captain can be selected from Matt, Rebecca and Helen.

Create a strategy Any of the three people can be chosen as captain. Once a captain has been selected, any of the two remaining people can be vice-captain.

Apply the idea There are three names to be selected from, so this means that there are 6 possible outcomes.

Captain

Vice-captain

Matt

Rebecca Helen

Rebecca

Matt Helen Matt

Helen

Rebecca

Example 3 Four cards numbered 1 to 4 are placed face down on a table. Two cards are drawn with replacement. a Construct a table array of this situation.

Create a strategy Since this is with replacement, our options will be the same for both draws.

Apply the idea For the ﬁrst selection, we have 1, 2, 3, 4 as possible options. Since each card is replaced, the options will stay the same for the second selection.

1 2 3 4

1 2 3 (1, 1) (1, 2) (1, 3) (2, 1) (2, 2) (2, 3) (3, 1) (3, 2) (3, 3) (4, 1) (4, 2) (4, 3)

4 (1, 4) (2, 4) (3, 4) (4, 4)

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b Calculate the probability that the two cards are both even.

Create a strategy Use the array table from part (a).

Apply the idea From the array table, we can see that there are 16 outcomes in the sample space. Four outcomes are our desired outcome with both numbers being even (2, 2), (2, 4), (4, 2), (4, 4). Therefore, the probability that the two cards are both even is .

1 2 3 4

1 2 3 (1, 1) (1, 2) (1, 3) (2, 1) (2, 2) (2, 3) (3, 1) (3, 2) (3, 3) (4, 1) (4, 2) (4, 3)

4 (1, 4) (2, 4) (3, 4) (4, 4)

Reﬂect and check A good check is that the size of the sample space is the number of possibilities in the first step times the number of possibilities in the second step. Sample space size = 4 × 4 = 16 which matches what we saw in the array table.

c Construct an array table that lists the sample space if the experiment is done without replacement.

Create a strategy Since this is without replacement, the first card is eliminated in the second selection.

Apply the idea When the experiment is done without replacement, the options for the second selection will change. The table array will show only valid outcomes where the first and second draws are different: In this array, the diagonal elements are excluded because a card cannot be drawn twice when there is no replacement. The sample space without replacement consists of the remaining 12 pairs.

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1 2 3 4

2 (1, 2)

3 4 (1, 3) (1, 4) (2, 1) (2, 3) (2, 4) (3, 1) (3, 2) (3, 4) (4, 1) (4, 2) (4, 3)

Example 4 Qian packed three shirts in his suitcase for a holiday. Two are pink and one is white. He selects two at random to bring on a day trip. a Construct a tree diagram of this situation.

Create a strategy Since he is bringing two distinct shirts, this is with out replacement.

Apply the idea For his first choice, he will have three shirts to choose from. For his second choice, he will only have two shirts to choose from. The options for the second shirt will change based on the first choice.

First shirt

Second shirt

Sample space

b Find the probability that Qian selects at least one pink shirt.

Create a strategy Use the tree diagram from part (a).

Apply the idea

Sample space

We can highlight the desired outcomes on the tree diagram. P

Start with the probability formula

Substitute

Simplify

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Reﬂect and check He must bring at least one pink shirt as he only has one shirt that isn’t pink. A probability of 100% means we are certain this will happen.

c Find the probability that Qian selects at most one pink shirt.

Create a strategy Use the tree diagram from part (a).

Apply the idea

Sample space

We can highlight the favourable outcomes. P

P(at most 1 pink shirt)

Start with the probability formula

Substitute

Simplify

Reﬂect and check This would be the same as the probability of him choosing one pink shirt and one white shirt.

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Example 5 Noive selects two letters from the word ONTO without replacement to see if a new word is formed. a Is the outcome NN possible?

Create a strategy Look at the number of letter Ns which remain after one N has been selected.

Apply the idea No, the outcome NN is not possible. There is only one N and letters are being selected without replacement, so once one N is selected first, there are none available for the second selection.

Reﬂect and check It is possible to have a double letter as an outcome, but there must be a double in the original word, for example OO. b Find the probability that the word ON is formed, in that order.

Create a strategy Create a tree diagram showing each of the possible two letter outcomes. Then using the tree diagram, we can divide the number of outcomes with the word ON by the total number of outcomes.

Apply the idea

Sample space

For the ﬁrst letter selection, it can be any of the letters of ONTO.

For the second letter, we will have to choose from the remaining three letters as this is without replacement.

Start with the probability formula

Substitute

Simplify

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c Find the probability that the word ON or NO is formed.

Create a strategy Use the tree diagram from part (b).

Apply the idea

Sample space

From the tree diagram, there are four outcomes which spell ON or NO.

Start with the probability formula

Substitute

Simplify

Reﬂect and check Another strategy is to ﬁnd the probability of ON and the probability of NO and add them together. There is no overlap between these outcomes, so we don’t need to subtract it. Sum the probabilities of the outcomes

Substitute

Add the fractions

Simplify

Idea summary Selecting items with replacement means putting each selected item back before you select another item. Selecting items without replacement means that we don’t put each selected item back before selecting another. When constructing tree diagrams for experiments without replacement, we must only list the options that are still available after the ﬁrst selection.

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10.04 Practice questions Understanding 1

Explorer 1–4

Adventurer 1–4

Trailblazer 1–4

Identify whether each experiment is with replacement or without replacement. a

A card is randomly selected from a normal deck of cards, and then returned to the deck. The deck is shuﬄed and another card is selected.

Each of four students is allowed to randomly pick an item from the teacher’s prize bag (and keep it).

A butterﬂy is captured and its gender is recorded. It is released back into the garden before another butterﬂy is captured.

Without looking, a child selects a chocolate from a box of milk and dark chocolates. They eat it, then select another.

For the same experiment done with and then without replacement, will the sample space change or stay the same?

A bag has 3 pink marbles and 2 blue marbles. Two marbles are drawn without replacement. Is each statement true or false?

The sample space will have 5 items in it.

The sample space will have 20 items in it.

The sample space is {PP, PB, BP, BB} and each outcome is equally likely.

It is more likely to draw two pink marbles than two blue marbles.

Three cards labelled 2, 3, and 4 are placed face down on a table. Two of the cards are selected randomly to form a two-digit number. The outcomes are displayed in the tree diagram. List the sample space.

4 2

Fluency 5

6–9, 11, 14–17

With replacement

Without replacement

A container holds four counters coloured red, blue, green and yellow. Draw a tree diagram representing all possible outcomes when two draws are done, and the ﬁrst counter is: a

Ex 1

5–9, 11, 12, 14, 15, 17

A bag has 2 blue marbles, 1 white marble, and 1 green marble. Draw a tree diagram for drawing two marbles. a

5–10, 12–14, 17

Replaced before the second draw

Not replaced before the next draw

Aditya has three lollies. A blue one, a green one, and a yellow one. They take one, eat it, and then eat one more. a

Is this experiment with or without replacement?

List the sample space for the possible lollies they eat. 10.04 With and without replacement mathspace.co

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Ex 2

A project team consists of four members: Sarah, David, Olivia, and Michael. The team needs to designate a leader and a coordinator. Construct a tree diagram to demonstrate all the possible ways a leader and a coordinator can be chosen.

Three cards labelled 7, 8 and 9 are placed face down on a table. Two of the cards are selected randomly without replacement to form a two-digit number.

Ex 3

Construct a tree diagram of this situation.

List the sample space for the possible two-digit numbers formed.

What is the probability that both cards are odd?

A bucket contains 3 green buttons and 1 black button. Two buttons are selected from the bucket. The ﬁrst button is replaced before the second button is selected. a

Construct a tree diagram of this situation.

Find the probability of selecting two black buttons.

A bag has ﬁve counters labelled 1 to 5. Two counters are drawn and the larger number is recorded as the outcome. a

Complete this array that lists the sample space if the experiment is done with replacement. 1 1

1 2 3 4 5 b

1 2 3 4 5

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3 3 3 3

4 4 4 4 4

5 5 5 5 5 5

1 3 4

Complete this array that lists the sample space if the experiment is done without replacement. 1

2 2 2

2 2

3 3 3

4 4 4 4

5 5 5 5 5

If the experiment is done without replacement, ﬁnd the probability of the larger number being: i

Even and prime

Odd

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iii

Even or prime

Ex 4

A container holds three cards coloured red, blue and green. a

Draw a tree diagram representing all possible outcomes when two draws occur, and the card is not replaced before the next draw.

Find the probability of drawing the blue card ﬁrst.

Find the probability of drawing a blue card in either the ﬁrst or second draw.

While playing a card game, Xander has a 9, a 6, a 5, and a 3. Two cards are drawn without replacement. a

Construct a tree diagram for the situation.

Calculate the probability that:

At least one of the cards is a multiple of 3

Construct a tree diagram of this situation.

Find the probability that Elliott-Rose selects: i

At least one cotton dress

Two cotton or two linen dresses

iii

At most one cotton dress

No cotton dresses

Two dresses of the same material and at least one linen dress

A band has three guitarists and one bassist. Two different musicians are chosen for solos at random. Find the probability of the two soloists being: Guitarists

Bassists

A number game uses a basket with 6 balls, all labelled with numbers from 1 to 6. Two balls are drawn at random. Find the probability that the ball labelled 3 is picked once if the balls are drawn: a

Ex 5

Neither of the cards are odd

Elliott-Rose has five dresses. Three are cotton and two are linen. She selects two for a weekend away.

a 16

With replacement

Without replacement

Two letters are randomly selected in order and without replacement from the word MONEY. a

List the sample space.

Y is considered a vowel here. Find the probability of: i

The word MY

The word ME or MY

iii

The letters having at least one vowel

Starting with an M or an N

Starting with an M and ending with a vowel

Starting with an M or ending with a vowel

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Reasoning

18, 20

18–20

Explain how selection with or without replacement affects the number of outcomes and branches in a tree diagram.

In a bag, there are 3 red marbles and 2 yellow marbles. Two marbles are drawn. This tree diagram shows the situation. a

Are the marbles drawn with or without replacement? Explain how you know.

Why do you think it is written as R1, R2, and R3 instead of writing them all as just R? Explain.

R1 R2 R3 Y1 Y2 R1 R2 R3 Y1 Y2 R1 R2 R3 Y1 Y2 R1 R2 R3 Y1 Y2 R1 R2 R3 Y1 Y2

Two letters are randomly selected in order and without replacement from the word YUMMY. a

Is the outcome YY possible? Explain.

Give an example of an outcome that is more likely than UM. Justify your answer.

Explain how the probability of getting MY would change if this was done with replacement.

Problem-solving 21, 23 21

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22–25

In a bakery, there are 6 wrapped pastries. Three are chocolate, two are custard, and one is strawberry. A customer buys two pastries at random. What is the probability that the customer gets: a

21–24

Two chocolate pastries

At least one chocolate pastry

A bag has counters with different numbers on them. Two counters are drawn without replacement. The smaller number is recorded as the outcome. a

Describe an event that would have a probability of 0.

Describe an event that would have a probability of 1.

Describe an event that would have a probability of less than 0.5.

Mathspace WA Curriculum – Year 9 mathspace.co

4 6 7

At the start of the week, Milana has three bananas and two apples in her fruit bowl. Each morning before school, she selects a piece of fruit at random to take and eat at lunchtime.

Is this experiment with replacement or without replacement?

Determine the probability that: i

She eats apples on Monday and Tuesday.

She eats an apple on Tuesday.

iii

There are no bananas left on Thursday morning.

A hand has ﬁve cards numbered from 3 to 7. Two cards are drawn and their product is the outcome. a

If the experiment is with replacement, show how an array could be used to list the sample space.

If the experiment is without replacement, show how an array could be used to list the sample space.

If the hand had 20 cards, how many ﬁlled cells would there be in the array: i

With replacement

Without replacement

There are four numbered cards. Two of the cards are randomly chosen and the possibilities for the sum of their numbers is: {15, 10, 8, 11, 9, 4}

Find the missing number on the fourth card.

If two cards are chosen at random, ﬁnd the probability that the sum of their numbers is: i

Even

10.05

At least 10

Multi-stage probability

After this lesson, you will be able to… • understand and apply the concept of tree diagrams with probabilities for experiments with and without replacement. • calculate probabilities for multi-stage experiments using tree diagrams. • construct tree diagrams for various multi-stage probability problems.

Tree diagrams with probabilities We have seen that the probabilities of outcomes are different depending on if the experiment was done with or without replacement. The more outcomes, the harder it can be to list the outcomes in 10.05 Multi-stage probability mathspace.co

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a simple tree diagram.

Exploration One tree diagram does not have probabilities on the branches, while another does. Sample space C

C C L L

CC CC CL CL

C C L L

CC CC CL CL

C C L L

CC CC CL CL

C C C L

LC LC LC LL

C C C L

LC LC LC LL

2 4

Without probabilities on the branches

With probabilities on the branches

C 3 5

2 5

2 4

3 4

1 4

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Could these tree diagrams represent the same situation? Explain.

For the tree diagram without probabilities on the branches, how could you ﬁnd the probability of CC?

Knowing the probability of CC from question 2, how could you ﬁnd the probability of CC from the other tree diagram?

What are the beneﬁts and drawbacks of each of the diagrams?

Mathspace WA Curriculum – Year 9 mathspace.co

When the outcomes are not equally likely, the probability will be written on the branches. In this example, there are probabilities on the branches.

W 60% 30%

If you add the probabilities in each trial, the sum of the branches equals to 1 (or 100%). This also helps us to check that all the outcomes are listed.

10% D

Here is an example of a multi-stage experiment, where the probabilities are written on the branches. D

0.7 0.7

0.3

0.7

0.3

0.7

0.3

0.7

0.3

0.7

0.3 N

This tree diagram shows that the probability that a child gets home during the day is P (D) = 0.7 and the probability that a child gets home during the night is P (N ) = 0.3. This is looked at over three days.

By multiplying across the branches of the tree diagram, from left to right, we get the probability of that outcome. So, P(DDD) = 0.7 × 0.7 × 0.7 = 0.343. Next, we get that the probability of P(DDN) = 0.7 × 0.7 × 0.3 = 0.147. When we add the probabilities of all the possible outcomes, it should equal to 1 (or 100%). When constructing tree diagrams for experiments without replacement, we must think carefully about the number of each item left after each selection, and change our probabilities accordingly.

Example 1 A standard deck of 52 cards has four queens. Ilham shuffles the deck then selects two cards, with replacement, noting whether or not they got a queen for each selection. a Construct a tree diagram of this situation with the correct probability on each branch.

Create a strategy Since each card is replaced, the probabilities will stay the same for the second selection. The total number of outcomes will be 52. The number of favourable outcomes will be the number of queens in the deck.

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Apply the idea For the ﬁrst and second selections, there are 52 cards in total, with 4 queens and 48 cards that are not queens. Let Q represent the event of drawing a queen and N represent the event of not drawing a queen.

4 52

Q 48 52 N

4 52

48 52 N

Q 48 52 N

b Calculate the probability that Ilham selects only one queen.

Create a strategy Use the tree diagram from part (a).

Apply the idea Let Q represent the outcome of drawing a queen. Let N represent the outcome of not drawing a queen. Ilham selects only one queen in the outcomes QN and NQ. Add the probabilities of each outcome Multiply along the branches of the tree diagram to calculate probabilities

Evaluate the multiplications

Evaluate the sum

Simplify

Reﬂect and check Notice that N is the same as Q′ since P (Q) + P (N ) = 1.

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Example 2 Noive selects two letters from the word PEPPER, without replacement. a Is the outcome RR possible?

Create a strategy

Apply the idea

Look at the number of Rs left after one R has been selected.

No, the outcome RR is not possible. There is only one R, and letters are being selected without replacement, so once one R is selected first, there are none available for the second selection.

b Find the probability that the letters P and R are selected, in that order.

Create a strategy Create a tree diagram with the correct probabilities on each branch, by considering the number of each item (and the total number of items) at each stage of the experiment.

Apply the idea For the first selection, there are 6 letters in total, with 3 of the letter P, 2 of the letter E and 1 of the letter R. For the second selection, there are 5 letters remaining: • If a P was selected first, there are 2 of the letter P, 2 of the letter E and 1 of the letter R. • If an E was selected first, there are 3 of the letter P, 1 of the letter E and 1 of the letter R. • If an R was selected first, there are 3 of the letter P, 2 of the letter E and none of the letter R.

2 5

P 2 5

1 5

3 6 2 6

3 5

1 5 3 5

1 6 R

E R P

1 5

E R P

2 5

Multiply along the branches of the tree diagram

Evaluate

Simplify

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c Find the probability that the letters P and R are selected, in any order.

Create a strategy Use the tree diagram from part (b).

Apply the idea Probability = P(PR) + P(RP ) From part (b) we know that P(PR) =

Sum the probabilities of the outcomes .

Multiply along the branches of the tree diagram

Evaluate

Simplify

Substitute the probabilities

Evaluate

Simplify

Reﬂect and check In general, order does not matter when we are making selections, so P ( AB) = P (BA).

Example 3 Luke plays three tennis matches. In each match, he has a 60% chance of winning: first match

second match

60%

40%

60%

40%

60%

40%

60%

40%

W W 60%

60%

40%

60%

40% L

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third match

a Find the probability that he will win all his matches.

Create a strategy Multiply the probabilities along the branches of the tree that lead to the event that he will win all three matches.

Apply the idea Based on the tree diagram, the probability that he will win his ﬁrst match is 0.6. The probability that he will win his second match is 0.6. The probability that he will win his third match is 0.6. 60%

40%

60%

40%

60%

40%

60%

40%

W 60% W 60%

40%

60%

40% L

The probability that Luke will win all his matches is: Probability = 0.6 × 0.6 × 0.6

Multiply all the probabilities

= 0.216

Evaluate the multiplication

b Find the probability that he will lose all his matches.

Create a strategy Multiply the probabilities along the branches of the tree that leads to the event that he will lose all three matches.

Apply the idea Based on the tree diagram, the probability that he will lose his ﬁrst match is 0.4. The probability that he will lose his second match is 0.4. The probability that he will lose his third match is 0.4. 60%

40%

60%

40%

60%

40%

60%

40%

W 60% W 60%

40%

60%

40% L

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The probability that Luke will lose all his matches is: Probability = 0.4 × 0.4 × 0.4

Multiply all the probabilities

= 0.064

Evaluate the multiplication

Reﬂect and check This is the least likely outcome as it is the product of all of the least likely outcomes at each stage.

c Find the probability that he will win more matches than he loses.

Create a strategy Use the fact that Luke needs to win more matches than he loses, so he must win either 2 or 3 matches.

Apply the idea Based on the tree diagram, we need to get the sum of the probabilities of the branches that result in 2 or three wins. i.e. WWW, WWL, WLW, LWW. 60%

40%

60%

40%

60%

40%

60%

40%

W 60% W 60%

40%

60%

40% L

The probability that Luke will win more matches than he loses is: Probability = P(WWW) + P(WWL) + P(WLW) + P(LWW) = (0.6 × 0.6 × 0.6) + (0.6 × 0.6 × 0.4) + (0.6 × 0.4 × 0.6) + (0.4 × 0.6 × 0.6) = 0.648

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Idea summary A tree diagram is useful in tracking multi-stage experiments. To find the probability we can: 1. Multiply along the branches to calculate the probability of individual outcomes. 2. Add down the list of outcomes to calculate the probability of multiple options. 3. The final percentages should add to 100%, or the final fractions should add to 1. When constructing tree diagrams for experiments without replacement, we must think carefully about the number of each item left after each selection, and change our probabilities accordingly.

Multiply probabilities We have seen that we can multiply along the branches of a tree diagram to ﬁnd the probability of an outcome in a multi-stage probability experiment. Instead of drawing the whole tree diagram, we can use what the probability will be for each step, based on the previous step, and then multiply them.

Exploration These two tree diagrams show drawing two cards from a standard and well shuﬄed deck of cards. The colour of the card is noted. 26 52

R R

R 26 52

26 52

25 51

26 52

26 51

26 52

26 51

25 51

26 52 B

B 26 52

With replacement

Without replacement

Without drawing another tree diagram, how could you ﬁnd the probability of drawing three red cards?

For the experiment with replacement, which outcomes have the same probabilities? Why do you think that is?

For the experiment without replacement, which outcomes have the same probabilities? Why do you think that is?

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We can find the probability of multi-stage events by multiplying the probabilities at each stage. The probabilities at each stage might stay the same or change.

Example 4 Nino randomly selects three cards from a standard deck of 52 cards. Find the probability that he selects a jack, then a queen of the same suit, then a king of the same suit, if his selections are made: a With replacement

Create a strategy Consider the number of each item at each stage in the experiment.

Apply the idea There are 4 jacks in the deck, so the probability that Nino selects a jack ﬁrst is

No matter which suit of jack is selected on the ﬁrst pick, there is only 1 queen of the same suit, so the probability he selected this is

. The probability he selected a king of the same suit is also

Substitute

Evaluate

b Without replacement

Create a strategy Consider the number of each item at each stage in the experiment.

Apply the idea As in part (a), the probability that Nino selects a jack ﬁrst is

If that card isn’t replaced, there are now only 51 cards remaining, so the probability that he selects the queen of the same suit next is

After that selection, there are only 50 cards remaining, so the probability that he then selects the king of the same suit is

Substitute

Evaluate

Reﬂect and check You may notice that our answer here is larger than the answer to (a). This makes sense, because when each selected card is not replaced, the probability of picking any other card increases slightly.

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Example 5 A bag contains ten pieces of paper with the numbers 1 to 10. Three pieces of paper are drawn without replacement. Find the probability of: a All numbers being even.

Create a strategy Since this is without replacement, we need to ﬁnd the probability of an even number at each stage. Then we can multiply them all together.

Apply the idea Initially, there were 10 numbers and 5 of them are even: 2, 4, 6, 8, 10. For the ﬁrst draw, the probability of drawing an even would be

The second draw will only have 4 even numbers and a total of 9 numbers left. The probability of drawing an even on the second draw is . The third draw will only have 3 even numbers and a total of 8 numbers left. The probability of drawing an even on the third draw is .

Substitute

Multiply

Simplify

The probability of drawing three even numbers is

Reflect and check The actual number we select is not important, we only need to know if it is even or odd. Another way to look at this would be: We have five even and five odd numbers to start: EEEEEOOOOO, so P (Even first)

b At least one number being even.

Create a strategy There are many ways for this to happen, so adding up all of those possibilities will take a long time. Instead, we can recognise that this is the same as there being not all odd.

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Apply the idea The probability of all odds, will be the same as the probability of all evens. We have ﬁve even and ﬁve odd numbers to start: EEEEEOOOOO, so P (Odd first)

Complementary events

Substitute

Multiply

Subtract

The probability of selecting at least one even number is

Reﬂect and check This outcome is highly likely.

Idea summary We can multiply the probabilities of each step of a multi-stage experiment without drawing a tree diagram. The probabilities for the second and third stages must consider what happened in the previous stages.

10.05 Practice questions Understanding 1

584

Explorer 1–4

Adventurer 1–4

Trailblazer 1–4

Are these statements true or false? a

If there are no probabilities shown on the branches of a tree diagram, then each branch is equally likely.

To ﬁnd the probability of an outcome on a tree diagram, we can add along the branch to ﬁnd its probability.

The sum of the probabilities of all outcomes is 100% or 1.

We must draw a tree diagram to determine multi-stage probabilities.

Each outcome in a sample space is always equally likely.

Alexis uses a tree diagram to represent an experiment where a fair six-sided die is rolled three times. A “win” is rolling a 6 and a “loss” is not rolling a 6. a

What probability should go on the “win” branch?

What probability should go on the “loss” branch?

Mathspace WA Curriculum – Year 9 mathspace.co

A fair coin is tossed and then the spinner is spun: a

Construct a tree diagram representing the situation.

Will each outcome be equally likely?

R Y

Y G

A bag has 3 green marbles, 5 pink marbles, and 2 yellow marbles. Two marbles are drawn without replacement. a

How many marbles are left after the ﬁrst draw?

How many pink marbles are left if a green was drawn ﬁrst?

How many pink marbles are left if a pink was drawn ﬁrst?

Fluency Ex 1

5, 6, 8–10, 12, 14–16

5, 6, 8, 9, 11–13, 15, 16

5, 7–9, 11–13, 15, 17

A bucket contains 5 green buttons and 7 black buttons. Two buttons are selected one after another from the bucket. The ﬁrst button is replaced before the second button is selected. a

Construct a tree diagram of this situation with the correct probability on each branch.

Find the probability of selecting two black buttons.

A remote community does not always have access to fresh fruit. For two days in a row, they have fresh berries. On both days, the probability being able to get raspberries (R) is 0.6 and being able to get strawberries (S) is 0.4. a

Find the probability that someone was able to get strawberries at least once.

Find the probability that the fruit available on the last day were raspberries.

Find the probability that strawberries were only available for one day.

0.6 R 0.6

0.4

S R

0.6

0.4 S

0.4

Students are given a choice of activity for free time. Of the 7 students in the room, four chose a board game (G) and three chose a book (B). Two students are selected to deliver a message to the oﬃce. a

Construct a tree diagram of this situation with the correct probability on each branch.

Find the probability that both of the students selected were playing a game.

Find the probability at least one book reader is selected.

10.05 Multi-stage probability mathspace.co

585

Ex 2

Two marbles are randomly drawn without replacement from a bag containing 1 black, 3 green and 5 pink marbles. a

Construct a tree diagram to show the sample space.

Find the probability of drawing: i

A black marble and a pink marble, in that order.

A green marble and a black marble, in that order.

iii

2 green marbles.

No green marbles.

2 black marbles.

A pink marble and a green marble, in that order.

vii A pink and a green marble, in any order. Ex 3

An archer has three arrows that each have a probability of arrows are shot at a target: a

Construct a tree diagram with the probability on each branch.

Find the probability that all three arrows will hit the target.

Find the probability that at least one arrow will miss the target.

Find the probability that at least one arrow will hit the target.

From a standard pack of cards, one card is randomly drawn and then put back into the pack. A second card is then drawn. Calculate the probability that: a

Ex 4

Both cards are red

Both cards are of different colours

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At least one of the cards is a diamond

Both cards are the same colour

A number game uses a basket with 6 balls, all labelled with numbers from 1 to 6. Two balls are drawn at random. Find the probability that the ball labelled 3 is picked once if the balls are drawn: With replacement

Without replacement

A standard deck of cards is used and 3 cards are drawn out. Find the probability, in fraction form, that: a

All 3 cards are diamonds if the cards are drawn with replacement.

All 3 cards are diamonds if the cards are drawn without replacement.

Three letters are randomly selected from the word MATHEMATICAL. Find the probability that the letters M, A, and T are selected, in that order, if the selection is made: a

Ex 5

Neither of the cards are diamonds

Amelia randomly selects two cards, with replacement, from a normal deck of cards. Find the probability that:

a 13

of hitting a target. If all three

With replacement

Without replacement

There are 4 green counters and 8 purple counters in a bag. Two counters are drawn without replacement. a

Find the probability of choosing a green counter, then choosing a purple counter.

Find the probability of choosing a purple counter ﬁrst or choosing two greens.

Three marbles are randomly drawn with replacement from a bag containing 6 red, 4 yellow, 3 white, 2 black and 4 green marbles. Find the probability of drawing: a

Three white marbles

No green marbles

At least one red marble

At least one white marble

Mathspace WA Curriculum – Year 9 mathspace.co

Beth randomly selects three cards, with replacement, from a normal deck of cards. Find the probability that: a

The cards are queen of diamonds, king of spades, and king of diamonds, in that order.

The cards are all black.

The ﬁrst card is a 9, the second card is a heart and the third card is red.

The cards are all hearts.

None of the cards is a 9.

Reasoning 18

18, 20

18–21

In the black or red game, you guess whether the next card is black or red. The deck of cards is re-shuﬄed every round. In this particular game, the previous 11 cards drawn were black. a

What is the probability that the next card is also black?

Would this be the case if the cards were not re-shuﬄed and just put into a discard pile? Explain your answer.

Explain how selection without replacement affects the probabilities in a tree diagram. In your answer, refer to the numerator and the denominator of the fractions for the probabilities.

4 gardeners and 3 bakers are at a craft show. Two winners are randomly selected for the large and small prize baskets. The large prize is selected first. We want to find the probability of a gardener winning the large basket. Atanasio says that we can just look at the first branch and

Who is correct? Explain.

4 7

1 2

2 3

3 7

1 3

Five six-sided dice are rolled. Explain how to ﬁnd the probability that the results are all less than 5.

Problem-solving 22, 24, 25 22

see that it is . Brook says that we can calculate P (GB) + P (GG), since those are the two favourable outcomes.

1 2

23, 25, 26

23, 26–29

In a pie shop, there are 15 pies. Five are vegetarian, six are beef, and four are chicken. A customer buys two pies at random. a

What is the probability that the customer gets two vegetarian pies?

What is the probability that the customer gets at least one beef pie?

10.05 Multi-stage probability mathspace.co

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Based on historical data, it rains 19 days out of 31 in July in Bangkok, Thailand. a

Find the chance it will rain in Bangkok on the 7th July 2027.

Find the probability, to the nearest percent, that: i

It will rain two days in a row.

From a Monday to a Wednesday in July, it only rained on the Monday.

iii

From a Monday to a Wednesday in July, it rained on only one of the days.

In a school, 25% of students ride skateboards and 20% of students have dark hair. One student is selected at random. Find the probability that the student: a

Has dark hair and rides a skateboard

Has light hair and does not ride a skateboard

Has dark hair and does not ride a skateboard

Has light hair and rides a skateboard

A coin is tossed, then the spinner is spun and either lands on A, B or C:

B C

Construct a probability tree diagram that correctly shows all possible outcomes and probabilities.

Find the probability of: i

Landing on tails and the spinner landing on C

Landing on tails, or the spinner landing on C, or both

In a basketball free throw contest, a player has two attempts to make a free throw. If they succeed on the ﬁrst attempt, they do not take the second. A player shoots with probabilities: • First attempt: 0.6 • Second attempt: 0.8

588

Construct a tree diagram showing the probability of the ﬁrst two attempts either being a success or fail.

Find the probability that the player needs to make a second attempt.

Find the probability that the player misses on both attempts.

Mathspace WA Curriculum – Year 9 mathspace.co

A sailor has four meal packets to choose from on her last day but only wants to eat three meals. Two of the packets are oatmeal (O) and two of them are toasted muesli (T ). She will eat three of them throughout the day, chosen at random for breakfast, lunch and dinner. The probability tree shows the options for the day. O A

? T

Fill in the probabilities matching the edges of the probability tree: A

Find the probability that she has the same type of meal for breakfast and lunch.

The probability of Mary hitting the bullseye in a game is 0.97. If Mary plays 13 games in a tournament, what is the probability that Mary misses the bullseye at least once in the whole tournament? Round your answer to the nearest percent.

A gardener has a rectangular ﬂower bed divided into two sections, A and B. In section A, the probability of a rose growing is , while the probability of a daisy growing is . In section B, the probability of a rose growing is , while the probability of a daisy growing is . The gardener randomly picks a ﬂower from anywhere in the garden. a

Let x be the probability of drawing a ﬂower from section A. Draw a tree diagram to represent the situation.

If the probability that a daisy is picked is

, and the total area of the garden is 100 m2,

determine the areas of sections A and B.

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MODELLING INVESTIGATION

Fish sampling Explore how mathematics can help you model fish populations in a Western Australian river, like the Swan River. Simulate catching fish to estimate the proportion of bream and mullet, using the stratified sampling method and determine what it means for the ecosystem or fishing rules.

Objectives • To analyse the fish sampling situation and decide how to investigate it. • To determine reliable datasets and use statistics to model fish populations. • To explain your findings and what they mean for the river.

Analyse the fish sampling situation You are modelling fish populations in a river by simulating catches of bream and mullet. Pick your total fish population, initial bream proportion and sample size to estimate species proportions. Choose a total fish population (for example 100 to 200), an initial bream proportion (for example 30% to 50%) and a sample size per method (for example 10 to 20 catches). Sketch your plan manually to outline it.

Investigate 1.

Choose your total fish population, bream proportion and sample size. Write down your numbers.

What questions do you want to answer? Will your probabilities be exact or approximate? Why?

What might affect your simulation? List one assumption and one thing that limits you.

Determine your dataset and analysis Simulate fish catches to create reliable datasets using the stratified sampling method, then analyse the proportions and distributions.

Interactive exploration

Discover this concept in action online

mathspace.co

Simulate stratified sampling by splitting the river into two strata: 40% downstream and 60% upstream. Within each area, sample according to specie: 40% bream and 60% mullet. Use the applet or calculate manually (for example 10 samples per area) to observe the distribution. 590

Mathspace WA Curriculum – Year 9 mathspace.co

Investigate 1.

Simulate your stratified sampling dataset. Record outcomes. How did you ensure it is reliable?

Simulate your stratified dataset. Demonstrate a method of how to calculate the probability of catching two bream with this method.

Make a two-way table and a histogram to show the distribution of catches.

Interpret your plan Look at your datasets and graphs to estimate the river’s fish population and explain what it means. Scale your stratified results to 500 fish, then adjust one factor, for example increase sample size by 100 or modify the proportion between zones or species.

Investigate 1.

Predict the number of bream in 500 fish using stratified sampling. How do your results change with different sampling choices (like zone sizes, species ratios, or sample sizes)? Why?

Change one variable, such as sample size or the proportions of zones or species. How does this affect your estimate?

Explain your fish population model to the class. What assumptions did you make, and how do they influence your results?

Discussion 1.

What was the trickiest part of modelling fish populations with mathematics? Why?

How could your model be improved? Could you take more samples or change how the river or fish types are divided?

Talk with a classmate about their method. Was their estimate of the bream proportion different from yours? What could explain the difference?

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10.06 Optional: Independent and dependent events After this lesson, you will be able to… • differentiate between independent and dependent events. • recognise the role of replacement in probability scenarios. • calculate probabilities of events occurring with and without replacement. • apply mathematical formulas to determine the probability of combined events. • analyse real-world examples to identify whether events are independent or dependent. • use tree diagrams to visualise sequences of events and calculate probabilities.

Independent and dependent events Two events are independent if the outcome of each event does not affect the outcome of the other event. For example, landing on heads and landing on tails when tossing a fair coin are independent events because the first toss has no effect on the second toss. If two events are independent, the probability of both events occurring is: P ( A ∩ B) = P ( A) × P (B) The converse statement is also true. This means that if P ( A ∩ B) = P ( A) × P (B), then A and B are independent. Two events that are not independent are called dependent. For example, receiving a first place prize and receiving a second place prize in a race are dependent events. Once someone has won the first place prize, the probability of winning the second place prize will increase for those who are still in the race.

Example 1 A fair standard six-sided die is rolled 691 times. a If it lands on a six 108 times, what is the probability that the next roll will land on a six?

Create a strategy We can use the formula:

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Apply the idea The previous results do not affect the outcome of the next roll. There are 6 numbers on a die and 1 of them is a six.

Substitute the number of outcomes

b Is the outcome of the next roll independent of or dependent on the outcomes of previous rolls?

Apply the idea The outcome of the next roll is not affected by the outcomes of previous rolls. So the outcome of the next roll is independent.

Example 2 Two events A and B are such that P ( A ∩ B) = 0.3 and P ( A) = 0.5. Calculate P (B) if events A and B are independent.

Create a strategy Use the fact that if the events A and B are independent, then P ( A) × P (B) = P ( A ∩ B).

Apply the idea Write the equation

Substitute the values

Divide both sides by 0.5

Evaluate

Idea summary Two events are independent if the outcome of each event does not affect the outcome of the other event. Two events that are not independent are called dependent. To check if two events are independent you can use: P ( A ∩ B) = P ( A) × P (B)

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Replacement A probability tree diagram is useful in tracking two events that happen one after the other. At each point, it branches out to all the possible events that could occur from that point. The probabilities of each event is written on the branches.

Probability H Branch

For example, selecting two cards from a deck of cards, one after the other, are events that occur in sequence. There are two ways this can happen: • Draw a card, then put it back in the deck before selecting another card. • Draw a card and keep it, then select another card from the deck.

0.1 Outcome 0.9 T

The ﬁrst method is described as “with replacement” because the card is placed back into the deck. The second method is called “without replacement” because the card is not placed back into the deck.

Exploration The jar of marbles shown contains 7 red marbles, 3 blue marbles, and 10 green marbles. 1. A red marble is drawn from the jar, then placed back into the jar. How many marbles are in the jar? 2. A green marble is removed from the jar. How many marbles are left in the jar? 3. Suppose one marble is drawn and placed back into the jar, then a second marble is drawn. Let event A represent drawing a red marble and event B represent drawing a green marble in the second draw. Are these events independent or dependent? Explain. 4. Suppose one marble is drawn, then a second marble is drawn. The ﬁrst marble was not replaced. Are events A and B independent or dependent? Explain.

When selecting objects from a group “with replacement”, selections are independent. Each time you select a card, you have the same probabilities. R R

will be

B R

B B

594

For example, selecting a red card from a standard deck

Mathspace WA Curriculum – Year 9 mathspace.co

every time if you replace the card.

When selecting objects from a group “without replacement”, selections are dependent. Each time you select a card, you change the probabilities for the next selection. R R

For example, the probability of drawing a red card from a standard deck is

B R

If we draw a red card and do not replace it, there are only 25 red cards left and only 51 cards left in the deck. The probability of selecting a second red card is

. This can be seen in the top branches of the tree

diagram.

B B

This shows the probability of the second event is dependent on what colour card was drawn ﬁrst.

Example 3 A number game uses a basket with 10 balls, all labelled with numbers from 1 to 10. One ball is drawn at random, then another ball is drawn at random. a What is the probability that the ball labelled 2 is picked once if the balls are drawn with replacement?

Create a strategy Find the probability for each way that this could happen, then add them together.

Apply the idea This could happen in two ways: 1. P (2) × P (2′): The ball labelled 2 is picked ﬁrst and a ball not labelled 2 is picked second. 2. P (2′) × P (2): A ball not labelled 2 is picked ﬁrst and the ball labelled 2 is picked second. Find the probability of picking the ball with 2

Find the probability of not picking the ball with 2 Add the probabilities of the two ways

Substitute the probabilities

Evaluate and simplify

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b What is the probability that the ball labelled 2 is picked once if the balls are drawn without replacement?

Create a strategy Find the probability for each way that this could happen, then add them together. For the second event, the total number of balls will need to decrease.

Apply the idea Similarly to part (a) this could happen in two ways. However, for the probability of the second ball, the total number of outcomes will be 10 − 1 = 9 since the ball was not replaced. Add the probabilities of the two ways

Substitute the probabilities

Evaluate and simplify

Example 4 Three marbles are randomly drawn without replacement from a bag containing 6 red, 6 yellow, 4 white, 6 black and 4 green marbles. Find: a The probability of drawing 3 white marbles.

Create a strategy Remember that the number of white marbles and the total number of marbles in the bag decrease by 1 every time a white marble is drawn.

Apply the idea There are 4 white marbles and a total of 26 marbles inside the bag. Multiply the probabilities

Evaluate

b The probability of drawing 3 black marbles.

Create a strategy Remember that the number of black marbles and the total number of marbles in the bag decrease by 1 every time a black marble is drawn.

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Apply the idea There are 6 black marbles and a total of 26 marbles inside the bag. Multiply the probabilities

Evaluate

c The probability of drawing zero green marbles.

Create a strategy Remember that the number of non-green marbles and the total number of marbles in the bag decrease by 1 every time a non-green marble is drawn.

Apply the idea There are 22 non-green marbles inside the bag. Multiply the probabilities

Evaluate

d The probability of drawing at least one red marble.

Create a strategy Use the fact that P (at least one red marble) is the complement of P (no red marbles).

Apply the idea There is a total of 20 non-red marbles inside the bag. Multiply the probabilities

Evaluate

We now calculate P (at least one red marble): Subtract P (at least one red marble) from 1

Evaluate

10.06 Optional: Independent and dependent events mathspace.co

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Example 5 Bart is purchasing a plane ticket online. He notices there are only 4 seats remaining, 1 of them is a window seat (W) and the other 3 are aisle seats ( A). Bart purchases the ticket first. His friend, Joh, gets on the computer and purchases a ticket immediately after. The seats are randomly allocated at the time of purchase.

A W

W d

b A

a Fill in the probabilities matching the edges of the probability tree for the seat Bart receives and the seat his friend receives:

Create a strategy Use the fact that the ﬁrst event is Bart receiving his seat allocation and the second event is Bart’s friend (Joh) receiving a seat allocation.

Apply the idea

There are 3 aisle seats and 1 window seat. This means that when Bart is assigned a seat, the probability of getting a window seat is , and the probability of getting an aisle seat is .

d A

A A

d A

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If Bart gets a window seat, there are only aisle seats left when Joh purchases his ticket. So the probability that Joh gets an aisle seat is 1.

If Bart gets an aisle seat, there are 2 aisle seats and 1 window seat left when Joh purchases his ticket. The probability that Joh gets a window seat is , and

the probability that he gets an aisle seat is . A A

So, the probabilities matching the edges of the probability tree for the seats are , and

b What is the probability that Bart’s friend, Joh, has an aisle seat?

Create a strategy Find the probability for each way that this could happen, then add them together.

Apply the idea A W

A A

The second set of branches that finish on a node labelled A correspond to Joh having an aisle seat. That is, WA and AA. The probability of each event occurring is found by multiplying previous probabilities along each branch.

This means that the probability that Joh has an aisle seat is given by:

Evaluate the multiplication

Evaluate the addition

10.06 Optional: Independent and dependent events mathspace.co

599

c What is the probability of Joh receiving an aisle seat if Bart has a window seat?

Create a strategy

Apply the idea

Use the probabilities found in part (a).

If Bart receives a window seat, only aisle seats are left. So, the probability of Joh receiving an aisle seat is 1.

Idea summary Events “with replacement” occur when the item drawn is placed back into the group before each selection. Each selection is independent of the others. Events “without replacement” occur when the item remains outside the group after selection. Each selection is dependent of the others. The probabilities of each selection will change depending on previous selections.

10.06 Practice questions Understanding 1

Explorer 1−3

Adventurer 1, 2, 4, 5

Trailblazer 1−5

Identify whether these events are independent or dependent: a

A card is randomly selected from a normal deck of cards, and then returned to the deck. The deck is shuﬄed and another card is selected.

Each student is allowed to randomly pick an item from the teacher’s prize bag.

A coin is tossed, the result noted, and then it is tossed again.

Ursula takes a bus to the station and then immediately gets on a train to work. Is the probability of her missing the train independent or dependent on her missing her bus?

Dean has a box that contains ﬁve blue marbles and four red marbles. He takes a marble from the box, then takes a second marble from the box. Determine whether these events are dependent or independent if:

600

He replaces the ﬁrst marble before taking the second.

He does not replace the marble before taking the second.

A bag contains 3 pink balls and 2 orange balls. You draw one ball, note its colour, then draw another ball without putting the first ball back. a

Find the missing probability for drawing a second pink ball on the tree diagram.

Which outcome can be found by multiplying

Which outcome(s) has a probability of

Are the events of drawing the balls independent or dependent?

Mathspace WA Curriculum – Year 9 mathspace.co

O O

P O

by ?

OO OP

PO PP

On the island of Guam, the probability that it rains during rainy season is P (R). This is shown in the tree diagram. a

What do the two sets of branches represent?

Does the weather on the first day affect the chance of rain on the second day?

Which two events have the same probability?

R 0.6

0.4

0.6

0.4 S

Fluency Ex 1

Ex 2

6, 7, 9, 11−13, 15−17, 19, 22

6, 7, 9−13, 15−17, 19, 22

0.6

0.4

S R

6, 8−10, 12, 14−16, 18, 20, 21, 23

On a roulette table, a ball can land on one of 18 red or 18 black numbers. a

If it lands on a red number on the ﬁrst go, ﬁnd the probability that it will land on a red number on the second go.

Are the successive events of twice landing on a red number dependent or independent?

From a standard pack of cards, 1 card is randomly drawn and then put back into the pack. A second card is then drawn. a

Find the probability that neither of the cards are spades.

Find the probability that at least 1 of the cards is a spade.

Are the two draws dependent or independent?

Two standard dice are rolled. One is red and one is white. a

Are the results from the red die and the white die dependent or independent?

Calculate the probability that: i

The same number is rolled.

The sum of the two numbers exceeds 9.

iii

The red die is 4 and the white die is 6.

The red die is even and the white die is odd.

The sum of the two numbers is less than 2.

Two events A and B are such that P ( A ∩ B) = 0.1 and P ( A) = 0.2. Calculate P (B) if events A and B are independent.

Two events A and B are such that P ( A ∩ B) = 0.04 and P ( A) = 0.4. If P (B) = 0.1, state whether the events A and B are independent.

A number game uses a basket with 8 balls, all labelled with numbers from 1 to 8. Three balls are drawn at random, with replacement. Find the probability that the ball labelled 4 is picked: a

Exactly once

Exactly three times

Exactly twice

10.06 Optional: Independent and dependent events mathspace.co

601

Ex 3

Christa randomly selects two cards from a normal deck of cards. Calculate the probability, in each of the following, in the case that the cards are selected:

With replacement

Without replacement

The ﬁrst card is a Queen of diamonds and the second card is a 10 of spades.

Both cards are 10s.

The ﬁrst card is a Queen and the second card is a King

The ﬁrst card is red and the second card is not a spade.

Mae deals two cards from a normal deck of cards, without replacement. a

Are the events of drawing Card 1 and drawing Card 2 dependent or independent?

Calculate the probability that she deals: i

Two 10s

Two red cards

iii

Two diamonds

A 10 of spades and an Ace of diamonds, in that order

A 10 of clubs and an Ace of spades, in any order

Valentina randomly selects three cards from a normal deck of cards. Calculate the probability if the cards are selected:

Ex 4

With replacement

The cards are ﬁve of clubs, King of clubs, and Jack of spades, in that order.

The cards are all red.

The ﬁrst card is the 2 of clubs, the second card is a spade and the third card is red.

The cards are all spades.

None of the cards is an 8.

Without replacement

Three marbles are randomly drawn without replacement from a bag containing 5 red, 4 yellow, 5 white, 4 black and 5 green marbles. Find the probability of drawing: Three white marbles

Three black marbles

Zero green marbles

Zero yellow marbles

At least one red marble

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In a game of cards, suppose a hand contains a 10, a Jack, a Queen, a King and an Ace. Two cards are drawn from the hand at random, in succession and without replacement. Find the probability that: a

The Ace is drawn.

The King is not drawn.

The Queen is the second card drawn.

A number game uses a basket with 5 balls, all labelled with numbers from 1 to 5. Three balls are drawn at random, without replacement. Find the probability that: a

The ball labelled 4 is picked.

The ball labelled 4 is picked and the ball labelled 1 is also picked.

From a set of 10 cards numbered 1 to 10, two cards are drawn at random, in succession and without replacement. Find the probability that: a

Both numbers are even.

One is even and the other is odd.

The sum of the numbers is 12.

Three fair coins are tossed. Find the probability of obtaining: H H T H H T T H H T T

H T T

At least 1 head.

TTH in that order.

THH in that order.

HHH HHT HTH HTT THH THT TTH TTT

10.06 Optional: Independent and dependent events mathspace.co

603

A basketball player is practicing his free throws. He makes the shot with a probability of and misses with the probability of . The tree diagram shows all the possible outcomes and probabilities. Find the probability that he will: Hit Hit Miss Hit Hit Miss Miss Hit Hit Miss Miss

Hit Miss Miss

Make all three free throws.

Miss at least one free throw.

Make at least one free throw.

James owns four green jackets and three blue jackets. He selects one of the jackets at random for himself and then another jacket at random for his friend. Find the probability that:

G G

James selects a blue jacket for himself.

Both jackets James selects are green.

B B Ex 5

Han owns four green ties and three blue ties. He selects one of the ties at random for himself and then another tie at random for his friend. a

Write the probabilities for the outcomes on the edges of the probability tree diagram.

Calculate the probability that: i

Han selects a blue tie for himself.

Han selects two green ties.

GG GB

BG BB

G G

B G

B B

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An ice cream shop offers one flavour of ice cream at a discounted price each day. There are 6 flavours for the owner to choose from that could be discounted. 3 of the flavours are sorbet and the other 3 are gelato. The three sorbet flavours are raspberry (R), lemon (L) and chocolate (C). The three gelato flavours are vanilla (V ), mint (M ) and chocolate (C). Each decision has equal probability. a

Write the sample space for which ice cream type and ﬂavour may be chosen.

Find the probability that the lemon sorbet is selected.

Determine the probability that the ﬂavour will be chocolate.

Reasoning 24

24, 25

R S

C V M G C

24, 26, 27

In a lottery there are 37 balls. a

Find the probability of the 22 ball being drawn ﬁrst.

A ball is discarded after it has been drawn. If ball number 22 is drawn on the ﬁrst go, ﬁnd the probability of ball number 15 being drawn next.

Is the probability of each successive ball drawn the same as the probability of the ﬁrst ball drawn?

Are the draws dependent or independent events? Explain your reasoning.

Two cards are to be chosen from a pack of cards numbered 1 to 11. Explain how to ﬁnd the probability of drawing a 1, 2 or 3 ﬁrst, and after replacing the card, drawing a 9, 10 or 11.

Recall that for mutually exclusive events, P ( A ∩ B) = 0. For two events A and B, where P ( A) ≠ 0 and P (B) ≠ 0, explain why the events cannot be both independent and mutually exclusive.

Using the Venn diagram shown, explain why events A and B are independent. A

6 4

Problem-solving 28−30, 33 28

29, 31, 33

28, 29, 31, 32, 34

Consider the word WOLLONGONG. If three letters are randomly selected from it without replacement, ﬁnd the probability that: a

The letters are W, O, L, in that order.

The letters are O, N, G, in that order.

All three letters are O.

None of the three letters is an O.

10.06 Optional: Independent and dependent events mathspace.co

605

Sarah has a box of Favourites chocolates. In this box there are 30 chocolates, 5 of which are Turkish Delights. Sarah takes and eats a chocolate without looking until she gets a Turkish Delight. a

Find the probability she only eats one chocolate.

Find the probability she eats only two chocolates, rounded to the nearest percent.

Find the probability she eats ﬁve chocolates, rounded to the nearest percent.

As she eats more and more chocolates, state whether the probability of the next chocolate being a Turkish Delight is getting higher or lower.

How many chocolates must Sarah have to eat to be certain that the next chocolate will be a Turkish Delight?

Tom has shuﬄed a standard pack of 52 playing cards. He draws one card from the pack at a time, and then puts it aside. He will stop when he draws the 9 of diamonds. a

Find the probability he has to draw exactly 1 card.

Find the probability he has to draw exactly 2 cards.

Find the probability he has to draw exactly 3 cards.

Find the probability Tom will stop after the 44th card.

As more cards are drawn that aren’t the desired card, does the probability that the next card selected is the 9 of diamonds increase or decrease?

A bag contains 3 green marbles and 4 purple marbles. Two marbles are drawn at random from the bag. a

If the ﬁrst marble is replaced before the second marble is drawn, complete the table with the probabilities matching the edges of the probability tree. If the ﬁrst marble is not replaced before the second marble is drawn, complete the table with the probabilities matching the edges of the probability tree.

G A

P G

B P

GG GP

Determine the impact that replacing the marble before drawing a second has on the probability of these events: i

Drawing two green marbles

Drawing two purple marbles

iii

Drawing marbles of different colours

Edge

Probability

Sureya has three races to swim at her school swimming carnival. The probability that she wins a particular race is dependent on whether she won the previous races, as summarised: • The chance she wins the first race is 0.7 • If she wins the first race, the chance of winning the second is 0.8 • If she loses the first race, the chance of winning the second is 0.4 • If she wins the first two, the chance of winning the third race is 0.9 • If she lost the first two, the chance of losing the third race is 0.9 • If she won only one of the first two races, the chance of winning the third is 0.6

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Construct a probability tree to represent all outcomes in this situation.

Calculate the probability Sureya won all three races, rounded to three decimal places.

Calculate the probability Sureya won the third race, rounded to three decimal places.

Mathspace WA Curriculum – Year 9 mathspace.co

For breakfast, Moriah has something to eat and drinks a hot drink. She will either eat toast or cereal and then will drink tea or Milo. • The chance of Moriah making toast is 0.7 • The chance of Moriah drinking Milo is 0.4

Construct a tree diagram illustrating all possible combinations of food and drink Moriah can have for breakfast and their associated probabilities.

Find the probability Moriah drinks tea and eats toast.

Find the probability Moriah drinks tea or eats toast.

The proportion of scholarship recipients at a particular university is

. The number of

students at the university is so large that even if a student is removed, we can say that the proportion of scholarship recipients remains the same. If three students are selected at random: a

Construct a probability tree showing all the possible combinations of recipients and non-recipients.

Find the probability that at least one of the students is a scholarship recipient.

Find the probability that at least one of the students is a non-recipient.

Find the probability that there is at least one recipient and one non-recipient in the selection.

Investigation: Simulation of compound events

Investigate online

mathspace.co

10.06 Optional: Independent and dependent events mathspace.co

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10 Chapter review 1

Jarvis rolled a 6-sided die 60 times. a

How many times would he expect a 4 to appear?

After rolling the die 60 times, he recorded that 4 had appeared 20 times. What is the experimental probability of rolling a 4?

Do you think this is a fair die? Explain.

A French bakery tracks the number of each type of macaron sold in one week: a

How many macarons were sold in total?

Calculate the experimental probability of a randomly selected customer purchasing a: i

Raspberry macaron

Macaron type Chocolate Lemon (dairy-free) Raspberry Salted caramel Vanilla (dairy-free) 3

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Frequency 204 51 72 191 82

Anju has a blank 12-sided die. On each side, she writes a 1, 2, 3, or 4. She gives the die to four friends who each roll it several times. The results are presented in this table: Briana Clinton Desi Elijah

Dairy-free macaron

1 5 9 22 25

2 6 4 19 27

3 1 2 3 5

4 2 1 4 6

How might Anju have labelled her die? Explain.

The die has regular pentagonal faces, each with an area of 208.2 mm2. Using your answer from part (a), what surface area of the die has numbers that are perfect squares?

A manager at a marketing agency needs to send one employee from the team to a conference in Paris, France. She creates a Venn diagram showing different skills that her team has. A suitable team member to send will be able to speak French and code with React JSX.

208.2 mm2

Speaks French

Codes with React JSX

How many team members are there?

What is the probability that a randomly selected team member can speak French or code with React JSX?

How many suitable team members are there?

The manager is told that a team member was selected by her supervisor at random. What is the probability that the selected team member is suitable?

Mathspace WA Curriculum – Year 9 mathspace.co

A group of teachers were asked if they have a pet and if they have ever gone out of state. A partially completed two-way table of the results is shown. a

Find the missing values.

If a random teacher is randomly selected, ﬁnd the probability that:

6 50

The teacher does not have a pet and has never been out of state.

iii

The teacher has a pet or has been out of state.

The teacher has a pet if we know they have been out of state.

This Venn diagram and two-way table are both used to represent the same sample space. Complete the Venn diagram and the two-way table. Likes yeast spread

Likes yeast spread

Does not like yeast spread

Australian resident Not Australian resident Total

16 68

1 2 3 4

What is the probability that the difference is: i

less than 2

iii

Total 49

A player rolls 2 four-sided dice and the total positive difference is recorded in an array. a

Total 54

The teacher has a pet.

Pet No pet Total

Australian resident

Has been Has not out of been out state of state

more than 1

Find the sum of parts ii and iii. Explain that sum.

1 0 1 2 3

2 1 0 1 2

3 2 1 0 1

4 3 2 1 0

One fair 4-sided and one fair 6-sided die are rolled at the same time. Their product is recorded. a

Construct an array to represent all possible outcomes.

Find the probability of an even product.

Find the probability of rolling a 3 on the 4-sided die and an even product.

Find the probability of rolling a multiple of 3 or an even number.

Two dice are rolled, and the two numbers are recorded. Is each statement true or false? Justify your answer.

1 2 3 4 5 6

1 1, 1 2, 1 3, 1 4, 1 5, 1 6, 1

2 1, 2 2, 2 3, 2 4, 2 5, 2 6, 2

If the two numbers are added, a sum less than or equal to 5 and a sum greater than 5 are equally likely.

If the two numbers are multiplied, an even product is more likely than an odd product.

If this was drawn as a tree diagram, it would have 36 branches.

3 1, 3 2, 3 3, 3 4, 3 5, 3 6, 3

4 1, 4 2, 4 3, 4 4, 4 5, 4 6, 4

5 1, 5 2, 5 3, 5 4, 5 5, 5 6, 5

Chapter 10 review mathspace.co

6 1, 6 2, 6 3, 6 4, 6 5, 6 6, 6

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Demetria teaches dance and gives students a sticker after class. She has one star, one unicorn, and one tractor left. Three students take a sticker, one after the other. a

Is this experiment with or without replacement?

List the sample space of the possible ways the stickers were handed out.

Four cards labelled 3, 5, 6, and 8 are placed face down on a table. Two of the cards are selected randomly without replacement to form a two-digit number. a

Construct a tree diagram of this situation.

List the sample space for the possible two-digit numbers formed.

What is the probability of the resulting number being: i

Even or a multiple of 5

A calming corner has 6 cuddly toys and 4 fidget spinners. One toy is selected, used, and replaced. Then another toy is selected and used. a

Construct a tree diagram of this situation with the correct probability on each branch.

Find the probability of selecting one of each type of toy.

A laundry basket has 3 black, 2 white and 6 red shirts. There is only room for two more shirts in the washing. Two shirts are selected at random. a

Construct a probability tree diagram to show the sample space.

Find the probability of drawing: i

2 red shirts.

2 white shirts.

iii

Two shirts of the same colour.

Two different coloured shirts.

A red shirt, then a white shirt in that order.

Exactly one red shirt.

Alice is hatchet throwing. Each of her three hatchets has a probability of bullseye. If all three hatchets are thrown at a target: a

Construct a tree diagram with the probability on each branch.

Find the probability that all three hatchets will hit the bullseye.

Find the probability that at least one hatchet will miss the bullseye.

Find the probability that at least two hatchets will hit the bullseye.

of hitting the

Hasan randomly selects three cards from a normal deck of cards. a

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Even and a multiple of 5

If he draws the cards with replacement, ﬁnd the probability that: i

The cards are the ace of diamonds, the two of hearts, and the three of spades in that order.

The cards are all red.

iii

The cards are all diamonds.

None of the cards are queens.

An ace, a king, and a queen, all of the same suit, in that order.

Mathspace WA Curriculum – Year 9 mathspace.co

If he draws the cards without replacement, ﬁnd the probability that: i

The cards are the three of hearts, the four of diamonds, and the ﬁve of spades, in that order.

The cards are all black.

iii

At least one card is black.

The cards are all black or diamonds.

Five six-sided dice are rolled. Explain how to ﬁnd the probability that less than two 4s were rolled. Give your answer as an exact value.

When Lila goes out to dinner, she needs to choose a restaurant and a meal, chicken or duck. For restaurants, she is half as likely to choose Arjun’s as Benny’s. If she goes to Arjun’s, she is four times more likely to order duck than chicken. If she goes to Benny’s, the probability she orders chicken is x. a

Draw a tree diagram to represent the situation

At both restaurants, the chicken costs $20 and the duck costs $30. The probability of her spending $20 is

c 18

. Find the probability that Lila chooses duck at Benny’s.

What is the expected cost of Lila’s meal to the nearest cent?

(Optional) Lucas picks three tiles randomly from a bag containing tiles numbered from 1 to 10. Calculate the probability if the tiles are selected:

With replacement

Without replacement

The tiles are number 3, number 7, and number 9, in that order.

The tiles are all even-numbered.

The first tile is the 4, the second tile is an odd number, and the third tile is greater than 5.

The tiles are all less than 6.

None of the tiles is a 10.

(Optional) Three balls are randomly drawn without replacement from a jar containing 6 blue, 3 orange, 4 purple, 5 gray, and 2 pink balls. Find the probability of drawing: Three purple balls

Three gray balls

Zero pink balls

Zero orange balls

At least one blue ball

(Optional) Sophia has shuffled a standard deck of 52 playing cards. She draws one card at a time and sets it aside. She will stop when she draws the Queen of hearts. a

Find the probability she has to draw exactly 1 card.

Find the probability she has to draw exactly 2 cards.

Find the probability she has to draw exactly 3 cards.

Find the probability Sophia will stop after the 35th card.

As more cards are drawn that are not the Queen of hearts, does the probability that the next card selected is the Queen of hearts increase or decrease?

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Big ideas • Understanding and analysing data, along with knowing the right measures of centre and spread help us make better conclusions and informed decisions. • Different tools like stem and leaf plots, frequency tables, and histograms help us organise and better understand data.

Curriculum Outcomes • Explore, choose and create graphical or visual representations and justify choice with regards to context, purpose, data type and intended audience • Interpret and compare multiple datasets represented in back-to-back stem and leaf plots and histograms with consideration of shape, spread and centre • (Optional) Produce and organise accurate and valid, ungrouped continuous data to construct histograms and frequency polygons. Determine summary statistics and analyse the distribution in terms of centre, shape and spread • Describe different sampling methods and analyse how the different methods can affect the results of surveys. Identify and explain how chance variation impacts on the data validity, reliability and conclusions drawn from surveys • Critically analyse statistics in the media and other real-life situations relating to data samples, including the effect of chance variation on sample analyses • Analyse data with multiple variables represented in tables, describe using statistical measures and relative frequencies to make inferences

11 Statistics Chapter outline 11.01 Types of data 11.02 Data displays Investigation: Opinion Polls 11.03 Back-to-back stem and leaf plots 11.04 Grouped histograms 11.05 Optional: Frequency histograms and polygons 11.06 Shape of data Investigation: Statistical reports in the media 11.07 Compare data sets Investigation: Plan and conduct statistical investigations Chapter 11 review

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Statistics help scientists study animal populations. They tag animals to monitor their movements and patterns to determine the health of the whole population.

11.01

Types of data

After this lesson, you will be able to… • distinguish between the types of data. • understand census and sampling in data collection. • identify the importance of representative data and the risks of bias. • apply strategies to avoid bias in data collection and interpretation.

Types of data Univariate data is the word commonly used in statistics to refer to data that consists of observations on a single characteristic. Data can be classified into two types, categorical data and numerical data. Numerical data is data that can be counted, ordered and measured. Categorical data is non-numerical, it describes the qualities or characteristics of a data set. Data that can be broken into groups or categories is called categorical data or qualitative data. Categorical data can be divided into two categories: • Nominal data involves categories that have no natural order. For example, if you asked someone what their favourite subject is, they might say “Maths” or “English”. • Ordinal data involves categories that have a natural order. For example, if you were tracking the position of people coming in after a race you would have “first”, “second”, etc. Things that have a natural order. Data that can be counted or measured is called numerical data or quantitative data. Numerical data can be divided into two categories: • Discrete data involves distinct or separate values. The data cannot take “in between” values. For example, if you were looking at the numbers of people that had entered a building over time, the responses could be 5 or 6, but we would not say 5.5, as we can’t have half a person. If you have to count to collect each piece of data, the data is discrete. • Continuous data is measured, so the values are not distinct. Within the range of possible values a piece of data could be any decimal or fractional number. If we measure the length of a person’s foot, we might get any decimal value, limited only by the precision of our ruler. The flowchart shows us the different types of data and sub-categories: Data

Numerical Made of numbers

Continuous Measurable

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Discrete Countable

Categorical Made of categories

Nominal Doesn’t have any ordering

Ordinal Elements are ordered

Example 1 Which one of the following data types is discrete? A Your height

The time it takes to swim 200 metres

C Daily temperature

The number of pets in your family

Create a strategy

Apply the idea

Choose the option which can be counted (involves distinct or separate values).

The correct answer is option D: The number of pets in your family.

Example 2 Classify each data as ordinal, nominal, continuous or discrete. a A class survey about where they went on their most recent holiday

Create a strategy

Apply the idea

Determine whether the answers would be words Names of places are words. So it’s categorical, or numbers. If they would answer with a word, and as there is no natural order, it is nominal. the data is categorical. If they would respond with a number, the data is numerical. b Weights of kittens

Create a strategy

Apply the idea

Determine whether the answers would be words The weight of a kitten can be measured. So or numbers. If they would answer with a word, it is numerical. Weight is a measurement that the data is categorical. If they would respond can have any number of decimal places, so it is with a number, the data is numerical. continuous.

Idea summary The ﬂowchart summarises the different types of data and sub-categories: Data

Numerical Made of numbers

Continuous Measurable

Discrete Countable

Categorical Made of categories

Nominal Doesn’t have any ordering

Ordinal Elements are ordered

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Data collection A census is when data is collected from a whole population. A sample is when data is collected from a part of a population. For example, if we wanted to know whether or not students at the school liked their uniform, we could survey every student in the school (a census). However, since this would be difficult and time consuming, we may choose to survey a smaller group of students (a sample). When we take a sample it is important that it is representative of the population. This means that the sample should reflect the characteristics of the entire group. For example, if we only survey Year 10 boys, this wouldn’t represent all students, since the school has different year levels and both boys and girls. If a survey gives unreliable results, it’s often because the data is biased. To avoid bias in a survey, here are some things to keep in mind: • Who is in the sample? Make sure it’s a good representation of the whole population. • How many people are surveyed? The more people you survey, the more accurate the results will be. After thinking about who you ask and how many you ask, the next thing to consider is how you ask the questions. • A leading question suggests a particular answer or influences how someone responds. For example, asking “Do you watch the most popular sport, soccer?” may push someone to answer “yes” because the question makes it seem like soccer is the correct choice. A better question would be, “Do you watch soccer?” • A loaded question is a question that contains an assumption or forces someone to answer in a way that might not be fair. For example, asking “How often do you skip practice?” assumes the person skips practice regularly, which may not be true and is emotive as skipping practice sounds bad. A better approach would be, “Do you attend practice regularly?” By avoiding leading and loaded questions, we can get more reliable and unbiased results.

Example 3 Yvonne, Lachlan and Tricia are investigating the favourite sports of students at their school. In each case, consider the sample and the survey question to determine whether the results are likely to be biased or not. a Yvonne asks students on the school soccer team, “What’s your favourite sport?”

Create a strategy Consider the following: • Who is being surveyed? • How many people are being surveyed? • Whether the question being asked actually addresses the question at hand. • Whether the question is leading.

Apply the idea The members of the soccer team will be more likely to answer soccer since they play it (the sample is not representative of the population). So the results are likely to be biased.

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b Lachlan randomly selected students from the school and asked, “What’s your favourite sport?”

Apply the idea Lachlan randomly selected students so the sample should be representative of the population. The question is not leading. The results are not likely to be biased.

c Tricia randomly selected students from the school. To each student she said, “The local AFL team is donating money to our school this term. What’s your favourite sport?”

Apply the idea The question is leading – because it mentions the donation from the local AFL team, students may feel pressured to choose AFL. So the results are likely to be biased.

Idea summary A census is when we collect data from a whole population. A sample is when we collect data from part of a population. Bias can affect the results of the data collection, or appear in the presentation of the data after collection. Consider: • • • •

Who is being surveyed? How many people are being surveyed? Whether the question being asked actually addresses the question at hand. Whether the question is leading.

11.01 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

True or false: a

Nominal data is a type of numerical data.

Numerical data can be discrete or continuous.

Qualitative data involves words or categories.

Quantitative data is another word for categorical data.

What is the difference between a sample and a census?

What are 3 ways we can reduce bias when collecting data?

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Fluency 4

Ex 1

Ex 2

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4–13

Classify each data set as numerical or categorical. a

Amount owing on layby

Types of vegetables

Brands of tablets

Types of folders

Colours of folders

Brands of phones

Maximum snowfall

Daily UV index

Types of dogs

Favourite colours

Cost of a computer game

Types of fruits

For each of these categorical data types, identify whether they are nominal or ordinal: a

Ice cream flavour

Smartphone application

Clothing size

Blood type

Customer satisfaction rating

Pain level

Membership rank

Cuisine preference

For each of these numerical data types, identify whether they are discrete or continuous: a

Number of classrooms at a school

Daily humidity

The ages of a group of people

The time taken to run 200 metres

The number of people taller than you

The number of coins in a coin collection

The amount of snowfall in a city

For each survey question, classify the data as ordinal, nominal, discrete or continuous. a

How far do you travel to school each day?

How many siblings do you have?

What modes of transport do you use to get to school?

What kinds of pets do you own?

How many pets do you own?

How tall are you?

How much does your schoolbag weigh?

Are these survey questions biased or fair? a

Do you want soup for lunch or a sandwich?

What do you do on a Sunday morning?

Do you think the government should be allowed to cut down some of the oldest trees in the area to construct a metro railway line in the city?

Do you eat at least the recommended number of servings of fruits and vegetables to ensure a healthy and long life?

Identify the type of bias that could be present in these biased survey questions: a

Do you prefer the natural beauty of hardwood floors in your home?

Do you feel that the TV news is an inaccurate portrayal of life’s problems?

Don’t you think this newspaper is biased?

Do you prefer the look and feel of thick lush carpet in your living room?

Mathspace WA Curriculum – Year 9 mathspace.co

Ex 3

Determine whether each question is likely to give biased results or not. If you think results will be biased, state why. a

Damon asked four of his friends, “What is your favourite subject at school?”

Lulu is interviewing people outside of a train station. She asks, “Should the government spend more money on public transportation?”

Lys is interviewing people about how shark nets are affecting shark attack numbers. He says, “Shark nets cost $1 000 000 per year and barely help. Do you support shark nets?”

After the government decided to increase the minimum retirement age, a survey is to select a group of people to ask their opinions on the changes. Should each of these groups be included in the survey?

Politicians and policy makers

Members in the community that are directly impacted by the changes

Members in the community not impacted by the changes

A radio station conducts a poll asking its listeners to call in to say if they are for or against restrictions on scalpers selling tickets for gigs at a higher price. State whether these are reasons why this is not an appropriate way to conduct a poll:

A large variety of people are likely to call.

A person can call more than once, so they could be counted more than once.

People with stronger views are more likely to call than those who don’t have a strong view.

A question is inappropriate for a survey given it fulﬁlls any of the following criteria: • Question is unclear • Emotional language • Question is too personal Using the criteria list, determine why each of the questions are inappropriate: a

Many people have worked incredibly hard and even died making this bridge. Do you like the bridge?

How many large electronic devices are in your home?

How many people have died in your family?

Reasoning 14

14–16

14–17

Explain why each survey question would give unreliable results. Rewrite each question to provide more reliable results. a

The Prime Minister believes that taxes are too high, do you think taxes are too high?

Do you have a Snapchat or Twitter account?

Do your siblings ever drive you to school?

Do you often waste time on social media?

What brand of shoe polish do you use every day?

Explain why it might be important to collect both categorical and numerical data in a survey about dietary habits.

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A city council wants to understand the commuting habits of its residents. What types of data should they collect and why?

A study claims that 75% of students prefer online learning. However, the survey was conducted by interviewing only 30 students from a single school. Comment on the reliability of this study and how the study might be improved.

Problem-solving 18–20 18

18, 20, 21

18, 20–22

Patricia surveys her class about their favourite music. Her results are shown in the table: a

According to the survey, which genre is most popular among her class?

This was her survey question:

Genre No. of students

Country 15

Pop 2

“The coolest kids like country music, and nobody likes pop. Do you like country music or pop music?” Do you trust the results of Patricia’s survey? Explain your answer. 19

You are part of a school council and you are tasked with ﬁnding out the most popular afterschool activities among students. a

Describe how and where you would carry out this survey, including which groups of people you would focus on.

What type of data would you be collecting?

How would you ensure that your data collection methods are unbiased and accurate?

You are a researcher studying the relationship between hours of sleep and academic performance among high school students. a

What type of data would you need to collect to investigate this?

How would you go about collecting this data?

A clothing company wants to understand their customer’s shopping habits better to improve their sales. a

What kind of data should they collect?

How could this data help improve their business strategy?

A company is conducting a survey to determine the average income of people in a city. They randomly select 50 employees from a local technology company and 50 employees from a small retail store to participate in the survey. The company ﬁnds that the average income of employees from the technology company, T, is $60 000, while the average income of employees from the retail store, R, is $30 000. However, the technology company mainly employs professionals with higher salaries, while the retail store employs a mix of part-time and full-time workers.

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Write an equation for the overall average income, A, of all 100 employees surveyed, in terms of T and R. Simplify if possible.

Explain why using the overall average income, A, in this case might lead to a biased representation of the city’s average income.

Mathspace WA Curriculum – Year 9 mathspace.co

11.02

Data displays

After this lesson, you will be able to… • understand the concept of univariate data and its representation through frequency tables, stem and leaf plots and dot plots. • apply knowledge of measures of centre and spread to univariate data displays. • calculate the mode, mean, median and range from frequency tables, stem and leaf plots and dot plots.

Frequency tables Frequency tables display univariate data. We can ﬁnd the mode, mean, median and range from a frequency table. The frequency table shows the number of goals scored in a football game, and the number of games that had that score (the frequency). Number of goals scored (x) 0 1 2 3 4 5 Total

Frequency ( f ) 4 11 5 1 3 1 25

The range is a measure of the spread of scores. We find the range by subtracting the lowest score from the highest. In the table, the range is the highest – lowest = 5 – 0 = 5 The mode is the most frequently occurring score in the data. We can find the mode by identifying the score with the highest frequency in the table. In the table, the mode is the most frequent score which is 1 goal, with a frequency of 11. The mean is the average score of the data set. We find the mean by summing all the scores and dividing by the total number of scores. Let’s add a column to the frequency table for fx, the frequency ( f ) multiplied by the score (x) for each row. Number of goals scored (x) 0 1 2 3 4 5 Total

Frequency ( f ) 4 11 5 1 3 1 25

fx 0×4=0 1 × 11 = 11 2 × 5 = 10 3×1=3 4 × 3 = 12 5×1=5 41

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The total for the fx column gives us the total number of goals scored in the 25 games played. To ﬁnd the mean, we divide the total number of goals scored (total of the fx column) by the amount of games played (total of the frequency column).

In this case, the mean is

= 1.64.

The median is the middle score when all the scores are arranged in order. To help with the median we can add a column called the cumulative frequency. This column is the sum of the frequencies of the score and each of the scores above it. The first row is the frequency of the first row. The second row is 11 more than the one before it. The third row is 5 more than the one before that, and so on for the full table. Number of goals scored (x) 0 1 2 3 4 5 Total

Frequency ( f ) 4 11 5 1 3 1 25

Cumulative frequency 4 4 + 11 = 15 15 + 5 = 20 20 + 1 = 21 21 + 3 = 24 24 + 1 = 25

The last row of the cumulative frequency should be the same as the total frequency. If we look at the score of 2 goals row, we’ll see that the cumulative frequency is 20. That means 20 games scored 2 goals or less. In a list of ordered scores, the median is found by selecting the middle score. In a frequency table, the position of the median can be found by adding 1 to the total frequency and then halving it.

In this example, the total frequency is 25, so the position of the median is median is the 13th score.

, that is, the

We can see that scores of 0 took up the first 4 positions, then scores of 1 took up the 5th to 15th positions. As the 13th position is found in this range, the median is 1 goal. If the median is between 2 positions, we would need to find the mean of the 2 scores. For example, if the median position is 12.5 then we need to calculate the mean of the 12th and 13th scores. We call the mean, median, range, mode measures of centre and spread.

Example 1 For the following data set: 27, 25, 24, 24, 24, 24, 24, 25, 23, 24, 26, 23, 27, 23, 24, 27, 23, 23, 27, 25, 24, 24, 27, 25, 23

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a Complete the frequency distribution table: Frequency ( f )

Score (x) 23 24 25 26 27 Total

Create a strategy Calculate frequency by counting score occurrences, multiply score (x) and frequency ( f ) for the fx column, and evaluate column totals for ﬁnal results.

Apply the idea Here’s the complete table: Frequency ( f ) 6 9 4 1 5 25

Score (x) 23 24 25 26 27 Total

fx 23 × 6 = 138 24 × 9 = 216 25 × 4 = 100 26 × 1 = 26 27 × 5 = 135 615

b Calculate the mean, rounded to one decimal place.

Create a strategy Use the formula

Apply the idea Substitute the values in part (a)

Evaluate

c Find the range.

Create a strategy Subtracting the lowest score from the highest score.

Apply the idea The highest score is 27 and the lowest score is 23. Range = 27 – 23 =4

Subtract the scores Evaluate

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d Find the mode.

Create a strategy

Apply the idea

Look at the table in part (a) and identify the score with the highest frequency.

24 has a frequency of 9, which is the highest frequency of all the scores. So the mode is 24.

e Find the median.

Create a strategy To ﬁnd the median we need to add a cumulative frequency column, then use the formula for median position.

Apply the idea Score (x) 23 24 25 26 27 Total

Frequency ( f ) 6 9 4 1 5 25

Cumulative frequency 6 6 + 9 = 15 15 + 4 = 19 19 + 1 = 20 20 + 5 = 25

Write the formula

Substitute the total frequency

Evaluate

The median will be 13th score. Looking at the cumulative frequency column of the table, the 13th score is 24 so the median is 24. f

Construct a column graph of the data.

Create a strategy Use the values we found in part (a).

Apply the idea

Frequency ( f )

Based on the frequency distribution table, the column graph should look like:

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10 9 8 7 6 5 4 3 2 1 0

23 24 25 26 27 Score (x)

Idea summary When ﬁnding measures of centre and spread from a frequency table: • The range is found by subtracting the lowest score from the highest score. • The mode is the score with the highest frequency. • The cumulative frequency is the sum of the frequencies of the score and each of the scores above it. Adding a cumulative frequency column to a frequency table helps us ﬁnd the median. • The position of the median score can be found with the formula:

• Adding an fx ( frequency times score) column to a frequency table is helpful for ﬁnding the mean. The mean can be found with the formula:

Stem and leaf plots A stem and leaf plot is a univariate data display, where scores are split using place value. The ﬁrst digit or digits in each value become the ‘stem’ and the last digit becomes the ‘leaf’. • The stems are arranged in ascending order, to form a column, with the lowest value at the top • The leaf values are arranged in ascending order from the stem out, in rows, next to their corresponding stem (each stem can have more than one leaf) • There are no commas or other symbols between the leaves, only even spacings

Score of 10

Score of 56

Stem

Leaf

Key: 2 | 1 = 21

8 9 All of these are in the 40s

When ﬁnding measures of centre and spread from a stem and leaf plot: • The range for the stem and leaf plot is the smallest score subtracted from the largest score. In the above example, it is 58 − 10 = 48 • The mode is the most common score, not to be confused with the most common leaf. In this case, there are two modes, 35 and 41. • The median is the middle score. In the stem and leaf above, the median is 36. • The mean is the sum of all the scores divided by the number of scores. The number of scores can be found by counting the number of leaves (don’t count the stems). In the above, the mean is

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Example 2 A cyclist measured his heart rate immediately after ﬁnishing each event in which he competed. The results are recorded in a stem and leaf plot. Stem

Leaf

4 4 4 6 8 8

0 0 4 6 8

0 2 3 3

2 3 4 7 7 8

Key: 16|1 = 161 a What was the difference between his slowest and fastest heart rate?

Create a strategy Subtract the smallest leaf on the lowest stem from the largest leaf on the highest stem.

Apply the idea The slowest recorded heart rate is 161 and the fastest recorded heart rate is 198. Difference = 198 – 161 = 37 beats per minute

Subtract the heart rates Evaluate

Reﬂect and check As a reminder, always remember to add the appropriate unit to your ﬁnal answer.

b For each extra heartbeat per minute, it takes 7 seconds longer for him to recover. How much longer would it take to recover when he ﬁnishes with the fastest heart rate than when he ﬁnishes with the slowest heart rate?

Create a strategy Multiply the result from part (a) by 7.

Apply the idea Recovery time = 37 × 7 = 259 seconds

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Multiply the heart rates Evaluate

c Find the median.

Create a strategy The median is the middle score in an ordered set of scores. In a stem and leaf plot, the scores are already ordered.

Apply the idea There are 23 scores in the data (the number of leaves gives the total count of scores). To find the middle score, we add 1 and then halve. This is the 12th position. Total scores = 23 Add 1 to the total scores and then halve

Median position

= 12 Evaluate Count from the bottom to find the 12th score. The median is 178. Stem

Leaf

0 0 4 6 8 8th 9th 10th 11th 12th

7 scores

Key: 16 | 1 = 161

Idea summary In a stem and leaf plot, the ﬁrst digit or digits in each value become the ‘stem’ and the last digit becomes the ‘leaf’.

Dot plots In a dot plot, each dot represents one data point. The dot plot displays the number of goals scored in a game, where each game is one dot.

1 2 3 4 Goals scored

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We can use a dot plot to ﬁnd measures of centre and spread: • The mode is the score with the most dots. In the above, there are two modes, 1 and 4. • The range is the smallest score with a dot above, subtracted from the highest score with a dot above. In this case, it is 5 − 0 = 5 • The median is the middle score. However, you should remember that each dot represents a score. In the above dot plot, the middle dot is above the score of 3, so the median is 3. • The mean is the sum of all the scores divided by the number of scores (the number of dots). In this case it is

(to one decimal place).

Example 3 The number of ‘three-pointers’ scored by a basketball team in each game of the season is represented in the dot plot. A ‘three-pointer’ is worth 3 points. 0

1 2 3 4 5 Three-point shots

a What was the total number of points scored from three-pointers during the season?

Create a strategy Multiply the number of dots with their respective three-point shots and sum the products. Then, multiply the sum by 3.

Apply the idea Looking at the dot plot, we have: Total no. of points = (0 × 4 + 1 × 3 + 2 × 1 + 3 × 4 + 4 × 1 + 5 × 2 + 6 × 4) × 3

Multiply the number of dots by threepoint shots, add them, then multiply by 3

= 55 × 3 Evaluate inside the brackets = 165

Evaluate

b What was the average number of points scored from three pointers each game of the season? Round to two decimal places if necessary.

Create a strategy Divide the total number of points by the total number of games.

Apply the idea We have the 19 dots representing the total number of games, and we found 165 as the total number of points in part (a). Remember to include the games with zero ‘three-pointers’ scored. Average no. of points = = 8.68

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Divide 165 by 19 Evaluate

c Find the median.

Create a strategy The median is the middle score in an ordered set of scores. In a dot plot, the scores are already ordered.

Apply the idea There are 19 scores in the data (the number of dots gives the total count of scores). To find the middle score, we add 1 and then halve. This is the 10th position. Total scores = 19 Median position

Add 1 to the total scores and then halve = 10 Evaluate

Count from the bottom to find the 10th score. The median number of three-point shots is 3. 8 scores

10th 9th 0

4 2 3 Three-point shots

Idea summary In a dot plot, each dot represents one data point, belonging to the value that it is placed above.

11.02 Practice questions Understanding

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

When might a frequency table help in simplifying the calculation of the mean of a data set, compared to working only with individual data points?

What type of data is best represented using a stem and leaf plot?

For a dot plot, describe how we determine the frequency of each score.

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4ac, 5, 6ab, 7, 9, 10, 12, 13

Fluency 4 Ex 1a

Ex 1d

4cd, 5, 6ac, 8, 9, 11, 13, 14

Complete the following frequency tables: a

Score Frequency Cumulative fx (x) (f ) frequency 1 7 2 4 3 7 4 8 5 8 6 8 Total

Score Frequency Cumulative fx (x) (f ) frequency 5 3 10 2 15 7 20 3 25 5 30 4 Total

Score Frequency Cumulative fx (x) (f ) frequency 34 119 36 125 38 160 40 143 42 185 44 137 Total

Score Frequency Cumulative fx (x) (f ) frequency 150 12 160 15 170 11 180 10 190 16 200 14 Total

State the mode of the following data sets: a

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4bc, 5, 6a, 7, 9, 10, 12–14

Score 3 4 5 6 7 8

Frequency 3 3 4 2 5 8

Mathspace WA Curriculum – Year 9 mathspace.co

Score 1100 1200 1300 1400 1500 1600

Frequency 125 129 140 151 153 149

Score 250 260 270 280 290 300

Frequency 28 27 26 28 26 24

6 Ex 1

For each scenario: i

Construct a frequency table for the results.

Calculate the range.

iii

Find the mode.

Calculate the mean, rounded to two decimal places.

Find the median.

Construct a column graph of the data.

Mr. Rodriguez recorded the number of pets owned by each of the students in his class. He found that 15 people had no pets, 19 people had one pet, 3 people had two pets and 8 people had three pets.

Mrs. Brown asked her students how many bedrooms they have in their house. Their responses were as follows:

Mr. Smith asked his students how many siblings they have. Their responses were as follows: 1, 1, 1, 2, 2, 1, 0, 3, 0, 1, 2, 1, 1, 2, 2, 3, 1, 2, 3, 3, 1, 2, 2, 1, 2

4, 4, 4, 3, 4, 3, 3, 5, 5, 4, 4, 3, 3, 4, 4, 4, 3, 4, 5, 4

The set of marks for a class of students is: 81, 81, 71, 81, 81, 61, 71, 93, 71, 58, 71, 58, 61, 93, 93, 71, 61, 61, 81, 58, 93

9 Ex 2

Organise the data into a frequency table.

Find the total number of students in the class.

Find the number of students who will get a Distinction grade (80 < Score ≤ 90).

Find the number of students who will get a High Distinction grade (Score > 90).

Find the mean mark, rounded to one decimal place.

A cyclist measured his heart rate immediately after ﬁnishing each event in which he competed. The results are recorded in a stem and leaf plot, but two values are missing: a

He can remember no number appears twice. What is the smaller missing number?

If both missing numbers sum to 367, what is the second missing number?

A particular genre of electronic dance music has a beat per minute (BPM) range of between 160 and 200. The BPM’s of 23 songs are shown in the stem and leaf plot. a

Calculate the mean BPM of the track selection, rounded to the nearest whole number.

Find the median BPM.

Find the modal BPM.

Stem

Leaf

3 5 9

0 3 ?

2 4 7 8

0 1

5 8

Key: 12|3 = 123 Stem

Leaf

4 4 4 6 8 8

0 0 4 6 8

0 2 3 3

2 3 4 7 7 8

Key: 1|2 = 12 beats per minute

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For this data set: 49, 49, 49, 51, 48, 47, 50, 49, 51, 50, 51, 49, 52, 47, 50, 49 51, 48, 51, 51, 50, 48, 49, 52, 49, 47, 47, 52, 52, 49, 47, 48 52, 48, 51, 51, 49, 52, 48, 48, 52, 50, 48, 47, 52, 52, 52, 48 a

Construct a dot plot that represents the data:

Summarise the data using measures of centre and spread.

A group of people were surveyed on the number of websites they visit on a daily basis. The results are displayed in the dot plot.

Number of websites visited daily

Number of websites visited daily (x) 0 1 2 3 4 5 6 Total

13 Ex 3

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Complete the frequency table based on the dot plot. Frequency ( f )

Cumulative frequency

What is the mean? Round your answer to two decimal places.

What is the median?

Consider the dot plot showing students scores: a

Complete a frequency table for the data.

How many students scored above 20?

How many students scored below 30?

What was the mode for the scores?

Find the mean score. Round your answer to two decimal places.

10 20 30 40 50 Score

The number of ‘three-pointers’ scored by a basketball team in each game of the season is represented in the dot plot. A ‘three-pointer’ is worth 3 points. a

What was the total number of points scored from threepointers during the season?

What was the mean number of points scored from threepointers each game of the season? Round your answer to two decimal places.

What was the median number of points scored from three-pointers each game of the season?

Mathspace WA Curriculum – Year 9 mathspace.co

1 2 3 4 5 Three Point Shots

The given dot plot shows the number of students in each section of the school band: a

Which section has the least members?

Find the number of people in the band.

How many more people are there in the Strings section than in the Brass section?

What is the mean number of musicians per group?

Reasoning 15

Band members

Strings Woodwind Brass Percussion

16, 17

15–18

The number of hours spent on homework per week by 30 students was collected. 2, 3, 2, 4, 3, 3, 4, 4, 3, 2, 4, 5, 6, 5, 4, 3, 2, 3, 4, 5, 6, 5, 4, 3, 2, 2, 4, 5, 6, 4. Explain how a frequency table could help to understand this data.

Alison and Sarah both found the average distance travelled in Sydney from the stem- and-leaf plot. Alison found the average distance travelled was 38.24 km, while Sarah found the average distance travelled was 41 km. Who is correct and why?

Sydney 1

0 3 6

5 5 6

0 3 6 8 8 9

6 7 8 1

5 6 9

Key: 1|2 = 12 kilometres traveled 17

Determine whether the statements are true or false in regards to dot plots. If false, explain your reasoning. a

A dot plot can effectively illustrate a correlation between two variables.

Dot plots can reveal multiple modes in the data set.

The mean, median and mode can all be determined directly by looking at a dot plot.

The range of the data cannot be determined from a dot plot.

Dot plots are only helpful for small data sets.

Here is a set of results: 10, 11, 11, 27, 28 Three new results are added to the data: 15, 15, 15 Without calculating the answer, describe the impact these new results will have on the range, mode, and median.

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Problem-solving 19–22 19

19–22

20, 21, 23

The level of mercury (ng/L) in 40 ﬁshing waters was tested and recorded. The results are given in the following stem and leaf plot: Stem

Leaf

3 5 5 6

2 4 4 5 5 6 8 9 9

2 2 2 2 2 3 4 6 9

2 3 3 3 5 6 7 7 7 8 8 8 8 9 Key: 1 |2 = 12

What was the second highest reading?

In how many places was a reading of 117 or higher recorded?

What was the 11th lowest score?

The safe level of mercury is recommended to be 109 ng/L or lower i

How many ﬁshing waters have safe levels of mercury?

What percentage of the places have safe levels of mercury?

A group of adults are asked: “How old were you when you passed your driving test?”. The responses were: 22, 17, 17, 17, 19, 21, 17, 22, 21, 18, 18, 17, 18, 22, 18 a

Plot the responses on a dot plot.

Find the following: i

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Range

Mode

iii

Median

How many people passed their driving test on or after their 19th birthday?

There was originally an outlier of 45. What would be the effect on the range, mode and median if it had not been removed?

Christa is a casual nurse. She used a dot plot to keep track of the number of shifts she did each week for a number of weeks: a

Over how many weeks did Christa record her shifts?

For how many weeks did she work 5 shifts?

How many weeks did she work less than 6 shifts?

If Christa works at least 6 shifts a week, she buys a weekly train ticket. What proportion of the time did she buy a weekly train ticket?

Mathspace WA Curriculum – Year 9 mathspace.co

4 5 6 7 8 Shifts per Week

At a national painting competition, the total number of paintings each of the 25 participating artists completed in a week is summarised in the frequency table: a

Determine the median number of paintings completed.

Determine the mode.

Find the mean number of paintings completed.

Total number of paintings completed

Number of artists

1 2 3 4 5 6 7 8 Total

3 4 3 2 5 3 4 1 25

The frequency table shows the number shown when a die is rolled. Carl rolls the die one more time and the new mean is 3. What was the number that Carl rolls? Length of hold (minutes) 1 2 3 4 5 6

Frequency 14 7 6 12 8 3

Investigation: Opinion Polls

Investigate online

mathspace.co

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11.03 Back-to-back stem and leaf plots After this lesson, you will be able to… • understand what back-to-back stem and leaf plots are and how they differ from regular stem and leaf plots. • construct back-to-back stem and leaf plots. • interpret and compare two sets of data using back-to-back stem and leaf plots. • solve problems involving back-to-back stem and leaf plots.

Back-to-back stem and leaf plots Back-to-back stem and leaf plots are very similar to stem and leaf plots. However, they compare and display 2 sets of data. We call this a bivariate data display because it displays two sets of data that we want to analyse together. Group A

6 8

This is a back-to-back stem and leaf plot.

Group B 7

Key: 3|1 = 13

Both Group A and Group B share a common stem.

8 9

Key: 1 |0 = 10

Group A’s data is read in reverse. The stem (in the centre) is always first, and the leaves are oriented in increasing order from the stem outwards. So the lowest number in group A is 13, and the largest number in group A is 58. Group B’s data is read as per a usual stem and leaf from left to right.

Back-to-back stem and leaf plots make it easier to identify similarities, differences and patterns between two sets of data.

Example 1 The test scores of 12 students in music and French are listed: • Music: 79, 59, 74, 94, 51, 71, 93, 84, 69, 61, 86, 86 • French: 62, 71, 64, 82, 83, 99, 87, 89, 66, 73, 59, 76 Display the data as a back-to-back stem and leaf plot.

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Create a strategy Start by sorting both data sets in ascending order. Then, create a stem and leaf plot with the stem representing all the digits before the ones place, and the leaf representing the ones place.

Apply the idea Let’s arrange the data sets in ascending order: • Music: 51, 59, 61, 69, 71, 74, 79, 84, 86, 86, 93, 94 • French: 59, 62, 64, 66, 71, 73, 76, 82, 83, 87, 89, 99 Here’s the stem and leaf plot for both subjects: French

Music 9

Key: 9|5 = 59

Key: 5 |1 = 51

Reﬂect and check Sorting the data in ascending order can help prevent errors when constructing the stem and leaf plot.

Example 2 The back-to-back stem and leaf plot shows the batting scores of two cricket teams, the Grasshoppers and the Cicadas. Grasshoppers

Key: 2|1 = 12

Cicadas 5

Key: 1 |6 = 16

a What is the highest score for the Grasshoppers?

Create a strategy

Apply the idea

Find the largest leaf of the largest stem in the Grasshopper’s scores.

To ﬁnd the highest score for the Grasshoppers, look for the largest stem, which is 5, and the largest leaf with it, which is 7. So, the highest score for the Grasshoppers is 57.

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b What is the highest score for the Cicadas?

Create a strategy

Apply the idea

Look for the largest leaf with the largest stem in the Cicada’s scores.

To ﬁnd the highest score for the Cicadas, look for the largest stem, which is 6, and the largest leaf with it, which is 5. So, the highest score for the Cicadas is 65.

Example 3 The back-to-back stem and leaf plot shows the length (in minutes) of a random sample of phone calls made by Sharon and Tricia. Sharon

Tricia 3

Key: 5|1 = 15

Key: 1 |2 = 12

Determine if each statement is true. a Sharon did not make a 13-minute phone call.

Create a strategy

Apply the idea

Sharon’s stem and leaf plot is on the left, so look for 3∣1 on her plot.

We can see from her stem and leaf plot that she has 3∣1 which means she made a 13-minute phone call. So, the statement is incorrect.

b Sharon’s median is higher than Tricia’s median.

Create a strategy Since both of them made 12 phone calls, ﬁnd the average of their 6th and 7th values.

Apply the idea Sharon’s 6th value is 27, and her 7th value is 38. So her median is Tricia’s 6th value is 32, and her 7th value is 34. So her median is

Sharon’s median (32.5) is lower than Tricia’s median (33), which makes the statement incorrect.

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c The mean is greater than the median of both groups.

Create a strategy Find the mean or the average minutes of Sharon and Tricia.

Apply the idea Get the average

Evaluate Get the average

Evaluate

Sharon’s mean (34.33) is greater than her median (32.5). Tricia’s mean (35.67) is greater than her median (33). So, the statement is correct.

Idea summary Back-to-back stem and leaf plots are constructed similarly to a typical stem and leaf plot. They are used to compare the measures of central tendency and spread between 2 sets of data.

11.03 Practice questions Understanding

Explorer 1–3

Adventurer

Trailblazer

1–3

How many sets of data does a back-to-back stem and leaf plot display?

What is the actual value of the 3 numbers circled?

Lemomade sales Week 1

Week 2 8

Key: 5|1 = 15 3

Key: 2 |1 = 21

For what purposes is a back-to-back stem and leaf plot useful?

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Fluency Ex 1

4–8, 10, 12

4, 6, 7, 9, 11

The test scores of 12 students in history and maths are listed: • History: 81, 65, 72, 92, 57, 75, 95, 88, 77, 63, 80, 86 • Maths: 60, 79, 68, 90, 85, 97, 89, 91, 64, 71, 55, 82 Display the data as a back-to-back stem and leaf plot.

Two friends have been growing sunﬂowers. They have measured the height of their sunﬂowers to the nearest centimetre, with their results shown: • Oliver: 15, 25, 19, 15, 34, 44, 38, 27 • Valentina: 7, 45, 14, 22, 43, 16, 21, 26, 29, 49, 8, 35, 2 Display the data as a back-to-back stem and leaf plot.

Ex 2

Ex 3

The back-to-back stem and leaf plot shows the batting scores of two cricket teams, the Wildcats and the Thunderbolts:

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Thunderbolts

What is the highest score for the Wildcats?

What is the highest score for the Thunderbolts?

Which team has the higher median?

Key: 7|3 = 37

Key: 3 |2 = 32

The back-to-back stem and leaf plot shows the length (in centimetres) of a random sample of paper chains made by Layla and Alina:

Layla

Alina

Wildcats

Who made the 14-centimetre paper chain?

Who has the higher median?

Is Layla’s mean greater than her median?

Is Alina’s mean greater than her median?

The weight (in kilograms) of a group of people aged below 30 and over 30 were recorded and presented in the back-to-back stem and leaf plot as shown: a

Find the mean weight of the group of 30 years and over.

Find the mean weight of the group of under 30 years of age.

Which group is heavier? Explain your answer using data.

Mathspace WA Curriculum – Year 9 mathspace.co

Key: 7|1 = 17

Key: 1 |5 = 15

Under 30 years of age

30 years and over

Key: 6|5 = 56

Key: 6 |3 = 63

The back-to-back stem and leaf plot shows the number of books read in a year by a random sample of university and high school students: a

Find the lowest score for the University students.

Compare the medians of both groups of students.

For which group is the mean greater than the median?

University students

High school students 7

5 6 3

Key: 4|1 = 14 10

The back-to-back stem and leaf plot shows the number of pieces of paper used over several days by Charlie and Dylan’s students: a

Did Charlie’s students use 32 pieces of paper on any day?

Who’s class had the higher median?

For which group is the mean greater than the median?

Find the lowest score for Hotel A.

Which hotel’s median is higher?

For which group is the mean greater than the median?

Charlie’s students

3 4

The back-to-back stem and leaf plot shows the number of desserts ordered at Hotel A and Hotel B over several days:

Key: 1 |2 = 12

2 3

Dylan’s students

Key: 4|0 = 4

Key: 1 |3 = 13

Hotel A

Hotel B

Key: 2|1 = 12

Key: 1 |5 = 15

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The back-to-back stem and leaf plot shows the amount of cash (in dollars) carried by a random sample of students and teachers:

Teachers

Students 7

Which group carried more cash?

Find the median amount of cash that the students carried.

5 4 1

2 6 8

Find the median amount of cash that the teachers carried.

8 5 4

3 4 4 6 6 8 9

9 8 2 2 2 1

3 4 6

9 7 4 3

8 5 2

3 1

Which group has a greater range in the amount of cash?

Key: 7|0 = 7

Reasoning 13

Key: 2 |6 = 26

13, 14

Elizabeth is analysing the test scores of two classes, Class A and Class B, using a back-to-back stem and leaf plot. Based on the plot, which class appears to have a higher average test score? Justify your answer.

13, 14

Test Score Class A

Class B 3

4 0 1 5 6 8 9 4

5 0 2 4 6 8

8 7 1 0 6 0 1 3 4 5 6 7 9 6 5 3 0

0 1 2 4 5 7

8 7 6 4 2 0 8 7 4 2 0 9 Key: 9|3 = 39

Adal and Dado are arguing over who performed best overall in their exams this year. They have drawn up a back-to-back stem and leaf plot to compare their results. Who has performed better? Justify your answer using data.

Key: 3 |5 = 35

Adal’s results

Mathspace WA Curriculum – Year 9 mathspace.co

Key: 2|1 = 12

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Dado’s results

Key: 0 |4 = 4

Problem-solving 15–17 15

Find the median score for: i

b c

India

Sri Lanka

Sri Lanka 9

Find the mean score of: i

15–19

The back-to-back stem and leaf plot shows the batting scores of two cricket teams, India and Sri Lanka: a

15–18

India

Sri Lanka

If a score of 95 were to be added to Sri Lanka:

Key: 4|3 = 34

Key: 3 |1 = 31

Is 95 within Sri Lanka’s usual score range? Why or why not?

How does the added score affect the mean and median for Sri Lanka?

iii

Which measure of centre was least affected by the addition of the outlier score of 95?

The back-to-back stem and leaf plot shows the test scores of a school’s two Year 7 classes, A and B.

Test score

What is the highest score in Class A?

Class A

What is the highest score in Class B?

Find the mean score of class A, rounded to two decimal places.

Find the mean score of class B, rounded to two decimal places.

Calculate the overall mean of the Year 7 students, rounded to two decimal places.

The back-to-back stem and leaf plot shows the results of a survey conducted on the price of concert tickets locally and the price of the same concerts at an international venue: a

What was the most expensive ticket price at the international venue?

At the international venue, what percentage of tickets cost between $90 and $110 (inclusive)?

Class B

Key: 9 |6 = 69

Key: 6 |4 = 64

Local

International

Key: 4|6 = 64

Key: 6 |1 = 61

At the local venue, what percentage of tickets cost between $90 and $100 (inclusive)?

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A basketball coach is analysing the scores, of players, on his team, Team Amazing, and an opposing team.

Calculate the probability that a randomly selected player from Team Amazing has a score greater than 70. Leave your answer as a fraction.

Team Amazing

Team Incredible 5

8 7 5 Compare the probability of selecting a 9 4 1 player with a score greater than 70 from Team Amazing to the probability of selecting Key: 0|6 = 60 a player with a score greater than 70 from Team Incredible. Which team has a higher probability?

9 Key: 5|1 = 51

A local library conducted two reading challenges, one in summer and one in winter, to encourage children to read. The back-to-back stem and leaf plot displays the number of books each child read during the two challenges.

The goal of the challenge was to read 47 books. a

What percentage of the children in summer achieved the goal?

What percentage of the children in winter achieved the goal?

Was the summer or winter challenge more successful?

If the goal for next year is to be adjusted to the overall median of the number of books read across the summer and winter challenge, calculate this value, rounded to the nearest whole number.

What is the mode of the number of books read across the summer and winter challenge, and what does it represent in the context of the reading challenge?

11.04

Reading challenge Summer

Winter

Key: 9|1 = 19

Grouped histograms

• understand grouped frequency tables and how they are used with continuous data. • construct and interpret grouped histograms. • calculate the measures of centre and spread in a grouped frequency table. • understand the advantages and disadvantages of grouped frequency tables. • solve problems related to grouped frequency tables and histograms.

Mathspace WA Curriculum – Year 9 mathspace.co

Key: 1 |2 = 12

After this lesson, you will be able to…

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Grouped frequency tables Grouped frequency tables can be helpful when collecting continuous numerical data or when the data is more spread out. It shows the number of values within a class interval or group. We can ﬁnd the frequency of a group by adding all the frequencies of the results contained in that group. Each class interval must be the same size. This grouped frequency table shows the time (minutes) taken, x, for 126 people to complete a 5 kilometre walking track: Class Interval 45 ≤ x < 50 50 ≤ x < 55 55 ≤ x < 60 60 ≤ x < 65 65 ≤ x < 70

Frequency 9 7 20 30 60

This is continuous data, so we use a class interval that captures the full range of scores possible.

The ﬁrst class interval includes the walking times for 9 different people. Each of their times fall within 45 to 50 minutes, including 45 but not including 50. The class centre is the midpoint or mean of each class interval. The class centre for the ﬁrst class interval would be:

Class Interval 45 ≤ x < 50 50 ≤ x < 55 55 ≤ x < 60 60 ≤ x < 65 65 ≤ x < 70

Class center 47.5 52.5 57.5 62.5 67.5

Frequency 9 7 20 30 60

To ﬁnd the measures of centre and spread in the grouped frequency table above: • The modal class in a grouped frequency table is the group that has the greatest frequency. If there are multiple groups that share the greatest frequency, then there will be more than one modal class. The modal class is 65 to 70. • The range, in grouped frequency tables, can be found by calculating the difference between the highest and lowest possible values. The range is 70 − 45 = 25. • The median is the middle value of ordered data. In this case, the position of the median is between the 63rd and 64th position when the numbers are arranged in ascending order. The median would be found in the 60–65 class interval. The median would be the class centre of this interval, which is 62.5. • We also use the class centres to calculate the mean. The mean is found by ﬁrst multiplying the class centre of each group of scores by its corresponding frequency, then adding our results, and dividing it by the total frequency. The mean is approximately 62.46, rounded to two decimal places. The disadvantage of a grouped frequency table is that the data becomes less precise, since we are observing grouped multiple data points together rather than individual data points.

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Example 1 Consider the table below. Score 1–4 5–8 9–12 13–16 17–20

Frequency 2 7 15 5 1

Note: This is discrete data so we display the intervals as discrete values, 1–4, then 5-8. It implies that there are no scores possible between 4 and 5.

a Use the class centre (midpoint) of each class interval to determine an estimate for the mean of the sample distribution. Round your answer to one decimal place.

Create a strategy To ﬁnd the midpoint of each class, ﬁnd the average of the two end points.

Apply the idea For the ﬁrst class, we get the midpoint (or class centre) to be for each class, we get the following midpoints: Score 1–4 5–8 9–12 13–16 17–20 Total

Class center 2.5 6.5 10.5 14.5 18.5

Frequency 2 7 15 5 1 30

. Using the same process

Class center × Frequency 5 45.5 157.5 72.5 18.5 299

To ﬁnd the mean, we multiply each class centre by the frequency, add them together, and divide by the number of scores. The number of scores can be found by adding the frequencies.

Use the formula

Substitue the values

Evaluate and round

b Find the modal class of the data.

Create a strategy Choose the class that has the highest frequency.

Apply the idea Looking at the table, we can see that the modal class is the group 9–12, since it has the highest frequency.

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Example 2 The scores of a basketball team are collected over a season. 65, 41, 53, 76, 37, 48, 68, 34, 59, 72, 45, 50, 32, 78, 62, 69, 56, 76, 75, 47 a Determine a set of ﬁve class intervals that could be used to separate this data.

Create a strategy

Apply the idea

Identify the maximum and minimum values in the The minimum value is 32 and the maximum data set and ﬁnd appropriate class intervals. value is 78. Each class interval must be the same size.

Five class intervals could be: 30–39, 40–49, 50–59, 60–69 and 70–79.

b Construct a grouped frequency table for the data.

Create a strategy

Apply the idea

Sort the data, and determine which class For 30–39, there are 3 values: 32, 34, 37 intervals they belong to. Then, count the number For 40–49, there are 4 values: 41, 45, 47, 48 of values in each class interval. For 50–59, there are 4 values: 50, 53, 56, 59 Note: This is discrete data, it is not possible to For 60–69, there are 4 values: 62, 65, 68, 69 achieve a score of 39.5, so the classes are set For 70–79, there are 5 values: 72, 75, 76, 76, 78 as 30–39, 40–49, etc. Arrange these into a grouped frequency table. Scores 30–39 40–49 50–59 60–69 70–79

Frequency 3 4 4 4 5

Idea summary A grouped frequency table shows the number of values within a set class interval. We can ﬁnd the frequency of a group by adding all the frequencies of the data contained in each group. Each class interval must be the same size. The class centre is the mean of the highest and lowest possible scores in the group. The measures of centre and spread of grouped frequency tables include: • The modal class is the group that has the greatest frequency. • The mean can be found by ﬁrst multiplying the class centre of each group of scores by its corresponding frequency, then adding our results, and dividing it by the total frequency. • The median is the middle value of ordered data.

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Grouped histograms A histogram is a visual representation of a frequency table. It displays the frequency distribution of continuous data. In a histogram, the class frequency is the height of the rectangle.

Exploration Histograms and column graphs look very similar. The ﬁrst graph is a histogram, the second is a column graph. 16 14 Frequency

12 10 8 6 4 2 0

100

110

120

130 140 150 Speed (km/h)

160

170

Waste generated in Australia by industry 2016/17 financial year

Amount of waste (millions of tonnes)

25 20 15 10 5 0

Households Construction Commercial Coal powered and local and and electricity councils demolition industrial generation

Compare the data type used for each graph.

What do you notice about the axes for the histogram and column graph?

Are you able to change the order of the columns in either of the graphs?

The key features of a histogram are: • The horizontal axis is a continuous numerical scale (like a number line), where the class intervals are placed on either side of the columns. • The vertical axis is the frequency of each data value or group of values. • There are no gaps between the columns because the horizontal axis is a continuous scale. It is possible for a class interval to have a frequency of zero, but this is not the same as having gaps between each column. • When creating a histogram, leave a half-column-width gap between the vertical axis and the ﬁrst column. 648

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This frequency distribution table and histogram shows the time taken, for 73 runners, to complete a 10 kilometre race: Class interval

Class centre

Frequency

45 ≤ x < 50

47.5

50 ≤ x < 55

52.5

55 ≤ x < 60

57.5

60 ≤ x < 65

62.5

65 ≤ x < 70

67.5

Frequency

Note: This is continuous data so we use inequality notation to capture the full range of scores.

35 30 25 20 15 10 5 0

The median, which is the middle score, is the 37th in the list of 73 runners. Adding the frequencies for the first three intervals (9 + 7 + 20 = 36) shows that the 37th score falls within the 60–65 minute interval. Since the data is grouped into intervals, the exact median value cannot be determined directly, we call this the median class.

Distribution of running times for a 10 km race 73 runners

The mode is the most common class interval (tallest column). The modal class is 60–65.

50 55 60 65 Running time (minutes) 70 – 45 = 25 Range

The mean can be calculated in a similar way as we did for the frequency table. First, multiply the class centre of each group of scores by its corresponding frequency. Then, sum those up and divide by the total frequency. The mean is approximately 58.80. The range is the difference between the highest and lowest score.

The class centre (midpoint) is not typically used in the histogram itself but can be useful for statistical calculations, like estimating means.

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Example 3 Some people were asked how many of their high school friends they remained in contact with after high school. The results were grouped into categories, presented in the frequency distribution table: Note: The score is a count of people which is discrete. It is not possible to have a score of 9.5 people.

Score 0–9 10–19 20–29 30–39 40–49

Frequency 7 10 7 1 2

a Construct a histogram for the data.

Create a strategy Examine and use the frequency table to determine how many times each range of scores occurred.

The vertical axis as the frequency of each data value or group of values.

Frequency

Apply the idea

Half-column-width gap between the vertical axis and the first column.

10 9 8 7 6 5 4 3 2 1 0

There are no gaps between the columns.

0 10 20 30 40 50 No. of friends

The horizontal axis is a continuous numerical scale, where intervals are placed on either side of the columns.

b Find the modal class of the data.

Create a strategy

Apply the idea

Examine the graph or use the frequency table to Based on the given frequency table and the assess which range of scores occurs the most graph from part (a), the modal class is 10–19 as (has the highest frequency). this is the range of scores that occurs the most.

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Example 4 The heights of students in a class were recorded. The frequency table shows the ﬁndings. Height (in cm) 140.9 < h ≤ 150.9 150.9 < h ≤ 160.9 160.9 < h ≤ 170.9 170.9 < h ≤ 180.9 180.9 < h ≤ 190.9

Frequency 7 3 10 2 3

Note: Heights are continuous data, so we write the classes using inequalities to capture the full range of scores.

a What could be the height of the tallest student in the class?

Create a strategy

Apply the idea

Identify the highest class interval value in the last row.

190.9

b How many students in the class are taller than 160.9 cm?

Create a strategy

Apply the idea

Identify the total frequency in the corresponding We need to sum the frequencies from the class intervals. following class intervals: 160.9 < h ≤ 170.9, 170.9 < h ≤ 180.9 and 180.9 < h ≤ 190.9 10 + 2 + 3 = 15 So there are 15 students who are taller than 160.9 cm. c Construct a histogram for the data.

Create a strategy Examine and use the frequency table to determine how many times each range of height (in cm) occurred.

Apply the idea

Frequency

0. 15 9 0. 16 9 0. 17 9 0. 18 9 0. 19 9 0. 9

Half-column-width gap between the vertical axis and the first column.

There are no gaps between the columns.

The vertical axis as the frequency of each data value or group of values.

10 9 8 7 6 5 4 3 2 1 0 Height (in cm)

The horizontal axis is a continuous numerical scale, where intervals are placed on either side of the columns.

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d If an additional column is added to the right side of the histogram, what would the new class interval be?

Create a strategy Recognise from previous rows that the lower class interval value uses a “less than” symbol (<) while the higher class interval value uses a “less than or equal to” symbol (≤). This is because we want to include the value from the previous class interval, but exclude it in the next class interval. Also notice that the end values have a difference of 10.

Apply the idea 190.9 < h ≤ 200.9

Idea summary Histograms differ from column graphs: • They are for numerical continuous data, not categorical data. • They have a numerical scale on the horizontal axis, where the class intervals are on either side of the columns. • They do not have gaps between each column. Measures of centre and spread for data presented in a histogram can be calculated in a similar way to those from frequency tables.

11.04 Practice questions Understanding 1

Explorer 1–2

Adventurer 1–3

Complete the frequency table based on the data set: 36, 50, 66, 46, 78, 63, 58, 39, 84, 40, 81, 45, 86, 51, 68, 43, 64, 67, 37, 68, 48, 65, 60, 82, 75

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Trailblazer 1–3

Class 30–39 40–49 50–59 60–69 70–79 80–89

Frequency

For each survey, state whether the data found should be grouped or ungrouped when constructing a frequency table: a

A survey conducted of 1000 people, asking them how many languages they speak.

A survey conducted of 1000 people, asking them how many different countries they know the names of.

A survey conducted of a class of 30 students, asking them how many pets they have.

A survey conducted of a class of 30 students, asking them how many minutes they spend on their phone each week.

Identify if the statements are true or false. a

The modal class is the one with the least frequency.

The class centre is found by calculating the range of the class interval values.

The median in a grouped frequency table is found by identifying the middle position and ﬁnding the corresponding class centre.

The mean in a grouped frequency table is found by ﬁrst multiplying the class centre of each group of scores by its corresponding frequency, then adding our results, and dividing it by the total frequency.

Fluency 4 Ex 1a

4–6, 8a, 9, 11–16

4–7, 8b, 10, 11, 13–17

For the frequency table: a

Complete the table.

Calculate an estimate for the mean. Round your answer to two decimal places.

Class 1–9 10–18 19–27 28–36 37–45

Class centre

f 8 6 4 6 8

State the modal class of the data set.

Class 30–39 40–49 50–59 60–69 70–79 80–89

Frequency 8 5 4 3 3 2

For the frequency table:

Score 1–5 6–10 11–15 16–20 21–25 26–30

Frequency 20 15 8 4 3 2

Ex 1b

4–7, 8b, 10, 11, 13–17

Estimate the mean by calculating the class centre for each row. Round your answer to one decimal place.

State the modal group of scores.

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Ex 2

For the frequency table: a

Estimate the mean. Round your answer to one decimal place.

State the modal group of scores.

Score 1≤x<6 6 ≤ x < 11 11 ≤ x < 16 16 ≤ x < 21 21 ≤ x < 26 26 ≤ x < 31

Frequency 18 14 9 5 3 1

Construct a grouped frequency table for the data. a

46, 54, 35, 23, 24, 28, 26, 11, 19, 17, 32

83, 68, 39, 42, 86, 66, 64, 76, 63, 43, 65, 83, 63, 67, 49, 51, 32, 55, 38, 65, 41, 73, 35, 36, 74

For the set of scores: 30, 67, 24, 51, 49, 53, 36, 24, 57, 66, 26

Determine a set of ﬁve class intervals that could be used to analyse this data.

Construct a grouped frequency table for the data.

For the set of scores: 12, 59, 61, 27, 58, 18, 76, 27, 52, 19, 13, 56, 71, 31, 73, 60, 41, 17, 22, 68, 57, 15, 40, 19, 76, 44, 60, 55, 36

Ex 3

Determine a set of seven class intervals that could be used to analyse this data.

Construct a grouped frequency table for the data.

A survey was conducted to collect data on the number of hours a group of students spent studying per week, rounded to the nearest hour. The collected data was then organised into a frequency distribution table.

Create a histogram to represent the data.

Find the modal class of the data.

In product testing, the number of faults detected in producing a certain machinery is recorded each day for several days. The frequency table shows the results. a

Construct a histogram to represent the data.

What is the lowest possible number of faults that could have been recorded on any particular day?

Hours 0−4 5−9 10−14 15−19 20−24

Number of faults 0–3 4–7 8–11 12–15

Frequency 12 20 15 8 5

Frequency 10 14 20 16

If the range of a set of continuous data is 28, ﬁnd the size of each class interval if we group the data into: a

4 equal width class intervals.

7 equal width class intervals.

If one of the class intervals for height is 10 < height ≤ 19, ﬁnd the next equal-width class interval.

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As part of a fuel watch initiative, the price of petrol, p, at a service station was recorded each day for 21 days. The frequency table shows the ﬁndings.

Frequency 4 6 5 6

Find the highest price that could have been recorded.

How many days was the price above 130.9 cents?

Create a histogram for the petrol prices over the 21 days.

If an additional column is added to the right side of the histogram, what would the new class interval be?

The histogram shows the number of hours that students in a particular class had slept for the night before. 12 11 10 9 8 7 6 5 4 3 2 1 0

6 8 Time (hours)

Construct a histogram to display the data shown in the frequency table.

Which groups represent peaks in the data set?

18, 19

How many students are in the class?

How many students had at least 8 hours of sleep that night?

What percentage of students had less than 6 hours of sleep?

The frequency table shows the price (to the nearest dollar) of the most recent book that seventy-two university students bought.

Reasoning 18

Price (in cents per litre) 120.9 < p ≤ 125.9 125.9 < p ≤ 130.9 130.9 < p ≤ 135.9 135.9 < p ≤ 140.9

Frequency

Ex 4

18, 19

Price ($) 0–19 20–39 40–59 60–79 80–99 100–119 Total

Frequency 6 20 12 6 19 9 72

18–20

The frequency table shows the data distribution for the length of leaves (in millimetres) collected from a species of tree in the botanical gardens. a

How many leaves less than 60 mm were collected?

What was the most common interval of leaf length collected?

Are leaves more likely to be at least 60 mm in length?

Can we conclude from the table that there were no leaves collected with a length less than 5 mm? Explain your answer.

Leaf length 0 ≤ x < 20 20 ≤ x < 40 40 ≤ x < 60 60 ≤ x < 80 80 ≤ x < 84

Frequency 5 11 19 49 43

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The frequency table displays the number of books read by a group of readers in a month. a

Create a histogram to represent the data from the frequency table.

Determine whether each of the statements about the data is accurate and justify your reasoning.

The majority of readers in the group read between 5 and 10 books in a month.

Most readers in the group read less than 15 books in a month.

iii

Only a small number of readers in the group read 20 or more books in a month.

Salvia and Lee create two different histograms for the same data set. The data collected shows the years of life expectancy after patients were diagnosed with lung cancer. Salvia’s Histogram

60 Frequency

Frequency

15 10 5 0

a b

Lee’s Histogram

40 30 20 10 0

5 10 15 20 25 30 Years

20 30 Years

Explain which histogram is more useful to analyse the data set. Lee claims that their histogram shows that 51 patients lived 30 years after being diagnosed with lung cancer. Is this correct? Explain your answer.

Problem-solving 21, 22

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Frequency 12 18 8 5 2

Number of books read 0≤x<5 5 ≤ x < 10 10 ≤ x < 15 15 ≤ x < 20 20 ≤ x < 25

21–23

The frequency table shows the average time spent travelling to work for 50 people: a

Construct a histogram to display the data shown in the frequency table.

State whether the statements about the data are accurate.

21–23

Commute time (minutes) 0 ≤ Time < 20 20 ≤ Time < 40 40 ≤ Time < 60 60 ≤ Time < 80 80 ≤ Time < 100 Total

Frequency 14 16 10 7 3 50

The data shows that most people travel to work by car or by walking, since most travel times are fairly short, and only a few people travel by bus or train.

The data suggests that people prefer a shorter commute to work. A majority live within 40 minutes travel, and in general the longer the commute the less people there are in that category.

iii

The data suggests that people don’t care too much about how far away from work they live. Roughly equal portions of people live less than 40 minutes away and more than 40 minutes away.

The data shows that everyone lives within an hours travel from their work, with the peak amount of people living between 20 and 40 minutes away.

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The histogram shows the speed of balls delivered by cricket bowlers. 16 14

Frequency

12 10 8 6 4 2 0

110

120

130 140 Speed (km/h)

150

160

170

Complete a frequency table for this set of data.

What percentage of balls were bowled at less than 130 km/h? Round your answer to one decimal place.

A battery manufacturer tests their batteries in a portable music player, to ﬁnd out how long the batteries last under constant use. The histogram shows the data from their tests. 35 30 25 Frequency

100

20 15 10 5 0

400 500 600 700 800 900 1000 1100 1200 1300 Battery Life (mins)

Create a frequency table for this set of data.

What percentage of batteries last longer than 10 hours? Round your answer to one decimal place.

If the average length of a song is 4 minutes, what percentage of batteries will last for an average of 250 songs or more? Round your answer to one decimal place.

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11.05 Optional: Frequency histograms and polygons After this lesson, you will be able to… • interpret the data represented in a histogram. • interpret the data represented in a frequency polygon. • construct a frequency table from a given histogram or frequency polygon. • solve problems involving histograms and frequency polygons.

Frequency histograms and polygons A frequency polygon is like a histogram in that it displays the scores on the horizontal axis and the frequencies on the vertical axis. To create the frequency polygon, the centre of the top of each column is joined together and the endpoints joined to the horizontal axis. 10

0 2 4 6 8 10 12 14 16 1820

Histogram

1 3 5 7 9 11 13 15 17 1921

Frequency polygon with the same data

Notice that the vertices of the frequency polygon are in the same place as the centre of the tops of the bars of the histogram. We can use frequency polygons in the same way that we use histograms.

Example 1 People in a city responded to a survey that asked how many minutes, m, they spent on average using public transportation to and from work each day. The results are displayed in the frequency polygon. a Complete the frequency table for the given frequency polygon. Minutes 140 ≤ m < 150 150 ≤ m < 160 160 ≤ m < 170 170 ≤ m < 180 180 ≤ m < 190 190 ≤ m < 200 658

Class centre

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Frequency

18 16 14 12 10 8 6 4 2

Minutes 135 145 155 165 175 185 195 205 215

Create a strategy The horizontal values of the vertices of the frequency polygon represent the class centres of each interval. The frequency is the height, or vertical value, of each vertex.

Apply the idea Minutes 140 ≤ m < 150 150 ≤ m < 160 160 ≤ m < 170 170 ≤ m < 180 180 ≤ m < 190 190 ≤ m < 200

Class centre 145 155 165 175 185 195

Frequency 2 6 14 20 10 6

Reﬂect and check

Frequency

This image shows what the frequency histogram would look like. Notice that each vertex of the frequency polygon is in the centre of each column of the histogram, and the endpoints of the frequency polygon lie on the horizontal axis. 22 20 18 16 14 12 10 8 6 4 2 0

130

140

150

160

170 180 Minutes

190

200

210

b Estimate the mean of the data set, rounded to two decimal places.

Create a strategy To estimate the mean, multiply each class centre by the frequency, ﬁnd the sum of all the products, then divide by the number of scores.

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Apply the idea Minutes 140 ≤ m < 150 150 ≤ m < 160 160 ≤ m < 170 170 ≤ m < 180 180 ≤ m < 190 190 ≤ m < 200 Total

Class centre 145 155 165 175 185 195

Frequency 2 6 14 20 10 6 58

Class centre × Frequency 290 930 2310 3500 1850 1170 10 050 Write the formula Substitute the values Evaluate and round

Example 2 Consider this frequency polygon:

Frequency

8 7 6 5 4 3 2 1

What is the modal class of this data set?

Score (x) 0 4 8 12 16 20 24 28 32 36

Create a strategy Look for the score with the highest point on the polygon. Then, ﬁnd the midpoints of the class centres to ﬁnd the end values of the class interval.

Apply the idea The score 28 has the highest frequency. Since 28 is the class centre, the end values of the class interval will be the midpoints of the class centres. The class centre below 28 is 24.

The class centre above 28 is 32.

The modal class is 26 ≤ x < 30.

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Example 3 A Year 9 class was surveyed on the approximate number of minutes they spent on homework last night. The raw data collected from 15 students was as follows: 35, 50, 25, 65, 40, 75, 30, 55, 45, 60, 20, 70, 50, 35, 60 a Organise this data into a grouped frequency table.

Create a strategy First, find the range of the data. Then, determine an appropriate class width to create 4 equal intervals starting from 20 minutes. Finally, tally the data into these intervals to find the frequencies and calculate the class centres.

Apply the idea The minimum value is 20 and the maximum value is 75. The range to cover starting from 20 up to 75 is 75 − 20 = 55. Desired number of intervals is 4. Estimated class width is convenient class width of 15 minutes. Time (minutes) 20 ≤ x < 35 35 ≤ x < 50 50 ≤ x < 65 65 ≤ x < 80

Class Centre 27.5 42.5 57.5 72.5

. Let’s use a

Frequency 3 4 5 3

The total frequency is 3 + 4 + 5 + 3 = 15.

b Construct a frequency polygon for this grouped data.

Create a strategy

Apply the idea

Plot points using the class centres as x-coordinates and frequencies as y-coordinates. Connect these points with lines. Extend the polygon to the horizontal axis one class width before the first class centre and one class width after the last class centre, both with frequency 0.

The class width is 15. Points for polygon: (27.5, 3), (42.5, 4), (57.5, 5) and (72.5, 3). Start point on axis: (27.5 − 15, 0) is (12.5, 0). End point on axis: (72.5 + 15, 0) is (87.5, 0). Frequency 5 4 3 2 1 Time (minutes) 20 30 40 50 60 70 80

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c From the grouped data, identify the modal class and estimate the mean homework time rounded to one decimal place.

Create a strategy The modal class is the class interval with the highest frequency. To estimate the mean, multiply each class centre by its frequency, sum these products, and divide by the total frequency.

Apply the idea The modal class is 50 ≤ x <65 minutes, as it has the highest frequency (5). Estimate the mean: Class Centre (m) 27.5 42.5 57.5 72.5 Total

Frequency ( f) 3 4 5 3 15

f×m 82.5 170.0 287.5 217.5 757.5

Write the formula

Substitute the values

Evaluate

The estimated mean homework time is 50.5 minutes.

d Describe the shape and spread of this distribution of homework times.

Create a strategy Examine the frequency polygon and the table to describe the shape (e.g., symmetrical, skewed, unimodal). For spread, consider the range of the data and how frequencies are distributed.

Apply the idea Shape: The distribution is unimodal, with its peak in the 50 ≤ x < 65 minutes class. It appears to be reasonably symmetrical, with frequencies decreasing on either side of the modal class at a similar rate (3, 4 rising to 5, then 5 falling to 3). There isn’t a strong skew to the left or right. Spread: The original data ranged from 20 to 75 minutes, a range of 55 minutes. The grouped data covers intervals from 20 minutes up to (but not including) 80 minutes. The frequencies are spread across all four intervals, but with a clear concentration in the middle two intervals (35 ≤ x< 65 minutes account for 4 + 5 = 9 out of 15 students, or 60%). This indicates a moderate spread, not extremely concentrated nor extremely dispersed.

Idea summary A frequency polygon is similar to a histogram in that it displays the frequencies on the vertical axis. However, the scores on the horizontal axis represent the class centres of each class interval.

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11.05 Practice questions Explorer

Understanding

1, 2

1–3

Trailblazer 1–3

Identify if each of these graphs are histograms or column graphs. Number of calls

10 9 8 7 6 5 4 3 2 1 0

2 3 4 5 Number of minutes

Number of students

Walkathon 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

Number of laps

Number of Boys

c 10 9 8 7 6 5 4 3 2 1 0

Preferred Sport

Football Tennis

Rugby Basketball Hockey

Sport

Frequency

Adventurer

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Match each of these frequency polygons to their corresponding frequency histogram. Graph A

Histogram 1

Frequency

7 6

Frequency

5 4 3 2

5 4 3 2 1 0

1 Score 0

10 20 30 40 50 60 70

Graph B

Histogram 2

Frequency 7

Frequency

5 4

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3 2

5 4

Score 0

0 10 20 30 40 50 60 80 Class intervals

When constructing a frequency polygon from a frequency histogram: a

What does the horizontal position (x-coordinate) of each main vertex (point) of the frequency polygon correspond to on the original histogram?

What does the vertical position ( y-coordinate) of each main vertex (point) of the frequency polygon correspond to on the original histogram?

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Fluency Ex 1a

4a, 5–7

4b, 5–8

4bc, 6–9

Complete these frequency tables using the given frequency polygons: a

Frequency 7 6 5 4 3

Class 4≤x<6 6≤x<8 8 ≤ x < 10 10 ≤ x < 12 12 ≤ x < 14

Frequency

Class 20 ≤ x < 30 30 ≤ x < 40 40 ≤ x < 50 50 ≤ x < 60 60 ≤ x < 70 70 ≤ x < 80

Frequency

Class 76 ≤ x < 84 84 ≤ x < 92 92 ≤ x < 100 100 ≤ x < 108 108 ≤ x < 116 116 ≤ x < 124

Frequency

2 1

Score (x) 3

Frequency 9 8 7 6 5 4 3 2 1

Score (x) 15 25 35 45 55 65 75 85

Frequency 9 8 7 6 5 4 3 2 1

Score (x) 72 80 88 96 104 112 120 128

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Ex 1b

Estimate the mean of this data set, rounded to one decimal place.

Frequency

8 7 6 5 4 3 2 1

Score 30 40 50 60 70 80 90 100

Ex 2

Ex 3

stimate the mean of the scores, rounded to E two decimal places.

Find the modal class.

20 18 16 14 12 10 8 6 4 2

Frequency

Score (x) 38 40 42 44 46 48 50

The following data represents the scores achieved by 15 students in a science quiz (out of 50): 28, 42, 35, 22, 48, 30, 38, 26, 45, 33, 29, 40, 36, 24, 39

Organise the data into a grouped frequency table.

Construct a frequency polygon for this grouped data.

Identify the modal class(es) and estimate the mean score, rounded to one decimal place.

Describe the shape and spread of the distribution of these science quiz scores.

Find the class interval in which the median of the data set lies.

Frequency

7 6 5 4 3 2 1

Score (x) 20 30 40 50 60 70 80 90 100

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Estimate the mean.

Find the class interval in which the median lies.

Find the modal class. Frequency

9 8 7 6 5 4 3 2 1

Score (x) 0

Reasoning

10–12

11–13

Explain why the lines of a frequency polygon are extended to meet the horizontal axis at both ends. What do these points on the axis typically represent in relation to the data’s class intervals?

A researcher has collected data on the test scores for two different classes. They want to compare the distribution of scores for both classes on the same set of axes. Explain why a frequency polygon might be a more suitable choice than a histogram for this specific purpose of comparison.

A student observes a frequency polygon where a single point is significantly higher than all other points, creating a sharp peak. The student concludes that this peak represents a single, very specific data value that occurred with the highest frequency in the dataset. Critically evaluate this student’s conclusion. Is it necessarily correct? Explain your reasoning, considering how frequency polygons for grouped data are constructed.

Imagine two frequency polygons plotted on the same axes, representing the results of two different surveys on daily screen time in hours. • Polygon X is tall and narrow, with its peak around 2 – 3 hours. • Polygon Y is shorter and wider, more spread out, with its peak also around 2 – 3 hours but with significant frequencies extending up to 6 – 7 hours. Based on these descriptions, what can you infer about the consistency or variability of daily screen time in the group represented by Polygon X compared to the group represented by Polygon Y? Which group is more likely to have individuals with very high screen times? Justify your answer.

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Problem-solving 14–16

15–19

The frequency polygon shows the frequency of calls made during each day of the week:

Number of calls

15–17

15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

Mon

Tue

Wed

Fri

Thurs

Sat

Sun

Days

How many calls were made on Friday?

What was the maximum number of calls made on a single day?

What was the total number of calls made during the week?

Find the mean number of calls made during the week.

The ages of new mothers in a particular hospital are displayed on this histogram:

Frequency

Age Frequency 28 ≤ Age < 30 30 ≤ Age < 32 32 ≤ Age < 34 34 ≤ Age < 36 36 ≤ Age < 38 38 ≤ Age < 40

New Mothers 20 16 12 8 4 0

668

29 31 33 35 37 39 Age

Complete the frequency table based on the histogram.

How many new mothers were there?

How many mothers were under 34 years old?

How many mothers were 36 years old or older?

Mathspace WA Curriculum – Year 9 mathspace.co

The Cancer Council conducted a study to determine how long people spend in the sun in a month. They asked 120 participants to wear a detector at all times. Their results are summarised in the given histogram: Hours in sun Frequency 0 ≤ Hours < 10 10 ≤ Hours < 20 20 ≤ Hours < 30 30 ≤ Hours < 40 40 ≤ Hours < 50 50 ≤ Hours < 60 60 ≤ Hours < 70 70 ≤ Hours < 80 80 ≤ Hours < 90

Cancer Council Study

16 14 Frequency

12 10 8 6 4 2 0

5 15 25 35 45 55 65 75 85 Hours in the sun

Complete the frequency table based on the histogram.

How many participants spent less than 30 hours in the sun?

What fraction of participants spent at least 60 hours in the sun?

A government agency records how long people wait on hold to speak to their representatives. The results are displayed in the given histogram:

Frequency

Time on hold in minutes Frequency 0.5 ≤ Time < 1.5 1.5 ≤ Time < 2.5 2.5 ≤ Time < 3.5 3.5 ≤ Time < 4.5 4.5 ≤ Time < 5.5

Time on hold 12 10 8 6 4 2 0

1 2 3 4 5 Length of hold (in minutes)

Complete the frequency table based on the data from the histogram.

How many phone calls were made?

How many people were on hold for at least 2.5 minutes?

How many people were on hold for less than 3.5 minutes?

This frequency table shows the number of hours students spent on social media each day. Draw a histogram and its polygon on the same graph.

Number of hours 0–1 2–3 4–5 6–7 8–9 10–11

Number of students 4 8 15 12 0 2

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What was the most common response?

Find the mean number of people who kept in touch with friends after high school. Round your answer to the nearest whole number.

11.06

Frequency

A group of people were asked approximately how many of their high school friends they remained in contact with after high school to the nearest 10 friends. The results are presented in the following frequency polygon:

20 18 16 14 12 10 8 6 4 2 0

10 20 30 40 No. of Friends

Shape of data

After this lesson, you will be able to… • identify and describe the features of symmetrical data distributions. • distinguish between positively and negatively skewed data distributions. • analyse the implications of skewness on the mean, median, and mode. • interpret different data representations to determine the skewness and clustering in data. • recognise and explain the presence of clusters in data sets. • identify clusters in data and distinguish between unimodal, bimodal and multimodal data.

Symmetry and skew Data may be described as symmetrical or skewed. This is the shape that symmetrical data would look like. It is similar to the shape of a bell. This is often referred to as a normal distribution.

There are a few features that are related to symmetrical data: • The data tends to be around a central value. • Roughly 50% of scores will be above the mean and 50% of scores will be below the mean. • The mean, median and mode should roughly coincide.

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This is the shape of data that is positively skewed. Notice the tail at the positive end. There are a few features that are related to positively skewed data: The mean is usually greater than the median. The median is usually greater than the mode. A negatively skewed graph has this general shape with the left side stretched out. There are a few features that are related to negatively skewed data: The mean is usually less than the median. The median is usually less than the mode.

Interactive exploration

Discover this concept in action online

This is the general shape of a frequency histogram with a symmetrical shape.

30 Percentage

mathspace.co

Even though the data is not exactly symmetrical, the dark line shows that a symmetrical curve can roughly be drawn over the histogram following the data.

20 15 10 5

Frequency

75 80 85 90 95 100 105 110 Weight

To describe shape we don’t need the data to be exactly the same, we are looking to describe what it is most like.

This is the general shape shown over a histogram of positively skewed data.

Frequency

Score

This is the general shape shown over a histogram of negatively skewed data.

Score

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Example 1 State whether the scores in each data display can be best described as positively skewed, negatively skewed or symmetrical.

Frequency

10 9 8 7 6 5 4 3 2 1 0

14 15 Score

Create a strategy Check where the bulk of the data sits and look at the general shape of the distribution.

Apply the idea The scores are roughly even in both the high and low end, so the distribution is symmetrical.

10 11 12 13 14 15 16 17 18 19 Score

Apply the idea The scores have higher-frequency dot stacks on the left (the lower scores), so the distribution is positively skewed.

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Stem

Leaf

0 4

2 7

3 8

0 4 7

0 0 3 4 6 7 7 9

0 1

2 2 5 6 8 9 2 4 5 5 6 7 9 9 2 2 3 5 6 8 8

Key: 1|2 = 12 score

Create a strategy

Apply the idea

If most of the scores have lower values, the data is positively skewed. If most of the scores are centred in the middle of the range, the data is symmetric. If most of the scores have higher values, the data is negatively skewed.

Most of the scores have higher values, so the distribution is negatively skewed.

Idea summary A distribution is said to be symmetric if its left and right sides are roughly mirror images of one another. A data set that has a positive skew has a longer tail of values to the right of the data set. The mass of the distribution is concentrated on the left of the ﬁgure. A data set that has a negative skew has a longer tail of values to the left of the data set. The mass of the distribution is concentrated on the right of the ﬁgure.

Clusters In a set of data, a cluster occurs when a large number of the scores are grouped together within a small range. Clustering may occur at a single location or several locations.

Frequency

If the data in a histogram has two clear peaks, then the shape is called bimodal.

Score

If the shape of data in a histogram has more than 2 peaks (modes), then it is described as multimodal. 11.06 Shape of data mathspace.co

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Example 2

Frequency

The percentage of faulty computer chips in 42 batches were recorded in the histogram.

a Which of these makes this statement true? The distribution is: A

Unimodal

Multimodal, but not bimodal

Bimodal

14 12 10 8 6 4 2 0

1 2 3 4 5 6 Faulty chips (%)

Create a strategy Count the number of peaks in the distribution.

Apply the idea Looking at the histogram, there are two peaks. So the distribution is bimodal. The correct answer is option B. b Which of these are the modal classes? Select all that apply. A

0–1

1–2

4–5

5–6

2–3

3–4

Create a strategy The modal classes are those with the highest frequency.

Apply the idea The highest frequency is 12, and the classes with this frequency are 1–2 and 3–4. So the correct answers are options B and D.

Example 3 For the data shown in this column graph: 45 40

Frequency

35 30 25 20 15 10 5 0 1

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6 Score

a Is there any clustering of data? A

Yes

Create a strategy

Apply the idea

Check to see if the data is ‘bunched’ up in any particular range.

Looking at the column graph, there are more data values (scores with higher frequencies) on the left side of the graph. The correct answer is option A.

b Where does the clustering occur? A

8 to 11

1 to 4

5 to 7

Create a strategy

Apply the idea

Which of the regions has a signiﬁcantly higher number of scores?

The highest frequencies are found in the classes 1 to 4. So the correct answer is option B.

c What is the mode?

Create a strategy

Apply the idea

Remember that the mode is the score that occurs most often.

Looking at the column graph, we can see that the highest frequency can be found in score 3.

d Describe the distribution of the data.

Create a strategy

Apply the idea

If the scores are even in both the high and low end, the distribution is symmetrical. If the right tail is longer, the distribution is positively skewed. If the left tail is longer, the distribution is negatively skewed.

The right tail appears to be longer than the left. So the data is positively skewed.

Idea summary To determine the modality of a data distribution: • If there is a single mode, the data is unimodal. • If there are two modes, the data is bimodal. • If there are more than two, the data is multimodal.

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11.06 Practice questions Understanding

Adventurer

Trailblazer

1–5

What is a feature of a distribution that is symmetrical?

Describe what positive skew and negative skew mean in the context of a data set.

How can clusters be identiﬁed within a data set?

How do we describe the shape of a histogram with two distinct peaks?

A set of data is strongly positively skewed. If the median is 60, is the mean less than, equal to or greater than 60?

Fluency

6–18

Identify if the graph is positively skewed, negatively skewed, or symmetrical.

8 9 10 11 12 13 14 15 16 17 Score

Mathspace WA Curriculum – Year 9 mathspace.co

10 9 8 7 6 5 4 3 2 1 0

6 10 14 18 22 26 30 34 38 42 46 Class

15 10 5

25 28 31 34 37 40 43 46 49 52 55 Class

d Frequency

9 8 7 6 5 4 3 2 1 0

0 5 10 15 20 25 30 35 40 45 50 Class

10 Frequency

9 8 7 6 5 4 3 2 1 0

Frequency

10 Frequency

676

6–18

f Frequency

Frequency

Ex 1

Explorer 1–5

15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

8 9 Score

4 5 Score

30 25 Frequency

Frequency

20 15 10

5 0

4 5 Score

130 120 110 100 90 80 70 60 50 40 30 20 10 0

Frequency

d Frequency

120 110 100 90 80 70 60 50 40 30 20 10 0

130 120 110 100 90 80 70 60 50 40 30 20 10 0

0 10 20 30 40 50 60 70 80 90 100110120 Length (cm)

130 Frequency

2 3 4 5 6 7 Score

Identify if each graph is unimodal or bimodal.

Frequency

0 10 20 30 40 50 60 70 80 90100110120 Length (cm)

20 30 40 50 60 70 80 90 100 110 120 Length (cm)

130 120 110 100 90 80 70 60 50 40 30 20 10 0

10 20 30 40 50 60 70 80 90 100 Length (cm)

The stem and leaf plot shows the ages of a group of people on a bus. a

Find the:

Leaf 1

Median age

Difference between the lowest age and the median

iii

Difference between the highest age and the median

The mean age, rounded to two decimal places

Key: 1|4 = 14 years old

Is the data positively or negatively skewed?

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The stem and leaf plot shows the number of hours worked per week by a group of people.

No. of hours 2

Is there any clustering of data? If yes, in what interval?

Find the mode(s).

Key: 5|2 = 52 hours 10

For the stem and leaf plot: a

Is there any clustering of data? If yes, in what interval?

Find the mode(s).

Describe the shape of the data.

Leaf 0

5 6 7 8 2

Key: 1|2 = 12 11

678

The percentage of faulty mobile phones in 118 batches were recorded. a

Is the data unimodal, bimodal, or multimodal but not bimodal?

Find the modal class(es) of the data.

Mathspace WA Curriculum – Year 9 mathspace.co

20 18 16 14 12 10 8 6 4 2 0

Frequency

Ex 2

1 2 3 4 5 6 7 8 9 10 Faulty mobile phones (%)

The number of peanuts in mixed nut packets were sampled and recorded in the stem and leaf plot.

Leaf

Describe the clustering of data.

3 6 8

2 2

0 0 3

4 6 8 8 9

0 2 4 5 6 9

0 4 5

7 1

8 8 4 5 9

2 3 5 5 6

Key: 5|2 = 52 13

For the given column graph: a

Is there any clustering of data?

Where does the clustering occur: 1 to 4, 5 to 7, or 8 to 11?

Find the mode.

Describe the shape of the distribution of the data. 45 40 35 Frequency

30 25 20 15 10 5 0

6 Score

The shoe sizes of all the students in a class were measured. a

Is there any clustering of data?

Where does the clustering occur?

Find the mode of the shoe sizes.

Describe the shape of the distribution of the data.

Frequency

Ex 3

10 5

8 9 Shoe size

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The dot plot shows the temperatures recorded over a period of time. a

Is there any clustering of data? If yes, in what interval?

Find the mode of the temperatures.

Describe the shape of the data. 40 41 42 43 44 45 46 47 48 49 Temperature

For this histogram representing students’ heights in centimetres:

Does the histogram represent grouped data or individual scores?

Estimate the value of the mean, rounded to one decimal place.

Describe the shape of the distribution.

Frequency

10 5

105 115 125 135 145 155 165 175 Score

For the dot plot, describe the shape of the distribution. 1 2 3 4 5 6 7 8 9 Score

For the dot plot: a

Is there any clustering of data? If yes, in what interval?

Find the mode(s).

Describe the shape of the data.

Reasoning

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Sample

19, 20

Explain how a positively skewed distribution might impact the mean, median, and mode.

The ages of people were collected across 5 suburbs in a certain town. What factors might lead to this data set having one or more clusters?

Problem-solving 21

21, 22

A company records the number of sales made by its salespeople over a month. The data is heavily skewed to the right. What can be inferred about the sales performance of the team?

A teacher wants to compare the test scores of two different classes. One class has a symmetrical distribution, while the other has a positively skewed distribution. How might the teacher interpret and compare the performance of the two classes based on these distributions?

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Investigation: Statistical reports in the media

Investigate online

11.07

mathspace.co

Compare data sets

After this lesson, you will be able to… • compare and contrast two or more data sets using measures of spread, central tendency, and data shape. • interpret histograms, stem and leaf plots, tables, and column graphs to draw comparisons between data sets.

Compare data sets When we compare data sets, we should ask a few questions: • How do the spreads of data compare? Is one set of data more spread out than the other? • How does the skew of data compare? Is one more symmetrical? • Is there a big difference in their measures of centre (mean, median or mode)?

Exploration Consider the following histograms that show the height of students in two basketball teams. We know that one graph represents a team made up of Year 12 students and the other represents a Year 8 team. Team A:

Team B: 12 Number of students

Number of students

10 8 6 4 2 0

155 160 165 170 175 180 185 190 195 200 Height (cm)

10 8 6 4 2 135 140 145 150 155 160 165 Height (cm)

Are there the same number of students in each team? Does it matter?

What are the similarities and differences in terms of measures of spread, central tendency and shape of data?

Which team do you think corresponds to the Year 12 team, and which team do you think corresponds to the Year 8 team?

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It is important to be able to compare data sets because it helps us make conclusions or judgements about the data. For example, suppose Jim scores 50% in a geography test and 60% in a history test. Based on those marks alone, it makes sense to say that he did better in history. But what if everyone else in his geography class scored 40%, while everyone else in his history class scored 80%? Now we know that Jim had the highest score in the class in geography, and the lowest score in the class in history. With this extra information, it makes more sense to say that he did better in geography.

Example 1 The weights (in grams) of female and male fish caught in a tag and release program were recorded and presented in the stem and leaf plot shown. Female

Stem

Male

Key: 1|6 = 61

Key: 6|2 = 62

a What is the mean weight of the group of male fish?

Create a strategy Use the formula: Mean =

Apply the idea We are focusing on the stems and the leaves under “Male” on the right side of the stem and leaf plot. Substitute the values

Evaluate the addition = 70.3 g

b What is the mean weight of the group of female fish?

Create a strategy Use the formula: Mean =

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Evaluate the division

Apply the idea We are focusing on the stems and the leaves under “Female” on the left side of the stem and leaf plot. Substitute the values

Evaluate the addition = 62.9 g

Evaluate the division

c Based on the mean, which group is heavier? A

Female fish

Male fish

Create a strategy Compare the mean weights of the two groups.

Apply the idea The group with the higher mean weight will also, on average, have individual fish that are heavier. The male fish had a mean of 70.3 g which were heavier than the female fish who had a mean of 62.9 g. So the correct answer is option B.

Example 2 The table shows the scores of Student A and Student B in 5 separate tests. Test 1 2 3 4 5

Student A 79 74 86 89 98

Student B 77 83 72 94 97

a Find the mean of the scores of Student A.

Create a strategy Use the formula: Mean =

Apply the idea We add the scores obtained by Student A divided by the number of tests. Substitute the values

Evaluate the addition

Evaluate the division

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b Find the mean of the scores of Student B.

Create a strategy Use the formula: Mean =

Apply the idea We add the scores obtained by Student B divided by the number of tests. Substitute the values

Evaluate the addition

Evaluate the division

c Which student did better overall on their tests?

Create a strategy

Apply the idea

We can compare the students by using the mean values calculated in parts (a) and (b).

Student A has a higher mean, 85.2, and did slightly better than Student B.

d What is the highest score overall? Which student (A or B) obtained that score?

Create a strategy

Apply the idea

Refer to the score columns and ﬁnd the highest score. Then, ﬁnd the student that obtained that score by referring to the column heading.

Based on the table, the highest score is 98, which was obtained by Student A.

e What is the lowest score overall? Which student (A or B) obtained that score?

Create a strategy

Apply the idea

Refer to the score columns and ﬁnd the lowest score. Then, ﬁnd the student that obtained that score by referring to the column heading.

Based on the table, the lowest score is 72, which was obtained by Student B.

Do the answers to parts (d) and (e) impact the answer to part (c)?

Create a strategy Determine the difference between what is required in part (c) and what is required in parts (d) and (e).

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Apply the idea In part (c), we are asked to determine which student did better overall, meaning their average score across all five tests. In contrast, parts (d) and (e) ask us to find the highest and lowest individual scores among all 10 tests. Therefore, the answers to parts (d) and (e) do not impact the answer to part (c).

Example 3 The following column graphs show the season results of two soccer groups, Group A and Group B, and the number of games (frequency) in which they scored a certain number of goals (scores). Group A

4 Frequency

Group B

3 2

3 4 Score

a What is the mode for Group A?

Create a strategy

Apply the idea

Examine the column graph for Group A and assess which score occurred the most.

Based on the column graph for Group A, the score that occurred the most is 3.

b What is the mode for Group B?

Create a strategy

Apply the idea

Examine the column graph for Group B and assess which score occurred the most.

Based on the column graph for Group B, the score that occurred the most is 4.

c What is the range for Group A?

Create a strategy Use the fact that the range is the difference between the highest and lowest scores.

Apply the idea Based on the column graph for Group A, the highest score is 5 and the lowest score is 1. Range = 5 – 1 =4

Subtract the lowest score recorded from the highest Evaluate

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d What is the range for Group B?

Create a strategy Use the fact that the range is the difference between the highest and lowest scores.

Apply the idea Based on the column graph for Group B, the highest score is 6 and the lowest score is 0. Range = 6 – 0 =6

Subtract the lowest score recorded from the highest Evaluate

e Which group scored the lowest total number of goals during the season? A

Group A

Group B

Create a strategy To ﬁnd the total number of goals scored by each group, we need to multiply each score by its frequency and then add the values.

Apply the idea We can create frequency tables for both Group A and Group B to easily ﬁnd and compare the total number of goals scored between the two groups. Frequency table for Group A: Scores (x) 1 2 3 4 5

Frequency ( f ) 2 4 5 4 2 Total

fx 1×2=2 2×4=8 3 × 5 = 15 4 × 4 = 16 5 × 2 = 10 51

Frequency table for Group B: Scores (x) 0 1 2 3 4 5 6

Frequency ( f ) 3 3 0 1 4 3 3 Total

fx 0×3=0 1×3=3 2×0=0 3×1=3 4 × 4 = 16 5 × 3 = 15 6 × 3 = 18 55

Comparing the total number of goals scored between the two groups, the group that scored the lowest during the season is Group A. So, the correct answer is option A.

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Which group had more varied results? What might be a reason for this?

Create a strategy

Apply the idea

Examine both column graphs and determine which data set is more spread out.

The range of goals scored for Group B is larger than for Group A, indicating a wider spread of results. There are many possible reasons why Group B has more varied results. For example, Group B might have a wider range of skill levels among its players, leading to a larger spread of scores.

Idea summary By comparing the measures of central tendency in a data set (the mean, median and mode), the measure of spread (the range) and the shape of the data, we can make comparisons between different groups and draw conclusions about our data.

11.07 Practice questions Understanding

Explorer

Adventurer

1–4

Trailblazer

1–4

What are three statistical measures we can use to compare data sets?

Can two data sets have the same mean but different medians?

Which data set has the lowest score?

9 8 7 6 5 4 3 2 1 0

Graph 2 5 4 Frequency

Frequency

Graph 1

3 2 1

3 4 Score

6 7 Score

Give two reasons using a data visualisation like a graph can make comparisons of data sets easier.

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Fluency 5 Ex 1

Ex 2

Key: 1|6 = 61

Which group takes less time on average to get to school?

The back-to-back stem and leaf plot shows the length (in minutes) of a random sample of phone calls made by Jenny and Joanne.

Jenny did not make a 13-minute phone call.

Jenny’s median is higher than Joanne’s median.

iii

The mean is greater than the median in both groups.

Is Jenny’s mean greater than her median?

Is Joannes’s mean greater than her median?

Find the total number of goals scored by both teams in Match C.

Find the total number of goals scored by Team 1 across all the matches.

Calculate the mean number of goals scored by: Team 1

Find the mean of the scores of: i

Alice

Key: 6 |2 = 62

Match A B C D E

Team 1 2 4 5 3 3

Team 2 3 2 2 5 4

Subject Maths Science History English

Alice 88 90 85 82

Bob 72 75 80 85

Team 2

The table shows the test scores of two students, Alice and Bob, in 4 different subjects. Bob

Find the mean of the combined scores of Alice and Bob.

Find the highest score overall and which student obtained that score.

Find the lowest score overall and which student obtained that score.

Mathspace WA Curriculum – Year 9 mathspace.co

Joanne

Key: 1|6 = 61

The number of goals scored by Team 1 and Team 2 in a football tournament are recorded in the table.

Key: 6 |2 = 62

Jenny

688

High school

Calculate the mean time taken to get to school for the group of high school students.

i 8

Calculate the mean time taken to get to school for the group of primary students.

Determine whether the following statements are true or false.

5, 6, 8, 10–12, 14–16

Primary

5, 6, 8, 10–13, 15–17

The time taken to get to school (in minutes) by a group of students was recorded and presented in a stem and leaf plot. a

5–11, 13, 15, 16

The points scored by each player in a basketball game are displayed: • Team X: 10, 12, 15, 18, 20, 15, 12, 10, 10, 8, 5 • Team Y: 14, 16, 10, 20, 24, 12, 15, 8, 5, 5, 4 a

Which team scored more points?

Find the mean points of the teams, rounded to one decimal place: i

Team X

Team Y

In a Mathematics test out of 100, the students in two classes scored the results displayed: • Class A: 72, 66, 92, 76, 77, 63, 67, 91, 66, 77, 66, 73, 80, 68, 68, • Class B: 84, 70, 53, 82, 65, 85, 81, 87, 64, 81, 86, 79, 69, 82, 77 a

Find the median mark for: i

b 11

Class A

Class B

Using the medians, which class did better on the test?

The monthly rainfall for two cities across a year is displayed: • City A: 122, 120, 121, 67, 79, 24, 8, 46, 27, 67, 80, 128 • City B: 51, 66, 33, 78, 79, 92, 78, 99, 41, 60, 26, 56 a

Find the range in monthly rainfall for: i

b 12

City A

City B

Which city had more consistent rainfall throughout the year?

Harry is tracking the growth of two different types of sunflower plants. He records the number of sunflowers that he picks from each plant for 10 days. • Plant A: 15, 7, 8, 10, 9, 15, 6, 6, 12, 15 • Plant B: 11, 10, 9, 9, 12, 10, 11, 10, 9, 10 a

Find the mean daily growth for Plant A. Round your answer to one decimal place.

Find the mean daily growth for Plant B. Round your answer to one decimal place.

Find the range of daily growth for Plant A.

Find the range of daily growth for Plant B?

Which plant shows more growth on average?

Which plant shows a more consistent daily growth?

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The stem and leaf plot shows the running times (in minutes) of two track teams, Team A and Team B. a

Find the fastest running time in: i

Team B

Key: 1|6 = 61

Key: 6|2 = 62

Perth

Adelaide

9 8 7 6 5 4 3

Perth

1 0

5 6 7 8 9

0 1

Key: 1|6 = 61

2 5 8

Key: 6|2 = 62

Adelaide

Find the mean temperature of: Perth

Adelaide

Which city would be described as being warmer? Is this supported by your answer to part (b)?

The back-to-back stem and leaf plot shows the number of books read in a year by students from two different schools, Kinderhaus and Learning Tree: a

Did any student from Kinderhaus read 12 books in a year?

Which school’s students had the higher median number of books read?

690

Team A

Find the highest temperature recorded in:

i c

Which team is quicker? Is this supported by your answer to part (b)?

i b

Team B

The stem and leaf plot shows the temperature (in degrees Celsius) recorded in two cities, Perth and Adelaide, over 10 days. a

Team B

Find the mean running time of: i

Team A

Is the median number of books read greater than the mean number of books read in both groups?

Mathspace WA Curriculum – Year 9 mathspace.co

Kinderhaus

Learning Tree

Key: 1|6 = 61

Key: 6|2 = 62

The graphs show the performance of two hockey teams, Team X and Team Y, in various games, along with the number of goals they scored in each game. 5

5 Frequency

Frequency

Ex 3

3 2 1

3 4 5 Score - Team X

2 3 4 Score - Team Y

Find the mode for Team X.

Find the mode for Team Y.

Find the range for Team X.

Find the range for Team Y.

Which team scored the lowest total number of points during the season?

Which team has the most varied results?

The pulse rates of two groups are given: • Group 1: 82, 85, 88, 65, 73, 89, 79, 90, 76, 68, 88, 65, 63, 62, 88, 82 • Group 2: 75, 88, 74, 73, 80, 76, 67, 81, 71, 83, 89, 62, 63, 80, 71, 78 a

Find the mean pulse rate of Group 1, rounded to two decimal places.

Find the mean pulse rate of Group 2, rounded to two decimal places.

What is the range for Group 1?

What is the range for Group 2?

Which group has the greater spread?

Reasoning 18

19, 20

18–20

18, 21

The weights of two groups of turtles are measured, in kilograms, to determine whether they might belong to the same species. The results are presented: • Group 1: 56, 60, 58, 59, 62, 55, 57, 61, 56, 59 • Group 2: 62, 64, 60, 65, 66, 62, 61, 64, 67, 63 Do you think the two groups of turtles are from the same species? Explain your answer.

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691

Two Biology classes, each with 15 students, sit a 10-question multiple choice test, each with four possible answers (only one of which is correct). Their class results, out of 10, are: • Class 1: 3, 2, 1, 1, 2, 5, 2, 2, 2, 3, 2, 1, 5, 5, 4 • Class 2: 10, 6, 9, 8, 10, 9, 7, 7, 6, 8, 8, 10, 8, 9, 8 a

Complete the following table, rounding your answers to one decimal place when necessary: Mean

Median

Mode

Range

Class 1 Class 2 b 20

Which class was more likely to have studied for their test? Explain your answer.

The median salary in Company Alpha is $60 000 with a mean salary of $65 000 and the median salary in Company Beta is $60 000 with a mean salary of $55 000. Which company is most likely to have some very high paying jobs? Explain your answer.

The table shows the ages of employees working in two different departments of the same company. Department A Department B a

25 45

30 50

Copy and complete the table, rounding your answers two decimal places where appropriate:

45 45

35 50

Department A

40 55 Department B

Mean Median Mode Range

23–26

24–27

The prices of sales at two locations of a shop are recorded in the histograms. 13 12 11 10 9 8 7 6 5 4 3 2 1 0

Location B

Sales

Location A

Sales

13 12 11 10 9 8 7 6 5 4 3 2 1 0

692

40 60

Which department seems to have older employees? Justify your answer.

Problem-solving 22, 23, 25 22

35 55

5 10 15 20 25 30 35 40 45 50 55 Price

What is the shape of the distribution of the histogram of Location A?

Which location made more sales?

Which location had a higher average sale price?

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An online shopping website records the user ratings for two similar products in the column graph. 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0

30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0

0.0

0.5

1.0

1.5

2.0 2.5 3.0 User rating

3.5

4.0

4.5

5.0

3.5

4.0

4.5

5.0

Product B

0.0

0.5

1.0

1.5

2.0 2.5 3.0 User rating

Find the median user rating for: i

Product A

Product B

According to the website’s rating system, a perfect rating is 5.0. Which product scored more perfect ratings?

According to the website’s rating system, a positive rating is greater than 3. Which product scored more positive ratings?

The ages of employees at two competing fast food restaurants on a Saturday night are recorded. Some statistics are given in the table. Barry’s Burgers Frank’s Fries

Mean 18 18

Median 17 19

Range 8 4

If the data for each restaurant was represented using a histogram, what would the likely shape of the histogram for Barry’s Burgers be?

Which restaurant likely has the oldest employee on the night the data is recorded?

Which restaurant has the most employees of a similar age?

Which statistical piece of evidence supports your answer to part (c)?

Which restaurant has an older workforce?

Which statistical piece of evidence supports your answer to part (e)?

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693

The hours of sleep per night for two people over a two week period are given: • Person A: 10, 9, 6, 6, 7, 8, 5, 6, 9, 9, 7, 5, 10, 9 • Person B: 7.5, 7.5, 8, 7.5, 7, 8, 7.5, 7, 7, 7, 7, 7, 8, 8 The table shows the measures of central tendency and spread for Person A and Person B. Mean 7.6 7.4

Person A Person B

Median 7.5 7.5

Mode 9 7

Range 5 1

Which person seems to be the least consistent in their sleep habits?

Which statistical piece of evidence supports your answer in part (a)?

Which person had the most sleep over the 14 nights?

Which statistical piece of evidence supports your answer in part (c)?

A stationery shop kept a weekly record of the number of pens and notebooks sold. 25

20 Number sold

Number sold

20 15 10 5 0

694

Mon Tue Wed Thu Fri Pens

15 10 5 0

Mon Tue Wed Thu Fri Notebooks

What is the shape of the histogram of the number of pens sold?

Find the percentage of pens sold on Thursday. Round your answer to two decimal places.

Calculate the mean number of notebooks sold per day. Round your answer to one decimal place.

Calculate the mean number of pens sold per day. Round your answer to one decimal place.

Which is the better selling product? Justify your answer.

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A basketball coach is analysing the free throw shooting performance of two basketball teams, Team X and Team Y. The data shows the number of successful free throws each player made during a season: • Team X free throws: 8, 9, 7, 6, 9, 8, 7, 10, 7, 6 • Team Y free throws: 9, 8, 10, 8, 7, 9, 8, 10, 7, 8 a

What is the lowest and highest successful free throws for both Team X and Team Y?

Calculate the mean number of successful free throws for both Team X and Team Y.

Calculate the median number of successful free throws for both Team X and Team Y.

Calculate the probability that a randomly selected player from Team X made 8 or more successful free throws.

Calculate the probability that a randomly selected player from Team Y made 8 or more successful free throws.

Based on the calculated means, modes, and probabilities, as well as the spread of data, what can you infer about the performance of Team X and Team Y? Which team’s players appear to have a more consistent performance in terms of free throw success?

Investigation: Plan and conduct statistical investigations

Investigate online

mathspace.co

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695

11 Chapter review 1

For each survey question, classify the data as ordinal, nominal, discrete or continuous. a

How long did you sleep for last night?

How many people live in your house?

How many languages do you speak at home?

How long did you spend on homework last night?

After the government decided to increase the minimum retirement age, a news poll selected a group of people to ask their opinions on the changes. Should each of these groups be included to avoid sampling bias?

Only people who are employed

People in the community who have a wide variety of views, excluding politicians and policymakers

People in the community who have a wide variety of views, even if they are not directly affected by the changes

Members in the community not impacted by the changes

A local doctor’s clinic is trying out a new way to help patients get better quicker. a

What type of info should the clinic collect to see if this new way works?

How can the clinic make sure the info they get is not biased?

The goals scored by a football team in their matches are represented in the dot plot: a

Complete a frequency table for this data.

Calculate the mean, median, mode, and range of the goals scored. Round your answer to two decimal places.

Make a statement for each of the measures of centre and how it connects visually to the dot plot.

1 2 3 4 Goals scored

Consider the frequency distribution table: Score (x) 20–24 25–29 30–34 35–39 40–44 45–49 50–54

696

Frequency ( f ) 9 18 37 14 9 6 3

Cumulative frequency (cf )

Complete the table.

Calculate the total frequency.

Describe the shape of the distribution.

Approximately one third of the scores recorded are greater than what score?

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The data shows the running times (in minutes) of two track teams, Ace Rockets and Thunders: • Ace Rockets: 36, 37, 41, 41, 43, 47, 49, 50, 52, 56 • Thunders: 30, 32, 34, 36, 38, 42, 44, 55, 57, 62

Find the mean score of each team.

Display the data as a back-to-back stem and leaf plot.

The back-to-back stem and leaf plot shows the number of tasks completed over several days by Selma’s and Sirin’s students: a

Did Selma’s students complete 7 tasks on any day?

Whose class had the higher median?

For which group is the mean greater than the median?

Selma’s students

2 4

2 3

Sirin’s students

Key: 2|0 = 2 8

Key: 2|3 = 23

This histogram shows the speed of balls delivered by cricket bowlers: 16

Construct a grouped frequency table for this set of data.

What percentage of balls were bowled at less than 130 km/h? Round your answer to one decimal place.

14 Frequency

12 10 8 6 4 2 0

100

110

120

130 140 150 Speed (km/h)

160

170

Consider the grouped frequency table showing the length of the phone calls made by Pauline in a month: Length of call (minutes) 0<t≤1 1<t≤2 2<t≤3 3<t≤4 4<t≤5 5<t≤6 6<t≤7

Number of calls 5 7 2 9 5 9 8

Pauline’s phone company charges phone calls as if they were at the upper boundary of the interval. For example, a phone call lasting 1 minute and 48 seconds would be charged as if it were a 2 minute call. a

Create a histogram for this set of data.

Find the total number of minutes for which Pauline will be charged for phone calls this month.

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697

(Optional) For the frequency table: a

Estimate the mean by calculating the class centre for each row. Round your answer to one decimal place.

State the modal group of scores.

Class 5–10 11–15 16–20 21–25 26–30

Frequency 18 12 9 5 4

(Optional) For the set of scores: 15, 42, 78, 23, 56, 34, 67, 19, 45, 71

Determine a set of ﬁve class intervals that could be used to analyse this data.

Construct a grouped frequency table for the data.

(Optional) In a quality control check, the number of defects found in batches of handmade pottery is recorded over several weeks. The frequency table shows the results. a

Construct a histogram to represent the data.

The median number of defects is in which class interval?

What is the lowest possible number of defects that could have been recorded in any batch?

Number of defects 1–3 4–6 7–9 10–12

Frequency 8 12 15 10

(Optional) The histogram shows the speed of balls delivered by cricket bowlers. 16 14

Frequency

12 10 8 6 4 2 0

698

100

110

120

130 140 Speed (km/h)

150

160

170

Complete a frequency table for this set of data.

What percentage of balls were bowled at less than 130 km/h? Round your answer to one decimal place.

Mathspace WA Curriculum – Year 9 mathspace.co

(Optional) Complete these frequency tables using the given frequency polygons: a

Frequency 9 8 7 6 5 4 3

Class 20 ≤ x < 30 30 ≤ x < 40 40 ≤ x < 50 50 ≤ x < 60 60 ≤ x < 70 70 ≤ x < 80

Frequency

Class 76 ≤ x < 84 84 ≤ x < 92 92 ≤ x < 100 100 ≤ x < 108 108 ≤ x < 116 116 ≤ x < 124

Frequency

2 1

Score (x) 15 25 35 45 55 65 75 85

Frequency 9 8 7 6 5 4 3 2 1

Score (x) 72 80 88 96 104 112 120 128

(Optional) The ages of new mothers in a particular hospital are displayed on the following histogram: Age 28 ≤ Age < 30 30 ≤ Age < 32 32 ≤ Age < 34 34 ≤ Age < 36 36 ≤ Age < 38 38 ≤ Age < 40

Frequency

New Mothers

16 Frequency

12 8 4 0

Complete the frequency table based on the histogram.

How many new mothers were there?

How many mothers were under 34 years old?

How many mothers were 36 years old or older?

29 31 33 35 37 39 Age

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699

(Optional) The table shows the number of suspense novels in a bookshop for different price ranges: a

Plot this data in a histogram.

Describe the shape of the distribution of the data.

Price of suspense novel 0≤p<5 5 ≤ p < 10 10 ≤ p < 15 15 ≤ p < 20 20 ≤ p < 25 25 ≤ p < 30 30 ≤ p < 35

The histogram shows the number of crime novels in a bookshop for different price ranges rounded off to the nearest $5: Novel genres

Frequency

15 10 5 0

5 10 15 20 25 30 35 Price

Describe the shape of the distribution of the data. 18

Consider the data shown in the graph: 45 40 Frequency

35 30 25 20 15 10 5 1 2 3 4 5 6 7 8 9 10 11 Score

700

Are there any outliers?

Is there any clustering of data? If so, in what interval?

State the mode.

Describe the shape of the distribution.

Mathspace WA Curriculum – Year 9 mathspace.co

Frequency 5 10 17 8 17 10 5

Consider the two histograms which show the grade distributions in two university courses:

20 18 16 14 12 10 8 6 4 2 0

Grades in a university course (Section 1, N = 100)

Number of individuals

80 Section 1

20 18 16 14 12 10 8 6 4 2 0

Which distribution has a clear mode?

Which distribution would be described as “uniform”?

Which distribution has the highest range?

Compare the mean and median of Section 1.

The stem and leaf plot shows the ages of people on a bus to enter through the gates of a concert in the ﬁrst 5 seconds. a

Find the mean age, rounded to two decimal places.

After the data was collected, one more person boards the bus. After they board, the new mean age becomes 25 years.

Grades in a university course (Section 2, N = 100)

4 5

6 Key: 1 |4 = 14 years old

Is the data positively or negatively skewed?

Ten participants had their pulse measured before and after exercise with results shown in the stem and leaf plot: a

80 85 Section 2

Leaf

How old is the person that just boarded the bus? c

Pulse rate before exercise

Pulse rate after exercise

Find the mean pulse rate:

Before exercise

After exercise

11 12

What is the mode pulse rate after exercise? Find the range of pulse rates: i

Before exercise

After exercise

What can you conclude about exercise and pulse rate from the measures of centre and spread?

Key: 6|1|2 = 12 and 16

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701

A battery manufacturer tests their batteries in a portable music player, to ﬁnd out how long the batteries last under constant use. The frequency table shows the data from their tests: a

Construct a histogram to represent the data.

What percentage of batteries last 10 hours or longer? Round your answer to one decimal place.

If the average length of a song is 4 minutes, what percentage of batteries will last for an average of 250 songs or more? Round your answer to one decimal place.

Battery life (mins) 400 ≤ x < 500 500 ≤ x < 600 600 ≤ x < 700 700 ≤ x < 800 800 ≤ x < 900 900 ≤ x < 1000 1000 ≤ x < 1100 1100 ≤ x < 1200 1200 ≤ x < 1300 Total:

Frequency 3 9 13 32 22 27 6 4 1 117

A pet store recorded the sales of different breeds of dogs over a year. The data showed a symmetric distribution for the sales of Labradors, but a positive skew for Beagles. How might the store owner interpret the popularity and sales trends of these two breeds based on this information?

The following table shows the number of units of blood donated per week over 10 weeks in two states: VIC SA

702

3986 1355

3738 1213

3949 1275

3909 1397

4130 1181

4079 1252

3894 1372

4079 1247

3711 1501

3871 1175

What was the average number of units donated per week in Victoria and South Australia?

Using the average units donated per week for each state, what is an estimate for the number of units collected in Victoria and South Australia over a year? Assume that there are 52 weeks in a year.

A rough estimate of the units of blood currently required annually per state is 3% of the population. If the population of Victoria is 5 640 900 and the population of South Australia is 1 659 800. Will each state meet the required number of blood donations?

One study suggests that demand for blood is expected to increase by 25% in four years time due to factors such as prolonged ageing. How many units of blood annually will Victoria require in four years’ time?

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“The world cannot be understood without numbers. But the world cannot be understood with numbers alone.” Hans Rosling

Answers

16 a

1.01 Rational and irrational numbers

b ii

c i

b π

d iii

d 8

e 6.2 × 10−1

f 2.5 × 100

h π3

Reasoning

4 a Rational b Rational c Rational d Rational e Rational f Irrational g Irrational h Rational

19 Yes. For example

i Irrational j Rational k Irrational l Rational

20 Yes. For example

Fluency

21 a Irrational

5 a True

b False

c True

d False

π is the exact value, so it is the most accurate.

6 a 4, Rational

, Rational

, Irrational

b Rational

c Irrational

d Irrational

e Irrational

f Rational

g Rational

h Irrational

9 a b c d

−3

−2

−1

−2

−1

Since the difference between π and smaller,

0 −1 0

10 a 7

b 11

c 8

d 15

11 a C

b C

c A

d B

12 a 49 < 59 < 64

b 7<

13 a 16 < 22 < 25

14 a

15 4

Mathspace WA Curriculum – Year 9 mathspace.co

23 a

3 3

is closer to π than 3.14.

is between

and

b Her answer should be less than 5.5 as 26 is much closer to 25 than it is to 36.

, and

−1

−3

We can ﬁnd the difference between π, 3.14 to ﬁgure out which one is closer.

d 0.08, Rational

7 a Rational

−4

b Rational

e True

704

d 2

18 a

c 5

17 a 12

Understanding 1 a iv

Problem-solving 24 a c = 13

25 26 a 12 square units b 13.0384 square units c 7.96% 27 a 2.6 cm b It’s diﬃcult to measure small lengths like 1 cm accurately. It’s also diﬃcult to draw an angle of exactly 90°. So, your answer to (a) might have been incorrect due to inaccuracies in the drawing. c Since it’s easier to accurately measure larger lengths, we could enlarge the spiral. For example, if we enlarge the spiral by a factor of

2, we would start with a right triangle with sides of 2 cm. Then the hypotenuse of the cm. We would sixth triangle would be 2 × measure the hypotenuse of this triangle and divide by 2 to estimate the value of . d

11 The mistake occurred in step 1, where addition was performed before multiplication. The correct working step should be:

2 2

2 4 2 3 2 2

2 5 2 6

2 1 2 17

2 7 2 16 2 8 2 15 2 9 2 10

= 34 − (−4)

= 34 + 4

= 38

12 The order of operations has not been applied properly. The division should have been performed before the addition. The correct answer is −9. 13 Yes. The distributive property of multiplication over addition is applied in both expressions.

2 14 2 11 2 12

(−2) + 12 × 3 − (−4) = (−2) + 36 − (−4)

2 13

14 The side lengths must be rational number.

is not a

1.02 Operations with integers Problem-solving Understanding 1 a False

15 60 minutes b True

c False

d False

2 A

16 14 people 17 a 11th ﬂoor

3 C

b 4

18 $156

4 A

Chapter 1 review Fluency 5 a −1

1 a Rational b Rational c Rational d Irrational b −8

c 21

d −19

e −25

f 6

g 4

h 59

6 a −34

b 63

c 42

d 53

e −36

f 120

g −85

h 87

7 a −15

b 6

c 72

d −9

e 5

f −5

g 30

h −5

3 a D

b B

8 a 17

b −2

c 18

d −10

4 a

f −9

g 42

h 1

e −58

e Rational f Irrational g Irrational h Rational 2 a b

−4

−3

−2

−1

c 5 a 4

9 a Reasoning

10 Orion because he evaluated the expression from left to right.

b 7

6 a The perimeter is

c 2 , so it is irrational.

b The area is 7, so it is rational. 7 a 7 × 4 = 28 m2

b 27.3861 m2

8 a −2

b 9

c −12

d 20

9 a −23

b 65

c −36

d 80

10 a 2

b −6

c 7

d 17

11 a 18

b −96

c −29

d 20

Answers mathspace.co

705

12 a

13 The error is in step 2. After the ﬁrst step, 15 was added to 7 to get 22, but multiplication should be done before addition. 15 − (−7) × 2 + (−9) = 15 + 7 × 2 − 9

b 20c7 d7

4 9

d −6u10 v11

11 6

e 12t u

f −25a5 b6 c10

g −15w6 x8 y10

h 27a5 b5 c8

c 14a b

8 a

= 15 + 14 − 9

2 3

= 29 − 9

= 20 Reasoning

14 $600

9 The multiplication law requires the bases to be the same. Since the bases in the expression are different, we cannot simplify it using the multiplication law.

15 $25 000

2.01 Multiplication law Understanding

10 Marvin’s solution is correct. When multiplying terms with the same base, we add the indices: (b2) × (b3) × (b7) = b(2 + 3 + 7) = b12.

1 a 3 =3×3×3×3×3 b 34 = 3 × 3 × 3 × 3 c 35 × 34 = (3 × 3 × 3 × 3 × 3) × (3 × 3 × 3 × 3)

11 a (−2)3 = −2 × −2 × −2 = −8. −23 = −(2 × 2 × 2) = −8.

d When you multiply terms that have the same bases you add the indices. 2 a 5×5×5 c a×a×a×a×a

So these expressions are equal. b (−2)4 = −2 × −2 × −2 × −2 = 16.

b 24

−24 = − (2 × 2 × 2 × 2) = −16.

d b6

So these expressions are not equal.

3 a Trent’s statement is incorrect because the expanded forms of the terms do not match. To show this, let’s expand each term: 22 = 2 × 2 = 4, 32 = 3 × 3 = 9. So, 22 + 32 = 4 + 9 = 13, while 52 = 5 × 5 = 25. Therefore 22 + 32 is not equal to 52 since 13 is not equal to 25. b No, because multiplication index law applies to multiplication with the same base. In 22 + 32, the operation used is addition and the bases are different.

c • If a negative number is raised to an odd power, the result will be negative. • If a negative number is raised to an even power, the result will be positive. Problem-solving 12 a a2 × a9 = a11 b c1 × c7 = c8 c b5 × b6 = b11 d 2a2 b2 × 3a2 b = 6a4 b3 e 3mc2 × −1bm6 × 2b3 c3 = −6b4 c5 m7 f

Fluency 4 a 27

b 38

c 26

d 55

e 34

f 215

g 39

h 710

i (−2)8

j 29

k 38

5 a y5

b x9

c a12

d a5

e b

f y

g u

i a3

j x3

k d3

l e−2

6 a 5y5

b 3y4

c 4x9

d 6a10

e b12

f 21a6

g 72u10

h 15y4

706

7 a 6x3 y5

h b

Mathspace WA Curriculum – Year 9 mathspace.co

a1 b2 c4 ×

a1 b1 c2 =

a2 b3 c6

13 a5 cm2 14 a 5, x = 9, y = 1, x = 8, y = 2, x = 7, y = 3, x = 6, y = 4, x = 5, y = 5. b There would be an inﬁnite number of possible solutions, x = 11, y = −1, x = 12, y = −2 etc.

2.02 Division law

Reasoning 10 Jane’s solution is incorrect. According to the division law, when dividing terms with the same base, we subtract the indices. Also, when she divided the coeﬃcients, she took the reciprocal, but shouldn’t have. The answer should be

Understanding 1

2 a False, when dividing like bases, you subtract the indices. b False, the bases must be the same. c True

12 Becky is correct. We cannot simplify using the index laws, but we can still evaluate using the

d True e True f F alse, different bases and we don’t subtract the bases. g True h False, the coeﬃcients should be divided, not subtracted.

order of operations 13 Using the division law, we know that when dividing two numbers with the same base and different indices, we subtract the indices:

Now, let’s consider dividing m6 by itself:

Fluency 7

3 a 3

b 7

d 3

c 9

e 6

f 2

g (5)

h 22

i (−8)7

4 a 51 = 5

d (−2)1 = −2

e 3 =9

f (−1)5 = −1

g (−2)5 = −32

which simpliﬁes to: m0 Now, we know that any non-zero number divided by itself is equal to 1:

5 a a4

b x2

c y6

d b4

h s9

f x

g f

i p

j h

k c

l d6

6 a 4b4

b 3y11

c 6m11

d 4c5

e 5b8

f 5y3

g 4k5

h 8t2

i 2k5

j 8g7

k 3p3

l 5l6

7 a x6

b y6

c t5

d k14

h w2

e a

According to the division law, we subtract the indices:

b 24 = 16

c 43 = 64

e j

11 The division law is am ÷ an = am−n. The base has to be the same.

f b

g u

So, we have: m0 = 1 This shows that, using the division law of indices, m0 is equal to 1 for any non-zero value of m. Problem-solving 14 a b9 ÷ b4 = b5

8 a f 2 g8

b c4 d7

c m7 p10

d wv7

e 5h4 j 3

f 3xy

g 20r3 s3

h 4k4 l9

3 4

3 3

9 a x y

b a b

3 2

e n p

f c d

c jt 3 2

g a z

d k m

b x5 ÷ x3 = x2

h q5 w6

Answers mathspace.co

707

15 Many possible solutions. For example:

Reasoning 8 If we apply the index law (am)n to each expression, we get w6.

16 4a4 cm

If we go even more basic, (w2)3 = w2 × w2 × w2 = w6, and (w3)2 = w3 × w3 = w6

9 The equation simpliﬁes to a2x = ax , so we need a value that satisﬁes 2x = x2. 2

18 a Answers may look like:

x = 2 is one possible value (x = 0 is another).

b Answers may look like:

2.03 Power of a power law

10 The student correctly simpliﬁed the numerator, (z4)3 to get z12, but incorrectly divided by z3. Instead of dividing 12 and 3, the student should have subtracted the powers. The correct simpliﬁcation is z12−3 = z9.

Understanding

Problem-solving

1 a Yes

b Yes

c Yes

d No

e No

b (2m2 n3)3 = 8m6 n9

2 a False

b True

3 a ( p3)2 = p6 mn

11 a (abc)3 = a3 b3 c3

b (p ) = p

d Fluency 4 a 221

b 316

c 718

d 56

h 560

e 9

f 11

i 742

j (−3)27

k (−2)28

l 1330

5 a j 18

b w8

c t12

d a21

e h42

f p18

g a25

h b28

i c30

j e44

k gxy

l wab

6 a a 3 b6

b w4 x12

e t10 u15

f a28 b8

g 4h10

h 27a6

i 25c14

j a5 b20 c15 k

m −32a15

n −27a6

7 a 222

b 314

c a43

d x12

e y5

f f 19

g 9t11

708

g 2

e f (4m2)3 × (2b2)2 = 256m6 b4 g (0.1a3)2 × (10b4)1 = 0.1a6 b4 h [(2a2)3]2 = 64a12 12 216p 13 33a−2c × 52b−3d 14 125a12 cubic units

2.04 Zero and negative indices Understanding

Mathspace WA Curriculum – Year 9 mathspace.co

1 a 2 b

14 a −2

b −5

c −3

d −3

Power: −4

e −11

f −3

g −4

h −5

c Base: 13

Power: −10

b 10

−1

−7

d 35

d Base: −5

Power: −8

e 9−12

f 822

g 7−12

h 100−1

−2

−23

l −824

2 a Base: 10

Power: −7

b Base: 2

15 a 5

c 2

3 C

i 2

j 6

4 B

16 a 1

b 16

c 27

5 A

i 49

k 125

b 28a2

e 8h7

h 25m3

k 3x8

n 3x8

18 a

6 Less than 1

k 19

Fluency 7 a 1

b 27

c 1

d 1

e 1

f −1

g 1

h 7

i −1

j −9

k 32

l 1

n 1

o 1

p 1

8 a 0

b 7

c 10

d 0

9 a

104

103

102

105

100 000 10 000 1000 100 c

33 27

17 a 20y6

−1

101 100 10−1 10

3−1

3−2

3−3

19 a 32y8

b 60y5

e 6y

10 a 7

b 16

c 3

d 2

f 16

11 a

12 a

e an

f −4y

b 37−1 −3

c 5−1

e −15

−5

f 10

g (−24)

i 13−11

j −7−8

k 16−12

d 4−7 −10

h 25−3 l (−45)−7

g 2x

−1

20 a 36x−4

b 100x6 y−8

d 250xy−6

e 81x−10 y16

d 72x−7 h 21x−3

Reasoning 21 a 0 b

13 a 3−1

c 40x−2

and dividing by zero is

impossible. 22 The expression 2x0 represents the number 2 times x raised to the power of 0. Any non-zero number raised to the power of 0 is equal to 1. This means that 2x0 simpliﬁes to 2. The expression (2x)0 represents the entire quantity 2x raised to the power of 0. When a non-zero base is raised to the power of 0, the result is always 1. This means that (2x)0 simpliﬁes to 1.

Answers mathspace.co

709

23 The correct value is −1 because the negation applies to the base, not to the exponent. 24 The −1 index only applies to the term directly in front of it. In this case it only applies to the a,

5 a 4.5 × 101

b 1.28 × 102

d 8.84 × 105

c 2 × 10

not the 5. The correct working is

25 The expression x−n is the same as

due to

the negative index law. This law can be applied to any value of x, including negative values.

e 8.4626 × 10

f 2.3456 × 104

g 6.14 × 100

h 9.7 × 109

6 a 6 × 10−1

b 2 × 10−3

c 7.4 × 10−4

d 1 × 10−4

−4

e 6.76 × 10

−5

g 5.932 × 10

Problem-solving

7 a 3 000 000 000

26 a

c a−1 b3 × 2a2 b−4 = 2ab−1 2 −4 7

f L argest: 4.7 × 109 × 108 Smallest: 4.7 × 10−8 × 108

3 −8

d ( gh ) g = g h

f b 2200

27 a

c 4y8

d 36x7

d 897 000 f 3014

g 8004.5

h 13 008 000 b 0.007

c 0.000 004 7

d 0.000 000 213

e 8.29

f 0.000 0347

g 0.0972

h 0.000 064

9 a 2

b 0.000 73

c 7

d 4

e 734 000

f −5 b 8 × 10−5

c 1.8 × 10

d 4 × 105

f 5 × 10−5

e 3 × 10

b 4.21 × 10−2

11 a 8.3 × 105

cm2

c 5.39 × 10 30 4

12 2 × 10−6 m

2.05 Scientific notation

13 5 000 000 m

Understanding

14 7.12 × 1020 litres 15 a 3.8 × 10−3 seconds

1 C

2 A 3 a 1011

b 10−8

c 103

d 102

d 7.49 × 10−4

b 2.2 × 10−8 g

c 1.67 × 10 L

d 6.3 × 107 L

e 3.4 × 10−10 m

f 8 × 1011 bytes

−2

g 5 × 10

bels

h 1.2 × 105 pascals

Reasoning

Fluency 4 a Largest: 7.27 × 10−7

Smallest: 2.22 × 10−7

−9

b Largest: 8.31 × 10

Smallest: 5.13 × 10

c Largest: 1.25 × 102

Smallest: 1.25 × 10−6

d Largest: 9.37 × 10−4

Smallest: 6 × 10−9

e Largest: 7.88 × 10

710

b 41 300

e 503 000

10 a 6 × 109

h 9.2 × 10−2

c 9000

8 a 0.0528

f 3 × 10−5

Smallest: 2.93 × 10−2

Mathspace WA Curriculum – Year 9 mathspace.co

16 Extremely large and small numbers have many zeros so are diﬃcult to read and write. When given in scientiﬁc notation these numbers are much shorter to write, and a scientist can see at a glance the magnitude of the number by looking at the index of ten.

17 a i 4.75 × 107

ii 4.75 × 107

b The answers are both 4.75 × 107 because 1.35 × 103 is very small compared with 4.75 × 107. 18 a When a number is correctly expressed in scientific notation, the number part, a, must be such that 1 ≤ a < 10. Here a > 10. b When a number is correctly expressed in scientific notation, it is of the form a × 10n, where n is an integer. Here n is not an integer. c When a number is correctly expressed in scientific notation, the number part, a, must be such that 1 ≤ a < 10. Here a < 1. d When a number is correctly expressed in scientific notation, it is of the form a × 10n. Here the base is 8, not 10.

9 a x10

d 3t6

10 a 16

11 a 2.34 × 102 −1

c 8 × 10

d 8.9 × 10−4

12 a 60 000 000

b 38 500

c 1.4

d 0.004

e 0.0336

f 0.000 003 005 b 6 × 10−7

13 a 8 × 109

d 3 × 10−8

c 2 × 105

b 2.35 × 10−2

14 a 3.51 × 10

−3

15 a 5.1 × 10

seconds

c 5 × 1010 bytes

d 8.9 × 10−8 g

e 2.3 × 10 bels

19 a 2 300 000

b 1.38 × 108

17 P6 ÷ P2 = P4

20 a 3 × 10−1 km/s

b 3 × 105 km/s

18 315 m

c 106 times

−23

20 3 × 104 mm

21 a 8.63 × 10

b 7.2%

22 a 3.14 × 10−4 kg

b 2.76 × 10−3 cm

Chapter 2 review 8

d b11

8+h

b (−2)

c 3

e c2

f 5x15

2 a 4 = 64

b (−6) = 36

c 26 = 64

d b (−5)20−t c x3 y5

4 a 115 = 1

b (−2)4 = 16

5 a 914

b j 26

c x5 y10 z15

d 2a

6 a 25

b −1

c 8

d 1

7 a

8 a

3.01 Algebra review Understanding

1 a 7

b 3.04 × 103 m

16 a8 m3

Problem-solving

3 a a9

b 1.13 × 105

1 a 6 e −9

b x and y c 5

d −1

f −3y and 5y

2 a True

b False

c True

e False

f False

g False

d False

Fluency 3 a 13x

b 3y

c 9c

d −2x

e 9c + 2d

f 5x + 5y

g 5x + 7y + z

h 4x + 7 − 2y

i x2 + 11x + 10

j 3ab + 5bc

k x2 − 16

l 7a2 b + 4ab2

m 8x + 5

n 5ab + 10cd

o 4xy + 2x2 y − 2xy2 − 7 p −4x + x2 4 a 45u

b 28r

c 6xy

d 24rs

e 15u3 v5

f 60pq2

g 15cd

h −25ef

Answers mathspace.co

711

j 36x2 y2 z2

i 30xy 3

k −54a

l 20a2 b2

m 20a7 b5

n −18w4 x5 y5

o 42c7 d8

p −15e14f 7

both the coeﬃcients and the powers when she should have divided the coeﬃcients and subtracted the powers. Problem-solving

5 a

c 3

d −4

13 (6b + 2c) cm

g 4

14 (10x − 2y) cm

i 3r

j −2

6 a 15

15 a Middle box 7a + 2b. Top box 11a + 3b. b Middle box 2x − 4y. Bottom left box 3y. Bottom right box −x − 3y. 16 36a2 cm 17 $(10x + 5y + z) b 3xy2

18 a xy e

j 35fg

d 0

f 18ab

i 3

k 9c

m 3a3

n 3u6 v5

o 15x5 y4

p 7a6 b5

19 (8.5 − 2x) + x + x + (8.5 − 2x) + x + x + (11 − 2x) + x + x + (11 − 2x) + x + x = 39

3.02 Distributive law 7 a 8c2

Reasoning 8 a 3a and 4b are not like terms. We can only add or subtract like terms. b Yes, if a = 1 and b = 1 the expression is true. 9 In algebra, “like terms” refer to terms that have exactly the same variable factors, i.e., the same variables raised to the same power. Even though x2 and x have the same variable, x, they are raised to different powers. 10 This would only be the case for a = 0. 11 In 7a + 5a × 2, if we do the multiplication first we get the correct answer of 10a. If we do the addition first the answer will be 24a, which is incorrect. By following the order of operations, we ensure that we do not get two different answers to the same problem. 12 Lawrence subtracted both the coefficients and the powers of the variable. However, when dividing by a monomial, the coefficients should be divided and the powers should be subtracted. Marika, on the other hand, divided

712

Mathspace WA Curriculum – Year 9 mathspace.co

Understanding 1 Dortha is correct because the rule of exponents dictates that the proper way to multiply a3 and a3 is by adding the exponents, which gives a6. 2 Commutative property of multiplication states that the order of numbers doesn’t matter when multiplying two or more numbers together. As long as we multiply every term inside the bracket by the term outside the bracket, then we maintain equivalence of the expression. 3 a 3a(2a + b) = 3a × 2a + 3a × b b 6m2 v(3 − 2m2 + v) = 6m2 v × 3 − 6m2 v × 2m2 + 6m2 v × v c 4ab(3b + 2c − 6a) ≠ 4ab × 3 + 4ab × c − 4ab × 6 Fluency 4 a 3 (2x + 4) = 6x + 12 b 2 (−4s + 5) = −8s + 10 c 5 (3p + (−6)) = 15p − 30 d −2 (3g − 7) = −6g + 14 5 a ab + 7a 2

b m2 + 7m

c y+y

d 2y + 12

e 4m2 + 40

f m2 − 3m

g 4y2 − 10y

h 8m − m2

j 4b2 + 32b

i 5t + 70t

k −15a2 + 20a

l −w2 − w

m −16t − 12t

n −11mn − 19mp

o 56s3 t5 − 63s3

p 6wy2 + 6w2 y

6 a 7y3 + 28y

b −15y3 + 18y2

14 a If w metres represents the width of the garden, then the area will be w(2w − 1) = 2w2 − w square metres. b 8w + 12 c 11 × 21 metres

c 18a10 + 36a8

d 24u5 + 30u7

e −15x3 + 6x6

f −2s4 − 5s13

15 −2a4 + 8a2 + a

16 a A(B + C) = (x + 1) (2x + 3)

7 a 2f − 16fg − 14fh

b 6a − 30ab + 18ac

c 20x2 + 36xy − 8x

d 2a4 + 8a3 − 14a2 b h2 + 14h + 81

8 a 11y + 75 c x2 + 6x + 10 3

d 27c3 + 6c2 + 21c

b A(B + C) = 2x2 + 5x + 3 17 a 40x2 − 14x b The range of possible surface area is between 26 cm2 and 3860 cm2.

3.03 Binomial expansion

f 17y2 − 52y

g 10y − 57y

h 39x − 54y

i 26x + 58xy

j 8x − 14

k 6x − 9

l 16u2 + 20uv

1 a 64

b 25

c 9x2

d 144y2

m −7n2 − 147n

n −5n2 − 50n

2 a Yes

b No

c No

d Yes

e 32c − 12c − 20c

Reasoning 9 a i Leon did not multiply both terms. ii Jill used addition for the ﬁrst term instead of multiplication. iii Claire did not change negative and positive signs. iv Beryl forgot the negative when multiplying the second term. b −6(4x − 25) = −24x + 150 10 a i x + 1 ii x + 2 b x + 2(x + 1) = 15 + 2(x + 2)

Understanding

3 a i A = y2

ii B = 3y

iii C = 6y

iv D = 18

b y2 + 9y + 18 4 a False

b True

c True

Fluency 5 a x2 − 6x − 72

b x2 + 8x + 12

c v + 21v + 110

d x2 − 11x + 28

e m2 + m − 30

f 10m2 − 61m − 63

g 21w + 53w + 30

h 30y2 − 3y − 27

i −28y2 − 64y − 36

6 a a2 − 9

b x2 − 4

c 17, 18, 19

c y − 25

d 4x2 − 25

d No, because the equation used to solve for x in part (a) has only one unknown, x, to the power of one. Therefore there is only one solution.

e 16x2 − 36

f 25x2 − 4

h 9y2 − 16

11 15π + 10

g 9x − 49 2

i 81m − 64

j 144x2 − 1

7 a x2 + 14x + 49

b x2 + 8x + 16

Problem-solving

c x − 6x + 9

d x2 − 10x + 25

12 2 (x − 12) + 9 = 3x − 8

e x2 + 12x + 36

f x2 − 4x + 4

13 a A = 2 + (x + 1) (x + 5) A = x + 7 + x (x + 5)

b A = x + 6x + 7 c The variable x represents a length. Lengths in geometry cannot be negative, as they measure size or distance.

8 a 9a2 + 24ab + 16b2

b 4a2 − 12a + 9

c 4x2 − 28xy + 49y2 2

d 36p2 + 36pq + 9q2

e 16m − 40mn + 25n f x2 − 4xy + 4y2 g

Answers mathspace.co

713

17 a x3 + 9x2 + 27x + 27

9 a ab − 12a + 6b − 72 2

b 8x3 − 48x2 + 96x − 64

b k + 5k − 2kr − 10r 2

c 2x2 + 10xy − 8xz + 3x + 15y − 12z

c q t + 9q − 7t − 63 d x3 + 5x2 y + xy + 5y2 3

d 6t3 + 25t2 + 16t − 15

e 3m − 3m p − 4mp + 4p

18 4x2 + 20x + 25 square units

f 24g3 + 20g2 h3 + 18gh + 15h4

19 x2 + 7x + 12 cm2

g 21a4 − 18a3 b2 − 7ab2 + 6b4 h 35x2 y4 − 60x7 y + 28xy3 − 48x6

20 n + 8 21 4x2 + 48x + 140 m3

Reasoning 10 The difference in the number of terms arises because in (x + 2) (x + 2), the middle terms (2x and 2x) combine to give 4x, resulting in three terms. In contrast, in (x − 2) (x + 2), the middle terms (2x and −2x) cancel each other out, leaving only two terms.

3.04 Algebraic factors

11 a We can determine Jane’s bedroom by multiplying (x − 1) by (x + 1). This means that the area of Jane’s bedroom is (x − 1)(x + 1) = x2 − 1. Since x2 > (x2 − 1) for all values of x, then Marvin has the larger bedroom.

2 a −12x + 5, No

b Marvin’s bedroom is larger by 1 m . 12 Yes, the results of each expansion are x2 − 8x + 16 and 16 − 8x + x2 respectively. This occurs because when a negative number is multiplied by itself, it becomes positive; and when expanding a perfect square, where the two expressions are identical, the negative signs negate each other and yield the same result. 13 Not correct, as (a + b)2 does not equal a2 + b2. The square of a binomial has three terms. Rahul forgot to include twice the product of x and 9 in his answer. The answer should be x2 + 18x + 81. 14 Anette made a mistake when using the area model calculate the area of the top-left square. (10x) (10x) = 100x2, but Anette put 100x instead. Her ﬁnal answer should have been 100x2 + 20x + 30x + 6 = 100x2 + 50x + 6 Problem-solving 2

15 a 3y + 24y + 45 2

1 a 2

b 5

c 4

d 6

e 8

f 6

g 9

h 20

c 20x − 12x2, No

b −12x + 20x2, Yes d −12x + 20x2, Yes

3 a 10ab = 2 × 5 × a × b = 2a × 5b After crossing out 5b, the remaining factor is 2a. b 15a2 b = 3 × 5 × a × a × b

= 3a2 × 5b

After crossing out 5b, the remaining factor is 3a2. c 5ab = 5 × a × b = a × 5b After crossing out 5b, the remaining factor is a. d 5a2 b = 5 × a × a × b = a2 × 5b After crossing out 5b, the remaining factor is a2. e 5ab2 = 5 × a × b × b = ab × 5b After crossing out 5b, the remaining factor is ab. f 120a2 b2 = 2 × 2 × 2 × 3 × 5 × a × a × b × b

b 8x − 64x + 96 2

c −x − 11x − 30

d −4y − 40y − 36

e −9y2 + 9y + 378

f −48x2 + 72x − 27

g 24x2 + 40x − 16

= 24a2 b × 5b

After crossing out 5b, the remaining factor is 24a2 b. Fluency

16 a (t + 2)

b (t + 11)

c (t − 5)

e (t + 12)

f 2 (t + 9)

g 2 (v − 3) h 2 (t + 8)

714

Understanding

d (t + 6)

Mathspace WA Curriculum – Year 9 mathspace.co

4 a x e 4m

b x f bx

c b2

d y2

5 a y ( y + 5)

b 2t (t + 1)

15 a

c 3y ( y + 2)

d −m (m − 19)

e −y ( y + 2)

f v (8 − v)

6 a 4(a + 3)

b 5(a − 2)

c 5(3 + 4a)

d −2 (9a − 8)

b x(2x + 16)

e 2a(2b + 3)

f m (11k − n)

c 2x(x + 8)

g −2 (6s − 5)

h h (8j − 9g)

i kt (16j − 5)

j −7 (5m − 7n)

7 a −3(a + 2)

b 2(2 − 3a)

c −5(c + 6)

d −4(2 + 3d)

e −b(4 + 5a)

f −a(bc + 5a)

g −w (w + 6)

h −9mq (5n + 8)

8 a y ( y + 4)

b k (k − 9) 2

d z (1 + 6z )

c x (3 − x) 2

e z (1 + 8z )

f 5u (u − 3)

g w2 (3y − 8)

h −7 (3r − 2) (3r + 2) 2

16 a 4x2 y − 4xy2 b x−y

i 4uv (11 − 2u)

j 5k t (1 + 8t)

k 7pq (7p − 4q)

l −2a (8 + 9b)

c i xy(x − y)

m −5w (6 + 5y)

n uv (9v − 10u)

17 a 60a3 b2 − 30a2 b3

9 a 4(x + 3 + 4xy)

b 3(x + 3 + 4yx)

c 2x(1 + 5xy + 4y)

d 4x(xy + 2y2 + 3z2)

e 5bc(ab + 5c2 + 20a) f 10c(3b2 + ab + 2c) g 2ab(ab − 1)

h 5y(x2 y + 3z2 + 5y + 12)

Reasoning 10 The factor of −5 was not factored out of the −15x term correctly. The correct answer is −5x (x + 3). 11 They are both correct. If you expand the answer each gave you will see that they both produce −5x + 10y. 12 Although he took out 2 as a common factor, he forgot to take out x as a common factor. The fully factorised form is 2x(x2 + 4x − 2) Problem-solving 13 x − 4 14 a 5n(4 + n), so the dimensions are 5n and 4+n b Length = 5n, Width = 4 + n c Each section is now n × (4 + n) and to fence this the perimeter is 8 + 4n units of fencing.

c 40%

ii 4 (x − y) b 60a2 b2(2a − b) d

18 (t − 5) (3t + 4)

3.05 Factorise by grouping Understanding 1 x+7 2 18, 9 3 a p = 9, q = 10

b p = −9, q = 7

c p = −6, q = −4 Fluency 4 a (x + 2) (x + 3)

b (x + 2) (x + 5)

c (x + 3) (x + 3)

d (x + 2) (x + 6)

e (x − 1) (x − 2)

f (x − 2) (x − 4)

g (x − 2) (x − 3)

h (x − 3) (x − 3)

i (x + 8) (x + 3)

j (x + 9) (x + 8)

k (x − 3) (x − 1)

l (x − 5) (x − 12)

m (x + 5) (x − 6)

n (x + 5) (x − 4)

o (x + 6) (x − 9)

p (x + 7) (x − 10)

q (x + 5) (x + 8)

r (x − 5) (x − 7)

s −(x + 4) (x + 2)

t −(x − 3) (x − 4)

Answers mathspace.co

715

5 a ( y + 5) ( y + 7)

b (a − 4) (a − 3)

c ( p − 3) ( p + 6)

d (q + 4) (5 − q)

e (r − 3) (5 − r)

f (2r − s) (6r − rs)

g (t + u) (7t + 2u)

h ( y − z) (x − w)

i (4w + 3x) (5y − z)

j ( y − 4) (8y − 3)

6 a (x − 2y) (8 + z)

b (3 + x) (8 + y)

c (x + z) (7y + w)

d (x + 2) (x + 5)

e (x − 3) (x + 8)

f (5 + x) (10 + y)

g (12y + w) (x + z)

h (x + 5) (x + 8)

i (x − 5) (x + 10)

j ( p + 2) (2m + 3)

k ( p + 3) (5m + 2)

l (2x + 3z) (1 + 6y)

m (4x + 3z) (1 + 8y)

n (a + 8) (a2 + 1)

b True

c True

d False

Fluency 4 a 20

5 a

e 4p

k 0

6 a

j 0

7 a

8 a

b (x + 5)

7 a (x + 4)

c (x − 2)2

d (x − 10)2

e (8 + x)2

f (6 − x)2

Reasoning 8 ( y − z) ≠ (z − y) so either a factor of −5x needs to be taken out of the ﬁrst group or a factor of −7 needs to be taken out of the second group. 9 We know that the factor pairs of 24 are (1, 24), (2, 12), (3, 8), (6, 4), (−1, −24), (−2, −12), (−3, −8) and (−6, −4). When we add these pairs up, the largest possible integer we can get is 25, so, k = 25. Problem-solving 10 a 2 (x + 2) (x + 5)

b 4 (x + 2) (x + 4)

c 3 (x − 5) (x − 2) 2

h −4 (x − 5) (x + 2)

11 a x (x + 3) (x + 4)

b 2x (x + 5) (x + 3)

b (3x2 − 5y) (x + 4y)

13 a t + 7

We get:

f 4 (x − 9) (x + 8)

c ( g + h) (3f + g + h)

9 Since the denominators are the same, we just need to add the numerators. The numerators are like terms, so we can add their coeﬃcients.

10 Example answers:

e −3 (x − 2)

12 a (2x + 3z) (2y − 1)

Reasoning

d −2 (x − 2) (x + 11)

g 5 (x − 2) (x + 3)

d (2c − d) (c + 5d + 3) b 4 (t + 7)

14 (4 + x) (3 − x) units

3.06 Optional: Add and subtract fractions with numeric denominators Understanding 1 a

c 1

2 a 6

b 6

c 6

d 12

716

3 a False

Mathspace WA Curriculum – Year 9 mathspace.co

11 Simplifying fractions before adding or subtracting them in algebraic equations can make the process easier and the calculations less complex. It also reduces the chance of errors. Additionally, it can help with understanding the problem and the relationship between the numbers. 12 No, we cannot directly add or subtract fractions with different denominators. The denominators represent the total number of equal parts the whole is divided into. Different denominators mean the fractions are divided into different sized parts. Therefore, to add or subtract fractions, we ﬁrst have to ﬁnd a common denominator, which changes the fractions so that they are divided into the same sized parts.

Problem-solving

Fluency

13 a

d 2k

14 a

4 a

5 a

d ab

16 a

6 a

l −3c2

7 y 6 5 4 3 2 1 −15 −10 −5 −1 −2 −3 −4 −5 −6 −7

x 5

Reasoning 7 a

was not multiplied by the reciprocal of

17 b

3.07 Optional: Multiply and divide algebraic fractions with numeric denominators Understanding 1 a

2 a True

b False

c False

d True

3 a

Start with the equation

Multiply by the reciprocal of

Simplify

8 Both Amarion and Dean could go on to get the correct answer. Amarion needs to simplify the two fractions and then multiply them. Dean needs to simplify the fraction by removing common factors from the numerator and denominator to get the ﬁnal answer. Amarion:

Answers mathspace.co

717

7 a 6f 3 + 18f

Dean:

b −16a3 + 8a6 or

c 2

d 5g − 20fg + 40g 8 a −5j 2 − 70j

b 23n + s

c 15y + 5

d 7c2 − 40c

9 4 (x − 8) + 13 = x + 14 10 a b2 + 10b + 21 9 a Sometimes true.

c 3d2 − 13d − 30 a numeric

For example, fraction, while

11 a h2 + 6h + 9

c Never true. We cannot divide by 0, so this expression is undeﬁned, not 1. Problem-solving b

c 16c2 + 40cd + 25d2

b Never true. An algebraic fraction has at least one variable in the numerator or denominator. Multiplying two fractions with no variables, cannot result in any variables.

10 a

d 15e2 − 2e − 24 f 144g2 − 100

e f − 81

an algebraic fraction.

b c2 − 19c + 88

11 a

12 a rs + 7r − 8s − 56 b p3 + 13p2 w + pw + 13w2 13 14 a t

b 2y2

15 a 3 (x + 3)

b a (16b − c)

c −4 (2x + 1)

d −7x (5y + 3)

16 a 7c (c − 4)

b −f 2 g (3 − g)

c 5 (k + 7jk + 3)

d 3t (st + 3s2 + 10)

17 a 3x (6x + 1) b Length = 6x + 1 Width = 3x c Fencing = 12x + 1

12 a 14k

18 a (x + 5) (x + 7)

b (x − 3) (7x − 2)

19 a ( g + 2) ( g + 6)

b (a − 2c) (3 + b)

c ( y + 5) (3x + 2) 20 a (x + 4) (x + 7)

c (x + 4) (x − 8)

Chapter 3 review 1 a 5y

21 a 3 (x + 5) (x + 2) b 8a d −10a2 + a

c 4x + 7y

b 42u2 v4

2 a 6rs 2 2

d −6c14 d5

c 80xy z 3 a

4 a −2

d 5t

b (x − 8) (x − 6) d (x − 9)2 b (7a − 13b) (a2 − 3b)

22 a

23 a

24 25 a

5 Total cost = 2000 + 50x + 75y

6 a 2x2 − 5x

c −4g2 − 8g

718

b −11ab − 9ac d 21c3 d4 − 63c3

Mathspace WA Curriculum – Year 9 mathspace.co

26 a

c v

c i M (−2, −3) ii

2 A −12−10−8 −6 −4 −2 −2 M −4

4.01 Midpoint Understanding 1 One

−6

2 a M (2, 5)

b M (10, 6)

3 a Yes

b No

d i M (5, −2) ii

4 a M (5, 7)

8 6 4 2

c M (−3, 2)

d M (−3, −2)

e M (−2, 1)

f M (2, −4)

5 a M (8, 6)

c M (−5, 9)

e M (−9, −4)

f M (−5, 2)

g M (2, −6)

h M (−9, −7)

−2

−2 −4 −6 −8 −10 −12

y B x 2

6 M

e i M (5, 7) ii

6 a i M (9, 7) 11 10 9 8 7 6 5 4 3 2 1

−8

Fluency

x 2 4 6 8

6 A

5 4

3 M

2 1

x 0

x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

b i M (−3, 4) ii

y 8

6 M

4 2 x

−12−10−8 −6 −4 −2 A −2

f i M (−3, 2) ii

5 4 3 M 2 1 −10 −8 −6 −4 −2 −1 A −2 −3 −4

x 2

2 4 6 8

Answers mathspace.co

719

Problem-solving

g i M (−3, −2) ii

14 a i B (7, 7)

4 2

y 8

−7 −6 −5 −4 −3 −2 −1 M −2

1 2 3 4

−4

−6

−8

M x

−10

−2 −2

2 A

h i M (−4, −5) ii

1 −12 −10 −8 −6 −4 −2 −1

b i B (−5, −7)

y x 2

ii A

−2

−8

−3 −4 M A

4 −6

−4

2 x

−2

2 −2

−5 −6

−4 −6

−7

−8

7 $29 million 8 63 years

15 a A (5, 6)

b A (6, −5)

c A (5, 6)

d A (5, −6)

e A (−5, 6)

f A (4, −3)

g A (−4, −5)

h A (6, 22)

i A (−3, 3)

j A (−7, −4)

9 10 a 207 203

b 268 291

Reasoning 11 a The diagonals bisect one another. b i

c In a parallelogram, the diagonals bisect each other, meaning the midpoints of the diagonals coincide (are the same point). Since the midpoints of AC and BD are both , ABCD is a parallelogram. 12 a T (8, 8)

b S (6, 1)

c Parallelogram, because the diagonals bisect each other. 13 The mean of two numbers is halfway between the two numbers. The mean of the x-coordinates is the midpoint of the x-coordinates and the mean of the y-coordinates is the midpoint of the y-coordinates.

720

Mathspace WA Curriculum – Year 9 mathspace.co

16 The centre of a circle is always the midpoint of its diameter. So we can use the midpoint formula to ﬁnd the coordinates of point B. So B (8, −1). 17 (4x, 3y) 18 Q (−6, 1), R (−8, 4), S (−10, 7) 19 a = −6 and c = 0 or a = 18 and c = 12 20 a M (3, 7)

b k = −11

21 (0, 5.5) 22 23 a

b b=0

4.02 Gradient

c i

−1

Understanding

−1

1 A

−2

2 a Negative gradient

b Positive gradient

−3

3 a Zero gradient

b Undeﬁned

−4 −5

Fluency 4 a Increasing

b Decreasing

5 a Negative

b Positive

ii −5 d i

6 5 4 3 2 1

6 a The line has a gradient of zero. b The line has an undeﬁned gradient. 7 C, B, A, D

−6−5−4−3−2 −1 −1 −2 −3 −4 −5 −6

x 1 2 3 4 5 6

8 a i 2

ii 6

iii

b i 12

ii 6

iii 2

c i 14

ii 7

iii 2

d i 12

ii 4

iii 3

e i −12

ii 4

iii −3

f i −6

ii 2

iii −3

11 a 0

b −4

c −3

d 3

f 0

g −2

i 1

j −1

m −10

9 a 2

10 a i

y 6 5 4 3

12 Undeﬁned

13 As x increases, the value of y decreases.

1 −5 −4 −3 −2 −1

x 1 2 3 4 5

ii −3 b i

y 3

14 As x increases, the value of y increases. 15 a

16 a y = 27

b 5

Reasoning

1 −2 −1 −1 −2 −3

x 1 2 3 4 5 6 7 8

17 a

c Yes, the points A, B, and C lie on a straight line and the gradients of AB and BC are the same. Since both gradients are equal to

, the points

A, B, and C do indeed lie on a straight line.

Answers mathspace.co

721

18 Because a vertical line has no run, the formula for the gradient

point increases as time passes. In the context of the problem, this implies that Jay is moving forward towards his school along the straight track.

will have a denominator

of zero, and any fraction with a zero on the denominator is undeﬁned. 19 They will get the same gradient. One characteristic of a line is that its gradient is constant all the way along it. So, you can choose any 2 points along a straight line and their gradients will be equal.

Jay’s average walking speed during his journey to school. The rise corresponds to the change in distance, while the run corresponds to the change in time. A steeper gradient indicates a faster walking speed, and a ﬂatter gradient indicates a slower walking speed.

20 Line 2 is steeper (increases faster) because it has a larger gradient value than Line 1. The gradient of a line represents its steepness, and a larger gradient means that the line increases more rapidly in the vertical direction as it moves horizontally. In this case, for every unit increase in the horizontal direction, Line 2 increases by 5 units vertically, while Line 1 increases by only 3 units. Therefore, Line 2 is steeper and increases faster than Line 1. Problem-solving 21 a 3

b 15

c 20

22 m = 1 23 A line with a gradient of 0 must have the same y-values, therefore the pair of points that fit is A(3, 5) and D(7, 5).

in this context represents

b The gradient

26 i a Positive

b 2

c In the context of the graph, this positive gradient means the object is moving away from the starting point at an average speed of 2 metres per second. ii a Positive

b 5

c In the context of the graph, this positive gradient means the object is moving away from the starting point at an average speed of 5 metres per second. b −3

iii a Negative

c In the context of the graph, this negative gradient means the object is moving back towards the starting point at an average speed of 3 metres per second. iv a Neither

b 0

A line with a positive gradient increases from left to right, which means that as the x-value increases, the y-value also increases. So, the possible pair of points are:

c In the context of the graph, this zero gradient means the object is not moving, and its distance from the starting point remains constant over time.

• E(−3, 10) and F (4, 12)

27 a

• C(−1, 8) and F (4, 12) A line with a negative gradient decreases from left to right, which means that as the x-value increases, the y-value decreases. So, the possible pair of points are: • C(−1, 8) and B (4, 6) • E(−3, 10) and B (4, 6) 24 a y = 1

b y=1 c i 1

ii −1

d −1 25 a Since the line has a positive gradient, it means that Jay’s distance from the starting

722

100 Distance (metres) 90 80 70 60 50 40 30 20 10 Time (seconds)

Mathspace WA Curriculum – Year 9 mathspace.co

b Positive c The gradient is 20 metres per second. d The gradient, which is 20 metres per second, indicates that the car maintains an average speed of 20 metres per second throughout this period.

4.03 Distance between two points

9 a 20 units

b 25 units

c 11.40 units Understanding 1 2 a AB = 10 units

b AB = 5 units

c AB = 3 units

d AB = 5 units

3 a AB = 14 units

b AB = 8 units

c AB = 6 units

d AB = 3 units

4 a True

b False

c False

d True

Fluency 5 a i PQ = 3 units

ii QR = 4 units

b i PQ = 7 units

ii QR = 4 units

iii PR = 8.06 units c i PQ = 3 units

ii QR = 8 units

iii PR = 8.54 units ii QR = 4 units

iii PR = 5.66 units b AC = 10 units

c AC = 13 units

d AC = 13.04 units b AC =

units

8 a i

4 A

units

c AC =

units

d AC =

units

11 a 10 units

b 15 units

units

e 8 units

units

12 a 4.47 units

b 6.71 units

c 13.89 units

d 10.20 units

e 5.10 units

f 9.49 units

13 a 6.40 units

b 3.09 units

c 10.11 units

d 12.36 units

e 8.67 units

f 0.91 units

14 a 9.90 units

b 11.18 units

c N

units

b AM = 1.80 units

Reasoning 16 a

units

17 a AC =

units

units is smaller than

b BC =

units

18 AK = 7.2 units

19 a i AB = 5 units

1 −6 −5 −4 −3 −2 −1 −1

x 1

20 a i AB =

units

−3

ii AC =

units

−4

iii BC =

units

b • To show it has a 90° angle, check if the sides satisfy Pythagoras’ Theorem.

• To show it has two 45° angles, check if two sides are equal since the triangle must be isosceles.

4 3

c Yes

2 1

−2

iii BC = 8 units

−2

−8 −6 −4 −2 −1

ii AC = 5 units

b Isosceles

units

b i

d No. If we plot points A, B, and C in the Cartesian plane, we see that the points do not lie on the same line. Problem-solving

ii AC =

b AC =

c C is equidistant from A and B.

6 a AC = 3.61 units

units

c N is closer because

d i PQ = 4 units

ii AC =

10 a AC =

15 a AB = 3.61 units

iii PR = 5 units

7 a AC =

d 6.40 units

x 2

b (7x, −2x)

21 a 10x 22 a i MN =

units

ii NR =

units

b 2:5 B

23 23.1 units

units

Answers mathspace.co

723

24 a i PQ =

units

iii PR =

units

ii QR =

units

b Yes 25 a i

iii

b i

units

ii BD =

units

b 60 units2 27 a 50 units2

28 a 17 units

b 34π units

units

c d= c 17 units

d Yes, because the distance between the centre and the point (20, 4) is the same as the radius.

4.04 Gradient-intercept form

2 a The graph of the line will have an upward incline, moving from left to right. This means that as the x-coordinate increases, the y-coordinate also increases. b The graph of the line will have a downward incline, moving from left to right. This means that as the x-coordinate increases, the y-coordinate decreases. c The graph of the line will be horizontal, having no incline. This means that regardless of the x-coordinate, the y-coordinate remains constant. b m = 7, c = 0 d m = −1, c = 6

4 a m = −4

b −4

Fluency 5 a i (0, 3)

ii m = 2

b i (0, 3)

ii m = 4

c i (0, 5)

ii m = −2

d i (0, 2)

ii m = −6

724

e y = 8x − 4

7 a i m = −2

ii c = 9

b i m = −4

ii c = −8

c i m=8

ii c = 6

d i

ii c = −1

8 a i

ii c = 7

b i m=9

ii c = −8

c i m = 12

ii c = −9

d i m = 18

ii c = −12

e i m=4

ii c = −5

f i m = −1

ii c = 4 b y = −2x + 3

c y = −51

1 m represents the gradient of the line and c represents the y-intercept of the line.

9 a m = −2, c = 3

Understanding

3 a m = 15, c = −10

b y = −6x + 9

g y = −8x − 3

c Parallelogram 26 a i AC =

6 a y = 2x + 5

Mathspace WA Curriculum – Year 9 mathspace.co

10 a y = 2x + 1

b y = 59 kilometres

11 a y = 3x – 4

b y = 146 metres

12 a y = 3x − 4

d y = 3x + 1

e y = −2x + 5

13 a y = 5x + 3

b y = 5x − 6

c 14 a y = 4x + 3

c y = 0.6x + 3 15 a y = −3x – 2

b Yes

16 a y = −3x + 8

b No

17 a i 2

ii y = 2x + 2

b i 2

ii y = 2x − 9

c i -6

ii y = −6x + 2

d i 7

ii y = 7x − 3

18 a c y = −x – 1

b d

19 a

4 2

1 x −2

−1

−10 −8 −6 −4 −2 −2

2 4 6 8 10

−4

−1

−6

−2

−8 −10

b i

ii y = 12

3 2

iii x = −15

1 −4 −3 −2

−1

x 1

−1

y 20

−2

12 8

−20−16 −12 −8 −4 −4

8 12 16 20

−8

1 x −2

x 4

−1

−12

−16

−1

−20

−2 −3

Reasoning d

21 a The constant term is the same.

b All the graphs intersect the y-axis at the same point.

2 1 x −1

1 −1 −2

20 a i y = −2x − 2 ii y = −2 iii x = −1

c Lines of varying gradient will each pass through same y-intercept of (0, 1). 22 a The coeﬃcient of x is the same. b All the graphs have the same gradient. c The lines would be parallel on a Cartesian plane because they have the same gradient but different y-intercepts. This means they will not intersect each other, because they are always the same distance apart. 23 a b Line A. The gradient of line A is which has a larger magnitude than the gradient of line B.

, ,

Answers mathspace.co

725

24 Line B. The gradient of line A is

, which has a

smaller magnitude than −1, the gradient of line B. 25 The gradient is −3. A negative gradient tells us that we draw a line going down or decreasing from left to right. 26 The equation y = 2x represents a line with a gradient of 2 and a y-intercept at (0, 0). The equation y = −2x represents a line with a gradient of −2 and a y-intercept at (0, 0).

which gives the point (3, 6), then draw a line passing through the points (0, 4) and (3, 6). 3 You let y = 0, then you solve for x. Fluency 4 a i x −1 0 1 2 y 8 9 10 11 ii

10 9 8 7 6 5 4 3 2 1

The difference between the lines is y = 2x is increasing from left to right while y = −2x is decreasing from left to right. 27 No, it is not possible for a vertical line to be represented by an equation in the gradientintercept form y = mx + b. This is because a vertical line has an undeﬁned gradient, and m in the gradient-intercept form represents the gradient of the line. Vertical lines can be represented by an equation in the form x = k, where k is the x-coordinate of any point on the line.

y=x+9

−9 −8 −7 −6 −5 −4 −3 −2 −1 −1

b i

−1

x 1

y 4

Problem-solving

3 y = 3x + 3

29 a i m = 1

ii c = −13

b i

ii c = −3

c i

ii c = −7

d i m=1

ii c = 3

e i

f i

1 x −2

−1

30 a y = −x + 2

c i

−1

−4

y 4

b a=8

31 y = x + 15

32 y = 28x + 85

4.05 Linear graphs

1 B 2 To graph

, plot (0, 4) on the y-axis,

then rise 2 units up and run 3 units to the right,

726

x −2

−1

1 −1

Understanding

Mathspace WA Curriculum – Year 9 mathspace.co

y = 4x 2

d i

−1

5 a i

−3

−1 −10 −19

4 y = −x + 1

y = −3x − 1

x −2

−1

x −2

−1

−2

−1

e i

−1

−7 −4

−1

b i

−9 −7 −5 −3

y 1

−1

y 1 1

−4 −3 −2 −1

−1

x 1

−1

−2

−2 y = 3x − 4

−3

−4

−5

−5 −6

f i

−1

10 9 8 7 6 5 4 3 2 1 −1

−7

y = 2x – 9

−8 −9

c i y = −2x + 8

−3

y 3

x 1

2 1 x −7 −6 −5 −4 −3 −2 −1 x y= +2 −1 3

Answers mathspace.co

727

d i

−2

−2 −1

−1

3 1 2

−3 −4 x

c i (4, 0)

ii (0, 4)

iii

4 3

2 1

−4 −3 −2 −1 −1

−2 x

−2

−3

−4 −3 −2 −1 −1

−1 −2 −3 −4

4 2

y 9 8 7 3x 6 y=− 2 +6 5 4 3 2 1 −3 −2 −1

6 a

iii

−1

−3

−4

−1

ii (0, −4)

d i

−2

iii

−3

4 3 2

c (0, −2)

d (−2, 0)

e y = −5

f x=2

7 a i (−1, 0) iii

ii (0, 1) 4

−2

−3 −4

3 2

ii (0, −3)

e i (1, 0)

1 −4 −3 −2 −1 −1

−4 −3 −2 −1 −1

x 1

iii

2 1

−2 −3 −4

−2

−1

1 −1 −2

b i

ii (0, −1)

−3 −4

f i (2, 0)

728

Mathspace WA Curriculum – Year 9 mathspace.co

ii (0, 4)

iii

c (0, −6)

5 4

9 a i

ii (0, 1)

iii

−1

g i (0, 0)

e y = −2

d (9, 0)

−2

−1

−2

−1

ii (0, 0)

iii

3 2

ii (0, −9)

b i (−2, 0)

iii

−4 −3 −2 −1 −1

−2 −3 −4

h i (2, 0) iii

−1

8 a

−3

−8 −6 −4

2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −11

1 −2

ii (0, −9)

c i (3, 0) iii

2 4 6 8 10

−4 −6 −8 −10 −12

−1 −2

−10 −8 −6 −4 −2 −2

ii (0, 6) 7 6 5 4 3 2 1

12 10 8 6 4 2

d i (−2, 0) x

1 2 3 4 5 6 7 8 9 10

iii

−1

x 1

−1 −2 −3 −4 −5 −6 −7 −8 −9 −10

ii (0, 8) 10 8 6 4 2

−10−8 −6 −4 −2 −2 −4 −6 −8 −10

x 2 4 6 8 10

Answers mathspace.co

729

10 a

7 6 5 4 3 2 1 −3

−2

−1

4 3 2 1

−3 −4

−2

−1

−2

−3

1 −4 −3 −2 −1

11 a

−1

ii (0, −3)

12 a i 3

iii

4 3 2 1

x −1

−1 −2

−2

−3 −4

x 1

−1 −2 −3 −4 −5 −6

b i −2 iii

2 1 −4 −3 −2 −1 −1

−1

x 1

−2 −3 −4

ii (0, −1) 4 3 1

−4 −3 −2 −1 −1 −2 −3

Mathspace WA Curriculum – Year 9 mathspace.co

−4

730

−2

−4 −3 −2 −1 −1

x 1

−1 −2

x 1

ii (0, −1)

c i 1 iii

3 2 1

−4 −3 −2 −1 −1

16 The graphs of y = 7x − 1 and y = 7x + 3 would be parallel lines on a Cartesian plane because they have the same gradient but different y-intercepts. This means they will not intersect each other, because they are always the same distance apart.

−2 −3 −4

ii (0, 3)

d i iii

15 No, the graph of y = −4x + 1 does not rise from left to right. It falls from left to right because the gradient of the line is −4, which is a negative number. In a graph, a negative gradient means the line goes downhill as you move from left to right.

Problem-solving

17 y = 10x + 50 Money saved

3 2 1

150

−4 −3 −2 −1 −1

100

−2 −3

−4

Weeks

13 a

5 4 3 2 1

5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

2 3 4 5 6 7 8 9

It would take 10 weeks for Mary to save $150. 18 a i 2 x

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

iii

1 2 3 4 5

ii (0, 6) 10 8 6 4 2

−5 −4 −3 −2 −1 −2 −4 −6 −8 −10

x 1 2 3 4 5

Reasoning 14 Elisabeth was correct. She remembered to go downhill for her rise, as the gradient was negative.

1 2 3 4 5

ii (0, −8)

b i 2 iii

10 8 6 4 2 −5 −4 −3 −2 −1 −2 −4 −6 −8 −10

x 1 2 3 4 5

Answers mathspace.co

731

19 a W = 5 − L b

6 a i y = 2x + 7

W 5

0 −1 −2 −3 −4 −5

10 y 9 8 7 6 5 4 3 2 1 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9

9 10

y = 2x + 7

−4 −3 −2

x 1

−1 −2

1 2 3 4 5 6 7 8 9 10

b i ii

Understanding

5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

4.06 Equation of a straight line

1 2 3 4 5 y=−

7 a i −4

1 a

b −4

f −1

2 a No

b Yes

c −3

c Yes

d No

3 B 4 a i m = −3

ii (1, 6)

b i

ii (−4, 5)

c i m = −9

d i m=1

ii (−8, −3)

5 a y = −4x − 24

b y = 4x + 19

c y = −3x + 15

d y = −9x − 33

e y = −2x − 15

h j

Mathspace WA Curriculum – Year 9 mathspace.co

3x −1 2

ii y = −4x + 9

b i

c i

8 4x − 7y + 40 = 0 9 a i y = −3x ii

−3

−6

−9 −12 −48

−28

−4

−2

−21

−9

−8

−7

−6

b i ii

Fluency

732

−1

d Yes, the graph should have endpoints. For the shape to be a rectangle, L needs to be greater than 0 but less than 5. That means W must also be greater than 0 but less than 5.

8 7 6 5 4 3 2 1

c i y = −8x + 90 ii

−15

210

d i y = −2x + 5 ii

−8

−4

−1

−7

−29

4.07 Direct proportion

e i y = −8x − 6 ii

−3

−1

−14

−30 −646

−9

−6

−45 −30

f i y = 5x ii

Understanding 1 a Gradient = 3

b Gradient =

2 a Yes

b No

c No

d No

3 a True

b True

c False

d True

c No

d Yes

e False 10 a y = −5x + 40 c y = −6x + 66

b y = 2x Fluency

4 11 a Yes

b No

c Yes

d No

Reasoning

12 Yes, it is possible. You can ﬁnd the gradient using the coordinates of the two intercepts using the formula

. Once you have

the gradient, you can use the point-gradient formula or the gradient-intercept form to ﬁnd the equation of the line. 13 If the line is horizontal, the gradient m is 0. The formula becomes y − y1 = 0(x − x1), which simpliﬁes to y = y1. Since y1 is a single value, this is the equation of a horizontal line.

15 Example answers:

10 20

8 a

r 10

40 30

b y = −4x + 16

18 a (1, 1)

b 3x − 10y + 7 = 0

19 a p = 17

b q = 12

20 18 square units

e r = 44

16 5x − 4y + 44 = 0

c 9x − 2y − 129 = 0

d q = 15

• y = −x + 4

17 a (4, 0)

• y = 3x

Problem-solving

b y = 2x − 10

All lines have different gradients and y-intercepts.

b No

7 a b = 9a

9 a

• y = 2x + 1

6 a No

14 To sketch the graph of the line from the equation, begin by plotting the point (x1, y1 ) which was given in the equation. Then, using the gradient from the equation, use the rise and the run to sketch more points on the line. Connect the points to get the full graph of the line.

b 0

b 6 c a = 6b d a = 42 e b = 10

Answers mathspace.co

733

10 a 280

b 70

b i

b 5

11 a

12 10

12 a 300 Water

270 240 210 180 150 120 90 60 30

6 4 2 10 20 30 40 50 60 70 80 90 100 110 120 130 140

b 2.1 c For every 100 g of rice William uses, he adds 210 mL of water.

0 2 4 6 8 10 12 14 16 18

ii c i

The unit rate of water with respect to rice is 2.1 mL/g. 9 8 7 6 5 4 3 2 1

Rice

13 a

y 120 100 80

Fish caught

60 40 20

x 0

ii Hours 0 1

d i

2 3 4 5 6 7 8 9

y 15

b c 9 fishes are caught every 4 hours 14 a i

y 5 15

x 0

ii 5

15 a

x 0

Reasoning 16 It increases by a factor of 7. 17 a

734

Mathspace WA Curriculum – Year 9 mathspace.co

Days

Amount saved

21 28

22 a

Amount saved

22 20 18 16 14 12 10 8 6 4 2

Jimmy

Number of books read

Robots built

Ben 1

Days 3

4 3 2

Time (hours) 2 3 4 5 6 7 8 9

b Derek

35 30 25

Online

5 6

25 a k = 0.82

b h = 0.82s b 21 grams

4.08 Optional: Indirect proportion

x 4

b 84 cm

26 a A = 3C

Newspaper

24 a 6 c h = 1.64 cm

15 10

Yuri

1 0 1

Derek

18 a y = 4x

c There were more online enquiries each day than there were newspaper enquries. d The total number of enquiries, both online and newspaper, are directly proportional to the number of days advertised. Problem-solving 19 m = kp

1 a False

b True

c True

d False

2 a Direct

b Inverse

3 a Inverse

b Inverse

c Neither

d Neither

e Neither

f Neither

g Inverse

h Inverse

Fluency

20 r = kn

4 a k = 200

Hours worked

Plates painted

18 16 14 12 10 8 6 4 2

c By comparing the constant of proportionality or unit rate. The higher unit rate would be the steeper graph, which represents the person who is saving more quickly.

b Carl 23 a

21 a

Number of weeks

c y = 40

5 a k = 12

b k = −6 d

c 6 a Yes

b Yes

c No

d No

7 a

−5

−4

−3

−2

−1

b Increasing x 0

c As x increases, y increases at a faster and faster rate.

Answers mathspace.co

735

8 a k = −144 c y = −18 9 a

c No, the pressure cannot ever reach zero. When dividing 6000 by a very large number we can get closer to zero, but it will never actually get to zero.

b d x = −9

Problem-solving

16 a k = 7.35 20

17 a = 9, b = 42 18 a k = 1220

10 r 15

b k = 200

11 a k = 9

19 32 metric tons

4.09 Model linear equations Understanding 1 a x+4=6

Time (min) 32 30 28

b x − 10 = 9

c 11x = 88

e 4x − 13 = 21

f 8x + 11 = −5

g 6(x − 20) = −10

2 The independent variable.

26 24

Fluency

Rice (g) 100 110 120 130 140

b k = 3000 c • The cooking time decreases as the amount of rice increases. • The product of the rice amount and cooking time is constant. Reasoning 13 The other quantity decreases by a factor of 7. 14 a b There were more online enquiries on the first day, so the number of enquiries decreased more quickly over time, making the line steeper than that for newspaper enquiries. 15 a 6000 kg/cm2 b The pressure decreases. Since we are dividing 6000 by the volume, an increase in volume means we divide by a larger number, which results in a lower pressure.

736

b 40 minutes

20 d

10 a k = 1

12 a

c h = 3.68 cm

Mathspace WA Curriculum – Year 9 mathspace.co

3 a y = 6.8x + 0.8

b $41.60

4 a 0.2 seconds

b y = −0.2x + 6

c 5.1 seconds

d x = 2.5 hours

5 a

Number of minutes passed (x)

Amount of fuel left in 150 125 100 75 50 the tank ( y) b y = 150 − 5x

c 30 minutes

6 a 1.4 m/min

b y = 1.4x

c 8.4 m

d 9 minutes

7 a 0.5x + 1.5y = 5

b 4 sheets

8 a 135 mg

b 70 hours

9 a y = 45x + 150

b $420

10 a y = 16x – 1300

b $3660

11 a y = 100 − 9x

b 59.5 km

12 a y = 75x + 50

b 218.75 km

4.10 Multi-step equations

Reasoning 13 a The cost of each text message. b The amount of credit Paul purchased. c $21.64 14 The baseball is moving downwards instead of upwards at T = 4 seconds.

Understanding 1 5 − 2(3x − 4) = 3(2x + 1) − 7x Start with the equation 5 − 6x + 8 = 6x + 3 − 7x

Expand the brackets

13 − 6x = −x + 3

Combine like terms

13 = 5x + 3

Add 6x to both sides

10 = 5x

Subtract 3 from both sides

b 28

2=x

Divide both sides by 5

c The water level before any water is pumped out.

2 a Incorrect

b Correct

c Incorrect

d Correct

15 a The rate at which the water level is falling per minute.

16 a 49 represents the initial value of the heartbeat (i.e. resting heartbeat). b The change in one minute of Anri’s heartbeat. Problem-solving

b 48 ( g − 2) + 9 = 10 ( g + 3) − 4 c 32 (x + 5) − 6 = −28 (x − 1) + 5 Fluency

17 a y = 10x + 340

b $700

18 a $63.50

b $5.50

c A ﬁxed cost such as a toll charge b S = 8 − 0.5(D − 1)

19 a 3.5 megabytes c 15th day

e d = 10

g q = −2

i y=8

j b

7 a Inﬁnitely many solutions

c Inﬁnitely many solutions

e Inﬁnitely many solutions

4 a

5 a

20 a C = 2920R + 50 b $0.21

c If the smart speaker is left on for 24 hours a day, the total running cost would be 24 × 365 × R = 8760R. Adding the initial cost of the smart speaker, we get the new total cost: Cnew = 50 + 8760R. To show that the cost does not triple, we can compare the new cost with three times the original cost. Substituting the expressions for Cnew, we have: Cnew = 50 + 8760R = 50 + 8760(0.21) = 1889.60 Three times the original cost is 3 × 650 = 1950. From the computation, we can conclude that the cost does not triple when the smart speaker is permanently left on. 21 a 188 bpm

3 a 60 ( f + 4) + 10 = −12 ( f + 1) − 7

b m = 220 − a

c Between 94 bpm and 159.8 ≈ 160 bpm.

6 a e

h x = −8

Reasoning 8 a Clara provided the correct solution. Here’s why: • Clara started with the given equation: 2z + 6 = 3z − 5 • She subtracted 3z from both sides: −z + 6 = −5 • She multiplied both sides by −1: z − 6 = 5 • She added 6 to both sides: z = 11

Answers mathspace.co

737

b David made a mistake in his working:

• David started with the given equation: 2z + 6 = 3z − 5 • He subtracted 3z from both sides: −z + 6 = −5 • He should have subtracted 6 from both sides: −z = −11

(14p − 28) = −8p + 16

Start with the equation

4(14p − 28) = −56p + 112

Multiply both sides by 7 to clear the fraction

56p − 112 = −56p + 112

Expand the brackets

112p − 112 = 112

Add 56p to both sides

112p = 224

Add 112 to both sides

• He switched the right-hand side and the left-hand side and then changed the sign of z: −11 = z (this is the mistake) The mistake made by David was in the fourth step, where he should have multiplied both sides by −1. 9 a His mistake was in Step 4. Instead of subtracting 5 from both sides, he added 5. The correct step should be subtracting 5 from both sides to get 2x = 5. b

4(2x + 5) − 3 = 37 4(2x + 5) = 40

Add 3 to both sides

2x + 5 = 10

Divide both sides by 4

2x = 5

12 a

Divide both sides by 2

10 a Alice provided the correct solution. She correctly expanded the equation, simpliﬁed it, and solved for x, resulting in x = 5.

11 a

a=3

Start with the answer

a+2=5

Add 2 to both sides

2a + 4 = 10

Multiply both sides by 2

2a + 4 + a = 10 + a

Add a to both sides

Subtract 5 from both sides

b Bob made a mistake in the third step of his working. After reaching −x + 17 = x + 7, he should have added x to both sides of the equation to get 2x = 10, just like Alice did. Instead, he subtracted x from both sides, resulting in −2x = 10, which led to the incorrect solution of x = −5. Start with the equation Expand the brackets Simplify Add 8p to both sides Add 16 to both sides Divide both sides 16

Divide both sides by 112

c Preference for either method is subjective. Method 1 may be considered easier due to fewer steps and simpler arithmetic, while Method 2 involves larger numbers and more calculations. It depends on individual comfort with fractions and arithmetic.

Start with the equation

p=2

2a − 7 + 4 + a = 10 + a – 7 Subtract 7 from both sides

3a − 3 = a + 3 b

Simplify

Start with the answer Add 1 to both sides Double both sides Add 2b to both sides Simplify

Problem-solving 13 x = 18 14 a Let r represent the cost of a rugby ball: Price of a cricket ball = r + 1.36 b 8(r + 1.36) + 9r = 119 c $6.36 d $7.72 15 a Let y cm represent the length of the longer side: Length of shorter side = y − 4 b 2( y − 4) + 2y = 12 c 5 cm × 1 cm

738

Mathspace WA Curriculum – Year 9 mathspace.co

16 x = 5

Reasoning

17 y = 8

9 a Dylan made an error in Step 1. He missed the constant term and he should have made sure had the same denominator. This that

18 y = 18 years old

should be the Step 1 before proceeding to cross-multiplication. By writing the binomial

4.11 Optional: Equations with algebraic fractions

1 a 2

b 42

c 6

d 12

2 a True

b False

c True

d False

with the same denominator we have Next, should have multiplied 2 by the binomial 32 − x, giving us the answer of 64 − 2x.

Understanding

3 a

Write the right-hand side as a fraction

Cross multiply

Expand the brackets and simplify Add 2x to both sides and subtract 48 from both sides

Fluency 4 a 5 a

c b y = −1

c x = −1

h c=1

Divide both sides by 26 and simplify

10 11 One example is

. Multiply both sides

by 12 (LCM of 6 and 4): 6x = 6x + 12. Subtract 6x from both sides: 0 = 12, which is a contradiction. Thus, no value of x satisfies the equation.

6 a

12 a

g q = −1

13 x = 6

Problem-solving b x = 28

14 k = 60 7 a

e n = −1

8 a

15 r = 0 16 The equation for the total funds raised is . Solving for x gives x = $1153.85.

4.12 Optional: Rearrange formulae Understanding 1 The purpose of rearranging formulae is to be able to evaluate for different variables when doing repeated calculations. 2 We should make c the subject

Answers mathspace.co

739

3 C 4 a False

b True

c True

d False

b m = 4, y = −16 y 2

−2 −1 −2 −4 −6 −8 −10 −12 −14 −16 −18

Fluency 5 a x=y−z

b x = −8y

f x = −7y

6 a m = ghy

c m = −2, y = 4

7 a

b y = 4 − 3x2

d y = 2x3 + 3

8 a

d m = 3, y = 6

d y = 4x −5

e y = 3x − 5

f y = −2x + 3

g y = 2x + 4

h y = 2x + 3

12 a

14 a

−4 −3 −2 −1 −1 −2 −3

740

x 1

x 1 2 3 4 5

,y=5 6 5 4 3 2 1

7 6 5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3

b y = −2x + 3

c y = −2x + 3

5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

11 a y = x − 1

2 3 4

Mathspace WA Curriculum – Year 9 mathspace.co

15 a

g z = x2

i c = sn2 + 6

23 Ruﬁno has an error in his work.

The equation for KE as given

Multiplying both sides by 2 Now he should have divided by v2

18 19

Problem-solving

Reasoning 20 Both equations can be rearranged for x. When 3x + 8 = 2 is rearranged for x the answer is numerical. Whereas when 3x + y = z is rearranged for x the answer has unknowns, y and z, in it. However if you substitute y = 8 and z = 2, you get the same numerical answer as solving 3x + 8 = 2. 21 Alice is correct in her rearrangement, while Bob made a mistake. Alice rearranged the equation to make a2 the subject by subtracting b2 from both sides: 2

c =a +b

a2 = c2 − b2

Bob, on the other hand, made a mistake in his rearrangement. He should have subtracted a2 from both sides to make b2 the subject, like this:

25 a

b r = 2.79 cm

26 a

b PV = $2027

27 a

b r = 3 mm

Chapter 4 review 1 a M (−1, 2)

b M (−2, −2)

2 a (−2, 0)

3 a A (−6, 6)

b A (8, 3)

4 a Positive

b Undeﬁned

c Negative

c2 = a2 + b2

5 a −2

c2 − a2 = a2 + b2 − a2

b2 = c2 − a2

6 −9

8 a

b2 = c2 + a2

d 0

units

b 9.43 units b

units

9 28.4 units

So, Alice is correct, and Bob made a mistake in his rearrangement. 22 Pamela’s mistake was in Step 2 of her algebraic working. When she divided both sides of the equation by L, she incorrectly wrote the right-hand side of the equation as

d Zero b Undeﬁned

7 a 8.06 units

However, he added a2 instead, yielding an incorrect equation:

b h = 3 cm

c − b = a + b − b

24 a

10 a i m = −4

ii c = 12

b i m = 12

ii c = 4

c i

ii c = −7

d i

instead

. The correct equation should be

11 a m = −3, c = 1

b y = −3x + 1

c y = −59

Answers mathspace.co

741

12 a

−3

x y

−1

16 a (2, −3)

b 6x − 7y − 33 = 0

−1

17 a Yes, y = 2x

b No

18 a

y 12

19 a For an inverse relationship, the product of the coordinates gives us the constant of proportionality, so k = 150.

10 8 6 4

−3 −2

−1

5 4 3 2 1

x 1

−2

c y = 15

20 a

−5

−4

−3

−2

−1

b Increasing

c As x increases, y increases at a faster and faster rate. 21 a k = 20 x

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

b 5 hours

1 2 3 4 5

22 Yes, there is an inverse variation because as the number of winners increases, the prize money decreases. The constant of proportionality is 275. 23 a $280

iii

2 1

112

93.33

200 100

0 1 2 3 4 5 6 7 8 9 10

x 1

e x is inversely proportional to y. 24 a $0.04 25 a

−3 −4 −5 −6

742

300

−2

15 a y = 4x + 15

400

1 −2 −1 −1

560 280 186.67 140

500

ii (0, −3)

b i (5, 0)

x y

−4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7

iii

ii (0, −6)

14 a i

Number of hours passed (x)

Mathspace WA Curriculum – Year 9 mathspace.co

c 281 g 0.5

1.5

Number of kilometres 150 110 left to drive ( y)

b y = 150 − 80x b y = −3x − 22

b $12.60 0

c 112.5 minutes

Fluency

b a = −4

26 a c

27 a b x=2 c Infinitely many solutions d 28 a z = 6

c No

d No

f Yes

g Yes

h No

i No

j No

k Yes

l No

m Yes

n No

o No

p Yes

5 a Quadratic

b Linear

c Quadratic

d Neither

e Neither

f Quadratic

6 a Quadratic

b Linear

c Quadratic

d Neither

b Yes

e No

7 a

4 a No

b d

31 a

b r = 2.52 cm

−3 −2 −1 0 1 2 3 4 5 y 2.25 1 0.25 0 0.25 1 2.25 4 6.25 x

16 36 64 100 144 196 256 324 400

−2

−1

3 0.75 0 0.75 3 6.75 12 18.75 27

−3 −2 −1 0 1 2 3 4 5 y −3.6 −1.6 −0.4 0 −0.4 −1.6 −3.6 −6.4 −10

5.01 Quadratic relationships

8 a i T he first figure has 7 squares. The second figure has 10 squares. The third figure has 13 squares and so on.

Understanding

The number of squares in each ﬁgure is increasing by 3.

−2

−8

−18

4 −32

First Difference −2 − 0 = −2

Second Difference −6 − (−2) = −4

−8 − (−2) = −6 −18 − (−8) = −10 −32 − (−18) = −14

−10 − (−6) = −4 −14 − (−10) = −4

2 a Linear

b Quadratic

c Linear

d Quadratic

e Linear

f Linear

g Quadratic

h Quadratic

i Quadratic

j Quadratic

3 a

ii Number of squares = 19 iii Linear b i T he first figure has 1 hexagon. The second figure has 4 hexagons. The third figure has 9 hexagons and so on. At each step, the increase goes up by 2. Example: +3, +5, +7, and so on. ii Number of hexagons = 25 iii Quadratic c i T he first figure has 3 squares. The second figure has 6 squares. The third figure has 11 squares and so on.

Step number

Number of squares

b To go from one step to the next, we add the next whole number. First, we add 1, then 2, then 3, and so on.

At each step, the increase goes up by 2. Example: +3, +5, +7, and so on. ii Number of squares = 27 iii Quadratic

c 15

Answers mathspace.co

743

Reasoning

11 10 9 8 7 6 5 4 3 2 1

9 To conﬁrm this, we could calculate the ﬁrst and second differences and check if the second difference is a non-zero constant. The second differences are zero so the table of values cannot have a quadratic relationship. 10 a y1 is a linear relationship as the highest power of x is 1. y2 is a quadratic relationship as it has an x-squared term and no higher powers. b y1 is a linear relationship because it increases by 3 for each increase in x by one unit. y2 is a quadratic relationship because the increase in y increases by 2 each time for each change in x by one unit (example: +1, +3, +5...). This would make the second differences all 2, which is a non-zero constant. c y1 has a minimum and is U-shaped, so it is a quadratic relationship. y2 has no minimum or maximum output and has a constant rate of change, so it is a linear relationship.

−3

−2

−1

11.25 5 1.25 0 1.25 5 11.25

−3

−2

−1

11.25

1.25

First differences Second differences

1.25 5

3 11.25

−6.25 −3.75 −1.25 1.25 3.75 6.25 2.5

2.5

2.5 2.5 2.5

Since the second differences are all the same non-zero number, this relationship must be quadratic. If we were to plot these points on a graph and join them with a curve, we get a U-shaped graph with a minimum, so this is a quadratic relationship.

744

Mathspace WA Curriculum – Year 9 mathspace.co

2 3 4

12 a Sometimes true. For example, the ﬁrst table of values has all the y-values positive, but has non-zero constant second differences, so it is quadratic. The second table has all the y-values positive, but has constant ﬁrst differences, so it is linear.

If we ﬁnd the ﬁrst and second differences, we get:

−4 −3 −2 −1 −1

11 Many possible answers. For example: x

−2

−1

13 3.25 0 3.25 13 29.25

−2 −1

−4

−1

−4 −9

−1

b Always true. A change in signs like this means there is a turning point. Linear relationships do not have turning points. c Never true. If all of the y-values are the same, then it would be graphed as a horizontal line, so it is not a quadratic relationship. Also, it would have zero ﬁrst and second differences, so it is not quadratic. Problem-solving 13 Any pattern that has 2, 8, 18, and 32 elements in the ﬁgures in the correct order.

14 a Side length of small square (m)

Total area (sq m)

Perimeter of large square (m)

100

144

b The large square has a perimeter of 40 m, so its side length is 40 ÷ 4 = 10 m, and its area is 102 = 100 square metres. Set 4s2 = 100: s2 = 100 ÷ 4 = 25, so s = 5 metres. The total area is 100 square metres. c The large square’s side length is 10 m, with an area of 100 square metres. For s = 5, the total area is 4(5)2 = 4 × 25 = 100 square metres, matching the large square. This could represent four smaller squares (each 25 sqm) arranged in a 2 × 2 grid within the 10 m × 10 m space, confirming the design.

5 a i The turning point is (0, 0). ii Minimum iii The equation of the axis of symmetry is x = 0. iv The x-intercept and y-intercept are both at (0, 0). b i The turning point is (0, 0). ii Maximum iii The equation of the axis of symmetry is x = 0. iv The x-intercept and y-intercept are both at (0, 0). 6 a

−2

−1

−28

−7

−28

−1

b −2

−5 −10 −15 −20

15 a To ﬁnd the y-values, ﬁnd the square of x-values, and then multiply the result by 4.

−25

b Quadratic

−30

c y = 4x2 16 Many possible answers. For example: x

−2 −1

10 2.5 0 2.5 10 22.5

c x = ±1

−2

−1

−5 (1, −7)

5.02 Graphs and features of quadratics

(1, −7)

−10 −15

Understanding

−20

1 a Concave up

b Concave down

2 a True

c False

b False

d True

3 The turning point lies right on the axis of symmetry. Knowing the axis of symmetry makes it possible to ﬁnd the turning point, and vice versa. Both elements help in understanding the shape and position of the graph. Fluency 4 a i Concave down

−25 y = −7x2 −30

7 a i

−2

−1

−44

−11

−44

y −2

−1

−5

x 1

−10

ii Maximum point

b i Concave up

ii Minimum point

c i Concave down

ii Maximum point

d i Concave up

ii Minimum point

−15 −20 −25 −30 −35 −40

Answers mathspace.co

745

b i

−2

−1

ii 35

8 7 6

5 4

−2

10 5 −2

c i

f i

−1

−2

−1

−64

−16

−64

ii −1 −5

−10 −15 −20 −25 −30 −35 −40 −45 −50 −55 −60

d i

−2

−1

−2

−1

−52

−13

−52

y −2

g i

−1

y −2

−5 −10 −15 −20 −25 −30 −35 −40 −45 −50

−2

−1

−4

−1

−4

y x −2

−1

−2

−3

30 25

−4

20 15

h i

10 5 −2

e i

−1

x 1

−2

−1

−2

−1

45 40 35 30 25 20 15 10 5 −2

746

Mathspace WA Curriculum – Year 9 mathspace.co

−1

x 1

8 a

−3

−2

−1

−18

−8

−2

−8

−18

b Maximum

c Negative

9 a Concave up c

b Turning point: (0, 0) 70 y 65 60 55 50 45 40 35 30 25 20 15 10 5

−1

and the range of the function will be all y-values less than or equal to this maximum value. 13 a i C

ii B

iii A

iv D

b i When a is positive, the parabola opens upward, and solutions to ax2 = 18 exist if y = 18 intersects the graph. Larger a values make the parabola steeper, reducing the x-values of the solutions for a = 9). (e.g., x = ±3 for a = 2 vs. When a is negative, the parabola opens downward, and since y = 18 is positive, there are no real solutions as the graph never reaches positive y-values above the vertex (e.g., a = −7). 14 a D

b A

c B

d C

Reasoning

Problem-solving

10 Answers may vary. An explanation may be: The square of any number (positive or negative) is always greater than or equal to 0. Therefore 0 would be the minimum

15 a The parabola is concave up because the coefficient of the x2 term is positive (22). It has a minimum value.

11 Always true. All parabolas have a turning point, which has either the maximum value (if the parabola opens downwards) or the minimum value (if the parabola opens upwards):

3 maximum

−2

c The minimum value is 0. representing the profit when no items are produced (x = 0). d x2 = 4, so x = ±2. These represent producing 200 items (since x = 2 hundred) or a theoretical negative production, yielding a profit of 88 thousand dollars. In context, only x = 2 is practical. 16 a

−4 −3 −2 −1 −1

b (0, 0)

Number of ﬂoors

x 1

minimum

−3 −4

12 The y-value of the turning point is signiﬁcant in determining the range of a quadratic function because it represents the minimum or maximum value of the function, depending on whether the parabola opens up or down. For an upwardopening parabola, the y-value of the turning point is the lowest point on the graph, and the range of the function will be all y-values greater than or equal to this minimum value. For a downward-opening parabola, the y-value of the turning point is the highest point on the graph,

Amount charged

$1000

$4000

$9000

$16 000

$25 000

$36 000

Amount charged

35 000 30 000 25 000 20 000 15 000 10 000 5000

Number of floors 0

1 2

c y = 1000x

d $64 000

Answers mathspace.co

747

e 20 ﬂoors high 17 a i

Fluency

207 368

5 a 8

7 6

y = x2

C 350

4 y = 3x2 3

300

250

200

−4 −3 −2 −1

150

y = 3x2 is vertically enlarged.

100

x 1

y −4 −3 −2 −1 −1

x 1

−2

b Solving: x2 = 16, so x = ±4. Only x = 4 makes sense, indicating that producing 4 items costs $368.

−3 −4 y = −2x2 −5

y = −x2

−6

c The vertex at (0,0) represents the minimum cost of $0 when no items are produced.

−7 −8

5.03 Basic quadratics Understanding

y = −2x2 is vertically enlarged. 6 a

y 8

1 Any equation where a > 1

7 y = x2 6

Examples: y = 2x2, y = 1.5x2, y = 5x2 2 Any equation where 0 < a < 1

5 4

Examples: 3 The outputs of y = x2 are 0 and positive values, and the outputs of y = −x2 are 0 and negative values. The graph of y = x2 is concave up, and the graph of y = −x2 is concave down. The graph of y = x2 has a minimum value, but the graph of y = −x2 has a maximum value. The shape of graphs are the same, and they both have a turning point at the origin. The axis of symmetry for both graphs is the same.

3 2

1 2 x 3

−4 −3 −2 −1

is vertically compressed.

y −4 −3 −2 −1 −1

x 1

−2 −3

4 a Vertical enlargement b Vertical enlargement

−4

c Vertical compression d Vertical compression

−5 −6 −7 y = −x2 −8

748

Mathspace WA Curriculum – Year 9 mathspace.co

is vertically compressed.

13 For any value of x, the result of x2 will be positive. Since the result of x2 is multiplied by −3, and a negative times a positive is negative, the value of y will always be negative.

7 a y = 2x2, y = x2, y = 0.2x2 b 8 a y = −x2

b y = 2x2

c y = −3x2

Problem-solving

14 a V = 30s2 2

9 a y = −2x

c y = x2

8 7 6 5 4 3 2 1

d y = 2x2

10 a y = 4x2

Reasoning 11

x 2

y=x

−3

−2

−1

b s = 8 cm y

−3 −2 −1

x 1

Area ≈ 7 units2 16 a

b 2

The outputs of

are

of the outputs of

y = x2. This means the outputs of

are

closer to the x-axis, and the parabola is less steep (wider). 12 Similarities:

c y = −4x

d y = 10x2

f b 6 countries

17 a

5.04 Solve algebraically

• All three graphs are vertically enlarged. • All three graphs have the same axis of symmetry. • All three graphs have the same x-intercept, (0, 0). • The graphs of y = 2x2 and y = −2x2 are enlarged by the same factor. • The graphs of y = 2x2 and y = (2x)2 are concave up. Differences: • The graph of y = (2x)2 is narrower than the graphs of y = 2x2 and y = −2x2. 2

• The graph of y = −2x has been vertically reﬂected across the x-axis, and is concave down. The other two have not been reﬂected and are concave up. • The graph of y = −2x2 has a maximum value, but the other two graphs have minimum values. • The graph of y = (2x)2 has a a different enlargement factor than y = 2x2 and y = −2x2.

Understanding 1 Both have a coefficient a and a constant k, but a linear equation has one solution, unlike a quadratic equation, which can have up to two solutions. 2 a True

b False

c True

d True

3 x = −12 Fluency 4 a

c x = ±11

b x = ±5 d

5 a y = ±5

c k = ±2

d d = ±6

e c = ±8

f j = ±9

g b = ±3

h n = ±10

6 a x = ±7 e

b p = ±4

c z = ±6

d y = ±8

Answers mathspace.co

749

7 a x = ±11

b x = ±5

c x = ±3

17 Samantha made an error when she interchanged the numerator and the denominator on the right-hand side when dividing both sides by 36.

g x = ±8

h x = ±15

Problem-solving

18 2 seconds

9 x = 6 trays

19 After 3 weeks

10 p = 5 atm

20 a 2x2 = 338

b 13 or −13

21 a A = 3x

11 t = −2° C

b x = 5 or x = −5

12 b = 5 beats per second

c 5 cm

13 d = −8 dollars

22 a h = 0.8 metre

Reasoning

b t = 0.5 seconds

14 T he equation can be rewritten as 3x2 = 27, which simplifies to x2 = 9. This equation has two solutions because there are two values of x which, when squared, make 9. One is the positive square root of 9 (x = 3), and the other is the negative square root of 9 (x = −3).

For x = 3: Left-hand side = 3x2 Write the left-hand side of the equation

= 3 × 32 =3×9 = 27

Substitute x = 3

or approximately 0.83 metres

5.05 Optional: Null factor law Understanding 1 A 2 a a = 3, b = 5

b a = 6, b = −3

c a = 0, b = 4 3 a (x − 5)(x − 1) = 0

b (x + 3)(x − 8) = 0

Evaluate the power

c (x + 2)(x − 9) = 0

d x(x + 7) = 0

Evalute

e (4x + 1)(5x − 3) = 0

f (x − 4)2 = 0

For x = −3: Left-hand side = 3x2 Write the left-hand side of the equation

= 3 × (−3)2 Substitute x = −3 =3×9 Evaluate the power = 27 Evalute Each result is the same as the right-hand side of the equation, so the values x = 3 and x = −3 are correct. 2

15 The equation can be rewritten as 2x = −4, which simplifies to x2 = −2. Since the square of a real number is always non-negative, there is no real number x that satisﬁes the equation. Therefore, the equation has no real solutions. 16 Ted made an error when he only took the positive square root of both sides. To correct the error, he should have taken both the positive and negative square roots of both sides.

750

Mathspace WA Curriculum – Year 9 mathspace.co

Fluency 4 a j = 0 , −7

c m = 0 , −5

d m = 0 , −5

h e=4,2

i s = −6 , −7

5 a x = 3, x = −8

b y = −1, y = 5

c m = −4, m = 2

d p = 7, p = −3

e t = −6, t = 1

f q = −9, q = 3

6 a x = 6, x = −4

,k=5

c m = −12,

, p = −8

e t = 30,

7 a x = 0, x = −7

b x = 0, x = −3

c x = 0, x = −3

d x = 0, x = 7

e x = 0, x = −3

f x = 0, x = 5

8 a k = 0, k = −10

b p = 0, p = 7

c t = −7, t = −8

d w = 6, w = −5

e r=8

f n = −2, n = 9

g s = 6, s = 17

h m = 10, m = −2

i v = 4, v = −3

j u = 2, u = −4

16 T he quadratic equation has a common factor of x, so x = 0. The linear factor is cx + d. Solving cx + d = 0 gives the second solution

17 S asha is correct in this speciﬁc case but not universally. The null factor law states that if (2x − 5)(x + 3) = 0, then 2x − 5 = 0 or x + 3 = 0, giving

and x = −3. Since the factors are

different, there are two distinct solutions.

10 a b = 3t

b b = −t

c b = −2t

d b = 4t

However, if the factors were identical, such as in (x − 2)(x − 2) = 0, the null factor law gives x − 2 = 0 or x − 2 = 0, both yielding x = 2, a single solution. Thus, two factors don’t always guarantee two distinct solutions—it depends on whether the factors are distinct.

11 a k = 3

b k=2

18 a i x = 4 or x = 5

c k=9

d k = 15

9 a x = a, x = −2a

, x = −a

12 a L = 5, L = −2 Since length must be positive, L = 5 metres. b L = −3, L = 7 Since length must be positive, L = 7 metres. c L = 4, L = −1 Since length must be positive, L = 4 metres. d L = −6, L = 2 Since length must be positive, L = 2 metres. 13

ii x = −2 or x = 7

iii x = −1 b i (4, 0) and (5, 0)

ii (−2, 0) and (7, 0)

iii (−1, 0) c The solutions from part (a) are the x-intercepts of the graphs in part (b). Problem-solving 19 a a = −12

b a=3

20 c = 3 21 a x = 13

b Minimum

c x = 6.5 cm 22 6 hours

Reasoning 14 J ade’s answer is incorrect. When using the null factor law, the correct equations should be x − 4 = 0 and x + 6 = 0. Solving these equations, we get x = 4 and x = −6 as the correct solutions, not x = −4 and x = 6.

, w = −5; since w must be positive,

recommend

24 The drone reaches the target altitude at minutes and h = 1 minute. The earliest time is minutes.

15 M ilo is incorrect. The null factor law states that if a product equals zero, such as (x − a) (x − b) = 0, then x − a = 0 or x − b = 0, giving x = a or x = b. Changing the order to (x − b) (x − a) = 0 still gives x − b = 0 or x − a = 0, resulting in the same solutions: x = b or x = a.

Chapter 5 review

The order doesn’t matter because multiplication is commutative. For example,

iii x = 0

(x − 2)(x + 3) = 0 has solutions x = 2 and x = −3, and (x + 3)(x − 2) = 0 has the same solutions: x = −3 and x = 2.

1 a Yes

b No

c No

2 a Quadratic

b Neither

3 a i Concave down

ii (0, 0)

d No

iv x-intercept: (0, 0) y-intercept: (0, 0) b i Concave up

ii (0, 0)

iii x = 0 iv x-intercepts: (0, 0) y-intercept: (0, 0)

Answers mathspace.co

751

b At d = 0 days

4 a 0 °C

d i

d After d = 0 days

c 0 °C 2

b y = 0.5x

5 a y = 3x

c y = −0.5x2

d y = −x2

6 7 a i −2

−1

2 −4 −3 −2 −1

−5

ii Concave up −10

iii Minimum point

e i

−3 −2 −1

−15

2 3 4

x 1

−5

−20

−10

ii Concave down b i

iii Maximum point −15

y 8

ii Concave down

f i

iii Maximum point y

2 x −4 −3 −2 −1

ii Concave up

iii Minimum point

c i

y −4

2 3 4

−2

x 2

4 −2

−2 −4

−1

ii Concave up

iii Minimum point

8 a m=0

b y = ±2

−6

c x = ±15

d x = ±7

−8

ii Concave down

iii Maximum point

9 a p = 0, 4

b q = 0, −2

e t = −3, 5

10 d = 8 11 a w = 0, 9

752

Mathspace WA Curriculum – Year 9 mathspace.co

b w = 4.5 m

6.01 Cost, profit, revenue

iv For 10 items a loss of $1095

Understanding

ii The ﬁxed costs are $10 000

c i The revenue per item is $3500 iii For 3 items a loss of $3250

1 B 2 a True

b False

c True

d True

iv For 10 items a proﬁt of $12 500 13 a $1800

b $560

c $1240

3 Find the point of intersection of the lines for revenue and cost, or solve the equation: Revenue = Cost.

14 a $200

Fluency

15 a Proﬁt of $500

b Loss of $250

16 a Proﬁt of $400

b Loss of $100

4 a Proﬁt of $5340

b Proﬁt of $730

c Loss of $500

d Loss of $480

5 a $326.40

b $220

c Loss

d Loss of $20

c 4 17 a C = 25x + 200, R = 50x

b $108

b 8

c $1260

c Loss of $100

6 a R = 26x

b R = 1890x

7 a $2250

b $2610

Reasoning

8 a C = 170x + 220

b C = 520x + 440

9 a $400

b $300

18 a From the original break-even point where 10x + 120 = 12x and x = 60, if the cost increases to 11 this means that 11x + 120 = 12x and x = 120. So the number of units sold would have to double just to break-even, from 60 to 120.

c Profit of $100

d 10

e $100 10 a The cost is $10 000 and revenue is $0 b The cost is $30 000 and revenue is $40 000 c 4 11 a 900

b Calculating the number of items if the ﬁxed price rises from $120 to $200 or if the cost per item rises from $10 to $10.50, 12x = 10.5x + 120

Dollars

x = 80

800

12x = 10x + 200

700

x = 100

C = 40x + 380

600 500

This means that it’s harder for a business to break-even when the ﬁxed price increases from $120 to $200, as they need to sell 100 items instead of 80 to cover their costs.

R = 120x

400 300

x 2

b $120

c $380

d $600

12 a i The revenue per item is $150 ii The ﬁxed costs are $400 iii For 3 items a loss of $70 iv For 10 items a proﬁt of $700 b i The revenue per item is $25.5

e $860

19 a i N o, the business will never make a loss because the revenue is always greater than the cost. ii P = 15x b i T he cost will be a horizontal line at C = 500, where C is the y-axis. ii P = 20x − 500

ii The ﬁxed costs are $1250 iii For 3 items a loss of $1203.5

Answers mathspace.co

753

20 a 700

3 C

Dollars C = 9x + 600

600

Fluency

500

4 $17 264

400

5 $294

300

6 $8373.50

200

7 $2696.08

100 0

R = 8x x 10 12 14

b The weekly cost is always greater than the weekly revenue. This is because the revenue per smoothie is less than the cost per smoothie.

10 a $6333

b $2923

c $1462

d $38

11 $272

b $64 800

14 a $28.50

b $741

15 a $49 400

b $53 300

b 800 meals

16 a 65 hours

b $24.65

c $700

17 $5902

c $1725.50

18 $59 489.55

23 a 900

Dollars

19 a y = 0.02x + 1 700

800

C = 50x + 390

700 600

c $175 000 20 a y = 0.005x + 800 21 a $2500

R = 110x

400

c $3000

300

x 2

b $210

c 19.1%

d P = 60x − 390

6.02 Income

b $1225 b 10% d $15 000

Reasoning 22 No, Mbacke is incorrect. The conversion 1 month ≈ 4 weeks is not very accurate. To calculate the monthly pay for the manager job, he should multiply by 52 then divide by 12 which gives $2946.67. 23 Elna should choose Offer B.

Understanding 1 Wage is a rate of money earned per time period, so the total amount earned is based on how much time you end up working. Salary on the other hand means that you earn a set amount independent of the number of hours worked. 2 a False

b $6678

c $110 000

500

754

c $47 424 d $3952

13 a y = 36x

c $25 per meal b $300

b $1824

b 41.5 hours

Problem-solving

22 a $420

9 a $912

12 a y = 52x

c She needs to increase the revenue per smoothie, or decrease the cost per smoothie, so that the revenue per smoothie is greater than the cost per smoothie.

21 a C = 15x + 8000

8 a $54 840 b $53 040 c Offer 1

b True

c True

Mathspace WA Curriculum – Year 9 mathspace.co

Her predicted income under Offer B is 5% of $4000 plus $1000 which is $1200. Her predicted income under Offer A is 10% of $4000 plus $500, which is only $900. Offer B is also the safer offer because she will earn more even if she has overestimated the value of sales she will make.

12 a $4996.53

b $10 329.30

b 40a

c $1470.75

d $8835.44

c 100b

e $10 509.60

f $4980.36

24 a 20a + 40b, or 20(a + 2b)

25 Answers vary, but may include: Job A has an annual value of $34 200, so total income is greater per year with Job B. Job A only requires 40 weeks of work, but Job B is working 48 weeks. So maybe Job A is preferable to work less for still a reasonable amount of money.

b $7800

27 a 22.5 hours

14 Option 1 15 $1634 16 a 8%

b 9%

c 4%

17 $80 18 $78 585

Problem-solving 26 a $27.63

13 Option 2

c 18.1% b 15 hours

c 7 hours

19 Fortnightly Reasoning 20 Yarel has multiplied an annual interest rate by the number of months, so the units of time in his calculation are not the same. He must either

28 At least 36 hours 29 $240

convert 22 months to

30 2%

interest rate of 8% p.a. to

31 a Profit = Line C Cost = Line A Revenue = Line B

b 10 units

21 The interest from Option 1 will be 0.5 × 2 × 12 = 12% of the principal, while the of the principal, so Option 1 will earn Ariela the most interest.

Understanding 1 a $5400

b $1080

2 a Divide by 12

b Multiply by 12

3 a True

c True

Fluency 4 a $342

per month.

interest from Option 2 will be

6.03 Simple interest

b False

years or convert the

b $7942

5 $10 080

22 When calculating interest we multiply by the loan term, so a shorter loan term will result in less interest paid. When we calculate repayments, we divide by the number of repayments. If making monthly repayments, for example, the number of repayments is 12 × T. A smaller T value means a smaller number of monthly repayments, which means that the repayment amount may be greater. Problem-solving

6 a $2500

b 5 years

7 1.82 years 8 a 1.5 years

b 6 quarters

23 a $960

b $290

24 a $2400

b 6%

25 a $2784

b $5490 d $2490

9 a 7.56%

b 10.95%

c 7.84%

c $10 764

10 a 1.5%

b 0.5%

c 0.12%

26 a $1000

11 a $693.00

b $692.75

27 5 years

c $2164.80

d $657.99

b $200

c 2%

28 25.36% p.a. 29 $2857.14 30 Option 1

Answers mathspace.co

755

6.04 Re-investing with simple interest

8 a $1600

b 10 weeks

c 1% per fortnight Understanding 1 Starting with the opening balance, add the interest and subtract any repayments. 2 a True

b False

c True

9 a

Month

Opening Balance

Interest

Closing Balance

9400

9447

47.24

9494.24

47.47

9541.71

47.71

9589.42

d False

e True Fluency 3

4 a

Month

Opening Balance

Interest

Closing Balance

73 000

73 073

73 146

73 219

73 292

10 a $8754.88

73 292

73 365

73 438

11 a 0.5%

Quarter

Opening Balance

Interest

Closing Balance

880

44 880

Month

b 0.4%

Opening Closing Interest Repayment Balance Balance

868.48

4.34

250

622.82

3.11

250

375.93

44 000

44 880

880

45 760

375.93

1.88

250

127.81

45 760

880

46 640

127.81

0.64

128.45

46 640

880

47 520

Month

c 6 years

d $17 878.45

e $2878.45 12 a 0.75%

Opening Closing Interest Repayment Balance Balance

Month

Opening Closing Interest Repayment Balance Balance

2500

700

1875

146 26 452.84 198.40

5000

21 651.24

1875

700

1250

147

21 651.24 162.38

5000

16 813.62

1250

700

625

148

16 813.62 126.10

5000

11 939.72

149

11 939.72

89.55

5000

7029.27

150

7029.27

52.72

5000

2081.99

151

2081.99

15.61

2097.6

625

700

b 4 months c $300

c 12 years and 7 months Opening Closing Quarter Interest Repayment Balance Balance 1

27 800

417

3200

25 017

417

3200

22 234

417

3200

19 451

417

3200

16 668

b $12 800 7 a 15 years c $598 500

756

c Under simple interest Page would have earned $188 interest, so she has earned $1.42 more than with simple interest.

c $3520

6 a

b $189.42

b 3 quarters or 9 months

5 a

d $8400

e $400

b 0.5% per month d $283 500

Mathspace WA Curriculum – Year 9 mathspace.co

d $752 097.60

e $450 000

Reasoning 13 If the opening balance plus interest is less than the normal repayment amount, the borrower only needs to pay back that amount. If they made their normal repayment, they would be paying back more than they owe. 14 The equivalent monthly rate for Option A is 0.75% simple interest. Since they have the same rate,

Option B would earn Ibryam the most interest because the interest is calculated as a percentage of the total amount (principal plus interest previously earned) instead of just the principal. 15 When being charged simple interest, making higher repayments can decrease the number of months taken to pay off the loan, which decreases the total interest paid. When interest is calculated as a percentage of the total amount, making a higher monthly repayment also decreases the interest charged on the next month, so the effect is even greater. Problem-solving 16 a Year 1: $10 Year 2: $10 Year 3: $10 Year 4: $10 b Year 1: $10 Year 2: $11 Year 3: $12.10 Year 4: $13.31 c

Interest ($)

17 a i 6 months to repay the loan and total interest is $45.84 ii 4 months to repay the loan and total interest is $36.38 b $9.46 18 a i $112 b

38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2

ii $97.59

Amount owing on loan ($) Option B

Option A

Time (Months) 1

Interest on total amount 12

c Ashly should choose Option B since the amount will decrease over time.

Simple interest

8 6

Chapter 6 review 1 a Proﬁt of $6090

b Proﬁt of $780

c Loss of $500

d Loss of $480

2 a $200 c Loss

3 a $29.62

b $270 d Loss of $70 b $5200

c 13.51%

4 a The weekly retainer is $600 and the commission is 5%

2 Year 0

d Simple interest points lie in a horizontal line because the amount of simple interest stays the same each year. The interest calculated as a percentage of the total amount increases by a greater amount each year, so the points curve upward, rather than lie in a straight line.

b $57 200 c $20 000 5 a $476.92

b $787.50

c $4482.00

d $795.43

6 a $850

b $9350

Answers mathspace.co

757

Month

8 a

Opening Balance

Closing Balance

Interest

14 J aden has multiplied an annual interest rate by the number of months, so the units of time in his calculation are not the same. He must either

85 000

127.5

85 127.5

127.69

85 255.19

convert 18 months to

interest rate of 6% p.a. to

85 255.19

127.88

85 383.07

128.07

85 511.14

128.27

85 639.41

128.46

85 767.87

Month

Opening Balance

Interest

Closing Balance 30 600

30 000

600

30 600

612

31 212

624.24

31 836.24

636.72

32 472.96

b Year 1: $12 Year 2: $12.96 Year 3: $14.00 Year 4: $15.12 c

Interest ($)

18 16

c Simple interest rate of 2.06% per month

9 a 0.5%

Interest on total amount

Opening Repay- Closing Month Interest Balance ment Balance

945.60

4.73

280

670.33

3.35

280

393.68

1.97

280

115.65

0.58

116.23

10 a i C

d $16 636.23

e $2636.23

ii A

iii B

b 80 units 11 a C = 5x + 10 000

b 1000 cups

c $15 per cup 12 No, Aria is not correct. The conversion 1 month ≈ 4 weeks is not very accurate. To calculate the monthly pay for the tutoring job, they should multiply by 52 then divide by 12 which gives $3250. 13 Leo should choose the Executive package. His predicted income under the Executive package is 8% of $5000 plus $1800 which is $2200. His predicted income under the Strategist package is 15% of $5000 plus $1200, which is $1950. The Executive package is also the better offer because he will earn more even if he has overestimated the value of the projects he will complete.

Simple interest

c 5 years

Year 1

Mathspace WA Curriculum – Year 9 mathspace.co

d The scenario from part (b) returned the larger investment by approximately $6.08 16 a i $162.50 b

ii $132.02

Amount owing on loan ($) 45

Option 2

40 35

Option 1

30 25 20 15 10 5 1 c Option 2

758

per month.

15 a Year 1: $12 Year 2: $12 Year 3: $12 Year 4: $12

b $2472.96

years, or convert the

Time (Months) 4 5

Since each triangle has a right angle and shares angle x, the remaining angle in all the triangles will be equal to (90 − x)°. This shows that all the angles are equal, so all the triangles are similar.

7.01 Similar right-angled triangles Understanding 1 a i ∠Q

ii ∠R

iii ∠P

b QP c RP

Problem-solving

2 a Each pair of corresponding angles must be equal. b Each pair of corresponding sides must be proportional. 3 a False

b True

e False

f True

c False

12 564 cm 13 14.7 m 14 43.75 m

d False

15 The height of the roof is about 4.33 m.

7.02 Scale factors

Fluency

Understanding

4 a A and D

b A and B

c A and B 5 a No

b Yes

c Yes

d Yes

1 a NL

b NM

2 a RS

b ∠RUT

3 a True

b False

c False

e True

f True

g True

6 a y = 17.5 cm

b y = 5 mm

c y=6m

d y = 9 cm

e y = 7.6 m

f y = 12.025 m

4 Smaller

7 a r = 55, s = 35

b m = 3.4, n = 15

c x = 10, y = 45

d y = 10, x = 6

Fluency

,y=6

8 a

Reasoning 9 a Yes. Since a pair of corresponding sides on the right triangles is in the same ratio as the two hypotenuses, the remaining sides will also be proportional by Pythagoras’ theorem.

5 a x = 28, y = 89

b a = 64, b = 85

c p = 71, q = 38

d j = 112, k = 68

e r = 80, s = 118

f x = 150, y = 117

6 a f = 9, g = 5

b c = 15, d = 30

c j = 24, k = 22

d s = 4, t = 12.5

e a = 11, b = 26

f e = 27, f = 11

g a = 6, b = 25

h x = 16.5, y = 7

b Yes. Since a pair of corresponding angles in the right triangles has the same measure, all pairs of corresponding angles are equal.

7 a 3

b 3

10 All pairs of angles in the two triangles are equal. It is possible to ﬁnd the missing angle since the sum of the angles in a triangle are equal to 180°. The side lengths given of both triangles are in the same proportion.

8 a i 2

ii x = 8

b i

ii a = 32

c i

ii x = 12

d i

ii m = 4

11 All the triangles H

x°D x° A

9 4 10 a 2

x° CA

x° EA

d True

b 12.7 cm

11 a

Answers mathspace.co

759

• Corresponding angles must be equal.

16 cm 4 cm closet

12 cm

8 cm

2.4 cm 18.4 cm

Reasoning 12 She is incorrect. For a quadrilateral to be similar, the AAA similarity test does not hold. We also need to show that all corresponding sides are in the same ratio. But some sides have a scale factor of 6 while others have a scale factor of 2, so these are not similar shapes. 13 Dorothy is correct. If the picture wasn’t distorted, it would be in the ratio of 1 : 1, not 1 : 3. This makes it look taller and narrower that in reality. If he had enlarged it by a scale factor of 3, it would be 300 × 300 px. 14 The scale factor of

2 Congruent shapes have equal side lengths, but the side lengths of similar shapes are proportional. 3 a False

b True

c True

d False

e True

f True

g False

4 a Congruence

b Congruence

c Similarity

d Congruence

5 a Any rigid transformation including a reflection, a translation or a rotation b A dilation or any rigid transformation, including a reflection, a translation or a rotation c Any rigid transformation including a reflection, a translation or a rotation Fluency 6 a

would reduce it back to its

original size. For example, if a side had length x the enlarged length would be 3x, so if we use a scale factor of

we get

, the original length.

7 2

Problem-solving 15 a 30 cm

3 b 60°

16 3

7 a Yes

b Yes

Scale factor = 2 Scale factor =

17 L = 5.3 18 a 2 cm

b 1 cm

19 a i Display correctly ii Distorted or cropped

c No

d Yes

Scale factor = 8 a i Similar shapes

ii Dilation

iii Distorted or cropped

b i Congruent shapes

ii Rotation

iv Display correctly

c i Congruent shapes

ii Reﬂection

v Distorted or cropped

d i Similar shapes

ii Dilation

vi Display correctly b i 100 cm2

ii Taller

7.03 Transformations and similarity Understanding 1 • Corresponding sides must be proportional to each other.

760

Mathspace WA Curriculum – Year 9 mathspace.co

9 a i ii Reﬂection and dilation by a scale factor of b i 3 ii Dilation by a scale factor of 3 and a translation c i 3

ii Rotation, dilation by a factor of 3 and a translation OR a dilation by a factor of 3 and a horizontal and vertical reﬂection. d i

24 Not always, similar shapes are a scaled version of each other. Only if the scale factor is 1 are they also congruent. Problem-solving

ii Rotation and dilation by a scale factor of

25 3 cm

10 a N

26 2 m

b Each pair of angles must be equal.

L M c MN

d ML

11 a Yes, the scale factor is 3.

7.04 Optional: Congruent triangles

b No 12 a Scale factor =

b Scale factor = 3

c Scale factor =

d Scale factor = 6

13 a LN 14 a i ∠Q

ii ∠R

b PQ 15 a Similar

b LM iii ∠P

2 a Yes

b No

c No

d Yes

c PR

3 a AAS

b SSS

c ASA

d RHS

b EF

c ∠EHG

e SAS Fluency

b x = 26

4 a Yes, they satisfy RHS. b Unknown, there is not enough information.

18 a 2 b i s = 14

Understanding 1 All three pairs are corresponding sides are equal in length, and all three pairs of corresponding angles are equal in measure.

16 y = 17.5 17 a x = 30

27 The dimensions of the smaller image would be 222 pixels by 103 pixels as we can’t have half a pixel.

ii n = 122

19 a 4

b x = 20

20 a

iii m = 44

c Yes, they satisfy AAS. d No, they are not congruent. e Yes, they satisfy SAS. f Yes, they satisfy RHS. g No, they are not congruent.

Reasoning

h Yes, they satisfy SSS.

21 No, because even though all sides are in the same ratio, the matching angles are not equal.

i Yes, they satisfy AAS.

22 a The two hexagons are each equilateral, giving them proportional sides, and equiangular, giving them equal angles. b All pairs of angles in the two triangles are equal and all pairs of sides are in the same proportion. 23 Yes, the conditions for similarity are equal angles and proportional sides. All corresponding sides and angles will be equal. All corresponding sides will have a scale factor of 1, which makes them proportional.

j Yes, they satisfy SSS. 5 a Triangle A and B

b AAS

6 a Triangle B and C

b RHS

7 a Triangle A and C

b SSS

8 a Triangle A and B

b AAS

9 a Triangle B and C

b SAS

10 a Triangle A and C

b RHS

11 a Triangle A and C

b SAS

12 a Triangle A and D, Triangle B and C b ASA and AAS 13 SAS

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761

14 a △ ACD and △BCE

b SSS

15 a △JKL and △JMN

b AAS

16 a △PRN and △QRM b SSS and SAS 17 a SAS

b EF = EH Given ∠FEG = ∠HED Vertically opposite angles EG = ED Given 18 a SAS

b PS = RQ Given QS is common ∠PQS = ∠RSQ Alternate angles on parallel lines 19 a AAS

23 Consider the triangles shown in the diagram. With the given information there are two possible triangles, one of which is not congruent. So we do not have enough information to conclude that the triangles are congruent.

b Example Answer:

BX = CX Given ∠ ABX = ∠DCX Alternate angles on parallel lines

∠BXA = ∠CXD Vertically opposite angles

C Q

Reasoning 20 If two pairs of angles of two triangles are equal, then the third pair of angles must be equal as well because the angle sum of any triangle is 180°. So if we know two angles and the included side are equal to the corresponding parts of another triangle, then the third angle is also equal and we have the information required to show Angle-Angle-Side. 21 Two triangles can have equal angles but have different side lengths. Consider these two equilateral triangles (where all angles are 60°):

1 cm 60°

1 cm

8 cm

1 cm 60°

Problem-solving 24 a ∠QRP = ∠SRP

b QR = RS

c ∠RPQ = ∠RPS 25 a JN = MN b Possible Answers: ∠JKN = ∠MLN c Possible Answers: ∠KJN = ∠LMN 26 a They are all radii of the circle. b PQ = SR

c ∠QXP = ∠SXR

b ∠ AXB = ∠DXC Vertically opposite angles BX = CX Radii of the same circle

60°

60° 8 cm

22 Yes. The intersecting lines form a pair of vertically opposite angles which are equal. Because the segments bisect each other, the triangles will be congruent by side-angle-side congruence (SAS) no matter how long the segments are.

762

27 a SAS

60° 60°

Mathspace WA Curriculum – Year 9 mathspace.co

AX = DX Radii of the same circle 28 a i AAS or RHS

ii x = 3

b i SSS or RHS

ii x = 9

c i SAS or RHS

ii x = 11

d i SSS or RHS

ii x = 7

18 a SAS

Chapter 7 review

b AB = CD Given

1 a Yes

b Yes

2 a y = 19.5

b y=7

3 a r = 62, s = 28

b a = 65, b = 7.5

BD is common

4 a One triangle is the one with the woman as the ﬁrst side, the side length of 1.8 m as the second side, and the line from the woman’s head to point S as the hypotenuse. The second triangle is the one with the tree as the ﬁrst side, the side length of 5.4 m as the second side, and the line from the top of the tree to point S as the hypotenuse. These two triangles are similar because the share angle S as a common angle, and they both have right angles. This means they are similar by the AAA similarity test. b 4.8 m

19 a Triangle C and D b SSS

8.01 Pythagoras’ theorem Understanding 1 a True 2

6 a i

ii z = 36

b i

ii z = 6

7 a f = 83, g = 68 8 a r = 15, s = 30

9 Jay is incorrect. For two quadrilaterals to be similar, all corresponding sides must be in the same ratio. In this case, Jay only dilated the rectangle vertically, so only two pairs corresponding sides are in the same ratio. Two sides have a scale factor of 2, while the others have a scale factor of 1. So, the rectangles are not similar.

d True

3 a i △ ABC, △ ABD, △ BCD iii BC

iv BD

v AB

b i No

ii No

iii Yes

iv Yes

b Yes

c No

d No

Fluency c c = 29 m

b c = 16, d = 6

c True

ii AC

5 a c = 25

b a = 117, b = 63

b False

2 13 + x = 19

4 a Yes

5 27 m

b c = 15 d c = 50 cm

6 a c=

b c=

c c=

d c=

e c=

f c=

g c=

h c=

7 a y = 12

b x = 16

c b = 21

d a = 18

8 a a=

b b=

c k=

d m=

e q=

f r=

9 h = 10

10 4 11 a 21 cm 12 a 40 m

∠ ABD = ∠CDB Alternate angles on parallel lines

b 13.6 cm b 46 m

13 a Yes

c 52 m b 3

10 165 cm 11 25 cm 12 a 62 + 122 = 36 + 144 = 180 and 142 = 196. Since 62 + 122 ≠ 132, the triangle is not right-angled. b 31.4 cm

15 16 Possible Answers: ∠RPQ = ∠RPS, QR = QS, ∠QRP = ∠SRP

c 36 cm2 13 x = 70

17 Possible Answers: JN = MN, ∠JKN = ∠MLN, ∠KJN = ∠LMN

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763

Reasoning

Problem-solving

14 Using the converse of the Pythagoras’ theorem, if a2 + b2 = c2 where c is the longest side, then the triangle is right-angled. Let a = 10, b = 24, and c = 26. a2 + b2 = c2 2 10 + 242 = 262 676 = 676

19 a d = 105

Since 10 + 24 = 26 , the triangle is a right triangle. 15 a She made a mistake on step 4 as she simplified extracting

to c − b. This is

incorrect because the expression cannot be further simplified. b Since we are solving for a length, we know that it must be positive. If she did show the negative square root, it would need to be discounted in the final answer.

b 14 595 m2

c 3.6

20 x = 3.9 21 a 11.2 cm

b 12.8 cm

22 a x = 5.39 m

b y = 5.39 m

c 21 m

d $777

23 a 6

b x = 6.8

24 a x = 10 cm

b 96 cm

25 Using the converse of the Pythagoras’ theorem, if the sides a = 4 m and b = 7 m satisfy a2 + b2 = c2 with the measured diagonal c = 8 m, then the path is the true diagonal. Solving for the real diagonal:

c2 = a2 + b2

c2 − b2 = a2

=a 16 a 52 + 72 = 25 + 49 = 74 and 92 = 81 Since a2 + b2 ≠ c2 then it can not be a right-angled triangle. b 7.52 + 102 = 56.25 + 100 = 156.25 and 12.52 = 156.25 2

Since a + b = c then it is a right-angled triangle.

Since the measured length is 8 m, which is not exactly equal to ≈ 8.06 m, the path is slightly off from the true corner-to-corner diagonal. 26 1379 mm 27 a 1178 m

b 14 minutes

8.02 Trigonometric ratios

17 35.78 cm. Since opposite sides of a rectangle are equal, then using Pythagoras’ theorem to ﬁnd each diagonal will gave us the same answer of 35.78 cm.

Understanding

18 The new side lengths will be a = 6, b = 8 and c = 10 units after the enlargement. a2 + b2 = c2 62 + 82 = 102 36 + 64 = 102 100 = 100

3 a H

b A

c tan

4 a True

b True

c True

So, Pythagoras’ theorem still holds for the enlarged triangle.

1 a Z 2 a ∠C

d c

b ∠A

d True

5 a Adjacent: BC

Opposite: AC

b Adjacent: BC

Opposite: AC

c Adjacent: YZ

Opposite: XZ

d Adjacent: YZ

Opposite: XZ

c 20 cm

Mathspace WA Curriculum – Year 9 mathspace.co

c X

Fluency

6 a 52 cm

764

b X

b 48 cm d

7 a Sine

b Cosine

e Sine

f Cosine

c Tangent d Tangent

14 a

c Yes, the value of the ratio for sin θ and cos α

8 a i

iii

b i

iii

c i

iii

d i

iii

e i

iii

f i

iii

g i

iii

h i

iii

9 a i x = 12

b i x=4

c i x = 15

d i x = 20

is the same, which is . d Yes, this is true for all right-angled triangles. This is because in any right-angled triangle, the opposite side in reference to one angle is the adjacent side in reference to the other angle, and the hypotenuse remains the same for both angles. So, sin θ = cos α for all right-angled triangles. 15 From the diagram we have the following relationships: sin θ =

, cos θ =

tan θ =

which is equal to tan θ

as required. Problem-solving 16

10 a

11 a i Sine

b i Tangent

c i Cosine

d i Sine

ii sin(25°) =

12 a i

b i

17 a ∠BCA

18 a ∠BCA

19 a

20 tan θ = tan θ = a tan θ = QS a tan θ = 2a tan θ = QR

Reasoning

21 a 19 : 21

13 The opposite side is doubled in length, the adjacent side is unchanged and the hypotenuse will increase to compensate but not by double. So the sine ratio ratio

QR = 2a tan θ

will increase, the cosine

will decrease and the tangent ratio will exactly double.

c 21 : 28.32 22

≤ tan x ≤

b 19 : 21 d 57 : 84.96 will ensure that the roof height

is between 3.5 and 5 m. And then using Pythagoras’ theorem, min d2 = 252 + 3.52 and max d2 = 252 + 52 This means d comes out between 25.25 and 25.49 m.

Answers mathspace.co

765

14 Ethan has only checked for the cosine ratio assuming that 10 cm is the adjacent side to the 30° angle. Instead Ethan can use the sine ratio with opposite side to ﬁnd that:

8.03 Side lengths Understanding 1 Adjacent and Opposite 2 a Yes

b No

c Yes

d Yes

3 a Use the tangent ratio: tan 45° = b Use the sine ratio: sin 30° = So, we end up with the hypotenuse being 20 cm.

c Use the cosine ratio: cos 60° = Fluency 4 a 0.990

b 0.883

c 9.514

d 0.342

e 1.072

f 0.629

g 0.588

h 0.087

5 a 8.019

b 11.911

c 4.254

d 4.096

e 2.669

f 1.641

g 1.751

h 0.016

6 a 0.135

b 0.106

c 0.183

d 0.093

e 4.877

f 0.313

g 0.083

h 0.208

This shows that it is possible to have a rightangled triangle with an angle of 30° and side lengths of 10 cm and 20 cm. 15 a i 0.98 and 0.98

ii 0.92 and 0.92

iii 0.71 and 0.71

iv 0.53 and 0.53

b Sine of an angle is equal to cosine of the complementary angle. c The angles are complementary angles. That is, θ + α = 90°.

7 a f = 16.308

b y = 5.032

c f = 5.129

d y = 6.359

Problem-solving

e x = 10.037

f x = 10.941

16 a a = 6.26

b b = 3.77

8 a g = 9.465

b h = 15.243

c b = 8.260

d x = 81.144

17 a AD = 5.20

b BD = 12.85 c AB = 18.05

e x = 15.475

f f = 25.812

c d=5

18 l = 20.6 m

9 AC = 15.7

19 a r = 11.3 cm

10 a = 12 cm

20 a i y = 90.24 ii w = 65.67

iii x = 24.6

b i y = 77.80 ii w = 26.74

iii x = 51.1

11 a h = 21.03

b f = 21.53

b 12.7 cm

21 29 m

12 d = 3303.31 m

8.04 Angles

Reasoning 13 Since we are given the angle, the opposite side, and the adjacent side, we will use the tangent ratio to solve for b.

Understanding 1 a AC

b BC

c BC

d AC

e ∠ ACB

Use the tangent ratio Substitute tan θ =

Opposite = b, Adjacent = 10 Multiply both sides by 10

iii

b i

iii

c i

iii

d i

iii

e i

iii

Evaluate using a calculator

This shows b = 6 m

766

2 a i

Mathspace WA Curriculum – Year 9 mathspace.co

f i

16 sin θ = cos θ when θ = 45°. When the sine and the

iii

3 Inverse trigonometric functions (sin−1, cos−1, tan−1) help you ﬁnd the unknown angles in a right-angled triangle when you know the side lengths. On the other hand, regular trigonometric functions (sin, cos, tan) are used to work out the ratios of the side lengths of a right-angled triangle when one of the angles is already known. Fluency 4 a 45°

b 64°

c 27°

d 16°

e 50°

f 83°

g 55°

h 45°

5 a 0°

b 0°

c 12°

d 60°

e 90°

f 45°

g 73°

h 41°

i 0°

j 60°

k 26°

l 66°

m 66°

n 54°

o 24°

6 a 59.46°

b 61.15°

c 83.63°

7 a x = 74°

b x = 69°

c x = 74°

d x = 74°

e θ = 36°

f x = 62°

g x = 67°

h y = 47°

i θ = 67°

j x = 67°

cosine are equal, This means that the opposite and adjacent sides must be equal, which occurs in a right isosceles triangle with angle θ = 45°.

17 If the sine ratio of an angle is greater than the cosine ratio of the same angle, it means that the side opposite the angle is longer than the side adjacent to the angle. This occurs when the angle is greater than 45°. Angles in a rightangled triangle (which aren’t the right angle) must be less than 90°. Therefore, the angle is in the range (45°, 90°).

8 θ = 54°47′

18 False. When the gradient is doubled, it does not mean that the angle between the line and the positive x-axis is also doubled. The relationship between the gradient and the angle is not linear. Instead, the gradient is related to the angle through the tangent function: m = tan θ. So, when the gradient is doubled, the angle between the line and the positive x-axis will not necessarily double, but the tangent of the angle will be doubled.

9 tan 33°

Problem-solving

10 a 0.2

b 0.6

c 1.2

d 0.4

e 0.8

f 1.8

11 a 27°

b 45°

c 63°

d 76°

12 a

b θ = 29.05°

19 x = 26.57°, y = 50.19°, z = 13.24° 20 x = 21.9°, y = 48.8°

13 θ = 47.73° 14 θ = 25.02°

21 a

22 a x = 29.74°

b y = 150.26°

23 59° 24 θ = 21.80° b θ = 33.02°

Reasoning

25 a QR = 9.75 cm

15 a A = 66.4°

8.05 Applications of trigonometry

b To create a right-angled triangle from the isosceles triangle given in the question, we must add a vertical line which divides the isosceles triangle into two halves (the line of symmetry). Then we know that the hypotenuse is 10 cm and the adjacent side to angle A is 4 cm. The cosine function is the most appropriate choice as it uses the hypotenuse and adjacent sides to an angle.

Understanding 1 α 2 a σ

b θ and σ

3 Tangent

Answers mathspace.co

767

17 a A diagram to represent the situation is shown.

4 a

1.36 m

44°

26° 48’

b Sine

The shorter side is opposite the angle made by the diagonal and the longer side, and the diagonal is the hypotenuse. The sine ratio is used when we have the opposite side and the hypotenuse.

Fluency 5 h = 11.15 6 d = 285 m

b y = 4.51

7 θ = 41.81°

18 Since the distance from the base of the ladder to the wall is given in centimetres, and the ladder is given in metres, we ﬁrst convert the distances to metres by dividing by 100:

8 θ = 43.79° 9 y = 58.02 10 a

2.79 m

2.25 m

190 cm = 1.9 m

42 cm = 0.42 m

86 cm = 0.86 m

Drawing a diagram helps organise the known information and determine which trigonometric ratio is needed. Using the ﬁrst length given, a diagram can be drawn as shown.

θ° b θ = 53.75° 11 a 74.85 m

x°

b 28.31°

3.46 m

12 a 7m 1.9 m

Finally, we can set up equations for the angle based on each length and evaluate:

24 m b θ = 16.26° 13 θ = 23.9

14 d = 814 m

Reasoning

The only distance which gives an angle between 10° and 20° is 86 cm, so the ladder is safest to climb when it is 86 cm from the wall.

16 Luca is correct, because the two sides given are the opposite side and the adjacent side.

19 Vern is correct. As the ship drifts, the angle decreases. The angle, x, is approximately 38°.

15 d = 1857.11

768

Mathspace WA Curriculum – Year 9 mathspace.co

Problem-solving

9 a b h = 41 101

20 a 73 500 m

21 a

10 a AW = 11.31 cm

b VW = 23.41 cm

11 a r = 5 m

b 10 m

12 a 124.85 cm2

b 56.83 cm

13 13 cm Reasoning

65° 20 m

1.6 m

b 44.49 m

AG2 = c2 + a2 + b2 and then

23 Angle SPR = 33.61° 24 36.7 m

8.06 Optional: Pythagoras in 3D

1 No, the Pythagoras’ theorem can only be applied to right-angled triangles. It states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. 2 a No, we only know BF, which has a given length c. From the given information, we can work out x. c Yes

16 a = 5, b = 2, c = 1. This is the only integer set that works. None of the lengths can be more than 5, since 62 = 36, and it is easy to show that no other set will work by trying them. Problem-solving 17 a 2.4 m 4.2 m

4 a HF = 19.21 cm

b DF = 22.65 cm

5 a EG =

b AG =

c DG = b XC = 13.60 cm

c DF = 15.26 cm c DG = 11.70 cm

hypotenuse.

3 x = 7.07, y = 8.66

b AG = 12.08 cm d 17.55 cm2

0.94 m

b 4.9 m 18 13 cm 19 151 mm 20 a

b 6.18 m

21 a To show that △PSR is a right-angled triangle, PR must be equal to HG since they are opposite sides. We can see that PS = 12 since PA = PS and RS = 9 since RC = RS.

8 a b c AF = 55.77

. We can ﬁnd the

length EN using the △EFN, EN would be the

Fluency

7 a EG = 11.40 cm

15 We would need to know the heights of the triangular faces, EM and EN. We can ﬁnd the length EM using the △EFM, EM would be the hypotenuse.

Understanding

6 a EC = 15.26 cm

, leading to

b θ = 36.68

22 a 63.3 m

b △EFH

14 Use △EHG, to calculate that EG is Then use △AGE to get

d AF = 72.46

Answers mathspace.co

769

7 a ∠ AED = 11.31°

b CE = 18.03

c BE = 18.28

d ∠BEC = 9.45°

e CX = 15.81

f ∠BXC = 10.74°

8 a x = 20

b y = 22.36

c θ = 27°

9 a i α = 90°

ii β = 90°

iii γ = 90°

b x = 1.41 cm

Since PR and HG are both 15 cm, △PSR is a right-angled triangle. 2

b A = 54 cm

c 7.2 cm

d V = 288 cm

c y = 1.73 cm d θ = 35° b 109 cm2

10 a

22 a 26.5 cm

b 40.0 cm

c θ = 16°

c 14.1 cm

d 61.9 cm

11 a BC = 82 m

b 48 m

23 h = 7

12 a z = 8.49 cm

b θ = 68.34°

8.07 Optional: Trigonometry in 3D

13 a MD = 5.66 cm

b x = 61.87°

c h = 10.58 cm

d MN = 4 cm

e PN = 11.31 cm

f 69.30°

Understanding 2

g 70.53°

1 L = 8 + 11 2

14 a 40.76 m

3 a

Reasoning

b 58.21 m

15 a The cross-section of the cone is an isosceles triangle which can be cut into two right-

y cm

angled triangles with angle 5 cm

x cm

at the top.

θ 2

13 cm

5 cm

b x = 7.07 c

7 cm

b θ = 65°

y cm

5 cm

16 AE = 26.0 m Problem-solving

7.07 cm

d y = 8.66

4 a i EG = 9.90 cm ii 35°

ii AG = 12.12 cm iii 90°

iv 55°

5 a z = 14.32 cm

b θ = 11.83°

6 a z = 21.47 cm

b 75.15 cm2

770

c 11°

d 16°

18 a i V A = 8.49 cm

ii V C = 8.49 cm

iii AC = 8.49 cm

Fluency

b i 45°

17 a 91.92 m b 37.14 m

Mathspace WA Curriculum – Year 9 mathspace.co

b ∠V CA = 60°. Since V A = V C = AC we know that △V AC is equilateral, so all angles are equal and the sum of the interior angles of a triangle is 180°, so ∠V CA = 19 15x

= 60°.

8.08 Optional: The unit circle Understanding 1 a Quadrant 4

b Quadrant 1

c Quadrant 2

d Quadrant 3

e Quadrant 1

f Quadrant 4

g Quadrant 2

h Quadrant 4

sin

cos

12 a i

b i

13 a b i 135°

ii 225°

iii 315°

d −sin 45°

b i 120°

ii 240°

iii 300°

d tan 60°

b i 150°

ii 210°

iii 330°

d −cos 30°

b −1

g −1

tan 14 a

30° 45°

60° 15 a 3 a −1.732

b 0.423

c −0.309

d −0.158

e 0.961

f −0.719

g −1

h 1

4 a False

b False

c True

d True

b x

c y

e True 5 a 1

e (cos θ, sin θ ) Fluency

16 a

6 a i

iii

b i

iii

c i

iii

17 a Quadrant 2

b Quadrant 2

iii

c Quadrant 2

d Quadrant 1

Reasoning

d i

7 a (0.77, 0.64)

b (−0.94, 0.34)

8 a Positive

b Negative

c Negative

d Positive

e Negative

f Negative

g Positive

h Positive

0° intersects the unit circle at (1, 0) therefore sin 0° = 0.

9 a 50°

b 80°

c 20°

d 65°

e 12°

f 42°

g 76°

h 71°

10 a sin 215° ≠ sin 35°

18 a The points are the coordinates where the angles 0°, 90°, 180° and 270° intersect the unit circle. The y-coordinate is the sine of the angle.

b cos 300° = cos 60°

c tan 275°  ≠ tan 85°

d sin 150° = sin 30°

e −cos 110° = cos 70°

f tan 145° = −tan 35°

11 a sin 74°

b −cos 31°

c −tan 56°

d −sin 60°

e cos 45°

f −tan 40°

g sin 6°

h −cos 30°

i −tan 65°

j −sin 40°

k −cos 69°

l tan 84°

90° intersects the unit circle at (0, 1) therefore sin 90° = 1. 180° intersects the unit circle at (−1, 0) therefore sin 180° = 0. 270° intersects the unit circle at (0, −1) therefore sin 270° = −1. b 0° intersects the unit circle at (1, 0) therefore cos 0° = 1. 90° intersects the unit circle at (0, 1) therefore cos 90° = 0.

Answers mathspace.co

771

180° intersects the unit circle at (−1, 0) therefore cos 180° = −1. 270° intersects the unit circle at (0, −1) therefore cos 270° = 0.

4 Pythagoras’ theorem can only be used to ﬁnd missing side lengths in right-angled triangles. Since the angles in a triangle sum to 180°, we can ﬁnd the missing angle by subtracting the other two angles from 180°.

19 W e can draw a vertical line from the point to the x-axis to form a right-angled triangle with one side cos θ and the other side sin θ. Since the point is on the unit circle, the hypotenuse of the triangle is 1. Using the Pythagoras’ theorem, we have x2 + y2 = 12, or cos2 θ + sin2 θ = 1.

180 − 28 − 63 = 180 − 91

20 A represents all ratios being positive in the ﬁrst quadrant.

6 1020 cm

S represents the sine ratio being positive in the second quadrant.

= 89

This show that the triangle does not have a right angle, so Pyhtagoras’ theorem cannot be used. 5 d = 1844 m

7 d = 21.3 8 a Cosine

b Sine

T represents the tangent ratio being positive in the third quadrant.

9 a i sin θ =

ii cos θ =

iii tan θ =

C represents the cosine ratio being positive in the fourth quadrant.

b i sin θ =

ii cos θ =

iii tan θ =

10 a i y = 21

ii tan θ =

b i y = 30

ii tan θ =

11 a Since sin θ =

, the hypotenuse has a

Problem-solving 21 a P (0.41, 0.91)

b (−0.41, 0.91) d −0.41

c 0.91 e −2.25

, and the opposite side from θ length of also has a length of 1.

22 a 23 a True

b True

c True

d True

b False

c True

d True

e False 24 a False

Using Pyhtaogras’ theorem, we ﬁnd the third side to have a length of 1.

e True 25 a i Q(−a, b)

ii R(−a, −b)

iii S(a, −b)

b i − tan 62° ii tan 62°

iii − tan 62°

c i − tan x

iii − tan x

ii tan x

26 a i 60°

b i 0°

ii 0

c i 60°

d i 0°

ii 0

This means the triangle is an isosceles right-angled triangle. b Any triangle with side lengths in the form x, x, . For example: 1

Chapter 8 review

1 a

2 a

3 a 15 cm

θ 4

b Perimeter = 36 cm

c Area = 54 cm2 2

772

Mathspace WA Curriculum – Year 9 mathspace.co

θ 4

b There is more than one way. One option is to make a right-angled triangle with CX as the hypotenuse and sides CF and FX. We know that CF = 15, but to find FX, we need to use

that X is the midpoint of EF, so FX = 2

12 a 0.64

b 5.14

c 0.97

d 6.78

e 3.09

f 0.85

g 2.23

h 0.33

13 a v = 21.87

b s = 8.15

c y = 74.67

d g = 32.91

e t = 1410.93

f h = 40.45

14 a 30°

b 45°

c 30°

d 40°

e 37°

f 36°

g 75°

h 55°

15 a z = 70° c m = 36°

b w = 50° d j = 28°

c There is more than one way. One option is to make a right-angled triangle with BE as the hypotenuse and sides AB and AE. We know that AB = 12, but to find AE, we need to set up another right-angled triangle, △ADE. We can then set up AE2 = AD2 + DE2 and solve for AE using AD = 4 and DE = 15. Once we have AE, we can set up BE2 = AE2 + AB and solve for BE using AB = 12 and .

17 θ = 49.40°

24 L = 17 cm

b x = 2.60

25 a x = 15

19 H = 43 m 20 Ezekiel is correct. As Alena moves closer, the angle increases. At the beginning of Alena’s hike, the angle from the ground to the peak is about tan−1

≈ 15°. After moving toward the

mountain, the angle is about tan−1

≈ 18°.

This shows the angle is increasing as Alena moves toward the mountain. 21 θ = 37.7° 22 a

c 23 a Since we know it is a triangular prism, the two triangle faces are parallel, so all of the other faces are rectangles. There is more than one way. One option is to make a right-angled triangle with CE as the hypotenuse and sides DC and DE. Opposite sides of rectangles are the same size, so even though they aren’t labelled, we know that DC = 12 and DE = 15. We can then set up CE2 = DC2 + DE2 and solve for CE.

= 6.

Now we can set up CX = CF + FX and solve for CX.

16 71.6°

18 a θ = 30°

b y = 17

c θ = 28° 26 a BC = 86 m

b 52 m

27 a AE = 25.6 m

b θ = 18

28 a

c 29 a Negative

b Positive

c Positive

d Negative

30 a −cos 50°

b sin 65°

c −tan 9°

d −sin 42° b

31 a

9.01 Area Understanding 1 a False

b True

d False

e True

2 a i Incorrect b i Correct

c True ii Exact ii Approximate

c i Incorrect

ii Approximate

d i Correct

ii Exact

Answers mathspace.co

773

3 a 300 cm c 270 000 m2

b 0.7 cm

Reasoning

d 5 ha

14 a The area of a triangle with base b and height h is half of the area of a rectangle with length and width b and h.

Fluency 4 a 36 cm2

b 145 mm2

c 144 cm2

d 54 m2

e 121 cm

f 1539 cm2

g 30 m2

h 420 cm2

i 90 m

j 81 cm

k 276 mm2

l 34 m2

m 76 mm2

n 75 mm2

Base

Height

Area

5 mm

2 mm

10 mm2

15 cm

7 cm

105 cm2

10 cm

5 cm

50 cm2

7.5 m

37.5 m2

h b

b The area of a parallelogram with base b and height h is the same as the area of a rectangle with length and width b and h.

c The area of a kite with diagonals of x and y is half of the area of a rectangle with length and width of x and y.

a=5m h=8m

b=9m

A = 56 m

7 h = 16 cm

d The area of a rhombus with diagonals of x and y is half of the area of a rectangle with length and width of x and y.

b = 18 cm

A = 288 cm2

8 a 20 cm2

b 3.3 cm2

c 27 cm2

d 80.0 cm2 y

9 130 mm2 10 a 169π cm2 c 121π cm2 11 a 283.5 cm2 c 804.2 cm2

c 28.5 cm

π m2

d 153.9 m2

15 No. Doubling the side lengths does not double the area. The current area is 54 cm2. Doubling the side lengths would result in an area of 216 cm2, which is more than double the area.

b 17 mm

16 Nadine is correct. The new radius will be .

d 3.16 mm

If we substitute this into the area formula it

b 452.4 mm2

becomes area.

774

b 484π mm2

12 314.16 cm2 13 a 4 cm

Mathspace WA Curriculum – Year 9 mathspace.co

. This is a quarter of the original

For example:

separately, using their respective formulas, and then add the two areas together.

r = 6 cm; A = 36π cm2 r=

A = 9π cm2 =

= 3 cm;

cm2

17 The circumference of a circle is C = 2 × π × r, so we can divide the circumference by 2 × π to ﬁnd the radius,

Now, we can put the radius into the area formula A = π r2 to get A = π × 82 = 64π. We can leave this as an exact value or round to get the approximate area. 18 a a + b

b h(a + b)

19 a 150 000 m 20 a

b 15 ha

Farm

Length (m)

Width (m)

Area (m2)

200

100

20 000

350

7350

100

2400

400

16 000

100

10 000

b Farm 1 and Farm 4 c Farm 3 and Farm 2

4 a x=7m

b y=8m

c 42 m

5 a 52 m2

b 24 m2

c 76 m2

6 a 78.3 cm

b 97.1 cm

c 208.6 cm

d 46.6 cm

7 a a = 6 cm

b b = 6 cm

c x = 8.49 cm 8 a 96 cm2

d 36.49 cm b 143 mm2

c 463 mm

d 180 cm2

e 194.85 units2

f 513 cm2

9 a 144 cm2

b 120 cm2

d 63 cm2

c 136 cm

d Farm 5

10 22 mm

e 11 150 m2 21 a 4 m2

11 a i One semicircle and one triangle

b 5 plants

ii 98.87 cm2

22 42 m2

b i One rectangle and four semicircles

23 a 84 cm2

ii 233.30 mm2

b 24 tiles

24 150.92 m2

12 a 636.2 cm2 2

25 1884.96 m2 26 14.1 cm2

Understanding b False

c True

d True

2 a Rectangle and triangle b To calculate the total area of a composite shape composed of a rectangle and a triangle, you would ﬁnd the area of the rectangle and the area of the triangle

b 276.3 m2

c 44.6 cm

d 70.7 cm2

13 a 40.8 cm2

b 242.9 cm2

c 242.9 cm2

9.02 Composite shapes

1 a True

b To calculate the total area of a composite shape composed of a square and a triangle, you would ﬁnd the area of the square and the area of the triangle separately, using their respective formulas, and then subtract the area of the triangle from the area of the square. Fluency

Problem-solving 2

3 a Square and triangle

d 17 398.5 mm2

14 a 245.46 m2

b 461.81 mm2

15 a 414 m2

b 318 m2

Reasoning 16 a A = 2w2 − s2

b No

17 There are two triangles with base 18 cm and height 8 cm, and one smaller triangle with base and height each 8 cm.

Answers mathspace.co

775

Remember that the area of a triangle is given by Area =

× base × height.

23 a Example answer: 50 m

3.2 m

Evaluate each product ﬁrst

30 m

Evaluate the addition

18 Joshua’s approach is correct. In calculating the area of composite shapes, it’s important to break down the composite shape into its simpler components, ﬁnd the area of each component and then add these areas together. The composite shape in this case is made up of a rectangle and a semicircle, so it’s appropriate to calculate the area of the rectangle and the semicircle separately, and then add these areas together. Joshua correctly calculates the area of the rectangle as 10 cm × 6 cm = 60 cm2 and the area of the semicircle as . Adding these together gives the total area of the composite shape as approximately 74.13 cm2. On the other hand, Lila’s approach is incorrect because she calculates the area of a full circle instead of a semicircle. As a result, she overestimates the total area of the composite shape. 19 Ava is correct in this situation. The problem describes a composite shape made up of a square and a quarter-circle. The diameter of the circle is equal to the side length of the square, not the radius. Hence, the radius of the quarter-circle is half the side length of the square. Therefore, Ava’s calculation of the areas of the square and quarter-circle are correct, while Max’s calculation incorrectly uses the side length of the square as the radius of the quarter-circle. Problem-solving 20 144.62 cm2 21 a 780 m

b 21.84 km

22 96 pieces of the 15 cm tile, 126 pieces of the 30 cm tile

776

Mathspace WA Curriculum – Year 9 mathspace.co

b 1461.49 m2 24 a 18 m2

c 365.37 kg b $864

25 $3134.26 26 a 28.5 m2

b 7.13 m2 d 35.37 m2

c $427.56 e $3537 27 a 4025 cm2 2

c 6507 cm

b 3867.9 cm2 d $1440

9.03 Surface area of prisms Understanding 1 a 6

b 5

c 6

d 8

2 a False

b True

c True

d False

3 C Fluency 4 a

6 cm 6 cm

b 216 cm2 5 a 54 cm2

b 864 cm2

c 222 m

d 180 cm2

e 232 m2

f 484 mm2

6 a 756 cm2

b 487 cm2

c 434 cm2

d 264 cm2

7 a 306 cm2

b 510 cm2

8 a d = 18.0 cm

b 492 cm2

9 708 cm2

22 22b2 cm2

10 21 108 cm2

23 a 52x2 m2

b 156x2 m2

11 152.62 m2

24 a 3240 m2

b 3175.2 m2

12 a x = 10 cm and surface area is 432 cm2

25 318.8 cm2

b x = 5 cm and surface area is 1220 cm 13 a 295.99 cm2

9.04 Volume of prisms

b 153.30 m2

Understanding

Reasoning 14 No, a cube and a rectangular prism with the same volume do not necessarily have the same surface area. This is because the surface area depends on the dimensions (length, width, height) of the prism, not just the volume. A cube has equal dimensions, while a rectangular prism can have different dimensions, leading to different surface areas.

1 a Square, 21 cm c Rectangle, 3 cm 2

Volume

15 John is correct. The surface area of the rectangular prism is 2(lw + lh + 2wh) = 2 × (8 × 5 + 8 × 3 + 3 × 8) = 158 square units. 16 Vlad is incorrect. Even though they have the same base area and the same height, the surface areas will be different because the faces of the prisms are different. 17

Substituting into the area of a trapezium Substituting into the area of a rectangle Factorising out the common term, l

1 mL

50 cm3

50 mL

750 cm3

750 mL

19 11 cm 20 14 m 21 474 cm2

1 m3

1 kL

1 m3

1000 L

1 m3

1 000 000 mL

3 a True

b False

c False

d True

Fluency 4 a 126 cm3

b 156 m3

5 a 336 cm3 3

b 2250 cm3

c 560 mm

d 240 cm3

e 52.5 m3

f 240 cm3

6 a 35 m

b 5824 mm3 3

c 63 cm

d 360 cm3

7 16 m3 8 120 000 cm3 9 h = 6 cm 10 y = 9 11 l = 30 cm

Sum of all 6 sides

12 a 450 mL

Substitute

13 80 L

Simplify

14 a 40 mL

b 0.45 L

b 87 880 000 mm3

c 2197 containers

Problem-solving 18 a 172 cm2

d Trapezium, 3 cm

Capacity

1 cm3

For example, a cube with sides of length 3 units has a volume of 3 × 3 × 3 = 27 cubic units, and a surface area of 6 × (3 × 3) = 54 square units. A rectangular prism with dimensions 1 unit by 1 unit by 27 units has the same volume (1 × 1 × 27 = 27 cubic units) but its surface area is 2 × (1 × 1 + 1 × 27 + 1 × 27) = 110 square units.

b Triangle, 8 cm

b 13.11 cm

15 b = 18 m Reasoning 16 No, it is not possible because the volume can be found by multiplying base by height.

Answers mathspace.co

777

17 Locker A contains a volume of 192 000 cm3 and Locker B contains a volume of 195 750 cm3. Stuart should choose locker B as it contains a larger volume. 18 If the length, width and height of a rectangular prism are all doubled, the volume will increase by a factor of 23 = 8, so the volume will be 8 times the original volume. 19 If two prisms have the same volume, it doesn’t necessarily mean they have the same base area or height. It is possible for one prism to have a larger base area but a shorter height, or vice versa. An example is a prism with base area of 6 cm2 and a height of 4 cm. Its volume will be 24 cm3, same with a prism with base area of 8 cm2 and a height of 3 cm. 20 Yes, it’s possible. The volume of a prism is a product of its base area and height. For example, consider Prism A with a base area of 10 square units and a height of 3 units. Its volume would be 10 × 3 = 30 cubic units. Now, consider Prism B with a larger base area of 15 square units but a shorter height of 2 units. Its volume would be 15 × 2 = 30 cubic units. Despite Prism B having a larger base area, its volume is equal to Prism A due to its shorter height. However, if the height of Prism B was even less (for instance, 1 unit), its volume would be 15 × 1 = 15 cubic units, which is less than the volume of Prism A. Thus, a prism can have a larger base area but a smaller volume if its height is signiﬁcantly shorter. Problem-solving 21 1.245 cm

23 784 cm3 24 4112.4 mm3 b 9

26 There are many possible answers. Prism A: 2 by 8 by 9 and Prism B: 4 by 3 by 6. Another possible solution is Prism A: 8 by 6 by 1 and Prism B: 2 by 3 by 4. 27 a 240 m2 28 a y = 80 − 2x

778

b 8.8 minutes

9.05 Surface area of cylinders Understanding b A = π x2

1 a C = 2π x 2 a True

b False

3 a Yes

c False b No

Fluency 4 a 219.91 m2

b 169.65 m2

5 872.9 m2 6 a i 75.40 cm2

ii 100.53 cm2

ii 131.95 m2

ii 301.59 m2

d i 226.19 m2

ii 353.43 m2

b i 75.40 m

c i 201.06 m 7 144π cm2 8 a 565.49 cm2 2

b 188.50 m2 d 15 896.46 cm2

c 314.16 cm

e 9424.78 cm

f 66 853.09 unit2

9 a radius = 15 m

b height = 92 m

10 height = 73 m 11 height = 51 cm Reasoning 12 Matea is incorrect. Doubling the height of a cylinder will not double its surface area. The area of the curved part of the cylinder will be doubled but the circular bases will not be doubled. The overall surface area will increase but not be doubled. For example, the surface area of a cylinder with radius 3 cm and a height of 5 cm is 48π cm2. While the cylinder with radius 3 cm and a height of 10 cm will have a surface area of 72π cm2.

22 486 L

25 a 7605 cm3

29 a 3.52 m3

b 4800 L

c 100 mm 2

b 80x − 2x3

Mathspace WA Curriculum – Year 9 mathspace.co

13 If the surface area is ﬁxed and you increase the radius of the cylinder, its height must decrease. This is because the surface area of a cylinder is directly proportional to both its radius and its height. So if one increases and the total surface area is to remain the same, the other must decrease.

14 Yes. By increasing the total amount of paper, you will increase the diameter which will affect the total surface area of the toilet paper roll, although the height will remain the same.

9 a 5780π cm3

b 1408π mm3

c 2873π m3

10 a 137.4 cm3

b 527.8 mm3

Problem-solving

c 150.8 m3

15 593.76 cm2

11 a 1837.83 m3

d 7.6 cm3 or 7634.1 mm3 b 829.38 cm3

d 7037.17 cm3 or 0.01 m3

c 1633.63 m

16 125.66 cm2 17 742.11 cm2

π km3

12 81π mm3 2

18 5.2756π m

19 a 326.73 cm2

b 353.43 cm2

Reasoning

20 452.6 m2

13 The volume of the smaller can is 1152π cm3 and the volume of the larger can is 9216π cm3. The volume of the bigger can, compared to the smaller one, is 8 times larger. Lyn is incorrect

21 101.8 m2

14 a 1130.97 cm3

c Vincent’s cylinder has larger surface area by 26.7 cm².

c 360π cm3

22 1376.02 cm2

d 1 130 973 mm3 3

e To go from cm to mm3, we multiply by 1000. This means that any rounding that we make can be ampliﬁed. Since in part (a) we only rounded to two decimal places, that means we lost one decimal place of accuracy as when we multiply by 1000, the decimal places “moves” three positions. There is a difference of 3 mm3 due to rounding. Converting from an exact value instead of a rounded value gives a more accurate volume reading for mm3.

23 138.98 m2 24 a TSA = 4π r2 + 6π rh b 4.95 cm 25 1.81 m2

9.06 Volume of cylinders Understanding 1 Circle 2 a

3 a 201.06 cm2

b 153.94 cm2

4 a 256π

b 231π

5 a True

b 1 130 970 mm3

b False

c False

d False

Fluency 6 a 452.4 cm3

b 5598.3 m3 3

d 69 115.0 cm3

15 Bob should buy the second can because it has a larger volume. The volume of a cylinder is given by the formula V = π r2 h. Even though the height of the second can is less than the ﬁrst, the increase in the radius (from the increase in diameter) has a greater effect on the volume because the radius is squared in the formula.

f 2 413.7 cm3 or 2 413 748.5 mm3

Problem-solving

d 19 088.3 mm3 or 19.1 cm3

c 5674.5 cm 7 a 350π mm3

b 896π cm3

c 3610π km

d 307.2π m3

8 a 294.5 cm3

b 10 995.6 cm3

c 402.1 m3 3

e 23 605.9 cm

Depending on the context this may or may not matter. In a scientiﬁc experiment, this could make a big different, especially if this inaccurate measurement is being used repeatedly. If this was a measuring cup for cake batter, this would not make a signﬁcant difference as this is less than 1 mL off.

16 2.8 cm

Answers mathspace.co

779

17 16 286.02 L

17 a 527.79 mm2

b 5183.63 cm2

18 275.08π cm3

18 a Radius = 12 m

b height = 42 m

19 2387.61 cm

19 a 417.83 cm

b 537.21 cm2

20 72π cm3

20 a 2309.1 mm3

b 5654.9 m3

21 a 29.69 cm3

21 a 981.7 mm3

b 175.9 cm3

b 169.2219 g costs $84.61

Chapter 9 review 1 a 96 m2

23 Radius= 2.19 cm b 560 cm2

c 150 mm

d 107.5 mm2

2 a 12.25π m2

b 38.48 m2

4 a 180 m2

b 24 m b 563.5 mm2

3 True

7 a 229 cm2

b 108 cm2

8 a 197.92 cm2

b 196.77 mm2

9 54 stones are needed. 900 ÷ 60 = 15 stones to cover the length of the pool, so 30 stones are need for both lengths. 600 ÷ 60 = 10 stones are needed to cover the width of the pool, so 20 stones are needed to cover both widths. However, this leaves the four corners without any stone, so we need to add 4 more stones. This gives a total of 54 stones. c 412 cm2 11 a x = 8 m

b 478.22 cm2

c 114 mm3

Fluency 4

6 7 a 5 times

8 a 12 times

9 2% 10 a 59%

b 2032 m2

b 41%

12 a

13 a 240 desserts

b 45%

14 a

15 a

16 a

17 a

d 1 c

b 112 m3 d 10 200 mm3

14 h = 7 15 39.375 mL 16 574.6 L

Mathspace WA Curriculum – Year 9 mathspace.co

18 a 102 times b i

780

b 138

5 a

d 148 m2

12 The original surface area is 2 (3x2 + 4x2 + 12x2) = 2 (19x2) = 38x2. The new dimensions would be 6x × 2x × 8x, so the new surface area is 2 (12x2 + 16x2 + 48x2) = 2 (76x2) = 152x2. This means the surface area increased by 152x2 − 38x2 = 114x2 square metres. 13 a 480 cm3

Understanding

2 In situations which involve experiments such as having a “loaded die” or historical events such as the winners of a contest.

b 24 pieces

5 a 30.28 m

10 a 304 cm2

10.01 Experimental probability

1 C

3 19 mm

6 a 187.5 mm

22 219.91 m3

iii

Problem-solving

19 a 50 times

20 a

31 a Summing the results over the 3 days results in 120 total spins and 32 total “Yes”. Thus, a possible spinner could look like this

21 a

22 a

YES

23 60 commuters b 58.90 cm2

24 57 times 25 a 48 000 people

b 480 people

32 a

26 a

b 160 tablets

33 Player A

27 a 400 deliveries

10.02 Venn diagrams and two-way tables

c Bicycle 28 a 71%

d 150 deliveries b 73%

c 69%

c 14 months

Understanding 1 a iv: A ∪ B, the union of events A and B

Reasoning

b i: Sample space

29 a 0.52

c iii: A ∩ B, the intersection of events A and B

b The coin seems like it might be fair, but more tosses should be done to check. c 0.504 d The more ﬂips that are performed the closer the coin’s probability gets to the theoretical probability. It is probably a fair coin. 30 a

b 72 people

d ii: A′, the complement of event A 2 a 6 plants

b 2 plants

c 5 plants

3 a True

b True

c False

d False

e False

4 a 13 students

b 25 students

c 30 students

c The probability and theoretical probability are not the same. The experimental Probability is greater than the Theoretical Probability.

Fluency 5 a

This difference might be due to chance or randomness in the experimental results. In the long run, with a large number of trials, the experimental probability should approach the theoretical probability more closely. However, in a relatively small number of trials (like the 19 tosses in this case), there can be variations that cause the experimental probability to differ from the theoretical probability.

6 a

7 a 28 players

b 1

c 8 players

8 Dogs

Walks regularly

Does not walk regularly

Total

Cats

Total

108

9 a 60

Answers mathspace.co

781

10 a

Month

Took off on time Delayed

May

135

Total

180

June

140

200

Total

275

105

380

b 380 ﬂights

c 30.0%

d 35.5%

16 a

Laptop owners Tablet owners

17 a

11 a

Likes smooth Likes crunchy Total peanut butter peanut butter Tea

105

Coffee

Total

180

Freelancer

Physical book

b 180 individuals

12 a

Reasoning

Enjoy

Don’t enjoy

Total

Teenagers Seniors

Total

18 Joanna is correct. 38 people speak French. Trent forgot to include the people who could speak French and Spanish. 19 a 6 people

b 13 a

Likes pizza

Doesn’t like pizza

Total

Likes sushi

Doesn’t Like sushi

b 45 people 14 a

Not vegetarian

782

Total 26

Adults

Total

15 a

Vegetarian

Children

c Goran is correct. 14 people play only 1 sport. Goran added the people that played both tennis and football, then took away the players that played both sports twice as people that played in both sports are counted in football and in tennis. 13 + 11 − 5 − 5 = 14 d The new probability will be lower than the probability calculated in part (b).

d Yes

b Yes, because more students joined the basketball club.

Borrowed Available

Total

Problem-solving 21 a 27 movies

Fiction

Non-ﬁction

Total

c 50%

Mathspace WA Curriculum – Year 9 mathspace.co

20 a Samuel added the number of students that joined the soccer and basketball clubs 75 + 78 = 153. This is 11 more than the total number of students that were asked to join. This means that 11 students are found in both groups.

d i

b 43 movies iii

c 15 movies iv

22 a 70 plants c 23 a

28 a

Restaurant Bowling alley

d Sketching

Pottery

29 25

Theme park

b 128 people b 12 people

c i

iii

24 a i

iii

d i

29 a

b Example answers:

i Not a primary student ii Not a primary student that catches the bus to school

iii

3 1

iii Either a senior student that does not catch a bus or a primary students that catches a bus 25 a i

iii

b c $45.80

b Example answers: i Not a non-musician

ii A musician who is not a child 1

iii A child musician 26 a 7 employees b

Leadership development

Project management

1 0

Probability that a plant at random has aphids but not mites is 30 a

c 27 a

b i

Rectangular

Not Total rectangular

Detected Detected Total Pesticides as Organic Fruit has pesticides

Emerald

140

Not emerald

Fruit is organic

Total

120

200

Total

b 60%

c 66.67%

iii

Answers mathspace.co

783

10.03 Arrays and simple tree diagrams

Understanding

1 a True

b False

c True

d False

2 a Equal probability

9 a

b Not equal probabilities c Equal probability 3 Sample space = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

1, A

1, B

1, C

2, A 2, B 2, C

3, A 3, B 3, C

b 9 Fluency

10 a

Cyan (C)

4 a P (9) =

C, B

P, B

W, B

Grey (G)

C, G

P, G

W, G

Red (R)

C, R

P, R

W, R

Yellow (Y )

C, Y

P, Y

W, Y

b 12

d P (7 or higher) =

b P (more than 6) =

c i

c P (6 or less) + P (more than 6) = 1

d (a) and (b) are complementary events, so they cover all possible outcomes and have a sum of 1.

12 a

iii

13 a

6 a 8

Red Red

b i

iii

Heads

H,1

H,2

H,3

H,4

H,5

H,6

Tails

T,1

T,2

T,3

T,4

T,5

T,6

b 12

Red Black Red

Black

Red

Red Black Red

Black Black b

Black Sally Peter

Alvin

8 a

b {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

5 a P (6 or less) =

784

11 a 8

e P (an even number) =

7 a

Pink (P ) White (W )

Black (B)

b P (less than 5) = c P (more than 6) =

Mathspace WA Curriculum – Year 9 mathspace.co

Sally

Alvin

Peter Alvin Peter Sally

Peter Sally Peter Alvin Sally Alvin

22 a The 12-sided and 20-sided dice

Reasoning 14 An array would be more efficient as there are 36 possible outcomes. Drawing up that many possibilities on a tree diagram would take a long time. However, filling in a 6 × 6 grid would not take as long. An array presents all possible outcomes at once, making it easy to see every combination. This is particularly beneficial when you need to calculate probabilities for specific events or sums of the dice. 15 The probability the next flip is a heads is . This is because each flip is independent of the others, and for any given flip, there are only two possible outcomes: heads or tails. Since there are no specific factors or conditions mentioned that would affect the flip, the probability is the same. 16 The probabilities are not the same. For both dice, there is only one way this can happen, so .

the probability is

b The 4-sided and 6-sided dice c Any combination of dice

10.04 With and without replacement Understanding 1 a With replacement

b Without replacement

c With replacement

d Without replacement

2 The sample space changes between experiments with and without replacement. With replacement, the sample space remains constant, while without replacement, it decreases after each draw. 3 a True

and

c False

Fluency 5 a B

These are different, so the chance of rolling a 6 and then a 2, is not the same for Frankie and Myah.

Problem Solving 17 a

18 a

19 a

v b i True 21 a e 0

ii vi

iii vii

iv viii

ii False

iii True

iv False

d 1

B B W G

Sample Space BB BB BW BG

B B W G

BB BB BW BG

B B W G

WB WB WW WG

B B W G

GB GB GW GG

Sample Space

20 a i

d True

4 {23, 24, 32, 34, 42, 43}

If we made an array for Myah, we could see that the sample space has 36 outcomes. While for Frankie the sample space has 36 outcomes. So the probabilities would be

b False

Answers mathspace.co

785

6 a

Red Blue Green Yellow Red Blue Green Yellow Red Blue Green Yellow Red Blue Green Yellow

Red

Blue

Green

Yellow

9 a

Sample Space 8 78

9 8

b {78, 79, 87, 89, 97, 98} c 10 a

Sample Space G G G B G G G B G G G B G G G B

Blue

Red

Green

Yellow Red

Blue

Green

Yellow Red

Green

Blue Yellow Red Blue

Yellow

b 11 a

Green

7 a Without replacement b {BG, BY, GB, GY, YB, YG} 8

Sarah

Olivia

David

Micheal

786

Olivia David Micheal Sarah

Olivia David

Mathspace WA Curriculum – Year 9 mathspace.co

1 1

c i

David Micheal Sarah Olivia Micheal Sarah

5 5 iii 0

iv 1

GG GG GG GB GG GG GG GB GG GG GG GB BG BG BG BB

12 a

Red

Blue

15 a

b 0

Red

16 a

Green

17 a {MO, MN, ME, MY, OM, ON, OE, OY, NM, NO, NE, NY, EM, EO, EN, EY, YM, YO, YN, YE}

Green Blue

Red Green

Blue

13 a

Sample Space 9

3 b i 0

9 5

69 65

5 6

35 36

iii

18 When picking items with replacement, the number of choices (branches) and the chance of each outcome stay the same throughout. Without replacement, the number of choices decreases for each level, and the chance of picking each item changes.

C C L L C C L L C C L L

Sample Space CC CC CL CL CC CC CL CL CC CC CL CL

C C C L C C C L

LC LC LC LL LC LC LC LL

For example, imagine a bowl with 2 balls (A and B). If you pick balls with replacement, there are always 2 choices at each level, and you can have 4 different results (AA, AB, BA, BB). But if you pick balls without replacement, there are 2 choices in the beginning, and then only 1 choice afterward. So, you can only have 2 results (AB, BA). 19 a The marbles are drawn with replacement. You can tell because after a marble is drawn in the tree diagram, it does appear as an option in the next level of branches. b The labels R1, R2, and R3 are used instead of just R to show that there are three different red marbles. Even though the marbles are drawn with replacement, using these labels helps us see that there are multiple red marbles in the bag, and each one can be drawn separately. 20 a Yes, getting YY is possible when you select two letters one after the other without putting them back from the word YUMMY. There are two Ys in the word. The chance of selecting YY is

b i

iii

Reasoning

ii 1

14 a

b i

because you can select one Y ﬁrst

and then the other Y next. In fractions, that’s .

Answers mathspace.co

787

b MY. First, we calculate the chance of getting UM. That’s

24 a

. Now, considering the

word YUMMY, we notice there are two Ys and two Ms. This gives us a bigger chance of ending up with MY. We can work out this

. If we compare

probability as

these probabilities, we see that getting MY is more likely than UM.

second selection remains . So the overall

Problem Solving b

21 a 22 a Answers may vary.

An event with a probability of 0 would be drawing two counters with the same number, for example, drawing both counters with the number 4, which is impossible as each number appears only once in the bag. b Answers may vary. An event with a probability of 1 would be drawing two counters with a sum greater than 8, as any pair of counters taken from the bag would have a sum larger than 8 (like 4 + 5 = 9 and the sums only increase from there). c Answers may vary. An event with a probability of less than 0.5 would be drawing a smaller number like 4 as the outcome. The possible draws for this outcome are (4, 5), (4, 6), (4, 7) and (4, 8), which has a probability of

3 3

would be , and since we put M back, the .

c If we select with replacement, the probability of getting MY increases. The ﬁrst selection

probability of getting MY would be

5 6

42 42

c i T here would be 20 × 20 = 400 ﬁlled cells, as every combination of the 20 cards can occur. ii There would be 20 × 19 = 380 ﬁlled cells, as each card can only be paired with different cards, and the array would be symmetrical along the diagonal (eliminating duplicate combinations). 25 a 1 ii

b i

10.05 Multi-stage probability Understanding 1 a True

b False

d False

e False

c True

2 a 3 a

1 6 H 1 2

1 6

or 0.2,

1 6

1 2

23 a Without replacement ii

5 15

considering all the 20 possible pairs of counters.

b i

4 12

iii

1 6

1 3 1 3

1 3

R B Y G R

B 1 Y 3 G

b No 4 a 9

788

Mathspace WA Curriculum – Year 9 mathspace.co

b 5

c 4

Fluency

9 a

5 a B

7 12

7 12 5 12

1 5

7 12

5 12

Hit

4 5

5 12

1 5

4 5

b b 0.6

7 a G

4 7

3 7

c 0.48 1 2

1 2 2 3

10 a

11 a

1 3

B 12 a

13 a

14 a

15 a

4 5

1 5

Hit 4 5 Miss

Miss

6 a 0.64

1 5

4 5

Hit 4 5 Miss

1 5

Hit Miss Hit Miss

1 5 1 5

4 5

Hit Miss Hit Miss

b 8 a B 1 9 1 3

3 8

5 8

1 8

B 2 8

5 8

1 8

5 9 R

B 3 8

4 8

16 a

17 a

e Reasoning

18 a b i iv vii

iii

v 0

b No, the probabilities would change because there would be fewer black cards, and fewer cards in total, for the next draw. 19 Making a selection without replacement means that the number of remaining items will decrease by one. If non-simpliﬁed fractions are used, the denominator of the fractions for the probabilities will decrease by one at the next stage. Depending on what type of item was selected, the numerator of the fraction may also decrease by one, or may stay the same.

Answers mathspace.co

789

20 They are both correct. The sum of the probability on any set of branches must be 1, so adding up the P(GB) + P(GG) is the same as doing P(Gardener large prize) × 1. We can see this if we do:

28 33% 29 a A x

P (Gardener large prize)

Garden 1−x

Multiply across, add down

Common factor of

Rose

2 5

Daisy

1 4

Rose

3 4

Daisy

The probability of drawing a ﬂower from Section A is represented by x, and from Section B is 1 − x.

Add fractions

21 There are 4 numbers on a die less than 5. The probability of rolling a number less than 5 on any die is .

b Section A area is 14.29 m2 while Section B area is 85.71 m2

10.06 Optional: Independent and dependent events

Probability=

3 5

Problem Solving

Understanding 1 a Independent

22 a

b Dependent

c Independent 23 a

2 Dependent

b i 38% 24 a 0.05

ii 9%

iii 28%

b 0.6

c 0.15

25 a

–1 2

H –1 2

–1 2 T

3 a Independent d 0.2

4 a

–1 8

b Drawing an orange ball, then a pink ball c • Drawing a pink ball, then an orange ball

3 – 8

–1 2

–1 8

• Drawing a pink ball, then a pink ball d Dependent 5 a Two days

3 – 8

26 a

Fluency

Makes 0.6

6 a

0.4

Misses

b 0.4 27 a

0.8

Makes

0.2

Misses

b Independent

7 a b

c 0.08 B

b No

c Rain on the ﬁrst day, sun on the second day (RS) and sun on the ﬁrst day, rain on the second day (SR).

b i

b Dependent

c Independent 8 a Independent b i 6 possibilities out of 36:

790

Mathspace WA Curriculum – Year 9 mathspace.co

(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6). The answer is

e i ii

ii 6 possibilities out of 36: (4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6).

15 a

The answer is b iii

v 0

c 9 0.5 10 Yes, events A and B are independent.

d e

11 a b

16 a

12 a i

17 a

b i

c i

18 a

d i

13 a Dependent b i

iii

c 8 possibilities out of 90: (2, 10), (3, 9), (4, 8), (5, 7), (7, 5), (8, 4), (9, 3), (10, 2). The answer is 19 a

20 a

v 14 a i ii

21 a

22 a

G G

b i ii

B G

c i B

GG GB

BG BB

d i b i

Answers mathspace.co

791

23 a {SR, SL, SC, GV, GM, GC} b

e Increase 31 a

Reasoning 24 a

d The draws are independent if the probability of each successive ball drawn is the same as the probability of the ﬁrst ball drawn. The draws are dependent if the probability of each successive ball drawn is not the same as the probability of the ﬁrst ball drawn.

Edge

C D

Probability

c No

Edge Probability

c i Higher probability by

In this case, it is dependent.

ii Higher probability by

25 3 of the cards are labelled 1, 2 or 3.

iii Lower probability by

3 of the cards are labelled 9, 10 or 11. Since the card is replaced, there are always 11 cards to choose from.

32 a 1st race 0.8 W 0.7

0.2

2nd race W

0.9

0.6

0.1

0.4

26 Independent would mean that P (A ∩ B) = P (A) × P (B) and since P (A) ≠ 0 and P (B) ≠ 0, it follows that P (A ∩ B) ≠ 0. For mutually exclusive events P (A ∩ B) = 0. 27 If the events A and B are independent, then P (A ∩ B) = P (A) × P (B). Here

0.4

0.3 L

0.6

0.4 0.1

0.9

b 0.504

c 0.678

33 a

3rd race W WWW L WWL W WLW L WLL W LWW L LWL W LLW L LLL

Tea Toast

Milo

Tea

Problem-solving

Cereal Milo

28 a

29 a

b 0.42

c 0.88

34 a

Recipient

Recipient Not

Recipient Recipient Not

Recipient

d The probability is getting higher.

Recipient Not

e 25 30 a

792

Not

Mathspace WA Curriculum – Year 9 mathspace.co

Not

Recipient Not

Not

Australian resident

Likes yeast spread

c 4

Chapter 10 review 1 a 10 times

Likes yeast spread

Does not like yeast spread

Total

c The experimental probability showed that rolling a 4 is twice as likely as expected with

Australian resident

theoretical probability, . We expected to see a four 10 times, but it appeared 20 times. The data so far suggests that the dice is not fair. It is also possible that this high number of 4s is random and the die is fair, if you did more trials you would be able to test this theory.

Not Australian resident

Total

7 a i P (1) = iii P (more than 1) =

b P (less than 2) + P (more than 1) =

2 a 600 b i 12%

ii 22.17%

3 a Adding the results of all of her friends. The relative frequencies are: 1

Relative frequency 43% 40%

These two events are complementary, because less than 2 includes 0 and 1, while more than 1 includes 2 and 3. This makes up the whole sample space, so would have probability of 100%. 8 a

This is close to: Theoretical probability 1 2 b

3 4 This means she may have labelled the 12 sides with ﬁve 1s, ﬁve 2s, one 3, and one 4. b 1249.2 mm2 4 a 18

b 100%

c 3

5 a

b i

ii P (less than 2) =

Has been Has not been Total out of state out of state Pet

No pet

Total

iii

9 a False. There are 10 outcomes less than or equal to 5 and 26 outcomes greater 5. Since these are not the same, their probabilities would be different, so are not equally likely. b True. We can only make an odd product by multiplying two odd numbers. An even product can result from the product of two evens or an even and an odd. There are 27 even products and 9 odd products. This means P (Even) = and P (Odd) = , so an even product is more likely. c True. The ﬁrst step would have 6 branches for the ﬁrst die. The second step would split each of the 6 branches into 6 more branches, making 6 × 6 = 36 branches.

Answers mathspace.co

793

10 a Without replacement b Let S represent the star sticker, U represent the unicorn sticker, and T represent the tractor sticker. {SUT, STU, UST, UTS, TSU, TUS} 11 a

14 a

Sample Space 5 35 6 36 8 38

3 6 8

3 5 8

63 65 68

3 5 6

83 85 86

c i 0

Hit

Miss Hit

1 6 Miss

1 6 1 6

5 6

Hit

1 6

53 56 58

b {35, 36, 38, 53, 56, 58, 63, 65, 68, 83, 85, 86}

1 6

Miss

Miss Hit

5 6

1 6

5 6

Miss Hit

5 6

Miss Hit Miss

15 a i

iii

12 a C

6 10

4 10 B

6 10

Sample Space C CC

4 10

6 10

4 10

b i

iii

16 Less than two 4s, means one or none.

Probability of not rolling a four is

Multiply

b 13 a B

2 10

6 10 3 10

3 11

2 11

b i

2 10

1 10

6 10 3 10

6 11

794

iii

2 10

5 10

B W

Sample Space BB BW

Mathspace WA Curriculum – Year 9 mathspace.co

To find P (one four), we can find the probability of the one being in each possible spot, first, second, third, fourth, or fifth and add them all together. Probability of one four is Multiply

Probability of one four is Multiply

This probability can be repeated 3 more times, for a four in the third, fourth and ﬁfth spots. The probability of each of these will also be So P (one four) =

3 Answers vary, but may include: Substitute Add Simplify

17 a

–1 5 4 – 5

A –1 3 2 – 3

• Use random sampling: Ensure that the sample is representative of the entire population by selecting participants randomly. • Design neutral survey questions: Avoid leading or loaded questions that suggest a particular answer. Questions should be worded neutrally to avoid influencing the respondent’s answers. • Ensure anonymity and confidentiality: Allow respondents to remain anonymous and assure them that their responses will be confidential.

D C

x B

2 A census involves collecting data from an entire population, while taking a sample means collecting data from a group or subset from within the population.

Fluency 1−x

b c $25.33

4 a Numerical

b Categorical

c Categorical

d Categorical

e Categorical

f Categorical

g Numerical

h Numerical

i Categorical

j Categorical

k Numerical

l Categorical

18 a i

b i

5 a Nominal

b Nominal

c i

c Ordinal

d Nominal

e Ordinal

f Ordinal

g Ordinal

h Nominal

6 a Discrete

b Continuous

d i e i

19 a

c Continuous

d Continuous

e Discrete

f Discrete

g Continuous 7 a Continuous

b Discrete numerical 20 a

c Categorical nominal d Categorical nominal e Discrete numerical f Continuous numerical

e Increase

g Continuous numerical

11.01 Types of data

8 a Fair

Understanding 1 a False

b True

c True

d False

b Fair

c Biased

d Biased

9 a Leading Question

b Loaded Question

c Leading Question

d Leading Question

Answers mathspace.co

795

10 a Not biased, because the sample size is small and the sample is not representative of the population. b Biased, because the group of people being asked are already users of public transport and so will beneﬁt if the government spends more money on the public transportation. c Biased, because the question is leading and contains negative information about shark nets.

16 To understand commuting habits, they should collect both categorical and numerical data. Categorical data might include types of transport used (bus, train, car, bicycle, etc.), while numerical data could include distance travelled, time taken, and frequency of travel.

b Question is unclear: how large is “large”?

17 The reliability of this study is questionable due to the small sample size and the sample not being representative of all students (it only includes students from a single school). To make a claim about all students, a larger, more diverse sample would be needed.

c This question is too personal. Individuals might not be comfortable discussing this openly.

Problem-solving

11 a Yes

b Yes

c No

12 a No

b Yes

c Yes

13 a Emotional language

Reasoning 14 a The question uses emotive or leading language by including the Prime Minister’s beliefs. Question should be: “Do you think taxes are too high, too low, or just right?” b The question asks more than one question and makes a false assumption that a person has one of the accounts. Question should be: “Do you have a Snapchat account?” and “Do you have a Twitter account?” c The question makes a false assumption that the person has a sibling and that the sibling can drive. Question should be: “Do you have a sibling? If yes, how often do they drive you to school?” d The question uses emotive or leading language as it suggests social media is time wasting. Question should be: “How much time do you spend on social media each day?” e The question makes a false assumption that a person uses shoe polish. Question should be: “Do you use shoe polish? If yes, what brand do you use most often?” 15 Collecting both categorical and numerical data in a survey about dietary habits can provide a more comprehensive understanding of the habits. Categorical data can help identify types

796

of food or meals preferred, while numerical data can indicate quantities or frequencies, providing insight into the extent of these habits.

Mathspace WA Curriculum – Year 9 mathspace.co

18 a Country b No, because the question uses emotive or leading language and suggests that country music is cool. It also asks two questions rather than one, as it’s possible to like both types of music (or neither). 19 a I would conduct the survey during lunch breaks in common areas like the cafeteria, ensuring a diverse sample by including students from different grades and activities. The questionnaire would list various after-school activities for students to select their favourites, providing insights into the most popular options. b The data collected would be categorical (nominal) as it involves the classiﬁcation of activities into categories. c To ensure that your data collection methods are unbiased and accurate, you could use stratiﬁed sampling by selecting a representative sample of students from each grade level, ensuring that the survey is anonymous, and encouraging honest responses. 20 a To study the relationship between hours of sleep and academic performance, you would need to collect numerical data on the number of hours of sleep per night for each student and their academic performance (e.g., exam scores, GPA).

b You could collect this data by using selfreport questionnaires for hours of sleep and obtaining academic performance data from the school records.

Score (x)

20 105

12 15

125

120

Total

Score (x)

Frequency ( f )

Understanding 1 A frequency table would help when dealing with a large data set. It would help to reduce the time taken to add all the scores as well as reduce the possibility of errors due to the amount of scores.

119

4046

244

4500

160

404

6080

143

547

5720

185

732

7770

137

869

6028

Total

869

Score (x)

Frequency ( f )

Cumulative frequency

150

1800

160

2400

170

1870

180

1800

190

3040

200

Total

78 b 1500

Cumulative frequency

Total

48 156

34 144

2800 13 710

c 250 and 280

Number of pets

Frequency

iii 1

iv 1.09

Frequency

Frequency ( f )

125

vi Score (x)

Cumulative frequency

5 a 8 6 a i

445

ii 3

Fluency 4 a

3 2 7

3 By counting the number of dots above a particular score

5 10

b This case might lead to bias because the survey is not truly representative of the entire city’s population.

2 Stem and leaf plots are best used for smaller sets of data that are integers or decimal numbers. For example, student test scores, ages of people, etc.

22 a

11.02 Data displays

Cumulative frequency

21 a The clothing company should collect data on customer preferences, purchasing habits, frequency of visits to their store or website, and average spending per visit. They could also collect demographic data such as age, gender, and location. b This data could be used to tailor marketing campaigns, optimise inventory and product selection, and create targeted promotions to improve sales.

Frequency ( f )

20 18 16 14 12 10 8 6 4 2 0

v 1

1 2 3 Number of pets

Answers mathspace.co

797

b i

Number of bedrooms

Number of students

ii 2

iii 4

iv 3.85

b The data is bi-modal has modes of 49 and 52 and a range of 5. The least frequent score is 50. The median is 49 and the mean is 49.625

v 4

11 a

Number of students

12 10

Number of websites visited daily(x) 0

Total

4 2 3 4 5 Number of bedrooms

Number of siblings

Number of students

b 3.36 12 a

Frequency 9

Number of students

v 2

8 6

b 20

13 a 201

2 0

7 a

iv 1.6

0 1 2 3 Number of siblings

Score Frequency

58 61

81 93

b 21

c 5 students

d 4 students

e 73.8

8 a 174

b 193

9 a 179 beats per minute b 178 beats per minute c 164 beats per minute 10 a

c 4

iii 1

Score

ii 3 vi

c i

Frequency Cumulative ( f ) frequency

c 17

d 10

b 10.58

c 9

e 27.84

14 a Percussion

b 23 people

c 3 people

d 6 people

Reasoning 15 A frequency table would help to see the most common number of hours spent on homework (mode). You could quickly calculate the mean and median with less chance of making an error. 16 When calculating the “average”, Alison found the mean, while Sarah found the median. Both answers are correct as measures of centre, however the average is usually taken to be the mean, so this would mean that Alison is correct. 17 a False, a dot plot only shows the results from one variable not two. b True c False, while the mode can be easily determined from the highest peak in the dot

798

Mathspace WA Curriculum – Year 9 mathspace.co

plot, the mean and median can’t be directly observed but can be estimated by looking at the general distribution of the data.

Total number Number of of paintings artists completed

d False, the range can be determined by identifying the maximum and minimum values on the dot plot. e False, dot plots can be useful for both small and large data sets. 18 • The range is the difference between the highest and lowest values. Since the new values (15, 15, 15) are within the existing range (10 to 28), the range will remain the same. • The mode is the value that appears most frequently. Before adding the new values, the mode was 11. After adding three occurrences of 15, 15 will become the new mode, as it now occurs more frequently than any other value. • The median is the middle value when the data is ordered. Adding three new values of 15 will shift the middle of the dataset, likely bringing the median closer to 15. The exact median will depend on how the numbers are arranged, but it will shift towards the middle of the new dataset.

Cumulative frequency 3

Total

107

b 5

c 4.28

23 1

11.03 Back-to-back stem and leaf Understanding 1 2 2 18, 34, 40 3 Answers vary, but may include: • Displaying distributions • Comparing central tendencies and variability • Drawing conclusions

Problem-solving 19 a 126 ng/L

b 17 places

d i 15

c 105 ng/L

ii 37.5%

20 a

Fluency History 7

Responses

8 17

b i 5

ii 17

iii 18

c 6 d The range will increase while the mode and median remain the same. The new range is 28, new mode is unchanged and is still 17 and new median is also unchanged at 18. 21 a 21 22 a 5

b 4

c 10

Math

Key: 5|5 = 55

Key: 5|7 = 57

Oliver

Valentina

Key: 4|1 = 14

Key: 1|2 = 12

Answers mathspace.co

799

6 a 59 7 a Alina

b 67 b Layla

8 a 70.3 kg

c Wildcats c Yes

c i No, as it will be their scores’ outlier if added. d Yes

b 62.9 kg

c • Under 30 years of age: 53, 56, 57, 61, 61, 63, 66, 68, 71, 73 Mean: 62.9 • 30 years and over: 62, 63, 63, 68, 69, 71, 72, 74, 78, 83 Mean: 70.3 Hence, the group “30 years and over” appears to be heavier because their mean weight is higher. 9 a The lowest score for University students is 7 books. b Both groups have the same median, 24 books.

ii Since there are 11 scores now, the new median will move at the 6th position which is 42. The new mean will be which makes the new Sri Lanka’s mean higher now than India’s mean (45.5). iii First median is 40.5 while the new median is 42 which is still less than India’s median. First mean is 43.8 while the new mean is 48.45 which is now higher than India’s mean. So the measure of centre that was least affected by adding the outlier is the median. 16 a 87

c Both groups 10 a Yes

b Dylan’s

c Neither

17 a $109

12 a Teachers c $42

c 73.86

b 50%

c 25%

b The probability of selecting a player with a score of greater than 70 from Team Amazing

c Neither

b $36

. The probability of selecting a player

with a score of greater than 70 from Team

d Teachers

Amazing is Reasoning 13 Visually, Class A clearly has higher test scores overall. The mean for Class A is approximately 78.1. The mean for Class B is approximately 58.8. The median for Class A is 80. The median for Class B is 60.5. When you compare the mean and median for both classes, we notice that the measures of centre are higher for Class A, than for Class B. 14 • Adal’s results: 14, 15, 17, 17, 18, 21, 24, 26, 27, 28 Mean: 20.7 • Dado’s results: 7, 10 15, 18, 21, 22, 23, 26, 29, 30 Mean: 20.1

The coach’s team, Team Amazing, has a higher probability of selecting a player with a score that is greater than 70. 19 a 50%

b 44%

c Summer challenge

11.04 Grouped histograms Understanding 1

Class

Frequency

30–39

40–49

Hence, Adal appear to be the best performer because his mean result is higher.

60–69

70–79

Problem Solving

80–89

b i 45.5

800

d 46 books

e 39 books

50–59

15 a i 45

d 79.13

18 a

11 a The lowest score for Hotel A is 3 desserts. b Hotel B

b 95

e 76.67

ii 40.5

2 a Ungrouped

b Grouped

ii 43.8

c Ungrouped

d Grouped

Mathspace WA Curriculum – Year 9 mathspace.co

3 a False

b False

c True

d True

Score (x) 10–19

20–29

30–39

40–49

Fluency 4 a

Class 1–9

Class centre

10–18

50–59

19–27

60–69

28–36

70–79

37–45

11 a Number of hours 25

5 30–39

6 a 9.3

b 1–5 b 1≤x<6 Frequency ( f )

10–19

20–29

30–39

40–49

50–59

Score (x)

Frequency ( f )

30–39

40–49

50–59

60–69

70–79

80–89

9 a 20–29, 30–39, 40–49, 50–59 and 60–69 b

Score (x)

Frequency ( f )

20–29

30–39

40–49

50–59

60–69

10 a 10–19, 20–29, 30–39, 40–49, 50–59, 60–69 and 70–79

b 5–9

spent on studying

15 10 5 0

0 5 10 15 20 25 Hours spent on studying

12 a Faulty machinery

Frequency

Score (x)

20 18 16 14 12 10 8 6 4 2 0

b 0

12 16 4 8 Number of faults

13 a 7

b 4

14 19 < height ≤ 28 15 a 140.9 cents c

b 11 days

7 6 5 Frequency

7 a 9.9

Frequency

b 23

8 a

Frequency ( f )

4 3 2 1 0

120.9 125.9 130.9 135.9 140.9 Price (in cents per litre)

d 140.9 < p ≤ 145.9 16 a 25 students

b 13 students

c 16%

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801

20 18 16 14 12 10 8 6 4 2 0

Problem-solving 21 a

20 18 16 14 12 10 8 6 4 2

Frequency

17 a

0 20 40 60 80 100 120 Price ($)

0 20 40 60 80 100 Commute time (minutes)

b 20–39 and 80–99 Reasoning b 60 mm ≤ x < 80 mm

18 a 35 c Yes

b i No 22 a

d No. The table shows ranges of lengths and not individual leaves. We do not know if any of the leaves in the range 0 ≤ x < 20 are less than 5 mm.

Frequency

19 a

20 18 16 14 12 10 8 6 4 2 0

Number of books read

Speed (km/h)

iii No

iv No

Frequency

110 ≤ x < 120

120 ≤ x < 130

130 ≤ x < 140

140 ≤ x < 150

150 ≤ x < 160

Total

b 51.6% 23 a

5 10 15 20 25 Number of books

b i Y es, the highest frequency is in the range 5 ≤ x < 10, indicating that the majority of readers fall in this category. ii Yes, the sum of frequencies in the ranges 0 ≤ x < 5 and 5 ≤ x < 10 is greater than the sum of frequencies in the ranges 10 ≤ x < 15 and 15 ≤ x < 20, indicating that most readers read less than 15 books. iii Yes, the frequency in the range 20 ≤ x < 25 is only 2, indicating that a small number of readers read 20 or more books. 20 a Salvia’s is more useful, as the class intervals are smaller so we can describe the shape of the data with more accuracy and identify trends in more detail. b No, Lee is not correct as the histogram shows that 51 patients lived between 21–30 years. In fact, by using Salvia’s histogram, we can see that 0 patients lived 30 years after being diagnosed.

802

ii Yes

Mathspace WA Curriculum – Year 9 mathspace.co

Battery life (minutes)

Frequency

400 ≤ x < 500

500 ≤ x < 600

600 ≤ x < 700

700 ≤ x < 800

800 ≤ x < 900

900 ≤ x < 1000

1000 ≤ x < 1100

1100 ≤ x < 1200

1200 ≤ x < 1300

Total

117

b 89.7%

c 9.4%

11.05 Optional: Frequency histograms and polygons Understanding 1 a Histogram c Column graph

b Column graph d Histogram

2 • Graph A/Table 2 • Graph B/Table 1 3 a The class centre (or midpoint of the top) of each bar/column in the histogram.

b The frequency (or height) of the corresponding bar/column in the histogram.

Class

Frequency

4≤x<6

6≤x<8

8 ≤ x < 10

10 ≤ x < 12

12 ≤ x < 14

Class

Frequency

20 ≤ x < 30

30 ≤ x < 40

Spread: The scores range from 22 to 48. The grouped data is spread across all four intervals, indicating a moderate spread. The majority of scores (5 + 4 = 9 out of 15, or 60%) fall between scores of 27 and 41 (not including 41).

40 ≤ x < 50

50 ≤ x < 60

8 50 ≤ x < 60

60 ≤ x < 70

70 ≤ x < 80

9 a 12.7

Class

Frequency

76 ≤ x < 84

84 ≤ x < 92

92 ≤ x < 100

100 ≤ x < 108

108 ≤ x < 116

116 ≤ x < 124

Reasoning 10 The lines are extended to meet the horizontal axis to show that the frequency is zero before the first class interval and after the last class interval. These points on the axis are typically placed at the class centre of hypothetical class intervals with zero frequency, one half-class-interval width before the first actual class centre and one half-class-interval width after the last actual class centre, effectively closing the polygon.

b 43 ≤ x < 45

6 a 42.6 Score

Class Centre

Frequency

20 ≤ x < 27

23.5

27 ≤ x < 34

30.5

34 ≤ x < 41

37.5

41 ≤ x < 48

44.5

Frequency (Number of students)

b 10 ≤ x < 15

c 10 ≤ x < 15

5 a 66.5

7 a

Mean: Estimated mean score is 34.2. d Shape: The distribution is unimodal, peaking at the 27 ≤ x < 34 class. It appears slightly skewed to the right (positively skewed), as there’s a tail extending towards the higher scores from the modal class, and the frequencies drop off more gradually on the right side of the peak.

Fluency 4 a

c Modal Class: 34 ≤ x < 41 (Frequency 5).

11 Frequency polygons are more suitable for comparing multiple distributions on the same axes because: • Overlapping polygons are easier to distinguish than overlapping histograms. The lines of polygons can cross without obscuring each other as much as solid bars would. • Polygons highlight the shape and trend of the distribution more clearly, making direct visual comparison of these aspects between the two datasets simpler. With histograms, the blockiness of the bars can sometimes make subtle differences in shape harder to discern when overlaid.

5 4 3 2 1 20

Science quiz score 30 40 50

While histograms are excellent for showing the frequency of individual groups, polygons excel at showing the overall shape and facilitating comparisons between datasets.

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803

12 The student’s conclusion is not necessarily correct. Here’s why: • In a frequency polygon for grouped data (as discussed in the lesson), each vertex (point) represents the class centre of a class interval and its corresponding frequency. • A high peak indicates that the class interval associated with that class centre has the highest frequency. It doesn’t mean a single, specific data value is the most frequent. The individual scores within that modal class could be varied. • For example, if a peak is at a class centre of 25 for an interval “20 − 30”, it means the group of scores from 20 up to (but not including) 30 is the most frequent, not necessarily that the score ‘25’ itself is the most frequent single score. The only way the student’s conclusion would be directly correct is if the class intervals had a width of 1, effectively making each class centre a unique score. However, frequency polygons are typically used for grouped data with wider intervals.

Justification: The spread of a frequency polygon (like a histogram) is an indicator of the dispersion or variability of the data. A narrower polygon implies lower variability (data points are close together), while a wider polygon implies higher variability (data points are more spread out). The height of the polygon indicates frequency for those class intervals. Problem-solving 14 a 11 15 a

b 13 Age

c 63

Frequency

28 ≤ Age < 30

30 ≤ Age < 32

32 ≤ Age < 34

34 ≤ Age < 36

36 ≤ Age < 38

38 ≤ Age < 40

b 82 mothers c 44 mothers d 26 mothers 16 a

Hours in Sun

Frequency

0 ≤ Hours < 10

10 ≤ Hours < 20

13 From the descriptions:

20 ≤ Hours < 30

• Polygon X (tall and narrow): This shape suggests that the daily screen times for this group are more consistent and clustered around the peak (2−3 hours). The narrowness indicates less variability; most individuals in this group have screen times relatively close to the average or most common value.

30 ≤ Hours < 40

40 ≤ Hours < 50

50 ≤ Hours < 60

60 ≤ Hours < 70

70 ≤ Hours < 80

80 ≤ Hours < 90

• Polygon Y (shorter and wider, more spread out):This shape suggests greater variability in daily screen times for this group. While the peak might be similar, the wider spread and significant frequencies extending to 6−7 hours indicate that individuals in this group have a broader range of screen times. Inference on high screen times: The group represented by Polygon Y is more likely to have individuals with very high screen times. The “wider spread” and “significant frequencies extending up to 6−7 hours” explicitly indicate that higher values are more common in this group compared to Polygon X, which is described as narrow.

804

Mathspace WA Curriculum – Year 9 mathspace.co

b 45 people c 17 a

Time on hold in minutes

Frequency

0.5 ≤ Time < 1.5

1.5 ≤ Time < 2.5

2.5 ≤ Time < 3.5

3.5 ≤ Time < 4.5

4.5 ≤ Time < 5.5

b 40 calls

c 17 people

d 34 people

Number of students

18 16

9 a No

b 26 and 36

10 a Yes, 30s–40s

b 48

c Positively skewed

11 a Multimodal but not bimodal

b 4%–5% and 8%–9%

4 0

0.5

1.5

2.5 3.5 4.5 Number of hours

19 a 10

5.5

b 18

12 The data is clustered around 80–89 and around 100–109. 13 a Yes

b 1 to 4

c 3

d Positively skewed

14 a Yes

11.06 Shape of data

c 9 Understanding

d Negatively skewed

15 a Yes, 45–49

b 47

c Negatively skewed

1 Possible answers: • The data is evenly distributed around the mean. It means that the left side of the distribution (less than the mean) mirrors the right side (greater than the mean).

16 a Grouped data

• The left side of the distribution (less than the mean) mirrors the right side (greater than the mean).

18 a No

• The mean, median and mode are very similar. 2 Positive skew refers to a distribution where the tail is on the right side of the distribution. Negative skew refers to a distribution where the tail is on the left side of the distribution. 3 Clusters within a data set can be identiﬁed by looking at areas where data points are concentrated. These are typically groups of data points that are relatively close together compared to the rest of the data set. 4 Bimodal 5 Greater than 60 Fluency 6 a Negatively skewed

b Positively skewed

c Symmetrical

d Positively skewed

e Positively skewed

f Negatively skewed

g Positively skewed

h Symmetrical

7 a Unimodal

b Bimodal

c Unimodal

d Unimodal

8 a i 21 years

ii 11 years

iii 30 years

iv 23.32 years

b Positively skewed

b 9–10

b 151.5 cm

c Negatively skewed 17 Positively skewed b 15 and 21

c Symmetrical

Reasoning 19 In a positively skewed distribution, the mean is typically greater than the median, which is greater than the mode. This is because the mean is affected by the long tail of high values, pulling it further to the right. 20 C lusters in the age data collected across the 5 suburbs might form due to common characteristics or attributes shared by people within each suburb. For example, one suburb might have a cluster of younger people due to the presence of schools and playgrounds, while another suburb might have a cluster of seniors due to retirement communities or more affordable housing for older adults. Problem-solving 21 The heavy right skew in the sales data suggests that most salespeople have low sales numbers, while a few have very high sales numbers. 22 The teacher might consider that the symmetrical distribution indicates a more consistent performance across students, while the positively skewed distribution suggests that most students scored lower, with a few high achievers. The teacher should also consider

Answers mathspace.co

805

using measures of centre and spread to compare the two classes more accurately.

13 a i 36

Understanding 1 We can compare data sets using: • Measures of central tendency (mean, median and mode) • Measure of spread (range) • Shape of the data 2 Yes 3 Graph 1

14 a i 34

ii 38

b i 27.7

ii 30.1

c Adelaide is warmer if we use the average temperature to compare the two cities. This can be supported by the answer in part b. 15 a Yes

b Learning Tree

c Yes

16 a 4

b 3

c 5

d 6

c 28

d 27

e Team X f Team Y

4 Answers may vary.

17 a 77.69

Here are some of the reasons: • Reveals patterns: Graphs show relationships between variables, enhancing comparisons. • Simplify complex data: Graphs break down large data sets into a more understandable format, aiding comparisons. • Quick processing: Graphs help visualise data trends quickly, making comparisons easier. Fluency 5 a 21 minutes

b 25.7 minutes

c Primary students 6 a i False

ii False

b Yes

c Yes

7 a 7

iii True b 17

c i 3.4

ii 3.2

8 a i 86.25

ii 78

Reasoning 18 Looking at the data, although the ranges are identical (both are 7 kg). The mean for Group 1 is 58.3 kg. The mean for Group 2 is 63.4 kg. The mean values are quite different. Group 2 has a noticeably higher average weight than Group 1. This difference in mean weights might suggest that these two groups of turtles are not of the same species. However, it’s important to note that the determination of species is based on multiple factors, not just weight. Additional data and biological information would be needed to conclusively determine species. 19 a

d 72, which was obtained by Bob.

10 a i 72

ii 81

b City B c 9 cm

Class 2

8.2

Mean

ii 73

12 a 10.3 cm b 10.1 cm

Mode

21 a

b Class B 11 a i 120

Median

2.7

Range

20 Company Alpha, because the mean is higher than the median, which means the higher salaries must be very large.

9 a Team X ii 12.1

Mean Class 1

b Class 2, because their mean, median and mode were higher.

c 90, which was obtained by Alice.

b i 12.3

b 75.69

e Group 1

b 82.125

806

ii 43.8

c Team A is quicker if we use the average running time to compare the two teams. This can be supported by the answer in part b.

11.07 Compare data sets

e Plant A

ii 30

b i 45.5

d 3 cm

f Plant B

Mathspace WA Curriculum – Year 9 mathspace.co

Department A

Department B

35.71

51.43

Median

Mode

35, 40

45, 50, 55

Range

b Department B seems to have older employees as the mean, median and mode ages are all higher for Department B. Problem-solving 22 a Positively skewed

b Location A

c Location B 23 a i 3

ii 2.5

b Product B 24 a Positively skewed

c Product A b Barry’s Burgers

c Frank’s Fries

d Range

e Frank’s Fries

f Median

25 a Person A

b Range

c Person A

d Mean

26 a Symmetrical c 13.2

3 a Answers may vary. One possible answer: The clinic should note down how long (in days) it takes for patients to feel better using the new method compared to the old ways. They might also ask patients to rate how they felt during the treatment. b Answers may vary. One possible answer: The clinic could test the new method with a mix of different patients, make sure they compare it properly to the old methods, and double-check all the details they write down for mistakes. It’s also important to see if anything else might be making the patients feel better. 4 a

Goals scored

Frequency

b 20.17%

d 23.8

e Pens because the daily mean is higher. 27 a Team X lowest: 6 Team X highest: 10 Team Y highest: 10 Team Y: 8.4 c Team X: 7.5 Team Y: 8

= 0.8

The mode represents the most frequent score. On the dot plot, it appears as the column with the most dots or the highest peak. 5 a

Chapter 11 review 1 a Continuous numerical b Discrete numerical

Score (x)

Frequency ( f)

Cumulative frequency (cf )

20–24

25–29

30–34

35–39

40–44

45–49

50–54

b 96

c Discrete numerical c Yes

c Positively skewed

d 34

d Continuous numerical b No

The median is the middle value, so on the dot plot, it would be where the number of dots on the left and right are equal.

f T he mean and median is consistently higher for Team Y. The probablity that a randomly selected player will have made 8 or more successful free throws, is signiﬁcantly higher for Team Y. Team Y has a more consistent performance than Team X. Team X has a slightly more diverse range of successful free throws for each player.

2 a No

c The mean represents the average number of goals scored and is typically where the dots in the dot plot are most densely clustered.

b Team X: 7.7

= 0.5

b Mean: 2.62 goals, Median: 3 goals, Mode: 1 and 4 goals, Range: 5 goals

Team Y lowest: 7

d No

6 a Ace Rockets: 45.2 minutes Thunders: 43 minutes

Answers mathspace.co

807

Ace Rockets 7 9

Key: 1|6 = 61

8 a

7 a Yes

13 a

Thunders 4

b Sirin’s

130 ≤ x < 140

140 ≤ x < 150

150 ≤ x < 160

Total:

Class

Frequency

20 ≤ x < 30

20 ≤ x < 40

130 ≤ s < 140

40 ≤ x < 50

140 ≤ s < 150

50 ≤ x < 60

150 ≤ s < 160

60 ≤ x < 70

70 ≤ x < 80

Total: b 51.6% Number of Calls

4 12

14 a

Frequency

120 ≤ s < 130

10 9 8 7 6 5 4 3 2 1 0

b 196 minutes 10 a 14.2

Class

Frequency

76 ≤ x < 84

84 ≤ x < 92

92 ≤ x < 100

100 ≤ x < 108

108 ≤ x < 1160

116 ≤ x < 124

Score

Frequency

15 a b 5–10

11 a 0–9, 10–19, 20–39, 40–59, 60–79

12 a

110 ≤ x < 120

b 51.6%

c Neither

110 ≤ s < 120

Frequency

120 ≤ x < 130

Key: 6|2 = 62

Speed (km/h)

9 a

Speed (km/h)

28 ≤ Age < 30

30 ≤ Age < 32

32 ≤ Age < 34

24 ≤ Age < 36

36 ≤ Age < 38

10–19

38 ≤ Age < 40

20–39

40–59

60–79

Score (x)

Frequency (f)

0–9

b 82 mothers

16 a

c 44 mothers

d 26 mothers Novel genres

14 15 Frequency

Frequency

12 10 8 6

10 5

4 2 0

b 7–9

808

0 1-3 4-6 7-9 10-12 Number of defects

c 1

Mathspace WA Curriculum – Year 9 mathspace.co

0 5 10 15 20 25 30 35 Price

b Symmetrical

17 Symmetrical and bimodal 18 a No

b Yes, 1–4

c 3

d Positive skew/Right skew

19 a Section 1

b Section 2

c They have the same range. d The mean is equal to the median. 20 a 24.47 years

b 35 years

c Positively skewed

23 The symmetric distribution for Labradors suggests consistent sales throughout the year, indicating steady popularity. The positive skew for Beagles suggests that they were less commonly sold, with occasional periods (or perhaps promotions) where they sold in higher numbers. 24 a Victoria: 3935.6, South Australia: 1296.8 b Victoria: 204 651.2, South Australia: 67 433.6 c Yes

21 a 115 b i 30

ii 37

c i 65.6

ii 96.1

d 211 533.75 units

d The range of pulse rates and the mean pulse rate increase after exercise. 22 a

35 30 Frequency

25 20 15 10 5 0

400 500 600 700 800 900 1000 1100 1200 1300 Battery Life (mins)

b 89.7%

c 9.4%

Answers mathspace.co

809