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IL FOUNDATION SERIES

MATHEMATICS

A Reliable Companion for JEE | NEET | Olympiads

IL Foundation Series - Mathematics Class 9 Legal Disclaimer This book is intended for educational purposes only. The information contained herein is provided on an “as-is” and “as-available” basis without any representations or warranties, express or implied. The authors (including any affiliated organizations) and publishers make no representations or warranties in relation to the accuracy, completeness, or suitability of the information contained in this book for any purpose. The authors (including any affiliated organizations) and publishers of the book have made reasonable efforts to ensure the accuracy and completeness of the content and information contained in this book. However, the authors (including any affiliated organizations) and publishers make no warranties or representations regarding the accuracy, completeness, or suitability for any purpose of the information contained in this book, including without limitation, any implied warranties of merchantability and fitness for a particular purpose, and non-infringement. The authors (including any affiliated organizations) and publishers disclaim any liability or responsibility for any errors, omissions, or inaccuracies in the content or information provided in this book. This book does not constitute legal, professional, or academic advice, and readers are encouraged to seek appropriate professional and academic advice before making any decisions based on the information contained in this book. The authors (including any affiliated organizations) and publishers disclaim any liability or responsibility for any decisions made based on the information provided in this book. The authors (including any affiliated organizations) and publishers disclaim any and all liability, loss, or risk incurred as a consequence, directly or indirectly, of the use and/or application of any of the contents or information contained in this book. The inclusion of any references or links to external sources does not imply endorsement or validation by the authors (including any affiliated organizations) and publishers of the same. All trademarks, service marks, trade names, and product names mentioned in this book are the property of their respective owners and are used for identification purposes only. No part of this publication may be reproduced, stored, or transmitted in any form or by any means, including without limitation, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the authors (including any affiliated organizations) and publishers. The authors (including any affiliated organizations) and publishers shall make commercially reasonable efforts to rectify any errors or omissions in the future editions of the book that may be brought to their notice from time to time. Subject to Hyderabad jurisdiction only. Copyright © 2025 Rankguru Technology Solutions Private Limited. All rights reserved. ISBN 978-81-985385-2-9 Second Edition

Contents Module 1 1.

Number Systems

2. Polynomials

3. Coordinate Geometry

4. Linear Equations in Two Variables

5. Euclid’s Geometry

132

6. Lines and Angles

146

7. Triangles

183

8. Quadrilaterals

222

1.1

NUMBER SYSTEMS

RECALL RATIONAL NUMBERS

p A number, which is written in the form of , where 'p' and 'q' are integers and q ≠ 0 is called a q rational number. The set of rational numbers is denoted by Q. A rational number may be positive, zero, or negative. Example:

1 2 −2 0 −5 are rational numbers. , , , , and 2 2 3 1 11

Note: Every rational number, when expressed in decimal form, is either a terminating decimal or a non-terminating repeating (or recurring) decimal, and its converse is also true. Example:

1 = 0.5 (Terminating decimal) 2

15 = 5 (Terminating decimal) 3 1 = 0.333..(Non-terminating repeating decimal )= 0.3 3 1.1.1 Decimal representation of a rational number Here are some basic rules related to the representation of a rational number: Rule 1: A rational number has a terminating decimal representation if its denominator in its standard form has 2, 5, or both as factors and has no other factors. Rule 2: If the denominator in standard form has some factor other than 2 and 5, then the rational number is a repeating decimal or non-terminating recurring decimal. Rule 3: Between any two different rational numbers, a and b, there exists another rational number a+b a+b . That is, a < < b. 2 2

1.2

IRRATIONAL NUMBERS

p A number s is called an irrational number if it cannot be written in the form , where p and q are q integers and q ≠ 0 . A number whose decimal expansion is non-terminating and non-recurring is called an irrational number. 1

NUMBER SYSTEMS

Example:

2, 3, 5, 4 3 are irrational numbers.

Some more examples of irrational numbers: Type 1: i) Clearly, 0.010010001... is a non-terminating and non-repeating decimal and, therefore, it is an irrational number. ii) 1.41412413414... and 1.42442444244442... are irrational numbers. Type 2: If 'm' is a positive integer, which is not a perfect square, then m is an irrational number. Thus, 2, 3, 5, 7, 8, 10, 11 etc. are all irrational numbers. If 'm' is a positive integer, which is not a perfect cube, then 3 m is an irrational number. Thus, 3 2, 3 3, 3 4, 3 5, 3 7, 3 9 etc. are all irrational numbers. Type 3: π is an irrational number, 'e' is an irrational number. π and e are called transcendental numbers. Properties of irrational numbers: Irrational numbers satisfy the commutative, associative, and distributive laws under addition and multiplication, multiplication is distributive over addition. Note: 1) The sum of two irrational numbers need not be an irrational number. Example: Each one of (2 + 3) and (4 − 3) is irrational. But (2 + 3) + (4 − 3) = 6, which is rational. 2) The difference between two irrational numbers need not be an irrational number. Example: Each one of (5 + 2) and (3 + 2) is irrational. But, (5 + 2) - (3 + 2) = 2, which is rational. 3) The product of two irrational numbers need not be an irrational number. Example: 3 is an irrational. But

3× 3 = 3, which is rational.

4) The quotient of two irrational numbers need not be an irrational number. Example: Each one of 2 3 and 5) If a is a rational number and are irrational numbers.

3 is an irrational. But

2 3 = 2, which is rational. 3

b is an irrational, then each of (a + b ), (a − b ), a b and

3 Example: Each one of (4 + 3), (8 − 5), 5 3 , and is irrational. 2

a b

IL Foundation Series Class 9

Example: 'Product of two irrational numbers is always an irrational number'. Negate the statement by giving a counter-example. Solution: Take 2 − 5, and 2 + 5 . Both are irrational. Their product = (2 − 5)(2 + 5) =22 − ( 5)2 =4 − 5 =−1 , which is a rational number.

1.3

DECIMAL REPRESENTATION OF REAL NUMBERS

All rational numbers can be converted into decimal numbers. There are two types of decimals. Terminating decimals: Terminating decimal is the decimal number in which there is an end digit. There is a finite number of digits after the decimal point. 13 7 7 Example: = 0.104, = 0.28, = 0.875 125 25 8 These are examples of terminating decimals. Non-terminating repeating decimal: In a non-terminating and repeating decimal, a single digit or a block of digits repeats itself infinitely after the decimal point. 2 1 Examples: = 0.666 = … 0.6,= 0.0909 = … 0.09 3 11 These are examples of non-terminating repeating decimals. Note: 1) If the denominator of a fraction, in its simplest form, contains either 2 or 5 as prime factors, only then the fraction can be written as a terminating decimal. 2) A fraction can be written as a terminating decimal if the denominator can be written in the form of 10n, since 2 and 5 are the only prime factors of 10. Example: Find whether the following rational number is a terminating or non-terminating decimal. i)

−5

4 5

ii)

1 3

Solution: 4 −29 −58 (i) −5 = = = −5.8 5 5 10 Thus, the above rational number represents a terminating decimal.

NUMBER SYSTEMS

1 (ii) = 0.3333… 3 Thus, the above rational number represents a non-terminating recurring decimal.

1.4

DENSITY PROPERTY AND REAL NUMBER LINE

1.4.1 Density property If x and y are any two rational numbers such that x < y, then x < x+y lies between x and y. 2 1 1 Example: Let and be two rational numbers. 2 3 1 1 1 1 3+ 2  1 5 5 Then, ×  +  =  = × = 2  2 3  2  6  2 6 12

x+y < y or the rational number 2

1 1× 6 6 1 1× 4 4 We can write and = = = = 2 2 × 6 12 3 3 × 4 12 Now, clearly Again

5 1 4 5 6  1 5 1 1 lies between and  < < ∴ < < i.e 12 2 3 12 12 12  3 12 2

1  1 5  1  6 + 5  1 11 11  +  = × = × = 2  2 12  2  12  2 12 24

1 1 and and so on. 2 3 Containing the above process, we can find an infinite number of rational numbers between two given rational numbers. This property is called the density property of rational numbers. Which lies between

1.4.2 Representation of rational number on the number line: Procedure to represent rational numbers on the number line: • Draw a line and locate the point 'O'. This point is known as the origin. O

• If the given number is positive, mark it on the right side of the origin. If it is a negative number, mark it on the left side of zero. 4 • Divide each unit into values equal to the fraction's denominator. For example, to represent on 5 the number line, you need to divide each unit into 5 subunits.

IL Foundation Series Class 9

Example: Represent Solution:

2 on a number line. 3

2 is a positive rational number, and it is known that 2 is less than 1 and greater than 0. 3 3 2 Therefore, lies between 0 and 1 on the number line. 3 Here, the denominator is 3, so we will divide each unit length into 3 subunits between 0 and 1. 2 3 0 0 3

1 1 3

2 3

3 3

1.4.3 Representation of decimals on the number line: We know that natural numbers, whole numbers, and integers can be represented on the number line. Similarly, decimal expansion can be represented on the number line. Example: Represent 3.765 on the number line. We know that 3.765 lies between 3 and 4. The distance between 3 and 4 is divided into 10 equal parts. Then, the first mark to the right of 3 will represent 3.1 and the second 3.2 and so on. Now, 3.765 lies between 3.7 and 3.8. Next, we divide the distance between 3.7 and 3.8 into 10 equal parts. 3.76 will be on the right of 3.7 at the sixth mark, 3.77 will be on the right of 3.7 at the 7th mark, and 3.765 will be between 3.76 and 3.77 and so on. 3

3.70

3.760

3.1

3.2

3.3

3.4

3.71

3.72

3.73

3.74

3.763

3.764

3.761

3.762

3.5

3.6

3.7

3.8

3.9

3.75 3.76

3.77

3.78

3.79

3.765

3.767

3.766

3.768

3.80

3.770 3.769

To mark 3.765, we have to use a magnifying glass.

NUMBER SYSTEMS

1.4.4 Representation of irrational numbers on the number line: Method 1: Consider the number line, mark a point O on it, and let it represent zero. Let A represent a point on the number line such that OA = 1 unit. At A, draw AB ⊥ OA such that AB = 1 unit. Join OB. B

2 1

Here, OA = 1, AB = 1 By Pythagoras theorem, 2 OB = OA 2 + AB 2

⇒ OB 2 =+ 12 12 ⇒ OB = 2 Draw an arc with centre O and radius OB. The arc cuts the number line at P. Thus, P represents

2 on the number line.

The irrational number

2 finds a point on the number line.

Draw a right-angled triangle OBC such that BC = 1. D

C 3

-1

Here, = OB

= 2, BC 1

By Pythagoras theorem, OC 2 = OB 2 + BC 2 ⇒ OC 2 = ( 2) 2 + (1) 2 ⇒ OC 2= 2 + 1 ⇒ 6

OC = 3

B 1 A

IL Foundation Series Class 9

Draw an arc with centre O and radius OC = 3 . This arc cuts the number line at Q. Thus, Q represents

3 on the number line.

In this way we can show that there exist points on number line representing

8 etc.

Method 2: Example: Represent

9.3 on the number line.

Solution: Let's look into the steps below to represent

9.3 on the number line.

Step I: Draw a line and take AB = 9.3 units on it. Step II: F rom B, measure a distance of 1 unit and mark C on the number line. Mark the midpoint of AC as O. Step III: With 'O' as the centre and OC as the radius, draw a semicircle. Step IV: At B, draw a perpendicular to cut the semicircle at D. Step V: With B as a centre and BD as the radius, draw an arc to cut the number line at E. Thus, taking B as the origin, the distance BE = 9.3 . Therefore, point E represents

9.3 on the number line.

5.1

9.3 A

4.15

B 1 unit

9.3

Let's look at the proof shown below. AB = 9.3, BC = 1 AC = AB + BC = 10.3 OC =

AC 10.3 = = 5.15 2 2

NUMBER SYSTEMS

OC = OD = 5.15 OB = OC - BC = 5.15 - 1 = 4.15 In right-angled ∆OBD, using Pythagoras theorem, we have, 2 BD = OD 2 − OB 2

= (5.15)2 - (4.15)2 = ( 5.15 + 4.15 )( 5.15 − 4.15 ) [Using a 2 − b 2 = ( a + b )( a − b )  = 9.3 ◊ 1 = 9.3 Hence, = BD

= 9.3 BE [Since they are the radii of the same circle]

Thus, we can say that point E represents 9.3 on the number line.

SOLVED EXAMPLES

(

) (

)

Example 1: Solve: 2 2 + 7 7 + 13 2 − 4 7 . Solution:

( 2 2 + 7 7 ) + (13 2 − 4 7 ) = ( 2 2 + 13 2 ) + ( 7 7 − 4 7 ) = ( 2 + 13) 2 + ( 7 − 4 ) 7 = 15 2 + 3 7

(

) (

)

Example 2: Solve: 7 7 × −4 7 . Solution:

( 7 7 ) × ( −4 7 ) = 7 × −4 × 7 × 7 = −28 × 7 = - 196 Example 3: Divide 8 15 by 2 3 . Solution: 8 15 ÷ 2= 3 8

8 3× 5 = 4 5 2 3

IL Foundation Series Class 9

p Example 4: Show that 0.3333… = 0.3 can be expressed in the form , where p and q are integers q and q ≠ 0 . Solution: Since we do not know what 0.3 is, let us call it ' x ' and so = x 0.3333… Now, here is where the trick comes in. Look at, 10 x = 10 × ( 0.333…) = 3.333… Now, 3.3333…= 3 + x, since x = 0.3333… Therefore, 10x = 3 + x Solving for x, we get = 9 x 3,= i.e., x

1 3

Hence, 0.33333... =

1.5

1 3

CONCEPT OF INFINITY AND INTERVAL NOTATION

1.5.1 Concept of infinity What does "infinity" mean? Infinity is not a real number. Infinity is larger than any number that can be imagined. It is an idea that has no bounds. A line is an example of having no bounds. For instance, the number line has arrows at the end to represent this idea of having no bounds. The symbol used to represent infinity is ∞. On the left side of the number line is -∞ and on the right side of the number line is ∞ to describe the boundless behaviour of the number line. 1.5.2 Interval notation Interval notation is a way to describe a range of numbers using just two endpoints. Imagine you have a line with numbers on it, like a ruler. Interval notation helps us talk about all the numbers between two points on that line. We use brackets [ ] or parentheses ( ) to show if the endpoints are included or not. • Square brackets [ ] mean that the endpoint is included. For example, [3, 7] means all the numbers from 3 to 7, including 3 and 7. • Round parentheses ( ) mean that the endpoint is not included. For example, (2, 5) means all the numbers from 2 to 5, but not including 2 and 5. 9

NUMBER SYSTEMS

• We also use ∞ (infinity) and -∞ (negative infinity) in interval notation to show when a range goes on forever in one direction or the other. For example, t(- ∞, 4) means all numbers less than 4, but not including 4, and (- ∞, ∞) means all real numbers. Let us look at some of the important notations used in mathematics. () → Open Brackets [ ] → Closed Brackets o → Open Value( x cannot take this value)  → Closed Value ( x can take this value) (-1,1) → x cannot take values -1 and 1. [-1,1) → x can take value -1, but not 1. (-1,1] → x cannot take value -1, but it can take value 1. [-1,1] → x can take both -1 and 1 values.

1.6

ABSOLUTE VALUE

Absolute value of an integer: The absolute value of an integer is the numerical value of the integer regardless of its sign. If 'a' is an integer, then its absolute value is denoted by a and is defined as = a a if a ≥ 0, a = −a if a < 0 Example: 7 = 7 and −7 =− ( −7 ) =7. Here, i)

a = a if a is positive.

ii)

a = 0 if a is zero.

iii) a = −a if a is negative.

1.7

SURDS

1.7.1 Definition of surds Let 'a' be a positive rational number and 'n' be a positive integer ( ≠ 1) .

IL Foundation Series Class 9 1

If n a or a n is not a rational number (irrational), then n a is called a surd of order 'n' or nth order surd. • The other name of a 'surd' is 'radical'. • The symbol n

is called a radical sign.

• 'n' is called the order of the surd. • 'a' is called the radicand. •

a can be read as "nth root of a".

Example: Note:

2, 3 2, 4 9,

25 10 , etc., are surds. 3

1) n a n = a . 2) In general, 2 a is written as

3) Every surd is an irrational number, but every irrational number need not be a surd. Example:

2 is a surd as well as an irrational, but π is only an irrational, not a surd.

1.7.2 Types of surds The different types of surds are as follows: Simple surd or monomial surd: A surd consisting of a single term is called a simple or monomial surd. Example: 2 3, 3 4,5 3 9 , etc. Pure Surd: A simple surd expressed in the form a n b , where a = 1, is called an entire surd or a pure surd. Example:: 7, 40, 3 10, 4 27 … Mixed surd: A simple surd, which is not pure surd, is called mixed surd. Example: 2 3, 4 7 9, 6 9 18… Compound surd: A surd consisting of two or more terms is called a compound surd. Example: 3 + 2, 2 − 3, 5 + 7 − 3 4 , etc. Similar surds or like surds: If two surds are different multiples of the same simple surd, then they are called similar surds (or like surds); otherwise, they are called dissimilar surds or unlike surds. Example: 1) 2 3,3 3,5 3 are similar surds. 2) 2 3, 2 5, 7 2 are dissimilar surds. 11

NUMBER SYSTEMS

1.7.3 Rules of surds a×b=

Rule 1:

a× b 98 .

Example: Simplify Solution:

98 = 7 × 7 × 2 = 7 2 × 2 , since 7 is the greatest perfect square factor of 98. Therefore,

( 7 × 2) = 2

98 =

a a = b b

Rule 2:

Example: Simplify

72 × 2 = 7 2

12 . 121

Solution:

12 = 121

2× 2×3 2 3 = 11 11×11

ac bd Rule 3: a b × c d = Example: Simplify 2 3 × 4 3. Solution: 2 3 × 4 3 = ( 2 × 4 ) × 3 × 3 = 8 × 3 = 24 1.7.4 Addition, subtraction, multiplication, and division of surds Addition and subtraction of surds: When adding or subtracting surds, we need to ensure that the surds are like terms, meaning they have the same radical part. For example, 3 and 3 are like terms, but 3 and 5 are not. To add or subtract surds, simply combine the like terms and keep the radical part the same. Here are some examples: • 2 3 +3 3 = 5 3 • 4 5−2 5 = 2 5 Multiplication of surds: To multiply surds, we simply multiply the coefficients (the numbers outside the radical) and then multiply the radicals together. Remember that we can simplify the result if possible.

IL Foundation Series Class 9

For example: • 2 3 ×3 5 = 6 15 • 4 2×2 3 = 8 6 Division of surds: Dividing surds involves rationalising the denominator, which means eliminating the surd from the denominator. To do this, we multiply both the numerator and the denominator by the conjugate of the denominator (the same expression with the opposite sign between the terms). This process helps eliminate the radical from the denominator. 5 2 3 3 To rationalise the denominator, multiply both the numerator and the denominator by 3 3 to eliminate the radical from the denominator: Here's an example:

5 2 × 3 3 15 6 = 27 3 3 ×3 3

1.8

LAWS OF EXPONENTS

The laws of exponents are rules that help us simplify and solve problems involving expressions with exponents. Here are the basic laws of exponents typically used. Exponent Rules Product law

Quotient law

am × an = a m+n am = a m − n , if m > n n a am 1 = n − m , if m < n, where a ≠ 0 n a a

Power raised to a power

(a ) = a

Power of a product

(abc = …) n a n b n c n ……

m n

Power of a quotient

am a = , where (b ≠ 0)   bm b

NUMBER SYSTEMS

= a−n

1 , where (a ≠ 0) an

Power with negative exponents −n n a b =     , where (a, b ≠ 0) b a Zero exponent law

0 a= 1(a ≠ 0)

Equality of exponents

a m = a n ⇔= m n (a ≠ 0,1)

SOLVED EXAMPLES Example 1: Find the sum of 12 and

27 .

Solution: 12 + 27 2 × 2 × 3 + 3× 3× 3

= 2 3 +3 3 = (2 + 3) 3 =5 3 Example 2: Subtract 2 45 from 4 20 . Solution: The required difference is: = 4 20 − 2 45 = 4 2 × 2 × 5 − 2 3× 3× 5 = ( 4 × 2 ) 5 − (2 × 3) 5 = 8 5 −6 5 =2 5 Example 3: Find the product: (3 2)(5 2) . Solution: (3 2)(5 2) = (3 × 5)( 2 × 2) = 15 × 2 = 30

IL Foundation Series Class 9

Example 4: Simplify: 13 × 11 Solution. 13 × 11 =

13 ×11 =

143

Example 5: Write the surd Solution:

4 in the simplest form: 1+ 2 3

4 1+ 2 3 To simplify the given surd expression, change the sign in the denominator of the surd expression and simplify it. Given expression:

4 1+ 2 3 =

4 1− 2 3 × 1+ 2 3 1− 2 3

4(1 − 2 3) 1 − 4(3)

4−8 3 1 − 12

4−8 3 −11

− (4 − 8 3) 11

Hence, the simplification of the given surd is:

1.9

4 4-8 3 =11 1+ 2 3

COMPLEX NUMBERS

1.9.1 Introduction In the real number system, we have seen that the equation x2 + 1 = 0 has no real solution since it gives x2 = - 1, and the square of every real number is non-negative. So, we need to extend the real number system to a larger system so that we can find solutions to the equation x2 = - 1. In fact, the main objective is to solve the equation ax 2 + bx + c =, 0 for the case b2 - 4ac < 0, which is not possible in the real number system.

NUMBER SYSTEMS

It was Euler (1707-1783) who identified a root for the quadratic equation x2 + 1 = 0 with the symbol 'i', called it an imaginary root and termed the numbers of the form a + ib, a, b ∈ R , a, b ∈ R as complex numbers. Here, i= −1 , i.e., i2 = - 1. We commonly refer to 'i' as 'iota’. Note: All imaginary numbers can be expressed in terms of i. Examples: 1.

−2 =

−2 × −3 =i 2 × i 3 =i 2 2 × 3 =−1 6 =− 6

Note:

−1× 2 =

−1 × 2 = i 2

−a × −b ≠ ab , a, b ∈ R +

Complex numbers: If a and b are two real numbers, then a number of the form a + ib is called a complex number. Generally, a + ib is denoted by 'z'. Here, a is the real part of the complex number, and ib is the imaginary part. Example: 7 + 2i, - 1 + i, 3 - 2i, 0 + 2i, 1 + 0i etc., are complex numbers. If z1= a + ib and z2 = c + id are two complex numbers such that: • (a + ib) + (c + id) = (a + c) + i(b + d) • (a + ib ) - (c + id) = (a-c) + i(b - d) • (a + ib) ⋅ (c + id) = (ac - bd) + i(ad + bc) (a + ib ) (ac + bd ) (bc − ad ) +i 2 • = 2 2 (c + id ) c +d c + d2 1.9.2 Problems related to complex numbers Example 1: Express the following expression in the form of a + bi. (1 - i) - (- 1 + i6) Solution: (1 - i) - (- 1 + i6) = 1 - i + 1 - i6 = 2 - 7i Example 2: Find the value of

−16 .

Solution: = 16

16 × −1= i 4 × 4= 4i [i=

−1]

IL Foundation Series Class 9

Example 3: Find the addition of two complex numbers (2 + 3i) and (- 9 - 2i). Solution: (2 (2 + + 33ii )) + + ((− −99 − − 22ii )) == ( 22 − (3ii − − 99 ) + + (3 − 22ii )) =− −77 + + ii =

Example 4: Express the complex number (1 - i) + (-1 + 6i) in the standard form a + ib. Solution: (1 - i) + (-1 + 6i) = 1 - i - 1 + 6i = 1 - 1- i + 6i = 0 + 5i, which is the required form. Example 5: z = 2 - 3i then find z2. Solution: Given, z = 2 - 3i z2 = z ◊ z = (2 - 3i) (2 - 3i) = 2(2) - 2(3i) - (3i)(2) + (3i)(3i) =− 4 6i − 6i + 9i 2 {Since i 2 = −1} = 4 - 12i + 9 (-1) = 4-12i - 9 = - 5 - 12i Therefore, z 2 =−5 − 12i .

QUICK REVIEW p • A number s is called an irrational number if it cannot be written in the form , where p and q are q integers and q ≠ 0 . • The decimal expansion of a rational number is either terminating or non-terminating recurring. • The decimal expansion of an irrational number is non-terminating or non-recurring. • For positive real numbers a and b, the following identities hold: 17

NUMBER SYSTEMS

(i)

ab = a b

(ii)

a a = b b

a b (iii) ( a + b )( a − b ) =− (iv) (a + b )(a − b ) =a 2 − b a + 2 ab + b (v) ( a + b ) 2 =

• A number of the form a + ib, where a and b are real numbers, is called a complex number; a is called the real part and b is called the imaginary part of the complex number. Let z1 = a + ib and z2 = c + id. Then (i) z1 + z2 = (a + c ) + i(b + d) (ii) z1 z2 = (ac - bd) + i(ad + bc)

WORKSHEET - 1 I.

IRRATIONAL NUMBERS 1. Classify the following numbers as rational or irrational. 2 7 d) 2π 7 7 2. Which of the following numbers is an irrational number?

a) 2 − 5

b) (3 + 23) − 23

a) 0.25

b) 9

c) 4 5 − 2 5 = 2 5 d) 5

m . n 1 4. Rationalise the denominator of . 7 1 5. Rationalise the denominator of . 6− 5 2 6. Simplify ( 5 + 2) . 3. Represent 0.23 in the form of

1 . x 8. Add 2 2 + 5 3 and 2 − 3 3 . 7. If x = 2 +

3 , then find x +

9. Subtract 2 5 from

5 .

10. Multiply 6 5 by 2 5 . 11. Find the number of decimal places after which the decimal expansion of the rational number 14587 will terminate. 1250 p 12. Express 0.6 + 0.7 + 0.47 in the form , where p and q are integers and q ≠ 0 . q 18

IL Foundation Series Class 9

II. DENSITY PROPERTY AND REAL NUMBER LINE 1. Write the following in decimal form and say what kind of decimal expansion it has: 36 1 3 1 b) c) 4 d) 100 13 8 111 2. Find three rational numbers between 2 and 3.

3. Insert 5 rational numbers between 4. Represent

5 on the number line.

−3 8 and . 11 11

5. Represent 9.3 on the number line. 6. Find two rational numbers whose absolute value is 7. Solve the following equation: 6u =|1+3u|

1 . 5

8. Solve the following equation: 1 z + 4 = | 4 z − 6 | 2 III. CONCEPT OF INFINITY AND INTERVAL NOTATION 1. The set of all real numbers lying between 4 and 5 is represented as a) [4,5)

b) (4,5]

c) (4,5)

d) [4,5]

2. The set of all real numbers lying between 5 and 6, including 5, is represented as a) (5,6)

b) [5,6)

c) [5,6]

d) (5,6]

b) x

c) ± x

d) - x

3. When x < 0, then |x|= a) 0

4. If a < b and c < 0, then which of the following is true? a b a) ac < bc b) < c) ac > bc c c 5. All real numbers less than or equal to 6 are included in

d) None of these

a) (−∞, 6]

c) (−∞, −3)  (6, ∞)

d) (6, ∞)

c) 4 8

d) 2 3 2

c) 2 3 5

d) 3 5

c) 16

d) 5 5

b) (−∞, 6)

IV. SURDS 1. Which of the following is a mixed surd? a) 3 16

b) 4 16

2. The simplest form of 3 250 is: a) 3 3 2

b) 5 3 2

3. Which of the following is a surd? a)

b) 3 27

NUMBER SYSTEMS

4. Solve the equation 2 x = 6 for x. 5. If a = 3 2 and b = 2 6 , find the value of ab by performing multiplication of surds. 6. If = a

2 + 3 and= b

2 − 3 , find the value of (a + b).

7. What is the result of adding (2 3 + 3 2) and (5 2 − 3) ? 8. If

a− b= c , when a = 25 and b = 9, what is the value of c?

9. Simplify 18 − 8 as much as possible. 2 x , and y = 4, what is the value of x? 10. If 3 y − x = 11. If (2 3)(3 5) = a , what is the value of a? 12. Simplify the expression ( 3 + 5)( 3 − 5) . 13. Find the product of ( 5 + 2 7) × (2 7 − 5) . 14. Simplify the expression:

10 8 × 5 2

4 6 2 2 1 16. If= x 2 2 + 7 , then x + = x a) 2 2 b) 4 2 15. Simplify the expression:

c) 8

5 32 , then find the value of x. 17. If 25 x ÷ 2 x =

18. If 52 x −1 − (25) x −1 = 2500 , then find the value of x. 19. If= a x b= , b y c and c z = a, find the value of xyz. 20. The product 3 2 ⋅ 4 2 ⋅ 12 32 equals a) 2

b) 2

21. If 2 = 1.4142 , then

2 − 1 is equal to 2 +1

a) 2.4142

b) 5.8282

22. If x = 23.

d) 12 32

c) 0.4142

d) 0.1718

c) 1

d) 11

7+4 3 , then x 2 ( x − 14) 2 = 7−4 3

(32)0.2 + (81)0.25 = (256)0.5 − (121)0.5

a) 2

c) 12 2

b) 5

IL Foundation Series Class 9

24. Prove that:  xa   b x  

a +b

 xb  × c  x 

b+c

 xc  × a  x 

c+a

= 1

x y 25. If a= b= c z and b2 = ac, then the value of y is equal to

2xz x+z

xz 2( x − z )

xz 2( z − x )

2 xz ( x − z)

26. If 21998 − 21997 − 21996 + 21995 = k ⋅ 21995 , then the value of k is a) 1 27. Simplify:

b) 2 1 1+ x

b −a

c −a

c) 3 1

1+ x

9n × 32 × ( 3− n /2 ) − (27) n −2

28. If

3 ×2

a −b

c −b

1 1+ x

a −c

1 , then m - n = 27

29.= If 2 1.414, = 3 1.732 , then find the value of 30. Find the value of

4 2 − 3

(216) V. COMPLEX NUMBERS

1 (256)

3 − 4

d) 4

2 −

(243) 5

+ x b −c

4 3 + . 3 3−2 2 3 3+2 2

1. Convert the expression (1 - i) - (1 - i6) into the form a + bi. 2. Convert (5 - 2i ) - (1 + 3i) into the form a + bi. 3. Find the sum of the complex numbers (6 - 2i) and ( - 1 + 7i). 4. What is the product of the complex numbers (1 + 7i) and (- 4 + 9i)? 5. For the complex number b = 4 + 5i, find the value of b2. 6. Evaluate: (1 + i)6 + (1-i)3 i 4 n +1 − i 4 n −1 ? 2 8. What is the smallest positive integer n, for which (1 + i)2n = (1 - i)2n? 7. What is the value of

9. What is the reciprocal of 3 + 7i ? n

1 n 10. For a positive integer n, find the value of (1 − i ) 1 −  .  i 3 3  1+ i   1− i  x + iy , then find (x, y). 11. If   −  =  1− i   1+ i  12. If

(1 + i ) 2 = x + iy , then find the value of x + y. 2−i

NUMBER SYSTEMS 100

1− i  13. If   = a + ib , then find (a, b).  1+ i 

WORKSHEET - 2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1 A non-terminating but recurring decimal is

a) a rational number

b) an integer

c) a natural number

d) a whole number

p form of the number 0.9 is q 9 9 a) b) 100 10 2. The

1 d) 1 9

83 is a

a) A natural number

b) A rational number

c) An integer

d) An irrational number

4. 0.73457456 ... is a) An irrational number

b) A rational number

c) A natural number

d) A whole number

(5 − 11)(5 + 11) is equal to

a) -6

b) 6

c) 14

d) 36

6. The sum of 0.3 and 0.4 is 7 7 b) 10 9 7. Every point on a number line represents

7 99

a) A natural number

b) A rational number

c) A unique number

d) An irrational number

8. The simplest rationalising factor of a) 5 2

7 11

50 is

d) 50

9. Among the following, an irrational number is a) 0.15

b) 0.1516

c) 0.1516

10. π is

a) A rational number

b) An integer

c) An irrational number

d) A whole number

d) 0.501500150001 ..

IL Foundation Series Class 9

11. The rational part of a pure surd is a) 0

b) 1

c) -1

d) Not defined

−2

12. The value of 4 (81) is 1 1 a) b) c) 9 3 9 13. x is the answer to one of the following. Identify it. a) 14.

12 7

x −x

1 4 3

b) 12 ( x )

5 7

d) 2 3

1 81

( x)

d) 0

c) 43

d) 24

c) 7i

d) -7i

d) x3 - x2

81 − 8 3 216 + 15 5 32 + 225 = b) 15

 2 1  15. If x= 3 + 8 , then the value of  x + 2  is x   a) 31 b) 34 16. What is the square root of -49? a) 7

b) -7

17. Which of the following expressions is equivalent to (3 + 4i) - (2 - i)? a) 5 + 3i

b) 1 + 5i

c) 1 + 3i

d) 5 - i

c) 5 - 5i

d) 5 + 5i

c) -13

d) -12

18. What is the product of (2 + i) and (1 - 3i) ? a) 5 - i

b) 5 + i

19. If z = 12 - 13i, what is the real part of z ? a) 13

b) 12

20. The set of all real numbers lying between 3 and 9 is represented as a) [3,9)

b) (3,9]

c) (3,9)

d) [3,9]

II. FILL IN THE BLANKS 1. 15 15 ÷ 3 3 = ____________. 2.

2 is ____________ number.

( 3 + 7)( 3 − 7) is equal to ____________. 1

4. The value of 7 2 8 2 is ____________. p is ____________. q 1 6. The smallest rational number by which should be multiplied so that its decimal expansion 3 terminates after one place of decimal is ____________. 5.

0.235 in the form of

NUMBER SYSTEMS

7. The value of 4 2 + 2 32 = ____________. 8.

( 9 + 2 3)( 9 − 2 3) = ____________.

15 18 ____________. × = 3 2

10. If z = 5i, then the real part of z is ____________. III. SUBJECTIVE QUESTIONS 1. If x= 7 − 4 3 , then find the value of x +

1 ? x

3 −1 = a+b 3 . 3 +1 2, then find the value of 'x'? If 5 5 x + 2 = 3 Rationalise the denominator of . 7− 2 p Express 2.93 in the form of , where p and q are integers, and q ≠ 0 . q Check whether −5 + 2 5 − 5 is an irrational or a rational number.

2. Find the values of 'a' and 'b', if 3. 4. 5. 6.

2157 in decimal form. 625 8. If z = 7 - 4i, what is the value of z2? 7. Express

9. Express the complex number (2 - 7i) + (- 1 + 8i) in the standard form a + ib. 10. Find the product of the complex numbers (3 + 5i) and (-1+ 6i)?

POLYNOMIALS

2.1 POLYNOMIALS - TERMINOLOGY Variable: A symbol that can represent various numerical values is called a variable or literal. Constant: A symbol with a fixed value is called a constant. Term: The constants alone or the variables alone or their combinations by operation of multiplication or division are called terms. Constant term: A term of an expression without any literals is called a constant term. Algebraic expression: The combination of terms obtained by the fundamental operations, , , , is called an algebraic expression. Example: 2 x 3, 5 2y, 6a, 7 b . Coefficient: The coefficient of a polynomial is the number multiplied by the variable. Example: For polynomial, x 3 3 x 2 4 x 10 Term

Coefficient

−3 x 2

-3

Polynomial: A polynomial p x in one variable x is an algebraic expression of the form p x an x n an 1 x n 1 ... a2 x 2 a1 x a0 ,

where a0 , a1 , a2 ,¼, an are constants and an ≠ 0 . 0

a0 , a1 , a2 , ¼, an are respectively the coefficients of x , x , x ,¼, x , and n is called the degree of the polynomial. Each of an x n , an−1 x n−1 , …, a0, with an ≠ 0 , is called a term of the polynomial p ( x ). Or, An algebraic expression in which the power of variables are non-negative integral powers (i.e. whole numbers) is called a polynomial.

POLYNOMIALS

Example: i) 8 x 2 6 x 6 is a polynomial, ii) 2 x−3 + 7 is not a polynomial. Degree of a polynomial: The degree of a polynomial is the highest power of the term in the polynomial. Example: The degree of 3 x 2 y 2 + 5 xy + 6 is 4. 2.1.1 Types of polynomials (based on the number of terms) Monomial: A polynomial containing one term is called a monomial. Example: 5 x 2 ,

-2 xy, 7 y, 6 3

Binomial: A polynomial containing two terms is called a binomial. Example: 5 2 x, 7 x 2 3y Trinomial: A polynomial containing three terms is called a trinomial. Example: x 2 − 4 x + 1, 5 + 2y + z Multinomial: An algebraic expression containing one or more terms is called multinomial. Note: All polynomials are multinomials but all multinomials need not be polynomials. Example: x 2 xy 2 is a multinomial but not a polynomial. 2.1.2 Types of polynomials (based on degree) Linear polynomial: If the degree of the polynomial is '1', then the polynomial is called a linear polynomial. Example: x + y , 2x - 4y Quadratic polynomial: If the degree of the polynomial is '2', then the polynomial is called a quadratic polynomial. Example: 5 x 2 + 2 x + 1 Cubic polynomial: If the degree of the polynomial is '3', then the polynomial is called a cubic polynomial. Example: x 3 3 x 2 x 1 Zero polynomial: If all the coefficients of a polynomial are zero, then it is called a zero polynomial. Note: p x ax 2 bx c is a zero polynomial if and only if a= b= c= 0.

IL Foundation Series Class 9

Constant polynomial: The polynomial having degree zero is called a constant polynomial. Example: 5, 2 2 Note: The degree of the zero polynomial is not defined.

SOLVED EXAMPLES Example 1: Write the coefficients of x 2 in each of the following. 2 4 5 a) 3 x 2 − 4y b) x + x 2 + 7 y c) 3 x 4y z 3 3 2 Solution: a) The coefficient of x 2 is 3. 4 b) The coefficient of x 2 is . 3 2 c) The coefficient of x is 0. Example 2: Which of the following expressions are polynomials? 1 a) x + b) x 2 + 1 c) 3 + x 2 + x5 x

3x2 + 2x + 1

Solution: a) x +

1 = x + x -1 , as the exponent of x −1 is -1 , it is not a polynomial. x

b) x 2 + 1 is a polynomial. c) 3 + x 2 + x5 is a polynomial. d)

3 x 2 + 2 x + 1 is a polynomial.

Example 3: Find the degree of each of the following polynomials. a) 2 x 2 x 7 x 7 3 Solution:

b) 2

c) 0

a) 2 x 2 x 7 x 7 3 The highest exponent of the variable x is 7. ∴ Degree is 7.

POLYNOMIALS

b) 2 2 can be written as 2 x 0 . ∴ The exponent of the variable is 0. ∴ Degree is 0. c) The degree of zero polynomial is not defined. Example 4: Which of the following are monomials, binomials, or trinomials? x 6 x5 x 4 a) 1 b) x 3 x x 2 c) + + 4 5 4 Solution: a) The given polynomial, i.e. 1, contains only one term. So, it is a monomial. b) x 3 x x 2 x 3 2 Clearly, it contains two terms. So, it is a binomial. c)

x 6 x5 x 4 + + 4 5 4 Clearly, it contains three terms, so it is a trinomial.

Example 5: Give one example for each of a binomial of degree 35, and of a monomial of degree 100. Solution: i) A binomial of degree 35 is x 35 + 1 . ii) A monomial of degree 100 is 3 x100 .

2.2 ZERO OF A POLYNOMIAL 2.2.1 Value of a polynomial The value of a polynomial f x at x = a is f a , obtained on replacing x by a . Example: Consider f x x 2 x 1. The value of f x at x = 2 is f 2 22 2 1 4 2 1 5 2.2.2 Zero of a polynomial The number for which the value of a polynomial is zero is called zero of the polynomial. Example: Zero of the polynomial x − 4 is 4. Example: Find the number of zeroes of the polynomial q ( y ) = y 2 − 1.

IL Foundation Series Class 9

Solution: q y y 2 1 is a quadratic polynomial. It has at most two zeroes. Let q y 0 y2 1 0

y 1 y 1 0

y 1 or y 1 ∴ The zeroes of the polynomial are -1 and 1.

2.3 FACTORISATION OF A POLYNOMIAL 2.3.1 Factorisation by splitting the middle term In this section, we shall learn the factorisation of polynomial of the form ax 2 + bx + c where a, b and c are real numbers. The rule to factorise the polynomial ax 2 + bx + c, where a, b, and c are real numbers, is to split b (the coefficient of x) into two real numbers such that the algebraic sum of these two numbers is b and their product is c , then factorise by grouping method. It is not always possible to factorise a polynomial ax 2 + bx + c (i.e., a quadratic expression); the following rule can save a lot of time: For the expression ax 2 + bx + c , work out b 2 − 4ac . If it is a perfect square, then the given expression will factorise; otherwise, it will not. Example: Factorise, 2 x 2 − 7 x − 15 using middle term split. Solution: To factorise, 2 x 2 − 7 x − 15 , we want to find two real numbers whose sum is b 7 and the product is ac 30. By trial, we see that 10 3 7 and 10 3 30. Therefore, 2 x 2 7 x 15 2 x 2 10 x 3 x 15 2x x 5 3 x 5 x 5 2x 3 Example: Factorise x 2 - 8 x + 12.

POLYNOMIALS

Solution: The middle term is -8x. We must split −8x so that the obtained numbers when multiplied together, give 12x2. Thus, −8x can be written as -2 x - 6 x. Therefore, x 2 8 x 12 x 2 2 x 6 x 12 x x 2 6 x 2 x 6 x 2 2.3.2 Remainder and factor theorem Remainder theorem

If f x is a polynomial in x of degree ≥ 1, 'a’ is any real number and if f x is divided by x a , then the remainder is f ( a ). Proof: Let us suppose that, when f x is divided by x a , the quotient is q x and the remainder is r x . By division algorithm we have f x x a q x r x , where r x 0 (or) deg r x deg x a . Since degree of x a is 1 , either r x 0 or deg r x 0( 1). So, r x is constant for all values of x. f x x a q x r , where r is a constant. In particular for x = a

f a a a q x r f a r

∴ Remainder f a Hence proved. Example: Find the remainder when x 4 x 3 2 x 2 x 1 is divided by ( x − 1 ). Solution: Let f x x 4 x 3 2 x 2 x 1 and the zero of x −1 is 1. By the remainder theorem, remainder f ( a ) = f ( 1 ). Thus, f 1 (1)4 (1)3 2(1)2 1 1 2. ∴ 2 is the remainder, when x 4 x 3 2 x 2 x 1 is divided by ( x − 1 ).

IL Foundation Series Class 9

Example: Let R1 and R2 be the remainders when the polynomials f x 4 x 3 3 x 2 12ax 5 and g x 2 x 3 ax 2 6 x 2 are divided by (x -1) and (x + 2) respectively. If 3 R1 R2 28 0 , then find the value of a. Solution: When f x is divided by ( x − 1 ), the remainder f 1 R1 When g x is divided by x 2 , the remainder g 2 R2 R1 f 1 4 3 12a 5 2 12a R2 g 2 2( 2)3 a( 2)2 6 2 2 116 4a 12 2 4a 2 If 3 R1 R2 28 0 ⇒ 3( 2 − 12a ) + 4a − 2 + 28 = 0 ⇒ 6 − 36a + 4a + 26 = 0 ⇒ −32a + 32 = 0 ⇒ 32a = 32 ⇒a = 1 Factor theorem

Let f x be a polynomial of degree ≥ 1 and a be a real number, then: i) x a is a factor of f x if f a 0. ii) f a 0 if x a is a factor of f (x ).

Proof: Let f x be a polynomial of degree ≥ 1 and a be a real number. i) Let x a be a factor of f x . f x divided by x a gives remainder zero. ∴ Remainder = 0 But by remainder theorem, when f x is divided by (x - a ), the remainder is f ( a ).

f a 0

ii) Let f a 0 ---------- (i) Let q x be the quotient when f x is divided by ( x − a ). By the remainder theorem, remainder f a

POLYNOMIALS

∴ By division algorithm, f x x a q x f a

f x x a q x 0

 from (i)

f x x a q x x a is a factor of f x Hence proved. Example: Determine whether x 1 is a factor of x 3 + x 2 + x + 1. Solution: Let f x x 3 x 2 x 1 By factor theorem, x 1 will be a factor of f x if f 1 0. Put x 1 f 1 ( 1)3 ( 1)2 1 1 1 1 1 1 0 f 1 0 Hence, x 1 is a factor of x 3 + x 2 + x + 1. Note: 1) If x a is a common factor of polynomials f x and g x , then f a g a . 2) The maximum number of factors a polynomial can have is equal to its degree. 3) If the sum of coefficients of the terms in a polynomial is zero, then x 1 is a factor. 4) If the sum of the coefficients of even powers of x and the constant term is equal to the sum of the coefficients of odd powers of x , then x 1 is a factor. Example: If x 1 is a factor of ax 3 + bx 2 + cx + d , then a c b d . 5) If g x is a factor of f x , then g x is also a factor of f x . Generally, we take g x (or) g x as a factor in which the highest degree term has a positive coefficient. Example: For the polynomial x 2 - 5 x + 6, (x - 2 ) and 2 x are the factors.

IL Foundation Series Class 9

SOLVED EXAMPLES Example 1: Find the zero of the polynomial, p x x 2. Solution: Given that p x x 2 To find zero of the polynomials, p x 0 p x 0 x 2 0 x 2 ∴ 2 is a zero of the given polynomials. Example 2: Find the zeroes of the polynomial, p x x 2 3 x 2. Solution: Given that p x x 2 3 x 2 To find zeros of a polynomial, p x 0 p x 0 x2 3x 2 0 x2 2x x 2 0 x x 2 1 x 2 0 x 1 x 2 0 x 1 0 or x 2 0 x 1 or x 2. ∴ 1 and 2 are the zeroes of the polynomial. Example 3: If x = 2 is a zero of the polynomial f x 2 x 2 3 x 7 a, find the value of a? Solution: Given that f x 2 x 2 3 x 7 a . And x = 2 is a zero of the given polynomial.

POLYNOMIALS

So, f 2 0 f x 2 x2 3 x 7a f 2 2(2)2 3 2 7a 0 8 6 7a 0 2 7a 0 7 a 2 a

2 7

∴ The value of a is

−2 . 7

Example 4: Verify that x =

4 is a zero of the polynomial, p x 5 x 5

Solution: Given that, p x 5 x 4 Put, x = 5 p x 5x 4 4 p 5 5 5 4 0. 4 p 0. 5 4 ∴ is not a zero of the given polynomial. 5 Example 5: Find the remainder when x 3 + 3 x 2 + 3 x + 1 is divided by 2 x − 1 . Solution: 1 Here, zero of 2 x − 1 is . 2 3 2 Let, p x x 3 x 3 x 1. We have by the remainder theorem, if ' a ' is a zero of a polynomial p x , then remainder is p a .

IL Foundation Series Class 9

1 ∴ remainder = p    2  1  1 3  1 2 =   + 3  + 3  + 1  2   2   2  1 3 3 1 + 6 + 12 + 8 27 = + + +1 = = . 8 4 2 8 8

Example 6: Check whether the polynomial p x 3 x 4 4 x 3 3 x 4 is a multiple of (x − 1). Solution: We know that p x is a multiple of x 1 only if x 1 divides p x leaving the remainder zero. p x 3x4 4x3 3x 4 p 1 3(1)4 4(1)3 3 1 4 3 4 3 4 0 p x is a multiple of x 1 .

Example 7: Using factor theorem, show that x 1 is a factor of x10 1 . Solution: f x x10 1 By factor theorem, x 1 is a factor of f x if f 1 0. f 1 110 1 1 1 0 f 1 0 Hence, x 1 is a factor of x10 − 1 .

POLYNOMIALS

Example 8: Factorise: 12 x 2 7 x 1 Solution: 12 x 2 7 x 1 12 x 2 4 x 3 x 1 4 x 3 x 1 1 3 x 1 4x 1 3x 1 12 x 2 7 x 1 4 x 1 3 x 1 Example 9: Factorise:

2x2 + 9x + 4 2

Solution: 2x2 9x 4 2 2x2 x 8x 4 2

2x 1 4 2 2x 1 2x 1 x 4 2 x

Example 10: Find the value of k, for which x 1 is a factor of kx 2 2 x 1 . Solution: Let, f x kx 2 2 x 1. If x 1 is a factor of f x , then f 1 0 f 1 k(1)2 2 1 1 0 k 2 1 0 k 2 1 k 2 1 Example 11: Find the value of ' p ' for which the polynomial, 2 x 4 + 3 x 3 + 2 px 2 + 3 x + 6 is divisible by x 2 . Solution: f x 2 x 4 3 x 3 2 px 2 3 x 6 f x is divisible by x 2 f 2 0

IL Foundation Series Class 9

2( 2)4 3( 2)3 2 p( 2)2 3 2 6 0 2 16 3 8 8 p 6 6 0 32 24 8 p 0 8p 8 0 p 1

2.4 LONG DIVISION OF POLYNOMIALS 2.4.1 Division algorithm If f (x ), g (x ) are two non-zero polynomials, then there exist polynomials q x and r x uniquely such that: f x q x g x r x Where r x 0 or deg r x deg g x . Note: If f x is divided by g ( x ), we get quotient q x and remainder r x . Now, dividend = divisor × quotient + remainder. f x g x q x r x Example: Divide 2 x 3 7 x 2 2 x 5 by (x − 3) and find the remainder. Solution: By long division, x-3

2x3 - 7x2 + 2x +5 2x3 - 6x2 (-) (+)

2x2 - x - 1

-x2 + 2x -x2 + 3x (+) (-) -x + 5 -x + 3 (+) (-)

∴ The remainder is 2.

POLYNOMIALS

SOLVED EXAMPLES Example 1: Verify division algorithm when 4 x 3 + 16 x 2 + 19 x + 9 is divided by (x + 2) . Solution: By long division, x + 2 4x3 + 16x2 + 19x +9 4x2 + 8x + 3 4x3 + 8x2 (-) (-)

8x2 + 19x 8x2 + 16x (-) (-) 3x + 9 3x + 6 (-) (-)

3 Here, Dividend 4 x 3 16 x 2 19 x 9 Divisor x 2 Quotient 4 x 2 8 x 3 Remainder 3. Now finding, (Divisior)(Quotient) + (Remainder), we get:

x 2 4x2 8x 3 3 4 x 3 8 x 2 3 x 8 x 2 16 x 6 3 4 x 3 16 x 2 19 x 9 Dividend ∴ Dividend = (Divisor) (Quotient) + Remainder. Hence, division algorithm is verified. Example 2: Divide p x by q x , where p x 7 x 2 7 x 2 x 3 28 and q x 5 2 x. Solution: Given, p x 7 x 2 7 x 2 x 3 28 The standard form of given polynomials is:

IL Foundation Series Class 9

p x 2 x 3 7 x 2 7 x 28 and q x 2 x 5 Now, by long division, we get:

2x + 5 2x3 + 7x2 - 7x +28 2x3 + 5x2 (-) (-)

x2 + x - 6

2x2 - 7x 2x2 + 5x (-) (-) -12x - 28 -12x - 30 (+)

(+)

2 Example 3: Use factor theorem to determine whether g x is a factor of f x . f (x )= 2 x 3 + 4 x + 6, g (x )= x + 1 Solution: f x 2 x 3 4 x 6 By factor theorem, if x + 1 is a factor of 2 x 3 + 4 x + 6 then, f 1 0 Now f 1 2( 1)3 4 1 6 2 4 6 0 f 1 0 x 1 is a factor of 2 x 3 + 4 x + 6. Thus, g x is a factor of f ( x ).

Example 4: Determine if x 1 is a factor of x 3 x 2 2 2 x 2 . Solution:

f x x3 x2 2 2 x 2 If x 1 is a factor of f ( x ), then f 1 0

POLYNOMIALS

Now,

(

)

f ( −1 )= (−1)3 − (−1)2 − 2 + 2 ( −1 ) + 2 = −1 − 1 + 2 + 2 + 2 = 2 2≠ 0

x 1 is not a factor of x 2 x 2 2 2 x 2 .

2.5 ALGEBRAIC IDENTITIES From your earlier classes, you may recall that an algebraic identity is an algebraic equation that is true for all values of the variables occurring in it. You have studied the following algebraic identities in earlier classes: Identity I : (x y)2 x 2 2 xy y 2 Identity II : (x y)2 x 2 2 xy y 2 Identity III: x 2 y 2 x y x y Identity IV : x a x b x 2 a b x ab You must have also used some of these algebraic identities to factorise the algebraic expressions. You can also see their utility in computations. Identity V : (x y z)2 x 2 y 2 z 2 2 xy 2yz 2 zx Remark: We call the right-hand side expression the expanded form of the left-hand side expression. Note that the expansion of (x + y + z)2 consists of three square terms and three product terms. Identity VI: (x y)3 x 3 y 3 3 xy x y Also, by replacing y by −y in the Identity VI, we get: Identity VII: (x y)3 x 3 y 3 3 xy x y x 3 3 x 2 y 3 xy 2 y 3 Some more identities:

a b a b a ab b a 3 b 3 a b a 2 ab b 2 3

(a b c)2 a 2 b 2 c2 2ab 2bc 2ca

a 3 b 3 c3 3abc a b c a 2 b 2 c2 ab bc ca x a x b x 2 a b x ab a b c 0 then a 3 b 3 c3 3abc 40

IL Foundation Series Class 9

SOLVED EXAMPLES Example 1: Factorise: 49a2 + 70ab + 25b2 Solution: 49a2 + 70ab + 25b2 (7 a)2 2 7 a 5b (5b)2 (7 a 5b)2 (a b)2 a 2 2ab b 2 7a 5 b 7a 5 b

49a 2 70ab 25 b 2 7a 5 b 7 a 5 b Example 2: Factorise: 8a 3 + b 3 + 12a 2b + 6ab 2 Solution: 8a 3 + b 3 + 12a 2b + 6ab 2 (2a)3 (b)3 3 2a b 2a b (2a b)3 (a b)3 a 3 b 3 3ab a b 2a b 2a b 2a b

8a 3 b 3 12a 2b 6ab 2 2a b 2a b 2a b Example 3: Factorise: 27 x 3 y 3 z 3 9 xyz Solution: Using the formula:

a 3 b 3 c3 3abc a b c a 2 b 2 c2 ab bc ca

we have 27 x 3 y 3 z 3 9 xyz (3 x)3 (y)3 (z)3 3 3 x y z 3 x y z (3 x)2 (y)2 (z)2 3 x y y z z 3 x

3 x y z 9 x 2 y 2 z 2 3 xy yz 3 xz

Example 4: If a b c 7 and ab bc ca 20, find the value of a 2 + b 2 + c2 . Solution: Given a + b + c = 7, ab + bc + ca = 20 we know that, 41

POLYNOMIALS

(a b c)2 a 2 b 2 c2 2 ab bc ca 7 2 a 2 b 2 c2 2 20 49 a 2 b 2 c2 40 a 2 b 2 c2 49 40 a 2 b 2 c2 9 Example 5: Evaluate 103 × 107 by using identity. Solution: Given 103 × 107 Now 103 107 100 3 100 7 (100)2 3 7 100 3 7  x a x b x 2 a b x ab 10000 1000 21 11021 Example 6: If x y z 0 , then show that x 3 y 3 z 3 3 xyz. Solution: We know that,

x 3 y 3 z 3 3 xyz x y z x 2 y 2 z 2 xy yz zx

But given x y z 0

x 3 y 3 z 3 3 xyz 0 x 2 y 2 z 2 xy yz zx

x 3 y 3 z 3 3 xyz 0 x 3 y 3 z 3 3 xyz Hence proved.

2.6 HCF AND LCM OF A POLYNOMIAL 2.6.1 HCF The highest common factor (HCF) of two polynomials f x and g x is the common factor which has the highest degree among all common factors and in which the coefficient of the highest degree term is positive.

IL Foundation Series Class 9

Method of finding HCF

Step 1: Write f x and g x as products of simple factors. Step 2: If there is no common factor, then HCF is 1. If there are common factors, find the smallest exponent of these factors and write as HCF. Step 3: If there are more than one common factor, find the smallest exponent of them and write their product as HCF.

Example: Find HCF of x 1 x 2 4 and x 2 1 x 2 .

q x x 1 x 2 x 1 x 1 x 2

Solution: Let p x x 1 x 2 4 x 1 x 2 x 2 and 2

The common factor of p x and q x x 1 x 2 ∴ HCF of p x and q x x 1 x 2 .

Solution: Let p x 21 x x 1 3 7 x x 1 x 1 Example: Find HCF of 21x x 2 1 and 3 x 3 (x + 1 ). 2

q x 3x3 x 1 3 x x x x 1

The common factors of p x and q x are 3, x and x 1 . ∴ HCF of p x and q x 3 x x 1 . 2.6.2 LCM The least common multiple (LCM) of two or more polynomials is the polynomial of the lowest degree, having the smallest numerical coefficient, which is exactly divisible by the given polynomials and whose coefficient of the highest degree term has the same sign as the sign of the coefficient of the highest degree term in their product. Method of finding LCM

Step 1: Express each polynomial as a product of powers of prime factors. Express numerical factors, if any, as a product of powers of the primes. Step 2: Consider each of the irreducible factors occurring in the given polynomials only once. Find the greatest exponent of each of these simple factors in the factorised form of the given polynomials. Step 3: Raise each irreducible factor to the greatest exponent found in step 2, and multiply to get the LCM.

POLYNOMIALS

Example: Find the LCM of the polynomials 90 x 2 5 x 6 (2 x 1)2 and 140(x 3)2 2 x 2 15 x 7 . Solution:

Let p x 90 x 2 5 x 6 (2 x 1)2 2 32 5 x 2 x 3 (2 x 1)2

q x 140(x 3)2 2 x 2 15 x 7

22 5 7(x 3)2 2 x 1 x 7 LCM of p x and g x 22 32 5 7 x 2 (x 3)2 (2 x 1)2 x 7

1260 x 2 (x 3)2 (2 x 1)2 x 7 .

2.6.3 Relation between LCM and HCF If the LCM and HCF of two polynomials p x and q x are l x and h x respectively, then h x l x p x q x . Here, i) h x is the common factor of p x and q ( x ). ii) p x and q x are the factors of l x .

2.7 HORNER’S SYNTHETIC DIVISION Whenever a polynomial is of a higher degree, it is difficult to factorise by using the factor theorem. By using the synthetic division method, we can find factors, remainder and quotient of the polynomial. Synthetic division is also known as 'Horner's synthetic division'. Method of synthetic division Step 1: Write the coefficients of the terms in the first row. Step 2: Write the zero under the first coefficient and add them. Multiply this value with the zero value of the given factor of the form x a where a ∈ Z . Step 3: Write this product under the next coefficient and add them. Step 4: Again, multiply this value with the zero value of the given factor. Write this product under the next coefficient and add them. Continue this till the last coefficient. The value under the last coefficient is said to be the remainder, and the polynomial formed by using the values under the remaining coefficients is the quotient. Example: Factorise, x 3 4 x 2 5 x 2.

IL Foundation Series Class 9

Solution: Let f x x 3 4 x 2 5 x 2 Sum of the coefficients 1 4 5 2 0 x 1 is a factor.

x=1 1

-4

-2

-3

remainder

f x x3 4x2 5x 2 x 1 x2 3x 2

x 1 x2 2x x 2

x 1 x x 2 x 2 x 1 x 2 x 1 (x 1)2 x 2 Example: Find the quotient and the remainder when 4 x 3 3 x 9 is divided by ( 2 x − 3 ). Solution: Let f x 4 x 3 3 x 9 3 Remainder f 2 3

3 3 4 3 9 2 2

27 9 27 9 18 36 18 9 2 2 2 2

Remainder = 18 3 The divisor 2 x 3 2 x 2

POLYNOMIALS

3 4 = 2 0

-3

2 4

6 18

Remainder 18 Quotient 2 x 2 3 x 3

SOLVED EXAMPLES Example 1: Find HCF and LCM of two polynomials 2 x 2 − x − 1 and 4 x 2 + 8 x + 3 and prove that the product of polynomials is the product of their LCM and HCF. Solution: Given two polynomials are 2 x 2 − x − 1 and 4 x 2 + 8 x + 3 . By factoring, 2 x 2 − x − 1 , we get: 2x2 2x x 1 2 x x 1 1 x 1 x 1 2x 1 By factoring, 4 x 2 + 8 x + 3 , we get: 4x2 6x 2x 3 2 x 2 x 3 1 2 x 3 2x 3 2x 1 The common factor in both polynomials is 2 x 1 Therefore, HCF = common factor 2 x 1 LCM = common factor × remaining factors = ( 2 x + 1 )×( 2 x + 3 )( x − 1 ) = ( 2 x + 1 )( 2 x + 3 )( x − 1 )

The product of two polynomials is equal to the product of LCM and HCF.

IL Foundation Series Class 9

2x x 1 4x 8x 3 2x 1 2x 3 x 1 2x 1 2

x 1 2 x 1 2 x 3 2 x 1 (2 x 1)2 2 x 3 x 1 Hence proved. Example 2: Find the HCF of polynomials x 3 + y 3 and x 4 + x 2 y 2 + y 4 whose LCM is

x y x xy y . 3

Solution: Given two polynomials are x 3 + y 3 and x 4 + x 2 y 2 + y 4 .

x xy y .

LCM of polynomials is x 3 y 3

By factoring x 4 + x 2 y 2 + y 4 , we get:

x2 y2

(xy) x xy y x xy y 2

Product of two polynomials = Product of their LCM and HCF

x y x x y y x y x xy y HCF 3

2 2

Cancel the common term x 3 y 3 .

x x y y x xy y HCF 4

2 2

x xy y x xy y x xy y HCF 2

Therefore, HCF x 2 xy y 2

Example 3: Find the HCF and LCM of the expressions a 2 12a 35 and a 2 − 8a + 7 by factorisation. Solution: First expression a 2 12a 35 a 2 7 a 5a 35 a a 7 5 a 7 a 7 a 5

POLYNOMIALS

a 2 8a 7 Second expression a 2 7 a a 7 a a 7 1 a 7 a 7 a 1 Therefore, the HCF a 7 and LCM a 7 a 5 a 1 Example 4: Factorise 2 x 3 11x 2 17 x 6 using synthetic division method. Solution: f x 2 x 3 11x 2 17 x 6

f 3 2 33 11 32 17 3 6 0 f x is divided by x 3 11 17 17− xxx= 11 666 ==333222− −− 11 17 −− 000 666− 15 −− 1515 666 222 − 555 222 000 −− Quotient = 2 x 2 5 x 2 2x2 4x x 2 2 x x 2 1 x 2 x 2 2x 1 f x x 3 x 2 2x 1 Example 5: Factorise x 3 4 x 2 5 x 2 . Solution: Let f x x 3 4 x 2 5 x 2 Sum of the coefficients 1 4 5 2 0 x 1 is a factor.

IL Foundation Series Class 9

x 1 1 4

5 2

1 3 1 3

2 0 Remainder

f x x3 4x3 5x 2 x 1 x2 3x 2 x 1 x2 2x x 2

x 1 x x 2 1 x 2 x 1 x 2 x 1 (x 1)2 x 2

2.8 INTRODUCTION TO QUADRATIC EQUATION 2.8.1 Quadratic expression An expression of the form ax 2 + bx + c , where a, b, c Î C and a ≠ 0 , is called a quadratic expression. 2.8.2 Quadratic equation An equation of the form ax 2 bx c 0 , where a, b, c Î C and a ≠ 0 , is called a quadratic equation. Here, a, b, c are called coefficients. It is called the standard form of the quadratic equation in x . Example: 3 x 2 4 x 4 0 is a quadratic equation. Note: The quadratic equation ax 2 bx c 0 is called incomplete if at least one of the coefficients, either b or c is zero. Pure quadratic equation: The quadratic equation ax 2 bx c 0 is called a pure quadratic equation if b = 0 . Example: 2 x 2 4 0 is a pure quadratic equation. Monic quadratic equation: The quadratic equation ax 2 bx c 0 is called monic quadratic equation if a = 1 . Example: x 2 3 x 10 0 is a monic quadratic equation. 2.8.3 Roots of quadratic equation A complex number α is called a solution of a quadratic equation f x ax 2 bx c 0 if f a 2 b c 0. If a quadratic equation f x ax 2 bx c 0 is satisfied by more than two distinct roots, then f x 0 becomes an 'Identity’ (i.e., a= b= c= 0 ).

POLYNOMIALS

If , are the two roots of a quadratic equation ax 2 bx c 0 , then,

•

b a c Product of the roots a A quadratic equation cannot have more than two roots. Sum of the roots

•

Example: Check whether the following are quadratic equations or not. i) (x 2)2 1 2 x 3 3 x2 x

ii) x

iii) x x 1 8 x 2 x 2 Solution: i)

(x 2)2 1 2 x 3

x2 4x 4 1 2x 3 x2 4x 5 2x 3 0 x2 6x 8 0 It is of the form ax 2 bx c 0 Therefore, the given equation is a quadratic equation. ii) x

3 x2 x

x2 3 x2 x x2 3 x3 x3 x2 3 0 The degree of this equation is 3. So, the given equation is not a quadratic equation. iii) x ( x + 1 ) + 8 = ( x + 2 )( x − 2 ) x2 + x + 8 = x2 − 4 x2 + x + 8 − x2 + 4 = 0 x + 12 = 0

It is not in the form ax 2 bx c 0 So, the given equation is not a quadratic equation. Example: Find the sum and product of the roots of the equation 5 x 2 2 x 7 0 .

IL Foundation Series Class 9

Solution: Here a 5, b 2, c 7 ∴ The sum of the roots

b 2 2 a 5 5

c 7 a 5 Example: Find the quadratic equation whose roots are 3 and 5. The product of the roots

Solution: If , are the roots of quadratic equation, then the quadratic equation is x 2 x 0 Here, α = 3, β = 5 ∴ the required quadratic equation is x 2 3 5 x 3 5 0 x 2 8 x 15 0 Example: If , are the roots of the equation x 2 x 20 0, then find the values of and 2 2 . Solution: Sum of the roots 1 Product of the roots 20

( )2 4 ( 1)2 4 20 1 80 81 9

2 2 1 9 9

2.9 SOLVING QUADRATIC EQUATION BY FACTORISATION METHOD To solve a quadratic equation using factorisation: 1) Transform the equation using standard form in which one side is zero. 2) Factor the non-zero side. 3) Set each factor to zero (Remember: a product of factors is zero if and only if one or more of the factors is zero). 4) Solve each resulting equation. 51

POLYNOMIALS

Example: Solve the following quadratic equations by factorisation. i) 2 x 2 5 x 3 0 ii) x 2 6 x 5 0 iii) 3 x 2 2 6 x 2 0 Solution: i) 2 x 2 − 5 x + 3 = 0 ⇒ 2x2 − 2x − 3x + 3 = 0 ⇒ 2 x ( x − 1 ) − 3( x − 1 ) = 0 ⇒( x − 1 )( 2 x − 3 ) = 0 ⇒ x − 1 = 0 or 2 x − 3 = 0 3 ⇒ x = 1 or x = 2 x 1 and

3 are roots of the equation. 2

ii) x 2 + 6 x + 5 = 0 ⇒ x2 + 5x + x + 5 = 0 ⇒ x ( x + 5 ) + 1( x + 5 ) = 0 ⇒ ( x + 1 )( x + 5 ) = 0 ⇒ x + 1 = 0 or x + 5 = 0 ∴ x = −1 or x = −5 x 1 and -5 are the roots of the equation. iii)

3x2 − 2 6x + 2 = 0 ⇒ 3x2 − 6x − 6x + 2 = 0

( 3x − 2 )− 2 ( 3x − 2 ) = 0 ⇒( 3 x − 2 )( 3 x − 2 ) = 0 ⇒ 3x

⇒ 3x − 2 = 0 ⇒x=

2 2 = 3 3

IL Foundation Series Class 9

This root is repeated twice, one for each repeated factor Therefore, the roots of 3 x 2 2 6 x 2 0 are

( 3x − 2 ).

2 2 , . 3 3

2.10 SOLVING QUADRATIC EQUATION BY QUADRATIC FORMULA Consider the quadratic equation ax 2 + bx + c = 0, a ≠ 0 . We have ax 2 bx c 0 (Dividing throughout by 'a' on both sides) b c x2 x 0 a a

b c x a a

c b x2 2 x a 2a 2

b Adding on both sides, we get: 2a 2

c b b b (x)2 2 x a 2a 2a 2a

b c b2 x 2a a 4a 2

4ac b 2 b x 2a 4a 2

[Taking square root of both sides and assuming b 2 4ac 0 ]

POLYNOMIALS

b 2 4ac b 2a 4a 2

b b 2 4ac 2a 2a

b 2 4ac b 2a 2a

b b 2 4ac 2a

b b 2 4ac x 2a

(or) x

b b 2 4ac 2a

b b 2 4ac −b − b 2 − 4ac So, the roots of ax bx c 0 are (or) . 2a 2a 2

This formula is for finding the roots of a quadratic equation as the quadratic formula.

Note: Here b 2 − 4ac is known as its discriminant and is generally denoted by 'D' or ' ' b 2 4ac Nature of the roots of a quadratic equation: Nature of the roots of the quadratic equation ax 2 bx c 0 depends on the discriminant b 2 4ac .

Here, a, b, c are real numbers, 1) If 0, then the roots are real and distinct. Example: Find the nature of the roots of the equation x 2 3 x 2 0. Solution: Here, a = 1, b = -3, c = 2 b 2 4ac ( 3)2 4 1 2 9 8 1 0 ∴ The roots are real and distinct. 2) If 0, then the roots are real and equal. Example: Find the nature of the roots of the equation x 2 4 x 4 0. Solution: Here, a = 1, b = -4, c = 4 b 2 4ac ( 4)2 4 1 4 16 16 0 ∴ The roots are real and equal. 54

IL Foundation Series Class 9

3) If 0, then the roots are non-real or imaginary. Example: Find the nature of the roots of the equation x 2 2 x 3 0. Solution: Here,= a 1= , b 2= , c 3. b 2 4ac 22 4 1 3 4 12 8 0 ∴ The roots are imaginary. Example: Find the roots of the following quadratic equations using the quadratic formula. i) 3 x 2 2 x 1 0

ii) x 2 x 1 0

Solution: i) 3x2 + 2x -1 = 0 Here, a = 3, b = 2, c = -1 D = b 2 - 4ac = (2)2 - 4 (3 )(-1 ) = 4 + 12 = 16 > 0 So, the given equation has real roots, given by x

b D 2 16 2 4 6 2a 2 3

2 4 2 4 or 6 6

2 1 6 or x or 1 6 6 3

ii) x 2 x 1 0 Here, a = 1, b = 1, c = 1

D b 2 4ac (1)2 4 1 1 1 4 3 0

D 0 So, the given equation has no real roots.

POLYNOMIALS

SOLVED EXAMPLES Example 1: Find the roots of the following equations: Solution:

1 1 11 x 4 x 7 30

x 7 x 4 11 x 4 x 7 30

x 7 x 4 11 x 2 3 x 28 30

11 11 x 3 x 28 30 2

x 2 3 x 28 30 x 2 3 x 28 30 0 x2 3x 2 0 x2 2x x 2 0 x x 2 1 x 2 0 x 1 x 2 0 x 1 or 2 Example 2: A plane left 30 minutes later than the scheduled time, and in order to reach its destination 1500 km away in time, it increased its speed by 250 km/hr from its usual speed. Find its usual speed. Solution: Let the usual speed of the plane be x km / h, then the new speed of the plane x 250 km/h Distance = 1500 km 1500 hrs x 1500 Time taken to cover 1500 km with the new speed hrs x 250 Time taken to cover 1500 km with the usual speed =

IL Foundation Series Class 9

According to the condition, 1500 1500 1 x x 250 2

1500 x 250 1500 x

1 2

1500 x 375000 1500 x

1 2

x x 250 x 2 250 x

x 2 250 x 750000 x 2 250 x 750000 0 x 2 1000 x 750 x 750000 0 x x 1000 750 x 10000 0 x 750 x 1000 0 ⇒ x = -1000 or x = 750, speed cannot be negative, Hence, the usual speed of the plane = 750 km/h. Example 3: Find the nature of the roots of the following quadratic equations. i) 2 x 2 x 1 0 Solution:

ii) x 2 4 x 4 0

iii) x 2 x 1 0

i) The given quadratic equation is: 2 x 2 x 1 0 Here, a = 2, b = 1, c = -1 D b 2 4ac (1)2 4 2 1 1 8 9 Since, D > 0 ∴ The given equation has real and distinct roots, ii) x 2 4 x 4 0 Here, a = 1, b = -4, c = 4

D b 2 4ac ( 4)2 4 1 4 16 16 0

Since, D = 0 ∴ The given equation has real and equal roots. iii) x 2 x 1 0 Here, a = 1, b = 1, c = 1

POLYNOMIALS

D b 2 4ac (1)2 4 1 1 1 4 3 Since, D < 0 ∴ The given equation has no real roots. Example 4: Find the value of k for which the equation x 2 2 x k 0 has equal roots. Solution: The given equation x 2 2 x k 0 Here, a = 1, b = -2, c = k For equal roots, D b 2 4ac 0 (−2)2 − 4( 1 )( k ) = 0 4 − 4k = 0 ⇒ 4k = 4 ⇒ k = 1

Example 5: If -4 is a root of the quadratic equation x 2 px 4 0 and the quadratic equation x 2 + px + k = 0 has equal roots, find the value of k. Solution: Since -4 is a root of the equation x 2 px 4 0 (−4)2 + p ( −4 ) − 4 = 0 ⇒ 16 − 4 p − 4 = 0 ⇒ 12 − 4 p = 0 ⇒ 4 p = 12 ⇒ p=3 The given equation x 2 px k 0 has equal roots. Here x 2 3 x k 0 a 1, b 3, c k

D b 2 4ac 0 (3)2 4 1 k 0 9 4k 0 4k 9 9 k= 4

IL Foundation Series Class 9

QUICK REVIEW •

A polynomial p x in one variable x is an algebraic expression in x of the form p x an x n an 1 x n 1 a2 x 2 a1 x a0 , where a0 , a1 , a2 ,…, an are constants and an ≠ 0 . a0 , a1 , a2 ,…, an are respectively the coefficients of x 0 , x, x 2 ,…, x n , and n is called the degree of the polynomial. Each of an x n , an 1 x n 1 , , a0 , with an ≠ 0 , is called a term of the polynomial p x .

•

A polynomial of one term is called a monomial.

•

A polynomial of two terms is called a binomial.

•

A polynomial of three terms is called a trinomial.

•

A polynomial of degree one is called a linear polynomial.

•

A polynomial of degree two is called a quadratic polynomial.

•

A polynomial of degree three is called a cubic polynomial.

•

A real number 'a' is a zero of a polynomial p x if p a 0. In this case, a is also called a root of the equation p x 0.

•

very linear polynomial in one variable has a unique zero, a non-zero constant polynomial has E no zero, and every real number is a zero of the zero polynomial.

•

emainder theorem: If p x is any polynomial of degree greater than or equal to 1 and p x is R divided by the linear polynomial x − a, then the remainder is p a .

•

actor theorem: x − a is a factor of the polynomial p x , if p a 0 . Also, if x − a is a factor F of p x , then p a 0 .

•

(x y z)2 x 2 y 2 z 2 2 xy 2yz 2 zx

•

(x y)3 x 3 y 3 3 xy x y

•

(x y)3 x 3 y 3 3 xy x y

•

x 3 y 3 z 3 3 xyz x y z x 2 y 2 z 2 xy yz zx

•

elation between LCM and HCF: If the LCM and HCF of two polynomials p x and q x R are l x and h x respectively, then h x l x p x q x .

•

polynomial of degree 2 is called a quadratic polynomial. The general form of a quadratic A polynomial is ax 2 + bx + c , where a, b, c are real numbers such that a ≠ 0 and x is a real variable.

•

I f p x ax 2 bx c, a 0 is a quadratic polynomial and α is a real number, then p a 2 b c is known as the value of the quadratic polynomial p x .

POLYNOMIALS

•

I f p x ax 2 bx c is a quadratic polynomial, then p x 0 i.e., ax 2 bx c 0 , if a ≠ 0 is called a quadratic equation.

•

The roots of the quadratic equation ax 2 + bx + c = 0, a ≠ 0 can be found by using the quadratic b b 2 4ac , provided that b 2 4ac 0 . 2a Nature of the roots of quadratic equation ax 2 bx c 0, a 0 depends upon the value of D b 2 4ac, which is known as the discriminant of the quadratic equation. formula

• •

The quadratic equation ax 2 bx c 0, a 0 has: i) Two distinct real roots, if D b 2 4ac 0. ii) Two equal roots, i.e. coincident real roots if D b 2 4ac 0. iii) No real roots, if D b 2 4ac 0.

WORKSHEET - 1 I.

POLYNOMIALS TERMINOLOGY AND ZERO OF A POLYNOMIAL 1. Which one of the following is a polynomial? 2

x 2 − 2 ii) 2 x − 1 2 x 2. Find the degree of the polynomial

iii) x +

3 3x 2

iv)

x 1 x 1

3. Find the degree of the polynomial 4 x 4 0 x 3 0 x5 5 x 7 . 4. Find the degree of the zero polynomial.

5. If p x x 2 2 2 x 1, then find the value of p 2 2 . 2

6. Find the value of the polynomial 5 x 4 x 3 at x 1. 7. If p x x 3, then find the value of p x p x . 8. Find zero of the polynomial p x 2 x 5. 9. Find the zeroes of the polynomial 2 x 2 7 x 4 . 10. Which of the following expressions are polynomials? Justify your answer: 1 5x 7 i) 3 x 2 − 2 x ii) 1 − 5 x iii) 5 x 2 iv)

x 2 x 4 x

1 1+ x

vi)

1 3 2 2 a a 4a 7 7 3

IL Foundation Series Class 9

x3 2x 1 7 2 11. For the polynomial x x 6 , write; 5 2 i) The degree of the polynomial ii) The coefficient of x 3 iii) The coefficient of x 6 iv) The constant term 12. Classify the following as a constant, linear, quadratic and cubic polynomials: i) 2 x 2 x 3 ii) 3 x 3 iii) 5t − 7 2

iv) 4 − 5y v) 3

vi) 2 + x

13. Find the value of the polynomial 3 x 3 4 x 2 7 x 5, when x = 3, and when x 3. 1 14. If p x x 2 4 x 3, evaluate: p 2 p 1 p . 2 15. Find the zeroes of the polynomial in each of the following: i) p x x 4 ii) g x 3 6 x iii) q x 2 x 7 iv) h y 2y

16. Find the zeroes of the polynomial: p x (x 2)2 (x 2)2. 17. Verify whether the following are zeroes of the polynomial indicated against them. i) p(x) = x2 - 1, x = - 1

ii) p(x) = (x + 1) (x - 2), x = - 1,2

iii) p(x) = x2, x = 0 II. FACTORISATION OF POLYNOMIALS 1. Use the factor theorem to determine whether g(x) is a factor of f(x)in the following case: f(x)= 4x3 + 20x2 + 33x + 18, g(x) = 2x + 3

2. Determine whether x 1 is a factor of x 3 x 2 2 2 x 2 . 3. Factorise: x 3 23 x 2 142 x 120 4. Find the remainder when f x 4 x 4 3 x 3 2 x 2 x 7 is divided by: 2 i) 2 x − 1 ii) x 3 5. Find the value of 'p' for which the polynomial 2 x 4 + 3 x 3 + 2 px 2 + 3 x + 6 is divisible by x 2 . 6. Without actual division, prove that x 4 2 x 3 2 x 2 2 x 3 is exactly divisible by x 2 2 x 3. 61

POLYNOMIALS

7. If f x x 2 5 x p and g x x 2 3 x q have a common factor, then: ii) Show that, ( p q)2 2 3 p 5q .

i) Find the common factor.

8. Find the remainder when x3 + 3x2 + 3x + 1 is divided by: i) x + 1 ii) x + π iii) 5 + 2x 9. Find the remainder when x3 - ax2 + 6x - a is divided by x - a. 10. Check whether 7 + 3x is a factor of 3x3 + 7x. 11. Determine which of the following polynomials has x 1 as a factor. i) x 3 + x 2 + x + 1

ii) x 4 + x 3 + x 2 + x + 1

iii) x 4 + 3 x 3 + 3 x 2 + x + 1

iv) x 3 x 2 2 2 x 2

12. Use the factor theorem to determine whether g x is a factor of p x in each of the following cases: i) p x 2 x 3 x 2 2 x 1, g x x 1 ii) p x x 3 3 x 2 3 x 1, g x x 2 iii) p x x 3 4 x 2 x 6, g x x 3 13. Find the value of k, if x −1 is a factor of p x in each of the following cases: i) p x x 2 x k

ii) p x 2 x 2 kx 2

14. Factorise the following:

i) 12 x 2 7 x 1 ii) 2 x 2 + 7 x + 3 iii) 6 x 2 5 x 6

iv) 3 x 2 − x − 4

15. Factorise the following: i) x 3 2 x 2 x 2

ii) x 3 − 3 x 2 − 9 x − 5

iii) x 3 + 13 x 2 + 32 x + 20

iv) 2y 3 y 2 2y 1

16. If x + 1 is a factor of the polynomial 2 x 2 + kx , then find the value of k. 17. Using the remainder theorem, factorise 2 x 3 13 x 2 26 x 15. 18. If x − a is the factor of 3 x 2 - mx - na , then prove that a

m n . 3

19. Show that 2 x + 1 is a factor of 2 x 3 − 11x 2 − 4 x + 1. 20. If the polynomials az 3 4 z 2 3 z 4 and z 3 4 z a leave the same remainder when divided by z − 3 , find the value of a.

IL Foundation Series Class 9

21. The polynomial p x x 4 2 x 3 3 x 2 ax 3a 7 when divided by x + 1 leaves the remainder 19. Find the value of a. Also, find the remainder when p x is divided by x + 2. 1 22. If both x − 2 and x − are factors of px 2 + 5 x + r, show that p = r . 2 23. Without actual division, prove that 2 x 4 5 x 3 2 x 2 x 2 is divisible by x 2 3 x 2 . III. LONG DIVISION AND HORNER’S SYNTHETIC DIVISION OF POLYNOMIALS 1. Verify the division algorithm when: i) x 4 3 x 3 x 2 x 6 is divided by x 3. ii) 2 x 4 5 x 3 2 x 2 5 x 10 is divided by x 2. iii) 7 x 4 10 x 3 3 x 2 3 x 3 is divided by x 1. iv) 2 x 4 8 x 3 5 x 2 4 x 2 is divided by x 2 4 x 2. v) 3 x 4 x 3 8 x 2 5 x 3 is divided by x 2 x 3. 2. Use polynomial division to simplify each of the following quotients. 3x4 9x3 5x2 6x 2 i) 3x2 2 x3 4x2 9 x 3 3. Factorise the following using synthetic division method. iii)

x3 − 2x2 − 4 ii) x −2 iv)

x 4 − 13 x − 42 x2 − x − 6

i) 2 x 3 11x 2 17 x 6 ii) 2 x 3 x 2 15 x 18 iii) 2 x 4 3 x 3 4 x 2 3 x 2 IV. PROBLEMS BASED ON ALGEBRAIC IDENTITIES 1. If x y z 0 , show that x 3 y 3 z 3 3 xyz. 2. Factorise 27 x 3 y 3 z 3 9 xyz. 3. Factorise each of the following: i) 27 y 3 + 125 z 3

ii) 64m3 − 343n 3

4. Factorise each of the following: i) 8a 3 + b 3 + 12a 2b + 6ab 2

ii) 27 125a 3 135a 225a 2

iii) 64a 3 27 b 3 144a 2b 108ab 2

iv) 27 p3

1 9 2 1 p p 216 2 4

5. Evaluate the following using suitable identities: i) (99)3 ii) (102)3 iii) (998)3 63

POLYNOMIALS

6. Expand each of the following using a suitable identity. i) (x + 2y + 4 z)2 ii) (2 x y z)2 iii) ( 2 x 3y 2 z)2 7. Evaluate the following products without multiplying directly. i) 103 × 107 ii) 95 × 96

iii) 104 × 96

8. Use suitable identities to find the following products. i) x 4 x 10

ii) x 8 x 10

iii) 3 x 4 3 x 5

9. If a b c 7 and ab + bc + ca = 20, find the value of a 2 + b 2 + c2 . 10. Factorise: i) 25 x 2 10 xy y 2 z 2

ii) x 2 4y 2 4y 1

iii) x 2 y 2 z 2 2 xy

iv) x 2 z 2 y 2 p2 2 pz 2 xy v) (ac bd)2 (ad bc)2

11. Prove that (a b)3 (b c)3 (c a)3 3 a b b c c a 2 a 3 b 3 c3 3abc . 1 1 1 1 12. Find the product a a a 2 2 a 4 4 using a suitable identity. a a a a 13. Factorise: i) a 4 + 4a 2 + 3 ii) x 8 x 4 132 14. Prove that (a + b + c)3 − a 3 − b 3 − c3 = 3 ( a + b ) ( b + c ) ( c + a ). 15. If a b c 5 and ab + bc + ca = 10, then prove that a 3 b 3 c3 3abc 25. 16. If a, b, c are all non-zero and a b c 0, prove that

a 2 b 2 c2 + + = 3. bc ca ab

17. Simplify (2 x 5y)3 (2 x 5y)3. V. HCF AND LCM OF POLYNOMIALS 1. Find the highest common factor (HCF) and lowest common multiple (LCM) of the two polynomials: i) a 3 2a 2 3a and 2a 3 5a 2 3a ii) 4u 2 − 9v 2 and 2u 2 − 3uv

iii) 4u 2 25v 2 and 6u 2 15uv

2. Find the LCM of pq - np, pq - mq, q - 3nq + 2n , pq - 2np - mq + 2mn and pq np mq mn . 3. The HCF of two polynomials p x and q x is 2 x x 2 and LCM is 24 x(x + 2)2 ( x − 2 ). If p ( x ) = 8 x 3 + 32 x 2 + 32 x, then find q ( x ).

IL Foundation Series Class 9

4. If x 6 is the HCF of x 2 − 2 x − 24 and x 2 − kx − 6, then what is the value of k ? 5. Find the HCF of 4 x 3 3 x 2 y 9 xy 2 2y 3 and x 2 xy 2y 2 . 6. The HCF of the two numbers is 98 and their LCM is 2352. Find the sum of the numbers. VI. QUADRATIC EQUATION AND ITS SOLUTION

1 1. The sum of the reciprocals of Rehman’s ages (in years) 3 years ago and 5 years from now is . 3 Find his present age. 2. A piece of cloth costs ` 200. If the piece was 5 m longer and each metre of cloth costs ` 2 less, the cost of the piece would have remained unchanged. How long is the piece, and what is the original rate per metre? 3. Solve the quadratic equation by factorisation method.

4 5 −3 −3 = , x ≠ 0, x 2x + 3 2

4. Determine the values of ' k ' for which the quadratic equation kx 2 5 x k 0 has equal roots. 5. If -4 is a root of the quadratic equation x 2 px 4 0 and the quadratic equation x 2 + px + k = 0 has equal roots, find the value of k. 6. Find the value of k for which the equation x 2 5 kx 16 0 has no real roots. 7. Find the discriminant of the equation 3 x 2 4 3 x 4 0 and find the nature of its roots. 8. If b = 0, c < 0, is it true that the roots of x 2 bx c 0 are numerically equal and opposite in sign? Justify your answer. 9. Is 0.2 a root of equation x 2 0.4 0 ? Justify your answer. 10. At t minutes past 2 p.m., the time needed by the minute hand of a clock to show 3 p.m. was t2 found to be 3 minutes less than minutes. Find t. 4

WORKSHEET - 2 I. MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. Degree of polynomial p x 4 x 4 2 x 3 x5 2 x 7 is: a) 7

b) 4

c) 5

d) 3

2. If x 1 is a factor of p x x 2 x k , then the value of k is: a) 3

b) 2

c) -2

d) 1

3. Degree of zero polynomial is: a) 0

b) 1

c) any natural number d) not defined

POLYNOMIALS

4. The coefficient of x 2 in 3 x 2 5 a) 12

4 4x is: 2

b) 5

c) -8

d) 8

5. The remainder when x 31 + 31 is divided by x + 1 is: a) 30

b) 31

c) -1

x y 3 3 + = −1, x ≠ 0, y ≠ 0, then the value of x − y is: y x 1 a) 1 b) -1 c) + 2

d) 0

6. If

d) 0

7. One of the factors of 16y 2 1 (1 4y)2 is: a) 4 + y

b) 4 − y

8. Zero of the polynomials a) 0

c) 4y + 1

d) 8y

4 11

11x 1 is: 4

4 11

c) −

11 4

9. Among the following, the cubic polynomial is: a) x 3 3 x 2 4 x 3

b) x 2 4 x 7

c) 3 x 2 + 4

d) 3 x 2 x 1

1 10. If p t t 2 t 2, then p is: 3 a)

22 9

14 9

16 9

15 9

11. Among the following, a polynomial in one variable is: a) 3 x 2 x

3x + 4

c) x 3 + y 3 + 7

d) x +

1 x

12. a 2 b 2 c2 ab bc ca a) (a + b + c)2

b) (a − b − c)2

c) (a b c)2

1 (a b)2 (b c)2 (c a)2 2

13. Among the following, the polynomial that has -3 as a zero is: a) x 3

b) x 2 − 9

c) x 2 − 3 x

d) x 2 + 3

IL Foundation Series Class 9

x x3 2 14. If p x 2 x , then p 1 is: 2 3 a)

15 6

17 6

13 6

1 15. The coefficient of x 2 in 3 x x 3 x is: x a) 3 b) 1 1

1 6

c) 4

d) 2

16. If x 3 y 3 z 3 0, then: a) x y z 0

1 1 1 b) x y z 3 x 3 y 3 z 3

c) x y z 3 xyz

d) x 3 y 3 z 3 3 xyz

17. If x 2 is a factor of 2 x 3 5 x 2 x k , then k is: a) 6

b) -24

c) -6

d) 24

18. The product of factors of (2a b)3 (b 2c)3 8(c a)3 is: a) 2a b b 2c c a

b) 3 2a b b 2c c a

c) 6 2a b b 2c c a

d) 2a × b × 2c

19. x 2 is a factor of: a) 4 x 3 13 x 6

b) x 3 + x 2 + x + 4

c) 4 x 3 13 x 25

d) 2 x 3 x 2 x 19

c) x −1

d) x + 4

20. One of the factors of x 1 x 2 1 is: a) x 2 − 1

b) x + 1

21. The remainder obtained when the polynomial p x is divided by b ax is: a a) p b

b b) p a

b c) p a

a d) p b

22. If 3 x 1 is a factor of the polynomial 81x 3 45 x 2 3a 6 , then a is: a)

8 3

−7 3

−10 3

11 3

23. If x 2 kx 6 x 2 x 3 then, the value of k is: a) 1

b) 2

c) 5

d) 3

POLYNOMIALS

24. Among the following, the polynomial having the degree as zero is: a) x

b) 1

d) x 2

c) 0

25. If the volume of a cuboid is 3 x 2 − 27, then its possible dimensions are: a) 3, x 2 , −27 x

b) 3, x 3, x 3

c) 3, x 2 , 27 x

d) 3, 3, 3

26. If the equation x 2 4 x k 0 has real and distinct roots, then: a) k < 4

b) k > 4

c) k ≥ 4

d) k ≤ 4

27. If the equation x 2 ax 1 0 has distinct roots, then: a) a = 2

b) a < 2

c) a > 2

d) None of these

28. If the sum of the roots of the equation x 2 x 2 x 1 is zero, then a) -2

c) −

b) 2

1 2

29. The positive value of k for which the equation x 2 kx 64 0 and x 2 8 x k 0 will both have real roots, is _______. a) 4

b) 8

c) 12

d) 16

30. If the roots of the equation a 2 b 2 x 2 2b a c x b 2 c2 0 are equal, then: a) 2b a c

b) b 2 = ac

c) b

2ac a c

II. FILL IN THE BLANKS 1. The degree of 2020 is __________. 2. If x p is a factor of x 2 px 3 p , then p = ___________. 3. Zero of the polynomials p x ax, a 0 is ___________. 4. The zeroes of the polynomial 3 x 2 x 2 is ___________. 5. If p x ax 3 x 2 x 4 and p 1 0 , then a = ______________. 6. If p x x 3 , then the value of p x p x _____________. 7. The coefficient of x in (2 x − 3)3 is ____________. 8. If p x x a q x , then the deg of q x is ______________. 9. If f x 3 x 2 , then the value of f 2 is _____________.

d) b = ac

IL Foundation Series Class 9

10. Product of the roots of the equation 13 x 2 x 5 is __________. 11. If one root of x 2 p 1 x 10 0 is 5, then the value of p is _________. 12. If

x 9 4 , then the value of x is ______________. 3 x

13. The ratio of the roots of the a1 x 2 b1 x c1 0 be equal to the ratio of roots of a2 x 2 b1 x c2 0 , then __________. 14. The roots of the quadratic equation ax 2 bx 0 are __________.

15. x 2 1

x 0 has _______ real roots. 2

16. If the discriminant of the equation 6 x 2 bx 2 0 is 1 , then the value of b is _______. 17. The value of k for which roots of the quadratic equation kx 2 2 x 3 0 are equal is _________.

18. The roots of x 2 2 x r 2 1 0 are ____________. 19. If 3 is a solution of 3 x 2 k 1 x 9 0 , then k = __________. 20. If ax 2 bx c 0 has equal roots, then c is ____________. 21. The value of c for which the equation ax 2 2bx c 0 has equal roots is ________________. 22. If the quadratic equation kx x 2 6 0 has two equal roots, then the value of k is _______. 23. The common root of the equation x 2 7 x 10 0 and x 2 10 x 16 0 is ______________. 24. The quadratic equation whose one of the roots is 3 + 2 5 is ________________. III. SUBJECTIVE QUESTIONS 1. If a b 7 and ab = 12 , find the value of a 2 + b 2 . 2. If x

4 1 1 , then find the value of 4 x 2 + 2 . x 2 x

3. If a + b + c = 0 , then write the value of

a 2 b 2 c2 + + . bc ca ab

4. If a b c 9 and ab bc ca 40 , find a 2 + b 2 + c2 . 5. Find the value of k if x 3 is a factor of k2 x 2 − kx − 2 . 6. Factorise: 5 xy − 15 x 3 y 3 .

POLYNOMIALS

7. Evaluate: (999)2 − 12 . 8. Find the zeroes of the polynomial: x 1 x 2 x 3 x 4 . 3

3 9. Expand: x 1 . 2 10. Check whether, x 1 is a factor of x 3 + x + x 2 + 1 . 11. If x = 3 is one root of the quadratic equation x 2 2 kx 6 0 , then find the value of k . 12. If the quadratic equation px 2 2 5 px 15 0 has two equal roots, then find the value of p. 13. Check whether 2 x 1 x 3 x 5 x 1 is a quadratic equation or not. 14. The product of two consecutive positive integers is 306. Represent this situation in the form of quadratic equation. 15. Check whether -3 is a solution of the equation 3 x 2 + 5 x + 2 = 0. 16. Write the condition to be satisfied by a , b , and c so that the quadratic equation ax 2 bx c 0 has two distinct real roots. 17. If one root of the equation 3 x 2 px 4 0 is

2 , then find the value of p. 3

18. Write the discriminant for the quadratic equation 2 x 2 − 3 x + 5 = 0. 19. Determine the nature of the roots of the quadratic equation 2 x 2 − 6 x + 3 = 0. 20. If

1 5 is a root of the equation x 2 kx 0 , then find the value of k. 2 4

COORDINATE GEOMETRY

3.1 CARTESIAN COORDINATE SYSTEM A French mathematician, René Déscartes (1596-1650), made a Y significant progress by introducing 'Analytical Geometry'. He is 4 accredited as the father of analytical geometry. Analytical geometry 3 Q (+,+) Q2(-,+) is also called Cartesian geometry. It is a method of studying 2 1 geometry using the coordinate system. Coordinate geometry allowed 1 ’ X X geometrical shapes to be expressed in algebraic equations. -4 -3-2-1O 1 2 3 4 -1 We draw a horizontal number line and a vertical number line -2 Q3(-,-) -3 Q4(+,-) perpendicular to the horizontal line on the graph paper. The point of intersection of the two lines is called origin, denoted by 'O'.

-4

Y’

Note: Here, a unit of length is chosen as 1 cm; from the origin, the positive integers, 1, 2, 3, ... are marked towards the right side of the origin and negative integers, -1, -2, -3, ... are marked towards the left side of the origin on the horizontal line. Similarly, the positive integers 1, 2, 3, ... are marked above the origin, and the negative integers -1, -2, -3, ... are marked below the origin or on the vertical line. The horizontal line is known as the x-axis and is denoted by XOX'. The vertical line is known as the y-axis and is denoted by YOY'. Here, OX is known as the positive direction of the x-axis. OX' is known as the negative direction of the x-axis. OY is known as the positive direction of the y-axis. OY' is known as the negative direction of the y-axis. Thus, the x-axis and y-axis represent the real number line. Important points: Let XOX' and YOY' be the coordinate axes. These two axes divide the XY-plane into four parts, namely XOY, YOX', X'OY', and Y'OX , and each part is called quadrant Q1, Q2, Q3, and Q4 , respectively. In the first quadrant, the x-coordinate is + ve and the y-coordinate is + ve. In the second quadrant, the x-coordinate is - ve and the y-coordinate is + ve. In the third quadrant, the x-coordinate is - ve and the y-coordinate is - ve. In the fourth quadrant, the x-coordinate is + ve and the y-coordinate is - ve. 71

COORDINATE GEOMETRY

Abscissa: The abscissa refers to the horizontal distance of a point from the vertical axis or the y-axis in a Cartesian coordinate system. It is often denoted by the letter 'x'. In simpler terms, the abscissa tells us how far a point is located to the right or left of the origin along the x-axis. Ordinate: The ordinate, on the other hand, refers to the vertical distance of a point from the horizontal axis or the x-axis in a Cartesian coordinate system. It is often denoted by the letter 'y'. The ordinate tells us how far a point is located above or below the origin along the y-axis. Together, the abscissa and the ordinate of a point uniquely determine its location in the Cartesian plane. Example: In the point (3, 4), the abscissa is 3 (the distance from the y-axis) and the ordinate is 4 (the distance from the x-axis). The coordinates of any point on the x-axis are of the form (x, 0). The coordinates of any point on the y-axis are of the form (0, y).

3.2 DISTANCE FORMULA Distance formula: The distance between any two points in the plane is the length of the line segment joining them. • Distance between two points P ( x1 , y1 ) and Q ( x2 , y2 ) is given by | PQ |= =| • Distance of any point P(x, y) from the origin O(0, 0) given by | OP

( x2 − x1 ) + ( y2 − y1 ) . 2

x2 + y2 .

Points to remember: In order to prove the given points, form a figure of a: i)

Square, if the four sides are equal and the diagonals are also equal.

ii) Rhombus, if the four sides are equal and the diagonals are not equal. iii) Rectangle, if the opposite sides are equal and diagonals are equal. iv) Parallelogram, if opposite sides are equal and diagonals are not equal. v) Right-angled triangle, if the sum of squares of any two sides is equal to the square of the third side. Example: Show that A(6, 4), B(5, –2), and C(7, –2) are the vertices of an isosceles triangle. Also, find the length of the median through A.

IL Foundation Series Class 9

Solution: We have, A(6, 4)

B(5, -2)

D(6, -2)

C(7, -2)

(5 − 6) 2 + (−2 − 4) 2

AB =

(−1) 2 + (−6) 2 =

1 + 36

= 37 units = AC =

(7 − 6) 2 + (−2 − 4) 2 (1) 2 + (−6) 2 =

1 + 36

= 37 units AC ∴ AB = So, ∆ABC is isosceles. Let D be the mid-point of BC.  5 + 7 (−2) + (−2)   12 −4  = Then, the coordinates of D are  ,   , =  (6, − 2) 2 2   2   2 ∴ AD = (6 − 6) 2 + (−2 − 4) 2 2 = 02 + (−6)= = 6 units

Example: If P(2, –1), Q(3, 4), R(–2, 3), and S(–3, –2) are four points in a plane, show that PQRS is a rhombus but not a square. Solution: The given points are, P(2, –1), Q(3, 4), R(–2, 3) and S(–3, –2).

COORDINATE GEOMETRY

PQ =

(3 − 2) 2 + (4 + 1) 2 = 1 + 25 =

26 units

QR =

(−2 − 3) 2 + (3 − 4) 2 =

25 + 1 =

26 units

RS =

(−3 + 2) 2 + (−2 − 3) 2 =

1 + 25 =

26 units

SP =

(−3 − 2) 2 + (−2 + 1) 2 =

25 + 1 =

26 units

PR =

(−2 − 2) 2 + (3 + 1) 2 =

16 + 16 =

32 = 4 2 units

And QS =

(−3 − 3) 2 + (−2 − 4) 2 =

36 + 36 =

72 = 6 2

\ PQ = QR = RS = SP = 26 units and PR ¹ QS . This means that PQRS is quadrilateral, whose sides are equal, but the diagonals are unequal. Thus, PQRS is a rhombus but not a square.

3.3 SECTION FORMULA Section formula: Let A and B be two points in the plane, and P be a point on the segment joining A and B such that AP : PB = m : n. Then, the point P divides the segment AB internally in the ratio m : n. n

m A

The coordinates of the point which divides the line segment joining the points (x1, y1) and (x2, y2) internally in the ratio m : n are given by  mx + nx1 my2 + ny1  P( x, y ) =  2 ,  m+n   m+n

If P is a point on AB produced such that AP : PB = m : n, then point P is said to divide AB externally in the ratio m : n. m A

P n

The coordinates of the point that divides the line segment joining the points (x1, y1) and (x2, y2) externally in the ratio m : n are given by  mx − nx1 my2 − ny1  P( x, y ) =  2 ,  m−n   m−n

IL Foundation Series Class 9

Note: 1) If P is the midpoint of AB, then it divides AB in the ratio 1 : 1, so

 x +x y +y  P= 1 2 , 1 2  2   2 A

B (x2, y2)

(x1, y1)

m m : 1 or λ :1 where λ = . Then, n n  λ x2 + x1 λ y2 + y1  P= , λ + 1   λ +1

2) The ratio m : n can also be written as

 x + x + x3 y1 + y2 + y3  3) Centroid formula, G =  1 2 ,  3 3   (x1, y1)

(x2, y2)

(x3, y3)

Example: Find the coordinates of points which trisect the line segment joining (1, –2) and (–3, 4). Solution:

A (1, -2)

B (-3, 4)

Let the points of trisection be P and Q, then AP = PQ = QB = λ (say). ∴ ⇒

PB = PQ + QB = 2λ AQ = AP + PQ = 2λ AP : PB = λ : 2λ = 1: 2 AQ= : QB 2= λ : λ 2 :1

So, P divides AB internally in the ratio 1 : 2, while Q divides internally in the ratio 2 : 1. Thus, the coordinates of P and Q are

COORDINATE GEOMETRY

 mx + nx1 my2 + ny1  P= 2 ,  m n m+n  +   1× (−3) + (2 ×1) (1× 4) + (2 × −2)  = ,  1+ 2 1+ 2    −3 + 2 4 − 4   −1  , =   , 0 3   3   3  mx + nx1 my2 + ny1  , Q= 2  m+n   m+n  (2 × −3) + (1×1) (2 × 4) + (1× −2)  , =  2 +1 2 +1    −6 + 1 8 − 2   −5  , =   , 2 3   3  3   −1  −5 ∴ The two points of trisection are  , 0  and  , 2  .  3   3  Example: Determine the ratio in which the line 3x + y - 9 = 0 divides the line segment joining A(1, 3) and B (2, 7). Solution: Suppose the line 3x + y - 9 = 0 divides the line segment joining A(1, 3) and B(2, 7) in the ratio k : 1 at point C. Then, the coordinates of C are:  2k + 1 7 k + 3   k + 1 , k + 1  But, C lies on 3x + y - 9 = 0  2k + 1  7 k + 3  ∴3  + −9 = 0  k + 1   k + 1  0 ⇒ 3(2 k + 1) + (7 k + 3) − 9( k + 1) = 0 ⇒ 6k + 3 + 7 k + 3 − 9k − 9 = 0 ⇒ 4k − 3 = 3 ⇒ 4k = 3 ⇒k= 4

IL Foundation Series Class 9

3.4 AREA OF TRIANGLES AND QUADRILATERALS Area of a triangle: The area of a triangle, the coordinates of whose vertices are ( x1 , y1 ) , ( x2 , y2 ) and 1 ( x3 , y3 ) is x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) sq. units. 2 Note: Three given points will be collinear if the area of the triangle formed by these points is zero Area of quadrilateral: The area of the quadrilateral can be found by splitting up the quadrilateral into two triangles and the sum of their areas. Thus, area of the quadrilateral ABCD = area ( ∆ABC ) + area ( ∆ADC ) D

Note: The area of the quadrilateral formed by the vertices ( x1 , y1 ) , ( x2 , y2 )( x3 , y3 ) , and ( x4 , y4 ) is 1 given by the formula x1y2 − x2 y1 + x2 y3 − x3 y2 + x3 y4 − x4 y3 + x4 y1 − x1y4 sq. units. 2 Example: Find the area of a triangle whose vertices are A(3, 2), B(11, 8) and C(8, 12). Solution: Let = A

x1 , y1 ) (3,= 2); B ( x= (11,= 8); C ( x= (8, 12) . (= 2 , y2 ) 3 , y3 ) 1 x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) 2 1 = | 3(8 − 12) + 11(12 − 2) + 8(2 − 8) | 2 1 = | (−12 + 110 − 48) | 2 1 = × 50 =25 sq. units. 2

= Area of ∆ABC

Example: Prove that the points (a, b + c), (b, c + a), and (c, a + b) are collinear. Solution: Let A =( x1 , y1 ) =(a, b + c); B =( x2 , y2 ) =(b, c + a ); C =( x3 , y3 ) =(c, a + b ) .

COORDINATE GEOMETRY

Area of the triangle ABC =

1 x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) 2

1 x1y2 + x2 y3 + x3 y1 − ( x1y3 + x2 y1 + x3 y2 ) 2 1 | a (c + a ) + b (a + b ) + c(b + c) − {a (a + b ) + b (b + c) + c(c + a )}| = 2 1 = ac + a 2 + ab + b 2 + cb + c 2 − a 2 − ab − b 2 − bc − c 2 − ca 2 1 = | 0 | = 0 sq. units. 2 =

Hence, the points are collinear.

SOLVED EXAMPLES Example 1: If the points (p, q), (m, n), and (p – m, q – n) are collinear, then show that pn = qm. Solution: If the given points are collinear, then the area of the triangle formed by the given points is zero. 1 x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) = 0 2 ⇒ p {n − ( q − n )} + m ( q − n − q ) + ( p − m )( q − n ) = 0 i.e.

⇒ p ( n − q + n ) + m ( −n ) + ( p − m )( q − n ) = 0 0 ⇒ p ( 2n − q ) − mn + ( p − m )( q − n ) = 0 ⇒ 2 pn − pq − mn + pq − pn − mq + nm = ⇒ pn − mq = 0 ⇒ pn − mq = 0 qm ⇒ pn = Example 2: For what values of k are the points ( k, 2 − 2 k ) , ( − k + 1, 2 k ) , and ( −4 − k, 6 − 2 k ) collinear? Solution: The given points will be collinear if the area of the triangle formed by them is zero. Let A ( k, 2 − 2 k ) , B ( − k + 1, 2 k ) , and C ( −4 − k, 6 − 2 k ). 1 ( x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) 2 1 ABC k{2 k − (6 − 2 k)} + (− k + 1){(6 − 2 k) − (2 − 2 k)} + (−4 − k)(2 − 2 k − 2 k) ⇒ Area of ∆= 2

∴ Area of ∆ABC =

IL Foundation Series Class 9

Since A, B, and C are collinear, then the area of ∆ABC = 0. 1 2 8k + 4 k − 4 = 0 2 0 ⇒ 8k 2 + 4 k − 4 = 2 ⇒ 2 k + k − 1 =0 0 ⇒ 2 k ( k + 1) − ( k + 1) = 0 ⇒ ( k + 1)( 2 k − 1) = k + 1 0 or 2= k −1 0 ⇒ = 1 ⇒ k= −1 or k = 2 Example 3: If A ( 5, − 1) , B ( −3, − 2 ) , and C ( −1, 8 ) are the vertices of triangle ABC, find the length of the median through A and the coordinates of the centroid.

∴

Solution: Let AD be the median through the vertex A of ∆ABC . A(5, -1) 2 G 1 B

(-3, -2)

D(-2, 3)

C (-1, 8)

Thus, D is the mid-point of BC. −3 − 1 −2 + 8  So, the coordinates of D are  , . 2   2 That is, D = (-2, 3) ∴ AD= =

(5 + 2) 2 + (−1 − 3) 2 49 + 16 =

65 units

et G be the centroid of ∆ABC . Then, G lies on the median AD and divides it in the ratio 2: 1. So, L the coordinates of G are  2 × −2 + 1× 5 2 × 3 + 1× −1  ,   2 +1 2 +1    −4 + 5 6 − 1  , =  3   3 1 5 = ,  3 3 79

COORDINATE GEOMETRY

Example 4: The coordinates of one endpoint of the diameter of a circle are (4, -1), and the coordinates of the centre of the circle are (1, -3). Find the coordinates of the other end of the diameter. Solution: Let AB be the diameter of the circle having its centre at C(1, -3) such that the coordinates of the one end A are (4, -1). B (x, y)

C (1, -3)

A (4, -1)

Let the coordinates of B be (x, y). x + 4 y −1  Since C is the mid-point of AB, then the coordinates of C are  , . 2   2 We know that the coordinates of C are (1, -3). x+4 y −1 = = −3 1 and 2 2 ⇒ x + 4 =2 and y − 1 =−6 ⇒x= −2 and y = −5

∴

Hence, the coordinates of B are (-2, -5). Example 5: Point P divides the line segment joining the points A(2, 1) and B(5, -8) such that AP 1 = . If P lies on the line 2x - y + k = 0, find the value of k. AB 3 Solution: We have, AP 1 = AB 3 AP 1 ⇒ = AP+PB 3 ⇒ 3AP =AP + PB BP ⇒ 2AP = AP 1 ⇒ = BP 2 80

IL Foundation Series Class 9

1 A(2, 1)

2 P

B(5,-8)

So, P divides AB in the ratio 1 : 2. ∴, the coordinates of P are  1× 5 + 2 × 2 1× −8 + 2 × 1  = ,  1+ 2  1+ 2   5 + 4 −8 + 2   9 −6  , =   ,  3  3 3   3 = ( 3, −2 ) Since P(3, -2) lies on the line, 2x - y + k = 0 0 ⇒ 2 ( 3 ) - ( -2 ) + k = ⇒ 6+2+ k = 0 ⇒8+k = 0 ⇒k= -8

QUICK REVIEW • The coordinates of any point on the x-axis are of the form (x, 0). • The coordinates of any point on the y-axis are of the form (0, y). • The distance between points P(x1, y1) and Q(x2, y2) is given by: PQ=

( x2 − x1 ) + ( y2 − y1 ) 2

• Distance of a point P(x, y) from the origin O(0, 0) is given by: OP =

x2 + y2

• The coordinates of the point which divides the joining of points P(x1, y1) and Q(x2, y2) internally in the ratio m : n are  mx2 + nx1 my2 + ny1  ,   m+n   m+n • The coordinates of the mid-point of the line segment joining the points (x1, y1) and (x2, y2) are  x1 + x2 y1 + y2   2 , 2 .  

COORDINATE GEOMETRY

• The coordinates of the centroid of the triangle formed by the points A ( x1 , y1 ) , B ( x2 , y2 ), and  x + x2 + x3 y1 + y2 + y3  C(x3, y3) are  1 , . 3 3   • The area of the triangle formed by the points A ( x1 , y1 ) , B ( x2 , y2 ), and C ( x3 , y3 ) is 1 x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) 2 • If points A ( x1 , y1 ) , B ( x2 , y2 ) and C(x3, y3) are collinear, then 1 x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) = 0 2

WORKSHEET - 1 I.

CARTESIAN COORDINATE SYSTEM 1. Plot the following points. Join them in order and identify the figure, PQRS thus obtained: P (1, 1) , Q ( 4, 2 ) , R ( 4, 8 ) , S (1, 10 ) . Write the mirror image of point P on the x-axis and y-axis. 2. Plot the points A, B, C, and D from the table: Point

-5

-4

-5

-16

Answer the following: a) Write the coordinates of A, B, C, and D. b) Shade the triangle ABC 3. Write the coordinates of the vertices of a rectangle whose length and breadth are 4 units and 3 units, respectively, and has one vertex at the origin, the longer side is on the x-axis, and one of the vertices lies in the IV quadrant. Also, find its area. 4. Three vertices of a rectangle are ( −1, 1) , ( 5, 1) , and (5, 3). Plot these points on graph paper and find the coordinates of the 4 vertices. 5. Three vertices of a square PQRS are P(-4, 0), Q(1, -5), and R(1, -5). Plot the points. Also, find the coordinates of the missing vertex S. 6. Plot the following points, join them in order and identify the figure thus formed: A (1, 3) , B (1, − 1) , C ( 7, − 1) , and D ( 7, 3) Write the coordinates of the point of intersection of the diagonals. 7. Plot the points O ( 0, 0 ) , B ( 3, 0 ) and D ( 0, 3) on graph paper. Complete the square OBCD. Find the coordinates of point C. Also, find the area of the square OBCD.

IL Foundation Series Class 9

8. Plot the points A (1, 4 ) , B ( −2, 2 ) , and C ( 3, 2 ) on graph paper. Join them in order, and identify the figure so obtained. Also, plot the mirror images of A, B, and C on the x-axis. 9. Plot the points A, B, C, and D, where: i) A lies on the x-axis and is at a distance of 2 units to the left of the origin. ii) B lies on the y-axis and is at a distance of 4 units above the origin. iii) C lies on both the x-axis and y-axis. iv) D lies on the second quadrant at a distance of 3 units from the x-axis and 2 units from the y-axis. 10. Plot the points A(4, 4) and B(-4, 4). Join the lines OA, OB, and BA. Name the figure formed. Give the mirror image of A and B on the x-axis. 11. Three vertices of a rectangle are A (1, 3) , B (1, − 1) , and C ( 7, − 1) . Plot these points and find the coordinates of the fourth vertex. Also, find its area. 12. Plot the points A(2, 0), B(2, 2), C(0, 2), and O(0, 0), and draw the line segments OA, AB, BC, and CO. What figure do you obtain? II.

DISTANCE FORMULA

1. Find the distance between the following pairs of points: i) (2, 3), (4, 1) ii) (-5, 7), (-1, 3) iii) (a, b), (-a, -b) iv) {( a + b ) , ( a − b )} , {( a − b ) , ( −a − b )}

( a sin α , b cos α ) and ( −a cos α , b sin α ) 2. Determine if the points (1,5 ) , ( 2,3 ) , and (-2, -11) are collinear.

3. Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle? 4. Find the point on the y-axis which is equidistant from (2, -5) and (2, 9). 5. Find the point on the x-axis which is equidistant from the points (6, 5) and (-4, 3). 6. Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units. 7. Find the value of x such that PQ = QR where the coordinates of P, Q and R are (6, -1), (1, 3) and (x, 8) respectively. 8. Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4). 9. If Q(0, 1) is equidistant from P(5, 3) and R(x, 6), then find the value of x. Also, find the distances QR and PR.

COORDINATE GEOMETRY

10. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:

( −1, − 2 ) , (1, 0 ) , ( −1, 2 ) , ( −3, 0 ) ii) ( −3, 5 ) , ( 3, 1) , ( 0, 3) , ( −1, − 4 ) iii) ( 4, 5 ) , ( 7, 6 ) , ( 4, 3) , (1, 2 ) iv) ( 0, − 1) , ( 6, 7 ) , ( −2, 3) , ( 8, 3) v) (1, − 2 ) , ( 3, 6 ) , ( 5, 10 ) , ( 3, 2 ) i)

III. SECTION FORMULA

1. Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3. 2. Find the points of trisection of the line segment joining the points: i) (5, -6) and (-7, 5)

ii) (3, -2) and (-3, -4)

iii) (4, -1) (-2, -3)

3. Find the ratio in which the point (2, y) divides the line segment joining the points A(-2, 2) and B(3, 7). Also, find the value of y. 4. Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6). 5. If (1, 2), (4, y), (x, 6), and (3, 5) are the vertices of a parallelogram taken in order, find x and y. 6. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4). 7. If the coordinates of the mid-points of the sides of a triangle are (3, 4), (4, 6), and (5, 7). Find its vertices. 8. If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that 3 AP = AB, and P lies on the line segment AB. 7 9. Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts. 10. Find the area of a rhombus if its vertices (3, 0), (4, 5), (-1, 4), and (-2, -1) are taken in order. 11. Find the ratio in which the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis. Also, find the coordinates of the point of division. IV. AREA OF TRIANGLES AND QUADRILATERALS

1. Find the area of a triangle whose vertices are

( 6, 3) , ( −3, 5) , and ( 4, − 2 ) ii) ( 2, 3)( −1, 0 ) , and ( 2, − 4 ) iii) ( a, c + a ) , ( a, c ) , and ( −a, c − a ) iv) ( −5, − 1)( 3, − 5 ) , and ( 5, 2 ) i)

IL Foundation Series Class 9

2. In each of the following, find the value of 'k' for which the points are collinear.

( 2, 5) , ( 4, k ) , and (8, 8) ii) ( 7, − 2 ) , ( 5, 1) , and ( 3, k ) iii) ( 8, 1) , ( k, − 4 ) , and ( 2, − 5 ) i)

3. Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5),(3, -2), and (2, 3). 4. P(5, 2), Q(4, 7), and R(7, -4) are the vertices of a ∆ PQR. Q(4, 7), R(7, -4), and S(6, 5) are the vertices of another triangle on the same base QR. Find the ratio of the areas of two triangles. 5. D(7, 9), E(1, 1), and F(-3, -7) are the vertices of a ∆DEF . L(4, 5), M(-1, -3), and N(2, 1) are area of ∆DEF 4 the mid-points of DE, EF, and FD, respectively. Prove that = . area of ∆LMN 1

WORKSHEET - 2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. The distance between the points ( cosθ , sinθ ) and ( sin θ , − cos θ ) is ______.

c) 2

d) 1

2. If the points (k, 2k), (3k, 3k), and (3, 1) are collinear, then k = ______. 1 1 2 2 b) − c) d) − 3 3 3 3 3. If (-1, 2), (2, -1), (3, 1), and (a, b), taken in order, are vertices of a parallelogram, then _____.

a)= a 2,= b 0

b) a = −2, b = 0

c)= a 0,= b 4

4. If points ( a, 0 ) , ( 0, b ) , and (1,1) are collinear, then a) 1

b) 2

d) a = 6, b = 2

1 1 + = _____. a b

c) 0

d) -1

5. The line segment joining points (-3, -4) and (1, -2) is divided by the y-axis in the ratio _____. a) 1 : 3

b) 2 : 3

c) 3 : 1

d) 2 : 3

6. If the centroid of the triangle formed by the points (a, b), (b, c), and (c, a) is at the origin, then a3 + b3 + c3 = _____. a) abc

b) 0

c) a + b + c

d) 3abc

7. The distance of the point P(-6, 8) from the origin is ______. a) 8

b) 2 7

c) 10

d) 6

COORDINATE GEOMETRY

8. The perimeter of a triangle with vertices (0, 4), (0, 0), and (3, 0) is ______. a) 5

b) 12

c) 11

d) 7 + 5

9. The fourth vertex D of a parallelogram ABCD whose three vertices are A(-2, 3), B(6, 7), and C(8, 3) is _______. a) (0, 1)

b) (0, -1)

c) (-1, 0)

d) (1, 0)

10. The distance between the points ( a cos 25° , 0 ) and ( 0, a cos 65° ) is ______. a) a

b) 2a

c) 3a

d) None of these

11. The coordinates of the point on the x-axis which are equidistant from the points (-3, 4) and (2, 5) are ______. 2  c)  , 0  d) None of these 5  12. The ratio in which the x-axis divides the segment joining (3, 6) and (12, -3) is _____.

a) (20, 0)

b) (-23, 0)

a) 2 : 1

b) 1 : 2

c) -2 : 1

d) 1 : -2

13. If the centroid of the triangle formed by (7, x), (y, - 6), and (9, 10) is at (6, 3), then (x, y) = ______. a) (4, 5)

b) (5, 4)

c) (-5, -2)

d) (5, 2)

14. If the point (x, 4) lies on a circle whose centre is at the origin and radius is 5, then x = _______. a) ±5

b) ±3

d) ±4

c) 0

15. If points A(5, p), B(1, 5), C(2, 1), and D(6, 2) form a square ABCD, then p = _____. a) 7

b) 3

c) 6

d) 8

16. If P ( 2, 4 ) , Q ( 0, 3) , R ( 3, 6 ), and S(5, y) are the vertices of a parallelogram PQRS, then the value of y is ______. a) 7

b) 5

c) -7

d) -8

17. A line intersects the y-axis and x-axis at P and Q, respectively. If (2, -5) is the midpoint of PQ, then the coordinates of P and Q, respectively, are: a) (0, -5) and (2, 0)

b) (0, 10) and (-4, 0)

c) (0, 4) and (-10, 0)

d) (0, -10) and (4, 0)

18. The distance between the points ( cos θ , sin θ ) and ( − sin θ , cos θ ) is a) 1

b) 2

d) 0

19. The x-axis divides the line segment joining (2, -3) and (5, 7) in the ratio a) -3 : 7

b) 3 : 7

c) -2 : 5

d) 2 : 5

20. The y-axis divides the line segment joining (3, 5) (-4, 7) in the ratio a) 5 : 7 86

b) -5 : 7

c) -3 : 4

d) 3 : 4

IL Foundation Series Class 9

21. If (1, 2), (4, -3), and (-2, 4) are the midpoints of the sides of a triangle, then its centroid is a) (1, 1)

b) (1, 0)

c) (1, 2)

d) (2, 1)

22. The incentre of the triangle formed by the points (0, 8), (6, 0), and (0, 0) is a) (1, 1)

b) (2, 2)

c) (3, 3)

d) (4, 4)

23. The excentre of the triangle formed by the points (0, 3), (4, 0), and (0, 0) which is opposite to (0, 0) is a) (3, 1)

b) (6, 6)

c) (3, 3)

d) (4, 4 )

24. If the orthocentre and circumcentre of a triangle, respectively, are (2, -3), (5, 6), then the centroid is −4   b)  −3,  c) (4, 3) d) (-1, -3) 3   25. If (0, 1) is the orthocentre and (2, 3) is the centroid of a triangle, then its circumcentre is

a) (2, 7)

a) (3, 2)

b) (1, 0)

c) (4, 3)

d) (3, 4)

26. If G is the centroid of ∆ ABC, and if the area of ∆ AGB is 5 sq.unit, then the area of ∆ ABC is a) 20 sq.unit

b) 15 sq.unit

c) 10 sq.unit

(

)

27. The area of the triangle formed by (0, 0), a x , 0 , and ( 0, a 6 x ) is a) 1 or 5

b) -1 or 5

28. The slope of the line whose inclination

c) 1 or -5

d) 25 sq.unit 1 sq. unit. Then x = 2a 5 d) -1 or -5

2π is 3

1 d) − 3 3 29. The inclination of the line passing through the points (-2, 3) and (-1, 4) is

a) 1

4 3 6 30. The equation of the line parallel to the y-axis through the point (-3, 2) is

a) x + 3 = 0

b) x - 3 = 0

c) y = 2

3π 4

d) y + 2 = 0

31. The equation of the line passing through the points (a, 0) and (0, b) is a) ax + by - ab = 0

b) bx + ay - ab = 0

c) ax - by + ab=0

d) bx - ay + ab = 0

32. The vertices of a ∆ABC are A(1, 1), B(-3, 4), C(2, -5). The equation to the altitude through the vertex A is a) 5x + 9y + 4 = 0

b) 5x + 9y - 4 = 0

c) 5x - 9y + 4 = 0

d) 5x - 9y - 4 = 0

33. The equation of the perpendicular bisector of the line segment joining the points (1, 2), (3, 4) is a) 2x - y + 5 = 0

b) x + y - 5 = 0

c) 3x - 2y + 5 = 0

d) x + y - 4 = 0

COORDINATE GEOMETRY

34. If a line cuts the y-axis at (0, 3), then the y-intercept is a) 3

b) -3

c) 0

2 4 and x-intercept − is 5 3 b) 2x - 3y - 14 = 0 c) 8x + 12y - 9 = 0

d) 6

35. The equation of the line having a slope − a) 3x - 4y + 20 = 0

d) 20x + 15y + 8 = 0

36. Reduce the equation 9x - 3y + 5 = 0 into slope intercept form. 5 3x 5x −4 x 5 y +2 y +5 y b) = c) = d)= + 3 4 3 2 2 37. The area of the triangle formed by the line passing through the points (3, 4), (5, 6) with the coordinate axes is 49 49 1 a) 24 sq.unit b) sq.unit c) sq.unit d) sq.unit 8 12 2 y 3x + a) =

II.

MULTIPLE CHOICE QUESTIONS WITH MULTIPLE CORRECT ANSWERS

1. If the perpendicular distance of a point P from the x-axis is 5 units and the foot of the perpendicular lies in the negative direction of the x-axis, then the y-coordinate of P is a) -5

b) 5

c) 0

d) Can't say

c) x-axis

d) y-axis

c) Q3

d) Q4.

c) Q3

d) Q4

2. The graph of x = 10 is parallel to ______axis. a) x = 7

b) y = 5

3. If xy < 0, then the point (x, y) lies in _____. a) Q1

b) Q2

4. If x + y > 0, then the point (x, y) lies in ______. a) Q1

b) Q2

III. ASSERTION AND REASON

1. Assertion(A): The value of y is 6, for which the distance between the points P(2, -3) and Q(10, y) is 10. Reason(R): Distance between two given points A(x1, y1) and B(x2, y2) is given, AB =

( x2 − x1 ) + ( y2 − y1 ) 2

a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A). b) Both Assertion (A) and Reason (R) are true, but Reason(R) is not the correct explanation of Assertion (A). c) Assertion (A) is true, but Reason (R) is false. d) Assertion (A) is false, but Reason (R) is true.

IL Foundation Series Class 9

2. Assertion(A): The points which divide the line segment joining the points (0, -3) and (7, 1)  14 −7  in the ratio 2 : 3 is  ,  .  5 5   x +x y +y  Reason (R): The mid-point of line segment joining of (x1, y1) and (x2, y2) is  1 2 , 1 2  . 2   2 a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A). b) Both Assertion (A) and Reason (R) are true, but Reason(R) is not the correct explanation of Assertion (A). c) Assertion (A) is true, but Reason (R) is false. d) Assertion (A) is false, but Reason (R) is true IV. FILL IN THE BLANKS

 2 1. If A(2, 3) and B(-2, 1) are the two vertices of ∆ABC and G 1,  is the centroid of ∆ABC ,  3 then the third vertex is _____. 2. If the points (-5, 1), (1, p), and (4, -2) are collinear, then p =_____. 3. If two vertices of an equilateral triangle are (3, 0) and (6, 0), then the third vertex is ______. 4. The endpoints of the diameter of a circle are (2, 4) and (-3, -1), then the radius of the circle is ______. 5. The area of a triangle with vertices (a, b + c), (b, c + a), and (c, a + b) is ______. 6. If the point P(x, y) is equidistant from A(5, 1) and B(-1, -5), then the relation between x and y is ______________. 7. If the centroid of a triangle formed by (3, x), (-1, 5), and (y, - 6) is (0, 1), then (x, y) is ______.

(

)

8. The points (a, a), (–a, –a), and − 3a, 3a form a/an ______ triangle. 9. If AB = 10 units, the coordinates of A are (-2, 3), and the ordinate of B is 9, then the abscissa of B is ______.

a  10. If P  , 4  is the midpoint of the line segment joining the points Q(-6, 5) and R(-2, 3), then 3  the value of a is ______. 11. If the points A(4, 3) and B(x, 5) are on the circle with centre O(2, 3), then the value of x is ______. 12. If the midpoint of the line segment joining the points A(3, 4) and B(a, 4) is P(x, y) and x + y - 20 = 0, then the value of a is ______. 13. The ratio in which the x-axis divides the line joining the points A(x1, y1) and B(x2, y2) is ______. 14. The value of a when the distance between the points (3, a) and (4, 1) is 10 is ______. 15. The points of trisection of the line segment joining the points (5, -6) and (-7, 5) is ______. 89

COORDINATE GEOMETRY

16. The ratio in which (-4, 6) divides the line segment joining the points A(-6, 10) and B(3, -8) is ________. 17. The ratio in which the straight line x - y = 2 divides the line segment joining the points (3, -1) and (8, 9)is ______. 18. ABCD is a parallelogram vertices A ( x1 , y1 ) , B ( x2 , y2 ) and C ( x3 , y3 ), then the coordinates of fourth vertex D is ______. 19. The x-coordinates of a point P is twice its y-coordinate. If P is equidistant from Q(2, -5) and R(-3, 6), then the coordinates of P are ______. 20. A, B, and C are collinear points, then ar ( ∆ABC ) = ______. V. SUBJECTIVE QUESTIONS

1. Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units. 2. Verify the vertices (2, 3), (7, -1), and (-1, 3) form a right-angled triangle. 3. Check whether (5, -2), (6, 4), and (7, -2) are the vertices of an isosceles triangle. 4. Find the radius of the circle whose centre is (3, 2) and passes through (-5, 6). 5. Can you draw a triangle with vertices (1, 5), (5, 8) and (13, 14)? Give reason. 6. Find a relation between x and y such that the point (x, y) is equidistant from the points (-2, 8) and (-3, -5). 7. If P is a point on the y-axis whose ordinate is 5 units and Q is the point (-3, 1), then find the length of PQ. 8. Name the type of quadrilateral formed, if any, by the following points and give reasons for your answer. i) (-1, -2), (1, 0), (-1, 2), (-3, 0)

ii) (-3, 5), (1, 10), (3, 1), (-1, -4)

iii) (4, 5), (7, 6), (4, 3), (1, 2) 9. Find the points which divide the line segment joining the points (1, 7), (-6, -3) in the ratio 2 : 3. 10. Find the ratio in which the x-axis and y-axis divide the line segment joining the point (1, 4) and (4, -5). 11. Find the fourth vertex of the parallelogram whose consecutive vertices are (8, 4), (5, 7), (-1,1). 12. The coordinates of A and B are (−3, α ) and (1, α + 4) and the mid point of AB is P(-1, 1). Find the value of α. 13. If A(2, 2), B(-4, -4), and C(5, -8) are the vertices of ∆ABC , then find the length of the median through C.

IL Foundation Series Class 9

14. If D(3, -2), E(-1, -5), and F(-2, 2) are the midpoints of BC , CA, AB of ∆ABC , then find the vertices of ∆ABC . 15. If G is the centroid of a triangle formed by A(6, 1), B(3, 5), and C(-1, -1), then find the area of the triangle GAB. 16. Find the area of the quadrilateral whose vertices are (-1, 5), (2, -3), (5, 1), (6, 9) taken in order.  1  1  1 17. Find the area of the triangle formed by the points  a,  ,  b,  ,  c,  .  a  b  c 18. Find the harmonic conjugate of (4, -2) with respect to (2, -4), (7, 1). 19. Find the coordinates of the points of trisection of the line segment joining (5, -1) and (-2, 8). VI. CASE STUDIES

Case study: If P(x, y) lies on the line joining A(x1, y1) and B(x2, y2), then P divides AB in the ratio x1 – x : x – x2 or y1 – y : y – y2. The x-axis divides the line segment joining (x1, y1) and (x2, y2) in the ratio – y1 : y2. The y-axis divides the line segment joining (x1, y1) and (x2, y2) in the ratio – x1 : x2. 1. The x-axis divides the line segment joining the points (2, -5) and (1, 9) in the ratio _____. a) 9 : 5

b) 2 : 1

c) 5 : 9

d) 1 : 2

2. The y-axis divides the line segment joining the points P(-4, 2) and Q(8, 3) in the ratio _____. a) 3 : 1

b) 1 : 3

c) 2 : 1

d) 1 : 2

3. If the point P(2, 1) lies on the line segment joining the points A(4, 2) and B(8, 4), then ______. a) 3AP = AB

b) AB = 4AP

c) 3PB = AB

d) 2AP = AB

4.1

LINEAR EQUATIONS IN TWO VARIABLES

INTRODUCTION TO LINEAR EQUATIONS

4.1.1 Equation An equation is a statement of equality of two algebraic expressions involving one or more unknown quantities called the variables. 4.1.2 Linear equation An equation in which the highest index of the variables is one is called a linear equation. Example: x y z 1 Linear Equation in one variable

A linear equation that involves only one variable is called a linear equation in one variable or a simple equation. Example: 3 x − 5 = x + 2 Note: ax b 0 is the general form of linear equations in one variable, where a, b are real numbers. Solution of a linear equation

The value of the variable (unknown), which, when substituted for the variable, makes both sides of the given equation equal, is called a solution of the equation. It is also known as the root of the equation. Example: x = 3 is the solution of the equation 3 x + 11 = 20 because L.H.S. 3 3 11 9 11 20 R.H.S. 4.1.3 Linear equations in two variables An equation of the form ax by c 0, where a, b, c are real numbers is called linear equation in two variables x and y, where a ≠ 0, b ≠ 0 . Example: 2 x 3y 9 Ordered pairs

Listing the two components in the specified order, separating them by a comma and enclosing the pair in parenthesis ( ) is called an ordered pair. The ordered pair of a and b is written as ( a, b ), where a is called the first element or the first component, and b is called the second element or the second component. 92

IL Foundation Series Class 9

Note: 1. The ordered pair of a and b is not the same as the ordered pair of b and a, i.e., a, b b, a if a ≠ b. 2.

a, b b, a a b ; similarly a, b c, d a c, b d .

Solutions of a linear equation in two variables

Let the linear equation be ax by c 0, a 0 and b ≠ 0 . Any ordered pair of values of x and y which satisfies (L.H.S. = R.H.S.) the equation ax + by + c = 0 is called its solution. x 2= , y 3. Example: Show that the solution of 2 x + y = 7 is= Solution: L.H.S. = 2 ( 2 ) + 3 = 7 R.H.S. 7 L.H.S. R.H.S. The solution is x, y 2, 3 . Note: 1. Any ordered pair of values which satisfies the equation is a solution of that equation. 2. A linear equation in two variables has an infinite number of solutions. Example: Let x y 10. For this, the solutions are 9, 1 , 1, 9 , 2, 8 , 8, 2 , 1, 11 , 11, 1 , etc. The set of all these solutions is called a solution set, written as:

9, 1 , 1, 9 , 2, 8 , 8, 2 , 1, 11 , 11, 1 . SOLVED EXAMPLES Example 1: Express 5x = − y in the form of ax + by + c = 0. Solution: Given: 5x = − y 5x y 0 5x 1 y 0 ax by c 0 where= a 5= , b 1 and c = 0 Example 2: Express 2 x 3y 9.35 in the form ax by c 0 and indicate the values of a, b and c.

LINEAR EQUATIONS IN TWO VARIABLES

Solution: Given: 2 x 3y 9.35 2 x 3y 9.35 0 ax by c 0 So, a 2, b 3, c 9.35 Example 3: Express y − 2 = 0 in the form ax + by + c = 0 and indicate the values of a, b and c. Solution: Given: y 2 0 0.x 1 y 2 0 ax by c 0 So, a 0, b 1, c 2 Example 4: Frame a linear equation in the form ax by c 0 by using the given values of a, b and c. i) a 2; b 3; c 4 a 5= ; b 0; c = 7 ii)= Solution: i) The given equation is ax + by + c = 0. Putting a 2, b 3, c 4 , we get 2 x 3y 4 0. This is the required linear equation. ii) The given equation is ax + by + c = 0. Putting= a 5= , b 0, c = 7 , we get 5 x + 0.y + 7 = 0 5x 7 0 This is the required linear equation. x 2= , y 1. Example 5: Find the value of p from the equation 3 x + 4y = p if one of its solutions is= Solution: Given equation: 3 x 4y p 1 Substituting= x 2= , y 1 in eq (1), we get: 3 2 4 1 p p 6 4 p 10 94

IL Foundation Series Class 9

Example 6: Write four solutions of 2 x + y = 7. Solution: Given equation 2x y 7 y 7 2x Put x = 0, then y = 7 − 2(0) = 7−0 7 0, 7 is a solution Put x = 1, then y 7 2 1 7 2 5 1, 5 is a solution Put x 1, then y 7 2 1 = 7+2 9 1, 9 is a solution Put x = 2, then y 7 2(2) 7 4 3 2, 3 is a solution ∴ The four solutions of the given equation are 0, 7 , 1, 5 , 1, 9 , 2, 3 . Example 7: A man is five times as old as his son. After 2 years, the man will be four times as old as his son. Represent these two situations in linear equations in two variables. Solution: Let the present age of the man = x years Let the present age of the son = y years According to the given problem: x 5y x 5y 0 …(1) Man’s age after 2 years x 2 years Son’s age after 2 years = ( y + 2 ) years

LINEAR EQUATIONS IN TWO VARIABLES

According to the given problem:

x 2 4 y 2

x 2 4y 8

x 4y 6 0

…(2)

Example 8: The taxi fare in a city is as follows: for the first kilometre, the fare is ₹8, and for the subsequent distance, it is ₹5 per km. Taking the distance covered as x km and the total fare as ₹y, write a linear equation for this information. Solution: Distance covered = x km The fare for the first km = ₹8 The rate of fare for the rest of (x-1)km = ₹5 per km ∴ The fare for (x-1)km = ₹5(x-1) Total fare = ₹y According to the given problem: 5( x − 1) + 8 = y ⇒ 5x − 5 + 8 = y ⇒

5x + 3 = y

Example 9: Find the value of k; if x = 2, y = 1 is a solution of the equation 2 x + 3y = k. Solution: We have 2 x + 3y = k. Putting x = 2 and y = 1 in 2 x + 3y = k, we get: 2 2 3 1 k

7 k

k 7 Thus, the required value of k is 7.

IL Foundation Series Class 9

4.2 SOLVING SIMULTANEOUS LINEAR EQUATIONS 4.2.1 Simultaneous linear equations in two variables Any system of linear equations in the same two variables taken together is called a system of simultaneous linear equations in two variables. Example: The equations 3 x − 2y + 10 = 0 and 2 x + 7 y − 1 = 0, taken together are called a system of simultaneous linear equations in x and y. Note: A linear equation in two variables has infinitely many solutions. Solution of a system of two simultaneous linear equations in two variables

The ordered pair of numbers that satisfies both equations is called a solution of a system of two simultaneous linear equations. Example: x + y = 5, 2 x − y = −2 is the system of linear equations, satisfied by x = 1 and y = 4. 1, 4 is a solution to the above system. 4.2.2 Methods of solving simultaneous linear equations Substitution method

a) Express one variable in terms of the other variable from one of the two equations. b) Substitute that value of the variable in the other equation and solve it for the variable left. c) Substitute this value in either of the two original equations and solve it to find the value of the other variable. Example: Solve x + y − 5 = 0, y − 2 = 2 x using the substitution method. Solution: Given equations are x + y − 5 = 0 … (i) and y − 2 = 2 x …. (ii) From (i), y = 5 − x substituting this in (ii), we get:

5 x 2 2x

x 2 x 3

3 x 3

x 1

Substituting this x = 1 in (i) or (ii), we get the value of y. From (ii), y = 2 x + 2 = 2 ( 1 ) + 2 = 4 ∴ The solution is= x 1= , y 4.

LINEAR EQUATIONS IN TWO VARIABLES

Elimination (or) addition-subtraction method

a) Multiply one or both equations (if necessary) to transform them such that either addition or subtraction will eliminate one variable. b) Solve the resulting single variable equation. c) Substitute this value in either of the two original equations and find the value of the second variable. Example: Solve, 5 x − 3y + 3 = 0, 6 x 5y 2 0 using the elimination method. Solution: Given equations are 5 x − 3y + 3 = 0…... (i) and 6 x 5y 2 0……. (ii) From (i) we get, 5 x 3y 3 ……. (iii) From (ii) we get, 6 x 5y 2 ……. (iv) To eliminate the variable x, we try to get the coefficients of x as L.C.M. of 5 and 6, i.e., 30. Multiplying on both sides of the equation (iii) by 6 and equation (iv) by 5, we get: ⇒ 30 x − 18y = −18 ⇒ 30 x − 25y = −

10 −

⇒

0 + 7 y = −28

⇒

y =

−4

Substituting y 4 in (i), we get: 5 x 3 4 3 0 5 x 15 x 3. ∴ The solution is x 3, y 4 . Cross-multiplication method

Let the system of linear equations be a1 x + b1y + c1 = 0 and a2 x + b2 y + c2 = 0, where It has a unique solution. The coefficients of the above system should be written in the following order:

a1 b1 ≠ . a2 b2

IL Foundation Series Class 9

x y 1 = = b1c2 − b2 c1 c1a2 − c2a1 a1b2 − a2b1

x 1 y 1 , b1c2 b2 c1 a1 b2 a2 b1 c1a2 c2a1 a1 b2 a2 b1

b1c2 b2 c1 c a c a ,y 1 2 2 1 a1b2 a2b1 a1b2 a2b1

This is the required solution. Example: Solve the following using cross-multiplication. 5 x − 3y + 3 = 0 6 x − 5y − 2 = 0 Solution: The given equations are: 5 x − 3y + 3 = 0, 6 x − 5y − 2 = 0 The coefficients of the above system can be written in the following order:

x y 1 3 2 5 3 3 6 2 5 5 5 6 3

x y 1 6 15 18 10 25 18

x y 1 21 28 7

21 28 ,y 7 7

x 3, y 4.

LINEAR EQUATIONS IN TWO VARIABLES

Verbal (word) problems:

a) Read the problem carefully and identify the unknown. b) Represent two unknown quantities by two variables. c) Translate the problem into two equations using the assumed variables. d) Solve the pair of equations using any method discussed earlier. e) Check the answer by satisfying the conditions given in the problem and not by substituting the values of the variables in the equations. Note: Remember the following points while solving problems based on two-digit numbers (yx). a) A two-digit number (yx) is represented by 10y x . b) In the number 10y + x, x is the units digit, and y is the tens digit. c) The sum of the digits of the number 10y + x is ( x + y ). d) The number obtained by reversing (or interchanging) the digits of the number 10y + x is 10x + y. e) The sum of a two-digit number and the number obtained by interchanging its digits is a multiple of 11. f) The difference between a two-digit number and the number obtained by interchanging its digits is a multiple of 9.

SOLVED EXAMPLES Example 1: Solve the equations 3 x − y = 5, 3 x + 3y = 21 using the substitution method. Solution: 3x − y = 5 ⇒ y = 3x − 5 Substitute y = 3 x − 5 in 3 x + 3y = 21 ⇒ 3 x + 3 ( 3 x − 5 ) = 21 ⇒ 3 x + 9 x − 15 = 21 ⇒ 12 x = 36 ⇒ x = 3 ∴ y = 3( 3 ) − 5 = 4 Example 2: Solve the equations 5 x + 3y = 11, 3 x + 5y = 13 using the elimination method. Solution: 5 x 3y 11 1 3 x 5y 13 2

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IL Foundation Series Class 9

(1) × 3 ⇒ 3 ( 5 x + 3y ) = 3 × 11 ⇒ 15 x + 9y = 33

( 2 ) × 5 ⇒ 5 ( 3 x + 5y ) = 5 × 13 ⇒ 15 x + 25y = 65 15 x + 9y = 33 15 x + 25y = 65 - - 16y = - 32 y=

32 ⇒y=2 16

Put y = 2 in equation (1), 5 x 3 2 11 5 x 5 x 1. Example 3: O ne number is greater than thrice the other number by 2, and 4 times the smaller number exceeds the greater one by 5. Find the numbers. Solution: Let the greater number be x, and the smaller number be y. Then, according to the given conditions, we have: x − 3y = 2 4y − x = 5 Thus, we have: x − 3y = 2…. ( i ) − x + 4y = 5….. ( ii ) Adding (i) and (ii), we get y = 7. Putting y = 7 in (i), we get x − 3 ( 7 ) = 2 ⇒ x = 2 + 21 = 23. ∴ The greater number = 23 and the smaller number = 7. Now, 3 ( 7 ) + 2 = 21 + 2 = 23 ∴ Thrice the smaller plus two is equal to larger. And, 4 ( 7 ) = 28 = 23 + 5. ∴ Four times smaller is equal to five more than the greater number. Example 4: 2 tables and 3 chairs together cost ` 2000, whereas 3 tables and 2 chairs together cost `2500. Find the total cost of 1 table and 5 chairs. Solution: Let the cost of a table be `x, and that of a chair be `y. Then: 101

LINEAR EQUATIONS IN TWO VARIABLES

2 x + 3y = 2000….. ( i )

3 x + 2y = 2500…. ( ii ) Multiplying equation (i) by 3 and (ii) by 2, we get: 6 x + 9y = 6000 ....(iii) 6 x + 4y = 5000 ....(iv) Subtracting (iv) from (iii), we get: 5y = 1000 y = 200 Substituting the value of y in equation (i), we get: 2 x + 3 ( 200 ) = 2000 ⇒ ⇒

2 x + 600 = 2000 2 x = 1400 ⇒ x = 700

So, the cost of a table = `700. The cost of a chair = `200. Hence, the cost of 1 table and 5 chairs ` x 5y = `( 700 + 5 × 200 ) = `1700 Example 5: F ind the solution of the given linear equations using the cross-multiplication method: 3 x − 4y = 2, y − 2 x = 7 Solution: 3 x − 4y = 2 ⇒ 3 x − 4y − 2 = 0 −2 x + y = 7 ⇒ −2 x + y − 7 = 0 By using the method of cross-multiplication, we get: x y 1 = = b1c2 − b2 c1 c1a2 − c2a1 b2a1 − b1a2 Now, we must substitute the value in the equation above. 1 x y = = ⇒ 28 + 2 4 + 21 3 − 8

102

⇒

x y 1 = =− 30 25 5

⇒

x = −6, y = −5

IL Foundation Series Class 9

Example 6: Solve the following system of equations using the method of cross-multiplication. x+y =7 5 x + 12y = 7 Solution: The given system of equations is: x y 7 0 5 x 12y 7 By cross-multiplication, we get: x y 1 1 7 12 7 7 5 7 1 1 12 5 1

x y 1 7 84 35 7 12 5

x y 1 77 28 7

28 77 and y 7 7

x 11 and y 4 Example 7: Solve the following system of equations by the method of cross-multiplication. ax + by = a − b bx − ay = a + b Solution: By using cross-multiplication, we get: ⇒

b { − ( a + b ) } − ( −a ) × − ( a − b )

−y 1 = 2 2 a × − ( a + b ) − b × − ( a − b ) −a − b

⇒

−y x 1 = = 2 −b ( a + b ) − a ( a − b ) −a ( a + b ) + b ( a − b ) − a + b2

⇒

x −y 1 = 2 2 = 2 2 −b − a −a − b − a + b2

⇒

(

)

1 y = 2 a +b − a2 + b2

( ) ( ) −(a + b ) ( a + b ) = −1 x= = 1 and dy= −(a + b ) −(a + b ) − a2 + b2

∴ x = 1, y = −1 103

⇒

x −y 1 = 2 2 = 2 2 −b − a −a − b − a + b2

(

)

x EQUATIONS y IN TWO1VARIABLES LINEAR =

( ) a + b −(a + b ) −(a + b ) ( a + b ) = −1 = 1 and dy= x= −(a + b ) −(a + b ) − a2 + b2

∴ x = 1, y = −1 Example 8: T he sum of the digits of a two-digit number is 15. If the number formed by reversing the digits is less than the original number by 27, find the original number. Solution: Let the units place digit = x and tens place digit = y ⇒ The number is 10y + x and the sum of digits = 15 x y 15 1 The number formed by reversing the digits is less than the original number by 27,

( 10y + x ) − ( 10 x + y ) = 27 ⇒ 10 y + x − 10 x − y = 27 ⇒ −9 x + 9y = 27 ⇒ x − y = −3 .... ( 2 ) After solving (1) and (2), we get x = 6, y = 9. ⇒ So, the required number is 10 y + x = 10 ( 9 ) + 6 = 96.

4.3 GRAPH OF LINEAR EQUATIONS 4.3.1 Graph of a linear equation in two variables To draw the graph of a linear equation ax by c 0, where a, b, c ∈ R and a ≠ 0, b ≠ 0 , we may take the following steps. Step (i): Let the given equation be ax by c 0. Step (ii): Write y

ax c .

b Step (iii): Take any convenient value for x and find the corresponding value of y. Let these values be x = p1 and y = q1. Step (iv): By giving another convenient value to x, find the corresponding value of y. Let these values be x = p2 , y = q2.

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IL Foundation Series Class 9

Step (v): Put the above values in the following tabular form. x

Step (vi): Plot the points A p1 , q1 , B p2 , q2 on the graph paper. Join AB and extend this line in both directions. Then, line AB is the required graph of the given equation. Note: a) If x , y is the solution of ax by c 0, we say that the point , lies on the line ax by c 0, otherwise we say that the point , does not lie on the line ax by c 0. b) Every point on the line ax + by + c = 0 gives its solution. c) The equation of the x-axis is y = 0. d) The equation of the y-axis is x = 0. e) Any line parallel to the x-axis is y = ± k units.(k is any real number) f) Any line parallel to the y-axis is x = ± k units. g) Every line is parallel to itself. h) For drawing the graph of a linear equation, plotting two points is sufficient, yet for the purpose of ensuring correctness, plotting at least three points is preferred.

SOLVED EXAMPLES Example 1: Draw the graph of the equation 2 x + y = 8. Solution: 2 x + y = 8 ⇒ y = 8 − 2 x. When x = 0, we have y = 8 − 2 × 0 = 8. When x = 1, we have y 8 2 1 6. When x = 4 , we have y = 8-2×4 = 0 . Thus, we have the following table : x

Plot the points 0, 8 , 1, 6 , and 2, 4 on the same graph paper.

105

LINEAR EQUATIONS IN TWO VARIABLES

Y (0, 8)

8 7 6 5 4 3 2 1

(4, 0)

1 2 3

4 5 6

X 7 8

Y Join these points and extend the line on both sides to obtain the graph of 2 x + y = 8. Example 2: Draw the graph of y = 3 x. Solution: Given equation y = 3 x The tabular form of y = 3 x is: x

-1

-2

y = 3x

-3

-6

Plot the points A 1, 3 , B 0, 0 , C 1, 3 , D 2, 6 , E 2, 6 on a graph paper. By joining these points, we get a straight line.

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IL Foundation Series Class 9

Y 7

D (2, 6)

6 5 4

A(1, 3)

3 2 1

B (0, 0)

x -7 -6 -5 -4 -3 -2 -1 O

-1

-2

C (-1, -3)

-3 -4 -5

E (-2, -6)

-6 -7

Y Example 3: D raw the graph of two lines, whose equations are 3 x − 2y + 6 = 0 and x + 2y − 6 = 0 on the same graph paper. Solution: Given equations are: 3 x 2 y 6 0 1 x 2y 6 0 2 The tabular form of eq. (1) will be: x

-2

he points are A 0, 3 and B 2, 0 . Plot these points on a graph paper and join them; we get a T straight line. The tabular form of eq. (2) will be: x

The points are C 0, 3 and D 6, 0 . 107

LINEAR EQUATIONS IN TWO VARIABLES

Plot these points on graph paper and join them; we get a straight line.

Y 7 6 5 4

A(0, 3)

3 2

B(-2, 0)

D(6, 0)

x -9 -8 -7 -6 -5 -4 -3 -2 -1 O

-1

-2 -3 -4 -5 -6 -7

Y Example 4: Give the geometric representation of 2 x 9 0 as an equation: i) in one variable

ii) in two variables.

Solution: Given 2 x + 9 = 0 ⇒ 2 x = −9 ⇒ x=

−9 2

−9 s an equation in one variable, x = , the geometric representation is the number on the number A 2 line. x = -9 2

108

X -5 -4 -3 -2

-1

IL Foundation Series Class 9

ii) As an equation in two variables, 2 x + 9 = 0 can be written as 2 x + 0.y + 9 = 0. The value of y can be −9 9 any number, but x will continue to be . It is a line parallel to the y-axis and units to the left of 2 2 the origin. Y

5 4 3 2 1

x = -9 2 X

-5 -4 -3 -2 -1 O

1 -1

2 3 4

5 6 7

-2 -3 -4 -5 Y

4.4 CONSISTENCY/INCONSISTENCY 4.4.1 Consistent system A system of simultaneous linear equations is said to be consistent if it has at least one solution. Graphical representation

Let a1 x b1y c1 0 1 , a2 x b2 y c2 0 2 be the given system of linear equations. Draw a graph of each of the given linear equations. Each graph is a straight line. Let L1 and L2 represent graphs of (1) and (2), respectively. Now, the following cases may arise: x p= , y q is a unique solution of the given system Case 1: If L1 and L2 intersect at P p, q , then= of equations. 109

LINEAR EQUATIONS IN TWO VARIABLES

y L2

L1 P(p, q) x

Note: The system of linear equations a1 x + b1y + c1 = 0, a2 x + b2 y + c2 = 0 is consistent and has a a b unique solution if 1 ≠ 1 . a2 b1 Case 2: If L1 and L2 are coinciding, then the given system of equations has infinitely many solutions.

y L1 L2 L2 L1 O

Note: If the system of linear equations a1 x + b1y + c1 = 0, a2 x + b2 y + c2 = 0 is consistent and has a1 b1 c1 infinitely many solutions, then = = . This system is known as the dependent system. a2 b2 c2 Conclusion: If a system of two linear equations in x and y has: a) A unique solution if two lines intersect at a point. b) Infinitely many solutions if two lines coincide. c) Only one solution, then it is called independent. d) Infinitely many solutions, then it is called dependent.

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IL Foundation Series Class 9

Example: Show that graphically, the system of equations 3 x + y = 4, 6 x + 2y = 8 has an infinite number of solutions. Solution: The given system is: 3x y 4 6 x 2y 8 Now, 3 x y 4. y 4 3x When x = 0, we have y 4 0 4 . When x = 1, we have y = 4 − 3 ( 1 ) = 1. When x = 2, we have y = 4 − 3 ( 2 ) = −2. ∴ The points to be plotted for the graph of equation (i) are 0, 4 , 1, 1 , and 2, 2 . 6 x + 2y = 8 ⇒ 2y = 8 − 6 x ⇒ y=

8 − 6x 2

When x = 0, we have y

8 0 4. 2

8 6 1 7. 2 8 6 2 When x 2, we have y 10. 2 ∴ The points to be plotted for the graph of equation (ii) are 0, 4 , 1, 7 , and 2, 10 . When x 1, we have y

111

LINEAR EQUATIONS IN TWO VARIABLES

y 10 9 8 7 6 5 4 3 2 3x + y = 4 1

6x + 2y = 8

-4

-3

-2

-1

-1 1

-2 -3 -4 y From the graph, we observe that the given two lines coincide. ∴ The given system has an infinite number of solutions. 4.4.2 Inconsistent system of linear equations A system of linear equations in two variables is said to be inconsistent if it has no solution at all. Graphical representation

If we observe the graph, L1 and L2 do not have any points in common. These lines are called parallel lines and are represented by an inconsistent system of equations.

L2 L1 O

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IL Foundation Series Class 9

Note: If the system of linear equations a1 x + b1y + c1 = 0 and a2 x + b2 y + c2 = 0 are inconsistent, then a1 b1 c1 = ≠ . a2 b2 c2 Conclusion: A system of two linear equations in x and y has no solution when the two lines are parallel, and have no common points. 4.4.3 Homogeneous system of linear equations The system of linear equations is of the form a1 x + b1y = 0, a2 x + b2 y = 0 is called homogeneous system of linear equations, where constant terms are zero. Note: The system of equations a1 x b1y 0, a2 x b2 y 0 has i) a unique solution= x 0= , y 0 , if

a1 b1 ≠ . a2 b2

ii) an infinite number of solutions, if

a1 b1 = . a2 b2

Clearly, this system is always consistent. iii) Any homogeneous system of equations always passes through the origin. Graphical representation

Let L1 and L2 represent the graphs of a1 x b1y 0 and a2 x b2 y 0 , respectively. Here, two cases will arise.

Case - I y

Case - II y L1

L1 L2 O Unique Solution

L2 x

Infinite Solution

Example: Show that the system of equations x y 3, x y 2 has no solution graphically. Solution: The given system is: x+y =3 x+y =2

113

LINEAR EQUATIONS IN TWO VARIABLES

Now, x y 3 y 3 x When x = 0, we have y = 3. When x = 1, we have y = 2. When x = 2, we have y = 1. ∴ The points to be plotted for the graph of equation (i) are 0, 3 , 1, 2 , 2, 1 . x y 2 y 2 x When x = 0, we have y = 2. When x = 1, we have y = 1. When x = 2, we have y = 0. ∴ The points to be plotted for the graph of equation (ii) are 0, 2 , 1, 1 , 2, 0 .

y 4 3 2

y= 3

-1 O

-1

y= 2

y From the graph, we observe that the two lines are parallel to each other. ∴ They do not have any point in common. ∴ The given system has no solution at all. Note: Let L1 : a1 x b1y c1 0, L2 : a2 x b2 y c2 0 be the system of non-homogeneous linear equations. 114

IL Foundation Series Class 9

System

No. of solutions

Condition

Consistent

Unique solution

a1 b1 ≠ a2 b2

Graphical representation

Nature of lines

Intersecting lines

P(p

, q) x

O Consistent

Infinite solutions

a1 b1 c1 = = a2 b2 c2

Coincident lines

L1 L2 L2 L1 x

O Inconsistent No solution

a1 b1 c1 a2 b2 c2

Parallel lines

L1 L2

Note: Let L1 : a1 x b1y 0, L2 : a2 x b2 y 0 be the system of homogeneous linear equations.

115

LINEAR EQUATIONS IN TWO VARIABLES

No. of solutions

Condition

Graphical representation

Nature of lines

Unique solution passing through the origin

a1 b1 ≠ a2 b2

Intersecting lines

L1 L2 x

Infinite solutions passing through the origin

a1 b1 = a2 b2

L2 x

Coincident Lines L1

L1 y

Note: A homogeneous system of linear equations is always consistent.

SOLVED EXAMPLES Example 1: Ten students of class X took part in a mathematics quiz. If the number of girls is 4 more than the number of boys, represent this situation algebraically and graphically. Solution: Let the number of girls be x and the number of boys be y. Given that the number of girls + number of boys = 10. x y 10. It is also given that the number of girls is 4 more than the number of boys. x y 4 x y 4 The algebraic representation of the given situation is: x y 10 (i) x y 4 (ii) 116

IL Foundation Series Class 9

Graphical Representation: The solutions of equation (i) are: x

-4

-2

The solutions of equation (ii) are:

Now, we plot the points on a graph sheet.

Y (3, 7)

7 6

(5, 5)

(6, 4)

4 3

(7, 3)

2 1

-7 -6 -5 -4 -3 -2 -1 O

(4, 0) -1

X y = 10

-2 -3

(0, -4)

(2, -2)

-4 -5

-6 -7

Y We observe that the two lines representing the two equations are intersecting at one point. Example 2: Two lines are represented by the equations x + 2y − 4 = 0 and 2 x + 4y − 12 = 0. Represent this situation graphically. Solution: The given equations are: 117

LINEAR EQUATIONS IN TWO VARIABLES

x + 2y − 4 = 0 2 x + 4y − 12 = 0 The solutions of equation (i) are: x

-2

The solutions of equation (ii) are:

Now, we plot the points on a graph sheet. We observe that the lines are parallel, and they do not intersect anywhere.

Y 7 6 5 4

(-2, 3)

3 2 1

X -7 -6 -5 -4 -3 -2 -1 O

(0, 3) (0, (2, 2) 2) (4, 0) -1 -2 -3

(6, 0) 4

2x + x + 4y 2y - 4 12 = 0 =0

-4 -5 -6 -7

Y Example 3: Romila went to a stationary stall and purchased 2 pencils and 3 erasers for ₹9. Her friend, Sonali, saw the new variety of pencils and erasers with Romila, and she also bought 4 pencils and 6 erasers of the same kind for ₹18. Represent this situation algebraically and graphically.

118

IL Foundation Series Class 9

Solution: et the cost of 1 pencil be ₹x, and that of the one eraser be ₹y. Then, the algebraic representation is L given by the following equations: 2 x 3y 9 4 x 6y 18 Graphical representation: The solutions of equation (i) are: x

-3

The solutions of equation (ii) are:

Now, we plot the points on a graph sheet.

Y 4x

(-3, 5)

6 5 4 3

(0, 3)

(3, 1)

X -7 -6 -5 -4 -3 -2 -1 O

-1 -2 -3

-4 -5 -6 -7

119

LINEAR EQUATIONS IN TWO VARIABLES

QUICK REVIEW •

n equation of the form ax + by + c = 0, where a, b and c are real numbers, such that and b are A not both zero, is called a linear equation in two variables.

•

A linear equation in two variables has infinitely many solutions.

•

The graph of every linear equation in two variables is a straight line.

•

x = 0 is the equation of the y-axis, and y = 0 is the equation of the x - axis.

•

The graph of x = a is a straight line parallel to the y-axis.

•

The graph of y = a is a straight line parallel to the x-axis.

•

An equation of the type y = mx represents a line passing through the origin.

•

very point on the graph of a linear equation in two variables is a solution of the linear E equation. Moreover, every solution of the linear equation is a point on the graph of the linear equation.

•

n equation of the form ax by c 0, where a ≠ 0 and b 0, a, b, c R is called a linear A equation in two variables, x and y.

•

ny ordered pair of values of x and y which satisfies (L.H.S. = R.H.S.) the equation A ax + by + c = 0 is called its solution.

•

The solution of system of linear equations be a1 x + b1y + c1 = 0 and a2 x + b2 y + c2 = 0, where: a1 b1 b c b c c a c a ≠ is x 1 2 2 1 ; y 1 2 2 1 a2 b2 a1b2 a2b1 a1b2 a2b1

•

pair of linear equations in two variables x and y can be represented algebraically as: A a1 x b1y c1 0 a2 x b2 y c2 0 where a1 , a2 , b1 , b2 , c1 , c2 are real numbers such that a12 b12 0, a22 b22 0 .

•

Graphically or geometrically, a pair of linear equations: a1 x b1y c1 0 a2 x b2 y c2 0,

in two variables represents a pair of straight lines which are: i) intersecting, if

ii) parallel, if 120

a1 b1 ≠ a2 b2

a1 b1 c1 = ≠ a2 b2 c2

IL Foundation Series Class 9

iii) coincident, if •

a1 b1 c1 = = a2 b2 c2

A pair of linear equations in two variables can be solved by the:

i) Graphical method ii) Algebraic method •

o solve a pair of linear equations in two variables by the graphical method, we first draw the T line represented by them.

i) If the pair of lines intersect at a point, then we say that the pair is consistent, and the coordinates of the point provide us with a unique solution. ii) If the pair of lines are parallel, then the pair has no solution and is called an inconsistent pair of equations. iii) If the pair of lines are coincident, then it has infinitely many solutions, with each point on the line being a solution. In this case, we say that the pair of linear equations is consistent with infinitely many solutions. •

o solve a pair of linear equations in two variables algebraically, we have the following T methods: i) Substitution method ii) Elimination method iii) Cross-multiplication method

•

I f a1 x b1y c1 0, a2 x b2 y c2 0 is a pair of linear equations in two variables, x and y, such that:

a1 b1 ≠ , then the pair of linear equations is consistent with a unique solution. a2 b2

ii)

a1 b1 c1 = = , then the pair of linear equations is consistent with infinitely many solutions a2 b2 c2

iii)

a1 b1 c1 = ≠ , then the pair of linear equations is inconsistent. a2 b2 c2

121

LINEAR EQUATIONS IN TWO VARIABLES

WORKSHEET - 1 I.

INTRODUCTION TO LINEAR EQUATION 1. Write the equation x = 7 in two variables form. 2. The price of 1 kg of grapes and 2 kg of apples on a day was found to be ₹160. After a month, the price of 2 kg of grapes and 4 kg of apples is ₹300. Represent the situation algebraically.

3. For what value of c, the linear equation 2 x + cy = 8 has equal values of x and y for its solution? 4. y varies directly as x. If y = 12 when x = 4, then write a linear equation. 5. Show that the points A 1, 2 , B 1, 16 and C 0, 7 lie on the graph of the linear equation y = 9 x − 7. 6. The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation: C

5 F 160 9

(i) If the temperature is 86 F, what is the temperature in Celsius? (ii) If the temperature is 35 C, what is the temperature in Fahrenheit? (iii) If the temperature is 0 C, what is the temperature in Fahrenheit, and if the temperature is 0 F, what is the temperature in Celsius? (iv) What is the numerical value of the temperature that is the same on both scales? 7. If the temperature of a liquid can be measured in kelvin units as x  K or in Fahrenheit units as y  F, then find the relation between the two systems of measurement of temperature. 8. If the point 3, 4 lies on the graph of the equation 3y = ax + 7, find the value of a. 9. The taxi fare in a city is as follows: for the first kilometre, the fare is ₹8, and for the subsequent distance, it is ₹5 per km. Taking the distance covered as x km and the total fare as ₹y, write a linear equation for this situation. 10. The work done by a body on application of a constant force is directly proportional to the distance travelled by the body, Express this in the form of an equation in two variables. II. SOLUTION OF SIMULTANEOUS LINEAR EQUATION 1. Solve: 101x + 99y = 499, 99 x + 101y = 501 2. Use the method of substitution to solve each other of the pair of simultaneous equations: i) x + y = 15, x − y = 3

ii) x + y = 0, x − y = 2

iii) 2 x − y = 3, 4 x + y = 3

iv) 2 x − 9y = 9, 5 x + 2y = 27

3. Solve each pair of equations given below using the elimination method: i) x + 2y = −4, 3 x − 5y = −1 122

IL Foundation Series Class 9

ii) 4 x + 9y = 5, −5 x + 3y = 8 ⎛ 3⎞ ⎛ 7⎞ iii) 2y − ⎜ ⎟ = 12, 5y + ⎜ ⎟ = 1 ⎝ x⎠ ⎝ x⎠ ⎛ 3 ⎞ ⎛ 2 ⎞ ⎛ 9 ⎞ ⎛ 9 ⎞ ⎛ 4 ⎞ ⎛ 21 ⎞ iv) ⎜ ⎟ + ⎜ ⎟ = ⎜ ⎟ , ⎜ ⎟ + ⎜ ⎟ = ⎜ ⎟ ⎝ x ⎠ ⎝ y ⎠ ⎝ xy ⎠ ⎝ x ⎠ ⎝ y ⎠ ⎝ xy ⎠ 8 5 3 2 34, 13 , where x ≠ 0, y ≠ 0 . x y x y 5. 2 tables and 3 chairs together cost `2000, whereas 3 tables and 2 chairs together cost `2500. Find the total cost of 1 table and 5 chairs. 4. Solve

6. A fraction is such that if the numerator is multiplied by 3, and 18 the denominator is reduced by 3, we get . However, if the numerator is increased by 8, and 11 2 the denominator is doubled, we get . Find the fraction. 5 7. The units digit of a two-digit number is twice its tens digit. If the digits are reversed, the number is 27 more than the original number. Find the number. 8. The sum of the weights of Rita and Aman is 60 pounds, and the difference is 2. Find the weights of Rita and Aman. 9. Find a two-digit number whose units digit is thrice the tens digit, and if 36 is added to the number, the digits interchange their place. 10. If x = 2 and y = 3 is a solution of the equation 8 x − ay + 2a = 0, then find the value of a . 11. If

x y 2, ax by a 2 b 2 , then find the value of y. a b

12. Find the solution of system of equations 2 x 3y 9, 3 x 4y 5 . 13. The sum of the digits of a two-digit number is 8, and the difference between this number and the one formed by reversing the digits is 18. Find the number. 14. Find the value of x and y by using the using the cross-multiplication method: 3 x + 4y − 17 = 0 4 x − 3y − 6 = 0 15. Solve the system of linear equations using the cross-multiplication method. ax + by − c2 = 0 a 2 x + b 2 y − c2 = 0 16. If 2 is added to each of the two given numbers, their ratio becomes 1 : 2. However, if 4 is subtracted from each of the given numbers, the ratio becomes 5 : 11. Find the numbers.

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LINEAR EQUATIONS IN TWO VARIABLES

III. GRAPH OF LINEAR EQUATION 1. Draw the graph of each of the following linear equations: i) x + 2y = 3 ii) 2 x − 3y = 5 iii) x − y = 1 2. Give the geometric representation of x − 1 = 0 in one variable. Give the geometric representation of x + 5 = 0 in two variables. 3. Find the two linear equations in two variables whose graphs pass through 2, 14 . How many such equations are possible? 4. Ravish tells his daughter Aarushi, "Seven years ago, I was seven times as old as you were then. And, three years from now, I shall be three times as old as you will be." Represent the present ages of Ravish and Aarushi both in terms of linear equations as well as graphically. 5. Draw the graphs of the following linear equations on the same graph paper: 2 x 3y 12, x y 1 6. Draw the graph of 5 x + 3y = 4. Use the graph to find three solutions of the equation. From the graph, check whether x 1, y 3 is a solution of the given equation. 7. The work done by a body on application of a constant force is the product of the constant force and the distance travelled by the body in the direction of force. Express this in the form of a linear equation in two variables and draw its graph by taking the constant force as 3 units. What is the work done when the distance travelled is 2 units? Verify it by plotting the graph. 8. Solve x + y = 5 and x − y = 1 graphically. 9. The force exerted to pull a cart is directly proportional to the acceleration produced in the body. Express the statement as a linear equation of two variables and draw the graph of the same by taking the constant mass equal to 6 kg. Using the graph, find the force required when the acceleration produced is (i) 5 m / s2 , (ii) 6 m / s2. 10. Draw the graph of the linear equation 3 x + 4y = 6. At what points does the graph cut the x-axis and the y-axis? 11. Amit bought two pencils and three chocolates for `11, and Sumeet bought one pencil and two chocolates for `7. Represent this situation in the form of a pair of linear equations. Find the price of one pencil and that of one chocolate graphically. IV. CONSISTENCY/INCONSISTENCY OF LINEAR EQUATION 1. For what value of k, does the pair of equations 4 x − 3y = 9 and 2 x + ky = 11 have no solution? 2. Check if the equations are consistent or not. i) x + 2y = 5 and 3 x + 4y = 20 ii) x + y = 2 and 2 x + 2y = 4 iii) 7 x + y = 10 and x + 7 y = 10 124

IL Foundation Series Class 9

3. Are the following pair of linear equations inconsistent? Justify your answer.

−3 x − 4y = 12 4y + 3 x = 12

4. For the pair of equations px + 3y = −7, 2 x + 6y = 14 to have infinitely many solutions, the value of p should be 1. Is the statement true? Give reasons. 5. For which value(s) of λ , does the pair of linear equations λ x + y = λ 2 and x + λ y = 1 have (i) no solution (ii) infinitely many solutions (iii) a unique solution 6. For which value(s) of k will the pair of equations kx + 3y = k − 3, 12x + ky = k have no solution? 7. For which values of a and b will the following pair of linear equations have infinitely many solutions? x + 2y = 1 ( a − b ) x + ( a + b )y = a + b − 2 8. Two straight paths are represented by the equations x − 3y = 2 and −2x + 6y = 5. Check whether the paths cross each other or not. 9. By the graphical method, find whether the following pair of equations are consistent or not. If consistent, solve them. (i) 3 x + y + 4 = 0 and 6 x − 2y + 4 = 0 (ii) x − 2y = 6 and 3 x − 6y = 0 10. Show graphically that the system of equations 3 x y 4, 6 x 2y 8 has an infinite number of solutions. 11. Show graphically that the system of equations x + y = 3, x + y = 2 has no solution. 12. How many solutions do the pair of equations y = 0 and y = −5 have? 13. Find whether the following pair of linear equations is consistent or inconsistent: 3 x + 2y = 8, 6 x − 4y = 9 14. Solve the following pair of linear equations graphically: x + 3y = 6; 2 x − 3y = 12 Also, find the area of the triangle formed by the lines representing the given equations with the y-axis.

125

LINEAR EQUATIONS IN TWO VARIABLES

WORKSHEET - 2 MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. A linear equation in two variables, x and y, is in the form ax + by + c = 0, where: a) a ≠ 0, b ≠ 0

b) a 0, b 0

c) a 0, b 0

a 0= ,c 0 d)=

2. How many linear equations in x and y can be satisfied by x = 1 and y = 2? a) Only one

b) Two

c) Infinitely many

d) Three

3. The equation x = 5 in two variables can be written as: a) 1.x + 1.y = 5

b) 0 x + 1.y = 5

c) 0. x + 0.y = 5

d) 1. x + 0.y = 5

4. The linear equation 2 x − 5y = 7 has: a) A unique solution

b) Two solutions

c) Infinitely many solutions

d) No solution

5. The graph of the linear equation 2 x + 3y = 6 cuts the y-axis at the point: a) 2, 0

b) 0, 3

c) 3, 0

d) 0, 2

6. The equation in the form y = mx + c will always: a) Intersect the x-axis

b) Intersect the y-axis

c) Passes through the origin

d) Intersect both axes.

7. Among the following, which is not a solution of 2 x + y = 4 is: a) 0, 4

b) 1, 2

c) 2, 0

d) 2, 1

8. The graph of the linear equation x − y = 0 passes through the point: 1 1 a) , 2 2

3 3 b) , 2 2

c) 0, 1

d) 1, 1

c) a, 0

d) a, a

9. Any point on the line y = x is of the form: a) (a, a)

b) 0, a

10. The graph of the line y = −3 does not pass through the point: a) 2, 3

b) 3, 3

c) 0, 3

d) 3, 2

11. The point of the form a, -a will always lie on the line: a) x = a

b) y = −a

c) y = x

d) x + y = 0

12. The positive solution of the equation ax + by + c = 0 always lie in the: a) 1st quadrant

b) 2nd quadrant

c) 3rd quadrant

d) 4th quadrant.

13. If 2, 0 is a solution of the linear equation 2 x + 3y = k, then the value of k is: a) 4 126

b) 5

c) 3

d) 2

IL Foundation Series Class 9

14. The equation of the x-axis is: a) x = 0

b) y = 0

c) x + y = 0

d) x = y

c) x + y = 0

d) x = y

15. The equation of the y-axis is: a) x = 0

b) y = 0

16. If the points 1, 0 and 2, 1 lie on the graph of a) a = 1, b = 1

b) a 1, b 1

x y + = 1, then the values of a and b are: a b

c) a = −1, b = 1

d) a = 0, b = −1

17. If ( 4, 19 ) is a solution of the equation y = ax + 3, then a =? a) 3

b) 4

c) 5

d) 6

18. If ( a, 4 ) lies on the graph of 3 x + y = 10 , then the value of a is: a) 3

b) 1

c) 2

d) 4

19. The equation x − 2 = 0 on the number line is represented by: a) A line

b) A point

c) Infinitely many lines

d) Two lines

20. If ( 2 k − 1, k ) is a solution of the equation 10 x − 9y = 12, then k = ? a) 1

b) 2

c) 3

d) 4

21. The distance between the graphs of the equations x = −3 and x = 2 is: a) 1

b) 2

c) 3

d) 5

22. The distance between the graphs of the equations y = −1 and y = 3 is: a) 2

b) 4

c) 3

d) 1

23. If the graph of the equation 4 x + 3y = 12 cuts the coordinate axes at A and B, then the hypotenuse of a right triangle AOB is of length ________. a) 4 units

b) 3 units

c) 5 units

d) None of these

24. The two lines x = 5 and y = 6 intersect at the point: a) ( 6, 5 )

b) ( 5, 6 )

c) ( 0, 0 )

d) ( −5, 6 )

25. Which of the following is not a solution of the equation 2 x + y = 7? a) ( 1, 5 )

b) ( 3, 1 )

c) ( 1, 3 )

d) ( 0, 7 )

26. The pair of linear equations 4 x + 6y = 9 and 2 x + 3y = 6 has: a) No solution

b) Many solutions

c) Two solutions

d) One solution

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LINEAR EQUATIONS IN TWO VARIABLES

27. Which of the following pairs of equations represent an inconsistent system? a) 3 x − 2y = 8, 2 x + 3y = 1

b) 3 x − y = −8, 3 x − y = 24

c) lx − y = m, x + my = 1

d) 5 x − y = 10, 10 x − 2y = 20

28. The graphical representation of the pair of equations x + 2y − 4 = 0 and 2 x + 4y − 12 = 0 is: a) Intersecting lines

b) Parallel lines

c) Coinciding lines

d) All of these

29. If am ≠ bl, then the pair of equations ax + by = c; x + my = n : a) Has a unique solution

b) Has no solution

c) Has infinitely many solutions

d) May or may not have a solution

30. For what value of k will the pair of equations 4 x − 3y = 9, 2 x + ky = 11 have no solution? a)

9 11

1 2

−3 2

−2 3

31. Two lines are parallel. The equation of one of the lines is 4 x + 3y = 14. The equation of the second line can be: a) 3 x + 4y = 14

b) 8 x + 6y = 28

c) 12 x + 9y = 42

d) −12 x = 9y

32. The point of intersection of the lines y = 3 x and x = 3y is: a) 3, 0

b) 0, 3

c) 3, 3

d) 0, 0

33. If two lines are intersecting at a point, then the equations will have: a) One solution

b) Infinitely many solutions

c) No solution

d) Two solutions

34. If the pair of equations 2 x + 3y − 5 = 0 and 4 x + ky − 10 = 0 has infinite number of solutions, then: a) k =

3 2

b) k = 6

c) k ≠

3 2

d) k ≠ 6

35. If a pair of linear equations is consistent, then the lines represented by these equations will be: a) Parallel

b) Coincident always

c) Intersecting (or) coincident

d) Intersecting always

36. The system of equations x + ky = 0, 2 x − y = 0 has a unique solution if: a) k ≠

128

−1 2

b) k ≠

2 3

c) k ≠

1 2

d) k ≠

3 4

IL Foundation Series Class 9

37. One equation of a pair of dependent linear equations is −5 x + 7 y = 2. The second equation can be: a) 10 x + 14y + 4 = 0

b) −10 x − 14y + 4 = 0

c) −10 x − 14y + 4 = 0

d) 10 x − 14y = −4

38. The sum of the digits of a two-digit number is 9. If 27 is added to it, the digits of the number 8 reversed. The number is: a) 25

b) 72

c) 63

d) 36

39. If four angles of a cyclic quadrilateral are (2 x 1) , y 5  , (2y 15) and (4 x − 7), then: a) x = 33, y = 50 40. If

b) x = 30, y = 53

c) x = 40, y = 43

d) x = 50, y = 33

c) x = 7, y = 2

d) x = 4, y = 4

44 30 55 40 10, 13 , then: x y x y x y x y

a) x = 3, y = 8

b) x = 8, y = 3

41. There are ₹2 and ₹5 coins in a purse. If there are 60 coins of value ₹195, the number of ₹2 coins is: a) 25

b) 30

c) 35

d) 45

42. Two equations in two variables taken together are called: a) Simple equations

b) Quadratic equations

c) Simultaneous equations

d) None of these

43. The solution of a system of equations x y 9, 2 x y 14 is: a) x = 2, y = 7

b) x = 4, y = 5

c) x = 5, y = 4

d) x = 6, y = 3

FILL IN THE BLANKS 1. The equation of a line passing through the point 3, 5 and parallel to the x-axis is __________. 2. The equation of the line parallel to the x-axis and 2 units above the origin is _________. 3. The number of line(s) passing through a point 3, 4 is/are __________. 4. If the line represented by the equation 3 x + α y = 8 passes through the points 2, 2 , then the value of α is _____________. 5. The graph of the linear equation 4 x = 5 in a plane is parallel to __________. 6. The graph of the equation 2 x + 3y = 9 cuts x-axis at the point ___________. 7. The geometrical representation of a linear equation in two variables is _________. 8. The equation of a straight line passing through the points 2, 2 , 0, 0 and 3, 3 is__________.

129

LINEAR EQUATIONS IN TWO VARIABLES

9. The maximum number of points lying on the graph of a linear equation in two variables is _________. 10. The age of x exceeds the age y by 5 years. Express this statement as a linear equation. ____________. 11. If the pair of linear equations a1 x + b1y + c1 = 0 and a2 x + b2 y + c2 = 0 has an infinite number of solutions, then the relation among the coefficients is __________. 12. If k = 2, then the pair of linear equations 3 x − y + 8 = 0 and 6 x − ky = −16 represent ___________ lines. 13. If a pair of linear equations is_________, then the lines will be intersecting or coincident. 14. The graph of y = 5 is a line parallel to the ___________ axis. 15. The graph of 3 x + 2y = 11 is always a ___________ line. 16. A linear equation in two variables has ____________ solution(s). a b c 17. 1 = 1 ≠ 1 , the pair of linear equations is ______________. a2 b2 c2 18. A pair of linear equations which has ___________ solution is called a dependent pair of equations. 19. The pair of equations 6 x − 3y + 10 = 0 and 2 x − y + 9 = 0 represent ___________ lines. 20. The solution of x = 2010, y = 2020 is ___________. 21. If

a1 b1 ≠ , then the pair of linear equations is _________. a2 b2

22. When

a1 b1 c1 = = , there is/are ___________ solution(s). a2 b2 c2

23. The graph of x = 3 is a line parallel to the __________. 24. A system of two linear equations in two variables is inconsistent if their graphs are __________. 25. The pair of linear equations 6 x − 3y + 10 = 0, 2 x − y + 9 = 0, representing are _________ lines. 26. The pair of equations x = 0 and y = −7 has ___________ solution(s). SUBJECTIVE QUESTIONS 1. Write the equation 3 x = 7 in two variables x and y. 2. Express variable y in terms of x, in the equation 3 x − y = 5. 3. Check whether x = 1 and y = −3 is a solution of 2 x − 5y = 17. 4. Find the value of m, if x = −1 and y = 2 is a solution of x + 5y = 3m . 5. Write an equation (linear) in two variables to represent the statement: The cost price of a pencil is one-fourth the cost price of a pen.

130

IL Foundation Series Class 9

6. If 0, p is a solution of the equation 5 x − 3y = 0, then find the value of p. 7. Express 5 = 2 x in the form of ax + by + c = 0 and indicate the values of a, b and c. 8. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. 9. Express y in terms of x from the equation 3 x + 2y = 8 and check whether the point 4, 2 lies on the line. 10. Find the coordinates of the points where the line 2 x − y = 3 meets both axes. 11. Is it true that the pair of equations x + 2y − 4 = 0 and 2 x + 4y − 2 = 0 are consistent? Justify your answer. 12. Two straight roads are represented by the equations x − 2y = 3 and −3 x + 6y = 5. Check whether the roads cross each other or not. 13. Is the pair of linear equations 2 x + 3y − 9 = 0 and 4 x + 6y − 18 = 0 consistent? Justify your answer. 14. Is= x 3= , y 4 a solution of the linear equation 4 x + 3y − 30 = 0? Justify your answer. 15. What is the minimum number of points we need to draw the graph of ax + by + c = 0? 16. For all real values of c, the pair of equations x − 2y = 8 and 5 x − 10 y = c has a unique solution. Justify whether it is true or false. 17. Does the pair of linear equations 2 x − 3y = 1 and 3 x − 2y − 4 have infinite solution? 18. If x = a, y = b is the solution of the equations x − y = 2 and x + y = 4, then find the values of a and b. 19. The pair of linear equations 3 k 1 x 3y 5 0 and 2 x − 3y + 5 = 0 has infinite solutions. Then, find the value of k. 20. Find the point of intersection of the lines represented by 3 x − 2y = 6 and the y-axis.

131

EUCLID’S GEOMETRY

5.1 INTRODUCTION 5.1.1 History and importance of geometry The term geometry comes from the Greek words 'geo’ (earth) and 'metron’ (measure). It arose from the necessity to measure land, particularly in ancient civilisations. Geometry was essential for practical problems, such as land surveying and construction. For example, the ancient Egyptians developed methods for redrawing boundaries after the Nile river flooded, which led to the invention of basic geometric techniques for measuring areas. This knowledge was also used for building structures like pyramids and calculating volumes, such as the volume of a truncated pyramid (a pyramid with the top cut off). In the Indus Valley Civilisation (around 3000 BCE), geometry played a significant role in city planning, such as the arrangement of roads and drainage systems. The bricks used in construction had a ratio of length to breadth to thickness as 4:2:1, showing an understanding of geometric proportions. Additionally, the Sulbasutras of ancient India (around 800-500 BCE) provided geometric instructions for constructing altars used in Vedic rituals. These included shapes like squares, circles, and combinations of rectangles, triangles, and trapeziums. However, geometry at this time was largely practical and not systematic. It wasn’t until ancient Greece that geometry became a formal, logical study, focusing on reasoning and proof. A Greek mathematician, Thales, is credited with providing the first known geometric proof — that a circle is bisected by its diameter. The Greek mathematician Thales is often credited with providing the first known proof, showing that the diameter of a circle divides it into two equal halves. One of his most famous students, Pythagoras (572 BCE), along with his followers, made crucial contributions to geometry, significantly expanding the field. This progress continued until around 300 BCE, when Euclid, a mathematics instructor in Alexandria, Egypt, gathered the knowledge of the time and systematically organised it in his famous work 'Elements’. Euclid divided 'Elements' into thirteen distinct books, which had a lasting influence on the global understanding of geometry for many generations.

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5.2 EUCLID'S DEFINITIONS, AXIOMS AND POSTULATES 5.2.1 Euclid’s definition of basic geometric terms Euclid’s geometry is based on logical reasoning, where abstract concepts such as points, lines, and planes are used to describe the world. These geometric concepts were derived from everyday experiences. Point: A point is a position in space with no length, width, or thickness. It has no dimensions. Line: A line has only length and no breadth. It extends infinitely in both directions. Plane: A plane is a flat surface that extends infinitely in all directions. Euclid began his work with a list of 23 definitions. For example: Definition 1: A point is that which has no part. Definition 2: A line is breadthless length. Definition 3: The ends of a line are points. While these definitions serve to introduce geometric concepts, some terms like “breadth” and “length” require further clarification, leading to more complex definitions. For practical purposes, terms like point, line, and plane are often treated as undefined terms. Their meanings are intuitive, and symbols or models represent them. Along with definitions, Euclid introduced axioms and postulates. Axioms are general, self-evident truths in mathematics. Postulates are assumptions specific to geometry. Some examples of Euclid’s axioms include: (i) Things which are equal to the same thing are equal to one another. (ii) If equals are added to equals, the sums are equal. Some examples of Euclid’s postulates are: (i) A straight line may be drawn from any one point to any other point. (ii) A terminated line can be extended indefinitely. (iii)A circle can be drawn with any centre and any radius. These postulates are accepted without proof, and they form the foundation for geometric reasoning.

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Proofs and theorems in Euclid’s geometry

Euclid’s system relied on proving theorems - statements that are logically derived from definitions, axioms, and postulates. Example 1: Sum of two line segments If A, B, and C are three points on a line, and B lies between A and C, then Euclid’s axiom tells us that: AB + BC = AC. This can be understood geometrically by noting that the length of AC is just the combined length of AB and BC. Example 2: Constructing an equilateral triangle To construct an equilateral triangle on a given line segment AB: (i) Draw a circle with A as the centre and AB as the radius. (ii) Draw another circle with B as the centre and BA as the radius. (iii) The two circles will intersect at point C. (iv) Draw line segments AC and BC to form the triangle ABC. By Euclid's postulate, the radii of the two circles are equal. Therefore, AB = AC = BC, making the triangle equilateral. 5.2.2 Euclid’s axioms Euclid’s axioms (also called common notions) are universally accepted truths in geometry that do not require proof. These axioms serve as foundational principles from which other geometric properties and theorems are derived. Axiom 1: Equality of equals “If two things are equal to the same thing, they are equal to each other.” Example: If A = B and B = C, then according to this axiom, we can conclude that A = C. This is an application of the transitive property of equality in mathematics. Axiom 2: Adding equals “If equal quantities are added to equal quantities, the sums will be equal.” Example: Suppose AB = CD and EF = GH, adding EF to AB and GH to CD will give the same result: AB + EF = CD + GH. This axiom shows that equality is maintained when the same quantity is added to both sides. Axiom 3: Subtracting equals “If equal quantities are subtracted from equal quantities, the remainders will be equal.” 134

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Example: If AB = CD and EF = GH, subtracting EF from AB and GH from CD results in: AB - EF = CD - GH. This shows that subtracting the same quantity from both sides preserves equality. Axiom 4: Coincidence of equal things “Things that coincide with one another are equal to each other.” Example: If you take two shapes, such as two circles, and place them on top of each other so that every point on one circle matches exactly with the other, the circles are congruent and equal in size. This axiom suggests that congruent figures must be equal. Axiom 5: Whole is greater than a part “The whole is greater than any of its parts.” Example: Consider a line segment AB and a point C on the segment. If C divides AB into two parts, AC and CB, then the entire segment AB is greater than either of its parts AC or CB. This axiom reflects that a whole is always greater than any individual portion of it. Axiom 6: Doubling of equal things “Things that are double of the same thing are equal to one another.” Example: If you have two pieces of string, AB and CD, of equal length, then 2  AB and 2  CD must also be equal. This axiom demonstrates that two things, each twice the size of another, are equal in size. Axiom 7: Halving of equal things “Things that are halves of the same thing are equal to one another.” Example: If AB = CD, then AB = CD . 2 2 If two quantities are equal, then dividing them into halves also results in equal parts. 5.2.3 Euclid’s postulates Postulates are specific assumptions about geometry that Euclid assumed to be true. These postulates are the basis for all further constructions and proofs in geometry. Euclid’s five postulates are the foundation of classical (Euclidean) geometry. Here is a description of each: Postulate 1: A straight line can be drawn from any one point to any other point.

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Note: Given two distinct points, there is a unique line that passes through them. Postulate 2: A terminated line can be extended indefinitely in both directions.

Postulate 3: A circle can be drawn with any centre and any radius. Radius Centre

Circle

Postulate 4: All right angles are equal to one another. D

∠ABC = ∠DEF = 90

F o

Postulate 5: If a straight line falling on two straight lines makes the interior angles on the same side less than 180°, then the two lines will meet on that side if extended. B P A C

1 Q 2

∠1 + ∠2 < 180o

The fifth postulate was especially important because it lead to the study of non-Euclidean geometry, where the parallel postulate may not hold true. 136

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5.3 EQUIVALENT VERSIONS OF EUCLID'S FIFTH POSTULATES 5.3.1 Equivalent versions of Euclid’s fifth postulates Euclid’s fifth postulate, also known as the parallel postulate, is one of the most important and widely discussed postulates in Euclidean geometry. It states: “If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, then the two straight lines, if extended indefinitely, meet on that side on which the angles are less than two right angles.

A C

When they are not parallel C

When they are parallel

This postulate, however, is not as intuitive as the other four postulates, and over time, it has led to alternative formulations or equivalent versions. The Playfair’s axiom

This is a modern restatement of Euclid’s fifth postulate, formulated by the mathematician John Playfair in the 18th century. It states: “Through a point not on a given line, there is exactly one line parallel to the given line.”

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5.3.2 Relationship between lines and angles Euclid’s fifth postulate defines the relationship between lines and angles when a transversal intersects two lines. The postulate helps to understand the behavior of angles formed by parallel lines and a transversal. Here’s how the fifth postulate relates to lines and angles: t 1 3 5 7

6 8

Key relationships between lines and angles

Corresponding angles: When a transversal intersects two parallel lines, the angles formed on the same side of the transversal and in the same relative position are equal. Example: If two parallel lines, l and m, are intersected by a transversal t, then the corresponding angles formed at the points of intersection are equal, i.e., ∠1 and ∠5, ∠2 and ∠6, etc. Alternate interior angles: When a transversal intersects two parallel lines, the angles formed on opposite sides of the transversal but between the two parallel lines are equal. Example: If two parallel lines, l and m, are intersected by a transversal t, then alternate interior angles will be equal, i.e., (∠3 and ∠6) and (∠4 and ∠5). Alternate exterior angles: When a transversal intersects two parallel lines, the angles formed on opposite sides of the transversal but outside the two parallel lines are equal. Example: If two parallel lines are intersected by a transversal, then alternate exterior angles will be equal, i.e., (∠1 and ∠8) and (∠2 and ∠7). Consecutive interior angles (or same-side interior angles): When a transversal intersects two parallel lines, the interior angles on the same side of the transversal add up to 180°. Example: If two parallel lines are intersected by a transversal, then the consecutive interior angles will be supplementary (i.e., their sum is 180°), i.e., (∠3 and ∠5) and (∠4 and ∠6).

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QUICK REVIEW •

Geometry originated from practical needs like land measurement, construction, and solving real-life problems in ancient civilisations such as Egypt, India, and Greece.

•

Euclid’s contributions: Euclid formalised geometry into a logical system in his work Elements. He introduced key concepts like points, lines, and planes, though some of these are considered undefined and intuitive.

•

Euclid’s Definitions: A point has no part. A line is a breadthless length. A straight line lies evenly with the points on itself.

•

Euclid’s Axioms: Axioms are self-evident truths used universally in geometry. Equality of Equals (Axiom 1): If two things are equal to the same thing, they are equal to each other. Adding Equals (Axiom 2): If equal quantities are added to equal quantities the sums will be equal. Subtracting Equals (Axiom 3): If equal quantities are subtracted from equal quantities, their remainders will be equal. Coincidence of Equal Things (Axiom 4): Things that coincide with each other are equal to one another. Whole is Greater Than a Part (Axiom 5): The whole is greater than any of its parts. Doubling of Equal Things (Axiom 6): Things that are double of the same thing are equal to one another. Halving of Equal Things (Axiom 7): Things that are halves of the same thing are equal to one another.

•

Euclid’s Postulates: These assumptions are specific to geometry, providing a basis for proving theorems. Postulate 1: A straight line can be drawn between any two points. Postulate 2: A terminated line can be extended indefinitely. Postulate 3: A circle can be drawn with any center and radius. Postulate 4: All right angles are equal to each other. Postulate 5: (Parallel Postulate): If two lines are cut by a transversal, and the interior angles on the same side of the transversal are less than 180°, the lines will meet.

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WORKSHEET - 1 Introduction, Euclid’s definitions, axioms and postulates 1. In the given figure AC = DC, CB = CE, show that AB = DE. E

2. In a triangle PQR, X and Y are the points on PQ and QR respectively. If PQ = QR and QX = QY, show that PX = RY. 3. P and Q are the centres of two intersecting circles. Prove that PQ = QR = PR. 4. In the given figure, if ∠1 = ∠3, ∠2 = ∠4 and ∠3 = ∠4, write the relation between ∠1 and ∠2, using Euclid’s axiom. 3 4

1 2

5. In a triangle ABC, X and Y are the points on AB and BC such that BX = BY and AB = BC. Show that AX = CY. State the Euclid’s Axiom used. 6. Consider two postulates given below: i) Given any two distinct points A and B, there exists a third point C, which is in between A and B. ii) There exist at least three points that are not on the same line. Do these postulates contain any undefined term? Are these postulates consistent? Do they follow with Euclid’s postulates? Explain. 7. Why is Axiom 5, in the list of Euclid’s axioms, considered a universal truth? (Note that the question is not about the fifth postulate.)

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8. Write whether the following statements are true or false? Justify your answer. i) Pyramid is a solid figure, the base of which is a triangle or square or some other polygon and its side faces are equilateral triangles that converges to a point at the top. ii) In the vedic period, squares and circular-shaped altars were used for household rituals, while altars whose shapes were combination of rectangles, triangles and trapeziums were used for public worship. iii) In geometry, we take a point, a line and a plane as undefined terms. iv) If the area of a triangle equals the area of a rectangle and the area of the rectangle equals that of a square, then the area of the triangle also equals the area of the square. v) Euclid’s fourth axiom says that everything equals itself. vi) The Euclidean geometry is valid only for figures in the plane. 9. In the given figure, we have: AC = XD, C is the mid-point of AB and D is the mid-point of XY. Using Euclid’s axiom, show that AB = XY. B

10. In the given figure, we have ∠1 = ∠3 and ∠2 = ∠4. Show that ∠A = ∠C. B

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WORKSHEET - 2 I. MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. The number of dimensions, a point has a)

Definition

Postulate

A proof

2. Which of the following needs a proof? a)

Theorem

Axiom

3. 'Lines are parallel if they do not intersect' is stated in the form of a)

An axiom

A definition

A postulate

4. The basic facts that are taken for granted, without proof and which are specific to geometry are called a)

Axiom

Postulates

Concurrent lines

Definition

5. Three or more lines intersecting at the same point are said to be a)

Parallel lines

Intersecting lines

Concurrent lines

Straight line

6. Assertion (A): According to Euclid’s axiom, when equals are added to equals, wholes are equal. Reason (R): If Rita and Rivi are of the same age that is 10 years then after 6 years also they will have the same age. a)

Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)

Assertion (A) is true but reason (R) is false.

Assertion (A) is false but reason (R) is true.

7. Assertion (A): According to Euclid’s 1st axiom-Things which are equal to the same thing are also equal to one another. Reason (R): If AB = PQ and PQ = XY, then AB = XY. a)

Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

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Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).

Assertion (A) is true but reason (R) is false.

Assertion (A) is false but reason (R) is true.

8. John is of the same age as Mohan. Ram is also of the same age as Mohan. State the Euclid’s axiom that illustrates the relative ages of John and Ram a)

First axiom

Second axiom

Third axiom

Fourth axiom

9. The three steps from solids to points are: a)

Solids - surfaces - lines - points

Solids - lines - surfaces - points

Lines - points - surfaces - solids

Lines - surfaces - points – solids

10. The number of dimensions, a solid has: a)

11. Euclid divided his famous treatise “The Elements” into: a)

13 chapters

12 chapters

11 chapters

9 chapters

Curve

Lines

Points

Lines

Surfaces

Points

Euclid

Both A and B

Archimedes

12. Boundaries of surfaces are: a)

Surfaces

13. Boundaries of solids are: a)

Curve

14. Pythagoras was a student of: a)

Thales

15. It is known that if x + y = 10 then x + y + z = 10 + z. The Euclid’s axiom that illustrates this statement is: a)

First axiom

Second axiom

Third axiom

Fourth axiom

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II. FILL IN THE BLANKS 1. The axiom that says “Things which are __________ of the same things are equal to one another” is one of Euclid’s common notions. 2. Euclid’s fifth axiom states that the __________ is greater than the part. 3. A solid has __________dimensions. 4. A point has _______ dimension. 5. There are ________ number of Euclid’s postulates. III. SUBJECTIVE QUESTIONS 1. Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they, and how might you define them? i. parallel lines ii. perpendicular lines iii. line segment iv. radius of a circle v. square 2. Using Euclid’s axiom, compare length of AD and AF. State which axiom you used here. Also give two more axioms other than the axiom used in the above situation. A

1 2

3. In the figure, if OX = XY, PX = XZ and, OX = PX, Show that XY = XZ. State which axiom you used here. Also give two more axioms other than the axiom used in the above situation. X

4. In the given figure, we have ∠ABC = ∠ACB, ∠3 = ∠4. Show that ∠1 = ∠2.

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D 4 B

3 C

5. In the given figure: (i) AB = BC, M is the mid-point of AB and N is the mid-point of BC. Show that AM = NC.

(ii) BM = BN, M is the mid-point of AB and N is the mid-point of BC. Show that AB = BC. A

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6.1 BASIC TERMS AND DEFINITIONS Angle: An angle is the union of two non-collinear rays with a common initial point. The two rays forming an angle are called the 'arms' of the angle, and the common initial point is called the 'vertex' of the angle. C

Notation: The angle formed by two rays, AB and AC, is denoted by the symbol ∠BAC or ∠CAB or ∠A . 6.1.1 Types of angles 1. Acute angle: An angle whose measure is less than 90° is called an acute angle. Example: 30°,77°,70° , etc.

C Acute angle

2. Right angle: An angle whose measure is exactly 90° is called a right angle. C Right angle o

3. Obtuse angle: An angle whose measure is greater than 90° and less than 180° is called an obtuse angle. Example: 120°,135°,140° , etc. C Obtuse angle

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4. Straight angle: An angle whose measure is 180° is called a straight angle. 180o

5. Reflex angle: An angle whose measure is more than 180° and less than 360° is called a reflex angle. Reflex angle

6.1.2 Angle bisector A ray AX is said to be the bisector of ∠BAC . X is a point in the interior of ∠BAC such that ∠BAX = ∠XAC. C X

1 Thus, if AX is the bisector of ∠BAC , then ∠BAX = ∠CAX = ∠BAC. 2

6.2 PAIRS OF ANGLES Complementary angles: Two angles are said to be complementary if their sum is 90° . Example: 30° and 60° are complementary angles. Supplementary angles: Two angles are said to be supplementary if their sum is 180° . Example: 70° and 110° are supplementary angles. Explementary angles: Two angles are said to be explementary if their sum is 360°. Example: 140° and 220° are explementary angles. Adjacent angles: Two angles are said to be adjacent if i) they have a common vertex i) they have a common arm iii) their non-common arms are on different sides of the common arm

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C D

Here, ∠ABD and ∠DBC are adjacent angles. Linear pair of angles: Two adjacent angles are said to form a linear pair of angles if their noncommon arms are two opposite rays. And the angles in a linear pair are supplementary. C

Here, ∠AOC and ∠BOC form a linear pair. Vertically opposite angles: Two angles are called a pair of vertically opposite angles if their arms form two pairs of opposite rays. Here, i) ∠AOC and ∠BOD form a pair of vertically opposite angles. ii) ∠BOC and ∠AOD form a pair of vertically opposite angles. C

B O

SOLVED EXAMPLES Example 1: In the figure, AB and AC are opposite rays. i) If ∠BAD =° 85 , find ∠CAD . ii) If ∠CAD = 135° , find ∠BAD . D

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Solution: i) Since ∠BAD and ∠DAC form a linear pair. ∴∠BAD + ∠DAC = 180° ⇒ 85° + ∠DAC =180° ⇒ ∠DAC = 180° − 85° = 95° ii) ∠BAD + ∠DAC =180° ⇒ ∠BAD + 135° =180° ⇒ ∠BAD =180° − 135° = 45° Example 2: In the figure, ∠AOC and ∠BOC form a linear pair. Determine the value of x. C

Solution:

4xo 2xo O

Since ∠AOC and ∠BOC form a linear pair. ⇒ ∠AOC + ∠BOC =180° ⇒ 4 x° + 2 x° = 180° ⇒ 6 x° = 180° 180° x° = ⇒ = 30° 6 x° = 30° ⇒ Example 3: In the figure, ∠POR and ∠QOR form a linear pair. If a − b = 80° , find the value of a and b. R

ao bo P

Solution: Since ∠POR and ∠QOR form a linear pair. ∴∠POR + ∠QOR = 180° ⇒ a + b= 180° .....(1)

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80° .....(2) Given that a − b = By adding (1) and (2), we get, 2= a 260° 260° ⇒ a= = 130° 2 From (1), a+b = 180° ⇒ 130° + b = 180° ⇒ = b 180° − 130° ⇒ b = 50° = ∴ a 130 = ° and b 50°. Example 4: In the given figure, OA and OB are opposite rays. Find the value of x. Further, find ∠AOC and ∠BOD. C

(x + 20)o A

xo O

(x + 10)o B

Solution: Since ∠AOC , ∠COD , and ∠BOD form a linear pair. ∴ ∠AOC + ∠COD + ∠BOD = 180° ⇒ (x + 20)° + x° + (x + 10)= ° 180° ⇒ 3 x + 30= ° 180° ⇒ 3= x 180° − 30° ⇒ 3 x = 150° 150° ⇒x= = 50° 3 ⇒x= 50° Now, ∠AOC= (x + 20)°= (50 + 20)°= 70° ∠BOD =(x + 10)°= (50 + 10)°= 60° Example 5: In the figure given below, POQ is a line, ray OR is perpendicular to line PQ, and OS is 1 another ray lying between rays OP and OR. Prove that ∠ROS= (∠QOS − ∠POS) . 2

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Solution:

Given: Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. To prove: ∠ROS=

1 (∠QOS − ∠POS) 2

Proof: Consider, ∠QOS − ∠POS = (∠QOR + ∠ROS) − ∠POS

= 90° + ∠ROS − ∠POS

= 90° − ∠POS + ∠ROS

= (∠POR − ∠POS) + ∠ROS

= ∠ROS + ∠ROS

= 2∠ROS

⇒ ∠QOS − ∠POS = 2∠ROS ∴∠ROS=

1 (∠QOS − ∠POS) 2

Hence proved. Example 6: In the figure given below, PO, OQ, OR, and OS are four rays. Prove that ∠POQ + ∠QOR + ∠SOR + ∠POS =360° P

O R

Solution: Let us produce a ray OT from the point T opposite to the ray OQ so that TOQ is a line. P Q

T O S

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Now, the ray OP stands on a line TOQ. Therefore, ∠TOP + ∠POQ = 180°

…(1)

Similarly, the ray OS stands on TOQ. Therefore, ∠TOS + ∠SOQ = 180°

…(2)

But we have, ∠SOQ = ∠SOR + ∠ROQ So, equation (2) becomes, ∠TOS + ∠SOR + ∠ROQ =180°

…(3)

Now, adding (1) and (3), we get, ∠TOP + ∠POQ + ∠TOS + ∠SOR + ∠ROQ = 360°

…(4)

But ∠TOP + ∠TOS = ∠SOP Therefore, equation (4) becomes ∠POQ + ∠QOR + ∠SOR + ∠POS = 360° . Example 7: In the figure, lines p and r intersect at the point O. If x = 45° , then find y, z, and u.

p y z

x u r

Solution: Given lines p and r intersect at O. ∴∠z = ∠x [vertically opposite angles] But ∠x = 45° (Given) ⇒ ∠z =45° Now, ∠x + ∠y =180° ⇒ 45° + ∠y =180° ⇒∠ = y 180° − 45° ∴∠y = 135° Again, ∠u =∠y (vertically opposite angles) ⇒ ∠u =135° Hence, = y 135 = °, z 45° and u = 135°.

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6.3 PARALLEL LINES AND TRANSVERSALS Transversal: A line which intersects two or more given lines at distinct points is called a transversal of the given lines. L

P Q

In the above figure, AB and CD are given two lines, and a transversal LM intersects them at points P and Q, then eight angles are formed. B

2 1 P 3 4

6 5Q 7 8

Alternate angles: Two angles are said to be a pair of alternate angles if a) they are on either side of the transversal. b) both are interior angles or exterior angles. c) they are not adjacent angles. Note: If alternate angles are equal, then the lines will be parallel. Alternate interior angles: A pair of angles on opposite sides of the transversal but inside the two lines are called alternate interior angles. Alternate exterior angles: The pairs of angles on opposite sides of the transversal but outside the two lines are called alternate exterior angles. The pair of angles (∠2, ∠8);(∠1, ∠7);(∠3, ∠5);(∠4; ∠6) are called the pair of alternate angles. Pair of alternate interior angles: (∠3, ∠5) and (∠4, ∠6) Pairs of alternate exterior angles: (∠2, ∠8) and (∠1, ∠7) Corresponding angles: Two angles are said to be a pair of corresponding angles if a) they are on the same side of the transversal. b) one is the interior angle, and the other is the exterior angle. c) they are not adjacent angles.

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Note: If the corresponding angles are equal, then the lines are parallel. The pair of angles (∠1, ∠5);(∠2, ∠6);(∠4, ∠8);(∠3, ∠7) are called the pairs of corresponding angles. Co-interior (or) Allied angles: The angles that are interior and on the same side of the transversal are called allied angles or co-interior angles. From the figure, (∠3, ∠6) and (∠4, ∠5)are called allied angles. Note: •

If a transversal intersects two parallel lines, then each pair of corresponding angles are equal.

•

I f a transversal intersects two lines, making a pair of corresponding angles equal, then the lines are parallel.

•

If a transversal intersects two parallel lines, then each pair of alternate interior angles are equal.

•

I f a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel.

•

I f a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.

•

I f a transversal intersects two lines such that a pair of interior angles on the same side of the transversal are supplementary, then the two lines are parallel.

SOLVED EXAMPLES Example 1: In the figure, transversal p intersects two lines, m and n, ∠4= 110°, ∠7= 65°. Is m  n ? p 2 3 6 7

Solution: Since ∠5 =∠7

[ vertically opposite angles]

∠5 =65° Again, ∠4 + ∠= 5 110° + 65= ° 175° ≠ 180° Since ∠4 and ∠5 are interior angles on the same side of the transversal p, their sum is not equal to 180° . ∴m is not parallel to n. 154

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Example 2: In the figure, if AB  CD , ∠APQ = 50° and ∠PRD =,find 127° x and y. A

P 50

127o

x C

R D

Solution: We have, ∠APR = ∠PRD [Alternate angles] ⇒ 50° += y 127° ⇒ y= 127° − 50°= 77° ∴ y = 77° Also, ∠APQ = ∠PQR [ Alternate angles ] ⇒ 50° = x Hence, x = 50° and y = 77° . Example 3: In the figure, find the values of x and y and show that AB  CD.

50o A

y C

130o

Solution: We have x + 50° = 180° [Linear pair] = ⇒ x 180 = ° − 50° 130° ∴ x = 130° y = 130° [Vertically opposite angles] So, alternate interior angles are equal, i.e., x = y = 130° . Hence, AB  CD .

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Example 4: In the figure, PQ  RS , find x. P

135o xo

142o

Solution: Draw TA  PQ  RS P

Q 135

1420

Since PQ  TA and PT is the transversal. ∴∠1 + ∠QPT = 180° ⇒ ∠1 + 135° =180° ⇒= ∠1 180° − 135 = ° 45° ⇒ ∠1 =45° Again, TA  RS and TR is transversal. ∴ ∠2 + ∠TRS= 180° ⇒ ∠2 + 142= ° 180° ⇒ ∠2= 180° − 142°= 38° ⇒ ∠2= 38° But x° = ∠1 + ∠2 = 45° + 38° = 83° ∴ x° = 83° Example 5: In the figure, PQ  ST . Find ∠a . S P

130o

110o a

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Solution: Draw EF parallel to PQ and also parallel to ST. S P

130o

110o ao

Since, PQ  ER

∠ERQ + ∠RQP =180° ⇒ ∠ERQ + 110° =180° ⇒ ∠ERQ = 180° − 110 = ° 70° ∴∠ERQ = 70° Again, EF  ST and RS is the transversal ÐTSR +ÐFRS = 180° Þ 130° +ÐFRS = 180° Þ ÐFRS = 180° - 130° = 50° \ ÐFRS = 50° Also, ∠ERQ + ∠a + ∠FRS = 180° [sum of all angles in a straight line] ⇒ 70° + ∠a + 50° = 180° ⇒∠ = a 180° − 120 = ° 60° ∴∠a = 60° Example 6: In the figure, if AB  CD, CD  EF and y : z = 3 : 7 , then find x, y and z.

A x

B y

C E

D F

Solution: We have, y : z = 3 : 7 Let y = 3 k and z = 7 k . Clearly, x and z are alternate angles, and they are equal.

157

LINES AND ANGLES

Þx=z Þ x = 7k

[ z = 7 k ]

Since x and y form a pair of interior angles on the same side of the transversal. ∴x + y = 180° ⇒ 7 k + 3k = 180° ⇒ 10 k = 180° 180° ⇒= k = 18° 10 Hence, x= 7 k =× 7 18° = 126° y= 3 k =× 3 18° = 54° z= 7 k =× 7 18° = 126°

6.4 ANGLE SUM PROPERTY OF A TRIANGLE Triangle: A plane figure bounded by three line segments is called a triangle. Theorem: The sum of the angles in a triangle is 180° . A

A 4

Given: A triangle ABC To prove: ∠A + ∠B + ∠C =180° i.e., ∠1 + ∠2 + ∠3 =180° Construction: Through A, draw a line l parallel to BC. Proof: Since l  BC and AB is a transversal. ∠2 =∠4 ...(1)

[Alternate interior angles]

Since l  BC and AC is a transversal. ∠3 =∠5 ...(2)

[Alternate interior angles]

By adding (1) and (2), we get, ∠2 + ∠3 = ∠4 + ∠5 Now add ∠1 on both sides

158

IL Foundation Series Class 9

∠1 + ∠2 + ∠3 = ∠1 + ∠4 + ∠5 ⇒ ∠1 + ∠2 + ∠3 =180° [Sum of angles in a straight line] ∴∠A + ∠B + ∠C =180°

6.5 EXTERIOR ANGLE PROPERTY Exterior angle: If the side BC of a triangle ABC is produced to form ray BD, then ∠ACD is called an exterior angle of ∆ABC at C and is denoted by ext. ( ∠ACD ) . Theorem: If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles. A 1 2

Given: In ∆ABC, D is a point on BC produced to D such that an exterior angle is formed. To prove: ∠4 = ∠1 + ∠2 Proof: In ∆ABC , We have ∠1 + ∠2 + ∠3 =180°

… (1)

Also, ∠3 + ∠4 =180°

… (2)

[Linear pair]

From (1) and (2), we get, ∠1 + ∠2 + ∠3 = ∠3 + ∠4 ⇒ ∠1 + ∠2 = ∠4 So, ∠4 = ∠1 + ∠2 . Hence proved.

SOLVED EXAMPLES Example 1: In the figure, given that ∠AOC = ∠ACO and ∠BOD = ∠BDO . Prove that AC  DB . Solution: D A O C

159

LINES AND ANGLES

We have ∠AOC = ∠ACO and ∠BOD = ∠BDO but ∠AOC = ∠BOD [Vertically opposite angles] ∴∠ACO = ∠BOD and ∠BDO = ∠BOD ⇒ ∠ACO = ∠BDO Thus, AC and BD are two lines intersected by transversal CD such that ∠ACO = ∠BDO . That is, alternate angles are equal. Therefore, AC  DB . Example 2: In a triangle, two angles are equal, and the third angle is greater than each of the angles by 15° . Find all the angles of the triangle. Solution: Let each of the two equal angles be x° . ∴ Third angle = (x + 15)° Now x + x + ( x + 15° ) = 180°

[Angle sum property]

⇒ 3 x + 15= ° 180° ⇒ 3 x = 165° ⇒ x = 55° ∴ The required angles are 55°, 55°, 70°. Example 3: In the given figure, ∠ACD= 120°, ∠CAE= 130° . Find ∠ABC . E A 130o

120o C D

Solution: Given ∠ACB + ∠ACD = 180°

[Linear pair]

⇒ ∠ACB + 120= ° 180° ⇒ ∠ACB= 60° Also, ∠BAC + ∠CAE =180°

[Linear pair]

⇒ ∠BAC + 130= ° 180° ⇒ ∠BAC= 50° Now in ∆ABC ,

160

IL Foundation Series Class 9

∠ABC + ∠BCA + ∠CAB = 180° ⇒ ∠ABC + 60° + 50= ° 180° ⇒ ∠ABC + 110= ° 180° ⇒ ∠ABC= 70° Example 4: If the bisectors of ∠ABC and ∠ACB of a triangle ABC meet at O, then prove that 1 ∠BOC= 90° + ∠A. 2 A O

Solution: Given: ∆ABC such that the bisectors of ∠ACB and ∠ABC meet at a point O. 1 To prove: ∠BOC= 90° + ∠A 2 Proof: In ∆BOC, ∠1 + ∠2 + ∠BOC = 180° ...(1) Also in ∆ABC C 180° ∠A + ∠B + ∠= ⇒ ∠A + 2∠1 + 2∠2= 180° ⇒ 2(∠1 + ∠2) = 180° − ∠A ∠A ⇒ ∠1 + ∠2= 90° − ...(2) 2

From (1) & (2), we get, ∠A + ∠BOC = 180° 2 ∠A ⇒ ∠BOC = 180° − 90° + 2 1 ⇒ ∠BOC= 90° + ∠A 2

90° −

Example 5: The sides AB and AC of ∆ABC are produced to P and Q, respectively. If the bisectors of 1 ∠PBC and ∠QCB intersect at O, then prove that ∠BOC= 90° − ∠A . 2

161

LINES AND ANGLES

B 1 2

C Q

P O

Solution: Given: A ∆ABC in which sides AB and AC are produced to P and Q, respectively. The bisectors of ∠PBC and ∠QCB intersect at O. 1 To prove : ∠BOC= 90° − ∠A 2 Proof: Since ∠ABC and ∠CBP form a linear pair. \ ÐABC +ÐCBP = 180° Þ ÐB + 2Ð1 = 180° [ ÐCBP = 2Ð1] Þ 2Ð1 = 180° -ÐB 1 Ð1 = 90 - ÐB¼ (1) 2 Again ∠ACB and ∠QCB form a linear pair. Therefore, ∠ACB + ∠QCB =180° ⇒ ∠C + 2∠2 =180° ⇒ 2∠= 2 180° − ∠C 1 ⇒ ∠2 90° − ∠C… (2) = 2 In ∆BOC , we have ∠1 + ∠2 + ∠BOC =180° 1 1 180° ⇒ 90° − ∠B + 90° − ∠C + ∠BOC = 2 2 1 ⇒ 180° − (∠B + ∠C) + ∠BOC =180° 2 1 ⇒ ∠BOC= (∠B + ∠C) 2 1 ⇒ ∠BOC = (180° − ∠A ) 2 1 ⇒ ∠BOC= 90° − ∠A 2 162

IL Foundation Series Class 9

Example 6: In the given figure, find x and y if AB  DF and AD  FG . A

65ᵒ y 125

C B

ᵒ

Solution: ∠y + 125° = 180° (Linear pair) ⇒∠ = y 180° − 125° ⇒ ∠y =55° Now, AB  FD , transversal AD cuts then

∠D = ∠A (Alternate angles) ⇒ ∠D= 65° Again, AD  FG , transversal FD cuts, then ∠F =∠D ⇒ ∠F =65° In ∆EFG , ∠x + ∠F + ∠y =180° ⇒ ∠x + 65° + 55° =180° ⇒∠ = x 180° − 120° ⇒ ∠x =60° ∴∠ = x 60°, ∠ = y 55° Example 7: In the figure, sides QP and RQ of ∆PQR are produced to points S and T, respectively. If ∠PQT =135° and ∠PQT =110° find ∠PRQ. S P 135o

110o

163

LINES AND ANGLES

Solution: We have ∠TQP + ∠PQR =180° (Linear pair) ⇒ 110° + ∠PQR = 180° ⇒ ∠PQR= 180° − 110°= 70° Also ∠SPR + ∠QPR = 180° (Linear pair) ⇒ 135° + ∠QPR = 180° ⇒ ∠QPR= 180° − 135°= 45° In ∆PQR, ∠QPR + ∠PRQ + ∠PQR =180° ⇒ 45° + ∠PRQ + 70° =180° ⇒ ∠PRQ = 180° − 115° = 65° 65° ∴ ∠PRQ =

QUICK REVIEW •

I f a ray intersects a line, then the sum of the two adjacent angles formed is 180 degrees, and conversely, if the sum of two adjacent angles is 180 degrees, then the ray intersects the line. This fundamental property is known as the linear pair axiom.

•

If two lines intersect each other, then the vertically opposite angles are equal.

•

When a transversal intersects two parallel lines: i) Corresponding angles are congruent. ii) Alternate interior angles are congruent. iii) Interior angles on the same side of the transversal are supplementary.

•

When a transversal intersects two lines: i) If any pair of corresponding angles are congruent, ii) If any pair of alternate interior angles are congruent, iii) If any pair of interior angles on the same side of the transversal are supplementary, then the lines are parallel.

164

•

Lines which are parallel to a given line are parallel to each other.

•

The sum of the three interior angles of a triangle is 180°.

•

hen a side of a triangle is extended, the exterior angle formed is equal to the sum of the two W interior opposite angles.

IL Foundation Series Class 9

WORKSHEET - 1 I. BASIC TERMS AND DEFINITIONS, COMPLEMENTARY ANGLES, SUPPLEMENTARY ANGLES AND EXPLEMENTARY ANGLES 1. Write the complement of each of the following angles: (i) 20°

(iii) 90°

(ii) 35°

(iv) 77°

(v) 30°

2. Write the supplement of each of the following angles: (i) 54°

(ii) 132°

(iii) 138°

3. If an angle is 28° less than its complement, find its measure. 4. If an angle is 30° more than half of its complement, determine the measurement of an angle. 5. Two supplementary angles have a ratio of 4 : 5 . Determine the measures of the angles. 6. Two supplementary angles have a difference of 48°. Determine the measures of the angles. 7. An angle is 8 times its complementary angle. Find the angle. 8. If the angles (2 x − 10)° and (x − 5)° are complementary angles, find x. 9. If an angle differs from its complement by 10° , find the angle. 10. If the supplement of an angle is two-third of the angle itself, calculate the measures of the angle and its supplement. 11. An angle is 14° more than its complementary angle. What is its measure? 12. An angle measures twice its supplementary angle. Determine the measurement of an angle. 13. If the complement of an angle equals the supplement of thrice its measure, calculate the measurement of an angle. 14. When the supplement of an angle is three times its complement, determine the measurement of an angle. 15. Write the explementary angle of each of the following angles: (i) 147°

(ii) 259°

(iii) 330°

II. ADJACENT ANGLES AND LINEAR PAIR OF ANGLES 1. In the below figure, OA and OB are opposite rays: (i) If x = 25°, what is the value of y? (ii) If y = 35°, what is the value of x? C (2y+5)o A

3xo B 165

LINES AND ANGLES

2. In the figure below, write all pairs of adjacent angles and all the linear pairs. D

3. In the given figure, find x. Further, find ∠BOC, ∠COD, and ∠AOD. C

(x + 10)

(x + 20)

4. In the figure, rays OA, OB, OC, OD, and OE have the common endpoint O. Show that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA =360°. B

C O

A D

5. In the figure, ∠AOC and ∠BOC form a linear pair. If a − 2b =° 30 , then find a and b. C a A

b O

6. How many pairs of adjacent angles are formed when two lines intersect at a point? A

O C

7. How many pairs of adjacent angles, in all, can you name in the figure given? D

166

IL Foundation Series Class 9

8. In the figure, determine the value of x. B 3x

O 150

9. In the figure, AOC is a line. Find x. B A

70o

2xo O C

10. In the figure, POS is a line. Find x. Q P

4x O

40o

III. VERTICALLY OPPOSITE ANGLES 1. In the figure, lines l1 and l2 intersect at O, forming angles as shown in the figure. If x = 45°. Find the values of y, z, and u. l1 z

y O u

2. In the figure, three coplanar lines intersect at point O, forming angles as shown. Find the values of x, y, z, and u. E

z u

y x O

90° 50°

167

LINES AND ANGLES

3. In the figure, find the values of x, y, and z. l1

y x

4. In the figure, find the value of x. F

5. If one of the four angles formed by two intersecting lines is a right angle, then show that each of the four angles is a right angle. 6. In the given figure, rays AB and CD intersect at O. (i) Determine y when x = 60° . (ii) Determine x when y = 40° . C

y O

7. In the given figure, lines AB, CD, and EF intersect at O. Find the measure of ∠AOC , ∠COF, ∠DOE , and ∠BOF . F

y o

168

B o

IL Foundation Series Class 9

8. AB, CD, and EF are three concurrent lines passing through the point O such that OF bisects ∠BOD . If ∠BOF = 35° , then find ∠BOC and ∠AOD . C E O

F D

9. In the given figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, then find ∠BOE and reflex ∠COE . C

E O

B D

10. Fill in the blanks so as to make the following statements true. (i) If one angle of a linear pair is acute, then its other angle will be ___________. (ii) A ray stands on a line, then the sum of the two adjacent angles so formed is _________. (iii) If the sum of two adjacent angles is 180°, then the ____________ arms of the two angles are opposite rays. IV. PARALLEL LINES AND TRANSVERSALS 1. In the figure, AB || CD and ∠1 and ∠2 are in the ratio 3 : 2 . Determine all angles from 1 to 8 . l 1 4

A 5 C

2 B

6 D

2. In the figure, l, m, and n are parallel lines intersected by transversal p at x, y, and z, respectively. Find Ð1, Ð2, and ∠3 . x 1 y

l m

120°

169

LINES AND ANGLES

3. In the figure, if AB  CD and CD  EF , find ∠ACE. B

A 70o E

130o C

4. In the figure, state which lines are parallel and why? A o 78

B D E

100o C

100o

5. In the figure, if l  m, n  p and ∠1 =85° , then find ∠2 . n

p m

3 2

6. If two straight lines are perpendicular to the same line, prove that they are parallel to each other. 7. Two unequal angles of a parallelogram are in the ratio 2 : 3 . Find all its angles in degrees. 8. If each of the two lines is perpendicular to the same line, what kind of lines are they to each other? 2 9. In the given figure, ∠1 =60° and ∠2 =   3

of a right angle. Prove that l  m. n l

2 3

170

IL Foundation Series Class 9

10. In the given figure, if l  m  n and ∠1 =60° , then find ∠2. p 1 3

l m n

V. ANGLE SUM PROPERTY OF TRIANGLES 1. In a triangle ABC, if ∠A = 55° , ∠B = 40° , then find ∠C . 2. If the angles of a triangle are in the ratio 1 : 2 : 3 , determine the three angles. °

1  3. The angles of a triangle are (x − 40) ,(x − 20) , and  x − 10  . Find the value of x. 2  4. Two angles of a triangle are equal, and the third angle is greater than each of those angles by 30° . Determine all the angles of the triangle. °

5. If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right angle triangle. 6. The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10° , then find the three angles. 7. ABC is a triangle in which ∠A = 72° , the internal bisectors of angles B and C meet at O. Find the magnitude of ∠BOC . 8. If the bisectors of the base angles of a triangle enclose an angle of 135° , prove that the triangle is a right triangle. 9. In a ∆ABC , ∠ABC = ∠ACB and the bisectors of ∠ABC and ∠ACB intersect at O such that ∠BOC = 120° . Show that ∠A =∠B =∠C =60° . 10. If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled. VI. EXTERIOR ANGLE PROPERTY 1. The exterior angles obtained from producing the base of a triangle both ways are 104° and 136° . Find all the angles of the triangle. 2. In the figure, the sides BC, CA and AB of a ∆ABC have been produced to D, E, and F respectively. If ∠ACD = 105° and ∠EAF = 45° , find all the angles of the ∆ABC . 3. In the given figure, AC ⊥ CE and ∠A : ∠B : ∠C = 3 : 2 : 1 , find the value of ∠ECD .

171

LINES AND ANGLES

A E

4. In the given figure, AB  DE . Find ∠ACD . A

300

400

5. Fill in the blanks to make the following statements true: (i) Sum of the angles of a triangle is ______. (ii) An exterior angle of a triangle is equal to the two ____ opposite angles. (iii) An exterior angle of a triangle is always ____ than either of the interior opposite angles. (iv) A triangle cannot have more than ____ right angles. (v) A triangle cannot have more than ____ obtuse angles. 6. In a ∆ABC , the internal bisectors of ∠B and ∠C meet at P, and the external bisectors of ∠B and ∠C meet at Q. Prove that ∠BPC + ∠BQC =180° . 7. In the figure, AB divides ∠DAC in the ratio 1 : 3 and AB = DB . Determine the value of x. E A 108o

x D

8. ABC is a triangle. The bisector of the exterior angle at B and the bisector of ∠C intersect each 1 other at D. Prove that ∠D = ∠A . 2 9. In the given figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠B = 65° and ∠C = 33°, then find ∠MAN . A

65o

33o

10. In a ∆ABC , AD bisects ∠A and ∠C > ∠B . Prove that ∠ADB > ∠ADC.

172

IL Foundation Series Class 9

WORKSHEET - 2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER

1. Side BC of a triangle ABC has been produced to a point D such that ∠ACD = 120° . If 1 ∠B = A then A is equal to 2 a) 80° b) 75° c) 60° d) 90° 2. In

ABC, if ∠A =100 , AD bisects ∠A and AD ⊥ BC, then ∠B is

a) 50°

b) 90°

c) 40°

d) 100°

3. If the bisectors of the acute angles of the right triangle meet at O, then the angle at O between the two bisectors is a) 45°

b) 95°

c) 135°

d) 90°

4 In the figure, the value of y in terms of x is A

y C

4 3 x b) x c) x 3 2 5. In the figure, if BF  CQ and AC = BC , then the measure of x is

4 x 3

20o x B

a) 30°

b) 25°

70o

c) 20°

d) 35°

6. The base BC of triangle ABC is produced both ways and measure of exterior angles formed 94° and 126° , then ∠BAC is a) 94°

b) 54°

c) 40°

d) 44°

173

LINES AND ANGLES

7. In the figure, if ∠b − ∠a = 45°, then ∠AOD is B 90o

a) 60°



d) 150°

c) 45°

b) 180°

8. In ABC , the sides AB and AC are produced to P and E, respectively, the bisector of ∠PBC and ∠ECB intersect at O, then ∠BOC is A B

E O

1 a) 90 − ∠A 2

∠B ∠C + 2 2

c) ∠B + ∠C

d) ∠A + ∠C

9. In the figure, the value of x is D x

25o

55o

40o

a) 65°

10. In the figure, if

b) 80°

c) 90°

174

d) 120°

y z = 5 and = 4 , then the value of x is x x

x a) 90°

b) 72°

zo c) 18°

d) 36°

IL Foundation Series Class 9

11. In the figure, ∠AOC and ∠BOC form a linear pair. If a − 2b = 30°, then find a and b. C

a A

a) 150° ,30°



b O

b) 130° ,50°

c) 140° , 40°

d) 105° , 75°

12. In ABC, ∠A = 150° and BC is produced to a point D if the bisectors of ∠ABC and ∠ACD meet at E, then ∠E = ? a) 75°

b) 40°

c) 50°

d) 20°

13. If l, m, n are lines in the same plane such that l intersects m and n, n  m then l and n are: a) Parallel lines

b) Intersecting lines

c) Always perpendicular

d) Always intersecting at 60°

14. In the figure, ∠PQR is o

75 P

105 R

a) 40°

b) 50

d) 105°

c) 30°

15. In the figure, if ll  l2  l3 , then the value of x is 40°

l1 l2 l3

a) 40°

b) 140°

c) 50°

d) 80°

16. If the angles of a triangle are in the ratio 3 : 5 : 7 , then the triangle is a) Acute angled

b) Right angled

c) Obtuse angled

d) Isosceles

17. If one of the angles of a triangle is 130° , then the angle between the bisectors of the other two angles can be a) 50°

b) 65°

c) 145°

d) 155°

175

LINES AND ANGLES

18. In the given figure, if ÐABD = 106° and ∠ACF = 118°. Then, the value of ∠A is A

a) 44°

106o

118o

D B

C F

b) 54°

c) 34°

d) 36°

19. The sum of two angles of a triangle is equal to its third angle, then the third angle is a) 45°

b) 120°

c) 90°

d) 150°

20. An exterior angle of a triangle is 110° , and one of its interior opposite angle is 45°, then the other interior opposite angle is a) 45°

b) 65°

c) 25°

d) 135°

21. In the given figure, if l  m and x : y = 2 : 3 , then y - x is equal to l

a) 108°

b) 72°

c) 36°

d) 54°

22. What could be inferred about the triangle from the fact that its interior angles are in the ratio 1: 2 : 3 ? a) It is a scalene triangle

b) It is an obtuse-angled triangle

c) It is a right-angled triangle

d) It is an isosceles triangle

23. In the figure, if ÐQ = 30°, and ÐS = 50° , then what is the value of x? Q

P 30o

x 50o

176

a) 280°

b) 80°

c) 360°

d) Cannot be determined

IL Foundation Series Class 9

24. In the given figure, if BU  EK and TE  UC , then what are the values of x and y? B 110o

T U

60o

x 50°

y 110°

110°

60°

100°

50°

110°

50°

25. In the given figure, if PE  HO, HON = 105° and ∠ONE = 25°, then what is the value of ∠PEN ? P

E O N

a) 130°

b) 80°

c) 125°

d) 105°

26. In the given figure, each ∠AOC and ∠BOD is at a right angle. B

C Q O

Which of the following can be determined using this information? 1 c) ∠QOD =∠AOP 2 27. In the given figure, AP  IL . The value of ∠LIR is

∠QOC a) OB Bisects ∠AOC b) ∠AOB =

d) None of these

121o

P 132o

a) 73°

b) 90°

c) 107°

d) 110°

177

LINES AND ANGLES

28. If two supplementary angles are in the ratio 3 : 7, then the smaller angle measures a) 72°

b) 54°

c) 18°

d) 126°

II. ASSERTION AND REASON 1. Assertion (A): = θ 120° is a reflex angle. Reason (R): Reflex angle lies between 180° and 360° . a) Both A and R are correct, and R is the correct explanation for A b) Both A and R are correct, and R is not the correct explanation for A c) A is true, R is false d) A is false, R is true 2. Assertion (A): An angle 30° added to one-sixth of its complement is 40° . Reason (R): The sum of the measure of two angles is equal to 90° , then they are called complementary angles. a) Both A and R are correct, and R is the correct explanation for A b) Both A and R are correct, and R is not the correct explanation for A c) A is true, R is false d) A is false, R is true 3. Assertion (A): = If AX 0.3= cm, XB 5= cm, AB 5.3 cm , then A, X, and B are called collinear. Reason (R): The points which belong to the same plane are called non-coplanar points. a) Both A and R are correct, and R is the correct explanation for A b) Both A and R are correct, and R is not the correct explanation for A c) A is true, R is false d) A is false, R is true III. FILL IN THE BLANKS 1. If three or more points lie on the same line, then they are called _________otherwise they are called ________. 2. If l parallel m and n is a transversal, then the sum of co-interior angles is _________. 3. If two supplementary angles differ by 34° . Then the largest angle is ________. 4. If l, m, n are three lines such that l parallel m and n ⊥ l . Then n is ________m. 5. Two straight lines, AB and CD, intersect one another at the point O. If ∠AOC + ∠COB + ∠BOD = 274° , then ∠AOD = _______. 6. Two straight lines, AB and CD, cut each other at O. If ∠BOD = 63° , then ∠BOC = ________.

178

IL Foundation Series Class 9

7. Two complementary angles are such that two times the measure of one is equal to three times the measure of the other. The measure of the smaller angle is _______. 8. If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2 : 3 , then the measure of the larger angle is _______. 9. The number of pairs of adjacent angles formed when two lines intersect at a point is _________. 10. In a ∆ABC , if ∠A = 60°, ∠B = 80° and the bisector of ∠B and ∠C meet at O, then ∠BOC = _________. IV. SUBJECTIVE QUESTIONS 1. Write the supplement of an angle of measure 2y° . 2. If a wheel has six spokes equally spaced, then find the measure of the angle between two adjacent spokes. 3. Name the alternate angles and corresponding angles in the adjacent figure. n 1

5 6

l m

4. In the figure, if l1  l2 , what is x + y in terms of w and z?

wo x

zo yo

5. In the figure, if l1  l2 , what is the value of y ?

l1 y0

l3 3x

179

LINES AND ANGLES

6. In the figure, if lines l and m are parallel, then the value of x is l x 125o

7. If the supplement of an angle is three times its complement, find the angle. 8. The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10° . Find all the three angles. 9. In the figure, what is z in terms of x and y?

zo A

yo C

10. If the angles of a triangle are in the ratio 2 : 1 : 3 , then find the measure of the smallest angle. 11. In the figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40° . Find ∠BOE and reflex ∠COE . E

12. What value of y would make AOB a line in the figure, if ∠AOC = 4y and ∠BOC =(6y + 30)° ? C

(6y+30)

4yo B

180

IL Foundation Series Class 9

13. Given ∠XYZ = 64° and XY is produced to point P. From this data, draw the figure. If ray YQ bisects ∠ZYP , find ∠XYQ and reflex ∠QYP . Q

Z 64o

14. In the figure, BAC is a line in which x : y : z = 5 : 6 : 9 . Find x, y, and z. E

D yo xo

zo A

15. In the figure, AB is a mirror, PQ is the incident ray, and QR is the reflected ray. If ∠PQR = 100° , then find ∠PQA . R

P 100o A

16. In the figure, if ∠POS + ∠QOS + ∠QOR = 302° , then find all the four angles. S

17. In the figure, find the value of y. B

6yo F

4yo O 8yo

D E

181

LINES AND ANGLES

18. In the figure, if AB  CD, ∠APQ =° 50 and ∠PRD = 127° , then find x and y. A

x C

127o

19. In the figure, the side QR of ∆PQR is produced to a point S. If the bisectors of ∠PQR and 1 ∠PRS meet at T, then prove that ∠QTR = ∠QPR . 2 P T

20. In the figure, if PQ ⊥ PS,PQ  SR, ∠SQR = 28° and ∠QRT =° 65 , then find the values of x and y. P

28o

y 65o

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7.1

INTRODUCTION AND PROPERTIES OF TRIANGLES

A triangle is a geometric shape with three sides, three vertices, and three angles. Triangles can be classified based on the length of their sides and the measures of their angles. Vertex 3 P 3 Side 3

Side 2

1 R Vertex 1

2 Side 1

Q Vertex 2

7.1.1 Types of triangles (based on length of sides) 1. Scalene triangle: A triangle with all three sides of different lengths.

2. Isosceles triangle: A triangle with two sides of equal length.

3. Equilateral triangle: A triangle with all three sides of equal length. 60° 60°

60°

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7.1.2 Types of triangles (based on measurement of angles) 1.

Acute triangle: All angles are less than 90°.

Right triangle: One angle is exactly equal to 90°.

Obtuse triangle: One angle is greater than 90°.

7.1.3 Properties of triangles 1. Angle sum property: The sum of all interior angles in any triangle is always equal to 180°.

∠A + ∠B + ∠C =180o 2. Exterior angle property: The measure of an exterior angle of a triangle is equal to the sum of the measures of its opposite interior angles. A a

Interior

Exterior d c C

Measure of exterior angle, ∠d = ∠a + ∠b 3. Triangle inequality theorem: The sum of the lengths of any two sides of a triangle is greater than the length of the third side. This holds true for all sides of the triangle. a

b c

a + b > c, b + c > a, c+a>b 4. Median of a triangle: A median of a triangle is a line segment connecting a vertex to the midpoint of the opposite side. Each triangle has three medians, and they intersect at the centroid.

Median B

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IL Foundation Series Class 9

5. Altitude of a triangle: An altitude of a triangle is a line segment through a vertex and perpendicular to a line containing the side opposite the vertex. Each triangle has three altitudes, and the point of intersection of the altitudes is called the orthocenter. A

Altitude

7.2

CONGRUENCE OF TRIANGLES

7.2.1 Introduction to congruence of triangles Congruence of triangles: Two triangles are congruent if one of them can be made superimposed on the other so as to cover it exactly. Example:

C E

In the above two triangles, ∆ABC ≅ ∆DEF . Corresponding parts of congruent triangles are equal and we write it in short 'CPCT'. AB = DE BC = EF AC = DF i.e., corresponding sides are equal. ∠A =∠D ∠B =∠E ∠C = ∠F i.e., corresponding angles are equal.

185

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7.2.2 SAS congruence rule SAS congruence rule/SAS criteria: Two triangles are congruent if two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of the other triangle. Example:

B D

In the above two triangles, AC = DF ∠A = ∠D And AB = DE Therefore, ∆ABC ≅ ∆DEF . 7.2.3 ASA congruence rule ASA congruence rule/ASA criteria: Two triangles are congruent if any two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle. Example:

B In the above two triangles, ∠B = ∠E BC = EF And, ∠C = ∠F Therefore, ∆ABC ≅ ∆DEF .

186

C E

IL Foundation Series Class 9

7.2.4 AAS congruence rule AAS congruence rule/AAS criteria: If any two angles and a non-included side of one triangle are equal to the corresponding two angles and the corresponding side of another triangle, then the two triangles are congruent. Example: P

In the above two triangles, ∠A = ∠P ∠B = ∠Q And, BC = QR Therefore, ∆ABC ≅ ∆PQR . 7.2.5 SSS congruence rule SSS congruence rule/SSS criteria: If three sides of one triangle are equal to corresponding three sides of the other triangle, then the two triangles are congruent. Example:

R Y

In the above two triangles, PQ = XY QR = YZ And PR = XZ Therefore, ∆PQR ≅ ∆XYZ .

187

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7.2.6 RHS congruence rule RHS congruence rule/RHS criteria: If the hypotenuse and a side of one right-angled triangle are equal (respectively) to the hypotenuse and corresponding side of the other right-angled triangle then the two triangles are congruent. Example:

In the above two triangles, ∠N = ∠Y = 90° MP = XZ (Hypotenuse) And NP = YZ Therefore, ∆MNP ≅ ∆XYZ .

SOLVED EXAMPLES Example 1: In the given figure, O is the mid-point of AB and CD. Prove that AC = BD and AC  BD . A

D O

Solution: In ∆AOC and ∆BOD , AO = OB [ O is the midpoint of AB ] OC = OD [ O is the midpoint of CD ] ∠AOC = ∠BOD (Vertically Opposite angles) So, by SAS congruency, ∆AOC = ∆BOD By CPCT, 188

IL Foundation Series Class 9

AC = BD And, ∠CAO = ∠DBO Now AC and BD are two lines intersecting by a transversal AB such that ∠CAO = ∠DBO (i.e alternate angles are equal). Hence, AC  BD . Example 2: In quadrilateral ADBC, AC = AD and AB bisects ∠A, then show that ∆ABC ≅ ∆ABD . C

Solution: Given a quadrilateral ADBC in which AC = AD and AB bisects ∠A. To Prove: ∆ABC ≅ ∆ABD Proof: In ∆ABC and ∆ABD AB = AB (Common) ∠BAC = ∠BAD ( AB bisects ∠A) AC = AD (Given) ∴∆ABC ≅ ∆ABD ( By SAS axiom) Example 3: In given figure, AB = CF, EF = BD, and ∠AFE = ∠DBC, then prove that ∆AFE ≅ ∆CBD . D

E 189

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Solution: We have, AB = CF ⇒ AB + BF = CF + BF ⇒ AF = CB ...(1) In ∆AFE and ∆CBD , we have AF = CB ( from (1)) ∠AFE = ∠DBC (given) EF = BD (given) ∴∆AFE ≅ ∆CBD (By SAS axiom) Hence, proved. Example 4: AD is the altitude of an isosceles triangle ABC in which AB = AC. Show that i) AD bisects BC ii) AD bisects ∠A A

Solution: Given: An isosceles

ii) AD bisects ∠A

Proof: In ∆ABD and ∆ACD AB = AC (Given) AD = AD (Common) ÐADB = ÐADC ( AD ^ BC ) \ DABD @ DACD (By RHS axiom) By CPCT BD = CD

Hence, AD bisects BC. ⇒∠BAD = ∠CAD Hence, AD bisects ∠A.

190

ABC in which AB = AC and AD ⊥ BC .

To prove: i) AD bisects BC

IL Foundation Series Class 9

Example 5: In the given figure, BD and CE are two altitudes of a triangle ABC such that BD = CE. Prove that ∆ABC is isosceles. A E

Solution: In right triangles ∆BEC and ∆BCD , BC = BC BD = CE ∠D = ∠E = 90° ∴∆BEC ≅ ∆CDB (By RHS rule) By CPCT, ∠EBC = ∠DCB i.e., ∠B = ∠C Then, AB = AC ∴∆ABC is an isosceles.

7.3

GEOMETRIC POINTS OF A TRIANGLE, EULER LINE

1. Centroid: The centroid of a triangle is the point where the three medians intersect. It divides each median into two segments, with the segment closer to the vertex being twice as long as the one closer to the midpoint of the opposite side. A

G B

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2. Incenter: The incenter is the point of concurrency for the angle bisectors. It is equidistant from all three sides of the triangle.

Incenter

3. Circumcenter: The circumcenter is the point of concurrency for the perpendicular bisectors. It is equidistant from all three vertices of the triangle. A

O C B

AO = BO = CO

4. Orthocenter: The orthocenter is the point of concurrency for the altitudes. It is the point where the three altitudes intersect. A

F D

192

Orthocenter

IL Foundation Series Class 9

Euler line: It is a significant concept in geometry named after the Swiss mathematician Leonhard Euler. It refers to a special line that is associated with various points of concurrency in a triangle. The Euler line passes through several important points in a triangle, including the centroid, circumcenter, and orthocenter. Specifically, the Euler line connects the following three key points in a triangle:

Orthocenter (H) Centroid (G) Circumcenter (O)

Euler line The Euler line is collinear with the circumcenter, centroid, and orthocenter. It exists in every triangle, regardless of the triangle's shape or size. In short, the Euler line is a line that connects the circumcenter, centroid, and orthocenter of a triangle, forming a straight line that passes through these three points of concurrency.

7.4

SIMILARITY OF TRIANGLES

7.4.1 Introduction to similarity Similar triangles: Two triangles are said to be similar if i) Their corresponding angles are equal. ii) Their corresponding sides are proportional (in the same ratio). Example: ∆ABC and ∆PQR,

P A

B If

AB BC CA = = and ∠A =∠P , ∠B =∠Q, ∠C =∠R then ∆ABC ~ ∆PQR. PQ QR RP

193

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7.4.2 Basic proportionality theorem (Thales theorem) Statement: 'If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio'. Proof:

A N

l C

Let us consider that the straight line l  BC in ∆ABC and let 'l' intersects AB and AC at D and E, AD AE respectively. Then, it is desired to prove that = . DB EC Let us draw DM ⊥ AC and EN ⊥ AB . Also, join DC and BE. 1 ∴ ar (∆ADE ) = × AD × EN 2 1 ar (∆BDE ) = × BD × EN 2 1 × AD × EN AD ar (∆ADE ) 2 ∴ = = ar (∆BDE ) 1 × BD × EN DB 2

...(1)

1 Again, ar ( ∆ADE ) = × AE × DM 2 1 and ar (∆CDE ) = × EC × DM 2 1 × AE × DM AE ar (∆ADE ) 2 ...(2) ∴ = = 1 ar (∆CDE ) × EC × DM EC 2 Also, ∆BDE and ∆CDE lie on the same base DE and lie between the same parallels l, BC. ∴ ar (∆BDE ) =ar (∆CDE ) ar (∆ADE ) AD ∴ From (1) ⇒ = ar (∆CDE ) DB ∴ From (2) and (3); we get AD AE = DB EC Hence proved. 194

...(3)

IL Foundation Series Class 9

Note: The basic proportionality theorem is also applicable when a line is drawn parallel to one side of a triangle intersecting other two sides externally at distinct points. Case - (ii)

Case - (i) A B D

A E

In both the above cases, DE  BC in ∆ABC. AD AE AD AE DB EC . = ⇒ = ⇒ = DB EC AB AC AB AC Converse of basic proportionality theorem (Converse of Thales theorem): ∴ By BPT,

Statement: 'If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side'. 7.4.3 Angle-Angle-Angle (AAA) similarity Statement: 'If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar'. Proof: D A P

ABC and DEF be two triangles such that ∠A = ∠D, ∠B = ∠E and ∠C = ∠F We need to prove that  ABC ~DEF. Let

...(1)

Mark points P and Q on DE and DF such that DP = AB and DQ = AC. Now, in

DPQ and ABC ,

DP = AB (by construction) DQ = AC (by construction) ∠A =∠D [(From(1)] ∴ By SAS congruency,

DPQ ≅  ABC

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⇒ ∠DPQ = ∠B and ∠DQP = ∠C [ CPCT ] ∴ ∠DPQ =∠E and ∠DQP =∠F

[ from (1) ]

By BPT, DP DQ = DE DF AB AC = ...(2) DE DF Similarly, by marking points X, Y on ED and EF respectively such that EX = BA and EY = BC. ⇒

We can prove that From (2) and (3)

AB BC = DE EF

...(3)

AB BC AC = = DE EF DF From (1) and (4)

...(4)

ABC ~DEF.

Hence proved. Note: Angle-Angle (AA) similarity: If in two triangles, two angles of one triangle are respectively equal to two angles of the other triangle, then the two triangles are similar. 7.4.4 Side-Angle-Side (SAS) similarity Statement: 'If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar'. Proof: Let

ABC and DEF ,be two triangles such that ∠A =∠D and AB =AC . DE

A P B We need to prove that

 ABC ~DEF.

Q F

Mark points P, Q on DE and DF respectively such that DP = AB and DQ = AC. Join PQ.

196

...(1)

IL Foundation Series Class 9

∴ In

ABC and DPQ ,

∠A =∠D , AB =DP and AC =DQ ∴ By SAS congruency,

ABC ≅DPQ

∴∠B = ∠DPQ and ∠C = ∠DQP From (1), we get

...(2)

AB AC = DE DF

DP DQ = DE DF ∴ Applying the converse of BPT in ⇒

DEF , we get

PQ  EF ∴∠DPQ = ∠E and ∠DQP = ∠F (corresponding angles)

...(3)

∴ From (2) and (3) ⇒ ∠B = ∠E and ∠C = ∠F So, in

ABC and DEF ,

∠A =∠D (given) ∠B =∠E , ∠C =∠F (proved alone) ∴ By AAA similarity, we get

ABC ~DEF. Hence proved. 7.4.5 Side-Side-Side (SSS) similarity Statement: 'If in two triangles, the sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal, and hence the two triangles are similar'.

D A P B

Q F

BC CA ABC and DEF be two triangles such that AB = = DE EF FD We need to prove that  ABC ~DEF. Let

...(1)

197

TRIANGLES

Mark points P, Q on DE and DF respectively such that DP = AB and DQ = AC. Join PQ. Now, from (1), we get, AB AC = DE DF ⇒

DP DQ = DE DF

∴ Applying the converse of BPT, PQ  EF ⇒ ∠DPQ =∠E and ∠DQP = ∠F (corresponding angles)

...(2)

∴ Applying AA similarity, we get,

DPQ ~DEF

∴

DP DQ PQ = = DE DF EF

BC CA But, from (1), AB = = DE EF FD DP BC DQ ⇒ = = DE EF FD DP DQ BC ⇒ = = DE FD EF ∴ From (3) and (4), PQ BC = EF EF ⇒ PQ = BC

⇒

∴ In

DPQ and  ABC ,

DP = AB, DQ = AC (by construction) and PQ = BC (proved above) ∴ By SSS congruency, ∆DPQ ≅ ∆ABC ∴∠A = ∠D, ∠B = ∠DPQ, ∠C = ∠DQP ⇒ ∠A=∠D, ∠B = ∠E , ∠C = ∠F , (by(2)) Hence,

198

 ABC ~ DEF .

...(3)

...(4)

IL Foundation Series Class 9

SOLVED EXAMPLES



Example 1: In ABC , D and E are points on AB and AC, respectively, such that DE  BC . If AD = = ( 5 x − 3) cm , then find the value of x. ( 4 x − 3) cm, AE = (8x − 7 ) cm, BD = ( 3x − 1) cm , and CE Solution:

D B

E C

Given: DE  BC ∴ By BPT

⇒ ⇒ ⇒ ⇒ ⇒ ⇒

AD AE = DB EC 4 x − 3 8x − 7 = 3x − 1 5 x − 3 ( 4 x − 3)( 5x − 3) = ( 3x − 1)(8x − 7 ) 20 x 2 − 12 x − 15 x += 9 24 x 2 − 21x − 8 x + 7 0 − 4x 2 + 2x + 2 = 2 x 2 − x − 1 =0 2 x 2 − 2 x + x − 1 =0

⇒ 2 x ( x − 1) + 1( x − 1) = 0 ⇒ ( 2 x + 1)( x − 1) = 0 ∴ 2 x= + 1 0 or x= −1 0 −1 ⇒ = x or= x 1 2 −1 But, for x = ; AD , DB , AE , EC gives negative values. 2 ∴ The value of x is 1. Example 2: Prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side. Solution: Let D and E be the midpoints of sides AB and AC, respectively.

199

TRIANGLES

D B

AE EC = ∴ AD DB and = AD AE ⇒ = 1 and = 1 DB EC AD AE ⇒ = DB EC ∴ Applying the converse of BPT in

ABC , we get DE  BC .

Hence proved. Example 3: Prove that diagonals of a trapezium divide each other proportionally. Solution: Let ABCD be a trapezium with AB  CD and diagonals AC and BD meeting at O.

A E

B O C

D Draw EO  AB  CD In

 ADC , EO  DC

OA AE = OC ED

...(1)

AE OB = ED OD

...(2)

∴ By B.P.T., Also, in

ABD, EO  AB

∴ By BPT,

OA OB = OC OD Hence, diagonals of trapezium divide each other proportionally. ∴ From (1) and (2) ⇒

200

IL Foundation Series Class 9



70 . Find Example 4: In the figure, ODC ~ OBA, ∠BOC = 125° and ∠CDO = ∠DOC , ∠DCO and ∠OAB. °

C 70°

125°

Solution: Given,

ODC ~OBA

⇒ ∠ODC =∠OBA =70° ∠COD and ∠BOC are a linear pair. ∴∠COD + ∠BOC = 180° ° 55° ⇒ ∠COD = 180° − ∠BOC = 180° − 125 =

∴∠DOC = 55° In ∆ODC, ∠DCO = 180° − ( 70° + 55° ) = 180° − 125° = 55° (Angle sum property) ∠AOB = ∠COD = 55° (Vertically opposite angles) In ∆OBA ∴∠OAB = 180° − ( 55° + 70° ) = 180° − 125° = 55° (Angle sum property) Example 5: In the figure,

QR QT and ∠1 = ∠2. Show that ∆PQS  ∆TQR. = QS PR

T P

2 S

201

TRIANGLES

Solution: Given ∠1 = ∠2 in

PQR .

∴ PR = PQ ∴

QR QT QR QT = ⇒ = QS PR QS PQ

∴ In

PQS and TQR

∠Q = ∠Q and

QR QT = QS PQ

∴∆PQS ~ ∆TQR (by SAS similarity) Example 6: In the given figure,

ABE ≅ ACD . Show that ADE ~ABC . A

Solution:



Given ABE ≅ ACD ∴ By CPCT , AB = AC and AD = AE ⇒ AB − AD = AC − AE ⇒ BD = CE AD AE = DB EC DE  BC (by converse of BPT ) ∴∠ADE = ∠ABC ( corresponding angles) ∴

So, in

ADE and ABC

∠A =∠A ∠ADE = ∠ABC ∴ ADE ~ ABC (By AA similarity)



202



IL Foundation Series Class 9

QUICK REVIEW • Two figures are congruent if they are of the same shape and of the same size. • If two triangles ABC and PQR are congruent under the correspondence A ↔ P , B ↔ Q , and C ↔ R , then symbolically, it is expressed as ∆ABC ≅ PQR . • SAS congruence rule: If two sides and the included angle of one triangle are equal to two sides and the included angle of the other triangle, then the two triangles are congruent. • ASA congruence rule: If two angles and one side of one triangle are equal to two angles and the corresponding side of the other triangle, then the two triangles are congruent. • AAS congruence rule: If two angles and one side of one triangle are equal to two angles and the corresponding side of the other triangle, then the two triangles are congruent. • SSS congruence rule: If the three sides of one triangle are equal to the three sides of the other triangle, then the two triangles are congruent. • RHS congruence rule: If in two right triangles, the hypotenuse and one side of a triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent. • Two figures having the same shape but not necessarily the same size are called similar figures. • All congruent figures are similar, but the converse is not true. • Two polygons having the same number of sides are similar if i) their corresponding angles are equal. ii) their corresponding sides are proportional (i.e., in the same ratio). • Basic proportionality theorem states that 'if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio'. • Converse of BPT: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side of the triangle. • The line drawn from the mid-points of a triangle is parallel to another side and bisects the third side. • The line joining the midpoints of two sides of a triangle is parallel to the third side. • AAA similarity criterion: If in two triangles, three angles of one triangle are respectively equal to three angles of the other triangle, then the two triangles are similar. • AA similarity criterion: If in two triangles, two angles of one triangle are respectively equal to two angles of the other triangle, then the two triangles are similar. • SAS similarity criterion: If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. • SSS similarity criterion: If in two triangles, corresponding sides are in the same ratio, then the two triangles are similar.

203

TRIANGLES

WORKSHEET - 1 I.

CONGRUENCE OF TRIANGLES 1. In the figure, LMN is an isosceles triangle with LM = LN and LP bisects ∠NLQ. Prove that LP || MN. Q



2. D and E are points on the base BC of ABC such that AD = AE and ∠BAD = ∠CAE. Prove that triangle ABC is an isosceles triangle. A

 ABC is an isosceles triangle.

3. In the figure, AD = AE, BD = EC. Prove that

4. Suppose line segments AB and CD intersect at O in such a way that AO = OD and OB = OC. Prove that AC = BD, but AC may not be parallel to BD. C B O A D

204

IL Foundation Series Class 9

5. Two sides AB, BC and median AM of one median PN of PQR . Prove that:



 ABC are respectively equal to sides PQ, QR and

 ABM ≅ PQN ii)  ABC ≅  PQR i)

6. Prove that the sum of two sides of a triangle is greater than twice that the median with respect to the third side. 7. In the given figure, ABC is a triangle. = AB AC , BL ⊥ AC and CM ⊥ AB . Show that BL = CM. Also, prove AM = AL. A

8. In the given figure, ∠QPR = ∠PQR and M and N are respectively on sides QR and PR of PQR such that QM = PN. Prove that OP = OQ where O is the point of intersection of PM and QN.



9. AB is a line segment P and Q are points on opposite sides of AB such that each of them is equidistant from points A and B. Show that the line PQ is the perpendicular bisector of AB.

P A

Q 10. In the given figure, ∠BCD = ∠ADC and ∠ACB = ∠BDA. Prove that: i) AD = BC ii) ∠A = ∠B A

205

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11. In figure, AB = AD, ∠1 = ∠2 and ∠3 =∠4. Prove that AP = AQ. D Q 2

4 3 1

C P B

12. Prove that two triangles are congruent if any two angles and the included side of one triangle are equal to any two angles and the included side of the other triangle. 13. In figure, OA = OB, OC = OD and ∠AOB = ∠COD. Prove that AC = BD. O

C B

II. GEOMETRIC POINTS OF A TRIANGLE 1. In

ABC , E is the midpoint of AC and G is the centroid of the triangle, then find BE : GE.

2. In an Euler line, if the length of OS = 9 cm, then find OG and GS.



3. In ABC , BC = 5, CA = 8, and AB = 7. If G is the centroid of GA 2 + GB 2 + GC 2 .

 ABC , then find

4. Define concurrent lines and centroid with diagrams. 5. Explain orthocentre and circumcentre, excentre and incentre through diagrams. 6. ABC is a triangle, right angled at C. If 'p' is the length of the perpendicular from C to AB and AB = c, BC = a and CA = b, then prove that: i) pc = ab 1 1 1 + ii) = 2 p a2 b2 7. In an equilateral triangle with side 'a', prove that: a) the altitude of length = b) area of the triangle = 206

3 a 2

3 2 a 4

IL Foundation Series Class 9

III. SIMILARITY OF TRIANGLES 1. In the given figure, DE  BC. If AD = x, DB = x − 2, AE = x + 2 and EC = x -1, then find the value of x.

C E

2. In the given figure, LM  AB. If AL = x − 3, AC = 2 x, BM = x − 2 and BC = 2x + 3, then find the value of x. C

3. In the figure, PQ  RS . Prove that,

POQ ~ SOR . R P O Q S

4. In the given figure, OA ⋅ OB = OC ⋅ OD . Show that ∠A =∠C and ∠B =∠D . C A O D B

207

TRIANGLES

5. In the given figure, CM and RN are, respectively, the medians of ABC ~ PQR , Prove that:

  i)  AMC ~  PNR ii) iii)

 ABC and PQR . If

CM AB = RN PQ

CMB ~ RNQ. A

M C

6. In the figure, E is a point on CB produced by an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC . Prove that, ABD ~ ECF .



 A

ABC such that ∠ADC = ∠BAC. Show that CA = CB ⋅ CD. 8. Sides AB and AC and median AD of  ABC are respectively proportional to sides PQ and PR at the median PM of another  PQR . Show that,  ABC ~  PQR . 7. D is a point on the side BC of

P A

B 9. If AD and PM are medians of AB AD = prove that, . PQ PM 208

ABC and ΔPQR, respectively, where  ABC ~ PQR , then

IL Foundation Series Class 9

10. Two triangles, BAC and BDC, right-angled at A and D, respectively, are drawn on the same base B and on the same side of BC. If AC and DB intersect at P, then prove that AP ⋅ PC = DP ⋅ PB . 11. In the figure,

ABC is right-angled at A and AD ⊥ BC . Prove that, AD = BD.DC . 2

12. In the figure, PA, QB and RC are perpendicular to AC. Prove that,

1 1 1 + =. x z y

P x



13. D is the midpoint of side BC of ABC . AD is bisected at the point E and BE produced cuts AC at the point X. Prove that BE : EX = 3 : 1.

WORKSHEET - 2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. In

ABC , BC = AB and ∠B = 80 , then ∠A is equal to °

a) 80° 2. In

c) 50°

d) 100°

b) ∠A = ∠C

c) ∠A = ∠B

d) ∠A < ∠C

b) 6 cm

c) 11 cm

d) 7 cm

ABC , if AB = BC, then

a) ∠B > ∠C 3. In

b) 40°

and BC 6= cm, AC 5 cm . The length of AB is ABC , ∠C = ∠A =

a) 5 cm

4. D is a point on the side BC of a triangle ABC, such that AD bisects ∠BAC, then a) BD = CD

b) BA > BD

c) BD > BA

d) CD > CA

209

TRIANGLES

5. The criteria used in

OAP ≅ OBP is A

a) SAS 6. In given figure,

b) SSS

c) RHS d) ASA

ABC , if AB = AC, then the value of x is A

80°

a) 80°

b) 1000°

c) 130° d) 120°

7. In the given figure, AD is the median, then ∠BAD is

40°

B a) 55°

b) 50°

C c) 100°

d) 40°

8. Among the following, the false statement is a) The two altitudes corresponding to the equal sides of a triangle are not equal b) The bisectors of two equal angles of a triangle are equal c) If the altitude from one vertex of a triangle bisects the opposite side, then the triangle is isosceles d) Angles opposite to equal sides of a triangle are equal

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IL Foundation Series Class 9

9. In the given figure, if two segments, AC and BD, intersect at their midpoint 'O', then

O B a) AB = DC 10. In

b) AD ≠ BC

c) ∠BOC = ∠ABD

d) ∠CAB ≠ ∠ACD

b) PQ > PR

c) PQ < PR

d) QR < PR

PQR , if ∠R > ∠Q, then

a) QR > PR



11. ∠X and ∠Y are exterior angles of a ABC , at the points B and C, respectively. Also ∠B > ∠C. then the relation between ∠X and ∠Y is a) ∠X > ∠Y

b) ∠X < ∠Y

c) ∠X = ∠Y

 ABC , then BM =

12. If M is the midpoint of the hypotenuse AC of right a) AC 13. If in two triangles a)

PQR

b) 2AC

d) ∠X = ∠Y + 90°

1 1 AC d) AC 3 2 PQR, AB =QR, ∠A =∠Q and ∠B = ∠R, then ABC ≡ c)

ABC and  b)  RQP

QRP



QPR

14. In the figure, ABC is a triangle, having sides BC and CA produced to D and E respectively comparing the sides shows

E 100° A 140° C

B a) AB > BC

b) AB > AC

c) AC > BC

d) BC > AC

15. In the figure, ABC is a triangle. If AB = 3 cm and AC = 4 cm, then BC = A

a) 7 cm

b) Less than 7 cm

c) More than 7 cm d) 14 cm

211

TRIANGLES

16. In the given triangle, AC = AD , ∠CAD = 50° and ∠BAD = 23° , then which of the following is true?

A 23°

B a) AB = BC 17. In a

50°

b) AB < BC

C c) AB > BC

ABC , ∠B =∠C =45 , which is the longest side?

a) AB

d) AB = BC + 73°

b) BC

c) CA d) All equal

18. If two angles of a triangle are unequal, then the smaller angle has the side opposite to it will be a) Smaller

b) Medium

c) Greater

d) Very smaller

19. In figure, if AB = AC and AP = AQ, then by which congruence criterion  PBC ≅ QCB ?

A P

B a) SSS

b) ASA

c) SAS d) RHS

20. In the right triangle, ∠E = 90° , then

212

a) DF is the shortest side

b) DF is the longest side

c) EF is the longest side

d) DE is the longest side

IL Foundation Series Class 9

= BC and AD = DC. The measure of ∠BCD

21. In the figure, ABCD is a quadrilateral in which is

A B

108°

42°

C a) 150°

b) 30°

c) 105° d) 72°

22. In triangles ABC and PQR, AB =AC , ∠C =∠P and ∠B =∠Q . The two triangles are a) Isosceles but not congruent b) Isosceles and congruent c) Congruent but not isosceles d) Neither isosceles nor congruent

PQR such that PE bisects ∠QPR, then

23. If E is a point on side QR of a a) QE = ER 24. If

c) QE > QP

d) ER > RP

b) CA = RP

c) AC = RQ

d) CB = QP

ABC ≅ PQR, then which of the following is true?

a) AB = RP 25. In

b) QP > QE

ABC , ∠A= 50 , ∠B= 60 , arranging the sides of the triangle in ascending order, we get °

a) AB < BC < CA

b) CA < AB < BC

c) BC < CA < AB

d) BC < AB < CA

26. If the altitudes AD, BE and CF of a triangle ABC are equal, then the triangle ABC is: a) An isosceles right-angled triangle

b) An equilateral triangle

c) A scalene triangle

d) An acute-angled triangle

27. If the circumcentre of a triangle lies on one of its sides, then the triangle is: a) An acute-angled triangle

b) An obtuse-angled triangle

c) A right-angled triangle

d) An equilateral triangle

28. In

ABC , if AB + BC = AC , then its orthocentre is: 2

a) A

b) B



c) C

d) Centroid of

 ABC

29. If 'H' is the orthocentre of ABC and X, Y, and Z are respectively the midpoints of AH, BH, and CH, then the point H is:

XYZ c) Orthocentre of  XYZ a) Circumcentre of

 XYZ d) Incentre of  XYZ

b) Centroid of

213

TRIANGLES

30. If two medians of a triangle are equal, then the triangle is: a) A scalene triangle

b) An isosceles triangle

c) An equilateral triangle

d) A right-angled triangle

31. A vertical stick 30 m long casts a shadow 15 m long on the ground. At the same time, a tower casts a shadow 75 m long on the ground. The height of the tower is a) 150 m

b) 100 m

c) 25 m

d) 200 m

32. In the figure given below, if DE  BC , then x equals to 3c

4 cm

D B

a) 3 cm

b) 2 cm

c) 4 cm



d) 6.7 cm

33. In the given figure, in XYZ , DE  YZ , so that the lengths of sides XD, XE, and EZ (in cm) are 2.4, 3.2, and 4.8 respectively. Then, the length of XY (in cm) is

X 2.4

3.2

4.8

a) 8 cm

a) ∠B = ∠E 36. In the given figure, if

b) ∠A = ∠D

c) ∠B = ∠D

d) ∠A = ∠F

ABC ~ PQR , the value of x is

214

d) 5 cm

AB BC = , then they will be similar when DE FD

6 cm

a) 2.5 cm

c) 4 cm

ABC = 20 cm , perimeter of PQR = 40 cm and PR = 8

b) 6 cm

35. If in triangles ABC and EDF ,

d) 1.6

4 cm

b) 3.5 cm

5 cm C



34. If ABC  PQR , perimeter of cm, then the length of AC is

c) 6.4

4. 5



b) 6

3.75 cm

a) 3.6

c) 2.75 cm

d) 3 cm

IL Foundation Series Class 9

37. In the figure PQ  BC , if

PQ 2 AP = , then is BC 5 PB A P

3 2 PX PY 1 38. In the following figure, XY  QR and = = , then XQ YR 2

2 5

2 3

3 5

Y R

Q a) XY = QR



39. If ABC ~ true?



1 1 b) XY = QR c) XY2 = QR2 d) XY = QR 3 2 EDF and ABC is not similar to DEF , then which of the following is not



a) BC ⋅ EF = AC ⋅ FD

b) AB ⋅ EF = AC ⋅ DE

c) BC ⋅ DE = AB ⋅ EF

d) BC ⋅ DE = AB ⋅ FD

40. If

ABC is obtuse-angled at C, then

a) AB > BC

b) AB = BC

c) AB < BC

d) AC > AB

41. It is not possible to construct a triangle when the lengths of its sides are a) 3 cm, 4 cm, 5 cm

b) 3 cm, 5 cm, 15 cm

c) 3 cm, 2.8 m, 3.1 cm

d) 9 cm, 5 cm, 7 cm

II. FILL IN THE BLANKS 1. The sum of the three medians of a triangle is ______ the perimeter. 2. The bisector of the vertical angle of a triangle bisects the base of the triangle, then the triangle is ______. 3. In an isosceles triangle, if the vertical angle is twice the sum of the base angles, then the vertical angle is ______.

215

TRIANGLES

4. If two triangles have the same area, then they are ______. 5. In

PQR , = if ∠P 60 , and = ∠Q 50 , then the longest side in the triangle is ______. °

6. If one angle of a triangle is equal to the sum of the other two angles, then the triangle is _______. 7. If O is any point in the interior of

 ABC , then OA + OB + OC > _______.

8. ABC is an isosceles triangle such that AB = AC and AD is the median to base BC. Then, ∠BAD = ______. A

350

ABC ≅ DEF , ∠A= 60 , ∠B= 70 , then ∠F= ______. 10. In  ABC , ∠A > ∠B and ∠B > ∠C , then the smallest side is _______. 9. If

11. In a trapezium, the diagonals divide each other _______. 12. In a triangle, a line divides any two sides in the same ratio then it is ____ to the third side. 13. Name the regular quadrilateral, which is always similar to any other regular quadrilateral. _______. 14. Two circles with different radii are always _______. 15. The line segments joining the midpoints of the adjacent sides of a quadrilateral form a _______. 16. The constant ratio between the corresponding sides of two similar figures is called _______. 17. In a

ABC , DE  BC , D is the midpoint of the AB, then AE = ________.

18. In an equilateral triangle of side 3 3 cm, the length of attitude is _______. 2 19. In a rectangle ABCD, E is a point = on AE AB = , AB 6= m, AD 3 m , the value of DE = 3 ________. III. SUBJECTIVE QUESTIONS 1. Prove that each angle of an equilateral triangle is 60° . 2. ABC is a right-angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C. 3. In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.

216

IL Foundation Series Class 9

4. In figure AC > AB and D is the point on AC such that AB = AD. Prove that, BC > CD. A D

5. In figure, ABC is an isosceles triangle whose side AC is produced to E. Through C, a line is drawn parallel to BA. Find the value of x. A D 520 B

x0 E

6. AB is a line segment and line l is its perpendicular bisector. If a point P lies on l, then show that P is equidistant from A and B. 7. In

PQR, QS bisects ∠PQR, PQ = PR and m∠PQS = 30 . Find m∠R. °

8. AB and PQ bisect each other at R. Prove that AQ = BP. 9. In x.

ABC if DE  AB, AD =+ 8 x 9, CD = x + 3, BE =+ 3x 4 and CE = x , then find the value of

10. In the given figure, LM  CB and LN  CD . Prove that

AM AN . = AB AD

M A

L N

11. In the figure, DE  BC . Find x. A 1.5 cm

1 cm

3 cm B

x C

217

TRIANGLES

12. In the given figure, ∆ACB  ∆APQ . If AB = 6 cm, BC =8 cm , and PQ = 4 cm, then find AQ. P B

13. If triangle ABC is similar to triangle DEF such that 2AB = DE and BC = 8 cm, then find EF. 14. In the given figure, find the value of x (in cm).

P 2.4 cm 3.2 cm 2 cm B A 4.8 cm 3.6 cm x cm Q R 15. If

PQR ~ XYZ , ∠Q =50 and ∠R =70 , then find ∠X + ∠Y. °

16. Two poles of height 6 m and 11 m stand vertically upright on a plane ground. If the distance between their foot is 12 m, then find the distance between their tops. 17. PQ is drawn parallel to the base BC of a cutting AB at P and AC at Q of and CQ = 2 cm, then find AQ.



 ABC . If AB = 4BP

18. ABC is such = that AB 3= cm, BC 2 cm , and CA = 2.5 cm . If ∆DEF ~ ∆ABC and EF = 4 cm, then find the perimeter of DEF .



19. In the given figure, ABCD is a square, AOB is equilateral triangle. Find the measurement of ∠COD. D

C O

20. In the given figure, AD = AE and ∠ACD = ∠BCD. Find the measurement of ∠EAC. A D E 370 B 218

IL Foundation Series Class 9

21. Find the value of x in the given figure if AC = BC = BD. D C 60°

x A

22. In the given figure BC = CD and ∠ABC  ∠BAD = 30°. Find ∠ABD. B

23. Find the sum of the vertex angle of the 5-vertex and 7-vertex star as shown in the diagram:

B A

G C

E (i)

C E

(ii)

24. In the right triangle ABC, right-angled at C, M is the midpoint of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. A

M B

Show that:

 AMC ≅ BMD (iii) DBC ≅  ACB (i)

(ii) ∠DBC is a right angle. (iv) CM =

1 AB 2

219

TRIANGLES

25. ∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. AD is extended to intersect BC at P. A

Show that, (i)

 ABD ≅  ACD

(ii)

(iii) AP bisects ∠A as well as ∠D

 ABP ≅  ACD

(iv) AP is the perpendicular bisector of BC.

26. If the angle bisector of a triangle bisects the opposite side, then prove that the triangle is an isosceles. 27. ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. Prove that AC + AD = BC. 28. In the given figure, Q is a point on the side SR of PQ.

PSR such that PQ = PR. Prove that PS > P

29. In the given figure, AC > AB and AD bisects ∠BAC. Prove that ∠ADC > ADB. A

30. O is any point in the interior of a

 ABC . Prove that (OA + OB + OC)> 12 (AB + BC + CA).

31. Prove that the perimeter of a triangle is greater than the sum of its three altitudes.

220

IL Foundation Series Class 9

32. In the given figure, AB||DC and diagonals AC and BD intersects at O. If OA = 3x - 1 and OB = 2x + 1, OC = 5x - 3 and OD = 6x - 5, find the value of x. D 6x 3x -

C 5x -

2x +

33. In the given figure, DE  AC and DF  AE . Prove that

FE EC = . BF BE

A D

221

8.1

QUADRILATERALS

INTRODUCTION TO QUADRILATERALS

8.1.1 Quadrilateral Different shapes can be formed by joining 4 points in various ways: All 4 collinear points on the same line result in a line segment. If 3 points are non-collinear, joining them forms a triangle.

Joining 4 non-collinear points creates a closed figure with four sides, known as a quadrilateral. A

A quadrilateral is a closed, two-dimensional figure bounded by line segments. In the above figure, AB, BC, CD, and DA are line segments. It is represented as quadrilateral ABCD or ◻ABCD. A quadrilateral comprises four sides (AB, BC, CD, and DA), four angles (∠ABC, ∠BCD, ∠CDA, and ∠DAB), four vertices (A, B, C, and D), and two diagonals (AC and BD). A diagonal is a line segment joining the opposite vertices of a quadrilateral. C D A

222

IL Foundation Series Class 9

In our surroundings, many objects take the shape of a quadrilateral, such as windows, blackboards, study tables, computer screens, LCD TVs, mobile phone screens, and book pages. Consider the quadrilateral ABCD shown in the figure. Two sides of a quadrilateral are adjacent sides if they have a common point (vertex). Adjacent Side

Common Vertex

AB and BC

BC and CD

CD and DA

DA and AB

Two sides of a quadrilateral are opposite sides if they have no common endpoint (vertex). In ◻ABCD, AB and DC are opposite sides, and BC and AD are also opposite sides.

Two angles of a quadrilateral are said to be adjacent angles if they have a common arm. Adjacent Angle

Common Arm

∠ABC and ∠BCD

∠BCD and ∠CDA

∠CDA and ∠DAB

∠DAB and ∠ABC

Two angles of a quadrilateral are said to be opposite angles if they do not have a common arm. In ◻ABCD, ∠ABC and ∠CDA, and ∠BCD and ∠DAB are pairs of opposite angles.

The sum of the lengths of all sides of the quadrilateral is the perimeter of a quadrilateral. Example: The perimeter of the following ◻ABCD = D

2 A

AB + BC + CD + DA = 5 + 4 + 3 + 2 = 14 units. Quadrilaterals can be convex or concave. i) Convex quadrilateral: A quadrilateral in which each interior angle measures less than 180° is termed a convex quadrilateral. 223

QUADRILATERALS

<180°

Convex quadrilateral

ii) Concave quadrilateral: A quadrilateral in which one of the interior angles measures more than 180° is referred to as a concave quadrilateral.

>180°

Concave quadrilateral

Note: 1) The sum of all four angles of a quadrilateral is 360° . Example: In a quadrilateral ABCD, ∠A + ∠B + ∠C +∠D = 360° 2) Each diagonal divides the quadrilateral into two triangles. Example: The diagonal AC divides the quadrilateral ABCD into two triangles, ∆ABC and ∆ADC.

3) Generally, a convex quadrilateral is considered a quadrilateral. 4) In a convex quadrilateral, both diagonals lie in the interior, whereas in a concave quadrilateral, one diagonal lies in the interior and the other in the exterior. 8.1.2 Types of quadrilaterals Quadrilaterals are classified based on their properties. Certain quadrilaterals have two pairs of parallel sides, while others only have one. Several quadrilaterals share the same properties; therefore, some quadrilaterals can be classified under multiple categories. 224

IL Foundation Series Class 9

The below flow chart shows the classification of quadrilaterals.

Quadrilateral

Two parallel sides Trapezoid

Isosceles Trapezoid

Two pairs of parallel sides

No parallel sides

Parallelogram

Kite

Rectangle

Rhombus

Square

Venn diagram of quadrilaterals: The following Venn diagram shows the inclusions and intersections of the various types of quadrilaterals.

225

QUADRILATERALS

Quadrilateral

Parallelogram

Trapezoid

Isosceles Trapezoid

Rhombus

Square

Rectangle

Kite

SOLVED EXAMPLES Example 1: In a quadrilateral ABCD, the angles ∠A + ∠B + ∠C + ∠D are in the ratio 1 : 2 : 3 : 4. Find the measure of each angle of the quadrilateral. Solution: Given ∠A + ∠B + ∠C + ∠D =1: 2 : 3 : 4 ; Total parts = 1 + 2 + 3 + 4 = 10 We know ∠A + ∠B + ∠C + ∠D =360° 1 2 ∠A = × 360° = 36° ; ∠B = × 360° = 72° ; 10 10 3 4 ∠C = × 360° = 108° ; ∠D = × 360° = 144° 10 10 ∴∠A= 36° , ∠B= 72° , ∠C= 108° , ∠D= 144°

226

IL Foundation Series Class 9

Example 2: If the three angles of a quadrilateral measure 110° ,82° , and 68° , then find the measure of the fourth angle. Solution: Let the measure of the fourth angle be x ° . Then, the sum of the angles of a quadrilateral being 360° , we have: 110° + 82° + 68° + x° = 360° ⇒ 260° + x ° = 360° ⇒ x° = 100° Hence, the measure of the fourth angle is 100° .

8.2

PROPERTIES OF SPECIFIC QUADRILATERALS

8.2.1 Trapezium A 'trapezium’ is a quadrilateral in which one pair of opposite sides are parallel.

(

)

Note: The parallel sides AB , CD are called the 'bases’ of the trapezium, and the other two sides are called its 'non-parallel sides (legs)’ BD , AC .

(

)

Altitude of a trapezium: The perpendicular distance from the vertex to the opposite base is the altitude of a trapezium. Properties of a trapezium: 1) ABCD is a trapezium such that AB  CD , then ∠A + ∠D = 180° , ∠B + ∠C = 180° . 2) The diagonal of a trapezium divides each other proportionally.

(

)

Example: If the diagonals ( AC and BD ) of a trapezium ABCD AB  CD intersect at O, then DO CO . = OB OA C

D O A

B 227

QUADRILATERALS

3) If the diagonals are proportional in a quadrilateral, then it forms a trapezium. 4) Any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally. Example: In a trapezium ABCD ( AB  CD ) , if EF  AB , then b

C h

E A

Area of the trapezium: Area of the trapezium=

AE BF = . ED FC

F B

1 1 ( AB + CD ) × CI= ( a + b ) × h , 2 2

where a, b are the lengths of parallel sides and ' h ' is the distance between the parallel sides. Isosceles trapezium: A trapezium in which the non-parallel sides are equal is known as an isosceles trapezium. In the isosceles trapezium ABCD , AB  CD , and AD = BC . Properties of an isosceles trapezium D

C 4

∠1 + ∠4 =180° and ∠2 + ∠3 = 180° .

2) Base angles are equal ( ∠1 =∠2 and ∠3 =∠4 ). 3) The lengths of diagonals are equal ( AC = BD ) .

SOLVED EXAMPLES Example 1: In the given figure, PQRS is an isosceles trapezium. Find x and y. S

R 3x

2x P 228

y Q

IL Foundation Series Class 9

Solution: We know that PQRS is a trapezium with SR  PQ Therefore, ∠P + ∠S =180° 2 x + 3x = 180° 5 x = 180° 180° 5 x = 36°  ∠P =∠Q( PS =RQ) ⇒ 2x = y x=

⇒ 2 × 36 = y ⇒y= 72 Hence, the required value for x is 36°, and y is 72O. Example 2: In the given figure, ABCD is a trapezium. Find the values of x and y. D

C 2x + 10°

x + 20° A

92°

y B

Solution: We know that ABCD is a trapezium with AB  DC Therefore, ∠A + ∠D =180° It is given that ∠A = x + 20° and ∠D = 2 x + 10° . 180° (co-interior angles) ( x + 20° ) + ( 2x + 10° ) =

3 x + 30° = 180° = 3 x 180° − 30° 3 x = 150° x = 50° Similarly,

229

QUADRILATERALS

y + 92° = 180° (co-interior angles) y 180° − 92° = y = 88° Hence, the required values for x and y are 50° and 88°, respectively. 8.2.2 Kite A quadrilateral having two pairs of equal adjacent sides but unequal opposite sides is called a kite. D A

ABCD is a kite with AB = BC and AD = CD . Properties of kite: 1) The diagonals of a kite are perpendicular to each other, i.e. BD ⊥ AC . 2) OA = OC. 3) ∠A =∠C . 4) Diagonal BD bisects ∠B & ∠D. 5) Diagonal BD divides the kite into two congruent triangles. 8.2.3 Parallelogram A quadrilateral in which both pairs of opposite sides are parallel is termed a parallelogram. D

O A

Properties: 1) In a parallelogram ABCD, two pairs of opposite sides are equal, i.e., AB = CD; BC = AD. 2) Opposite angles are equal, i.e., ∠A = ∠C =∠B = ∠D. 3) The diagonals of a parallelogram bisect each other = = , BO OD ) . ( AO OC 4) Each diagonal of a parallelogram divides it into two congruent triangles. 230

IL Foundation Series Class 9

5) The sum of any two adjacent angles in a parallelogram is 180°, i.e., (∠A + ∠B = 180°; ∠A + ∠D = 180°; ∠C +∠D = 180°; ∠B + ∠C = 180°; ∠B + ∠C = 180°). 6) If one pair of opposite sides is parallel and equal in a quadrilateral, it forms a parallelogram. 7) The sum of the squares of the four sides of the parallelogram equals the sum of the squares of the diagonals, i.e., in parallelogram ABCD : AB 2 + BC 2 + CD 2 + DA 2 = AC 2 + BD 2 . Rectangle: A parallelogram in which one angle is a right angle is called a rectangle (or); if one of the angles of a parallelogram is a right angle, then all angles are right angles. Such a parallelogram is called a rectangle. C

D O A

A rectangle satisfies all the properties of a parallelogram. a) The lengths of the diagonals of a rectangle are equal. b) Opposite sides are equal. c) Opposite angles are equal. d) Each diagonal divides it into two congruent right-angled triangles. e) The diagonals of a rectangle bisect each other. Note: D d

d H E

a a

G F

C c

bb B

1) The bisectors of angles of a parallelogram form a rectangle. (Here, ABCD is a parallelogram, and EFGH is a rectangle)

2) In a rectangle ABCD, O is an interior point of the rectangle, then OA 2 + OC 2 = OB 2 + OD 2 3) In a rectangle, the square of a diagonal is equal to the sum of the squares of the sides, i.e., in a 2 rectangle ABCD , AC = AD 2 + DC 2 .

Rhombus: A parallelogram in which two adjacent sides are equal is called a rhombus. D

B 231

QUADRILATERALS

Properties: a) Each diagonal of a rhombus divides it into two congruent isosceles triangles. b) Opposite angles are equal, and the sum of any two adjacent angles is 180°. c) The diagonals bisect each other perpendicularly. d) The diagonal AC bisects ∠A and ∠C ; the diagonal BD bisects ∠B and ∠D .

Square: A rectangle in which adjacent sides are equal is called a square. (OR) A rhombus in which one of its angles is a right angle is called a square. D

Properties: a) All sides are equal. b) Each angle is equal to 90°. c) The diagonals are equal and are mutually perpendicular bisectors. d) Each diagonal divides the square into two congruent right-angled isosceles triangles. Note: 1) If the diagonals of a parallelogram are the equal and right bisectors of each other, then it is a square. 2) In a rhombus ABCD, the diagonal AC bisects ∠A and ∠C , and the diagonal BD bisects ∠B and ∠D. 3) The diagonals of a rhombus divide it into four congruent right-angled isosceles triangles.

8.3

THEOREMS RELATED TO QUADRILATERALS

Theorem 1: A diagonal of a parallelogram divides it into two congruent triangles. Given: A parallelogram ABCD with AC as its diagonal C

A 232

IL Foundation Series Class 9

To prove: ∆ABC ≅ ∆ADC Proof: The opposite sides of a parallelogram are parallel. So, AB  DC and AD  BC Since AB  DC and AC is the transversal ∠BAC = ∠DCA (Alternate angles)

...(1)

Since AD  BC and AC is the transversal ∠DAC = ∠BCA (Alternate angles)

...(2)

In ∆ABC and ∆ADC, ∠BAC = ∠DCA

(From (1))

AC = AC (Common) ∠DAC = ∠BCA

(From (2))

∴∆ABC ≅ ∆ADC (ASA congruency) Hence proved. Theorem 2: In a parallelogram, opposite sides are equal. Given: A parallelogram ABCD with AC as its diagonal. To prove: AB = DC and AD = BC Construction: Draw AC, the diagonal of parallelogram ABCD C

D 2

4 A

Proof: In ∆ABC and ∆CDA AC = CA ( Common side ) ∠1 =∠2 ( Alternate angles ) ∠3 =∠4 ( Alternate angles ) Therefore, ∆ABC ≅ ∆CDA(ASA rule) Hence AB = DC and AD = BC (Corresponding parts of congruent triangles are congruent) Theorem 3: In a parallelogram, opposite angles are equal. Given: A parallelogram ABCD with AC as its diagonal. 233

QUADRILATERALS

To prove: ∠A =∠C , ∠B =∠D D

C 2

4 1 A

Proof: In ∆ABC and ∆ADC AC = CA ( Common ) ∠1 =∠2 ( Alternate angles ) ∠3 =∠4 ( Alternate angles )

Thus ∆ABC ≅ ∆CDA (ASA rule) Hence ∠B =∠D Similarly, we can prove ∠A =∠C Theorem 4: If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram Given: ABCD is a quadrilateral in which AB = DC, AD = BC To prove: Quadrilateral ABCD is a parallelogram Construction: Join A and C. D

Proof: In ∆ABC and ∆CDA AB = DC ( given ) AD = BC ( given ) AC = CA ( common side )

Thus, ∆ABC ≅ ∆CDA (SSS rule). ∠BAC = ∠DCA (Corresponding parts of congruent triangles are congruent) But, these are alternate angles. AB  CD and AB = DC Hence, quadrilateral ABCD is a parallelogram. Theorem 5: If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram. Given: ABCD is a quadrilateral in which ∠A =∠C and ∠B =∠D 234

IL Foundation Series Class 9

To prove: A quadrilateral is a parallelogram Proof: In quadrilateral ABCD ∠A + ∠B + ∠C + ∠D = 360° But ∠A =∠C and ∠B =∠D (given) ⇒ ∠A + ∠B + ∠A + ∠B = 360° ⇒ 2 [ ∠A + ∠B ] = 360° ⇒ ∠A + ∠B =180° But, ∠A and ∠B are interior angles on the same side of transversal AB, which cuts lines AD and BC AD  BC . Similarly, we can prove AB || DC. Hence, ABCD is a parallelogram. Theorem 6: The diagonals of a parallelogram bisect each other. Given: ABCD is a parallelogram in which diagonals AC and BD intersect each other at O. D

C O

To prove: OA = OC and OB = OD Proof: In ∆AOB, ∆COD, ∆AOB = ∆COD (vertically opposite angles) AB = CD (opposite sides of the parallelogram) ∠1 =∠2 (Alternate angles) Thus ∆AOB ≅ ∆COD Theorem 7: A quadrilateral is a parallelogram if a pair of opposite sides are equal and parallel. Given: ABCD is a quadrilateral in which AB = CD and AB  CD

235

QUADRILATERALS

C 2 3

4 A

To prove: Quadrilateral ABCD is a parallelogram Construction: Join A and C Proof: In ∆ABC and ∆CDA AB = CD (Given) ∠1 = ∠2 (Alternate angles) AC = CA (Common) Thus, ∆ABC ≅ ∆CDA (SAS rule) So ∠3 =∠4 (CPCT) But ∠3 and ∠4 are alternate angles AD  BC AD  BC and AB  DC Hence, quadrilateral ABCD is a parallelogram. Hence, OA = OC and OB = OD (CPCT)

SOLVED EXAMPLES Example 1: In parallelogram ABCD. Compute the values of x and y. A

° 28 ° 60

(10

)

(4y

Solution: Since ABCD is a parallelogram, AB  DC and AD  BC Now, AB  DC and transversal BD intersects them ∴∠ABD = ∠BDC ( Alternate angles are equal )

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IL Foundation Series Class 9

⇒ 4y = 28 ⇒y= 7° AD  BC , and transversal BD intersects them. ∴∠DBC = ∠ADB ⇒ 10 x = 60° ⇒ x = 6° Hence, x = 6° and y = 7° Example 2: In a parallelogram ABCD , ∠D = 105° , then find ∠A and ∠B . D

5°

Solution: Since the sum of any two consecutive angles of a parallelogram is 180°,

∠A + ∠D =180° and ∠A + ∠B =180

Now, ∠A + ∠D =180° ⇒ ∠A + 105° =180° ⇒= ∠A 75° = ∠D 105° (Given)  ⇒ 75° + ∠B= 180° ⇒ ∠B= 105° Thus, ∠A = 75° and ∠B = 105° Example 3: In a parallelogram ABCD, diagonals AC and BD intersect at O and AC = 6.8 cm, and BD = 5.6 cm. Find the measures of OD and OD. C

D O A

Solution: Since the diagonals of a parallelogram bisect each other, O is the midpoint of AC and BD. 1 1 ∴ OC = AC = × 6.8 cm = 3.4 cm , and 2 2 1 1 OD = BD = × 5.6 cm = 2.8 cm 2 2

237

QUADRILATERALS

Example 4: The diagonals of quadrilateral ABCD bisect each other. If ∠A = 35° , determine ∠B . Solution:

Given: The diagonals of a quadrilateral ABCD bisect each other, so it is a parallelogram. ∠A = 35° As the sum of interior angles between two parallel lines is 180° ∠A + ∠B =180°

Substituting the values, 35° + ∠B =180° ∠B = 180° − 35° So we get ∠B = 145° Therefore, ∠B = 145° Example 5: If the measures of opposite angles of a parallelogram are (60 − x)° and (3 x − 4)° , then find the measures of angles of the parallelogram. Solution: Let ABCD be a parallelogram, with (60 − x )°, and ∠C = (3x − 4)°. We know that in a parallelogram, the opposite angles are equal. Therefore,

∠A = ∠C 60-= x 3x − 4 − x − 3 x =−4 − 60 −4 x = −64 x = 16 Thus, the given angles become, ∠A= (60 − x)° = (60 − 16)° = 44° Similarly, ∠C = 44°

Also, the adjacent angles in a parallelogram form the consecutive interior angles of parallel lines, which must be supplementary.

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IL Foundation Series Class 9

Therefore, ∠A + ∠B =180° 44° + ∠B =180° ∠B= 180° − 44° ∠B = 136° Similarly, ∠D = ∠B ∠D = 136° Thus, the angles of a parallelogram are 44° ,136° , 44° , and 136° . Example 6: Calculate all the angles of a parallelogram if one of its angles is twice its adjacent angle. Solution: Let the angle of the required parallelogram be "x". Then, its adjacent angle will be 2x. It is said that the opposite angles of a parallelogram are equal. So, all the required angles of a parallelogram will be x, 2x, x and 2x. The sum of the required interior angles of a given parallelogram = 360° , x + 2x + x + 2x = 360° x = 60° Therefore, all the necessary angles will be 60° ,120° , 60° , and 120° . Example 7: If the diagonals of a parallelogram are equal, then show that it is a rectangle. Solution: Given: A parallelogram ABCD in which diagonal AC = diagonal BD. D

To prove: ABCD is a rectangle Proof: In ∆ABC and ∆DCB AC = BD

(Given)

BC = CB

(Common)

AB = DC

(opposite sides of a parallelogram)

Therefore, ∆ABC ≅ ∆DCB

(SSS rule) 239

QUADRILATERALS

∠ABC = ∠BCD

(CPCT)

But, ∠ABC + ∠BCD =180° [Interior angles on the same side of transversal BC and AB  CD ] So, 2∠ABC = 180° ∠ABC = 90° ∠BCD ] [ ∠ABC = Hence, parallelogram ABCD is a rectangle. Example 8: Show that if the diagonals of a quadrilateral bisect each other at the right angles, it is a rhombus. Solution: Given: A quadrilateral ABCD in which diagonals AC and BD bisect each other at right angles at O. D

O A

To prove: Quadrilateral ABCD is a rhombus Proof: In ∆AOD and ∆AOB

OA = OA( Common ) OD = OB ( Given ) ∠AOD = ∠AOB = 90 (Given) Therefore, ∆AOD ≅ ∆AOB

(SAS rule)

so, AD = AB (CPCT) Similarly AB = BC = CD = DA Hence, quadrilateral ABCD is a rhombus. Example 9: ABCD is a parallelogram, and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that

i) ∆APB ≅ ∆CQD ii) AP = CQ

Solution: Given: ABCD is a parallelogram, and AP and CQ are perpendiculars from vertices A and C on BD. To prove: i)∆APB ≅ ∆CQD Proof: i) In ∆APB and ∆CQD, we have ∠ABP = ∠CDQ (Alternate angles) 240

ii) AP = CQ

IL Foundation Series Class 9

AB = CD (opposite sides of a parallelogram) ∠APB = ∠CQD (Each 90°) ∆APB ≅ ∆CQD (AAS rule) ii) So, AP = CQ (CPCT).

8.4

THE MIDPOINT THEOREM

We have proved several parallelogram theorems. Now, let’s use these theorems to prove a few interesting and helpful triangle facts. Theorem: The line segment joining the midpoints of two sides of a triangle is parallel to the third side and is equal to half of it. A

Given: ∆ABC in which D and E are the midpoints of sides AB and AC, respectively. 1 BC . 2 Construction: Produce the line segment DE to F, such that DE = EF. Join FC. To prove: DE  BC and DE =

Proof: In ∆AED and ∆CEF, we have AE = CE

( E is the midpoint of AC )

⇒ ∠AED = ∠CEF

(Vertically opposite angles)

And DE = FE

(By construction)

Therefore, ∆AED ≅ ∆CEF (By SAS-criterion of congruence) Using corresponding parts of congruent triangles ⇒ AD = CF

...(1)

and ∠ADE = ∠CFE

...(2)

D is the midpoint of AB. ⇒ AD = DB

...(3)

241

QUADRILATERALS

From equation (1) and (3), we get CF = DB

...(4)

Now, DF intersects AD and FC at D and F, respectively, such that, ∠ADE = ∠CFE

(From (2))

That is, alternate interior angles are equal. ∴

AD  FC

⇒

DB  CF

...(5)

From (3) and (5), we find that DBCF is a quadrilateral such that one pair of sides is equal and parallel. ∴

DBCF is a parallelogram.

⇒

DF  BC and DF = BC ( Opposite sides of a parallelogram are equal and parallel)

But, D, E, F are collinear and DE = EF. 1 BC . 2 Converse of the midpoint theorem: The line drawn through the midpoint of one side of a triangle, parallel to another side, bisects the third side. ∴

DE  BC and DE =

E’ E

Given: ∆ABC in which D is the midpoint of DE and DE  BC To prove: E is the midpoint of AC. Construction: Join DE and DE’. Proof: We must prove that E is the midpoint of AC. Suppose E is not the midpoint of AC. Then, let E' be the midpoint of AC. Now, in ∆ABC, D is the midpoint of AB and E' is the midpoint of AC. But the line drawn through the midpoint of one side of a triangle, parallel to another side bisects the third side. Therefore, we have DE '  BC

242

...(1)

IL Foundation Series Class 9

Also Given, DE  BC

...(2)

From (1) and (2), we find that two intersecting lines, DE and DE’, are both parallel to line BC. This is a contradiction. So, our assumption is wrong. Hence, E is the midpoint of AC. Note: 1) The quadrilateral formed by joining the midpoints of sides of a quadrilateral is itself a parallelogram. 2) Similarly, the quadrilateral formed by joining the midpoints of sides of a parallelogram is also a parallelogram. 3) The quadrilateral formed by joining successively the midpoints of sides of a square is also a square. 4) The quadrilateral formed by joining successively the midpoints of sides of a rectangle is a rhombus.

SOLVED EXAMPLES Example 1: In the figure, D, E, and F are, respectively, the midpoints of sides BC, CA, and AB of an equilateral triangle ABC. Prove that triangle DEF is also an equilateral triangle. A

Solution: Given: D, E, and F are the midpoints of sides BC, CA, and AB of an equilateral triangle respectively. To prove: DEF is an equilateral triangle. Construction: Join DE, DF, and FE. Proof: Since the segment joining the midpoints of two sides of a triangle is half of the third side. Therefore, D and E are the midpoints of BC and AC, respectively. 1 ⇒ DE = AB 2 E and F are the midpoints of AC and AB, respectively.

...(1)

243

QUADRILATERALS

1 BC 2 F and D are the midpoints of AB and BC, respectively. EF =

1 AC 2 Now, ∆ABC is an equilateral triangle. FD =

...(2)

...(3)

⇒ AB = BC = CA ⇒

1 1 1 AB = BC = CA 2 2 2

⇒ DE = EF = FD Hence, ∆DEF is an equilateral triangle. Example 2: l, m, and n are three parallel lines. p and q are two transversals intersecting the parallel lines at A, B, C, D, E, and F, as shown in the figure. If AB : BC = 1:1, find the ratio of DE : EF. A

C p

F q

m n

Solution: Given: AB : BC = 1:1 To find: DE : EF Construction: Join AF such that it intersects line m at G. In △ACF

AB = BC (1:1 ratio) BG ∥ CF (as m ∥ n)

Therefore, by the converse of the midpoint theorem, G is the midpoint of AF (AG = GF) Now, in △AFD

AG = GF (proved above) GE ∥ AD (as l ∥ m)

Therefore, by the converse of the midpoint theorem, E is the midpoint of DF (FE = DE) So, DE : EF = 1 : 1 (as they are equal)

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IL Foundation Series Class 9

Example 3: In the figure given below, L, M, and N are the midpoints of the sides PQ, QR, and PR, respectively, of triangle PQR. If PQ = 8 cm, QR = 9 cm and PR = 6 cm. Find the perimeter of the triangle formed by joining L, M, and N. P L

Solution: As L and N are midpoints. By midpoint theorem LN ∥ QR and LN =

1 × (QR) 2

1 × 9 = 4.5 cm 2 1 1 Similarly, LM = × (PR) = ×(6) = 3 cm 2 2 1 1 Similarly, MN = × (PQ) = × (8) = 4 cm 2 2 LN =

Therefore, the perimeter of ∆LMN = LM + MN + LN = 3 + 4 + 4.5 = 11.5 cm The perimeter is 11.5 cm. Example 4: ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse AB and parallel to BC intersects AC at D. Show that i) D is the midpoint of AC Solution:

ii) MD ⊥ AC

= MA = iii) CM

1 AB 2

Given: A triangle ABC, in which ∠C = 90° and M is the midpoint of AB and BC  DM To Prove: i) D is the midpoint of AC ii) DM ⊥ AC 1 AB 2 Construction: Join CM iii) CM = MA =

245

QUADRILATERALS

Proof: i) In ∆ABC M is the midpoint of AB , BC  DM By the converse of the midpoint theorem, D is the midpoint of AC ii) ∠ADM = ∠ACB (Corresponding angles) But, ∠ACB = 90° ∴∠ADM = 90° But, ∠ADM + ∠CDM =180° ⇒ 90° + ∠CDM =180° ⇒ ∠CDM =90° Hence, DM ⊥ AC iii) AD = DC (D is the midpoint of AC) Now in ∆ADM and ∆CMD ∠ADM = ∠CDM = 90° AD = DC DM = DM [common side] ∆ADM ≅ ∆CMD [ by SAS rule] By CPCT CM = MA Since M is the midpoint of AB 1 ∴ MA = AB 2 Hence, CM = MA =

246

1 AB 2

IL Foundation Series Class 9

Example 5: If quadrilateral ABCD is a rhombus, and P, Q, R, and S are the midpoints of AB, BC, CD, and DA , respectively, prove that quadrilateral PQRS is a rectangle. D

S A

Q P

Solution: Given: ABCD is a rhombus P, Q, R, S are the midpoints of AB, BC, CD, DA respectively. To prove: PQRS is a rectangle Construction: join AC Proof: In ∆ABC, P and Q are the midpoints of the sides AB and BC.

1 ∴ PQ  AC and= PQ AC − (1) (midpoint theorem) Similarly, in ∆ADC SR  SR and= SR

1 AC − ( 2 ) 2

From (1) and (2), we get

PQ  SR and PQ = SR − ( 3) Hence, PQRS is a parallelogram Since the sides of a rhombus are equal, AB = BC. 1 1 × AB = × BC 2 2 (P and Q are the midpoints of sides AB and BC, respectively) ∠QPB = ∠PQB (Sides opposite to equal angles are equal) ---------- (3) In ∆APS and ∆CQR, (P and Q are the midpoints of sides AB and BC, respectively) (S and R are the midpoints of sides AD and CD, respectively) PS = QR (Opposite sides of a parallelogram are equal) By SSS congruency, ∆APS ≅ ∆CQR (3) So, ∠APS = ∠CQR (By|CPCT) -----------------(4) Since AB is a straight line, ∠APS + ∠SPQ + ∠QPB =180. Since BC is a straight line, ∠PQB + ∠PQR + ∠CQR =180. 247

QUADRILATERALS

∠APS + ∠SPQ + ∠QPB = ∠PQB + ∠PQR + ∠CQR By equations (3) and (4), we get, ∠APS + ∠SPQ + ∠QPB = ∠QPB + ∠PQR + ∠APS ∠SPQ = ∠PQR ----------(5) Since ∠SPQ and ∠PQR are interior angles on the same side of the transversal PQ, they form a pair of supplementary angles. ∠SPQ + ∠PQR =180 ⇒ 2∠SPQ = 180 [From (5)] ⇒ ∠SPQ =90 Clearly, PQRS is a parallelogram with one of its interior angles being 90. Hence, PQRS is a rectangle.

QUICK REVIEW • The sum of the angles of a quadrilateral is 360°. • A diagonal of a parallelogram divides it into two congruent triangles. • In a parallelogram: 1) Opposite sides are equal. 2) Opposite angles are equal. 3) Diagonals bisect each other. • A quadrilateral is a parallelogram if: 1) Opposite sides are equal. 2) Opposite angles are equal. 3) Diagonals bisect each other. 4) A pair of opposite sides is equal and parallel. • Diagonals of a rhombus bisect each other at right angles. • Diagonals of a rectangle bisect each other and are equal. • Diagonals of a square bisect each other at right angles and are equal. • The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is half of it. • A line through the midpoint of a side of a triangle, parallel to another side, is parallel to the third side. • The quadrilateral formed by joining the midpoints of the sides of a quadrilateral in order is a parallelogram

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IL Foundation Series Class 9

WORKSHEET - 1 I.

INTRODUCTION TO QUADRILATERALS 1. The opposite angles of a quadrilateral ABCD are equal. If AB = 4 cm, determine CD. 2. Can the angles of a quadrilateral be in the ratio 1 : 2 : 3 : 6. If not, why? Give reasons. 3. In a quadrilateral, one angle is 120° , and the remaining three angles are equal. Find the measures of these three angles. 4. In a quadrilateral, if the three angles are 55° , 65° , and 150° , find the fourth angle. 5. Is it possible to draw a quadrilateral whose all angles are obtuse? 6. Calculate all the angles of a quadrilateral if they are in the ratio 2 : 5 : 4 : 1. 7. The angles of the quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral. 8. In a quadrilateral, the angles are x ° , ( x + 10)° , ( x + 20)° , and ( x + 30) . Find the measure of the greatest angle. °

II. PROPERTIES AND THEOREMS RELATED TO QUADRILATERALS 1. ABCD is a trapezium in which AB  CD . If AD = BC, show that ∠A =∠B and ∠C = ∠D .

(

)

2. In an isosceles trapezium ABCD AB  CD , if AC = 5a + 20, BD = 3a + 40, then find a. 3. In the figure, ABCD is a trapezium in which AB  DC . If ∠A = 55° and ∠B = 70°, find ∠C , and ∠D . D

55°

C 70°

4. PARK is an isosceles trapezium in which ∠A = 6y − 60 and ∠R = 3y + 30 . Find y and measure all the angles. 5. ABCD is a trapezium in which AB  DC and ∠A =∠B =45° . Find out the necessary angles C and D of the trapezium. 6. In a rectangle, one of the diagonals forms an angle of 25° with one of its sides. Calculate the acute angle formed between the two diagonals. 7. Prove that a quadrilateral whose diagonals bisect each other at right angles is a rhombus. 8. Prove that the diagonals of a square are equal in length and bisect each other at right angles. 9. Prove that a quadrilateral with equal diagonals that bisect each other at right angles is a square. 10. Find all the angles of a parallelogram if one of its angles is twice the size of its adjacent angle. 249

QUADRILATERALS

11. If the diagonal AC of a parallelogram ABCD bisects ∠A, show that it also bisects ∠C and that ABCD is a rhombus. 12. Given a rhombus ABCD, show that diagonal AC bisects ∠A and ∠C, and diagonal BD bisects ∠B and ∠D. 13. Given a rectangle ABCD where diagonal AC bisects ∠A and ∠C, show that ABCD is a square and that diagonal BD bisects ∠B and ∠D. 14. If ABCD and AEFG are two parallelograms and ∠C equals 55°, find ∠F. 15. If the angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°, find the angles of the parallelogram. 16. ABCD is a quadrilateral whose diagonals AC and BD intersect at O. Prove that i) AB + BC + CD + AD > AC + BD ii) AB + BC + CD + AD < 2 ( AC + BD )

17. In the fig., AB  DE , AB = DE , AC  DF and AC = DF. Prove that BC  EF and BC = EF. A

D E

18. PQ and RS are two equal and parallel line segments. Any point M not lying on PQ or RS is joined to Q and S, and lines through P parallel to QM and through R parallel to SM meet at N. Prove that line segments MN and PQ are equal and parallel to each other. 19. In a quadrilateral ABCD, CO and DO are bisectors of ∠C and ∠D, respectively. Prove that 1 (∠A + ∠B ) ∠COD= 2 20. The sides AB and CD of a quadrilateral ABCD are extended to points P and Q, respectively. Is ∠ADQ + ∠CBP = ∠A + ∠C ? Give the reason. III. MIDPOINT THEOREM 1. Given a quadrilateral ABCD, where P, Q, R, and S are the midpoints of AB, BC, CD, and respectively, and AC = BD. Prove that PQRS forms a rhombus.

(

)

2. If the median of a trapezium ABCD , AB  CD is 16 cm , and AB = 4 cm , then find CD. 3. The bases of a trapezium are 13 cm and 9 cm . Find the length of the median of the trapezium. 4. ABCD is a trapezium with AB  CD . P and Q are the midpoints of diagonals AC and BD. Prove that i) PQ  AB  CD 250

ii)= PQ

1 ( AB − DC ) 2

IL Foundation Series Class 9

5. Given a quadrilateral ABCD, where P, Q, R, and S are the midpoints of AB, BC, CD, and DA, respectively, and AC ⊥ BD . Verify that PQRS forms a rectangle. Verify that DA = AR and CQ = QR. 6. Given a quadrilateral ABCD, where P, Q, R, and S are the midpoints of AB, BC, CD, and DA, respectively, and AC = BD and AC ⊥ BD . Verify that PQRS forms a square. 7. Given a parallelogram ABCD, where P and Q are the midpoints of the opposite sides AB and CD respectively, and AQ intersects DP at S and BQ intersects CP at R, show that PRQS forms a parallelogram. 8. Given a trapezium ABCD, where E and F are the midpoints of the parallel sides AD and BC 1 respectively, prove that EF  AB and EF = (AB + CD). 2 9. Prove that the line joining the midpoint and the line diagonals of a trapezium are parallel to the parallel sides of the trapezium. 10. Given a parallelogram ABCD, where P is the midpoint of side CD, and a line through C parallel to PA intersects AB at Q and DA produced at R. Verify that DA = AR and CQ = QR. 11. ABC is a triangle right angled at B, and P is the mid point of AC. Prove that PB = PA =

1 AC . 2

12. The given figure shows a triangle ABC in which AD is a median and E is the midpoint of median AD. B and E are joined, and BE is produced to meet AC at point F. Prove that 1 AF = AC . 3 A F E B

13. ABC is a right triangle. D and E are the midpoints of the sides AB and AC, respectively. If = AB 8= cm, BC 6 cm and AC = 10 cm. A

251

QUADRILATERALS

Find i) the area of the triangle ABC ii) the perimeter of the trapezoid BCED iii) the area of the trapezoid BCED 14. ABCD is a quadrilateral in which P, Q, R and S are the midpoints of the sides AB, BC, CD and DA. AC is a diagonal. Show that: D

R C

i) SR  AC and SR = ii) PQ = SR

1 AC 2

iii) PQRS is a parallelogram.

WORKSHEET - 2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. The sum of angles of a concave quadrilateral is

a) More than 360°

b) Less than 360°

c) Equal to 360°

d) Twice of 360°

2. In a quadrilateral, if diagonals are equal in length, then it is a) Rhombus

b) Parallelogram

c) Rectangle

d) None

3. The perimeter of a quadrilateral is __________ sum of its diagonals. a) Greater than

b) Less than

c) Equal to

d) None

4. A quadrilateral formed by joining angular bisectors of any cyclic quadrilateral is always [ ] a) Parallelogram

b) Rectangle

c) Square

d) Cyclic quadrilateral

b) Kite

c) Rhombus

d) None

5. A square is a a) Trapezium

= cm, CD 10 cm, AB  CD , P and Q are the midpoints of AD 6. In a trapezium ABCD= with AB 20 and BC, then the length of PQ is

a) 10 cm

b) 5 cm

c) 2 cm

d) 15 cm

7. A quadrilateral formed by joining the midpoints of any quadrilateral is always a) Rectangle 252

b) Square

c) Rhombus

d) Parallelogram

IL Foundation Series Class 9

8. If M is the midpoint of hypotenuse PR of a right-angled triangle PQR right angled at Q, then a) MP = PR

b) MQ = PQ

c) MQ = MP

d) None

9. The parallel sides of a trapezium are x and y in length. The length of the line segment joining the midpoints of the non-parallel sides is 2 x + 3y x⋅y x+y a) b) x + y c) d) 2 2 2 10. Two consecutive angles of a parallelogram are in the ratio 1 : 3, then the smaller angle is a) 45°

b) 50°

c) 60°

d) 90°

11. The diagonals of a parallelogram bisect each other at right angles, then it is a a) Rhombus b) Rectangle c) Trapezium d) Quadrilateral 12. In the given figure, the measure of ∠COD = D

C O

29°

a) 90°

b) 118°

33°

31° B

c) 180°

d) 60°

13. In parallelogram PQRS, PT and QT are angular bisectors of ∠P and ∠Q , then ∠PTQ is. a) 100°

b) 180°

c) 90°

d) 60°

14. In ∆ABC, D, E, and F are the midpoints of the sides BC, CA, and AB, respectively. If AC = 8.2 cm , then DF is ______________. a) 8.2 cm

b) 4.2 cm

d) 12.3 cm

c) 4.1 cm

15. The diagonals of rhombus ABCD intersect at the point ' O '. If ∠BDC = 50° , then ∠OAB a) 50°

b) 40

c) 25°

d) 20°

16. In APB and CQD are two parallel lines, then the bisectors of ∠APQ, ∠BPQ, ∠CQP and ∠PQD enclose a a) Square

b) Rhombus

c) Rectangle

d) Kite

17. The diagonals AC and BD of a parallelogram ABCD intersect each other at the point 'O' such ° ° that ∠DAC =30 , ∠AOB =70 , then ∠ADO = ? D

C O

30°

a) 40°

b) 35°

70° B

c) 45°

d) 50° 253

QUADRILATERALS

18. If an angle of a parallelogram is 24° less than twice the smallest angle, then the largest angle is a) 68°

b) 102°

c) 112°

d) 136°

19. If the given figure ABCD is a rhombus, then D

C O

a) AC 2 + BD 2 = AB 2

b) AC 2 + BD 2 = 2 AB 2

c) AC 2 + BD 2 = 4 AB 2

d) 2 ( AC 2 + BD 2 ) = 3 AB 2

20. In a parallelogram ABCD, E and F are the midpoints of the sides AB and CD, respectively. AF and CE meet the diagonal BD of length 12 at P and Q, respectively. Now, the length of PQ is a) 6 cm

b) 4 cm

d) 5 cm

c) 3 cm

21. The diagonals AC and BD of parallelogram ABCD intersect at the point O. If ∠DAC = 34° and ∠AOB = 75° , then measure ∠DBC is A

D 34° 75° O

a) 34°

b) 38°

c) 41°

d) 17°

22. Figure ABCD is a rhombus such that ∠ADB is A

50° D

a) 40°

b) 50°

c) 90°

d) 60°

110° . The angles of the triangle formed by joining the midpoints 23. In ∆ ABC , ∠A = 40° and ∠C = of the sides of this triangle are a) 70° , 70° , 40°

254

b) 60° , 40° ,80°

c) 30° , 40° ,110°

d) 60° , 70° ,50°

IL Foundation Series Class 9

24. In an equilateral triangle ABC, D and E are the midpoints of sides AB and AC, respectively. Then, length of DE is 2 3 1 BC b) BC c) BC d) Not possible to find 3 2 2 25. PQRS is a quadrilateral. PR and QS intersect each other at O. In which of the following cases is PQRS a parallelogram?

a) ∠P= 100° , ∠Q= 80° , ∠R= 100° . ° ° ° b) ∠P= 85 , ∠Q= 85 , ∠R= 95

cm ,QR 7= cm , RS 8= cm , SP 8 cm = c) PQ 7 =

= = cm, OQ 6.5 = cm, OR 5.2 = cm, SP 8 cm d) OP 6.5

26. What is the ratio of AC and CF if AD and BE are the median and DF  BE ? A

E F B

a) 4 : 1

b) 1 : 1

c) 1 : 4

d) 1 : 2

27. Identify the correct relation from the given options if PQRS is a parallelogram, and V is the midpoint of PQ. U T

1 1 PU b) TU = PU c) TU = BC PU d) TU = 2PU 4 2 28. Find the ratio of ( AB + CD ) / EG if EG  AB , and AB  DC , and E and G are the midpoints.

a) TU =

B F

a) 1

1 BC 2

c) 2

d) 3 255

QUADRILATERALS

29. If ABCR and PQRS are rectangles, and B is the midpoint of PR, find the ratio of AC and PR. S

R C

a) 1 : 2

b) 2 : 1

c) 1 : 1

d) 5 : 2

30. In the given figure, ABCD is a rhombus. Diagonals AC and BD intersect at O. E and F are the midpoints of AO and BO, respectively. If AC = 16 cm and BD = 12 cm , then EF is: D

O E

a) 4 cm

b) 5 cm

c) 6 cm

d) 7 cm

31. ABCD is a parallelogram in which diagonals AC and BD intersect at O. If E, F, G, and H are the midpoints of AO, BO, CO, and DO, respectively, then the ratio of the perimeter of the quadrilateral EFGH to the perimeter of parallelogram ABCD is D

G O

a) 1 : 3

b) 1 : 2

c) 2 : 3

d) 3 : 4

II. MULTIPLE CHOICE QUESTIONS WITH MULTIPLE CORRECT ANSWERS 1. A quadrilateral has a) Four sides

b) Four diagonals

c) Four vertices

d) Four angles

c) BD

d) BC

2. In a quadrilateral ABCD, a diagonal is a) AB

256

b) AC

IL Foundation Series Class 9

3. Which among the following has all its angles equal? a) Square

b) Rhombus

c) Rectangle

d) Parallelogram

4. In an isosceles trapezium, a correct statement is a) All sides are equal

b) Non-parallel sides are equal

c) Diagonals are equal

d) All angles are equal

5. The diagonals of a kite are a) Perpendicular to each other

b) Parallel to each other

c) Equal to each other

d) Not equal

6. If ABCD is an isosceles trapezium such that AB  CD and AD = BC, then a) ∠A + ∠D =180°

b) ∠A =∠B

c) AC = BD

d) ∠B + ∠C =180

7. If ABCD is a kite with AB = BC and AD = CD, then a) OA = OC

b) ∠A =∠C

c) OB = OD

d) ∠B =∠D

III. FILL IN THE BLANKS 1. If the diagonals of a rhombus are 12 cm and 16 cm, then the length of the side of the rhombus is ______________________. 2. The bisectors of the angles of a parallelogram enclose a _____________________. 3. ∆ABC is a right triangle and angled at B side. If AB = 6 cm, side BC = 8 cm, and D is the midpoint of AC, then the length of BD is ____________________. 4. The perimeter of a parallelogram is 32 cm. If the shorter side is 6.5 cm, then the measure of the longest side is _____________________. 5. If ABCD is a parallelogram, then ∠A − ∠C is equal to _____________________. 6. In a parallelogram ABCD, if ∠A = 80° , then ∠B is ____________________. 7. Two adjacent angles of a rhombus are 3x − 40° and 2 x + 20° . The measurement of the greater angle is ______________________. 257

QUADRILATERALS

8. A trapezium is called ________________ if its non-parallel sides are equal. 9. The line segment joining the midpoints of non-parallel sides of a trapezium is called its _______________________. 10. A diagonal of a rectangle is inclined to one side of the rectangle at 25° . The acute angle between the diagonals is _________________. 11. ABCD is a rhombus such that ∠ACB = 40°, then ∆ADB is _______________. 12. The quadrilateral formed by joining the midpoints of the sides of a quadrilateral PQRS, taken in order, is a rectangle if ______________________________. 13. If angles A, B, C, and D of the quadrilateral ABCD taken in order are in the ratio 3 : 7 : 6 : 4, then ABCD is a ___________________. 14. If bisectors of ∠A and ∠B of a quadrilateral ABCD intersect each other at P, of ∠B and ∠C at Q, of ∠C and ∠D at R, and of ∠D and ∠A at S, then PQRS is a ____________________. 15. D and E are the midpoints of the sides AB, and AC of ∆ABC and O is any point on side BC. O is joined to A. If P and Q are the midpoints of OB and OC, respectively, then DEQP is ________________________. IV. SUBJECTIVE QUESTIONS 1. In a parallelogram, if two adjacent angles are (2 x + 35)° and (3x − 15)° , find the value of x. 2. Find the measure of each angle of a parallelogram if one of its angles is 30° less than twice the smallest angle. 3. PQRS is a parallelogram in which PQ = 12 cm and its perimeter is 40 cm . Find the length of each side of the parallelogram. 4. If an angle of a parallelogram is two-thirds of its adjacent angle, find the angles of the parallelogram. 5. State the condition for a parallelogram to be a rectangle. 6. State the type of quadrilateral if its diagonals bisect each other. 7. If in a quadrilateral ABCD , ∠B = 90° and AB = BC = CD = DA, then write the type of quadrilateral ABCD. 8. In ∆DEF, PQ is a line segment drawn through midpoints of DE and DF, respectively. If EF = 7.6 cm , then find the length of PQ. 9. JUMP is a square with ∠JUP = 45° . Determine ∠PUM . 10. In a parallelogram PQRS taken in order if ∠Q = 128° , determine the measures of its other angles. 11. In the given figure, ABCD and BDCE are parallelograms with a common base DC. If BC ⊥ BD , then find ∠BEC .

258

IL Foundation Series Class 9

300

12. ABCD is a trapezium in which AB  CD and AD = BC. Show that i) ∠A =∠B

ii) ∠C = ∠D

iii) ∆ABC ≅ ∆BAD

iv) Diagonal AC = diagonal BD

13. ABCD is a rhombus in which the perpendicular bisector of AB passes through D. Find the angles of the rhombus. 14. In the given figure, P is the midpoint of side BC of a parallelogram ABCD such that BAP = DAP. D

Prove that AD = 2CD. 15. ABCD is a rhombus, and AB is produced to E and F such that AE = AB = BF. Prove that ED and FC are perpendicular to each other. 16. Find x in the following figure. 90°

E 60° D

90° l B

x 40°

17. ABCD is a parallelogram. The diagonals AC and BD intersect at a point O. If E, F, G, and H are ( EF + FG + GH + HE ) the midpoints of AO, DO, CO and BO, respectively, then find the ratio of . ( AD + DC + CB + BA) 18. PQRS is a square whose length is 30 cm. M is a point on QR, and N is a point inside the square such that MN is perpendicular to QR, where NP = NS = NM - 5. Then what will be the value of MN?

259

ANSWER KEY 1. NUMBER SYSTEMS

Worksheet 1 I. Irrational numbers 1. (a), (d) – Irrational and (b), (c) - Rational 2. c 3.

23 99 √7 7

167

12. 90

II. Density property and real number line 1. NA 5 9 11 , , 2 4 4

3. NA 4. NA 5. NA

1 5

6. − and

1 5

1 9 20 4 𝑧𝑧𝑧𝑧 = , 7 9

7. 𝑢𝑢𝑢𝑢 = − and 𝑢𝑢𝑢𝑢 =

1 3

III. Concept of infinity and interval notation 1. c 2. b 3. d 4. c 5. a IV. Surds 1. d 2. b 3. d 4. 𝑥𝑥𝑥𝑥 = 3 5. 12√3

6. 2√2 7. 8√2 + √3 8. 4 9. √2

260

14. 2√2 15. 2√3 16. b 17. 4

5. √6 + √5 6. 7 + 2√10 7. 4 8. 3√2 + 2√3 9. −√5 10. 60 11. 4 decimal places

10. 4 11. 540 12. -2 13. 23

18. 𝑥𝑥𝑥𝑥 = 3 19. 1 20. b 21. c 22. 1 23. 1 24. NA 25. a 26. c 27. 1 28. 1 29. 2 30. 214 V. Complex numbers 1. 0 + 5𝑖𝑖𝑖𝑖 2. 4 − 5𝑖𝑖𝑖𝑖 3. 5 + 5𝑖𝑖𝑖𝑖 4. −67 − 19𝑖𝑖𝑖𝑖 5. −9 + 40𝑖𝑖𝑖𝑖 6. −2 − 10𝑖𝑖𝑖𝑖 7. 𝑖𝑖𝑖𝑖 8. 2 9.

3−√7𝑖𝑖𝑖𝑖 16 𝑛𝑛𝑛𝑛

12.

2 5

10. 2 11. (0, −2) 13. (1, 0)

Worksheet 2 I. Multiple choice questions with single correct answer 1. a 2. d 3. d 4. a 5. c 6. b 7. c 8. b 9. d 10. c 11. b 12. a 13. c 14. d 15. b 16. c 17. b 18. c 19. b 20. c

ANSWER KEY 12. i) Cubic polynomial ii) Cubic polynomial iii) Linear polynomial iv) Quadratic polynomial v) Constant polynomial vi) Linear polynomial 13. 61, –143

II. Fill in the blanks 1. 5√5 2. an Irrational 3. −4 4. 2√14

233 990 3 10

14. −

15. i) 4

7. 12√2 8. –3 9. 3√5 10. 0 III. Subjective questions 1. 14 2. 𝑎𝑎𝑎𝑎 = 2, 𝑏𝑏𝑏𝑏 = −1 3. 6 3 4. (√7 + √2) 5.

iii)

5. 6. 7. 8.

Worksheet 1 I. Polynomials terminology and zero of a polynomial 3. 4. 5. 6. 7.

3𝑥𝑥𝑥𝑥 2 √𝑥𝑥𝑥𝑥

4 Not Defined 1 6 6 5

8. − 2

9. –4, 2 10. NA

11. i) 6

iii) -1

iv) 0

ii) −

–1 NA NA i) 0 ii) −𝜋𝜋𝜋𝜋 3 + 3𝜋𝜋𝜋𝜋 2 − 3𝜋𝜋𝜋𝜋 + 1 27 iii) −

557 81

2. POLYNOMIALS

1 2

7 2

4. i) − 8

6. Irrational number 7. 3.4512 8. 33 − 56𝑖𝑖𝑖𝑖 9. 1 + 𝑖𝑖𝑖𝑖 10. −33 + 13𝑖𝑖𝑖𝑖

ii) 2

16. 0 17. NA II. Factorisation of polynomials 1. NA 2. NA 3. (𝑥𝑥𝑥𝑥 − 1) (𝑥𝑥𝑥𝑥 − 10) (𝑥𝑥𝑥𝑥 − 12)

5 97 33

1. iii) 𝑥𝑥𝑥𝑥 2 +

31 4

ii) 5

1 5

iv)

9. 5𝑎𝑎𝑎𝑎 10. NA 11. i) 12. NA 13. i) 𝑘𝑘𝑘𝑘 = − 2 ii) 𝑘𝑘𝑘𝑘 = −�2 + √2� 14. i) (3𝑥𝑥𝑥𝑥 − 1)(4𝑥𝑥𝑥𝑥 − 1) ii) (2𝑥𝑥𝑥𝑥 + 1)(𝑥𝑥𝑥𝑥 + 3) iii) (3𝑥𝑥𝑥𝑥 − 2)(2𝑥𝑥𝑥𝑥 + 3) iv) (3𝑥𝑥𝑥𝑥 − 4)(𝑥𝑥𝑥𝑥 + 1) 15. i) (𝑥𝑥𝑥𝑥 − 2)(𝑥𝑥𝑥𝑥 + 1)(𝑥𝑥𝑥𝑥 − 1) ii) (𝑥𝑥𝑥𝑥 − 5)(𝑥𝑥𝑥𝑥 + 1)(𝑥𝑥𝑥𝑥 + 1) iii) (𝑥𝑥𝑥𝑥 + 10)(𝑥𝑥𝑥𝑥 + 2) iv) (𝑦𝑦𝑦𝑦 − 1)(2𝑦𝑦𝑦𝑦 + 1)(𝑦𝑦𝑦𝑦 + 1) 16. k = 2 17. (𝑥𝑥𝑥𝑥 − 1)(2𝑥𝑥𝑥𝑥 − 5)(𝑥𝑥𝑥𝑥 − 3) 18. NA 19. NA 20. – 1 21. a = 5, remainder = 62 22. NA 23. NA 261

ANSWER KEY III. Long division and Horner’s synthetic division of polynomials 1. NA 2. i) 𝑥𝑥𝑥𝑥 2 + 3𝑥𝑥𝑥𝑥 − 1 4 ii) 𝑥𝑥𝑥𝑥 2 − 𝑥𝑥𝑥𝑥𝑥𝑥

iii) 𝑥𝑥𝑥𝑥 2 – 𝑥𝑥𝑥𝑥 – 3 iv) 𝑥𝑥𝑥𝑥 2 + 𝑥𝑥𝑥𝑥 + 7 3. i) (𝑥𝑥𝑥𝑥 𝑥 2)(𝑥𝑥𝑥𝑥 𝑥 3)(2𝑥𝑥𝑥𝑥 𝑥 1) ii) (𝑥𝑥𝑥𝑥 𝑥 2)(2𝑥𝑥𝑥𝑥 𝑥 3)(𝑥𝑥𝑥𝑥 + 3) iii) (𝑥𝑥𝑥𝑥 + 1)(𝑥𝑥𝑥𝑥 𝑥 1)(2𝑥𝑥𝑥𝑥 𝑥 1)(𝑥𝑥𝑥𝑥 + 2)

IV. Problems based on algebraic identities 1. NA 2.

(

(3 x + y + z ) 9 x 2 + y 2 + z 2 − 3 xy − yz − 3 xz

3. i) (3𝑦𝑦𝑦𝑦 − 5𝑧𝑧𝑧𝑧)(9𝑦𝑦𝑦𝑦 2 − 15𝑦𝑦𝑦𝑦𝑧𝑧𝑧𝑧 + 25𝑧𝑧𝑧𝑧 2 ) ii) (4𝑚𝑚𝑚𝑚 − 7𝑛𝑛𝑛𝑛)(16𝑚𝑚𝑚𝑚2 + 28𝑚𝑚𝑚𝑚𝑛𝑛𝑛𝑛 + 49𝑛𝑛𝑛𝑛2 ) ii) (3 − 5𝑎𝑎𝑎𝑎)3 4. i) (2𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏)3 iii) (4𝑎𝑎𝑎𝑎 𝑎 3𝑏𝑏𝑏𝑏)3

1 3 6

iv) �3𝑝𝑝𝑝𝑝 − �

5. i) 970299 ii) 1061208 iii) 994011992 6. i) (𝑥𝑥𝑥𝑥 2 + 4𝑦𝑦𝑦𝑦 2 + 16𝑧𝑧𝑧𝑧 2 + 4𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦 + 16𝑦𝑦𝑦𝑦𝑧𝑧𝑧𝑧 + 8𝑥𝑥𝑥𝑥𝑧𝑧𝑧𝑧) ii) (4𝑥𝑥𝑥𝑥 2 + 𝑦𝑦𝑦𝑦 2 + 𝑧𝑧𝑧𝑧 2 − 4𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦 − 2𝑦𝑦𝑦𝑦𝑧𝑧𝑧𝑧 + 4𝑥𝑥𝑥𝑥𝑧𝑧𝑧𝑧) iii) (4𝑥𝑥𝑥𝑥 2 + 9𝑦𝑦𝑦𝑦 2 + 4𝑧𝑧𝑧𝑧 2 − 12𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦 + 2𝑦𝑦𝑦𝑦𝑧𝑧𝑧𝑧 – 8𝑥𝑥𝑥𝑥𝑧𝑧𝑧𝑧) 7. i) 11021 ii) 9120 iii) 9984 8. NA 9. 9 10. NA 11. NA 1 12. 𝑎𝑎𝑎𝑎8 − 8 𝑎𝑎𝑎𝑎

13. i) (𝑎𝑎𝑎𝑎2 + 1) (𝑎𝑎𝑎𝑎2 + 3) ii) (𝑥𝑥𝑥𝑥 4 − 11)(𝑥𝑥𝑥𝑥 4 + 12) 14. NA 15. NA 16. NA 17. −120𝑥𝑥𝑥𝑥 2 𝑦𝑦𝑦𝑦 𝑦 250𝑦𝑦𝑦𝑦 3 V. HCF and LCM of polynomials 1. i) H.C.F = 𝑎𝑎𝑎𝑎(𝑎𝑎𝑎𝑎 + 3) L.C.M. = 𝑎𝑎𝑎𝑎(𝑎𝑎𝑎𝑎 𝑎 1)(𝑎𝑎𝑎𝑎 + 3)(2𝑎𝑎𝑎𝑎 𝑎 1) (ii) H.C.F = 2𝑢𝑢𝑢𝑢 𝑢 3𝑣𝑣𝑣𝑣 L.C.M. = 𝑢𝑢𝑢𝑢(2𝑢𝑢𝑢𝑢 + 3𝑣𝑣𝑣𝑣)(2𝑢𝑢𝑢𝑢 𝑢 3𝑣𝑣𝑣𝑣) 262

)

(iii) H.C.F = 2𝑢𝑢𝑢𝑢 + 5𝑣𝑣𝑣𝑣 L.C.M. = 3𝑢𝑢𝑢𝑢(2𝑢𝑢𝑢𝑢 + 5𝑣𝑣𝑣𝑣)(2𝑢𝑢𝑢𝑢 𝑢 5𝑣𝑣𝑣𝑣) 2. 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝(𝑝𝑝𝑝𝑝 𝑝𝑝𝑝𝑝𝑝)(𝑝𝑝𝑝𝑝 − 𝑛𝑛𝑛𝑛)(𝑝𝑝𝑝𝑝 − 2𝑛𝑛𝑛𝑛) 3. 6𝑥𝑥𝑥𝑥 3 − 24𝑥𝑥𝑥𝑥 4. 5 5. (𝑥𝑥𝑥𝑥 + 2𝑦𝑦𝑦𝑦)(𝑥𝑥𝑥𝑥 𝑥𝑥𝑥𝑥𝑥) 6. 1372 VI. Quadratic equation and its solution 1. 7 years 2. 20 m, ₹10 3. 𝑥𝑥𝑥𝑥 = 1 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 – 2 5

4. k = ± 2 9

5. k= 4 6.

−8 8 < 𝑘𝑘𝑘𝑘 < 5 5

7. D = 0, real 8. NA 9. NA 10. 𝑡𝑡𝑡𝑡 = 14 min Worksheet 2 I. Multiple choice questions with single correct answer 1. c 2. c 3. d 4. c 5. a 6. d 7. d 8. b 9. a 10. c 11. a 12. d 13. b 14. b 15. c 16. b 17. a 18. c 19. a 20. c 21. c 22. a 23. c 24. b 25. b 26. a 27. c 28. c 29. d 30. b II. Fill in the blanks 1. 0 2. 3 3. 0 2

4. -1 and 3

5. 4 6. 6 7. 54 8. [degree of 𝑝𝑝𝑝𝑝(𝑥𝑥𝑥𝑥)] − 1 9. 4 10. 6 11. 8 12. 3 and 9 𝑎𝑎𝑎𝑎

𝑏𝑏𝑏𝑏

𝑐𝑐𝑐𝑐

13. 𝑎𝑎𝑎𝑎1 = 𝑏𝑏𝑏𝑏1 = 𝑐𝑐𝑐𝑐1 2

ANSWER KEY 14. 0,

−𝑏𝑏𝑏𝑏 𝑎𝑎𝑎𝑎

15. no 16. ±7

17.

1 3

18. 1 − 𝑜𝑜𝑜𝑜, 𝑜𝑜𝑜𝑜 + 1 19. -11

𝑏𝑏𝑏𝑏2 4𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏2 21. 𝑎𝑎𝑎𝑎

20.

22. 6 23. 2 24. 𝑥𝑥𝑥𝑥 2 − 6𝑥𝑥𝑥𝑥 𝑥 11 = 0 III. Subjective questions 1. 25 2. 9 3. 3 4. 1 −1

2. a) A: (7, 10), B: (-5, 13), C: (13, -5), D: (-4, -16) b)

5. 𝑘𝑘𝑘𝑘 = 3 (or) 𝑘𝑘𝑘𝑘 = 3

6. 5𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦(1 − 3𝑥𝑥𝑥𝑥 2 𝑦𝑦𝑦𝑦 2 ) 7. 998000 8. 1,2,3 and 4

1 [27𝑥𝑥𝑥𝑥 3 + 54𝑥𝑥𝑥𝑥 2 + 36𝑥𝑥𝑥𝑥 + 8] 8

10. Yes 1

11. 2 12. P = 3, But 𝑝𝑝𝑝𝑝 𝑝 0 because if p = 0, then the given equation is not a quadratic equation. 13. 𝑥𝑥𝑥𝑥 2 − 11𝑥𝑥𝑥𝑥 + 8 = 0 14. 𝑥𝑥𝑥𝑥 2 + 𝑥𝑥𝑥𝑥 𝑥 360 = 0 15. no 16. 𝑏𝑏𝑏𝑏 2 > 4𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 17. -8 18. -31 19. Real and distinct 20. 2

(0, 0), (4, 0), (0, −3), (4, −3) ,

Area = 12 units 4. The fourth vertex of the rectangle is (−3,3) 5. Coordinates of point S are (−4, −5) 6.

3. COORDINATE GEOMETRY Worksheet 1 I. Cartesian coordinate system

Point of intersection of diagonals = (4,1) 263

ANSWER KEY 7.

11.

Area = 9 sq units

12.

Coordinates of D (7,3), Area = 24 sq. units

10.

264

II. Distance formula ii) 4√2 1. i) 2√2 2 2 iii) 2√𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏 iv) 2√𝑏𝑏𝑏𝑏 2 + 𝑎𝑎𝑎𝑎2 v) (sin α + cos α) √𝑎𝑎𝑎𝑎2 + 𝑏𝑏𝑏𝑏 2 2. Points are not collinear 3. Points are vertices of isosceles triangle 4. (0,7) 5. (1.8, 0) 6. 𝑦𝑦𝑦𝑦 = 3 or 𝑦𝑦𝑦𝑦 = −9 7. 𝑥𝑥𝑥𝑥 = −3 or 𝑥𝑥𝑥𝑥 = 5 8. 3𝑥𝑥𝑥𝑥 + 𝑦𝑦𝑦𝑦 − 5 = 0 9. 𝑥𝑥𝑥𝑥 = 4, QR= √41, PR= √82 10. i) square ii) no quadrilateral iii) parallelogram iv) rectangle v) rectangle

ANSWER KEY III. Section formula 1. (1,3)

7 4 3 3 4 8 10 ii) � , − � or �−3, − � 3 3 3 5 8 8 iii) �2, − � or � , − � 3 3 3

2. i) �1, − � or �−3, � 3. 4. 5. 6. 7.

Ratio 4: 1, 𝑦𝑦𝑦𝑦 = 6 2: 7 𝑥𝑥𝑥𝑥 = 6, 𝑦𝑦𝑦𝑦 = 3 (3, -10) (4,5) (2,3) (6,9) 2 20 � 7 7 13 �1, 2 �

10. 24 11. 1:1 IV. Area of triangles and quadrilaterals 49 21 1. i) sq. units ii) sq. units 2

iv) 3 sq. units iii) a2 sq. units 2. i) 𝑘𝑘𝑘𝑘 = 6 ii) 𝑘𝑘𝑘𝑘 = 4 iii) 𝑘𝑘𝑘𝑘 = 3 3. 28 sq. units 4. 1:4 5. NA Worksheet 2 I. Multiple choice questions with single correct answer 1. a 2. b 3. c 4. a 5. c 6. d 7. c 8. b 9. b 10. a 11. c 12. a 13. d 14. b 15. c 16. d 17. d 18. c 19. b 20. d 21. d 22. b 23. b 24. c 25. d 26. a 27. d 28. d 29. a 30. a 31. b 32. c 33. b 34. a 35. d 36. a 37. c II. Multiple choice questions with multiple correct answers 1. a, b 2. a, d 3. b, c 4. a, b, d III. Assertion and reason 1. d 2. b IV. Fill in the blanks 1. (3, −2) 2. -1 9 2

3√3

9 2

5√2 units 2

5. 0 6. 2𝑥𝑥𝑥𝑥 = 37 7. (−2, 4) 8. Equilateral 9. (-10 or 6) 10. -12 11. 2 32 12. 3

8. �− , −

13. 𝑦𝑦𝑦𝑦1 ∶ 𝑦𝑦𝑦𝑦2 14. 4 or-2 15. (1, −73) and (−3,43) 16. 2:7 17. 2:3 18. 𝑥𝑥𝑥𝑥1 + 𝑥𝑥𝑥𝑥3 − 𝑥𝑥𝑥𝑥2 , 𝑦𝑦𝑦𝑦1 + 𝑦𝑦𝑦𝑦3 − 𝑦𝑦𝑦𝑦2 19. (16, 8) 20. 0 V. Subjective questions 1. 𝑦𝑦𝑦𝑦 = 3 or 𝑦𝑦𝑦𝑦 = −9 2. NA 3. NA 4. 4√5 5. NA 6. 𝑥𝑥𝑥𝑥 + 13𝑦𝑦𝑦𝑦– 17 = 0 7. 5 8. i) square ii) no quadrilateral iii) parallelogram 9 13 � 5 5

9. �− ,

10. 4:1 11. (2, −2) 12. −1 13. √85 14. (−6, −1)(2, 5)(4, −9) 17 15. 3

16. 44 units

17.

(𝑎𝑎𝑎𝑎−𝑏𝑏𝑏𝑏) (𝑏𝑏𝑏𝑏−𝑐𝑐𝑐𝑐)(𝑐𝑐𝑐𝑐−𝑎𝑎𝑎𝑎) 2𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏

18. (-8, -14) 8 3

1 3

19. � , 2� or � , 5� VI. Case studies 1. c 2. d 3. b

3√3

3. � + 2 � 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 � − 2 �

265

ANSWER KEY 4. LINEAR EQUATIONS IN TWO VARIABLES

Worksheet 1 I. Introduction to linear Equation 1. 1. 𝑥𝑥𝑥𝑥 + 0. 𝑦𝑦𝑦𝑦 = 7 2. 2𝑥𝑥𝑥𝑥 + 𝑦𝑦𝑦𝑦 = 160 and 4𝑥𝑥𝑥𝑥 + 2𝑦𝑦𝑦𝑦 = 300 8−2𝑥𝑥𝑥𝑥 , 𝑥𝑥𝑥𝑥 𝑥 0 3. 𝑎𝑎𝑎𝑎 =

𝑐𝑐𝑐𝑐 2 𝑎𝑎𝑎𝑎(𝑎𝑎𝑎𝑎𝑎𝑎)

𝑦𝑦𝑦𝑦 = 𝑏𝑏𝑏𝑏(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)

16. Required numbers = 34, 70 III. Graph of linear equation 1. i)

𝑥𝑥𝑥𝑥

4. 𝑦𝑦𝑦𝑦 = 3𝑥𝑥𝑥𝑥 5. NA 6. i) 30∘ C ii) 95∘ F

iii) 32∘ F and

−160∘ C 9

iv) −40° 9 7. 𝑦𝑦𝑦𝑦 = (𝑥𝑥𝑥𝑥 𝑥 273) + 32 5

8. 𝑎𝑎𝑎𝑎 = −

5 3

9. 𝑦𝑦𝑦𝑦 = 5𝑥𝑥𝑥𝑥 + 3 10. 𝑦𝑦𝑦𝑦 = 5𝑥𝑥𝑥𝑥 II. Solution of simultaneous linear equation 1. 𝑥𝑥𝑥𝑥 = 2, 𝑦𝑦𝑦𝑦 = 3 2. i) 𝑥𝑥𝑥𝑥 = 9, 𝑦𝑦𝑦𝑦 = 6 ii) 𝑥𝑥𝑥𝑥 = 1, 𝑦𝑦𝑦𝑦 = −1 iii) 𝑥𝑥𝑥𝑥 = 1, 𝑦𝑦𝑦𝑦 = −1 iv) 𝑥𝑥𝑥𝑥 =

261 9 , 𝑦𝑦𝑦𝑦 = 49 49

3. i) 𝑥𝑥𝑥𝑥 = −2, 𝑦𝑦𝑦𝑦 = −1 ii) 𝑥𝑥𝑥𝑥 = −1, 𝑦𝑦𝑦𝑦 = 1 1 2

iii) 𝑥𝑥𝑥𝑥 = − , 𝑦𝑦𝑦𝑦 = 3 iv) 𝑥𝑥𝑥𝑥 = 3, 𝑦𝑦𝑦𝑦 = 1

4. 𝑥𝑥𝑥𝑥

1 = , 𝑦𝑦𝑦𝑦 3

5. ₹1700 6.

12 25

1 = − 2

7. 63 8. Weight of Rita = 31 Pounds, Weight of Aman = 29 pounds 9. 62 10. 16 11. 𝑦𝑦𝑦𝑦 = 𝑏𝑏𝑏𝑏 12. 𝑥𝑥𝑥𝑥 = − 21, 𝑦𝑦𝑦𝑦 = 17 13. 53 14. 𝑥𝑥𝑥𝑥 = 3, 𝑦𝑦𝑦𝑦 = 2

𝑐𝑐𝑐𝑐 2 (1−𝑏𝑏𝑏𝑏) 15. 𝑥𝑥𝑥𝑥 = 𝑎𝑎𝑎𝑎(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)

266

ii)

iii)

2. NA 3. 7𝑥𝑥𝑥𝑥 𝑥𝑥𝑥𝑥𝑥 = 0 and 𝑥𝑥𝑥𝑥 𝑥𝑥𝑥𝑥𝑥 + 12 = 0, Infinite lines

ANSWER KEY 4.

6. NA 7. NA 8.

10.

i) 30 N

ii) 36 N

11.

IV. Consistency/inconsistency of linear equation 3 1. 𝑘𝑘𝑘𝑘 = − 2 267

ANSWER KEY 2. 3. 4. 5.

NA NA NA i) λ = – 1 ii) λ = 1 iii) all real value of λ except +1. 6. 𝑘𝑘𝑘𝑘 = − 6 7. 𝑎𝑎𝑎𝑎 = 3, 𝑏𝑏𝑏𝑏 = −1 8. NA 9. NA 10. NA 11. NA 12. No solution 13. Consistent 14. NA Worksheet 2 I. Multiple choice questions with single correct answer 1. a 2. c 3. b 4. a 5. a 6. b 7. b 8. b 9. a 10. c 11. b 12. c 13. a 14. a 15. b 16. b 17. d 18. b 19. b 20. b 21. b 22. b 23. c 24. c 25. b 26. a 27. b 28. b 29. a 30. c 31. d 22. d 33. a 34. b 35. c 36. a 37. d 38. d 39. a 40. b 41. c 42. c 43. c II. Fill in the blanks 1. 𝑦𝑦𝑦𝑦 = 5 2. 𝑦𝑦𝑦𝑦 = 2 3. Infinite 4. 1 5. 𝑌𝑌𝑌𝑌-axis 6. (4.5, 0) 7. Straight line 8. 𝑥𝑥𝑥𝑥 + 𝑦𝑦𝑦𝑦 = 0 9. Infinite 10. 𝑥𝑥𝑥𝑥 − 𝑦𝑦𝑦𝑦 − 5 = 0 𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏 𝑐𝑐𝑐𝑐 11. 𝑎𝑎𝑎𝑎1 = 𝑏𝑏𝑏𝑏1 = 𝑐𝑐𝑐𝑐1 2

12. Co-incident lines 13. consistent 14. X-axis 15. straight line 16. infinite 17. inconsistent 18. infinite 19. parallel

268

20. (2010, 2020) 21. consistent 22. infinite 23. Y - axis 24. parallel 25. parallel 26. no solution III. Subjective questions 1. 3𝑥𝑥𝑥𝑥 + 0, 𝑦𝑦𝑦𝑦 = 7 2. 𝑦𝑦𝑦𝑦 = 3𝑥𝑥𝑥𝑥 − 5 3. Yes (1, −3) is a solution of 2𝑥𝑥𝑥𝑥 − 5𝑦𝑦𝑦𝑦 = 17 4. 𝑚𝑚𝑚𝑚 = 3 5. 𝑥𝑥𝑥𝑥 − 4𝑦𝑦𝑦𝑦 = 0, where 𝑥𝑥𝑥𝑥 represents the cost price of pen and 𝑦𝑦𝑦𝑦 represents the cost of price of pencil. 6. 𝑝𝑝𝑝𝑝 = 0 7. 𝑎𝑎𝑎𝑎 = 2, 𝑏𝑏𝑏𝑏 = 0, 𝑎𝑎𝑎𝑎 = −5 8. 2𝑥𝑥𝑥𝑥 − 𝑦𝑦𝑦𝑦 = 0, where 𝑥𝑥𝑥𝑥 represents the cost of a pen 𝑦𝑦𝑦𝑦 represents the cost of notebook. −3 9. 𝑦𝑦𝑦𝑦 = 2 𝑥𝑥𝑥𝑥 + 4, yes, the point (4, −2) lies on the line. 3 10. � , 0� and (0, −3) 2 11. False. Given equations forms parallel lines 12. No. Two straight roads do not cross each other. 13. Yes. Because given equations form coincident lines. 14. No. Because LHS ≠ RHS 15. 2 𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏 16. False. Because 1 = 1 17. No 18. 𝑎𝑎𝑎𝑎 = 3, 𝑏𝑏𝑏𝑏 = 1 19. 𝑘𝑘𝑘𝑘 = −1 20. (0, −3)

𝑎𝑎𝑎𝑎2

𝑏𝑏𝑏𝑏2

5. EUCLID’S GEOMETRY

Worksheet 1 I. Introduction, Euclid’s definitions, axioms and postulates 1. NA 2. NA 3. NA 4. ∠1 = ∠2 5. NA 6. NA 7. NA

ANSWER KEY 8. i) False ii) True iii) True iv) True v) True vi) True 9. NA 10. NA Worksheet 2 I. Multiple choice questions with single correct answer 1. a 2. a 3. b 4. b 5. c 6. a 7. a 8. a 9. a 10. d 11. a 12. b 13. c 14. a 15. b II. Fill in the blanks 1. equal 2. whole 3. three 4. zero 5. five III. Subjective questions 1. i) NA ii) NA iii) NA iv) NA v) NA 2. NA 3. NA 4. NA 5. i) NA ii) NA

6. LINES AND ANGLES

Worksheet 1 I. Basic terms and definitions, complementary angles, supplementary angles and explementary angles 1. i) 70° ii) 55° iii) 0° iv) 13° v) 60° 2. i) 126° ii) 48° iii) 42° 3. 31° 4. 50° 5. 80° and 100° 6. 66° and 114° 7. 80° 8. 35° 9. 50° 10. The required angles is 108° and its supplement is 72° 11. The required angle is 52° 12. 120°

13. .45° 14. 45° 15. i) 213° ii) 101° iii) 30° II. Adjacent angles and linear pair of angles 1. i) 𝑦𝑦𝑦𝑦 = 50° ii) 𝑥𝑥𝑥𝑥 = 35° 2. Adjacent angles: ∠AOC and ∠COB; ∠AOD and ∠BOD ∠AOD and ∠COD; ∠BOC and ∠COD Linear Pair: ∠AOD and ∠BOD; ∠AOC and ∠BOC 3. ∠AOD = 60°, ∠COD = 50° and ∠BOC = 70° 4. NA 5. 𝑎𝑎𝑎𝑎 = 130° and 𝑏𝑏𝑏𝑏 = 50° 6. 4 7. 10 8. 30° 9. 55° 10. 20° III. Vertically opposite angles 1. 𝑥𝑥𝑥𝑥 = 135°, 𝑢𝑢𝑢𝑢 = 135° and 𝑧𝑧𝑧𝑧 = 45° 2. 𝑥𝑥𝑥𝑥 = 40°, 𝑦𝑦𝑦𝑦 = 50°, 𝑧𝑧𝑧𝑧 = 90° and 𝑢𝑢𝑢𝑢 = 40° 3. 𝑦𝑦𝑦𝑦 = 25° and 𝑧𝑧𝑧𝑧 = 155° 4. 𝑥𝑥𝑥𝑥 = 18° 5. NA 6. i) 𝑦𝑦𝑦𝑦 = 160° ii) 𝑥𝑥𝑥𝑥 = 70° 7. ∠AOC = 35°, ∠BOF = 40o, ∠DOE = 105° and ∠COF = 105° 8. ∠BOC = 110°, ∠AOD = 110° 9. ∠BOE = 30° and reflex of ∠COE = 250° 10. i) obtuse ii) 180° iii) uncommon IV. Parallel lines and transversals 1. ∠1=108° ∠2=72° ∠3=108° ∠4=72° ∠5=108° ∠6=72° ∠7=108° ∠8=72° 2. ∠1=60°, ∠2=120° and ∠3=60° 3. ∠ACE = 20° 4. NA 5. ∠2 = 95° 6. NA 7. ∠A=72° ∠B=108° ∠C=72° and ∠D=108° 8. Parallel lines 9. NA 10. ∠2 = 120° 269

ANSWER KEY V. Angle sum property of triangles 1. ∠C = 85° 2. 30°, 60° and 90° 3. 𝑥𝑥𝑥𝑥 = 100° 4. 50°, 50° and 80° 5. NA 6. 50°, 60° and 70° 7. 126° 8. NA 9. NA 10. NA VI. Exterior angle property 1. ∠A = 60°, ∠B =44° and ∠C =76° 2. ∠BAC = 45° ∠ACD = 75° and ∠ABC = 60° 3. ∠ECD = 60° 4. ∠ACD = 70° 5. i) 180° ii) interior iii) greater iv) one v) one 6. NA 7. 𝑥𝑥𝑥𝑥 = 90° 8. NA 9. 16° 10. NA Worksheet 2 I. Multiple choice questions with single correct answer 1. a 2. c 3. c 4. a 5. a 6. c 7. a 8. a 9. d 10. c 11. b 12. a 13. b 14. c 15. b 16. a 17. d 18. a 19. c 20. b 21. c 22. c 23. a 24. d 25. a 26. d 27. c 28. b II. Assertion and reason 1. d 2. a 3. c III. Fill in the blanks 1. Collinear points, non - collinear points 2. 180∘ 3. 107∘ 4. Perpendicular 5. 86∘ 6. 117∘ 7. 36∘ 8. 108∘ 9. 4 10. 120∘ IV. Subjective questions 270

1. 180∘ − 2𝑦𝑦𝑦𝑦 ∘ 2. 60∘ 3. Alternate angles : (∠1, ∠8), (∠4, ∠6), (∠2, ∠7), (∠3, ∠5) Corresponding angles (∠1, ∠5), (∠2, ∠6), (∠4, ∠7), (∠3, ∠8) 4. 180∘ − 𝑤𝑤𝑤𝑤 + 𝑧𝑧𝑧𝑧 5. 135∘ 6. 35∘ 7. 18° 8. 108° 9. 360 − (𝑥𝑥𝑥𝑥 ∘ + 𝑦𝑦𝑦𝑦 ∘ ) 10. 30∘ 11. 30∘ and 250∘ 12. 5 13. 122° and 302° 14. 45°, 54° and 81° 15. 40° 16. ∠𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 and ∠𝑄𝑄𝑄𝑄𝑃𝑃𝑃𝑃𝑄𝑄𝑄𝑄 = 58° ∠𝑄𝑄𝑄𝑄𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 and ∠𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑄𝑄𝑄𝑄 = 122° 17. 10 18. 50° and 77° 19. NA 20. 65° and 53°

7. TRIANGLES

Worksheet 1 I. Congruence of triangles 1. NA 2. NA 3. NA 4. NA 5. NA 6. NA 7. NA 8. NA 9. NA 10. NA 11. NA 12. NA 13. NA II. Geometric points of a triangle 1. 3: 1 2. 12 cm 3. 46 4. NA 5. NA 6. NA 7. NA

ANSWER KEY III. Similarity of triangles 1. 4 2. 9 3. NA 4. NA 5. NA 6. NA 7. NA 8. NA 9. NA 10. NA 11. NA 12. NA 13. NA Worksheet 2 I. Multiple choice questions with single correct answer 1. c 2. b 3. b 4. b 5. a 6. c 7. b 8. a 9. a 10. b 11. b 12. c 13. c 14. a 15. b 16. b 17. b 18. a 19. c 20. b 21. b 22. a 23. b 24. b 25. c 26. b 27. c 28. b 29. c 30. b 31. a 32. d 33. b 34. c 35. c 36. d 37. b 38. b 39. c 40. a 41. b II. Fill in the blanks 1. less than 2. Isosceles triangle 3. 120∘ 4. Congruent triangles 5. PQ 6. Right angled triangle 1 7. (𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 + 𝐴𝐴𝐴𝐴𝐵𝐵𝐵𝐵 + 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵) 2 8. 55∘ 9. 50∘ 10. AB 11. Proportionally 12. parallel 13. square 14. similar 15. parallelogram 16. scale factor 17. EC 18. 4.5 cm 19. 5 cm III. Subjective questions

1. 60∘ , 60∘ , 60∘ 2. ∠B = 45∘ , ∠C = 45∘ 3. 30∘ , 30∘ , 120∘ 4. BC > CD 5. 104∘ 6. PA = PB 7. 60∘ 8. 𝐴𝐴𝐴𝐴𝑄𝑄𝑄𝑄 = 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 9. 2 units 10. By BPT 11. 𝑥𝑥𝑥𝑥 = 2 cm 12. AQ = 3 cm 13. 16 cm 14. 𝑥𝑥𝑥𝑥 = 3 cm 15. 110∘ 16. 13 m 17. 6 cm 18. 15 cm 19. 150o 20. 37o 21. 𝑥𝑥𝑥𝑥 = 20° 22. 15° 23. i) ∠𝐴𝐴𝐴𝐴 + ∠𝐴𝐴𝐴𝐴 + ∠𝐵𝐵𝐵𝐵 + ∠𝐷𝐷𝐷𝐷 + ∠𝐸𝐸𝐸𝐸 = 180∘ ii) ∠𝐴𝐴𝐴𝐴 + ∠𝐴𝐴𝐴𝐴 + ∠𝐵𝐵𝐵𝐵 + ∠𝐷𝐷𝐷𝐷 + ∠𝐸𝐸𝐸𝐸 + ∠𝐹𝐹𝐹𝐹 + ∠𝐺𝐺𝐺𝐺 = 540∘ 24. NA 25. NA 26. NA 27. NA 28. NA 29. NA 30. NA 31. NA 32. NA 33. NA

8. QUADRILATERALS

Worksheet 1 I. Introduction to quadrilaterals 1. 4cm 2. NA 3. 80∘ 4. 90∘ 5. No 6. 60∘ , 150∘ , 120∘, and 30∘ 7. 36∘ , 60∘ , 108∘ , 156∘ 271

ANSWER KEY 8. 105∘ II. Properties and theorems related to quadrilaterals 1. NA 2. 𝑎𝑎𝑎𝑎 = 10 3. ∠C = 110∘ and ∠D = 125∘ 4. 𝑌𝑌𝑌𝑌 = 30∘ , ∠A = ∠R = 120∘ and ∠P = ∠K = 60∘ 5. ∠C = 135∘ and ∠D = 135∘ 6. 50∘ 7. NA 8. NA 9. NA 10. 120∘ , 60∘ , 120∘ , 60∘ , 11. NA 12. NA 13. NA 14. 55∘ 15. 60∘ , 120∘ , 60∘ , 120∘ 16. NA 17. NA 18. NA 19. NA 20. NA III. Mid point theorem 1. NA 2. 28 cm 3. 11 cm 4. NA 5. NA 6. NA 7. NA 8. NA 9. NA 10. NA 11. NA 12. NA ii) 20 cm2 13. i) 24cm2 2 iii) 22 cm 14. NA Worksheet 2 I. Multiple choice questions with single correct answer 272

1. c 2. c 3. a 4. d 5. c 6. d 7. d 8. c 9. a 10. a 11. a 12. b 13. c 14. c 15. b 16. c 17. a 18. c 19. c 20. b 21. c 22. a 23. c 24. c 25. a 26. a 27. c 28. c 29. a 30. b 31. b II. Multiple choice questions with multiple correct answer 1. a, c, d 2. b, c 3. a, c 4. b, c 5. a, d 6. a, b, c, d 7. a, b III. Fill in the blanks 1. 10 cm 2. Rectangle 3. 5 cm 4. 9.5 cm 5. 0∘ 6. 100∘ 7. 100∘ 8. isosceles trapezium 9. median 10. 50∘ 11. 50∘ 12. diagonals of PQRS are perpendicular. 13. Trapezium 14. quadrilateral whose opposite angles are supplementary. 15. Parallelogram IV. Subjective questions 1. 32∘ 2. 70∘ , 110∘ , 70∘ , 110∘ 3. PQ = 12 cm, QR = 8 cm, RS = 12 cm, SP = 8 cm 4. 72∘ , 108∘ , 72∘ , 108∘ 5. In a parallelogram, if each angle is equal to 90∘ , then it is a rectangle. 6. parallelogram, rectangle, square, rhombus 7. square 8. 3.8 cm 9. 45∘ 10. 52∘ , 128∘ , 52∘ , 128∘

ANSWER KEY 11. 60∘ 12. NA 13. ∠𝐴𝐴𝐴𝐴 = ∠𝐵𝐵𝐵𝐵 = 60∘ and ∠𝐴𝐴𝐴𝐴 = ∠𝐷𝐷𝐷𝐷 = 120∘ 14. NA 15. NA 16. 80∘ 17. 1: 2 18. 22

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