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1 Foundations of Geometry Big ideas • Different representations of data highlight different characteristics of the data. • Geometry uses standard vocabulary and symbols to communicate facts and relationships about geometric figures. • All constructions are based on the properties of geometric figures.

Chapter outline 1.01 1.02 1.03 1.04 1.05

Introduction to the data cycle and Venn diagrams Venn diagrams and sets Introduction to geometric notation Line segments and constructions Angles and constructions

4 18 31 37 49

1.01 Introduction to the data cycle and Venn diagrams After this lesson, you will be able to... • write statistical questions and use them to collect binary bivariate data. • collect binary bivariate data using observations, measurements, surveys, and experiments. • create Venn diagrams from a two-way table, raw survey data, or secondary sources. • analyze data represented in a Venn diagram.

Venn diagrams A Venn diagram is a tool that is used to compare and contrast properties of two things or groups. They allow us to visualize the overlaps and differences between these two categories. For example, in the following Venn diagram, we can see that: Fish

Both ﬁsh and dolphins: • Have ﬁns • Have teeth • Swim

Dolphins

Scales Gills Lay eggs

Swim Have teeth

Breathe air

Have fins

Give live birth Two legs

Dolphins, but not ﬁsh: • Breathe air • Give birth to live young Neither dolphins, nor ﬁsh have two legs.

A Venn diagram will usually have two or three categories. Where the circles overlap, the characteristics there are shared, in the rectangle, but outside all the circles are the characteristics that neither have. A A

Venn diagram with two categories

Venn diagrams with three categories

Instead of listing all of the characteristics or members that belong in each region, we can just list the number of items in each region. Likes running Agostino Darla Foster Isabella Jerome

Plays an instrument

Hajime

Mei

Chauncey

Gareth

Esme

Linda Konrad

Venn diagram that sorts students by who like running and who play an instrument 4

Likes running

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A Venn diagram that shows the number of students that ﬁt in each of the four regions

Exploration With a partner or in a group of three, draw a Venn diagram with two or three circles and label each circle with one of your names. 1.

Sort each of the characteristics into the correct region of your Venn diagram: • Is taking Geometry class • Ate breakfast • Is a vegetarian • Likes walking • Plays video games • Traveled out of state last summer

Come up with six more characteristics to sort and ensure that there is at least one characteristic in each region of the Venn diagram.

Rewrite the Venn diagram with the number of characteristics that are in each region instead of listing them.

Example 1 A marine biologist creates this diagram of her favorite ﬁsh: Saltwater fish

Freshwater fish

Goldfish Largemouth Bass Betta Fish

Blue Tang Barramundi Salmon

Clownfish Seahorse

a How many of the ﬁsh can live in fresh water?

Create a strategy This will include any of the ﬁsh in the freshwater ﬁsh circle.

Apply the idea

Reﬂect and check

There are 5 fish that can live in fresh water.

Some of these ﬁsh can also live in salt water and some of them cannot.

b How many of the ﬁsh are both freshwater and saltwater?

Create a strategy These ﬁsh will be in the overlap of the two circles.

Apply the idea There are 2 ﬁsh that can live in fresh and salt water: Salmon and Barramundi.

1.01 Introduction to the data cycle and Venn diagrams mathspace.co

c How many of the ﬁsh are freshwater or saltwater, but not both?

Create a strategy There are no ﬁsh outside of the two circles, so this will be the total number of ﬁsh, minus those that are both.

Apply the idea

Reﬂect and check

There are 6 fish that are either freshwater or saltwater, but not both: Goldfish, Betta Fish, Largemouth Bass Clownfish, Blue Tang, and Seahorse.

This Venn diagram sorted 8 types of ﬁsh, but there are many more types of ﬁsh that could be sorted using this Venn diagram.

Example 2 An ecologist listed some features of rivers and lakes. Organize the given characteristics in a Venn diagram. River Flows across a slope Often supports navigation Can be used for recreation and human activity Can have rapids or waterfalls Support ecosystems with aquatic life Has a deﬁned riverbed Subject to pollution and environmental concerns

Lake Usually situated in a basin Can regulate local climate Can be used for recreation and human activity Often a habitat for diverse species Support ecosystems with aquatic life Contains still or standing water Subject to pollution and environmental concerns

Create a strategy The characteristics should be organized in a Venn diagram with one circle representing rivers and the other representing lakes. The overlapping section should list the shared characteristics, while the non-overlapping sections should list the unique characteristics.

Apply the idea We can go through the list of characteristics for rivers ﬁrst and check whether each one should go in the overlap or not. For ones that are in the overlap, we should cross them off from the lakes list so we don’t double up.

Lakes

Rivers

Flows across a gradient Often supports navigation Can have rapids or waterfalls Has a defined riverbed

Support Usually situated ecosystems in a basin with aquatic life

Subject to pollution and environmental concerns Can be used for recreation and human activity

Can regulate local climate Often a habitat for diverse species

Contains still or standing water

Reﬂect and check This Venn diagram could be helpful for someone trying to decide whether to vacation near a lake or river if they are looking to do calm canoeing and look for lots of wildlife.

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The data could be collected and organized in a table like this, where an X means yes and a blank means no. Harry Potter X X

Amir Bilal Chloe Denel Emil

Lord of the Rings X X

Magic Tree House

X This Venn diagram can then be used to organize the data about the book series students have read. Has read Lord of the Rings

Has read Harry Potter

Has read Magic Tree House

Notice that: • A student can ﬁt into none, one, two, or all three categories. • Previously, we used single-selection surveys, but now we need to allow multiple selections. • If we wanted to have four or more possible responses, we couldn’t draw a Venn diagram that allows for all possibilities, but they can be used in some cases.

In order to make further conclusions or in a second iteration of the data cycle, it can be helpful to organize data from Venn diagram into a table or vice versa. We can convert between a table and a Venn diagram with two circles by matching up their parts. Left-handed

This Venn diagram shows data for students who were asked:

Entered

1. Are you left-handed? 2. Did you enter the math contest? 2

15 Left

Right

Entered

Didn’t enter

By converting the Venn diagram into a table, we can more clearly see what each of the four regions represents.

Once the display is created, we can analyze the Venn diagram or table to make conclusions about the proportion of the sample in the different regions or formulate further questions involving probabilities. For example, once our data is organized, we could ask questions like: “If a student is right-handed, when is the probability they entered the math contest?” Probability The likelihood of an event occurring Probability of an event =

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Instead of comparing the proportions in the different regions, we can also compare and contrast the properties or characteristics using Venn diagrams with questions like: • Compare and contrast the properties of the following parallelograms. • Rectangle • Square • Parallelogram • Are squirrels and chipmunks or squirrels and rabbits more similar? • Compare and contrast cultural practices in the US and Canada.

Example 3 Owns a laptop Owns a smartphone

Consider this Venn diagram:

a Write two statistical questions that could be summarized with the given Venn diagram.

Create a strategy There are two categories, “Owns a laptop” and “Owns a smartphone”. The formulated questions should explore the relationship between these two categories.

Apply the idea Two possible statistical questions are: 1. Are laptops or smartphones more popular among 10th graders? 2. What proportion of students don’t have access to technology at home?

Reﬂect and check Since the data was already collected, we could draw some conclusions to help answers these questions.

b Write a statistical question that involves probability that could be summarized with the given Venn diagram.

Create a strategy A probability question would involve comparing a speciﬁc subgroup to the whole population.

Apply the idea A possible statistical question could be: “What is the probability or likelihood that a student owns both a smartphone and a laptop?”

Reﬂect and check This is different from the survey questions we would ask the sample like: “Do you own a laptop?” and “Do you own a smartphone?”

1.01 Introduction to the data cycle and Venn diagrams mathspace.co

Example 4 Alaia wants to know more about how ice hockey and ﬁeld hockey are played. a Formulate a statistical question that could be used to start the data cycle for Alaia and would require a Venn diagram to analyze.

Create a strategy In this case, we will be using a Venn diagram to compare and contrast features, not to collect data from a population.

Apply the idea

Reﬂect and check

“Compare and contrast the features and rules of ice hockey and ﬁeld hockey.”

If Alaia wanted to know more about the popularity of ice hockey compared the ﬁeld hockey, she could have asked a question like “In Virigina, what proportion of the population has been to ice hockey games compared to ﬁeld hockey games?”

b Collect and organize data for Alaia using a Venn diagram.

Create a strategy Alaia can focus on one particular aspect of the sports such as equipment, rules, safety precautions, or league setup. She could then do another iteration of the data cycle to look at another aspect.

Apply the idea She can start by looking at the equipment for each sport. Here is a list she could have collected. Ice Hockey equipment Helmet with face cage Ice skates Gloves Puck Goalie pads Mouthguard Shin pads Hockey stick with a curved blade Elbow pads

Field Hockey equipment Throat protector Cleats Gloves Ball Leg guards Mouthguard Shin pads Field hockey stick (straight stick) Elbow pads

She can then organize it in a Venn diagram. Ice hockey equipment Helmet with face cage Ice skates Puck Goalie pads

Field hockey equipment

Gloves

Mouthguard Elbow pads

Throat protector Cleats Ball Leg guard

Hockey stick Shin pads Field hockey with curved stick blade

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Reflect and check From this she might note that for both sports the main equipment is safety pads and lead her to do another iteration of the data cycle looking at rates of injury or types of injuries between the two sports.

Example 5 A group of tourists in Japan were asked whether they spoke Filipino or Spanish. Their results were recorded in this table: Abby Carlo Thomas Dean Kevin Pam Jenny Rose Hiro Keia Aurora

Can speak Filipino Yes No Yes No No Yes Yes Yes Yes Yes Yes

Can speak Spanish No No No Yes Yes Yes Yes No No Yes No

a Complete this table to summarize the survey results. Spanish

Not Spanish

Filipino Not Filipino

Create a strategy We can go through the data one item at a time and make a tally of which of the four regions each person would ﬁt using a tally. We could also go through and count how many people have Yes and Yes, then how many have Yes and No, then No and Yes, then No and No.

Apply the idea

Reﬂect and check

We can start with a tally:

Review the tally counts to ensure that each response has been accurately recorded in the table. Double-check the numbers for those who speak both languages to ensure they are not double- counted in the single-language categories.

Spanish Filipino

∣∣∣

Not Filipino

∣∣

Not Spanish ∣

We can then convert to numbers: Filipino Not Filipino

Spanish 3 2

Not Spanish 5 1

If there was a large amount of data, we could use technology and write spreadsheet formulas to do the calculations.

1.01 Introduction to the data cycle and Venn diagrams mathspace.co

b Create a Venn diagram to summarize the data.

Create a strategy The four regions in the table correspond to the four regions in the Venn diagram.

Apply the idea We can match from the table the number of people who speak both languages (3), only Filipino (5), only Spanish (2), and neither language (1). Can speak Filipino

Can speak Spanish

Reﬂect and check We can double check we didn’t miss anyone by adding up all the numbers in the Venn diagram and counting the total number of people who answered the survey. Both should be 11. c If a random tourist was selected from the sample, what is the probability that they speak neither language?

Create a strategy Determine the total number of tourists surveyed to establish the denominator for the probability calculation. Use the formula for probability: Probability of an event =

Apply the idea There is one tourist (Carlo) who speaks neither Filipino nor Spanish out of a total of 10 tourists surveyed. P (Speaks Neither) =

Reﬂect and check This was a very small sample, so may not be representative of the population. We also cannot extend any conclusions to tourists in general as these were all tourists in Japan.

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d If a random tourist who spoke Filipino was selected from the sample, what is the probability that they do not speak Spanish?

Create a strategy Use the formula for probability: Probability of an event = where the favorable outcomes are tourists who speak only Filipino, and the total outcomes are all Filipino-speaking tourists.

Apply the idea Given the survey data: • Tourists who can speak Filipino: 8 in total (Abby, Thomas, Pam, Jenny, Rose, Hiro, Keia, and Aurora) • Tourists who can speak Filipino, but not Spanish: 5 in total (Abby, Thomas, Rose, Hiro, and Aurora). Therefore, tourists who speak only Filipino (not both languages) are 5 out of the 8 who can speak Filipino. P (Speaks Filipino, but not Spanish) =

Reﬂect and check This means that

speak both Filipino and Spanish.

Idea summary Venn diagrams are helpful to organize data when we ask questions that involve multiple-selection surveys or comparisons. The data cycle requires us to: 1. Formulate a statistical question about a speciﬁc population 2. Collect data using a sample survey, an experiment, or secondary data 3. Organize the data into displays like Venn diagrams and tables 4. Analyze data to draw conclusions to answer the original question 5. Possibly repeat the cycle for a new question that came up during the process

1.01 Introduction to the data cycle and Venn diagrams mathspace.co

Practice What do you remember? 1

Enoch wants to explore the statistical question: “How do the proportions of dog and snake owners compare for students at my school?” a

Dog Owners

Snake Owners

Determine whether or not the responses to each survey question could be used to create the given Venn diagram. i

What kind of pets do you have?

Which type of pets do you have?

Select all that apply. A

Dog

Snake

C None of the above iii Firstly, do you own a dog? Secondly, do you not own a dog? iv Firstly, do you own a dog? Secondly, do you own a snake? b

Complete the sentence using the words ‘Dog’ or ‘Snake’: i ii

The two categories in the diagram are ⬚ owners and ⬚ owners. 2 is the number of people who have a ⬚ AND a ⬚.

A student creates this diagram of their favorite animals. Is each statement true or false? a

Zebras have four legs and stripes.

Four of their favorite animals have four legs.

Three of their favorite animals have stripes.

A clown ﬁsh has stripes, but does not have four legs.

A ring-tailed lemur would have the same classiﬁcation as a zebra.

A lion would have the same classiﬁcation as a tiger.

A gardener creates this diagram of her favorite plants: a

How many of the plants are ﬂowering plants?

How many of the plants are both ﬂowering and succulent?

How many of the plants are either ﬂowering or succulent, but not both?

Four Legs

Stripes

Turtle Hippo

Zebra

Lion

Tiger

Clown Fish

Crow

Kangaroo

Elephant

Flowering plants Succulent plants Orchid Sunflower Christmas cactus Daisy Tulip

Crown of Thoms

Aloe Vera

Bamboo Ivy

SOL

This Venn diagram represents the languages spoken at an international conference.

Languages

Shade the region in the Venn diagram for participants who speak only English and French.

English

Japanese

French

Mandarin

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The given Venn diagram shows the number of students in a school playing rugby league, rugby union, both or neither: Rugby League

Find the number of students who:

Play rugby league only.

Play rugby league.

Play rugby union.

Play rugby union only.

Do not play rugby league.

Do not play rugby union.

169

Rugby Union 138 124

Carl collects data from the local pet shelter and makes a Venn diagram. He is curious about what would happen if he selected a pet at random. a

Find the probability that his new pet will be a cat that isn’t ﬂuffy.

Find the probability that his new pet won’t be ﬂuffy.

Cats

Fluffy

10 5

The table and Venn diagram show the same information: Piano Not Piano Total

Guitar 20 90 110

Not Guitar 40 50 90

Piano

Total 60 140 200

Guitar

D A

Find the value of: a

A+B+C+D

B+C

B+D

Let’s practice 8

Write two statistical questions that could be summarized with the given Venn diagram. Owns a pet

Owns a car

For each given survey question(s), can the results be summarized in a Venn diagram? If so, sketch the Venn diagram. If not, explain why not. a

What is your favorite sport?

Which other languages can you speak? Select all that apply. A Spanish

French

Dutch

Dog

Other. Fill in the box: ⬚

None of the above

Which pets do you have? A Cat

1. Have you taken public transportation this year? 2. Do you have your driver’s license?

1.01 Introduction to the data cycle and Venn diagrams mathspace.co

The investigative question “Compare and contrast the characteristics of two state animals in Virginia.” The following information was collected using a secondary source. Two of Virginia’s state animals are the Red Salamander and Eastern Garter Snake. Organize the given characteristics in a Venn diagram. Red Salamander • Four legs • Carnivorous diet • Live in moist, forested habitats • Cold-blooded • No scales • Amphibian • Brightly colored red or orange • State animal in Virginia

Eastern Garter Snake • No legs • Carnivorous diet • Live in moist, forested habitats • Cold-blooded • Has scales • Reptile • Greenish, brown, or black • State animal in Virginia

A survey was conducted and students were asked if they have been out of state before and if they have a passport. The results are recorded in the table. a

Complete the table to summarize the survey results. Has a passport

Does not have a passport

Has been out of state Has not been out of state b

Convert the table to a Venn diagram.

Anaya Ben Connie Dai Enzo Farida Goldie Han Isabella Joy Khalil Lizzie Masumi

Has been out of state Yes No Yes Yes No Yes Yes Yes Yes Yes Yes Yes Yes

A librarian checks on the status of 30 books one day. The books are categorized based on whether they are ﬁction and whether they have been borrowed. a

Using the Venn diagram, complete the table: Borrowed

Available

Total

Fiction Non-ﬁction Total

Has a passport No No Yes No Yes Yes No Yes Yes No No Yes No

What percentage of books are ﬁction and available?

What percentage of books are ﬁction and have been borrowed?

State two more observations or conclusions you can make from the Venn diagram.

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Fiction

Borrowed

6 ?

Bree formulated the statistical question “Are there any patterns or trends that can be seen for hobby preferences between building models, crafting, and baking?” She collected data using this table. a

Make a Venn diagram to summarize the collected data.

How many people like building models?

What proportion of people like baking?

Which hobby was the most popular?

Which two hobbies were most commonly preferred together?

What other conclusions or observations can she make?

Student Naomi Oliver Prince Quienie Reagan Shiela Trina Ulrich Valerie Warren Xyza Yvette Zoey

Likes building models X X X X

Likes crafting X

Likes baking X X X

X X

X X X

X X X X

X X

Let’s extend our thinking 14

Bradyn is interested in learning more about the relationship between environmental allergies and hobbies. He wants to use the data cycle to investigate. a

Formulate a question which could be investigated using a Venn diagram.

Describe a method which could be used to collect data.

Collect data for your formulated question.

Organize your data using a Venn diagram.

How can we clearly communicate the results? Draw conclusions to answer your formulated question.

This Venn diagram shows the data for a sample of students. Are the results similar or different to the results from your sample? Explain.

Allergies

Indoor

3 10

Based on your data analysis, formulate another question that arose. This question does not need to be represented with a Venn diagram.

Lois wants to know more about the structure of names. She notices that some students have hyphenated last names and some do not, some have middle names while others do not, and some students wish they could change their name while others are happy with their names. Go through the whole data cycle at least once using a context that is related to names and where the data can be organized into a Venn diagram.

1.01 Introduction to the data cycle and Venn diagrams mathspace.co

1.02 Venn diagrams and sets After this lesson, you will be able to... • use Venn diagrams to represent set relationships, including union, intersection, subset, and negation. • interpret Venn diagrams, including those representing contextual situations.

Venn diagrams and sets A collection of items is called a set. There are many situations which we can describe and explore mathematically using sets, and to do so, we will need some new terminology. Set

Proper subset

A collection of items, which are usually called elements. Sets are usually denoted using capital letters.

A subset B of A, where there is at least one element in A that is not in B. Written as B ⊂ A.

Example: A = {1, 2, 3, 4, 5, …} A

Element

A single member of a set. Elements are usually denoted using lower case letters. Subset A set B is a subset of another set A, denoted B ⊆ A, if every element of B is also an element of A Example: If A = {1, 2, 3, 4, 5} and B = {1, 3, 5}, then B ⊆ A. A set can be described by listing its elements inside a pair of braces, and we call this set notation. For example, if the set A is “the set of positive integers smaller than nine”, we can write this as A = {1, 2, 3, 4, 5, 6, 7, 8} If there are too many elements to write out but there is a clear pattern to the elements, we can use three dots to indicate that a pattern continues. For example, if the set B is “the set of even whole numbers”, we can write this as B = {2, 4, 6, 8, 10, …} A special set, called the empty set, is the set which contains no elements. It is usually represented by the symbol ∅, but can also be expressed in set notation as ∅ = {} Venn diagrams can be used to visually represent relationships between sets. Recall that a Venn diagram has four regions or subsets. These include: d

Set 1 a

Set 2 b

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• a: The elements that belong only to the ﬁrst set and not to the second set • b: The elements that belong to both sets • c: The elements that belong only to the second set and not to the ﬁrst set • d: The elements that are included in the universal set but do not belong to either subset

Universal set

Intersection

A set from which all elements in a problem can be found.

The intersection of two sets A and B, denoted A ∩ B, is the set of all elements which belong to both A and B

This is sometimes called the conjunction. A

Negation The negation of a set A, denoted A′ or “not A” is the set of all elements of the universal set which are not elements of A. This is sometimes called the complement. A

Union The union of two sets A and B, denoted A ∪ B, is the set of all elements which belong to either A or B

This is sometimes called the disconjunction. A

Hockey

Gymnastics

We can describe each of the four regions using a combination of unions, intersection, and complements.

For example, the shaded region represents those who play Hockey (H), but don’t do gymnastics (G). This means that in the set H ∩ G′ has 5 elements. 15

Exploration The given Venn diagram shows the sets A, B, and C. Suppose that C is the set of all integers. With a partner, discuss these questions.

Describe a possible set B. Explain. Is there any other possible set?

Describe a possible set A. Explain.

Describe the elements in A ∩ B′, the subset of elements in A, but not in B.

1.02 Venn diagrams and sets mathspace.co

Example 1 P is the set of odd numbers between 2 and 16, and Q is the set given by Q = {1, 2, 3, 5, 8, 13, 21}. Determine the set given by P ∩ Q.

Create a strategy P ∩ Q is the intersection of P and Q. That is, P ∩ Q contains all of the elements that belong to both P and Q. In other words, we are looking for all the elements that are in Q and are also an odd number between 2 and 16.

Apply the idea

Reﬂect and check

We can see that the odd elements of Q are 1, 3, 5, 13, and 21.

We can construct a Venn diagram to see the relationship between sets P and Q, particularly the elements they have in common.

Of these, 3, 5, and 13 lie between 2 and 16 and are therefore also elements of P.

So, we have that

P ∩ Q = {3, 5, 13}

11 9

3 5

Example 2 In the universal set S = {square, triangle, rhombus, parallelogram, hexagon, circle, trapezoid, rectangle} the subset A is “quadrilaterals” and the subset B is “words beginning with the letter t”. a Describe the set “not A” using words, then express it using set notation.

Create a strategy The keyword “not” indicates that we want to take the negation of A. In order to think about this, it might be useful to ﬁrst list out the elements of A.

Apply the idea The set A is the set of “quadrilaterals” from the universal set S. So we have A = {square, rhombus, parallelogram, trapezoid, rectangle} The negation of A is the set of all elements of S that are not in A. We can describe this as “the set of shapes which do not have exactly four sides”. In set notation, this will be A′ = {triangle, hexagon, circle}

Reﬂect and check A

B Square Rhombus

Triangle Trapezoid

Parallelogram Rectangle Circle

Hexagon

This Venn diagram represents the universal set with A′ highlighted.

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b Describe the set “B or not A” using words, then express it using set notation.

Create a strategy We determined the set “not A” in the ﬁrst part of this question. The keyword “or” indicates that we want to take the union of this with set B. This means we need to include everything that is in set B as well as everything in “not A”. This time it might be useful to list out the elements of B.

Apply the idea The set B is the set of “words beginning with the letter t” from the universal set S. So we have B = {triangle, trapezoid}. A′ = {triangle, hexagon, circle} as we found in part (a). The union of two sets will include any element that is in at least one of those sets. We can describe the set “B or not A” as “the set of shapes that begin with the letter t or are not quadrilaterals.” In set notation, this will be B ∪ A′ = {triangle, hexagon, circle, trapezoid}

Reﬂect and check Taking the negation of A: A

B Square Rhombus

Triangle Trapezoid

Parallelogram Rectangle Circle

and combining it with set B:

We get the Venn diagram:

B Square Rhombus

Triangle Trapezoid

Hexagon

B Square Rhombus

Parallelogram

Rectangle

Circle

Hexagon

Circle

Triangle Trapezoid

Hexagon

Notice that the element “triangle” appears in both B and A′. This element is still included in the union, since appearing in both sets means that it appears in at least one of the two sets. We only need to list it once; we do not need to write the element twice in the union even though it appears in both sets.

1.02 Venn diagrams and sets mathspace.co

Example 3 Consider the sets shown in the Venn diagram:

2 3 A 5 4 1 13 16 14 15

a Identify the elements in the universal set.

17 B 18 6 7 11 8 9 10 12 19 20

Create a strategy The universal set contains all the possible elements that can appear in any set in the context. For a Venn diagram, this includes any element within the outer rectangle (whether or not it is inside any circles).

Apply the idea

Reﬂect and check

In this case, every number from 1 to 20 appears exactly once, so the sample space is

In any universal set, the sets A and B are subsets of the universal set. Using geometric notation, we can write A ⊆ S ( A is a subset of S) and B ⊆ S (B is a subset of S).

S = {1, 2, 3, …, 19, 20}

There are many other subsets in a universal set such as A′ and A ∪ A′ and A ∩ B, to name a few.

b Indicate the region which represents A ∩ B′ on the diagram, and determine the elements of this set.

Create a strategy B′ is the negation of B which is the region outside of circle B. A

We then want to take the intersection of that with set A. A

The intersection will be the region where the shading overlaps.

Apply the idea

Reﬂect and check

We want to include everything that is inside circle A but not inside circle B:

In part (a), we found that the universal set for this Venn diagram is “all integers from 1 to 20 inclusive”.

2 3 A 5 4 1 13 16 14 15

17 B 18 6 7 11 8 9 10 12 19 20

In set notation, this is A ∩ B′ = {1, 4, 16}

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Within this universal set, we can describe set A as “the set of square numbers”, and we can describe set B as “integers between 6 and 12 inclusive”. Within the universal set, we could also describe the set A ∩ B′ as “square numbers that are not between 6 and 12” The set A ∩ B′ is a subset of the universal set.

c Write the set notation that represents the set {6, 7, 8, 10, 11, 12}.

Create a strategy We can shade the area of the Venn diagram where these numbers are found, describe the set in words, and then use set notation to describe the set.

Apply the idea 17

2 3 5 1 13

A 4

14 15

18 B 6 7 8 11 10

These numbers are found in this shaded area of the Venn diagram. In words, we can describe this as in set B, but not in set A. In set notation we can say B ∩ A′.

19 20

Reﬂect and check We could also say A′ ∩ B.

Example 4 A group of students were asked why they skipped breakfast. The two reasons given were that they were “not hungry” and they were “too busy”.

Not hungry

Too busy

a How many of the students skipped breakfast because of exactly one reason?

Create a strategy

Apply the idea

Add the number of students that are in the “Not hungry” circle or in the “Too busy” circle but not in both.

Students with one reason = 8 + 13 = 21

Add the numbers Evaluate

b Let N represent the set of students who skipped breakfast because they were not hungry. Let B represent the set of students who were too busy. Use set notation to describe the region which contains all students who skipped breakfast.

Create a strategy Set notation means to use symbols like ∩, ∪, and ′.

1.02 Venn diagrams and sets mathspace.co

Apply the idea

Reﬂect and check

The students that skipped breakfast are contained within the two circles.

Since there isn’t a number outside of the two circles, we can assume that there aren’t any students who skipped breakfast for another reason.

This is the union or N ∪ B.

c Use set notation to describe the region which contains the students who only skipped breakfast because they were too busy.

Create a strategy The word “only” means excluding other options. We can rewrite this in words and then write using set notation.

Apply the idea

Reﬂect and check

Since we want the students who only said “too busy,” we should not include the intersection.

This subset would have 13 students in it. While the subset B of “all students who gave too busy as a reason” would have 24 students in it.

We can think of this as those who are in B, but not in N. This is the set B ∩ N ′.

Example 5 One hundred students in a school are asked about the subjects that they study. 58 of them are studying both math and science, 70 are studying math, and 23 are not studying math nor science. a Construct a Venn diagram that represents the given information.

Create a strategy There are two sets: students who study math and students who study science. We are also considering their negations: students who do not study math and students who do not study science.

Apply the idea The two sets will represent the two circles in the Venn diagram. Math

Science

We know that 58 students study both math and science. This is the intersection of the sets, so we can put this number in the center section where the two circles overlap: Math

Science 58

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We also know that 70 students are studying math, but this includes the 58 that are studying science. So, there are 70 − 58 = 12 that are studying math, but not science. These students belong in the left circle but not the right circle: Math 12

Science 58

Next, we have also been told that 23 students are not studying either of these subjects. This number should be in the rectangle that represents the whole set, but not in either of the circles: 23

Math 12

Science 58

To complete the Venn diagram, we need to determine how many students are studying only science. We can use the fact that there are 100 students in total to ﬁnd how many study only science. Since all four regions sum to 100, we can subtract the three that are ﬁlled in: 100 − 23 − 12 − 58 = 7 This number goes in the right side of the circle for science only and completes the Venn diagram. 23 Math 12

Science 58

b One student is chosen at random. Determine the probability of randomly selecting a student that is studying science.

Create a strategy To calculate probability, we can use that: Probability of an event = There are 100 students in total, so when calculating probability this will the number of possible outcomes. We need to ﬁnd the number of favorable outcomes.

1.02 Venn diagrams and sets mathspace.co

Apply the idea 23 Math 12

Science 58

This shaded region shows all students who are studying science, the favorable outcome. The number of students in this circle is 58 + 7 = 65.

The probability of a randomly selecting a student studying science is 65%.

c Let M represent students who study Math and let S represent students who study Science. One student is chosen at random. Determine the probability of randomly selecting a student in the set (M ∪ S)′.

Create a strategy We can translate this set notation into words and then use the Venn diagram to ﬁnd the number of favorable outcomes.

Apply the idea (M ∪ S)′ is the complement or negation of M ∪ S, which is the union of the two sets. M ∪ S can be shown by shading the union and represents all students who study math or science.

23 Math 12

Science 58

(M ∪ S)′ can be shown by shading everything outside of the union and represents all students who study neither math nor science.

23 Math 12

Science 58

There are 23 students in the set (M ∪ S)′.

The probability of a randomly selecting a student in (M ∪ S)′ who is not studying math nor science is 23%.

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Reﬂect and check Notice that:

This means that the probability of a set and its complement is 1 or 100%. These two sets make up the universal set, so the favorable outcomes are the same as the possible outcomes.

d Explain why P (M ) + P (M ′) = 1.

Create a strategy Remember that M is the set of students who are studying math and M ′ is the set of students who are not studying math.

Apply the idea

Reflect and check

A student in the universal set must be in M, studying math, or in M ′, not studying math. There are no other possibilities. For this particular case:

We can make a connection to logic notation and the idea that p∨ ∼p is always true. This is because if p is true then one of the whole statement is true and if p is false, then ∼p is true, so the whole statement is true.

Idea summary When working with sets, we often want to find new subsets of a universal set. Where the two subsets overlap is called the intersection, is written as A ∩ B and contains the elements in both A and B. Everything in the two subsets is called the union, is written as A ∪ B, and contains the elements in either A or B. Everything outside one of subsets is called the negation or complement, is written as A′, and contains the elements not in A. A

1.02 Venn diagrams and sets mathspace.co

Practice What do you remember? 1

Match each symbol with its meaning. a

∪

Intersection (conjunction)

∩

Union (disjunction)

iii

{}

Complete each statement. a b c

Empty set

B = {∅} means that ⬚ is the ⬚ set.

A = {⬚∣ − ∞ < x ≤ 6} means that A is the set of all x ⬚ − ∞ < x ≤ 6.

A = {⬚ : ⬚} means that A is the set of all y such that −∞ < y < ∞ or that ⬚ is any real number.

For each pair, determine if B ⊂ A, B ⊆ A, or neither. a

A = {5, 9, 14, 15, 22, 28}

A = {red, blue, green, yellow}

B = {15, 9, 28, 5, 14} B = {green, yellow, red, purple} c

A = {−2, −1, 0, 1, 2}

A = the set of all integers

B = {−2, −1, 0, 1, 2} B = the set of all even whole numbers 4

Determine the intersection (conjunction) for each pair: a

A = {0, 1, 4, 9, 16, 25, 36, 49}

P = {1, 3, 5, 7, 9, 11, 13}

B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} Q = {0, 2, 4, 6, 8, 10, 12} 5

Determine the union (disjunction) of each pair: a

A = {9, 99, 999, 9999}

J = the set of multiples of 5

B = {0, 1, 99, 100, 101} K = the set of multiples of 10 6

Match each of the Venn diagrams to the correct statement. a

A ∩ B = {∅}

B ∩ A′ has 12 elements

(B ∪ A)′ has 12 elements

B⊂A

A B

iii A

A 32

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List the elements in each of the subsets: a

A∩B

A∪B

A ∩ B′

( A ∪ B)′

B 7

Let’s practice 8

In a survey, a group of students were asked about their siblings. The two categories show if they have at least one brother and if they have at least one sister: a

Have a sister

Have a brother

Find the number of students that have: i

At least one sibling.

At least one brother.

iii No sisters. b

Let S represent the set students who have a sister. Let B represent the set students who have a brother. Use set notation to describe the region which contains the students who have: No siblings.

A brother and a sister.

iii Only have a sister.

A brother or a sister, but not both.

June is struggling to decide what movie to watch online. A Venn diagram of her options sorts movies into three categories based on their genre: Comedy, Action and Horror.

She decides to pick one movie at random from the streaming website. What is the probability that she selects: a

A horror movie?

A movie that only ﬁts into one exactly genre?

A movie that ﬁts into at least two genres?

Action

Comedy 4

5 3

6 13 Horror

A group of 60 people were surveyed on whether they enjoyed skateboard riding or bike riding.

Skateboard

Bike

• 16 people said they enjoyed skateboard riding but not bike riding • 25 people said they enjoyed bike riding only • 12 people said they enjoyed both activities • 7 people said they did not enjoy either activity Complete the Venn diagram with the given information. 11

Out of a group of 40 students, 12 of them enjoy singing but not dancing. 13 enjoy both singing and dancing, while 8 do not enjoy either activity. a

Create a Venn diagram including the given information.

One student from the group is randomly selected. Determine the probability of: i

The student enjoying dancing, but not singing.

The student enjoying singing.

iii The student enjoying singing or dancing.

1.02 Venn diagrams and sets mathspace.co

Consider the Venn diagram: a b c

Set P is given as A ∩ B. Highlight set P on the diagram and determine its elements.

16 A

Set Q is given as A′ ∩ B. Highlight set Q on the diagram and state its elements.

2 4 7 8

Express the set P ∪ Q in terms of A and B.

A group of people were asked about whether they liked watching basketball and/or baseball. The results are shown in the Venn diagram:

19 20

Eiichiro makes this claim: “In total 70 people like watching basketball, and 63 people like watching baseball. When added to the 31 people who do not like watching either sport, this means that 164 people in total were surveyed.”

Basketball

Baseball 49

Explain why Eiichiro’s reasoning is incorrect, and ﬁx his mistake(s). 14

A manager is assessing the participation of employees in two different training programs: leadership development and project management. Out of a team of 40 members, 32 attended the leadership development program and 15 attended the project management training. a

Given that everyone in the team attended at least one training, ﬁnd out how many employees attended both trainings.

Construct a Venn diagram that represents the given information.

100 students were asked whether they studied History or Geography. Let H represent History, G represent Geography, and S represent the universal set. Select all correct statements. A

H ⊂ (H ∪ G)

S ⊆ (H ∪ H′)

G⊂H

(H ∩ H′) = {∅}

H ⊂ (H ∩ G)′

(H′ ∩ G) ⊂ G

19 History 31

Geography 26

Let’s extend our thinking 16

S is the set of all polygons, A is the set of all triangle, B is the set of all non-triangles. Are these statements true? Use a diagram to justify your answers. a

A⊂B

A ∩ B = {}

A∪B=S

A = B′

Given a set A and a universal set S, determine an expression for ( A′)′, the negation of the negation of A. Explain your answer.

Given any two sets A and B, determine which would be greater, P ( A ∪ B) or P ( A ∩ B). Explain your reasoning.

Draw a Venn diagram that meets the given criteria. • A∩B=∅ • C∩B≠∅ • A⊂C • C⊂S

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These undefined terms can now be used to define other geometric figures. Line segment

Angle

A line segment consists of two endpoints and all the points between them.

Angles are formed wherever two lines, segments or rays intersect. Angles are measured in degrees. The point of intersection is called the vertex of the angle.

B A

Ray A ray has one endpoint and extends without end in one direction. B A

The intersection of two geometric figures is the set of all points they share in common. Points are considered collinear if they lie on the same line and geometric figures are considered coplanar if they lie on the same plane. We can then build the following definitions: Collinear

Parallel planes

Points that lie on the same line.

Planes that do not intersect.

Non Collinear Points

Collinear Points

Q P

R S

Coplanar Figures that lie in the same plane. B A D

B C

Coplanar Points

A D

Non-Coplanar Points

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p r

Example 1 Use the diagram to identify the following geometric ﬁgures.

a A plane

Create a strategy Planes are named with three noncollinear, coplanar points. Identify three points on the plane in the diagram that do not form a line.

Apply the idea

Reﬂect and check

A, B, and E is one possible solution.

There are four points labeled on the plane: A, B, C, and E. Any combination except ABC would work to name the plane.

b A line segment, ray, and line that contain point A and point C

Create a strategy The correct notation for a line includes two arrows, for a ray includes a right arrow, and for a segment includes a line with no arrows.

Apply the idea

Reﬂect and check

Line:

We know from the postulates that through any two points there exists exactly one line. Since segments and rays are both parts of a line the two points that deﬁne the line can also deﬁne a segment and rays in either direction.

Ray: Segment:

Example 2 E

Use the diagram to identify geometric ﬁgures. A G

H K

I M J D

a Three collinear points

Create a strategy

Apply the idea

Collinear points are those that lie on the same line. Since there are multiple lines with three points labeled, there are multiple correct answers.

E, J, and I is one possible answer.

1.03 Introduction to geometric notation mathspace.co

b The intersection of

and

Create a strategy The line intersection postulate states that “if two distinct lines intersect, then they intersect in exactly one point”, so we know the answer will be a single point. Which point do the two lines have in common?

Apply the idea H

c All angles with I as their vertex.

Create a strategy In the diagram, we can see there are 4 non-straight angles and 2 straight angles that contain I as their vertex.

Apply the idea ∠EIM ∠EIH ∠HIJ ∠JIM ∠EIJ ∠HIM

Idea summary We can define all figures based on the undefined terms point, line, plane, and distance along a line and use these ideas throughout geometry.

Practice What do you remember? 1

Complete the deﬁnition using one of the given terms, then draw an example. i

Point

Line

iii

Ray

Angle

vii Collinear

A ⬚ has one endpoint and extends without end in one direction.

b c d e f

Plane

Line segment

viii Plane ﬁgure

A ⬚ has two dimensions extending without end. It is often represented by a parallelogram. A ⬚ has no dimension. It is a location on a plane. It is represented by a dot.

An ⬚ is formed wherever two lines, segments or rays intersect. The point of intersection is called the vertex of the ⬚. A ⬚ consists of two endpoints and all the points between them.

A ⬚ has one dimension. It is an inﬁnite set of points represented by a ⬚ with two arrowheads that extend without end.

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Using proper notation, identify the following features of the diagram shown: a

A line segment with one endpoint at point D

A set of three collinear points

A line passing through point F

A ray starting from point A

Another name for

C D E

F G

Using proper notation, identify the following features of the diagram shown: a

A line that is contained in plane PQR

A line that intersects, but is not contained in, plane PQR

Four coplanar points

Three collinear points

An angle that is contained in the plane

An angle that is not contained in the plane

A pair of opposite rays

R S

P C

Name the geometric ﬁgure shown in two different ways.

X Y

State the intersection of the two geometric ﬁgures shown in each of the following diagrams: a

and

and Plane BCD

A B

and Plane XY Z

Planes PQR and QRS

Q Z R

1.03 Introduction to geometric notation mathspace.co

The given ﬁgure is a cube. Identify three pairs of skew lines.

C G

A F E

Let’s practice 7

Answer the following questions: a

Draw a ray that has point A as an endpoint.

Write a name for the ray you have drawn.

State how many points you needed to use to name the ray.

When naming the ray, determine whether the order of the points matters. Explain your answer. and

8 a 9

Draw a diagram that shows this.

Identify a pair of opposite rays in the diagram.

Name the full intersection of the two planes.

Planes DEF and EFG have common points of E and F. a

meet at point M.

Draw a diagram that shows this.

contained in plane P, with

Draw a diagram that shows

intersecting (but not contained in) the plane.

Let’s extend our thinking 11

Determine whether each statement is true or false. If true, draw an example. If false, draw a diagram that shows it is false. a

Through any two points, there is exactly one line.

Two distinct pairs of points always form two different lines.

If two points are both contained in a plane, then the line that they form is also entirely contained in that plane.

For any three collinear points, there is exactly one plane that contains them.

If three points are collinear, they cannot be contained in a single plane.

For any two distinct lines, there is a plane which contains them both.

For each geometric ﬁgure, determine a real-life example which is a model for the ﬁgure. State whether there any limitations or restrictions on the real-life example. a

A line

A plane

A line and a ray.

Explain the similarities and differences between: a

A ray

A line segment and a ray.

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1.04 Line segments and constructions After this lesson, you will be able to... • use constructions to copy a line segment. • use constructions to find the midpoint of a line segment. • identify and use midpoints and segment bisectors to solve problems.

Line segments There are two postulates in Geometry that allow us to measure and solve problems involving segment lengths. Ruler postulate

Segment addition postulate

Every point on a line can be paired with a real number. This makes a one-to-one correspondence between the points on the line and the real numbers. The real number that corresponds to a point is called the coordinate of the point. The distance between points A and B, written as AB, is the absolute value of the difference of the coordinates of A and B.

If B is between A and C, then AB + BC = AC.

If AB + BC = AC, then B is between A and C. A

AB = |x − y|

A segment can be bisected, which means it has been divided into two congruent segments. The midpoint, which bisects a segment, is the point exactly halfway between the two endpoints of a segment. Congruent segments

Segment bisector

Two segments whose measures are equal.

A point, ray, line, line segment, or plane that intersects the segment at its midpoint.

Labeled with congruent marks. 4 cm L

4 cm M

M A

C B

Midpoint Divides a segment into two congruent segments. A

Midpoint

1.04 Line segments and constructions mathspace.co

Example 1 Use the ruler postulate to ﬁnd the length of

. X

Create a strategy Since X lines up with the real number 2 and Z lines up with the real number 5, the length of between 2 and 5.

is the difference

Apply the idea

Reﬂect and check

XZ = 3

The length of a line segment will not include the line segment symbol: XZ = 3 but not = 3.

Example 2 Use the segment addition postulate to ﬁnd the length of the following:

a QS

Create a strategy

Apply the idea

We know that QR = 33 and RS = 30. Since R is between Q and S, by the segment addition postulate, we know that QS = QR + RS.

QS = QR + RS

Segment addition postulate

QS = 33 + 30

Substitute QR = 33 and RS = 30

QS = 63

Evaluate the addition

b PQ

Create a strategy We know that PR = 57 and QR = 33. Since Q is between P and R, by the segment addition postulate, we know that PQ + QR = PR.

Apply the idea PQ + QR = PR

Segment addition postulate

PQ + 33 = 57

Substitute QR = 33 and PR = 57

PQ = 24

Subtract 33 from both sides

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Example 3 Point B bisects

a Identify two congruent segments.

Create a strategy To bisect something is to divide it into two congruent parts. Since B bisects as shown in the diagram. A

, it creates two congruent segments

Apply the idea ≅

b If AB = 7, ﬁnd the length of

Create a strategy Using the segment addition postulate we know that AB + BC = AC. Since B bisects which tells us that BC = 7.

, we know that

≅

Apply the idea AB + BC = AC 7 + 7 = AC 14 = AC

Segment addition postulate Substitute AB = 7 and BC = 7 Evaluate the addition

Idea summary Line segments can be measured using the ruler postulate by aligning the segment with a number line and ﬁnding the difference between the values each point lines up with. We can solve problems involving line segments using the segment addition postulate. Congruent segments can be labeled with congruency marks. Midpoints and bisectors are two ways a segment can be divided into congruent segments.

1.04 Line segments and constructions mathspace.co

Geometric constructions A geometric construction is the accurate drawing of angles, lines, and shapes. The tools used for these constructions are a straightedge, compass, and pencil. To construct a copy of a segment, we will: 1. Identify the segment we want to copy. 2. Draw a point that will become the ﬁrst endpoint of the copied segment. 3. Open the compass width to measure the distance between endpoints. 4. Without changing the compass width, place the point end of the compass on the point we constructed for the copy and draw a small arc. Place a point anywhere on the arc. 5. Connect the two points using a straightedge.

Step 1

Step 2

Step 3

Step 4

Step 5

To construct the bisector of a segment, we will: 1. Identify the segment we want to bisect. 2. Open the compass width to just past half the segment’s length and draw an arc from one endpoint that extends to both sides of the segment. 3. Without changing the compass width, draw another arc from the other endpoint that intersects the original arc on both sides of the segment. 4. Label the intersection of the arcs with points. 5. Connect the points.

Step 1

Step 2

Step 3

Step 4

Step 5

Example 4 Construct a copy of

. G

Create a strategy To construct a copy of a segment, we can follow the steps detailed in the idea introduction. We can do it by hand or using technology in the form of GeoGebra.

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Apply the idea Using technology:

First, we use the point tool to add a new point, which will be one endpoint of the copied segment.

Next, we can use the Distance or Length measurement tool to ﬁnd the length of the original equivalent of measuring the segment with a compass.

. This is the

1.04 Line segments and constructions mathspace.co

Finally, we can use the Segment with Given Length tool to create a segment from our new point using the length that we measured.

Reﬂect and check Note that there are many other tools in GeoGebra that we could use to construct a copy of a segment. For example, after using the measurement tool to ﬁnd the length GH, we could then use the Circle: Center & Radius tool to create a circle with a radius of 4.9 units. Any radius of that circle would then be a copy of the original segment.

Example 5 Ursula has a rectangular fenced sheep enclosure, shown below. She wants to build an additional fence that will divide the enclosure into two congruent rectangular sections.

Construct a line to represent the additional fence, showing all steps.

Create a strategy A line that divides the sheep enclosure into two congruent rectangular sections must bisect a pair of opposite sides of the enclosure. In particular, the line will be a perpendicular bisector of both sides. So, we can draw a line to divide the enclosure into two congruent rectangular sections by constructing a perpendicular bisector of and extending it to reach . We will do so in this case by making use of technology.

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Apply the idea

We start by creating an arc centered at A that has a radius that is greater than half the length of

The easiest way to achieve this is to use the Circle: Center & Radius tool and manually input the radius. This way, we will be able to create an arc with an identical radius at the next step.

We now create an identical circle by using the Circle: Center & Radius tool and inputting the same radius, but this time centered at the other endpoint B.

1.04 Line segments and constructions mathspace.co

Next, we use the Point tool to mark both intersections of these two arcs.

Finally, we can draw a line that passes through these two points of intersection. To divide the rectangle into two smaller congruent rectangles, we need to make sure this line extends all the way to intersect , so we use the Line tool (rather than the Segment tool).

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Reflect and check We could also have divided the sheep enclosure into two congruent rectangles by bisecting the other sides, . The result of this construction would look like the following:

and

Idea summary Geometric figures can be constructed using a compass and straight edge. We can copy a line segment to any point using congruent circles and bisect a line segment using overlapping congruent circles.

Practice What do you remember? 1

Use the diagram to ﬁnd each segment length. A

−4

−3

−2

−1

Write an equivalence statement using the segment addition postulate, then ﬁnd the length of the indicated segment. a

9 19 C

E F

1.04 Line segments and constructions mathspace.co

Use the diagrams provided to ﬁnd the indicated length. a

10.2 mm

25.7 cm

Use the diagram and the given information to answer each question. • H is the midpoint of

List all pairs of congruent segments.

Identify two additional segments that have been bisected, and the line which bisects them.

Identify two additional midpoints.

Step 1

Step 2

The bisection of

A line segment congruent to

An angle congruent to ∠ABD

≅

Construct a line segment PC such that

. b

A P

B P

Use a construction to locate the midpoint of

. Describe each step of your construction. P

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H K

M J

Which type of construction is illustrated in the ﬁgure?

C A

bisects segment EJ.

•

SOL

Let’s practice 8

Consider the segment shown in the diagram:

−4

−3

−2

−1

Determine which segments are congruent to

. .

Determine which segments are congruent to

If a point G was added at −1, determine how many additional congruent segments you would ﬁnd in parts (a) and (b).

Identify all midpoints in the diagram and the segments they bisect.

Use the segment addition postulate to ﬁnd the unknown length.

Given: JK = 5x + 2, KL = 7x + 4 and JL = 42. Find KL. K

Given: JT = 5x + 5, CT = 76 and CJ = 4x + 8. Find JT. C

Suppose the point B lies on the segment

between the points C and D.

If CD = 4x, BD = 2x and CB = 12, ﬁnd the length of

If CB = 3x − 6, BD = 4x + 3 and CD = 18, ﬁnd the length of

In the diagram, M is the midpoint of

In the diagram, Y is the midpoint of

. .

. Find the length ML. 5x + 6

8x − 21 M

. Find the length XZ. X 5k − 2

Y 18 + 3k

Q is the midpoint of

, with PQ = 3x − 8 and QR = 2x + 3.

Determine the length PQ. 14

bisects

at W.

If UW = 8x + 5, ﬁnd an expression for the length UV.

1.04 Line segments and constructions mathspace.co

Construct a segment that is half the length of AB with an endpoint on X. Describe each step of your construction. A

Construct a new line segment whose length is equal to AB + DE. A

Let’s extend our thinking 17

Given that points X, Y, and Z are collinear, and that XY = 12 and XZ = 18, determine the possible lengths for Y Z. Draw diagrams that support your solution.

Given that AB = 10, BC = 5, and AC = 12 determine if points A, B, and C are collinear. Explain.

Two students use the ruler postulate to measure the length of A

in the diagram.

Student A: −10 −8

−6

−4

−2

10 B

QB = ∣ 9 − 3∣ = ∣ 6∣ = 6 Student B: −10 −8 A

−6

−4 P

−2

0 Q

QB = ∣ 6 − 0∣ = ∣ 6∣ = 6 Explain why both students have the same solution for the length of

LeeAnn wants to divide her paper into four columns of equal width without making any folds. Describe the compass and straight edge constructions LeeAnn could use to accomplish her goal.

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1.05 Angles and constructions After this lesson, you will be able to... • use constructions to copy an angle. • use constructions to bisect an angle. • identify and use vertical angles, linear pairs, complementary and supplementary angles to solve problems.

Using properties of angles to solve problems There are two postulates that allow us to measure and solve problems with angles. Protractor postulate Consider a ray

and a point A on one side of

. Every ray of the form can be paired one to one with a real number from 0 to 180. The measure of ∠AOB, written as m∠AOB, is equal to the difference between the real numbers matched with and

on a protractor.

Angle addition postulate If P is in the interior of angle RST, then m∠RSP + m∠PST = m∠RST. R

A O

The angle addition postulate only works for adjacent angles, or angles that share a common leg and vertex, but do not overlap. The measure of an angle is deﬁned using the protractor postulate. Congruent angles

Linear pair

Angles with the same measure.

Adjacent angles that form a straight line.

57° 57°

The linear pair postulate states that if two angles form a linear pair, then they are supplementary, which means the sum of their angles is 180°. Vertical angles The opposite angles formed when two lines intersect.

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Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 1.05 to answer these questions. 1.

Create the following types of angles and observe the reflex angle: right, acute, obtuse, straight.

What do you notice about each reflex angle?

Angles can be classiﬁed based on their measure: Acute angle

Straight angle

An angle whose measure is between 0 and 90 degrees.

An angle whose measure is exactly 180 degrees. 180°

90°

55°

Reﬂex angle An angle whose measure is between 180 and 360 degrees.

Right angle An angle whose measure is exactly 90 degrees.

90°

Obtuse angle An angle whose measure is between 90 and 180 degrees. 90° 120°

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210° 180°

Angle bisector A line, segment or ray that divides an angle into two congruent angles.

Example 1 Solve for x.

152° x°

Create a strategy

Apply the idea

The angles in the diagram are marked as congruent. That means they have equal measure.

x = 152

Example 2 Consider the diagram, where m∠PQR = 145°.

P (3x + 7)°

(2x − 2)° Q

a Write an equation and solve for x.

Create a strategy Using the angle addition postulate we know that m∠PQS + m∠SQR = m∠PQR. Now, we can substitute and solve.

Apply the idea m∠PQS + m∠SQR = m∠PQR 3x + 7 + 2x − 2 = 145 5x + 5 = 145

Angle addition postulate Substitute m∠PQS = 3x + 7, m∠SQR = 2x − 2 and m∠PQR = 145 Combine like terms

5x = 140

Subtract 5 from both sides

x = 28

Divide both sides by 5

b Find m∠SQR.

Create a strategy Now that we know the value of x we can substitute it back into the expression for m∠SQR.

Apply the idea m∠SQR = 2x − 2

Expression for m∠SQR

m∠SQR = 2(28) − 2

Substitute x = 28

m∠SQR = 54

Evaluate the multiplication and subtraction

1.05 Angles and constructions mathspace.co

Example 3 The angles in the diagram are complementary. Find the value of x. x° 39°

Create a strategy Complementary angles have a sum of 90°. Use this to write an equation that includes the two angles in the diagram knowing that the angles are complementary. Then we want to solve for x.

Apply the idea x + 39 = 90 x = 51

Deﬁnition of complementary angles Subtract 39 from both sides

Reﬂect and check If the angles were supplementary, then the sum of the angles would equal 180° instead of 90°.

Example 4 Use the diagram to identify an example of each angle pair.

B A

28° F 62° E

a Vertical angles

Create a strategy

118°

62° D

Apply the idea

Vertical angles are formed by intersecting lines. ∠AFC and ∠EFD or ∠AFE and ∠CFD There is only one pair of intersecting lines in the diagram, and . Identify a pair of opposite angles formed by this intersection.

b Linear pair

Create a strategy Linear pairs are adjacent angles that form a line. First, we need to identify a line such as identify adjacent angles that form this line.

and see if we can

Apply the idea

Reﬂect and check

∠AFE and ∠EFD form a linear pair.

There are multiple linear pairs in the diagram, including ∠EFD and ∠CFD, ∠AFC and ∠CFD, and ∠AFC and ∠AFE.

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Idea summary We can use protractors and algebra to measure angles and solve problems involving angles. We can use deﬁnitions and postulates for supplementary, complementary, vertical, and adjacent angles to solve problems.

Angle constructions Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 1.05 to answer this question. 1.

Describe what is happening at each step of the construction.

To construct a copy of an angle, we will: 1. Identify the angle we want to copy. 2. Draw a ray that will form one of the legs of the copied angle. 3. With the compass point on the vertex of the original angle, use the compass to draw an arc that intersects both legs. 4. Copy the arc in Step 3 by placing the point end of the compass onto the endpoint of the ray. 5. On the original angle, use the compass to measure the distance between the points where the legs of the angle meets the arc drawn in Step 3. 6. Without changing the compass width, copy the distance by placing the compass point where the ray meets the copied arc and draw an intersecting arc. 7. Draw a ray that shares its end point with the ray from Step 2, and goes through the intersection found in Step 6.

Step 1

Step 2

Step 3

Step 4

Step 5

Step 6

Step 7

To construct the bisector of an angle, we will: 1. Identify the angle we want to bisect. 2. With the compass point on the vertex of the angle, use the compass to draw an arc that intersects both legs. 3. Label the intersections with points. 4. With the compass point on one of the points from Step 3, draw an arc that passes halfway through the interior of the angle. 5. With the compass point on the other point from Step 3, draw an arc that passes halfway through the interior of the angle and intersects the ﬁrst arc. 6. Label the intersection of the arcs drawn in parts 4 and 5 with a point.

1.05 Angles and constructions mathspace.co

7. Draw a line that connects the vertex of the angle and the point added in Step 6.

Step 1

Step 2

Step 3

Step 4

Step 5

Step 6

Step 7

Example 5 Construct a copy of the angle shown. 135°

Create a strategy To construct a copy of an angle, we will follow the series of steps detailed in the concept summary above, using technology in the form of GeoGebra.

Apply the idea

We start by drawing a ray nearby that will form one leg of the angle. We can do this by using the Vector tool as shown.

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Next, we want to create an arc on the original angle that intersects both legs, and then duplicate an arc with the same radius centered on the new ray. To do so, we can make use of the Circle: Center & Radius tool, to ensure that both arcs have the same radius.

We can then use the Point tool to create points at all of the intersections between an arc/circle and a ray.

1.05 Angles and constructions mathspace.co

We now want to get the radius of the circle centered at a point of intersection on the original angle (point F in this case) that passes through the other point of intersection (point G). Once again, we can make use of the Circle: Center & Radius tool, by clicking on F (to be the center) and then dragging to G before releasing. Once we have created the circle, we can check its radius in the algebra tab.

We can then duplicate this circle across to the new ray, centering it on the point of intersection (point H) and giving it the same radius using the Circle: Center & Radius tool.

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Finally, we can mark the point of intersection of the two circles we have created, and then use the Vector tool to create the other ray of our copied angle.

Reﬂect and check Note that the new copy of the angle doesn’t have to be drawn in the same orientation as the original angle - it can be rotated, as shown here. Also note that because we have used a circle tool to draw the arcs, there are two possible points of intersection to use at the last step. Using either of these will create an angle of the correct measure.

Example 6 A circle centered at O has radii

and

as shown. B

Find and label point C, which lies at the midpoint of minor arc

Create a strategy The ray which bisects ∠AOB will intersect the circle at the midpoint of and then mark the point of intersection with the circle.

. So, we can construct an angle bisector

We will do so by making use of technology in the form of GeoGebra.

1.05 Angles and constructions mathspace.co

Apply the idea

We start by drawing an arc centered at O that intersects both radii. The easiest way to do this is to use one of the Circle tools. We then mark both of the points of intersection, using the point tool.

Next, we draw an arc centered at one of these points of intersection that passes halfway between the interior of ∠AOB. It is easiest to do this using the Circle: Center & Radius tool, since the next step will be to create another arc (i.e. circle) of the same radius.

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We now repeat this, creating another circle of the same radius centered at the other point of intersection, so that it crosses the circle we just drew. We then mark the newly formed point of intersection.

We now construct the bisector of ∠AOB by drawing a line through this latest point of intersection and O.

1.05 Angles and constructions mathspace.co

Finally, we can label point C as the intersection of the bisector and

Reﬂect and check We can also ﬁnd the midpoint of minor arc

using string.

First, take a piece of string and tie it to the end of a pencil. Use a push pin to secure the other end of the string on point A and draw an arc.

Then, using the same string, place the push pin to secure the string on point B and draw another arc, intersecting the ﬁrst.

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Finally, draw a line from point O through the intersection of the two drawn arcs to the circle’s edge to find the . midpoint of minor arc

C A

Idea summary A compass and straightedge can be used to construct angles as well as the use of technology or string.

Practice What do you remember? 1

Name the angles shown in three different ways. a

X Y

For each angle shown: i

Identify the measure of the angle.

Use the measure to classify the angle as acute, obtuse, right, or straight.

1.05 Angles and constructions mathspace.co

Find the measure of each angle. a

If m∠IJK = 150°, ﬁnd m∠BJK

Find m∠KLM

B M

25° I

30°

60°

Use the diagrams to identify two pairs of congruent angles. a

b A

B J

40° G

50°

40°

Use the diagram to ﬁnd each angle measure. C B

m∠AEB

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m∠AED

m∠CED

m∠DEB

Determine whether each of the following diagrams shows the given angle pair. Explain. a

Complementary angles

Vertical angles

28° 68°

28°

12°

Linear pair

71° 109°

137° 43°

Identify the missing angle for the angle pair. a

⬚ forms a linear pair with ∠QOT.

⬚ forms a vertical pair with ∠BEH. C

N P

F H

Each statement shown has an error. Identify the error and rewrite the statement correctly. a

∠ABC = 90°

In ∠XZY, Y is the vertex.

180° is an obtuse angle since its measure is greater than 90°.

If point P lies in the interior of ∠JKL then m∠JKP + m∠JKL = m∠LKP.

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Determine the value of the variable in each of the following diagrams: a

B A

m° G

49°

80° D

63°

a°

J K

Let’s practice 10

Use the diagram to ﬁnd an example of each angle pair if it exists:

T U

Linear pair

Vertical angles

Complementary angles

Supplementary angles that are not also a linear pair

V X W Y Z

Use the angle addition postulate to ﬁnd the indicated angle. a

m∠PQR = 145°

m∠ADC = 120°

Find m∠SQR Find m∠ADB S P

(3x + 7)°

(2x − 2)°

(4x − 23)°

m∠ADC = (10x + 35)°

m∠ADC = 70°

Find m∠ADB Find m∠ADB D C

C (x − 9)°

D B

(5x + 1)°

A (5x − 15)° B A

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For each diagram: i

Write an equation that models the relationship shown in the diagram.

State the theorem or postulate that justiﬁes this equation.

iii

Solve for x.

(12x − 5)° (10x + 1)° (4x + 7)° (2x + 5)°

Use the given information to solve for the value of x. bisects ∠PRS

bisects ∠ADC and m∠ADC = 138°

(x + 40)° (3x − 20)° R

(4x − 23)° D

The measure of an angle’s complement is ﬁve less than half the measure of the angle’s supplement. Find the measure of the angle.

Use the given information to ﬁnd the measures of the indicated angles: • ∠1 and ∠3 form a linear pair • ∠1 and ∠2 are vertical angles

• m∠1 = 2x – 30 • m∠2 = 5x – 120 • m∠3 = 10x − 150 a

m∠1

Given that m∠ABC = 120°, ∠ABC is bisected by m∠EBC.

Construct ∠X so that ∠B ≅ ∠X.

m∠3

, and ∠ABD is bisected by

. Draw a diagram and ﬁnd

A C B

1.05 Angles and constructions mathspace.co

Construct

bisects ∠B.

so that

Use proper geometric notation to clearly describe each construction step shown. P

Step 1:

Step 2:

R S

Step 4:

Step 3:

P R

T O

N S

Let’s extend our thinking 20

Two students are debating over the answer to the following ﬂawed question: Find m∠Q Presley:

Qing:

The m∠Q should be 59°

The m∠Q should be 22° S

P 37°

22° Q

Determine the issue with this question and explain why the two students chose different answers. 21

Determine if the statement is always, sometimes, or never true. Create diagrams to justify your answers. a

Complementary angles are also adjacent.

A linear pair is supplementary.

An obtuse angle forms a linear pair with an acute angle.

Vertical angles form a linear pair.

Vertical angles are supplementary.

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Determine if the information given is enough to justify the conclusion. Draw a diagram to support your solution. Given: ∠MNP and ∠PNR form a linear pair and

bisects ∠MNR.

Conclusion: m∠MNP = m∠PNR = 90° 23

Construct an angle twice the size of ∠XY Z with a vertex at point P. Describe each step of your construction. X

Z Y

Construct an angle whose measure is equal to m∠ABC + m∠DEF. A

Describe a sequence of constructions that would create a 45° angle.

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2 Language of Logic Big ideas • Logic, in combination with facts and theorems can be used to draw conclusions about geometric figures.

Chapter outline 2.01 2.02 2.03 2.04

Translating logic statements Converse, inverse, and contrapositive Inductive reasoning Deductive reasoning

70 82 91 99

New statements can be formed from these logic statements. Negation is the opposite of a given statement. For the statement r: “an animal is a cat” the negation is “an animal is not a cat.” Negation The negative of a statement, often formed by adding the word “not” Example: The negation of “a shape is a square” would be “the shape is not a square” We can combine simple logic statements to form compound statements. One type of compound statement is a conjunction which connects two statements with the word and. The conjunction of p and q from the exploration is “an animal is a corgi and an animal is a dog.” Conjunction A conjunction connects two statements with the word “and” Example: The conjunction of “a shape is a square” and “the shape is green” would be “the shape is a square and the shape is green” A disjunction connects two statements with the word or. The disjunction of r and q from the exploration is “an animal is a cat or an animal is a dog.” In a disjunction, one or both statements may be true. Disjunction A disjunction connects two statements with the word “or” Example: The disjunction of “a shape is a square” and “the shape is green” would be “the shape is a square or the shape is green” A conditional statement, also called an if-then statement, is a type of compound statement that combines two simple statements in the form: If p, then q. Conditional statement A logical argument consisting of a set of premises, hypothesis ( p), and conclusion (q). Symbolically: If p, then q p → q Example: If p is the statement “a shape is a square” and q is the statement “a shape is a rectangle”, the conditional would be “if the shape is a square, then the shape is a rectangle” We call p the premise or the hypothesis. The hypothesis is the assertion that begins the statement of argument, and it typically starts with the word “if”. We call q the conclusion. It closes out the argument and is true if the premise is true. The converse of a conditional is formed by switching the place of the premise and conclusion of the conditional statement: If q, then p. Converse A logical statement formed by interchanging the hypothesis and conclusion of a conditional statement. Symbolically: If q, then p q → p Example: Given the conditional “if the shape is a square, then the shape is a rectangle”, the converse is “if the shape is a rectangle, then the shape is a square.” 2.01 Translating logic statements mathspace.co

A true conditional statement can have a false converse, and a false conditional can have a true converse. This pair has a true conditional and a false converse: Conditional Converse

If the shape is a square, then the shape is a rectangle. If the shape is a rectangle, then the shape is a square.

This pair has a false conditional and a true converse: Conditional Converse

If an animal is a dog, then it is a corgi. If an animal is a corgi, then the animal is a dog.

A biconditional statement is the conjunction of a conditional statement and its converse: If p, then q; and if q, then p. We may also see this written as: p if and only if q Biconditional statement A biconditional is the conjunction of a conditional and its converse. It may be written p iff q. Example: “A shape is a triangle if and only if it has three sides.”

Example 1 Consider the statement “If it is sunny, then Yasmin goes to the park.” a Identify the hypothesis of the statement.

Create a strategy The hypothesis is the assertion that begins the statement of argument, and it typically starts with the word “if”.

Apply the idea In the statement, the hypothesis is “it is sunny.”

b Identify the conclusion.

Create a strategy The conclusion closes out the argument and often follows the word “then”.

Apply the idea In the statement, the conclusion is “Yasmin goes to the park.”

c Write the converse of the statement.

Create a strategy The converse is formed by switching the place of the hypothesis and conclusion.

Apply the idea The converse is “If Yasmin goes to the park, then it is sunny.”

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d Use the statement and its converse to write a biconditional statement.

Create a strategy Use the converse from part (c). A biconditional is the conjunction of both the original statement and its converse.

Apply the idea

Reﬂect and check

A biconditional is “If Yasmin goes to the park, then it is sunny and if it is sunny then Yasmin goes to the park.”

The reverse of the biconditional has exactly the same meaning, so we could have written:

Another way to write the biconditional is “Yasmin goes to the park if and only if it is sunny.”

“It is sunny if and only Yasmin goes to the park.”

Idea summary For the statements p and q : • • •

Negation - not p Conjunction - p and q Disjunction - p or q

• • •

Conditional - if p, then q Converse - if q, then p Biconditional - p if and only if q

Translate logic statements We can use a variety of symbols when working with logical statements. This can make writing logical statements more eﬃcient. Negation This is the opposite of a statement. We write the negation of p as ∼ p. p: ∼p :

The plate is dirty. The plate is not dirty.

Note that the negation of a negation becomes the original proposition. In symbols we could write that ∼ (∼ p) is the same as (or equivalent to) p. Conjunction To join two logical statements together with the word and, we use a conjunction which is often symbolized as ∧. p: q: p∧q:

The plate is dirty. Dinner is over. The plate is dirty and dinner is over

Disjunction To join two logical statements together with the word or, we use a disjunction which is often symbolized as ∨. p∨q:

The plate is dirty or dinner is over

Conditionals The symbol used for the conditional connective is a right arrow →. Recall a conditional is in the form: If p then q. p→q:

If the plate is dirty, then dinner is over.

The connective, as a right arrow, tells us that the statement can only be read one way.

2.01 Translating logic statements mathspace.co

b “If the lemonade is sour, then the rice is not hot.”

Create a strategy

Apply the idea

The statement can be shortened as “if p then not q”, where → is “if-then” and ∼ is “not”.

In symbols, the statement is: p → ∼q

c “The lemonade is not sour, but the rice is hot.”

Create a strategy

Apply the idea

The statement can be shortened as “not p but q”, where ∼ is “not” and ∧ is “but”. Notice ∧ can mean both “and” and “but”.

In symbols, the statement is: ∼p ∧ q

d “Neither is the lemonade sour nor is the rice hot.”

Create a strategy

Apply the idea

The statement can be rephrased as “not p and not q”, where ∼ is “not” and ∧ is “and”.

In symbols, the statement is: ∼p ∧ ∼q

e “It is false that the lemonade is sour or the rice is hot.”

Create a strategy

Apply the idea

“False” negates the whole statement “p or q” so it must be grouped by parentheses.

In symbols, the statement is: ∼ ( p ∨ q)

Use ∼ for the word “false” and ∨ for the word “or”.

“It is false that if the rice is not hot, then the lemonade is sour.”

Create a strategy

Apply the idea

“False” negates the whole statement “if not q then p” so it must be grouped by parentheses.

In symbols, the statement is: ∼ (∼ q → p)

Use ∼ for the words “false” and “not”, and → for the words “if-then”.

2.01 Translating logic statements mathspace.co

Example 4 Consider the following propositions: p: q: r:

John is having roast beef for dinner. John is having Yorkshire pudding for dinner. John is having dessert for dinner.

Determine the meaning, in words, of the following compound proposition: ( p ∨ q) ∧ r

Create a strategy The expression is a conjunction of two propositions. The ﬁrst (in parentheses) is a disjunction. Remember that the disjunction is inclusive in that John could end up eating the roast beef, the Yorkshire pudding and the dessert.

Apply the idea It would read: “John is having roast beef or Yorkshire pudding, and having dessert for dinner.”

Reﬂect and check If we moved the parentheses so the compound proposition is p ∨ ( q ∧ r) it means something different This new statement would read: “John is having roast beef, or he is having Yorkshire pudding and dessert.” The placement of parenthesis may change the meaning of a statement.

Example 5 Consider the following statements: p: q:

A triangle is a right triangle. The sum of the squares of the legs is equal to the square of the hypotenuse.

Write a biconditional using those statements in both words and symbols.

Create a strategy Remember a biconditional means “if p, then q” and “if q, then p”.

Apply the idea In symbols, we can write this as p ↔ q. In words, we can. say “a triangle is a right triangle if and only if the sum of the squares of the legs is equal to the square of the hypotenuse.”

Reﬂect and check Notice, that a biconditional can be written both ways. The statement p ↔ q is exactly the same as q ↔ p.

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Idea summary We can use a variety of symbols to make writing logic arguments more eﬃcient. ∼ ∧ ∨ → ↔

Negation Conjunction Disjunction Conditional Biconditional

“not” “and” “or” “if - then” “if and only if”

Practice What do you remember? 1

For each conditional statement, identify the hypothesis and conclusion: a

If two angles are vertical, then they are congruent.

If the sum of a polygon’s interior angle measures is 180°, then the polygon is a triangle.

Two triangles are congruent if they have the same side lengths.

When two lines are parallel, the consecutive interior angles formed by a transversal are supplementary.

Determine whether each statement is conditional, biconditional, or neither. a

If two lines are parallel then they will never intersect.

Dogs are mammals.

A triangle is equilateral if and only if it is equiangular.

If a number is divisible by four then it is even.

A number is composite if and only if it has at least three factors.

Right angles measure 90°.

Consider the statement: “A ray is an angle bisector if and only if it divides an angle into two congruent angles.” State the two conditional statements that make up this biconditional statement.

Consider the two theorems: “If two sides of a triangle are congruent, then the angles opposite them are congruent.” “If two angles of a triangle are congruent, then the sides opposite them are congruent.” Write a biconditional statement combining these two theorems.

Match term to its relevant symbol. a

Not

→

And

∨

iii

∼

If… then…

↔

…if and only if…

∧

Therefore

∴

2.01 Translating logic statements mathspace.co

Complete the given table: Negation ⬚ Disjunction Conditional ⬚ ⬚ Biconditional

Read as Not p p⬚q p⬚q If p, then q p implies q p if and only q p⬚q

Symbolically ⬚ p∧q p∨q p⬚q p⬚q p⬚q p iff q

Let’s practice 7

For the given diagrams: • The hypothesis, p, is “m∠1 + m∠2 = 180°” • The conclusion, q, is “∠1 and ∠2 are supplementary.” a

Write the conditional statement p → q in words.

Write the converse statement q → p in words.

Write the biconditional statement p ↔ q in words.

1 2 Fig 1 1 2 Fig 2

Let c represent “∠A and ∠B are congruent.” Let d represent “∠A and ∠B have the same measure.” Translate the following argument into symbolic form. Statement ∠A and ∠B are congruent if and only if they have the same measure. ∠A and ∠B have different measures. Therefore, ∠A and ∠B are not congruent.

SOL

Symbolic representation

Let p represent “Two angles are complementary.” Let q represent “Two angles form a straight angle.” What is the symbolic representation of the following statement? If two angles form a straight angle, then they are not complementary. A

p→q

q→p

p → ∼q

Let • p: A zebra has a mane. • q: A dog can bark. Write each statement in symbolic form:

A zebra does not have a mane.

A zebra has a mane and a dog can bark.

A dog cannot bark or a zebra does not have a mane.

A dog cannot bark if, and only if, a zebra does not have a mane.

If a zebra does not have a mane, then a dog cannot bark.

A dog cannot bark, but a zebra has a mane.

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q → ∼p

Let • p: Dave has a laptop. • q: James has a smartphone. Write each statement in words:

∼q

p∧q

p∨q

∼p → q

∼p ↔ ∼q

∼ p∨ ∼ q

∼ ( p ∧ q)

∼ ( p ∨ q)

∼ p → (q ∨ r)

Let • p: The temperature is 90°. • q: The heater is working. • r: The apartment is cold. Write each statement in symbolic form:

The temperature is 90° and the heater is not working, and the apartment is cold.

The temperature is not 90° and the heater is working, but the apartment is cold.

The temperature is 90° and the heater is working, or the apartment is cold.

If the temperature is 90°, then the heater is working or the apartment is not cold.

Let • p: The water is 80 °F. • q: The sun is up. • r: We go snorkeling. Write each statement in words:

SOL

( p ∧ q) ∨ r

q → ( p ↔ r)

(q → p) ∨ r

(q ∧ r) → p

Let p represent “Angela bakes cookies.” Let q represent “Her children will eat cake.” Symbolically represent the following argument. p→q q→p

p → ∼p ∼p → q

p q

∴p ∴q

∴ ∼p ∴ ∼q

Statement If Angela bakes cookies, then her children will not eat cake. Angela’s children eat cake. Therefore, Angela did not bake cookies. 15

Symbolic representation

For each statement: i

Write the original statement in symbolic form and clearly identify the two premises.

Select the term(s) that best describe the statement: A Negation

Conjunction

Disjunction

Conditional

E Biconditional a

“Vincent started the bonﬁre and Valerie did not forget the marshmallows”

“It is false that if a building is up to code, then it has no windows.”

2.01 Translating logic statements mathspace.co

For each statement: i

Write the original statement in symbolic form and clearly identify the three premises.

Select the term(s) that best describe the statement: A Negation

Conjunction

Disjunction

Conditional

E Biconditional

SOL

“If the meeting is in Fairfax, then we can see a ﬁlm or we can go to the concert.”

“If a quadrilateral has four right angles and four congruent sides, then it is a square.”

“The movie theater is closed if, and only if, it is a public holiday, or it is before 7 a.m.”

Let p represent a polygon is a regular polygon Let q represent a polygon has at least one line of symmetry Determine which Venn diagram accurately represents p → q. A

Polygons

At least one line of symmetry

Regular polygons

Polygons

Regular polygons

Polygons

Regular polygons

At least one line of symmetry

Regular polygons

Let’s extend our thinking 18

Kane was given that: • p: the food contains gluten • q: the food contains peanuts • r: the food contains dairy For each statement, identify and correct his error.

∼ (q ∧ r) translates to “The food contains peanuts and dairy.”

p ∧ r translates to “The food contains gluten or dairy.”

∼ p → (q ∨ r) translates to “If food does not contain gluten, then it contains peanuts and dairy.”

Let p be “you are in Geometry class” and let q be “you are a mathematician.” i

Write each statement using symbols.

Determine if the statement is true or false.

If you are in Geometry class, then you are a mathematician.

If you are a mathematician, then you are in Geometry class.

If you are not in Geometry class, then you are not a mathematician.

If you are not a mathematician, then you are not in Geometry class.

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Let p represent “a2 + b2 = c2.” Let q represent “a + b > c.” The statement p → q is true and the statement q → p is false.

Write each statement using symbols.

Determine if the statement is true or false.

If a2 + b2 = c2, then a + b ≯ c.

If a + b ≯ c, then a2 + b2 ≠ c2.

If a + b > c, then a2 + b2 = c2.

If a2 + b2 = c2 if and only a + b > c.

The Venn diagram represents groups of tradespersons. Let p represent a tradesperson is a plumber

Tradespersons

Let q represent a tradesperson is a woodworker Let r represent a tradesperson is a carpenter Write three conditional statements in if-then form describing the relationships between the various groups of tradespersons. Give each statement in words and using symbols.

Plumbers

Wood workers Carpenters

2.01 Translating logic statements mathspace.co

2.02 Converse, inverse, and contrapositive After this lesson, you will be able to... • identify and determine the truth of the converse, inverse, and contrapositive of a conditional statement. • identify and determine the truth of a statement using the law of contrapositive. • identify counterexamples to disprove a given statement or conjecture.

Converse, inverse, and contrapositive Conjecture An unproven statement that is believed to be true but has not yet been proven. A conjecture may be true or false until proven. A conjecture is a statement that may be true or false until proven. There are many ways to relate, combine, and change conjectures that may change the truth value of statements. Inverse A statement formed by negating the hypothesis and conclusion of a conditional statement. Consider the following conditional statement and its inverse: Statement (If p then q): If an animal is a dog, then it has four legs. Inverse (If not p then not q): If an animal is not a dog, then it does not have four legs. Symbolic: If ∼ p then ∼ q Converse A statement formed by switching the hypothesis and conclusion of a conditional statement. Consider the following conditional statement and its converse: Statement (If p then q): If an animal is a dog, then it has four legs. Converse (If q then p): If an animal has four legs, then it is a dog. Symbolic: If q then p The contrapositive of the statement “If an animal is a dog, then it has four legs” is “If an animal does not have four legs, then it is not a dog.” Note that a contrapositive statement is logically equivalent to the original conditional statement. Contrapositive A statement formed by interchanging and negating the hypothesis and conclusion of a conditional statement.

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Consider the following conditional statement and its contrapositive: Statement (If p then q): If an animal is a dog, then it has four legs. Contrapositive (If not q then not p): If an animal does not have four legs, then it is not a dog. Symbolic: If ∼ q then ∼ p

Example 1 Consider the conditional statements. a “If a number ends in zero, then it is divisible by ten.” State the contrapositive of the statement.

Create a strategy

Apply the idea

The contrapositive of a statement negates both parts of a conditional statement and also switches the order.

The contrapositive of “If a number ends in zero, then it is divisible by ten” is “If a number is not divisible by ten, it does not end in zero.”

b “If c then d.” Where c and d are both conjectures. State the converse of the statement.

Create a strategy

Apply the idea

The converse of the conditional statement is formed by switching the hypothesis and conclusion.

The converse of “If c then d” is “If d then c.”

c “If b then d.” b and d are both conjectures. State the inverse of the statement.

Create a strategy The inverse of the conditional statement is formed by negating the hypothesis and conclusion.

Apply the idea

Reﬂect and check

The inverse of “If b then d” is “If ∼ b then ∼ d.”

The inverse statement is read as “if not b, then not d”.

Idea summary • • •

The converse of “if p then q” is “if q then p” and is logically independent of the original. The inverse of “if p then q” is “if ∼ p then ∼ q” and is logically independent of the original. The contrapositive of “if p then q” is “if ∼ q then ∼ p” and is logically equivalent to the original.

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Law of contrapositive and counterexamples Exploration • Truth is a property of statements (premises and conclusions). Truth is the complete accuracy of each individual statement. • Validity requires logical consistency between statements, but it does not require true statements. Validity is a property of the conclusion itself. Determine whether each statement is true and whether the conclusion can be valid. 1.

Statement 1: If I skip breakfast, then I am hungry. Statement 2: If I am hungry, I want pizza. Conclusion: Therefore, if I skip breakfast, I want pizza.

Statement 1: Only animals live on farms. Statement 2: Mary lives on a farm. Conclusion: Therefore, Mary must be an animal.

Can the statements be valid? Can the conclusions be true?

In the language of logic, there are some things that are always true: Law of contrapositive The law of contrapositive states that if p → q is true and ∼ q is true, then ∼ p is true. For example, consider the statements: p: ∼p : q: ∼q :

Two angles are vertical. Two angles are not vertical. They are congruent. They are not congruent.

Then, in words, by the law of contrapositive: If two angles are vertical, then they are congruent. ∠A ≠ ∠B, therefore ∠A and ∠B are not vertical. We can use the symbol ∴ instead of the word therefore. So, the conclusion of the law of contrapositive could be written: ∠A ≠ ∠B ∴

∠A and ∠B are not vertical

Statements can also be proven false. A counterexample is used to show a statement is false. A counterexample of a statement conﬁrms the hypothesis but negates the conclusion. Consider the statement: All triangles are equilateral. We could interpret this statement as “If a shape is a triangle, then it is equilateral.” 55°

This shape conﬁrms the hypothesis since it is a triangle, but it negates the conclusion since it is not equilateral. 35°

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Therefore, this is a counterexample which proves our original statement to be false.

Example 2 Consider the true conditional statement: If a shape is a square, then it is a rectangle. a Write the converse and determine whether it is true. Provide a counterexample if it is false.

Create a strategy The conditional is in the form “if p, then q”. Write the converse in the form “if q, then p”.

Apply the idea If a shape is a rectangle, then it is a square. This statement is false. Consider this rectangle. It is a rectangle, but it is not a square. 5m

10 m

b Write the inverse and determine whether it is true. Provide a counterexample if it is false.

Create a strategy The conditional is in the form “if p, then q”. Write the inverse in the form “if ∼ p, then ∼ q”.

Apply the idea If a shape is not a square, then it is not a rectangle. This statement is false. Consider this rectangle. It is not a square, but it is a rectangle. 5m

10 m

c Write the contrapositive and determine whether it is true. Provide a counterexample if it is false.

Create a strategy The conditional is in the form “if p, then q”. Write the contrapositive in the form “if ∼ q, then ∼ p”.

Apply the idea If a shape is not a rectangle, then it is not a square. This statement is true. Any shape that is not a rectangle cannot be a square since all squares are rectangles.

Reflect and check We could confirm this by the law of contrapositive. Since the conditional is true and ∼ q is true, the contrapositive is also true.

2.02 Converse, inverse, and contrapositive mathspace.co

Example 3 Consider the following statements p and q. Consider the conjecture: Every even number is divisible by 2. a Write the contrapositive.

Create a strategy

Apply the idea

First, rewrite the statement in the traditional “if, then” format. Then, write the contrapositive.

We can rewrite the conjecture as: “If a number is even, then it is divisible by 2.” So the contrapositive would be: “If a number is not divisible by 2, then it is not an even number.”

b Use the law of contrapositive to determine if 11 is an even number.

Create a strategy The law of contrapositive states that if p → q is true and ∼ q is true, then ∼ p is true.

Apply the idea

Reﬂect and check

Since it is true that if a number is even then it is divisible by 2, p → q is true.

We may see the word “therefore” replaced by the symbol ∴

Since 11 is not divisible by 2, we know ∼ q is true. By the law of contrapositive, ∼ p is also true.

Using this symbol, the last line could have been written: ∴ 11 is not even.

Therefore, 11 is not even.

Idea summary The law of contrapositive states that if p → q is true and ∼ q is true, then ∼ p is true. A single counterexample can prove that a statement is false.

Practice What do you remember? 1

For each pair of conditional statements, determine if they are converse, inverse, or contrapositive statements for each other: a

If two numbers are even, then their sum will be even. If the sum of two numbers is even, then the two numbers are even.

If two lines do not meet at a right angle, then they are not perpendicular. If two lines are perpendicular, then they meet at a right angle.

If two triangles are similar, they must have three congruent angles. If two triangles are not similar, they cannot have three congruent angles.

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Complete the sentences. a

b c

When a conditional ( p → q) and its converse (⬚ → ⬚) are true, the statements can be written as a ⬚

• ⬚ iff ⬚ • p if and ⬚ if q • p⬚q

A biconditional statement is the ⬚ of a conditional statement and its ⬚. For example, if p ↔ q is true, then ( p → q) ∧ (⬚ → ⬚) is true.

The law of contrapositive states that if p → q is ⬚ and ∼ q is true, then ⬚ is true.

For example, if two angles are ⬚, then the sum of the measures is 180° is true. 3

m∠A + m∠B ⬚ 180°, therefore ∠A and ∠B are not supplementary.

Suppose this statement is true.

“A quadrilateral is a parallelogram if and only if its diagonals bisect each other.” Write the two conditional statements that must be true. 4

Write the negation of each statement. a

“The ball is blue”

“The dog is not brown”

“The water is hot”

“Some turtles do not have claws”

Let’s practice 5

Which is the converse of the following statement? If 2a = b, then a = .

If 2a ≠ b, then a ≠ .

If a = , then 2a = b.

If a ≠ , then 2a ≠ b.

If 2a = b, then a ≠ .

Which is the inverse of the following statement? If the measure of an angle less than 90°, then it is an acute angle.

If the measure of an angle is not less than 90°, then it is not an acute angle.

If the measure of an angle is not less than 90°, then it is an acute angle.

If an angle is not an acute angle, then its measure is not less than 90°.

If an angle is an acute angle, then its measure is less than 90°.

The Pythagorean theorem says that “If a triangle is a right triangle, then the sum of the squares of the legs is equal to the square of the hypotenuse.” Which of the following is its converse? A

If the sum of the squares of the legs is equal to the square of the hypotenuse, then the triangle is a right triangle.

If the sum of the squares of the legs is equal to the square of the hypotenuse, then the triangle is not a right triangle.

If the sum of the squares of the legs is not equal to the square of the hypotenuse, then the triangle is a right triangle.

If the sum of the squares of the legs is not equal to the square of the hypotenuse, then the triangle is not a right triangle.

2.02 Converse, inverse, and contrapositive mathspace.co

For each of the statements, write the: i

converse

If an animal is a bird, then it has feathers.

If a dessert is listed in menu B, then it is also listed in menu C.

The square root of an even square number is also an even number.

inverse

iii

contrapositive

The conjecture a ﬁve-sided polygon is a pentagon is true. a

Write the conjecture in if-then form.

Write the contrapositive of the if-then statement.

If a polygon is not a pentagon, what conclusion can be made? Justify your answer.

All given statements are true. For each statement: i

Write the converse.

If an angle measures 90°, then it is a right angle.

If a shape has rectangle, then it is a quadrilateral.

If two angles are adjacent and supplementary, then they form a straight angle.

Is the converse is true or false?

All given statements are true. For each statement: i

Write the contrapositive.

If a number ends in zero, then it is divisible by ten.

If I have a Golden Retriever, then I have a dog.

Is the contrapositive true or false?

Is the inverse is true or false?

All given statements are true. For each statement: i

Write the inverse.

If two angles are congruent, then they have the same measure.

If I can paint a picture, then I am an artist.

Given: ∼ p → q. a

What is the converse of this statement?

What is the inverse of this statement?

What is the contrapositive of this statement?

Which value for x is a counterexample to the following argument? For all positive values of x, x3 ≥ x A

−1.0

−0.1

0.1

Determine if each argument is true or false. If it is false, give a counterexample: a

All triangles are equilateral.

All squares are rectangles.

If I have a part-time job, then I am part of the workforce.

If a person goes into a school, then they are a student.

Consider the statement “If x = 2, then x2 = 4”. a

Determine whether this statement true. If not, explain why.

Determine whether the converse of this statement is true. If not, explain why.

Determine whether the inverse of this statement is true. If not, explain why.

Determine whether the contrapositive of this statement is true. If not, explain why.

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1.0

Given statements: If a cat is in the room, then I will sneeze. I do not sneeze. a

b 18

Which is a valid conclusion that can be drawn from these statements? A There is no cat in the room.

There is a cat in the room.

C Cats are the only thing that make me sneeze.

I’m not allergic to cats.

Explain whether or not the statements and conclusions are true.

Given statements: If a shape has three sides, then it is not a quadrilateral. ABCD is a quadrilateral. a

b 19

Which is a logical conclusion from the given statements? A ABCD has three sides.

ABCD has ﬁve sides.

C ABCD does not have three sides.

ABCD is a quadrilateral.

Explain whether or not the statements and conclusions are true.

The true conditional statement, “If an animal is a shark, then it is a ﬁsh” can be represented using Venn diagram 1. a

Does Venn diagram 2 represent the inverse, converse or contrapositive of the original?

Write the conditional statement for the Venn diagram in part (a)

Determine if the conditional statement in part (b) is true or false. If false, provide a counterexample. Venn diagram 2

Venn diagram 1 Animal

Animal Fish

Not a fish

Shark

Not a shark

The true conditional statement, “If a food contains dairy, then it is not vegan” can be represented using Venn diagram 3. a

Does Venn diagram 4 represent the inverse, converse or contrapositive of the original?

Write the conditional statement for the Venn diagram in part (a)

Determine if the conditional statement in part (b) is true or false. If false, provide a counterexample. Venn diagram 4

Venn diagram 3 Food

Food Not vegan

Does not contain dairy

Contains dairy

Vegan

2.02 Converse, inverse, and contrapositive mathspace.co

Let’s extend our thinking 21

Determine whether each argument is true or false. If the argument is false, rewrite it as a true statement using the converse, inverse, or contrapositive. a

If two rectangles have equal area, then they are congruent.

If a quadrilateral is a parallelogram, then it is a rhombus.

If two polygons are similar, then their corresponding angles are congruent.

Can the statement “If m2 − 8 = m + 5, then m = 3” be combined with its converse to form a true biconditional statement? Explain your answer.

Write a conditional statement which is true but has a converse that is not true. Give an example of why the converse is not true.

Felix claims the statement “If it is sunny, then it is not winter” can be written as a true biconditional statement. Identify his error. Explain your reasoning.

Use the Venn diagram to determine whether the statement is true or false. Justify your answer. Assume that no region of the Venn diagram is empty.

If a number is a whole number, then it is also a rational number.

If a number is a rational number, then it is also a whole number.

If a number is a rational number, then it is also a decimal number.

Some rational numbers are not whole numbers.

If a number is a decimal number, then it is not a whole number.

Some numbers are both rational and decimal numbers.

Some numbers are both whole and decimal numbers.

Decimal Numbers

Whole Numbers

Determine if each statement is always, sometimes, or never true for a conditional statement p → q. Explain your answer, using examples where relevant. a

Rational Numbers

If p → q is true, then q → p, is true.

If p → q is true, then ∼ p → ∼ q, is true.

If p → q and ∼ q are true, ∼ p is also true.

If p → q is false, then the ∼ p → q is false.

If the converse is true, then the inverse is true.

If q is true, then ( p → q) ∨ (∼ p → q) is true.

If ( p → q) ∧ (q → r) is true, then p ↔ r is true.

The statements describe three true characteristics of mammals and monotremes. i

An animal that gives birth to live young is a mammal.

All mammals produce milk.

iii

Monotremes are mammals that lay eggs.

Write each statement in if-then form.

Write the converse of each of the statements in part (a). Is the converse of each statement true? Explain your reasoning.

Write a true if-then statement about mammals, that uses the given information, that is different from the ones in parts (a) and (b). Is the converse of your statement true or false? Explain your reasoning.

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2.03 Inductive reasoning After this lesson, you will be able to... • use inductive reasoning to draw conclusions from observations. • make and test conjectures using inductive reasoning. • use counterexamples to show a conjecture is false.

Inductive Reasoning Recall that a conjecture is a statement that may be true or false until proven. Conjecture An unproven statement that is believed to be true but has not yet been proven. A conjecture may be true or false until proven.

Exploration Consider the pattern:

Write a conjecture about the pattern.

Use your conjecture to draw the tenth ﬁgure in the pattern.

In the exploration, we used inductive reasoning. Inductive reasoning uses patterns and observations to write a conjecture. These conjectures are often helpful in identifying and generalizing patterns. Inductive reasoning A method of drawing conclusions from a limited set of observations. Consider the scenario: On Monday and Tuesday, after all the students sat down, the teacher started math class. We can use inductive reasoning to write the conjecture: Therefore, every day, the teacher will start class after all the students sit down. To show that a conjecture is true, you must show that it is true for all cases. However, you can show that a conjecture is false by ﬁnding just one counterexample. If, next week, there is a ﬁeld trip and once all the students sit down, they line up for the bus, this would be a counterexample to our conjecture. 2.03 Inductive reasoning mathspace.co

Example 1 Consider these appointment times: 11:00 a.m., 11:45 a.m., 12:30 p.m., … a Write a conjecture for the given pattern.

Create a strategy

Apply the idea

First, ﬁnd the pattern. We can use inductive reasoning to write a conjecture based on the pattern.

11:45 a.m. = 11:00 a.m. + 0:45 12:30 p.m. = 11:45 a.m. + 0:45 Each time is 45 minutes later than the previous time. A possible conjecture about this pattern is “There is a new appointment every 45 minutes.”

b Determine the next appointment time.

Create a strategy

Apply the idea

Use the conjecture from part (a) to predict the next time.

In part (a), we predicted a new appointment every 45 minutes. 1:15 p.m. = 12:30 p.m. + 0:45 Based on our conjecture, the next appointment will be at 1:15 p.m.

Example 2 Test each conjecture. a The sum of an odd integer and an even integer is odd.

Create a strategy First, select test cases that make the hypothesis true. Then, check to see if the conclusion is true.

Apply the idea To make the hypothesis true, we need an odd integer and an even integer. Test case 1: 3+4=7 This example confirms our hypothesis since an odd integer (3) and even integer (4) sum to an odd integer (7). Test case 2: 1+2=3 This example confirms our hypothesis since an odd integer (1) and even integer (2) sum to an odd integer (3). Test case 3: 11 + 20 = 31 This example confirms our hypothesis since an odd integer (11) and even integer (20) sum to an odd integer (31). All three of these tests confirm our conjecture.

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b The numbered day of every Wednesday in 2024 is a multiple of 7. FEBRUARY

2024

SUN MON TUE WED THU FRI SAT 1

9 16 23

10 17 24

4 11 18

5 12 19

6 13 20

7 14 21

8 15 22

Create a strategy Find the Wednesdays in 2024 which make the hypothesis true. Then, check to see if the numbered day is a multiple of 7 which would conﬁrm the conclusion.

Apply the idea The ﬁrst Wednesday in February has the numbered day “7” which is a multiple of 7. The second Wednesday in February has the numbered day “14” which is a multiple of 7. The third Wednesday in February has the numbered day “21” which is a multiple of 7. The fourth Wednesday in February has the numbered day “28” which is a multiple of 7. These all conﬁrm the conjecture that the numbered day of every Wednesday in 2024 is a multiple of 7.

Reﬂect and check Notice that testing a conjecture does not prove it to always be true. If we were to extend the calendar into March, the ﬁrst Wednesday in March has the numbered day “6” which is not a multiple of 7. This single counterexample disproves the conjecture.

Example 3 For each conjecture, ﬁnd a counterexample that shows it is false. a The square of any integer is even.

Create a strategy Find a counterexample where the square of an integer is odd.

Apply the idea 32 = 3 ⋅ 3 = 9 Since 9 is odd, this is a counterexample to the conjecture because the square of 3 is odd.

2.03 Inductive reasoning mathspace.co

b If ∠A and ∠B are complementary angles, then they share a common side.

Create a strategy

Apply the idea

Find a counterexample with two complementary angles that do not share a common side.

Draw any two angles that are not connected, but still sum to 90°. For example:

35° Angle A

Angle B

55°

Since ∠A and ∠B sum to 90°, they are complementary. Since they do not share a common side, this is a contradiction to the conjecture.

Idea summary Inductive reasoning is a method of drawing conclusions from a limited set of observations. To show that a conjecture is true, you must show that it is true for all cases. You can show that a conjecture is false by ﬁnding just one counterexample.

Practice What do you remember? 1

Match each term with its deﬁnition and an example. a

Inductive reasoning

Conjecture

Counterexample

Deﬁnitions: i

A speciﬁc case for which a conjecture is false.

A method of drawing conclusions from a limited set of observations.

iii

An unproven statement based on patterns or observations.

Examples: I

Given the pattern in the table, for term 11, there will be 2 points. Term number Number of points

6 12

7 10

8 8

9 6

10 4

For the conjecture “all roads are paved”, we ﬁnd some roads that are gravel.

III The sum of two odd numbers is even. 2

Determine if each conditional statement is true or false. If false, provide a counterexample. a

If a number is prime, then it has exactly two factors.

If a number is rational, then it can be simpliﬁed to an integer.

If shape has six sides, then it is a regular polygon.

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Use words to describe the pattern, then write the next two numbers or letters. a

1, 3, 5, 7, …

10, −8, 6, −4, …

H, G, F, E, …

A, E, I, …

Use words to describe the pattern, then draw the next two ﬁgures. a

Figure 1

Figure 2

Figure 3

b Month 1

Month 2

Month 3

Match each term with its deﬁnition: a

Prime number

One right after the other

Composite number

A number with exactly two factors

Even number

iii

The result of two numbers being added together

Odd number

A number that is divisible by two

Sum

A number with at least three factors

Consecutive

A number that is not divisible by two

Triangular number

vii A number that is the sum of the ﬁrst n whole numbers

Let’s practice 6

Write a conjecture about each pattern. a

−2, 3, 8, 13, …

1, 3, 6, 10, …

1, −1, 2, −2, …

1, 5, 13, 29, …

2 4

3 8

Write a conjecture about the given tables. a

n 1 2 3 4 Value 1 4 9 16

5 25

… …

n Value

0 1

1 2

4 16

… …

2.03 Inductive reasoning mathspace.co

Write a conjecture about the given quantities. a

The product of any two odd integers

The GCF of any two prime numbers

The sum of two odd integers

The sum of two consecutive integers

Draw Step 10 in the pattern.

For each pattern: i

Write a conjecture.

Step 1

Step 2

Step 3

Step 4

Step 2

Step 1

Step 4

Step 3

Step 1

Step 2

Step 3

Step 4

Step 6

Step 5

For each quantity: i

Write a conjecture.

The difference of two consecutive integers.

The product of two unique prime numbers.

The sign of (−2)n for whole number values of n.

The number of non-overlapping triangles that polygons with n sides can be split into using the original vertices as the triangle vertices.

Test the conjecture ﬁve times.

For each conjecture, test for n = 9 and n = 10, showing your work, and draw a conclusion based on the testing. > 1.

For a positive integer n, we have that

For an integer n, where n > 4, we have that 2n > n2.

For a positive integer n, we have that 1 + 2 + 3 + 4 + … + n =

The ﬁrst ten triangular numbers are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55. Jacques notices a pattern when ﬁnding the sums of the reciprocals of the triangular numbers.

Jacques makes the conjecture that the sum of the reciprocals of the triangular numbers approaches 2. Is Jacques’ conjecture accurate? Justify your answer.

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Determine whether you can make each conjecture from the graph regarding future elective vehicle (EV) sales? Explain your reasoning. Sold (in thousands) 30 25 20 15 10 5 Year 1

There will be an increase in EV sales in Year 8 compared to Year 7, assuming the trend continues.

Given the rate of increase in sales, EV sales might outpace the sales of traditional gasoline-powered vehicles by Year 9.

Each hexagon in this beehive is the face of a hexagonal prism with volume of 216 mm3.

Day 1

Day 2

Day 3

Day 4

Day 5

On Day 6, estimate the volume of the beehive. Explain your answer. 20

Use inductive reasoning to write a formula for the sum of the ﬁrst n positive odd integers.

A conjecture says that “for an integer a

Show that this is true for n = 1.

Rose shows that for n = 10, it is true that

Using this and laws of exponents, show that the conjecture is true for n = 11.

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2.04 Deductive reasoning After this lesson, you will be able to... • identify whether an argument uses inductive or deductive reasoning. • use deductive reasoning to draw conclusions. • use the law of syllogism, law of detachment, and law of contrapositive to make a logical argument or determine the validity of an argument. • use Venn diagrams to test the validity of an argument. • write two column and paragraph proofs. • use counterexamples to show an argument is false.

Deductive reasoning In mathematics, we have postulates which are accepted facts. If we apply deductive reasoning to postulates and other starting hypotheses, we can construct logical arguments for new facts of mathematics. Deductive reasoning A method that uses logic to draw conclusions based on definitions, postulates, and theorems.

Exploration Consider the Venn diagram: Polygons

Polygons

Parallelograms

Rhombus

Trapezoids Pair of opposite sides that are not parallel

Consider whether each conclusion is valid or not. Explain your reasoning. 1.

A polygon is a parallelogram. Therefore, it is not a trapezoid.

A polygon has two sides that are not parallel. Therefore, it is a parallelogram.

If a shape is a rhombus, then it must be a parallelogram.

There are several laws that can be used to construct an argument: Recall the Law of Contrapositive: Law of contrapositive The law of contrapositive states that if p → q is true and ∼ q is true, then ∼ p is true.

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The Law of Detachment says if the hypothesis of a true conditional statement is true, then the conclusion is also true. Law of detachment If p → q is true and p is true, then q is true. For example: p: q: p→q:

A number is odd. It is the sum of an even and odd number. If a number is odd, then it is the sum of an even and odd number.

Consider the number 11. Since p → q is true and p is true since 11 is odd, by the law of detachment, 11 is the sum of an even number and an odd number. The Law of Syllogism says if both p → q and q → r are true, then p → r is the logical conclusion. Law of syllogism If p → q is true and q → r is true, then p → r is true. For example: p→q: q→r:

If a shape has three sides, it is a triangle. If a shape is a triangle, the angles sum to 360 degrees.

Since p → q is true and q → r is true, by the law of syllogism, p → r is true which means “If a shape has three sides, the angles sum to 360 degrees”. Deductive reasoning uses facts, deﬁnitions, and the laws of logic to form an argument. Inductive reasoning uses observations and patterns to form a conjecture. Deductive reasoning can prove an argument is true, while we can only use inductive reasoning to prove a conjecture is false.

Example 1 Determine the law of logic that is used in each argument. a If the sum of the angles in a polygon is 360°, then it is a quadrilateral. You see the following shape and know it is a quadrilateral.

88°

85°

108° 79°

Create a strategy Determine whether this represents the law of contrapositive, law of detachment, or law of syllogism.

Apply the idea Let p and q represent the following statements. p q

The sum of the angles in a polygon is 360°. The shape is a quadrilateral.

Since we are told that “if the sum of the angles in a polygon is 360°, then it is a quadrilateral”, we know that p → q is true.

100

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Apply the idea Let p represent We win the game. Let q represent We win the tournament. Let r represent We go out for pizza. If we win the game, we will win the tournament. If we win the tournament, we will go out for pizza.

p→q q→r

Using the law of syllogism, we get: If we win the game, we will go out for pizza

p→r

The law of syllogism conﬁrms that the statement “If we win the game, we will go out for pizza” is true. So, the correct answer is B.

Reflect and check Notice that A and C could be true, but aren’t necessarily true. For choice A and C, we could go out for pizza even if we lose, so these do not follow from the original statements.

Example 3 Write the logical argument symbolically, then determine whether the argument is valid. Justify using deductive reasoning. If a quadrilateral has two sets of parallel sides, then it is a parallelogram. If a quadrilateral is a parallelogram, then its opposite angles are congruent. A quadrilateral does not have two sets of parallel sides. Therefore, its opposite angles are congruent.

Create a strategy Consider whether the law of syllogism, law of contrapositive, or law of detachment, or a few laws combined apply here.

Apply the idea We can start by translating to symbols to see if the argument is valid. Let p represent a quadrilateral has two set of parallel sides. Let q represent a quadrilateral is a parallelogram. Let r represent a quadrilateral has opposite angles that are congruent. Reason Given Given Law of syllogism Given Invalid conclusion

Statement If a quadrilateral has two sets of parallel sides, then it is a parallelogram. If a quadrilateral is a parallelogram, then its opposite angles are congruent. If a quadrilateral has two sets of parallel sides, then its opposite angles are congruent. A quadrilateral does not have two sets of parallel sides Therefore, its opposite angles are congruent.

Symbolic representation p→q q→r p→r ∼p ∴r

The first and second statements are true, and the third statement can be created using the law of syllogism. However, the conclusion contradicts the law of detachment, making the argument invalid and the conclusion false.

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Idea summary These laws can help determine the validity of arguments: Law of contrapositive Law of syllogism Law of detachment

If p → q is true and ∼ q is true, then ∼p is true. If p → q is true and q → r is true, then p → r is true. If p → q is true and p is true, then q is true.

Direct proof In mathematics, we have postulates which are accepted facts. If we apply deductive reasoning to postulates and other starting hypotheses, we can construct logical arguments for new facts of mathematics. A new fact is called a theorem and its argument is called its proof. Theorem

Proof

A true statement that follows as a result of other true statements.

A justification that is logically valid and based on initial assumptions, definitions, postulates, theorems, and/or properties.

We can use deductive reasoning to write proofs in a variety of ways. Formal proofs have a more rigid structure and can be helpful when we want to present our justiﬁcation in a very organized way. Paragraph proof

Two column proof

A type of proof written as in the form of a paragraph. Each statement must follow from a row above or be given. We can use the laws we know to form an argument.

A type of proof written as numbered rows which have the statement in one column and the reason in the other column. Each statement must follow from a row above or be given.

The structure of a two column proof follows: To prove: Proof goal Statements

Reasons

Statement

Given

Statement

Given

Statement

Justiﬁcation

Statement

Justiﬁcation

Statement

Justiﬁcation

Statement

Conclusion

Every statement in a proof must be show to be true. In geometry, an undefined term is a term or word that does not require further explanation or description. The include a point, set, line, and plane. A defined term is a term that has a formal definition and can be defined using other geometrical terms.

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We have learned many properties and laws that can be used to make arguments. Recall the properties of equality: Symmetric property of equality

If a = b, then b = a

Transitive property of equality

If a = b and b = c, then a = c

Addition property of equality

If a = b, then a + c = b + c

Subtraction property of equality

If a = b, then a − c = b − c

Multiplication property of equality

If a = b, then ac = bc

Division property of equality

If a = b and c ≠ 0, then

Substitution property of equality

If a = b, then b may be substituted for a in any expression

Additive identity

If a = b, then a + 0 = b and a = b + 0

Multiplicative identity

If a = b, then a ⋅ 1 = b and a = b ⋅ 1

We also have the properties of congruence: Properties of segment congruence tell us that segment congruence is reﬂexive, symmetric, and transitive Reﬂexive property of congruent segments

Any segment is congruent to itself, so

Symmetric property of congruent segments

, then

Transitive property of congruent segments

and

, then

Properties of angle congruence tell us that angle congruence is reﬂexive, symmetric, and transitive Reﬂexive property of congruent angles

Any angle is congruent to itself, so ∠A ≅ ∠A

Symmetric property of congruent angles

If ∠A ≅ ∠B, then ∠B ≅ ∠A

Transitive property of congruent angles

If ∠A ≅ ∠B and ∠B ≅ ∠C, then ∠A ≅ ∠C

Example 4 Consider the vertical angles theorem, which states that vertical angles are congruent.

Use transformations to prove the following: Given: ∠ABC and ∠DBE are vertical angles Prove: ∠ABC ≅ ∠DBE

B A

Create a strategy We can use translations, reﬂections, rotations, and dilations to demonstrate congruence between ﬁgures.

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Apply the idea

Reﬂect and check

We are given that ∠ABC and ∠DBE are vertical angles. Since ∠ABC and ∠DBE are vertical angles, they share vertex B. We can use tracing paper to rotate ∠ABC clockwise 180° around vertex B and verify that its image is ∠DBE. Therefore, ∠ABC ≅ ∠DBE.

This is an example of an informal paragraph proof that uses an annotated diagram.

B A

180° D

Example 5 Consider the congruent supplements theorem, which states that if two angles are supplementary to the same angle then they are congruent. Construct a two column proof to prove the following:

Given: • ∠A and ∠C are supplementary • ∠B and ∠C are supplementary Prove: ∠A ≅ ∠B

Apply the idea

1. 2. 3. 4. 5. 6. 7.

To prove: ∠A ≅ ∠B Statements Reasons ∠A and ∠C are supplementary Given ∠B and ∠C are supplementary Given m∠A + m∠C = 180° Definition of supplementary angles m∠B + m∠C = 180° Definition of supplementary angles m∠A + m∠C = m∠B + m∠C Transitive property of equality m∠A = m∠B Addition property of equality ∠A ≅ ∠B Definition of congruent angles

Reﬂect and check We can see that the proofs for the congruent supplements theorem and the congruent complements theorem are very similar. The ﬂow chart proof and two column proof use the same steps but display them differently.

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Idea summary Both paragraph and two column proofs are ways to use deductive reasoning to show an argument must be true. A proof shows that a conjecture is true for every case.

Practice What do you remember? 1

Determine if each statement is true or false. a

Postulates must be proved true before we can use them.

Theorems must be proved true before we can use them.

A postulate is assumed to be true.

A point is an undeﬁned term.

An angle does not have a deﬁnition, as it is considered an undeﬁned term.

In order to move from one statement to the next in a proof, the hypothesis of a theorem needs to be true, then the conclusion can be stated.

The law of detachment says that if the conclusion of a true conditional statement is true then the hypothesis is also true.

Complete each statement: a b c

The law of detachment says that, using deductive reasoning, if the ⬚ of a true conditional ⬚ is true, then the conclusion is ⬚. The law of syllogism says that, using deductive reasoning, we can draw a new conclusion from two ⬚ statements when the ⬚ of one is the hypothesis of the other. The law of contrapositive says that, that if a conditional statement is true and the negation of the conclusion is ⬚, then ⬚ is true.

State the law of logic that is used in each argument. a

If a member does not attend three consecutive meetings, then they are asked to provide a written explanation.

You did not attend the last two meetings and will not be attending the next one. You are asked to provide a written explanation. b

If you clear off the table, then you will start a new painting. If you start a new painting, then you will paint a portrait. If you clear off the table, then you will paint a portrait.

If ∠1 and ∠2 form a linear pair, then ∠1 and ∠2 are supplementary. If ∠1 and ∠2 are supplementary, then m∠1 + m∠2 = 180°. If ∠1 and ∠2 form a linear pair, then m∠1 + m∠2 = 180°.

d 4

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If a is an integer, then f (a) = 2a + 5. We have that a = −10, so f (a) = 2a + 5.

Is each statement true or false? Provide a counterexample to support your answer if the statement is false. a

For any two integers a and b, if a2 + b2 is even, then both a and b must be even.

If a ﬂight is delayed, it is always due to bad weather conditions.

If an angle has a measure of more than 90° and less than 180°, then it is obtuse.

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The process for writing a proof has been written in the wrong order. Reorder the steps correctly. 1.

Say “therefore, the proof goal is true”.

2. Write a true statement based on the given information. 3. Decide on the type of proof to do. 4. Read the given information and proof goal. 5. Use deductive reasoning to move from one true statement to another using postulates, theorems, properties, or deﬁnitions. 6

Copy and complete the table for the properties of congruent segments and angles. Congruence property ⬚

Symmetric property of congruent segments Transitive property of ⬚

Reflexive property of congruent angles ⬚ ⬚

Example and If ⬚, then ⬚

and ⬚, then . ⬚ If ∠A ≅ ∠B, then ∠B ≅ ∠A. If ∠A ≅ ∠B and ∠B ≅ ∠C, then ∠A ≅ ∠C.

Recall the properties of equality. For each of the following pairs of steps, determine which property or converse property of equality is used to get from Step 1 to Step 2: a

1 m∠A = m∠B 2

m∠A + m∠C = m∠B + m∠C

1 p+r=q+r 2

p=q

3x = 24

x=8

m∠X − m∠Z = m∠Y − m∠Z

m∠X = m∠Y

Let’s practice SOL

Which is a valid conclusion that can be drawn from these statements? If a triangle is a right triangle, then it has a 90° angle. If a triangle has a 90° angle, then it is not equilateral.

SOL

Every triangle is a right triangle.

No sides of a right triangles are congruent.

The angles of a triangle sum to 180°.

No right triangle is equilateral.

Given statements: If a shape is a rectangle, then opposite sides are congruent. A square is a rectangle. Which is a logical conclusion from the given statements? A

A square has opposite sides that are congruent.

The opposite angles of a square are congruent.

The diagonals of a square are congruent.

A square is a quadrilateral.

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Write each logical argument symbolically, then determine whether the argument is valid. Justify using deductive reasoning. a

If a triangle is equilateral, then it has three congruent sides. Triangle ABC does not have three congruent sides, therefore, triangle ABC is not an equilateral triangle.

If a quadrilateral is a rhombus, then it is a parallelogram. If a quadrilateral is a parallelogram, then it has opposite sides that are parallel. Therefore, a rhombus does not have opposite sides that are parallel

If a ray divides an angle into two congruent angles, then the ray is an angle bisector. bisector of ∠MNO. Therefore m∠MNL = m∠LNO.

If b = 5, then 7b = 35. If 7b = 35, then

If x4 = 256, then x = 4. If x = 4, then x3 = 64. x4 = 256, therefore, x3 = 64.

b = 2 . Therefore, if b = 5, then

is not an angle

b=2 .

Determine what conclusion, if any, can be drawn from the following statements. Indicate whether the law of detachment, law of syllogism, or law of contrapositive was used. a

If a parallelogram is a rhombus, then it has four congruent sides. Parallelogram ABCD does not have four congruent sides.

If a line divides a line segment into two congruent line segments, then the line is a segment bisector. divides into two congruent line segments.

If a figure is a trapezoid, then the figure is a quadrilateral. A figure is a trapezoid.

If ∣x∣ > 8, then f (x) = 2x − 3. ∣x∣ ≤ 8.

Given the Venn diagrams: Animals Fish

Animals Mammals

Has gills

Mammals Marsupials

Determine if each conclusion is valid or not.

An animal is a marsupial. Therefore, the animal is a mammal.

An animal is a marsupial. Therefore, the animal is not a ﬁsh.

An animal has gills. Therefore, the animal is not a marsupial.

An animal is a ﬁsh. Therefore, the animal is a mammal.

Determine if the argument is valid. If so, indicate the law(s) of logic used. If a figure has four sides, then it is a quadrilateral. If a figure is a quadrilateral, then the sum of the interior angles is 360°. Therefore, if the sum of the interior angles is not 360°, then the figure does not have four sides.

Laura is trying to classify a kite and shows this reasoning. If a shape is a parallelogram, then the shape has two pairs of congruent sides. A kite has two pairs of congruent sides. Therefore, using the law of detachment, a kite is a parallelogram. Identify her error.

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Vivienne and Bella are given the argument: • If you are flexible, then you do yoga. • If you do yoga, then you like green smoothies. • Therefore, if you are flexible, then you like green smoothies. Vivienne says if you are flexible, then you like green smoothies must be a true statement. Bella says it is a valid argument, but not true. Who is correct? Explain.

Decide whether inductive reasoning or deductive reasoning is used to reach the conclusion. Explain your reasoning. a

Every Saturday, your brother goes cycling. So, this coming Saturday, your brother would go cycling.

Rational numbers can be written as terminating decimals. Irrational numbers cannot be written as terminating decimals. So, 0.5 is a rational number.

Monotremes lay eggs. A platypus is a monotreme, so a platypus lays eggs.

Each time you get a high test score, you are allowed to play video games. So, the next time you get a high test score, you will be allowed to play video games.

Using the list of definitions, postulates, and theorems, complete the reasoning for each statement given the diagram. Each reason is used exactly one time. • Angle addition postulate • Linear pair postulate • Vertical angles theorem • Definition of complementary angles • Definition of supplementary angles • The measure of a right angle is 90°

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

B C A

D E

To prove: ∠APB and ∠CPD are complementary Statements Reasons ∠BPE is a straight angle Given ⬚ ∠BPC and ∠CPE are supplementary ⬚ m∠BPC + m∠CPE = 180° ⬚ m∠BPC = 90° 90° + m∠CPE = 180° Substitution m∠CPE = 90° Subtraction property of equality ⬚ m∠CPD + m∠DPE = m∠CPE m∠CPD + m∠DPE = 90° Substitution ⬚ ∠APB ≅ ∠DPE m∠APB = m∠DPE Deﬁnition of congruence m∠CPD + m∠APB = 90° Substitution ⬚ ∴ ∠APB and ∠CPD are complementary

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Copy and complete the proof of the linear pairs postulate by giving a reason for each step. Given: ∠AXB and ∠BXC are a linear pair

Prove: ∠AXB and ∠BXC are supplementary

1. 2. 3. 4. 5. 6. 19

To prove: ∠AXB and ∠BXC are supplementary Statements Reasons ⬚ ∠AXB and ∠BXC are a linear pair ⬚ ∠AXC is a straight angle ⬚ m∠AXC = 180° ⬚ m∠AXC = m∠AXB + m∠BXC ⬚ m∠AXB + m∠BXC = 180° ⬚ ∠AXB and ∠BXC are supplementary

Copy and complete the proof of the vertical angles theorem by giving the statement for each step, matching the reasons provided. Given: • •

is a straight line segment is a straight line segment

1. 2. 3. 4. 5. 6. 7.

Prove: ∠AXB ≅ ∠CXD

To prove: ∠AXB ≅ ∠CXD Statements Reasons ⬚ Given ⬚ Given ⬚ Deﬁnition of a linear pair ⬚ Deﬁnition of a linear pair ⬚ Linear pairs postulate ⬚ Linear pairs postulate ⬚ Congruent supplements theorem

Let’s extend our thinking 20

Determine if each conclusion is valid. Justify your reasoning. • Shenandoah is a National Park in Virginia. • You and your friend went hiking at Shenandoah National Park. • When you go hiking, you take pictures. • If you go on a hike, your friend goes with you. • You go on a hike. • There is a short hike to Dark Hollow Falls.

110

You went hiking in Virginia.

Your friend took pictures.

Your friend went on a hike.

You and your friend hiked to Dark Hollow Falls.

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The given two statements are true: If a shape is a regular polygon, then it has congruent sides and congruent interior angles. If a shape has congruent sides and congruent interior angles, then it is a regular polygon. a

Write a biconditional which combines the two conditional statements.

Write the biconditional statement as a deﬁnition.

For each hypothesis and conclusion, determine if it is a valid conclusion or not. i

A rhombus has congruent sides. This means that it is a regular polygon.

An equilateral triangle has congruent sides and congruent interior angles. This means that it is a regular polygon.

iii An isosceles trapezoid has two congruent angles. This means that it is a regular polygon. d 22

Explain whether, in this case, the valid conclusions in part (c) are true or not.

For the given diagram, identify all errors in logic for each of the following solutions. You do not need to correct the errors. a

AC = AB + BC

Segment addition postulate

AC = 6 + 4

Substitution

AC = 10

Evaluate the sum

A 6

101°

C 4

Priya is asked to ﬁnd m∠DBE. Her solution is: 1

m∠ABC = m∠ABD + m∠DBE

101° = 90° + m∠DBE

101° + 90° = m∠DBE

191° = m∠DBE

Angle addition postulate Substitution Converse of the addition property of equality Evaluate the sum

Construct a two column proof to prove the congruent complements theorem. Given: • ∠A and ∠C are complementary • ∠B and ∠C are complementary Prove: ∠A ≅ ∠B

Write a paragraph proof for the theorem “If two angles are congruent and supplementary, both are right angles”.

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3 Parallel & Perpendicular Lines Big ideas • Geometric figures are bound by properties that can be verified.

Chapter outline 3.01 3.02 3.03

Parallel lines and transversals Proving lines parallel Perpendicular lines

114 123 137

Same-side (consecutive) exterior angles

Alternate exterior angles

Angles that are on the exterior of two lines on the same side of the transversal.

Angles that are on the exterior of two lines on different lines and opposite sides of the transversal.

Alternate interior angles Angles that are on the interior of two lines on different lines and opposite sides of the transversal.

Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 3.01 to answer these questions. 1.

What relationship do each of the types of angle pairs have?

Uncheck the parallel lines box and drag the lines. Are the relationships stated in the previous question still true? How do you know?

Check the parallel lines box again and drag one of the parallel lines to create a translation. How can you use a translation to verify the relationships you found?

Parallel lines Two lines that never intersect. Lines are denoted as being parallel by the symbol ∥.

When lines cut by the transversal are parallel, the angle pairs created have special relationships. They will either be congruent or supplementary. Corresponding angles theorem If a transversal intersects two parallel lines, then corresponding angles are congruent.

We can use our knowledge of translations to show this theorem is true. We can imagine translating one of the angles along the traversal until it meets the second parallel line. It will match the corresponding angle exactly. 3.01 Parallel lines and transversals mathspace.co

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The corresponding angles theorem can be used as a basis for proving relationships between other angle pairs, as given in the following theorems. Consecutive interior angles theorem

Alternate interior angles theorem

If a transversal intersects two parallel lines, then same-side (consecutive) interior angles are supplementary.

If a transversal intersects two parallel lines, then alternate interior angles are congruent.

Alternate exterior angles theorem Consecutive exterior angles theorem If a transversal intersects two parallel lines, then same-side (consecutive) exterior angles are supplementary.

If a transversal intersects two parallel lines, then alternate exterior angles are congruent.

Example 1 For each of the following angle pairs, state the type of angle pair they are and state the relationship between the angles: A B D

C E

a ∠A and ∠C

Apply the idea

Reﬂect and check

The angles ∠A and ∠C are corresponding angles formed by a transversal crossing a pair of parallel lines.

The angles ∠A and ∠C are not considered same-side (consecutive) interior or exterior angles because ∠A is exterior while ∠C is interior.

By the corresponding angles theorem, they are congruent. b ∠B and ∠C

Apply the idea

Reﬂect and check

The angles ∠B and ∠C are same-side (consecutive) interior angles formed by a transversal crossing a pair of parallel lines.

Recall that supplementary angles have a sum of 180°.

By the consecutive interior angles theorem, they are supplementary.

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c ∠C and ∠D

Create a strategy

Apply the idea

Note that ∠C and ∠D are on opposite sides of the transversal, so we know they are alternating.

The angles ∠C and ∠D are alternate interior angles formed by a transversal crossing a pair of parallel lines. By the alternate interior angles theorem, they are congruent.

d ∠A and ∠E

Apply the idea

Reﬂect and check

The angles ∠A and ∠E are alternate exterior angles formed by a transversal crossing a pair of parallel lines.

∠A and ∠E would only be considered same-side (consecutive) if they were on the same side of the transversal.

By the alternate exterior angles theorem, they are congruent.

Example 2 The ﬁgure shows two intersecting pairs of parallel lines. 63°

y° x°

a Find the value of x and explain your answer.

Create a strategy

Apply the idea

We can see that angle labeled with a measure of x° forms a same-side (consecutive) exterior angle pair with the given angle. Since they lie on the transversal of two parallel lines, we can use the consecutive exterior angle theorem to relate their measures.

By the consecutive exterior angle theorem, the two angles are supplementary. This means that: x + 63 = 180 Solving this equation tells us that x = 117.

b Find the value of y and explain your answer.

Create a strategy We can see that the angles labeled with measures of x° and y° form a pair of alternate interior angles. Since they lie on the transversal of two parallel lines, we can use the alternate interior angles theorem to relate their measures.

Apply the idea

Reﬂect and check

By the alternate interior angles theorem, the angles labeled with measures of x° and y° are congruent. This means that:

When we have two intersecting pairs of parallel lines, we can use the theorems introduced in this topic to relate the measures of all the angles formed by their intersection.

x=y Using the value of x found in the previous part, this tells us that y = 117.

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Example 3 a

Determine if the information given is enough to justify the conclusion. Given: a ∥ b and ∠1 ≅ ∠3

Conclusion: ∠2 and ∠3 are supplementary

4 2

Create a strategy

Apply the idea

Use postulates and theorems about angles to determine if there is enough information to justify the conclusion.

Yes, we can conclude ∠2 and ∠3 are supplementary.

Since a ∥ b, ∠1 ≅ ∠4 by the corresponding angles theorem and ∠4 and ∠2 are supplementary by the linear pair postulate. So using the previous two statements, ∠1 and ∠2 are supplementary. Since ∠1 ≅ ∠3, and ∠1 and ∠2 are supplementary, we can conclude that ∠2 and ∠3 are supplementary.

Example 4 Construct a proof of the following: Given:

∥

Prove: m∠5 = m∠3 and m∠1 = m∠7

6 7

1 4

5 8

Create a strategy We can write a proof using the structure of a two column proof, paragraph proof, or flow chart proof.

Apply the idea

Reﬂect and check

One approach to prove the conclusion is with a ﬂow chart proof, as shown:

Notice that we just proved the alternate interior angles theorem and the alternate exterior angles theorem.

(Given) ∠1 ≅ ∠5 Corresponding angles theorem) m∠1 = m∠5 (Definition of congruent angles) m∠1 = m∠3

m∠5 = m∠7

(Vertical angles theorem)

m∠5 = m∠3

m∠1 = m∠7

(Transitive property of equality) (Transitive property of equality)

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Idea summary When a transversal cuts through two parallel lines, • • •

Corresponding angles are congruent Same-side (consecutive) interior angles and same-side (consecutive) exterior angles are supplementary Alternate interior angles and alternate exterior angles are congruent

Practice What do you remember? 1

In each of the following diagrams of parallel lines cut by a transversal: i

Name the type of angles highlighted.

State whether the two highlighted angles are congruent, complementary or supplementary.

For each of the following diagrams: i

Identify the types of angles formed by the transversal.

Describe how the highlighted angles are related to each other.

iii

Find the value of x.

x°

125°

x°

75°

125°

x°

f x°

x°

x° 107° 96° 104°

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SOL

In the ﬁgure of the parking lot shown, the lines from the parking spaces are parallel.

Curb Distance

Given that m∠3 = 51°, a

What is the value of m∠1? A 39°

45°

C 51°

129°

Determine m∠2. Explain your reasoning.

Actual Width

Offset

Review the following diagrams and statements. Determine whether the statements are correct. Explain your reasoning. a

m∠1 = 84° by the same-side (consecutive) exterior angles theorem

84°

m∠1 = 75.25° by the alternate interior angles theorem

1 104.75°

Let’s practice 5

Given the following diagrams: i

Find the value of the unknown variable(s).

Justify your answer using angle relationships.

c (−16 + 12m)° (3x − 15)° ( y + 25)°

( y + 12)° x°

105°

114°

2m°

(2x + 43)° (2x − 3)°

2(71.5 − a)° (6a − 9)°

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(3x − 4)° 113°

(3x − 7)°

(5x + 3)°

148°

(84.6 − 0.5k)° (6y)°

(x + 5)°

(5k + 12)°

In the ﬁgure, there are three pairs of parallel lines. Find the values of x, y, and z in the ﬁgure. x° 133° y°

53°

47°

z°

SOL

The ﬁgure represents a map of some major streets. Segments and are parallel to . Segments and are parallel to Segments and are parallel to . If m∠JFE = 120°, what is m∠FED? A C

60° 120°

B D

90°

150°

H F

120°

Determine if the information given is enough to justify the conclusion: a

Given: a ∥ b and ∠1 ≅ ∠9 Conclusion: ∠5 ≅ ∠9

Given: a ∥ b and ∠7 ≅ ∠11

Consider the following diagram, where Could she be correct? A

11 12 16 15

9 10 13 14



7 8

5 6

Conclusion: a ∥ c

3 4

1 2

. Sandy claims that ∠1 and ∠7 are always supplementary.

Yes, because ∠1 is supplementary to ∠2 because they form a linear pair, and ∠2 ≅ ∠7 by corresponding angles theorem.

No, because ∠1 and ∠7 are congruent by alternate exterior angles theorem. So they can only be supplementary when they are both 90°.

Yes, because ∠7 and ∠5 are congruent by vertical angles theorem, and ∠5 and ∠1 are supplementary by corresponding angles theorem.

No, because ∠1 and ∠7 are congruent by corresponding angles theorem. So they can only be supplementary when they are both 90°.

2 4

1 3 6 5 7 8

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In the given diagram, a ∥ b and ∠2 ≅ ∠3. Henry claims that ∠1 ≅ ∠8. Select the statement that proves or disproves his claim. A

There is not enough information to prove the validity of the statement. ∠1 and ∠8 are formed by different transversals and so we cannot make any valid conclusion on the relationship of the two angles. The statement is not true. The two transversals that cuts the lines a, b, and c are not parallel and so ∠1 and ∠8 can never be equal.

The statement is true. ∠1 ≅ ∠4 since they are supplementary to congruent angles. ∠4 ≅ ∠8 since they are corresponding angles formed by a transversal and two parallel lines a and b.

The statement is true. ∠1 ≅ ∠8 are alternate interior angles of parallel lines and so they must be equal.

3 4

1 2

7 8

5 6

11 12 16 15

9 10 13 14

Given m ∥ n, complete the proof that ∠2 ≅ ∠5.

1. 2. 3.

Statements ∠2 ≅ ∠3 ∠3 ≅ ∠5 ⬚

To prove: ∠2 ≅ ∠5 Reasons ⬚ ⬚ Transitive property of congruence

7 5

2 8

Let’s extend our thinking 12

Using the given diagram: a

Prove the alternate exterior angles theorem: G

∥

Given:

2 4

Prove: ∠2 ≅ ∠8 b

Prove the same-side exterior angles theorem:

1 3

6 5 7 8

∥

Given:

Prove: ∠1 and ∠8 are supplementary 13

Consider the given ﬁgure. a

Given: a ∥ b and c ∥ d Prove: ∠1 ≅ ∠3

Prove part (a) in a different way.

c a

Given:

∥

and ∠BHJ ≅ ∠HJE

Prove: ∠HBE ≅ ∠BEJ

C B

A G

H I

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D E

F J K

Using properties of parallel lines, we can relate three parallel lines given one line in common. Transitivity of parallelism

a b

If a ∥ b and b ∥ c, then a ∥ c.

Example 1 Determine whether or not there is a pair of parallel lines in the ﬁgure.

124°

46°

Create a strategy We can see that the two marked angles form a pair of consecutive (same-side) interior angles. This means that we can use the converse of consecutive interior angles theorem to check whether or not we have parallel lines.

Apply the idea If we add the measures of the consecutive (same-side) interior angles together we get: 124° + 46° = 170° Since the sum of the measures is not 180°, the two angles are not supplementary. The converse of consecutive interior angles states that the lines are parallel if the consecutive (same-side) interior angles are supplementary. Since they are not supplementary, the lines are not parallel. Therefore, there is not a pair of parallel lines in the ﬁgure.

Example 2 Find the value of x required for the ﬁgure to contain a pair of parallel lines. (5x − 8)°

(3x + 40)°

Create a strategy We can see that the two marked angles form a pair of alternate exterior angles. For there to be a pair of parallel lines, the converse of alternate exterior angles theorem tells us that the two marked angles must be congruent.

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Apply the idea We know that the two marked angles must be congruent, so we can set their measures to be equal and solve for x. 5x − 8 = 3x + 40

Deﬁnition of congruent angles

5x − 3x = 40 + 8

Add 8 and subtract 3x from both sides

2x = 48

Combine like terms

x = 24

Divide both sides by 2

Therefore, the ﬁgure contains a pair of parallel lines when x = 24.

Reflect and check For any other value of x, the equality would not be true and the lines would not be parallel.

Example 3 a

Determine if the information given is enough to justify the conclusion.

Given: ∠1 ≅ ∠3 1

Conclusion: a ∥ c

2 3

Create a strategy

Apply the idea

Note that there are no markings on the diagram to indicate any relationship between the lines.

Given that ∠1 ≅ ∠3, we could conclude that a ∥ b, using the converse of corresponding angles theorem. However, we do not have enough information to the conclude that a ∥ c. We would need some information involving ∠4 to draw a conclusion.

Example 4 Consider the given diagram: B A

C D

Construct a two column proof to prove that

∥

Create a strategy It is given that ∥ and ∠ABC is congruent to ∠CDA. Those angles are consecutive (same-side) interior, this will help us choose which theorems to use in our proof.

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Apply the idea ∥

To prove: Statements 1.

Reasons Given

∥ ∠ABC ≅ ∠CDA ∠ABC and ∠BCD are supplementary ∠CDA and ∠BCD are supplementary

2. 3. 4.

Given Consecutive interior angles theorem Substitution Converse of consecutive interior angles theorem

∥

Idea summary We can use the relationships between angles to choose which theorems we need to prove that lines are parallel.

Parallel line constructions Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 3.02 to answer these questions. 1.

Describe what is happening in each step.

We can construct a set of parallel lines using dynamic geometry software. Another way to construct parallel lines using a compass and straightedge as follows: 1. Choose a point A through which to construct a line parallel to

2. Set the compass width to the distance AC.

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3. Construct an arc centered at B with the radius AC.

4. Set the compass width to the distance AB.

5. Construct an arc centered at C with the radius AB.

6. Construct a line through A and the intersection of the two arcs. This line is parallel to

Notice, here we did not draw the full circles as was done with the dynamic geometry software. We could have drawn the full circles but it is only necessary to draw enough of the arcs of each circle to make the necessary points of intersection. Drawing a construction this way prevents creating any unnecessary points of intersection that may cause confusion. No matter which construction tools we use, parallel line constructions use the Converse of the Corresponding Angles Theorem which states, “If two lines and a transversal form corresponding angles that are congruent, then the lines are parallel.” Through the process of this construction, we are intentionally creating congruent corresponding angles which makes sure the lines will be parallel.

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Example 5 Use constructions to construct two parallel lines.

Create a strategy We can use a compass and straightedge or technology to construct parallel lines.

Apply the idea One approach for constructing parallel lines follows: 1. Use the line segment tool to construct line

and an arbitrary point C not on the line.

2. Use the point tool to add point D anywhere on line

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and use the line tool to draw line

3. Use the point tool to place point E on equal to ED centered at D.

closer to C and use the compass tool to construct a circle with radius

4. With the compass tool again, construct a circle with radius equal to ED centered at C.

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5. Use the point tool to label the intersection of circle C and intersection of circle D and

with F and use the point tool to label the

with G.

6. Select the compass tool and create a circle with radius GE centered at F and use the point tool to label the intersection of circle F and circle C with H.

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7. Use the line tool to draw a line through points C and H. This line is parallel to

Idea summary We can use various methods to construct parallel lines including technology and a compass and straightedge.

Practice What do you remember? 1

For each of the diagrams, determine whether the diagram contains parallel lines. Justify your choice with a theorem or postulate. a

b 124° 46°

67° 113°

d 79°

77°

74° 77°

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102°

112°

102°

90°

l 105° m

94° n 105°

Describe the error in the given statement.

“If ∠1 ≅ ∠2, then by the converse of the same-side interior angles theorem, l ∥ m.”

1 2

Describe the error in the given proof. ∠1 ≅ ∠2 is given. By the vertical angles theorem, ∠1 ≅ ∠3. So by the transitive property of congruence, ∠2 ≅ ∠3. By the converse of the corresponding angles postulate, l ∥ m.

3 m

Let’s practice 4

Given a ∥ b, determine y such that a ∥ c. (80x)° a (40x + 60)° b (5y)°

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Consider the diagram:

Find the value of x such that c ∥ d.

Find m∠1 such that c ∥ d. Justify your reasoning.

b c

136°

3(x + 7)° d

1 9x°

Determine the value of x that makes e ∥ f. a

e 108°

(25x − 3)°

(3x + 30)° (5x − 20)°

(21x + 45)°

e f

d 5x°

2x° (3x − 2)°

f 55°

SOL

(5x + 40)°

Lines j and k are cut by transversals m and n. (Figure is not drawn to scale.) (x + y)° m

(x + 40)°

n j

(4x − y − 120)° k

Which relationship is suﬃcient to prove j ∥ k? A

x = 40

y = 80

y = 60 − 2x

y = 140 − 2x

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Consider the diagram for the statements. Determine if the information given is enough to justify the conclusion. Explain your reasoning. a

3 4

5 6

7 8

9 10

11 12

13 14

15 16

17 18

19 20

21 22

23 24

Given:

• ∠1 and ∠14 are supplementary • ∠7 and ∠10 are supplementary • ∠22 ≅ ∠1 Conclusion: b ∥ c Conclusion: a ∥ b c

Given: d Given: • b ∥ c • ∠1 and ∠11 are supplementary • ∠5 and ∠19 are supplementary Conclusion: a ∥ c Conclusion: a ∥ b

Given:

• ∠18 and ∠20 are supplementary • a∥b Conclusion: b ∥ c SOL

• ∠1 ≅ ∠13 • ∠3 ≅ ∠24 Conclusion: a ∥ b

Lines a and b intersect lines c and d.

Which of the statements could be used to prove that a ∥ b and c ∥ d?

∠1 ≅ ∠4, ∠5 ≅ ∠6

∠1 ≅ ∠4, ∠1 and ∠5 are supplementary

∠2 ≅ ∠3, ∠3 and ∠4 are supplementary

∠1 and ∠5 are supplementary, ∠2 and ∠4 are supplementary

1 2 4

c 6 d

Consider the diagram: a

Identify the angles that are supplementary to ∠BMN.

If ∠BMN and ∠QNR are supplementary, prove

∥

. A

N S

Given:

• ∠A and ∠D are supplementary • ∠B ≅ ∠D Prove: 134

∥

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Using a ruler and compass, construct a line through a point A that is parallel to some line

C B

Let’s extend our thinking 13

Prove the converse of the same-side interior angles theorem. Given: ∠1 and ∠2 are supplementary

Prove: m ∥ n

6 8

Given:

∥

and

∥

Prove: ∠CST ≅ ∠XQP

Given:

∥

Prove:

∥

and ∠NBA and ∠ADC are supplementary P

N M

B L

A C

D T

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Given:

∥

Prove:

∥

and ∠MAB ≅ ∠BCQ P

N M

B L

A C

D T

Given: • ∠ACB and ∠CGH are supplementary • ∠EFJ ≅ ∠FJK

B m

Prove: m ∥ o n o

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F J

G K

3.03 Perpendicular lines After this lesson, you will be able to... • prove two lines are perpendicular. • construct perpendicular lines. • use perpendicular line relationships to solve problems.

Proofs and constructions of perpendicular lines Perpendicular lines

Midpoint

Two lines that intersect at right angles. Lines are denoted as being perpendicular by the symbol ⊥.

A point exactly halfway between the endpoints of a segment that divides it into two congruent line segments. B

Midpoint

To construct a perpendicular line, we can use the following method: 1. Choose two points on a given line.

2. Construct an arc centered at one of the points.

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3. Construct an arc centered at the other point, such that this arc intersects the arc constructed in Step 2 at two distinct points.

4. Plot points where the two arcs intersect. Here we have used A and A′.

A′

5. Construct the line

. It is perpendicular to the line constructed in Step 1.

A′

If we want the perpendicular line to pass through a particular point, we can adjust the radius of the compass so that both arcs pass through that point.

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Example 1 Construct a proof of the following:

Given m∠1 = m∠2 Prove: l ⊥ m

2 m

Create a strategy We can use a ﬂow chart proof, two column proof, bulleted list, or paragraph proof to prove that l ⊥ m. For this proof, we will use a two column proof.

Apply the idea To prove: l ⊥ m 1. 2. 3. 4. 5. 6.

Statements m∠1 = m∠2 ∠1 and ∠2 are supplementary m∠1 + m∠2 = 180° m∠1 + m∠1 = 180° m∠1 = 90° ∠1 is a right angle and l ⊥ m

Reasons Given Linear pair postulate Deﬁnition of supplementary angles Substitution property of equality Division property of equality Deﬁnition of perpendicular lines

Example 2 Ludek is creating a shirt design on his computer. He wants to include some perpendicular lines as part of the design, but the program he is using doesn’t measure angles so he can’t construct them directly using a right angle. The program Ludek is using can construct lines and circles accurately through or centered at points. Describe how Ludek could construct the perpendicular lines for his design.

Create a strategy One approach Ludek can use to construct perpendicular lines is by using dynamic technology software.

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Apply the idea 1. Use the ‘Line’ tool to construct a line through two points.

2. Use the ‘Circle with Center’ tool to construct a circle about each point on the line, such that the circles intersect at two distinct points.

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3. Use the ‘Line’ tool once more to construct a line through the points of intersection of the circles. This line will be perpendicular to the ﬁrst line.

Idea summary We can use the relationships between angles to prove that lines are perpendicular. We can use various methods to construct perpendicular lines including technology and a compass and straightedge.

Solve problems with perpendicular lines A particular type of transversal is one that is perpendicular to the lines that it intersects. Using the converse of corresponding angles postulate, we can determine that the two lines intersected by the transversal are parallel (because both angles in any pair of corresponding angles will measure 90° making them congruent).

When the transversal is perpendicular to the lines it intersect, we can get the following theorems relating parallel and perpendicular lines. Perpendicular transversal theorem

Converse of perpendicular transversal theorem

If a ⊥ c and a ∥ b, then b ⊥ c.

If a ⊥ c and b ⊥ c, then a ∥ b. c

c a

a b

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Example 3 Find the value of x that makes the diagram valid.

a b

(x + 38)° c

Create a strategy Using the perpendicular transversal theorem and its converse, we can conclude that the diagram is valid when a and c are perpendicular to b and d. This means that the angle of measure (x + 38) ° needs to be a right angle.

Apply the idea To ﬁnd the value of x that makes the diagram valid, we let: x + 38 = 90 Solving this equation gives us x = 52, which is when the diagram will be valid.

Example 4 Consider the given relations between lines: • a∥b • b⊥d

• c⊥d

a Determine the relationship between the lines b and c. Justify your answer.

Apply the idea Using the converse of perpendicular transversal theorem, since b ⊥ d and c ⊥ d, we have that b ∥ c.

Reﬂect and check We can use the given theorems to justify, but we can also check our reasoning by sketching a diagram.

By extending some of these lines and using other theorems to show parallelism and perpendicularity, we can reach the same results.

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b Determine the relationship between the lines a and d. Justify your answer.

Apply the idea Since parallel relations are symmetric and a ∥ b, then we can write b ∥ a. Using the perpendicular transversal theorem, since b ⊥ d and b ∥ a, we have that d ⊥ a.

Reﬂect and check Since perpendicular relations are also symmetric, we can also write the relation as a ⊥ d.

Idea summary The perpendicular transversal theorem and its converse help us draw conclusions about relationships between lines and solve problems. • •

The perpendicular transversal theorem states that if a ⊥ c and a ∥ b, then b ⊥ c. The converse of perpendicular transversal theorem states that if a ⊥ c and b ⊥ c, then a ∥ b.

Practice What do you remember? 1

Determine if each image has enough information to say the lines are perpendicular. a

−2

−1

−1 −2

90°

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Find the missing angle if a

⊥

30° x A

Complete each statement. a b

Perpendicular transversal theorem: If two lines are perpendicular and a third line is ⬚ to the ﬁrst, then the third line must also be perpendicular to the second line. For example: If a⬚c and a⬚b, then b ⊥ c. Converse of ⬚ ⬚ theorem: If two lines are perpendicular and a third line is also ⬚ to the second line, the third line must be parallel to the ﬁrst line. For example: If line a⬚c and line b ⊥ c, then line a⬚b.

The diagram shows a section of a public transit map in Broward county.

Deerfield Beach

Sample Rd

Hillsboro Beach

N. Powerline Rd

Hillsboro Blvd

Copans Rd

Atlantic Blvd

144

If Sample Rd. is parallel to Copans Rd. and Copans Rd. is parallel to Hillsboro Blvd, identify the relationship between Sample Rd. and Hillsboro Blvd.

If Atlantic Blvd is perpendicular to N. Powerline Rd. and Copans Rd. is perpendicular to N. Powerline Rd, identify the relationship between Atlantic Blvd and Copans Rd.

If Atlantic Blvd. is parallel to Copans Rd. and N. Powerline Rd. is perpendicular to Copans Rd, identify the relationship between Atlantic Blvd and N. Powerline Rd.

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Let’s practice 5

Solve for x given that a

⊥

m∠AOB = 2x + 42

m∠BOC = x − 10

Solve for x given that a

bisects ∠CEB.

m∠CEF = 12x − 15 and m∠FEB = 3x + 30

m∠CEF = 2x + 5

Determine which lines, if any, are parallel. Explain your reasoning. a

b a c

b c

c a

a b

d e

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Consider the given diagram. a

Given a ∥ b, determine y such that a ∥ c.

Determine whether or not c ⊥ d.

(9x − 6)°

(87)°

(5y + 9)° f

(8x)°

e (9y)°

d a

Consider the given diagram.

Find x. Give your answer in simplest radical form.

Determine y such that m ⊥ t.

If m ⊥ t, identify what you can conclude about the relationship between lines m and n.

2 (x2 + 2y − 5)° (2x )°

Given:

• a∥b • b⊥d • c⊥d • b⊥e • c⊥f

(x2 + 9)° d (3x + 9y + 18)° e

(2z + 7y)°

Find the values of x, y, and z.

Determine whether the statements are true or false. Explain your reasoning. a

If a ∥ b, b ∥ c, c ∥ d, then a ∥ d

If a ⊥ b, b ∥ c, c ∥ d, d ⊥ f, then a ⊥ f.

If a ∥ b, b ⊥ c, c ⊥ d, then a ∥ d.

Suppose that L1 ⊥ L2. Determine what other lines L1 is perpendicular to. Justify your answer.

Which statement describes the construction being illustrated on the square shown? A

A bisector of

A line segment congruent to

A perpendicular to

through point E, not on

A perpendicular to

through point F, on

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Construct two lines that are perpendicular using compass and straight-edge constructions.

Lutgardo is using constructions to draw perpendicular lines. Identify the step where Lutgardo made an error and explain what they should have done: 1.

Lutgardo started by constructing one of his lines.

2. Then, he chose two points on the line and constructed circles with those points as endpoints on the circles such that the circles have two distinct points of intersection. 3. The line through the points of intersection of the two circles must be perpendicular to the starting line. 16

Construct a rectangle using compass and straight-edge constructions.

Let’s extend our thinking 17

Prove the perpendicular transversal theorem: Given: m ⊥ p and m ∥ n Prove: n ⊥ p

1 m 2

Nur wants to divide a square piece of origami paper into eight equal parts, but she does not want to use measurements or make any folds. Explain and demonstrate how Nur can divide the square into eight equal parts using compass and straight edge constructions.

Describe three real-world examples of perpendicular lines.

Design a bookshelf using at least two pairs of perpendicular lines. Include lengths, units, and a sketch in your work. Justify your design using mathematics.

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4 Triangles Big ideas • Geometric figures are bound by properties that can be verified. • Relationships exist between the angles and segments of triangles and these relationships can be proven.

Chapter outline 4.01 4.02 4.03 4.04

Triangles and angles Isosceles and equilateral triangles Triangle inequalities Angle and perpendicular bisectors

150 160 170 179

4.01 Triangles and angles After this lesson, you will be able to... • solve for interior and exterior angles of a triangle, when given two angles. • solve problems, including contextual problems, involving the interior and exterior angles of a triangle.

Interior angles of triangles Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 4.01 to answer this question. 1.

What do you notice about the relationship between the angles in the triangle?

Triangle Angles Sum Theorem The sum of the measures of the interior angles of a triangle is 180°. Using the parallel postulate we know that we can construct an auxiliary line through one of the vertices of a triangle that is parallel to the opposite side. The three marked angles that have the shared vertex on the auxiliary line can be used to help us prove that the sum of the measures of the interior angles of a triangle must be 180°.

Example 1 Prove the Triangle Angles Sum Theorem.

Create a strategy To prove this theorem, construct a line

through point B that is parallel to Q

We are given △ABC with

150

∥

and need to prove m∠BAC + m∠ABC + m∠BCA = 180.

Mathspace Virginia SOL Geometry mathspace.co

Apply the idea To prove: m∠BAC + m∠ABC + m∠BCA = 180 Statements 1. 2. 3. 4. 5. 6. 7. 8.

Reasons

Given

∥ ∠QBC and ∠RBC are supplementary m∠QBC + m∠RBC = 180 m∠QBC = m∠QBA + m∠ABC m∠QBA + m∠ABC + m∠RBC = 180 ∠QBA ≅ ∠BAC

Linear angle pair Deﬁnition of supplementary angles Angle addition postulate Substitution property Alternate interior angle theorem

∠RBC ≅ ∠BCA m∠QBA = m∠BAC

Congruent angles have the same angle measure

m∠RBC = m∠BCA m∠BAC + m∠ABC + m∠BCA = 180

Substitution property of equality

Example 2 Determine the measure of the third interior angle of the triangle.

72°

58°

Create a strategy The Triangle Angles Sum Theorem tells us that the sum of the measures of the triangle will be 180°. Let the measure of the third interior angle be x° and solve for it.

Apply the idea By the Triangle Angles Sum Theorem, we have that: x + 72 + 58 = 180 x + 130 = 180 x = 50

Triangle Angles Sum Theorem Evaluate the addition Subtract 130 from both sides

So, the measure of the third interior angle of the triangle is 50°.

Reﬂect and check Since any interior angle of a triangle is supplementary to the sum of the other two interior angles, we could also use the calculation x = 180 − (72 + 58) to reach the same result.

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Example 3 Consider the diagram shown:

x°

65° 33°

y°

77°

Solve for x and y.

Create a strategy Calculate the measure of x using the triangle sum theorem, then calculate the measure of y using the Triangle Angles Sum Theorem.

Apply the idea The sum of the angles in a triangle is equal to 180°, so we have: x + 65 + 77 = 180 x + 142 = 180 x = 38

Triangle Angles Sum Theorem Combine like terms Subtract 142 from both sides

Now, we can see that the large triangle made up of both smaller triangles will also have an interior angle sum of 180°. Note that one angle of the larger triangle is ( y + 77)°. So, we have: y + 77 + 38 + 33 = 180

Triangle Angles Sum Theorem

y + 148 = 180

Combine like terms

y = 32

Subtract 148 from both sides

Reﬂect and check We could also solve for y after ﬁnding the unknown angle in the smaller bottom triangle. We know that a straight line is 180°, so we know that the 65° angle and the unknown angle that is supplementary to it must have a sum of 180°: 180° − 65° = 115° Then, we have: y + 115 + 33 = 180 y + 148 = 180 y = 32

Triangle Angles Sum Theorem Combine like terms Subtract 148 on both sides

Idea summary The Triangle Angles Sum Theorem states that the sum of the measures of the interior angles of a triangle is 180°.

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Exterior angles of triangles Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 4.01 to answer this question. 1.

What do you notice about the relationship between the interior and exterior angles?

Using the Triangle Angles Sum Theorem, we can also relate the measures of exterior angles and opposite interior angles of a triangle. Exterior angle of a polygon

Opposite interior angles

The angle outside of a polygon, between one side of the polygon and the extension of an adjacent side. This angle forms a linear pair with the interior angle it is adjacent to.

The interior angles of a polygon that are not adjacent to a given exterior angle.

Exterior Angle Theorem The measure of an exterior angle of a triangle is equal to the sum of the measures of the two opposite interior angles of the triangle.

For this triangle, the Exterior Angle Theorem tells us that:

m∠PAB = m∠B + m∠C

Example 4 Prove the Exterior Angle Theorem.

Create a strategy To prove this theorem, extend

We are given △ABC with exterior angle ∠BCD and need to prove m∠BAC + m∠ABC = m∠BCD. We can use the Triangle Angles Sum Theorem and the deﬁnition of supplementary angles to help us prove the statement.

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Apply the idea We can prove the statement by using a two column proof.

1. 2. 3. 4.

To prove: m∠BAC + m∠ABC = m∠BCD Statements Reasons m∠BAC + m∠ABC + m∠BCA = 180 Triangle Angles Sum Theorem m∠BCA + m∠BCD = 180 Linear pair postulate m∠BAC + m∠ABC + m∠BCA = m∠BCA + m∠BCD Transitive property of equality m∠BAC + m∠ABC = m∠BCD Subtraction property of equality

Reflect and check We can also use a flow chart proof to prove the statement: with exterior angle ∠BCD (Given) m∠BAC + m∠ABC + m∠BCA = 180

m∠BCA + m∠BCD = 180

(Triangle sum theorem)

(Linear pairs)

m∠BAC + m∠ABC + m∠BCA = m∠BCA + m∠BCD (Transitive property of equality)

m∠BAC + m∠ABC = m∠BCD (Subtraction property of equality)

Example 5 Determine whether or not the angle measures given in the diagram are valid. 36°

73°

Create a strategy The Exterior Angle Theorem tells us that the measure of the exterior angle should be equal to the sum of the measures of the two opposite interior angles. If this is not the case, then the diagram cannot be valid.

Apply the idea The exterior angle of the triangle has a measure of 73°. The two opposite interior angles of the triangle have measures of 36° and 39°. Adding the measures of the two remote angles together gives us: 36 + 39 = 75 ≠ 73 Since the angle measures of the ﬁgure do not satisfy the triangle Exterior Angle Theorem, they are not valid.

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39°

Reflect and check Another way to show that the figure is not valid would be to find the measure of the third interior angle of the triangle, using the Triangle Angles Sum Theorem, and then showing that it is not supplementary to the exterior angle: 36 + 39 + x = 180

Triangle Angles Sum Theorem

75 + x = 180

Combine like terms

x = 105

Subtract 75 from both sides

Since the third angle in the triangle is 105°, we should expect that it creates a linear pair with the exterior angle: 105 + 73 = 178 ≠ 180 Since the angles are not supplementary, the angle measures are not valid.

Example 6 Solve for x in the diagram shown:

C (2x + 5)°

109° (2x − 6)°

Create a strategy Relate the known angle in the diagram to the other two angles with measures given in terms of the variable x. In a triangle, the measure of the exterior angle is equal to the sum of the measures of the two opposite interior angles. We will express this relationship using an equation and solve for x.

Apply the idea We have: (2x − 6) + (2x + 5) = 109

Exterior Angle Theorem

4x − 1 = 109

Combine like terms

4x = 110

Add 1 to both sides

x = 27.5

Divide both sides by 4

Reﬂect and check Conﬁrm that the angles satisfy the Exterior Angle Theorem by substituting 27.5 in each expression for x: (2x − 6)° + (2x + 5)° = 190° [2 (27.5) − 6]° + [2 (27.5) + 5]° = 109° (55 − 6)° + (55 + 5)° = 109° 109° = 109°

Exterior Angle Theorem Substitute x = 27.5 Evaluate the multiplication Evaluate the subtraction and addition

Idea summary The Exterior Angle Theorem states that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two opposite interior angles of the triangle.

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Practice What do you remember? 1

State the sum of the measure of the interior angles of any triangle.

Using the given diagram, or otherwise, explain the sum of the interior angles of a triangle is 180°.

A 1 2 5

4 B

Find the value of x in the following diagrams: a

b 30°

x°

110° 55°

x°

70°

38°

x°

x° 73°

Find the measure of the third angle of a triangle given the measures of the other two angles: a

156

34°

36° and 94°

45° and 45°

3° and 150°

75° and x°

Consider the following diagram: a

Identify the numbered angles that are exterior angles.

∠7 is an exterior angle for a speciﬁc interior angle. Identify the described interior angle.

Describe how exterior angles 5 and 6 are related.

Identify the two interior angle measures that have the same measure as angle 5. Explain your reasoning.

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8 7

3 6 5 4

Let’s practice 6

Solve for x: a

132°

P 52°

R 127°

121°

x R

D A

B E

x C

Find the value of x in the following diagrams: a

x°

135°

b 52°

112°

x° 108°

x°

82°

68° 68°

x° x°

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Let’s extend our thinking 12

Fill in the blanks to complete the proof.

1. 2. 3. 4. 13

To prove: m∠BAC + m∠ABC = m∠PAB Statements Reasons ⬚ Triangle Angles Sum Theorem ⬚ Linear pair postulate ⬚ Transitive property of equality m∠ABC + m∠BCA = m∠PAB Subtraction property of equality

Determine whether the following statements are true or false. If false, provide a counterexample and correct the statement so it is true. If true, explain why. a

The interior angles of only right triangles sum to 180°.

The measure of an exterior angle of a triangle is equal to the sum of the two opposite interior angles of the triangle.

The sum of the measures of the exterior angles of a triangle is 180°.

In a right triangle, the right angle is 90° and the remaining two angles are always congruent.

Miriam claims that each pair of opposite interior angles in a triangle there are two possible ways to draw the exterior angles. State whether this is correct and provide a diagram to support your answer.

A triangle has two angles with measures of 37° and 85°. Find the measure of the smallest exterior angle of the triangle. Explain your reasoning.

By constructing a line through vertex A that is parallel to the side , show that the sum of angle measures in a triangle is 180° (Triangle Angles Sum Theorem). Justify your reasoning.

Consider the following diagram: a

Identify the relationship between ∠JQR and ∠JNP.

Identify the relationship between ∠MNK and ∠JNP.

Using the relationships discovered in the previous parts, ﬁnd the value of x.

x° Q

C L

155° N P

117° R

Consider the following diagram: a

Identify the relationship between ∠JHM and ∠KMH.

Using this relationship, ﬁnd the value of x.

J 66°

92°

x° K

Given: ∥ • • ∠ACB and ∠ABC are complementary • ∠DEF and ∠EFD are complementary Prove:

A F

∥

D H

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Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 4.02 to answer these questions. 1.

What is true about the orange angles?

What else do you notice about this triangle?

We can explore constructing triangles with two congruent sides or with two congruent angles, and notice the relationship between the base angles and congruent sides of an isosceles triangle. This relationship is summarized in the following theorems: Base angles theorem If two sides of a triangle are congruent, then the angles opposite them are congruent. A

In this triangle, since and are legs of the triangle, the theorem tells us that the base angles ∠ABC and ∠ACB are congruent.

Converse of base angles theorem If two angles of a triangle are congruent, then the sides opposite them are congruent. In this triangle, since ∠ABC and ∠ACB are congruent, the theorem

tells us that

≅

While we can understand these two theorems are valid using constructions or dynamic geometric software, we will also formally prove these theorems in a later lesson.

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We can apply these two theorems repeatedly to show the corollaries that follow about equilateral triangles. A corollary is a proposition inferred immediately from something that has already been proven. Corollary to the base angles theorem

Corollary to the converse of the base angles theorem

If a triangle is equilateral, then it is equiangular.

If a triangle is equiangular, then it is equilateral.

Example 1 Prove the corollary to the base angles theorem: If ABC is an equilateral triangle, prove all angles are congruent and equal to 60°.

Create a strategy Draw a diagram with the characteristics given. B

First, we need to prove that all angles are congruent. Then we can determine the measure of each angle.

Apply the idea To prove: All angles of an equilateral triangle are congruent Statements Reasons 1. 2. 3. 4. 5. 6. 7.

≅ △ABC is isosceles ∠A ≅ ∠C

Given

≅ △ABC is isosceles ∠A ≅ ∠B ∠A ≅ ∠B ≅ ∠C

Given

Deﬁnition of isosceles triangles Base angles theorem Deﬁnition of isosceles triangles Base angles theorem Transitive property of congruence

Since all three angles of an equilateral triangle are congruent, it follows from the triangle sum theorem that their sum must be 180°. Therefore, the measure of each angle in an equilateral triangle is

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= 60°.

Example 2 Is △CDE isosceles, equilateral, or neither? Justify your answer in a proof. B

45° C

F G

Create a strategy To prove △CDE isosceles or equilateral, we need to show congruent angles or congruent sides. Since we have some angle measures in the diagram, let’s try to ﬁnd the angle measures in △CDE.

Apply the idea m∠ CAB = m∠ ABC = m∠ BCA (Given)

EG ≅ FG (Given)

△ABC is equilateral and m∠CAB = m∠ABC = m∠BCA = 60° (Equiangular triangle are also equilateral)

△EFG is isosceles and ∠GEF ≅ ∠EFG (Base angles theorem)

m∠BCA = m∠DCE = 60° (Vertical angles)

m∠GEF = m∠EFG (Definition of congruence)

m∠CDE = 180° − (m∠DCE + m∠DEC) (Triangle Sum theorem)

m∠EFG = 45° (Given)

m∠CDE = 180° − (60° + 45°) = 75° (Substitution)

m∠GEF = 45° (Transitive property of equality)

m∠DCE = 60°, m∠DEC = 45°, m∠CDE = 75° (△CDE is neither isosceles nor equilateral)

m∠GEF = m∠DEC = 45° (Vertical angles)

Example 3 ≅

Consider the triangles in the diagram shown and given

Q R P

60° T

a Identify all possible pairs of congruent angles.

Create a strategy Look at angles already marked congruent and use information from the triangles to determine other congruent pairs of angles.

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Apply the idea ∠TRS ≅ ∠RST is given from the diagram. ≅ (given), we know that △QPT is isosceles, so we have∠QPT ≅ ∠QTP by the base angles theorem Since and therefore m∠QPT = m∠QTP = 60°. This would also give us: m∠PQT = 180° − ∠QPT − ∠QTP

Angle sum of △QTP

= 180° − 60° − 60°

Substitute m∠QPT = m∠QTP = 60°

= 60°

Evaluate the subtraction

We have shown △QTP is equiangular. So, we can also state that ∠QTP ≅ ∠TPQ, ∠QTP ≅ ∠PQT, and ∠TPQ ≅ ∠PQT.

b If m∠QTS = (7x + 85)°, ﬁnd the value of x.

Create a strategy ∠QTP and ∠RTS are a linear pair because they lie on m∠RTS and ﬁnd the value of x.

, we know their sum is 180°. We can use this to solve for

Apply the idea m∠QTP + m∠RTS = 180 60 + m∠RTS = 180

∠QTP and ∠RTS are a linear pair Substitution

m∠RTS = 120

Subtract 60 from both sides

7x + 85 = 120

Transitive property of equality

7x = 35

Subtract 85 from both sides

x=5

Divide both sides by 7

Example 4 Consider the polygon shown: 12.5

C 72° D

a If m∠ACD = (2x + 56)°, ﬁnd the value of x.

Create a strategy Since △ACD is an isosceles triangle, ∠ADC ≅ ∠ACD by the base angles theorem and therefore m∠ADC = m∠ACD by deﬁnition of congruency. Use the fact that m∠ACD = 72° to solve for x.

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Apply the idea m∠ACD = 72

△ACD is isosceles

2x + 56 = 72

Transitive property of equality

2x = 16

Subtract 56 from both sides

x=8

Divide by 2 on both sides

b If AD = 3y − 9.25, ﬁnd the value of y.

Create a strategy

Apply the idea

Based on the diagram, ≅ and we are given that AB = 12.5, so write and solve an equation using AD and AB.

AB = AD

Deﬁnition of congruence

12.5 = 3y − 9.25

Substitution

21.75 = 3y

Add 9.25 to both sides

7.25 = y

Divide both sides by 3

Idea summary We can use the following theorems to solve problems: • • •

The base angles theorem states that if two sides of a triangle are congruent, then the angles opposite them are congruent. The converse of base angles theorem states that if two angles of a triangle are congruent, then the sides opposite them are congruent. An equilateral triangle is a sub-class of isosceles triangles that has three equal-length sides and three 60° interior angles. It may also be referred to as an equiangular triangle.

Practice What do you remember? 1

State the relationships between the sides and angles of an: a

Isosceles triangle

Equilateral triangle

Label the parts of the given isosceles triangle. A

Determine if each statement is true or false. a

All angles in an equilateral triangle measure 60°.

All angles in a right isosceles triangle measure 90°.

Base angles in a right isosceles triangle measure 45°.

All equilateral triangles are also isosceles.

All isosceles triangles have three congruent sides.

All equilateral triangles have three congruent sides.

All equilateral triangles are equilangular.

In a triangle, angles that are opposite congruent sides must also be congruent. 4.02 Isosceles and equilateral triangles mathspace.co

165

Determine if the statements are true or false for the corresponding ﬁgure: a

This triangle has two congruent sides.

7 12

73° 73°

This triangle is equilateral.

This triangle is isosceles. 12

60°

This triangle has no congruent angles.

This triangle has no congruent sides.

62°

53°

65°

Complete the statements using the given triangle: a b c d

If ∠EAD ≅ ∠EDA, then ⬚ ≅ ⬚ If ∠BEA ≅ ∠BAE, then ⬚ ≅ ⬚ If

, then ∠⬚ ≅ ∠⬚

Let’s practice 6

Find the value of x. a

x°

82°

68°

x°

For the given ﬁgure: a

What type of triangle is the outer triangle?

Find the value of a.

Find the value of b.

47° b°

a°

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For each of the following triangles: i

Find the value of x.

Justify your answer.

x° 70°

4 x° 4

32°

d 60°

x 60°

x°

f 56°

26°

10 x°

For each of the following diagrams: i

Find x.

y°

Find y.

3x (7x + 4)°

110°

5y − 1

x°

For the given diagram: a

Find the value of a.

Find the value of b.

Find BD.

D 8 in 61°

b° A

For the given diagram: a

Find the value of x.

Find the value of m.

Identify another side with length of 15 yd.

a° C

E 15 yd 58° 19° x° A

m° C

The lengths of the two congruent legs in an isosceles triangle are 12 inches and (5x − 8) inches. Find x.

An isosceles triangle has a base angle with a measure of 73° and a vertex angle with a measure of y°. Find y.

Two sides in an equiangular triangle have lengths of (2x + 4) inches and (3x − 7) inches. Find x.

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Out of the following eight triangles, identify all possible pairs of triangles that have all sides and angles congruent. i

iii

iv 4

4 60°

62°

60°

4 56°

62°

vii viii 56°

56°

60°

60° 5

Let’s extend our thinking 16

A community garden is building a large shed to store all of the materials and tools over the winter. The given image shows the design of the framing. The collar tie, ceiling joist, and auxiliary line are all horizontal. Auxiliary line 25°

2 ft y° 6 ft

Rafters Collar tie x° Ceiling joist

7 ft

Find x and y.

Explain why the triangle with the collar tie as its base and the triangle with the ceiling joist as base are both isosceles.

If the collar tie is 3.6 ft long, find the total length of wood needed for this frame.

Describe and correct the error. Because

≅

, ∠B ≅ ∠C.

So, ∠B = 65°

65° A

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Determine whether the following statements are true or false. If true, justify your answer. If false, correct the statement. a

All equilateral triangles are acute triangles.

All isosceles triangles are acute triangles.

A triangle is acute if and only if it is equilateral.

Sketch a possible triangle for each description. You do not need to label all sides and angles. If not possible, explain why not. a

An equilateral triangle with perimeter of 21 in.

An isosceles triangle with area of 2 in2.

A right isosceles triangle with base angles of 50°.

An isosceles triangle with base angles of 100°.

An isosceles triangle with base angles of 30°.

An obtuse equilateral triangle.

Prove the corollary to the converse of the base angles theorem: that if all the angles of △ABC are congruent then △ABC is an equilateral triangle.

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We can make a connection between the relative size of each angle and its opposite side length in a triangle. The smallest angle will be opposite the shortest side, while the largest angle will be opposite the longest side of the triangle. Opposite side-angle relationship theorem

Opposite angle-side relationship theorem

If one side of a triangle is longer than another side, then the angle opposite the longer side is larger than the angle opposite the shorter side.

If one angle of a triangle is larger than another angle, then the side opposite the larger angle is longer than the side opposite the smaller angle.

C B m∠B > m∠ A

C AC > BC

In this triangle, we can see that:

AC > AB > BC Using the theorems above, we can say that: m∠B > m∠C > m∠A B

Similarly, we could observe the order of the angle measures and determine the order of the side lengths.

Corollary to the Triangle Angles Sum Theorem 1

The acute angles of a right triangle are complementary.

m∠1 + m∠2 = 90° Since the hypotenuse of a right triangle is always the longest side and is opposite the right angle, the legs must be shorter and opposite acute angles using the side-angle relationship.

Example 1 Suppose that we have three sides of lengths 4, 7 and 12. Determine if these three sides can form a valid triangle.

Create a strategy We want to check whether the lengths of the three sides satisﬁes the triangle inequality theorem or not.

Apply the idea

Reﬂect and check

Compare the sum of each pair of sides to the third side. • 4 + 7 < 12 • 4 + 12 > 7 • 7 + 12 > 4

When testing whether a set of sides satisﬁes the triangle inequality theorem or not, it is suﬃcient to only check that the combined length of the two shortest sides is greater than the longest side.

The side of length 12 is longer than the combined lengths of the other two sides, so we do not satisfy the triangle inequality theorem. The three given sides cannot form a valid triangle.

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Example 2 For the triangle in the ﬁgure, state its angles in order of ascending measure, from smallest to largest.

Create a strategy

Apply the idea

Since a longer side is opposite a larger angle, we can order the angles in ascending measure based on the order of ascending side lengths.

In ascending order, the sides lengths of the triangle are AB, AC, BC. This means that the angles, written in order of ascending measure, are ∠C, ∠B, ∠A.

Example 3 For the triangle in the ﬁgure, order the sides from shortest to longest.

A 72°

55° 52° B

Create a strategy We can use the fact that the longest side is opposite the largest angle and the shortest side is opposite the smallest angle to order the sides from shortest to longest.

Apply the idea

Reﬂect and check

Since ∠B is the smallest angle (52°), will be the shortest side. ∠C is the next smallest angle (55°), so will be the next shortest side. Finally, ∠A is the largest angle (72°), so will be the longest side.

Recall that when we are referring to the side of the triangle, we draw a line over the name of the side, . But when referring to the length of the side, we do not draw a line over the side name, AB.

This means that the sides, written in order from shortest to longest, are AC < AB < BC.

Example 4 A valid triangle has side lengths of 4, 10 and x. Find the range of values for x.

Create a strategy We can determine the upper and lower bounds for x using the inequalities from the triangle inequality theorem.

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Idea summary Triangles are valid if they satisfy the triangle inequality theorem, which states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side. The smallest angle in a triangle will be opposite the shortest side, while the largest angle will be opposite the longest side of the triangle.

Practice What do you remember? 1

For the following triangle, state the triangle inequality theorem in words and using notation.

Three possible sides of a triangle are 5, 10, and 13: a

b 3

Complete the following statements, using =, >, or <: 5 + 10 ⬚ 13, 5 + 13 ⬚ 10, 10 + 13 ⬚ 5

Determine if this is a valid triangle.

Consider the triangle in the given ﬁgure. a

→

174

47°

Greatest

Sides Angles b

97°

Complete the table to arrange the side lengths and angle measures from least to greatest. Least

36°

Describe how the angle and side in each column are related.

Determine if each statement is true or false. a

Three positive integer side lengths will form a valid triangle.

If three sides have the same whole number lengths, then it will form a valid triangle.

For an isosceles triangle with with legs of length x, the third side must satisfy the inequality 0 < third side < 2x.

A side opposite an obtuse angle will be the shortest side.

If one side of a triangle is longer than another side, then the angle opposite the longer side is larger than the angle opposite the shorter side.

If one angle of a triangle is larger than another angle, then the side opposite the larger angle is longer than the side opposite the smaller angle.

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For each of the following triangles, order the angles from smallest to largest. a

C 11 cm A

10 cm

9 cm B

△ABC where,

• AB = 50 • AB = 7 • AC = 62 • AC = 3.2 • BC = 70 • BC = 4.8 SOL

Margaret, Naomi, and Oriana are standing in a triangle playing Hacky Sack. The perimeter of △MNO, the triangle they form, is 150 in. • The distance between Margaret (M ) and Naomi (N ) is MN = 45 in. • The distance between Naomi (N ) and Oriana (O) is NO = 43 in. Which statement is true about the interior angles of △MNO?

m∠O is the greatest of the interior angles

m∠O is the least of the interior angles

m∠N is the greatest of the interior angles

m∠N is the least of the interior angles

Determine the possible range of values for x using the given ﬁgure.

B 4x − 9 x−5

2x + 7

Consider the given triangle: a

Find the value of x.

Find the measure of ∠CAB.

Find the measure of ∠ABC.

Find the measure of ∠BCA.

Which side is the longest side of the triangle?

C (6x + 10)° (7x − 2)° (8x + 4)° A

Draw the following triangles using the given measures. a

176

Side lengths Angle measures 12 24° 6 120° 8 36°

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Side lengths 8 6 10

Angle measures 53° 37°

SOL

The ﬁgure shows two triangular sections of a garden with irrigation pipes along the edges and sprinklers at each of the vertices. The common irrigation pipe, , has length x. Which inequality shows all the possible values for x? J

20 x L

22 M

A 16

1 < x < 39

1 < x < 46

2 < x < 39

2 < x < 46

For △ABC, we are given: • BC = 10 cm • m∠A = 30°

If m∠B = 80°, order the side lengths from shortest to longest.

If AC = 15 cm, can we order the angle measures from greatest to least? Explain why or why not.

If m∠C = 19°, can we order the side lengths from greatest to least? Explain why or why not.

Given two side lengths, can we order the angles from least to greatest? Explain.

Given two angle measures, can we order the angles from least to greatest? Explain.

△ABC has a perimeter of 60 cm. AB = x, BC = ordered from greatest to least.

AC, and AC =

AB. List the interior angles of △ABC,

Let’s extend our thinking 18

Bela and Nikol are debating over which angle is the biggest in the diagram. Bela thinks ∠6 has the largest measure, while Nikol is convinced that ∠2 has the largest measure. Who is correct? Explain your thinking. (The ﬁgure is not drawn to scale.) 15 cm B 1 23 cm

C 2 6

20 cm 5

19 cm 3 4

25 cm

A contractor is creating the plans for a house and a hallway has two standard doors very close by. He wants to have all of the doors open in to the hallway, as seen in the given diagram. A standard door has a width of 36 inches.

4.5 ft 16 in

The contractor has approved the plans and said that the doors will never hit each other because the triangle inequality says there is no valid triangle using those lengths. Identify and explain his error.

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4.04 Angle and perpendicular bisectors After this lesson, you will be able to... • construct the perpendicular and angle bisectors of a segment. • construct the perpendicular and angle bisectors of a triangle and use them to solve problems. • apply theorems about perpendicular and angle bisectors to solve problems involving triangles.

Perpendicular bisectors Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 4.04 to answer this question. 1.

What relationships in the diagram are always true? Can you explain why?

A segment, line, or ray that is perpendicular to a line segment at its midpoint is a perpendicular bisector. When a perpendicular bisector cuts a line segment at a right angle and into two congruent segments, we can use the perpendicular bisector theorem and the converse of the perpendicular bisector theorem to solve problems in angles and triangles. Perpendicular bisector theorem

Converse of perpendicular bisector theorem

If a point is on the perpendicular bisector of a line segment, then it is equidistant from the end points of the line segment.

If a point is equidistant from the end points of a line segment, then it is on the perpendicular bisector of that line segment.

Equidistant

The same distance from two or more objects.

S T M

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To construct the perpendicular bisector of a segment, we will: 1. Identify the segment we want to bisect. 2. Open the compass width to just past half the segment’s length and draw an arc from one endpoint that extends to both sides of the segment. 3. Without changing the compass width, draw another arc from the other endpoint that intersects the original arc on both sides of the segment. 4. Label the intersection of the arcs with points. 5. Connect the points.

Step 1

Step 2

Step 3

Step 4

Step 5

Example 1 Recall that the Pythagorean theorem states that given a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of its legs lengths. Prove the perpendicular bisector theorem.

Create a strategy ⊥

To prove this theorem, construct a segment

such that M is the midpoint of

common to both we can use Since both △AMP and △BMP are right angled triangles with AM = BM (given) and the Pythagorean theorem and substitution to ﬁnd the relationship between PA and PB. P

Apply the idea To prove: PA = PB Statements 1.

AM = BM

△AMP is a right triangle

△BMP is a right triangle

4. 5. 6. 7. 8.

180

Reasons Given

PA = AM + PM PB2 = BM 2 + PM 2 PB2 = AM 2 + PM 2 PA2 = PB2 PA = PB

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⊥ ⊥ Pythagorean theorem in △AMP Pythagorean theorem in △BMP Substitution property of equality Transitive property of equality Square root property

Example 2 Construct perpendicular bisector

through

. A

Create a strategy One approach to constructing the perpendicular bisector through

is using a compass and straightedge.

Apply the idea 1. Construct an arc centered at point A so that the arc is longer than half of

2. Construct an arc centered at point B using the same compass setting, such that this arc intersects the arc constructed in Step 1 at two distinct points. Label the two points C and D.

B D

3. Draw

. Label the point M where

. M is the midpoint of

intersects

B M

⊥

at the midpoint of

, so

is the perpendicular bisector of

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Example 3 Find the value of x. 15

2x + 3

Create a strategy

Apply the idea

We can see from the diagram that the vertical line is the perpendicular bisector because it cuts the horizontal line in half and is perpendicular to it.

Since we have a perpendicular bisector, we can apply the perpendicular bisector theorem which tells us that the lengths with measures of 15 and 2x + 3 are equal. 15 = 2x + 3

Perpendicular bisector theorem

12 = 2x

Subtract 3 from both sides

6=x

Divide both sides by 2

Idea summary A segment, line, or ray that is perpendicular to a line segment at its midpoint is a perpendicular bisector.

Angle bisectors Recall that an angle bisector is a line, segment or ray that divides an angle into two congruent angles. When an angle bisector cuts an angle into two congruent angles, we can use the angle bisector theorem and the converse of the angle bisector theorem to solve problems in angles and triangles. Angle bisector theorem

Converse of angle bisector theorem

If a point is on an angle bisector, then it is equidistant from the two sides of the angle.

If a point is in the interior of an angle and is equidistant from the sides of the angle, then it lies on the angle bisector. Incenter The point of concurrency of the angle bisectors of a triangle.

Recall, to construct the bisector of an angle, we: 1. Identify the angle we want to bisect. 2. With the compass point on the vertex of the angle, use the compass to draw an arc that intersects both legs. 3. Label the intersections with points. 4. With the compass point on one of the points from Step 3, draw an arc that passes halfway through the interior of the angle. 5. With the compass point on the other point from Step 3, draw an arc that passes halfway through the interior of the angle and intersects the ﬁrst arc. 182

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6. Label the intersection of the arcs drawn in parts 4 and 5 with a point. 7. Draw a line that connects the vertex of the angle and the point added in Step 6.

Step 1

Step 2

Step 3

Step 4

Step 5

Step 6

Step 7

Example 4 Find x. Justify your answer. x° 23°

Create a strategy

Apply the idea

The vertical line segment is an angle bisector by the converse of angle bisector theorem which states that if a point is in the interior of an angle and is equidistant from the sides of the angle, then it lies on the angle bisector. Use the theorem to determine x.

Since the vertical line segment is an angle bisector, x = 23.

Example 5 P is the incenter of the triangle.

76°

a Determine m∠BAP.

62°

Create a strategy Since P is the incenter of the triangle, it is the point of concurrency of the angle bisectors. So, we know that bisects ∠BAC, and therefore m∠BAP =

m∠BAC.

We can use the measures of the two known angles to ﬁnd the measure of ∠BAC.

Apply the idea m∠BAC + m∠ABC + m∠ACB = 180° m∠BAC + 76° + 62° = 180° m∠BAC + 138° = 180° m∠BAC = 42°

Triangle interior angle sum Substitute m∠ABC = 76° and m∠ACB = 62° Combine like terms Subtract 138° from both sides

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Then we have bisects ∠BAC

Substitute m∠BAC = 42°

Evaluate the multiplication

b Determine m∠BPC.

Create a strategy Again, since P is the incenter of the triangle, it is the point of concurrency of the angle bisectors. We can use this to ﬁnd the measures of ∠PBC and ∠PCB. and then use those to determine the measure of ∠BPC.

Apply the idea Deﬁnition of angle bisector

Substitute m∠ABC = 76°

Evaluate the multiplication

and Deﬁnition of angle bisector

Substitute angle measure

Simplify expression

Then we have m∠BPC + m∠PBC + m∠PCB = 180° m∠BPC + 38° + 31° = 180° m∠BPC + 69° = 180° m∠BPC = 111°

Triangle interior angle sum Substitute m∠BPC = 38° and m∠PCB = 31° Combine like terms Subtract 69° from both sides

Example 6 Draw a triangle and construct the inscribed circle.

Create a strategy The point of concurrency of the angle bisectors of a triangle is called the incenter because it is the center of the inscribed circle of the triangle. To find the incenter, we need to draw at least two of the three angle bisectors. Two is sufficient as the three lines are concurrent, but three would be more precise if constructing by hand. Once we have the incenter, we need to find the radius of the circle which will be the shortest (perpendicular) distance to the triangle, so we can draw a perpendicular line from the incenter to one of the sides and use that as the compass width for the inscribed circle.

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Apply the idea 1. Draw a triangle and construct one of the angle bisectors.

2. Construct at least one more angle bisector.

3. Find the point of intersection of the two angle bisectors, this is the incenter. Then construct a line that is perpendicular to one of the sides of the triangle that goes through the incenter.

4. Set the compass width to the distance between the incenter and where the perpendicular line crossed the triangle side. Draw a circle with center at the incenter.

5. Erase all construction marks.

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Example 7 A landscaper used a coordinate plane to plan the layout of a garden. The plan includes the three paths shown, that pass through the points A (30, 0), B (30, 45) and C (5, 26.25). The landscaper wants to place a fountain at an equal distance from each path.

45 40 35 30 25 20 15 10 5

y (ft) B

x (ft) A 5 10 15 20 25 30 35 40 45

−5 −5

a How could they ﬁnd the coordinates of the fountain?

Create a strategy

Apply the idea

We need to identify whether perpendicular bisectors or angle bisectors will help us solve the problem. The goal in this problem is to ﬁnd a location equal distance from each path.

The landscaper can ﬁnd the location of the fountain by considering the triangle formed by the paths and ﬁnding the intersection of the angle bisectors of the vertices of that triangle.

b What is the location of the fountain?

Create a strategy

Apply the idea

Copy the diagram and use constructions to ﬁnd the location of the fountain. We only need two angle bisectors to ﬁnd the incenter of the paths.

45 40 35 30 25 20 15 10 5 −5 −5

y (ft) B

x (ft) A 5 10 15 20 25 30 35 40 45

The fountain is located at X(20, 25) on the coordinate plane.

Idea summary The incenter is the point of concurrency of the angle bisectors of a triangle. It is called the incenter because it is the center of the inscribed circle of the triangle. The incenter will be equidistant from each side of a triangle. To construct the inscribed circle of a triangle, we can construct angle bisectors on two angles and a perpendicular bisector. Then, we can draw a circle using the incenter and the point where the perpendicular line crosses the triangle’s side as the radius.

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Practice What do you remember? 1

≅

, describe the relationship between

and

V X

Identify the line segment which is the perpendicular bisector of ∠ABC.

Complete the following statements: a

A point is on the perpendicular bisector of a line segment if and only if …

A point is on the bisector of an angle if and only if …

Identify if the stated line segment shown is the angle bisector, the perpendicular bisector, both, or neither in △ABC in the related ﬁgure:

B D

F C

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A B D

E A

is the perpendicular bisector of

and FC = 8.6 cm,

E D

Let’s practice 6

For each of the following diagrams: i

Find x.

Justify your answer.

x 9

d x° 26°

For each of the following diagrams, ﬁnd the value of the variable: a

b 5y − 7

188

2x + 25

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3y + 7

(6x)°

(8x − 8)°

4x + 9 6x + 3

is the perpendicular bisector of

and AC =

x + 1.

Solve for x given AB =

Solve for x given AB = 5.3 (3x + 55) and AC = 4.9 (x + 55).

Determine whether each of the following diagrams is valid. If not, identify the contradiction. a

38° 13

40°

d 47° 43°

For each of the given triangles: i

Classify the triangle.

Construct a perpendicular bisector through the side

iii

Construct an angle bisector for ∠C on the same ﬁgure as the perpendicular bisector.

State one thing you notice about the constructions.

B A

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189

C B

Using the constructions from question 10, or with an additional construction, justify if each statement is true or false. a

For an equilateral triangle, the perpendicular and angle bisectors are the same line.

The perpendicular and angle bisectors are always the same line.

A perpendicular bisector, if extended, will always go through the opposite vertex.

An angle bisector will always break the triangle into two congruent triangles.

An angle bisector will always break isosceles triangle into two congruent triangles.

A perpendicular bisector will always break the triangle into two triangles.

A perpendicular bisector for a right triangle will create a triangle that has the same interior angle measures as the original. ≅

Julian says that you can construct an angle bisector for an isosceles triangle △ABC where following these steps: 1.

Drawing a circle centered at B with radius BC.

2. Drawing a circle centered at C with radius BC. A

3. Drawing a line through the intersection points of the two circles.

Is Julian correct? Explain why or why not. 13

A landscaper used a coordinate plane to plan the layout of a garden. The plan includes the three paths shown that pass through the points A (30, 0), B (30, 45), and C (5, 26.25). The landscaper wants to place a fountain somewhere in the region formed by the three paths. a

Construct the angle bisector for ∠ACB.

On the same Coordinate plane, construct the angle bisector for ∠ABC.

On the same Coordinate plane, construct the angle bisector for ∠BAC.

What do you notice about the three angle bisectors?

45 40 35 30 25 20 15 10 5

y (ft) B

x (ft) A 5 10 15 20 25 30 35 40 45

−5 −5

Let’s extend our thinking 14

Justify the converse of perpendicular bisector theorem.

Consider the following information: • PA = PB • ⊥ Show that: AM = BM

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Gary, Uma and Ingrid are deciding on a location to meet up at during the holidays. a

Copy the diagram and construct perpendicular bisectors for the segments between Gary and Ingrid, and Gary and Uma. They will meet at the point of intersection of these two lines.

What is the signiﬁcance of the point of intersection of the perpendicular bisectors?

8 7 6 5 4 3 2 1

x 1 2 3 4 5 6 7 8 Ingrid

Gary Uma

Using the coordinate plane, Curtis is making plans for a stand with a triangular top for an indoor plant. He wants to drill a hole for the central column of the stand which will sit equidistant from each edge.

He tries two different constructions as he isn’t sure which will work. First he constructs two angle bisectors and notes their point of intersection. Then he constructs two perpendicular bisectors and notes their point of intersection.

Which construction should he use? Explain. Two angle bisectors and their point of intersection

Two perpendicular bisectors and their point of intersection y

D A

A x

B x

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5 Rigid Transformations Big ideas • Geometric transformations are functional relationships that can be described in terms of angles, circles, parallel lines, perpendicular lines, and line segments. • Geometry is present throughout the physical world and can be applied to solve many real-world problems.

Chapter outline 5.01 5.02 5.03 5.04 5.05

Translations Reﬂections Rotations Symmetry Sequences of transformations

194 206 218 229 242

Riccardo has been creating a tower with building blocks.

State whether it is possible to use the same transformation mapping to place block E or block N at the highest point on the tower. If yes, describe the mapping. If not, explain how you know.

S K F I C H

L M O

E Q A

Consider the ﬁgure shown, where △ABC has been translated to create △A′B′C′. The lines formed by corresponding sides of the triangles are parallel, and a transversal line cuts through these parallel lines. Use the properties of parallel lines cut by a transversal and the nature of translations to prove that the angles of △ABC are congruent to the corresponding angles of △A′B′C′. Given the translation vector and the positions of the triangles, explain how the angle relationships prove congruency of corresponding angles in triangles ABC and A′B′C′.

4 3 2 1 −4 −3 −2 −1 A′ −1

G N

C′ A

2 B′

x 4

−2 −3 −4

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Example 1 has been rotated counterclockwise about the origin. y A

3 2 1 −3 −2 −1 −1 B′

B 1

x 3

−2 −3

A′

a Describe the rotation required to map

Create a strategy We can identify the rotation that has taken place by drawing connecting segments from the origin (center of rotation), O, to point B and to point B′, and then measuring the angle.

Apply the idea

Reﬂect and check y

A 180° rotation in the clockwise direction gives the same images as a 180° rotation in the counterclockwise direction. This is why we did not need to specify the direction.

3 2 1 −3 −2 B′

−1

B 1

x 3

−1 −2

A′

−3

In this case, ∠BOB′ creates a straight line. The segment has been rotated 180°.

b Write the coordinate mapping.

Create a strategy We can compare the coordinates of the preimage and image to determine what the coordinate mapping looks like.

Apply the idea We can see that A → A′ maps (3, 3) → (−3, −3) and B → B′ maps (2, 1) → (−2, −1). Both are of the form (x, y) → (−x, −y) which corresponds to a 180° counterclockwise rotation about the origin.

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Idea summary When rotated counterclockwise about the origin, the transformation mappings are as follows: • • • •

Degree of rotation: 90° Degree of rotation: 180° Degree of rotation: 270° Degree of rotation: 360°

Coordinate mapping: (x, y) → (−y, x) Coordinate mapping: (x, y) → (−x, −y) Coordinate mapping: (x, y) → ( y, −x) Coordinate mapping: (x, y) → (x, y)

Practice What do you remember? 1

Describe the result of rotating a ﬁgure 90° clockwise about the origin.

Determine the coordinates of B′, the image produced by rotating B 90° clockwise about the origin.

5 4 3 2 1

y B

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

Determine the coordinates of B′, the image produced by rotating B 180° counterclockwise about the origin.

5 4 3 2 1

1 2 3 4 5

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

Consider the given diagram. 4

Select the algebraic mapping that transforms A to A′. (x, y) → (−x, −y)

(x, y) → (x, −y)

(x, y) → ( y, −x)

(x, y) → (−y, x)

−4 −3 −2 −1 −1 −2 −4

Mathspace Virginia SOL Geometry mathspace.co

A −3

222

A′

1 2 3 4 5

x 1

Create a strategy

Apply the idea

A shape has line symmetry if half of the ﬁgure is a reﬂection of the other.

The shape has no lines of symmetry.

A shape has point symmetry if it looks the same upside down.

The shape does have point symmetry because each point has a matching point on the opposite side of the center.

Create a strategy

Apply the idea

Determine if there is a line of reﬂection that will create two equal halves of the shape.

The shape has line symmetry.

Create a strategy

Apply the idea

Determine whether a line can be drawn that can divide the shape into two mirror images of each other or if the shape could map to itself with a 180° rotation.

This ﬁgure has neither line symmetry nor point symmetry.

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Apply the idea

Reﬂect and check

The ﬁgure has both line symmetry and point symmetry.

Shapes with point symmetry are often radially symmetric. This creates shapes that have more than just point symmetry, such as having other types of rotational symmetry or line symmetry.

Example 4 Determine whether each of the following statements is true about the quadrilateral shown.

7 6 5 4 3 2 1

−8−7−6−5−4−3−2−1 −1 −2 −3 −4 −5 −6

1 2 3 4 5 6 7

a y = 0 is a line of reﬂection.

Create a strategy A line of reﬂection splits a shape into two equal halves. If the line y = 0 divides the quadrilateral into two equal halves when drawn, it is a line of symmetry.

Apply the idea

7 6 5 4 3 2 1

The line y = 0 is a horizontal line through 0 on the y-axis.

−8−7−6−5−4−3−2−1 −1 −2 −3 −4 −5 −6

The line splits the quadrilateral into two equal halves. The line y = 0 is a line of symmetry.

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x 1 2 3 4 5 6 7

Reﬂect and check All corresponding points in the preimage and image (halves of the quadrilateral) are equidistant from the line of reﬂection. As an example, the distance from the line of symmetry to point A is 2 units, which is similar to the corresponding point B on the other side of the line of symmetry. Points C and D also have the same distance from the line of symmetry at 3 units.

2 units A

7 6 5 4 3 2 1

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

3 units x

1 2 3 4 5 6 7 D

3 units

b The quadrilateral has point symmetry.

Create a strategy A shape has point symmetry if it looks the same after being rotated 180°.

Apply the idea Remember a line is a straight angle measuring 180°. If we think of each axis as a line we can rotate the points from one side of the axis to the other to create a 180° rotation. 6

5 4 3 2 1 −8 −7 −6 −5 −4 −3 −2 −1 −1

x 1

2 3 4 5 6 7

−2 −3 −4 −5 −6

The shape does not map back to itself. The quadrilateral does not have point symmetry.

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235

Let’s practice 6

Sketch the lines of symmetry on each shape: a

How many lines of symmetry do each shape have? a

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237

SOL

Which ﬁgure appears to have exactly two lines of symmetry? A

2 1

4 3

−2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9

2 1 −5−4−3−2−1 −1

x 1 2 3 4 5 6 7 8 9

−2 −3 −4

y 5 4

−4 −3 −2 −1 −1

−2

x 1 2 3 4 5 6 7 8 9

SOL

−3

−5 −4 −3 −2 −1 −1

−4

−2

x 1 2 3 4 5

For which polygon are both x = −1 and y = −2 lines of symmetry? A

4 3

2 1 −7−6 −5 −4−3−2 −1 −1

−3 −4

2 1 −9−8−7−6−5−4−3−2−1 −1 −2

x 1 2 3 4 5 6

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6

1 2 3 4 5

−2

4 3 2 1

1 2 3 4 5

3 2 1

−3

−4 −3 −2 −1 −1

−4

−2

−5

−3

−6

−4

x 1

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Determine whether each diagram have line symmetry, point symmetry, both, or neither. a

The toy fan pictured below can spin either clockwise or counterclockwise.

Does the direction of the spin affect whether the fan has point symmetry? Explain. 15

Do all regular polygons have point symmetry? Explain your reasoning.

Consider the logo of a famous company, ‘Symmetra Corp’, known for its unique architectural designs. The logo, often used in their building facades, is shaped like a combination of two geometric ﬁgures: an equilateral triangle and a circle. The triangle is positioned such that one of its vertices points downwards, and it is inscribed within the circle. Analyze the logo and answer the following questions:

240

Identify and count the number of lines of symmetry in the logo. Draw these lines on the logo if possible.

Determine whether the logo has point symmetry, line symmetry, both, or neither. Justify your answer with a brief explanation.

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Let’s extend our thinking 17

Consider the capital letters of the English alphabet: a

Which capital letters have point symmetry?

Identify the capital letters which have line symmetry.

For the letters with line symmetry identiﬁed in the previous part: i

Which have vertical symmetry?

Which have horizontal symmetry?

iii Which have both vertical and horizontal symmetry? 18

Consider the road sign shown: a

Describe the point symmetry of the road sign.

Does the road sign have line symmetry?

Explain how symmetry relates to the concepts of reflections and rotations. Provide examples to illustrate your points. Consider including: • How line symmetry is related to reflections. • The relationship between point symmetry and rotational symmetry. • Examples from everyday objects or geometric figures.

Create your own ﬁgure or design that has both line and point symmetry.

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6 Congruence & Triangles Big ideas • Geometric transformations are functional relationships that can be described in terms of angles, circles, parallel lines, perpendicular lines, and line segments. • Congruent triangles are used to derive many geometric relationships.

Chapter outline 6.01 6.02 6.03 6.04 6.05

Congruence transformations Corresponding parts of congruent triangles SSS and SAS congruence criteria ASA and AAS congruence criteria Right triangle congruence

254 265 275 303 325

Example 1 The triangles in the diagram are congruent. B

C Q

a Identify the transformations that map one triangle to its image.

Create a strategy There are three rigid transformations that preserve length and angle measure: translations, reﬂections, and rotations. We will draw a diagram to determine which transformations can be applied to map one triangle onto the other.

C C′

A′

B′

Apply the idea

Reﬂect and check

A reflection and a translation will map the first figure onto the other.

It looks like a rotation would map the triangles onto each other, but after rotating one triangle the congruency marks do not correspond correctly.

b Complete the congruency statement: △ABC ≅ △⬚.

Create a strategy Identify the vertices that correspond to each A, B, and C, then put them in the same order.

Apply the idea Since A mapped to P, B mapped to R, and C mapped to Q, we can say △ABC ≅ △PRQ.

Reﬂect and check The order of a congruency statement matters. The vertices in the pre-image need to be in the same order as the vertices to which they map in the image.

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255

Describe the rigid transformations that could map: a

△ABC onto △DEF

△ABD onto △CDB

C B

△MNR onto △PNQ

△PSR onto △RQP

M N

S P

△GHI onto △LMN

△PQR onto △STU

P I G

U T

Triangle M is to be rotated 90° clockwise about point O, then translated 3 units left. Identify which triangle corresponds to the result of this transformation.

M V

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6.02 Corresponding parts of congruent triangles After this lesson, you will be able to... • identify corresponding parts of congruent triangles. • identify congruent triangles using their corresponding parts. • solve mathematical and real-world problems using corresponding parts of congruent triangles.

Corresponding parts of congruent triangles Exploration Consider the diagram below. △ABC ≅ △DEF. C

B F A

What sequence of rigid transformations could map △ABC to △DEF ?

What segment is the image of know?

What angle is the image of ∠B after performing motions on the pre-image? How do you know?

Each segment and each angle in △ABC maps to another segment or angle in △DEF. What can you say about the relationship between each mapped pair?

after performing rigid transformations on the pre-image? How do you

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When a figure can be mapped to an image using translations, reflections, and rotations, we can state that the triangles are congruent by definition of rigid transformations. Once we’ve established congruency between two triangles by mapping transformations, we can then justify the congruence of any pair of corresponding parts. Corresponding parts of congruent triangles theorem (CPCTC)

Rigid transformation A transformation that does not change the size or shape of a geometric ﬁgure. It is a special kind of transformation that does not change the size or shape of a ﬁgure.

Corresponding parts of congruent triangles are congruent.

Corresponding parts Parts in congruent ﬁgures that occupy the same relative location in the ﬁgures.

We can reference this theorem with its acronym, CPCTC. When the figures are oriented in the same direction, it is easier to identify the corresponding parts. If the figures have been reflected or rotated try to find a reference point (such as a labeled pair, a shared side, or a right angle) to help us identify the corresponding parts. 5

−4 −3 −2 −1

These ﬁgures are congruent by translation. The corresponding parts are all congruent:

∠A ≅ ∠X

∠B ≅ ∠Y

−1

∠C ≅ ∠Z

Y 1

Likewise, if we know all corresponding parts of a figure are congruent, then we know there is a way to map one figure to the other. This means two figures are congruent if all corresponding sides and all corresponding angles are congruent.

Example 1 Prove corresponding segments of congruent triangles are congruent.

Create a strategy Draw two triangles such that △ABC ≅ △PQR. A

C Q

We can use what we know about rigid transformations in this proof.

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b Identify the congruent parts of the triangles and write a congruency statement.

Create a strategy

Apply the idea

The corresponding parts of the triangles are marked. We can list their congruency, then write an overall congruency statement for the triangles.

• • • • ∠J ≅ ∠Q • ∠K ≅ ∠R • ∠L ≅ ∠S △JKL ≅ △QRS

Example 3 Find the values of x and y that would prove these triangles are congruent. 2x + 13

3y + 23

5x − 20

5y − 12

Create a strategy We are given a pair of corresponding, congruent angles in the diagram. Start with the sides opposite those angles as these sides must also be congruent. Next, look at the side adjacent to the angle in the direction of the side labeled with the y expression. Those sides must also be congruent.

2x + 13

3y + 23

5x − 20

5y − 12

Set the expressions equal to each other and solve for the unknown variables.

Apply the idea We have: 5x − 20 = 2x + 13

Congruent sides

5x = 2x + 33

Add 20 to both sides

3x = 33

Subtract 2x from both sides

x = 11

Divide by 3 on both sides

Next, we have: 3y + 23 = 5y – 12

268

Congruent sides

23 = 2y − 12

Subtract 2y from both sides

35 = 2y

Add 12 to both sides

17.5 = y

Divide by 2 on both sides

Mathspace Virginia SOL Geometry mathspace.co

Reﬂect and check Not all pictures are drawn to scale. When determining what sides correspond, use a reference point based on the given congruences. We can also conﬁrm that relative side lengths should still match up proportionally: the shortest side of one triangle will be the same length as the shortest side of the other. 5x − 20 = 5(11) – 20 = 35

Substitute x = 11 Evaluate the multiplication and addition

2x + 13 = 2(11) + 13 = 35

Substitute x = 11 Evaluate the multiplication and addition

Example 4 Hamida is working on a building for her architecture class ﬁnal exam. A tower in her building is constructed from congruent triangles that meet at the top of the tower’s roof.

If ∠BCD ≅ ∠BDC, and m∠BCD = 67°, determine m∠DBE.

Create a strategy We know that △BCD ≅ △BDE, so we will use what we know about △BCD to solve unknown parts of △BDE.

Apply the idea m∠BCD = m∠BDC 67 = m∠BDC m∠BCD + m∠BDC + m∠CBD = 180 67 + 67 + m∠CBD = 180 134 + m∠CBD = 180

Deﬁnition of congruence Substitution Triangle sum theorem Substitution Combine like terms

m∠CBD = 46

Subtract 134 from both sides

m∠CBD = m∠DBE

CPCTC and deﬁnition of congruence

46 = m∠DBE

Substitution

m∠DBE = 46°.

Idea summary If two triangles are congruent, all corresponding segment and angle pairs will be congruent. If all corresponding sides and all corresponding angles between a pair of ﬁgures are congruent, then the two ﬁgures will be congruent.

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Practice What do you remember? 1

State the side corresponding to a

in each pair of congruent shapes:

X Y C

B A

In each diagram, the two shapes are congruent. State the angle that corresponds to the angle labeled x. M

x D

x E

S R

c V U

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Consider the ﬁgure consisting of rectangle ABDE, its diagonal, and △BCD: a

Find the triangle congruent to △EAB.

Find the angle that corresponds to: ∠EBA

i c

∠BEA

For each pair of congruent triangles, find the value of x. a

P 4 58°

7 5

32° Q

x° 7

7 M

T C

Find the side that corresponds to: i

Given that the following two triangles are congruent, ﬁnd the length of SU.

4 Q

4 T

Write two different congruence statements using the points D, E, F.

△ABC ≅ ⬚

List all corresponding sides and angles for each congruency statement. Explain why the order of the points in the congruence statement is important.

Let’s practice 7

Consider the following triangles:

Given that LM = ST, and ∠LMK = 79°, ﬁnd the value of x.

42°

x° M

42° K

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The triangles shown in the diagrams are congruent. a

Find NM.

Find SU.

63°

28°

63°

28°

4 T

Find m∠DEF.

Find m∠NLM.

C D A 39°

84°

37°

57°

118° D

84°

7 M

5 4

Assume each pair of triangles is congruent. Find the value of x. a

E 15

L 2x − 1

11.6 G

13 x

9.5

x+6 X

272

5x − 3

2x + 5

Mathspace Virginia SOL Geometry mathspace.co

3x + 1

Bridge engineers use congruent triangles to distribute weight. The bridge in the diagram shows two congruent triangles.

Identify the corresponding parts of the congruent triangles that complete the congruency statements: i iii

b 11

∠ADB ≅ ⬚

≅ ⬚

= 15 meters, ﬁnd

∠BCD ≅ ⬚ ≅⬚

A hang glider’s design has congruent triangles.

Identify the corresponding parts of the congruent triangles that complete the congruency statements: i iii

∠RQT ≅ ⬚

≅ ⬚

= 10 feet, ﬁnd

∠RTS ≅ ⬚ ≅⬚

Let’s extend our thinking SOL

Complete the proof to show △ABC ≅ △CDA. A

To prove: △ABC ≅ △CDA Statements 1.

, ∠D ≅ ∠B

2. D

3. 4.

∠DAC ≅ ∠BCA, ∠DCA ≅ ∠BAC △ABC ≅ △CDA

Reasons Given ⬚

⬚ ⬚

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Farah and Kate are standing outside the Icon Brickell complex such that their angles of elevation to the top of the building are equal, indicated by the congruent angle markings in the diagram.

It takes Farah 20 strides to reach the base of the building, while it takes Kate 18 strides to reach the base of the building. If Kate’s stride length is 30 inches, ﬁnd Farah’s stride length. 14

274

Pegboards are used to create geometric ﬁgures with rubberbands. A triangle can be formed by connecting three pegs as shown in the image:

Determine the number triangles with different size or shape that can be formed on a 3 × 3 pegboard. Justify the uniqueness of each triangle.

Explain how you can prove that two triangles found on the pegboard are congruent.

Mathspace Virginia SOL Geometry mathspace.co

Sometimes, congruent parts are not given to us directly and instead have to be concluded from the diagram.

For example, we know that any segment is congruent to itself by the reﬂexive property of segments. We can use this fact when proving triangles congruent. In the diagram shown, segments. J

by the reﬂexive property of

Example 1 Consider the following initial step in the proof of the SSS congruency theorem, which states if the three sides of one triangle are congruent to the three sides of another triangle, then the triangles are congruent. Given: • • • Prove: △ABC ≅ △DEF A

C B F E D

Step 1. onto

There is a rigid transformation that will map

because

C A′

B C′ F

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B′ E

a Give an example of a sequence of rigid transformations that would map △ABC onto △A′B′C′ as shown in Step 1.

Apply the idea A reﬂection across a vertical line of symmetry, followed by a translation along a vector from A to A′ will map △ABC onto △A′B′C′. A

C C

A′

B C′ F D

A′

C′ F

B′ E

Reﬂect and check There exists more than one sequence of rigid transformations that could map △ABC onto △A′B′C′, such as a rotation, followed by a reﬂection, and then a translation.

b Complete the proof of SSS congruency theorem using a rigid transformation mapping.

Create a strategy We can use the converse of perpendicular bisector theorem, which states if a point is equidistant from the end points of a line segment, then it is on the perpendicular bisector of that line segment.

Apply the idea Step 2. and we have that Since the perpendicular bisector theorem.

must be the perpendicular bisector of

using the converse of

A′ C′ F

B′ E

Step 3. is the perpendicular bisector of Since on to point D by a reﬂection across .

, we have that:

. And this means that point A′ can be mapped

Step 4. Since △ABC maps to △DEF using a sequence of rigid transformations, △ABC ≅ △DEF.

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Example 2 This two-column proof shows that △DEH ≅ △FEG as seen in the diagram, but it is incomplete. Fill in the blanks to complete the proof. Given: E is the midpoint of E

To prove: △DEH ≅ △FEG Statements E is the midpoint of

1. G

Reasons Given

Given

Given ⬚ △DEH ≅ △FEG

4. 5.

⬚ ⬚

Create a strategy The best way to approach a proof is to label the given information and any information that can be concluded based on the given information. In this example, we can label because E is a midpoint. Take a look at the labeled diagram: E

We can see that these triangles have all three corresponding sides labeled as congruent so the triangles will be congruent by SSS congruence.

Apply the idea To prove: △DEH ≅ △FEG Statements 1.

E is the midpoint of

Reasons Given

Given

Deﬁnition of midpoint

△DEH ≅ △FEG

SSS congruence

Reﬂect and check For the most part, the order of the statements in a proof is up to us. Just make sure that any statements based on a given piece of information (such as the deﬁnition of midpoint in this proof) come after that given statement.

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Example 3 Find the value of x. K M

Create a strategy We know by SSS congruency that △LKM ≅ △RQS so by CPCTC, , so by the deﬁnition of congruence, KM = QS. We will use this information to write and solve an equation to ﬁnd the unknown variable.

Apply the idea

Deﬁnition of congruent segments

Substitution

Add 2 to both sides

Subtract

Multiply both sides by

from both sides

Example 4 Construct a triangle that is congruent to △ABC. Use the Side-Side-Side congurence criteria to justify your construction. B

Create a strategy To construct a triangle that is congruent to △ABC, we will use geometric constructions to make a copy of each of its sides and angles.

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Apply the idea We start by marking a point D that will become one vertex of the triangle. We can do this by using the Point tool as shown.

Next, we want to create an arc with a radius of length AC. This will ensure the side we create is congruent to To do this, we can make use of the Compass tool. Select vertices A and C ﬁrst then select point D.

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We can then use the Point tool to create a new point E on the arc of the circle we just drew. We know that the distance between points D and E is the same as the length of .

We now want to create an arc with a radius of length AB. This will give us a corresponding congruent side to

Once again, we can make use of the Compass tool.

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We won’t place a point yet because we need to make sure we place it so that the third side will be the correct length. So, ﬁrst we will create an arc with a radius of length BC using the Compass tool. This time, select vertices B and C ﬁrst to set the length. Then select point E. This will make sure the distance between point E and the new point is equal to the distance between points C and D while also making sure the distance between the new point and point D is the same as BA.

Next, we can mark the point of intersection of the two circles we have created using the Point tool.

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Finally, we can create △DFE using the Polygon tool. △DFE ≅ △ABC because all three pairs of corresponding sides are congruent.

Reﬂect and check If we want to conﬁrm that the constructed triangle is a copy of the original triangle, we can look at the side lengths of both triangles in the Algebra view.

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Example 5 F

The construction of a line perpendicular to a given line through point C on the line is shown. Write a proof using SSS. A D

Create a strategy

Apply the idea

We can use the fact that we used the same setting of the compass when drawing the last two arcs to justify that some segments are congruent. We will use SSS and

D and E lie on the same arc centered at C, so

CPCTC to prove that

and

are perpendicular.

E B

F lies on both the arc centered at D and the arc centered at E, and both arcs have the same radius. This means . F

A D

E B

is common to △DFC and △EFC. This means △DFC ≅ △EFC by SSS congruence. Using CPCTC, ∠DCF ≅ ∠ECF. Since ∠DCF and ∠ECF are also linear pair, then m∠DCF = m∠ECF = 90°. This means perpendicular to

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Example 6 The construction of an angle congruent to a given angle is shown. Write a proof to justify it. A I G

F C

H E

Create a strategy We can use the fact that we used the same setting of the compass when drawing the last two arcs to justify that some segments are congruent. We will use SSS and CPCTC to prove that ∠ABC ≅ ∠IDH.

Apply the idea G and F lie on the same arc centered at B and so

I and H lie on the same arc centered at D and so

. . Also,

Since, by construction the two arcs have the same radius, we know that construction. This means △GBF ≅ △IDH.

A I G

F C

H E

Using CPCTC, ∠GBF ≅ ∠IDH, so ∠ABC ≅ ∠IDH.

Reﬂect and check The method for constructing an angle congruent to a given angle is very similar to the method for copying a triangle using the three sides of the triangle. The difference is that when constructing an angle, we draw an isosceles triangle ﬁrst and copy that triangle.

Idea summary To show that two triangles are congruent, it is suﬃcient to demonstrate the following: •

Side-side-side, or SSS: The two triangles have three pairs of congruent sides

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SAS congruence criteria We can prove the congruence between two triangles when we are given two sides on each triangle with their included angle. Side-Angle-Side (SAS) congruency theorem

Included angle

If two sides of one triangle are congruent to two sides of another triangle, and the included angles are also congruent, then the triangles are congruent.

The angle between two sides of a polygon is known as the included angle of those two sides. A B

C D

In the diagram shown, △ABC ≅ △DEF by SAS congruency. F

E A

We can use the fact that vertical angles are congruent by the vertical angles theorem to help us prove triangles congruent.

In the diagram shown, ∠ACB ≅ ∠DCE by Vertical angle theorem.

Example 7 Prove the SAS congruency theorem using rigid transformations with the given diagram. Given:

• • • ∠BAC ≅ ∠EDF

Prove: △ABC ≅ △DEF D

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Apply the idea 1. There is a rigid transformation that will map onto because . First, we can rotate the triangle to get until . Then we can translate up or down and left or right until is on top of . Call this triangle △A′B′C′. If C′ and F are on the same side, reﬂect over until the ﬁgure is as shown. A

C′

B B′

A′ D

2. Since these transformations are all rigid transformations, we have that: and ∠BAC ≅ ∠B′A′C′ 3. Using the given information and the transitive property of congruence, we have that: and ∠B′A′C′ ≅ ∠EDF and ∠B′A′C′ ≅ ∠EDF we have that must be the angle bisector of ∠C′DF using the 4. Since deﬁnition of the angle bisector theorem since it breaks the angle into two congruent angles. C′

B′

A′

5. Since

is the angle bisector of ∠C′DF, we can use it as a line of reﬂection to map C′ onto F.

coincides with 6. Angle measures are preserved in reﬂections. We know that congruent, point C′ can be mapped on to point F by reﬂection across .

and since these segments are

7. We have now shown that: • A maps to D using rigid transformations • B maps to E using rigid transformations • C maps to F using rigid transformations So, we have that △ABC can be mapped onto △DEF, so △ABC ≅ △DEF.

Reﬂect and check Another approach for a rigid transformation transformation that could map diagonal line followed by a rotation and a translation.

onto

could be a reﬂection across a

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Example 8 Consider the triangles shown: A

R M

C E

a Identify the additional information needed to prove these triangles congruent by SAS congruence.

Create a strategy From the diagram we know that and . If we want these triangles to be congruent by SAS we will need to identify the corresponding angles that complete the congruency theorem.

Apply the idea

Reﬂect and check

∠R ≅ ∠A

Be sure that the angle identiﬁed is in between the given congruent sides.

b Suppose that the triangles are congruent by SAS and that ∠R = 57° and ∠A =

Create a strategy

. Solve for x.

Apply the idea

If the triangles are congruent by SAS, we know that ∠R ≅ ∠A and therefore m∠R = m∠A by the deﬁnition of congruence. We will use this to solve for x.

Deﬁnition of congruent angles

Substitution

Multiply both sides by 2

Subtract 90 from both sides

Divide both sides by 3

Example 9 Complete the proof of the base angles theorem. B

288

Start with an isosceles triangle where ∠ABC has been bisected by

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To prove: ∠A ≅ ∠C Statements

Reasons

Given

Reﬂexive property

3. 4. 5.

∠ABD ≅ ∠CBD △ABD ≅ △CBD ⬚

⬚ ⬚ CPCTC

Create a strategy Start by drawing any given information from the proof onto the diagram. We know that ∠ABD ≅ ∠CBD, so we can label it appropriately.

Now that we see a shared angle between two congruent corresponding pairs of side lengths, we will use SAS to prove congruency between the triangles and prove the base angles theorem.

Apply the idea To prove: ∠A ≅ ∠C Statements

Reasons

Given

Reﬂexive property

∠ABD ≅ ∠CBD

4. 5.

△ABD ≅ △CBD ∠BAD ≅ ∠BCD

is angle bisector of ∠ABC SAS congruency theorem CPCTC

Example 10 Using the truss bridge shown, the steel beam that makes up the base of the bridge is divided into 4 segments of equal length. The beams that appear horizontal and vertical in the diagram are perpendicular to one another.

Identify two triangles from the braces of the bridge that are congruent by naming their vertices and stating the correspondence.

Apply the idea △ABC ≅ △EBC because is the same in both triangles. We are given that . is a perpendicular bisector of since it is vertical to the base of the bridge, so m∠BCA = m∠BCE = 90°. Since ∠BCA is the shared angle of and , and ∠BCE is the shared angle of and , the triangles are congruent by SAS.

Reﬂect and check We could use the same justiﬁcation for △EFG and △HFG.

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Is there enough information to determine whether the triangles are congruent? If not, list a pair of congruent parts that would guarantee triangle congruence and name the postulate that justiﬁes it. a

A B

E A C

C D B

E M H

Let’s practice 4

For each of the following pairs of triangles, identify a congruence theorem or postulate that justiﬁes the congruence. If a theorem or postulate does not exist, determine whether the triangles are not congruent or if there is not enough information to justify the congruence. a

A C

84°

D 7

5 4

R S

8 9

84°

9 T

8 5 L

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SOL

What value of x makes △STW ≅ △XYZ? Z

T 6x + 3

A 6

6x − 7

4x + 7

2x + 15

8x − 5

5x − 4

Find the value of the variable(s) that would result in suﬃcient criteria to prove the two triangles are congruent. a

8 R

B 2.5 in

1.5 in

40°

60° 80°

2 in

2.5 in

x°

1.5 in K E

d 4m

30°

(2x + 8)°

(3x + 2)

12 in

4m (x + 12) in

12 in

(2y + 10)°

30°

3x − 54

Triangle A

Triangle B

7 5

Triangle C

Triangle D

4 9

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Which two triangles are congruent?

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Which congruence test is satisﬁed?

Triangle A

Triangle B

Triangle C

27°

Triangle D

27°

6 27° 5

a 9

Which two triangles are congruent?

Which congruence test is satisﬁed?

Use the given diagram to: i

State which two triangles are congruent.

State the congruence postulate the triangles satisfy.

6 R

E 5

If two line segments bisect each other, and their endpoints are joined, will a pair of congruent triangles always be formed? Explain.

Which congruence test do △ABD and △CDB satisfy?

D SOL

Given:

•

⊥

•

⊥

O P

Prove: △MPL ∼ △NPO

Use the options to fill in the blanks and complete the proof:

To prove: △MPL ∼ △NPO Statements Reasons ⊥

∥

2. 3. 4. A

Given

⊥

⬚

△MPL ∼ △NPO

∠OPN ≅ ∠MPL;

If two lines are perpendicular to a third line, then the two lines are parallel. If two parallel lines are cut by a transversal, alternate interior angles are congruent. ⬚

∠NML ≅ ∠LON;

∠NOL ≅ ∠MLO;

∠OLM ≅ ∠MNO ∠MLO ≅ ∠ONM ∠NML ≅ ∠MNO D

Angle-Angle (AA) Postulate

Side-Angle-Side (SAS) Postulate

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In the following ﬁgure EF is the perpendicular bisector of AD.

A′

Given: C′ F

• •

B′

Prove: △A′B′C′ ≅ △DEF by the Side-Angle-Side (SAS) congruency theorem.

E D

In the following ﬁgure,

Given:

• Prove: △ABC ≅ △CDA by the Side-Angle-Side (SAS) congruency theorem.

Use the diagram to prove that:

△SPQ ≅ △SRQ P X R

Given:

• E is the midpoint of • • H

Prove: △DEH ≅ △FEG 17

Construct a triangle that is congruent to △EFG using the SAS congruence theorem. Use a compass and straight edge.

Construct a triangle that is congruent to △QRS using SSS congruence theorem. a

Given: • ∠A ≅ ∠P • • Is there enough information to conclude △ABC ≅ △PQR? Draw a diagram to justify your conclusion.

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Let’s extend our thinking 20

Bowties look best when the two triangles formed on each side are congruent. State what measurements could be taken to ensure that the triangles formed by the bowtie are congruent by SAS congruence.

Consider △JKL with angle bisector JM of ∠KJL and KL = 6: a

What type of triangle is △JKL?

Are these statements correct? i

KM > ML

KM = ML

J 5

iii KM < ML

Is the line through J and M a line of symmetry for △JKL?

Are △KJM and △MJL congruent or not congruent?

Consider △PQR with midpoint X at side PR: a

What type of triangle is △PQR?

Are these statements correct? i

∠QXR > ∠QXP

∠QXR = ∠QXP

iii ∠QXR < ∠QXP

Is the line through Q and X a line of symmetry for △PQR?

Are ∠RQX and ∠PQX equal or not equal?

Consider △PQR with angle bisector QX at ∠PQR: a

What type of triangle is △PQR?

Are these statements correct? i

PX > RX

PX = RX

iii PX < RX

Is the line through Q and X a line of symmetry for △PQR?

Are △QXR and △QXP congruent or not congruent?

In construction technology class, Quentin has decided to construct a shelf with triangular openings. How should Quentin cut and assemble the pieces of wood to guarantee the triangles are congruent? Assume he will use no more than six planks to construct the shelf. Use a triangle congruence postulate to justify your solution.

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In the triangle △ABC we are given that and are congruent. By constructing to be the angle bisector of ∠BAC, give a proof for the base angles theorem.

If two line segments bisect each other, and their endpoints are joined, determine whether a pair of congruent triangles will always be formed. Explain.

Give an example of two triangles which have one pair of congruent angles, and two pairs of congruent sides, but are not themselves congruent triangles.

Give an example of two quadrilaterals in which all four pairs of sides are congruent, but the quadrilaterals themselves are not congruent.

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6.04 ASA and AAS congruence criteria After this lesson, you will be able to... • prove triangles are congruent using ASA and AAS triangle congruence. • solve problems using ASA and AAS triangle congruence. • use constructions to create a triangle congruent to a given triangle with the ASA and AAS congruence criteria.

ASA and AAS congruence criteria Exploration We are given two triangles with 2 pairs of congruent angles, and the corresponding sides between those angles are congruent. 1.

Sketch two triangles to ﬁt the description.

Label the triangles GHI and LMN, so that ∠G ≅ ∠L, ∠H ≅ ∠M, and

Formulate a plan for proving that there is a sequence of rigid transformations that will map △GHI to △LMN and explain how you know one or more vertices will align at each step.

If we are given two congruent corresponding angles and one congruent corresponding side, then we will be proving the triangles congruent by angle-side-angle or angle-angle-side congruency depending on the position of the given side. Included side The side between two angles of a polygon is the included side of those two angles. D

Angle-Side-Angle (ASA) congruency theorem If two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent. Angle-Angle-Side (AAS) congruency theorem If two angles and the non-included side of one triangle are congruent to the corresponding parts of another triangle, the triangles are congruent.

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X M A N

ASA

AAS

When proving triangles congruent, it can be diﬃcult to distinguish between ASA and AAS congruence. That’s usually due to a result of a corollary to the triangle sum theorem: Third angles theorem If two angles of one triangle are congruent to two angles of another triangle, then the third angles are also congruent.

Because of this theorem, any triangles that can be proven by ASA congruence can also be proven by AAS congruence and vice versa without any additional information.

Example 1 Use rigid transformations to prove the ASA congruency theorem: If two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent.

Create a strategy We can set up a ﬁgure so we can do the proof in terms of a particular diagram. Since we can’t use the ASA congruency theorem to prove itself, we must use another strategy. In this case, we need to identify a series of rigid transformations that will map △ABC to △DEF. Our rigid transformations are: • Translation • Rotation • Reﬂection

Apply the idea Given: • ∠ABC ≅ ∠DEF

•

• Prove: △ABC ≅ △DEF

B F

E D

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Apply the idea Both triangles are equiangular, which means that the two triangles share three common angle measures. Since they both also have a corresponding side of length 8, we can use the AAS test to justify that they are congruent.

Reﬂect and check We could also show that the triangles are congruent by stating that both triangles are equilateral, which means that both must have three sides of length 8. We can then use the SSS test to justify that they are congruent.

Example 3 In the following diagram,

and

are both straight line segments.

A X

Prove that △ABX ≅ △DCX.

Create a strategy C

We want to ﬁnd as much information as we can in order to satisfy one of the congruence tests. Since since

and

are straight line segments, we can ﬁnd vertically opposite angles, and we can ﬁnd alternate angles on parallel lines.

A X

Apply the idea To prove: △ABX ≅ △DCX Statements 1. 2. 3.

and are straight line segments ∠AXB and ∠DXC are vertically opposite angles ∠AXB ≅ ∠DXC

∠ABX and ∠DCX are alternate interior angles ∠ABX ≅ ∠DCX

306

is a transversal of

and

Alternate interior angles are congruent Given

7. 8.

Opposite angles between straight line segments Vertically opposite angles are congruent Given

5. 6.

Reasons Given

△ABX ≅ △DCX

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AAS congruence

Reﬂect and check We could also use ASA to prove that △ABX ≅ △DCX, ignoring vertical angles in a proof like the one that follows: To prove: △ABX ≅ △DCX Statements 1. 2.

and

Reasons Given

are straight line segments

∠ABX and ∠DCX are alternate interior angles

is a transversal of Given

3. 4. 5.

∠ABX ≅ ∠DCX ∠BAX ≅ ∠CDX

Alternate interior angles are congruent Alternate interior angles are congruent Given

6. 7.

and

△ABX ≅ △DCX

ASA congruence

Example 4 Find the value of x that makes the triangles congruent. D 23°

15 ft 64°

13.5 ft

F X 23°

Z 64°

(4x − 1) ft

Create a strategy The triangles are congruent by ASA, so we can identify corresponding parts and create an equation to solve for x. and the Based on the diagram, we see that ∠E ≅ ∠Y and ∠D ≅ ∠X, so the included side for the first triangle is included side for the second triangle is . By the definition of congruence, if , then the two segments will be equal in length. Therefore, we need to find x so that 15 = 4x − 1.

Apply the idea 15 = 4x – 1

DE = XY

16 = 4x

Add 1 to both sides

4=x

Divide by 4 on both sides

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Example 5 Construct a copy of the triangle shown using two angles and one side. B

Create a strategy We will use two angles and included side. To create a copy of △ABC, we will ﬁrst create a copy of . The endpoints of this new segment are the two vertices of the triangle. Then we will create copies of ∠A on one of these vertices and a copy of ∠C on the other vertex. The intersection of the sides of the angles will determine the third vertex.

Apply the idea We will use GeoGebra to implement the steps. We start by constructing a point D that will become one of the vertex of the triangles. We can do this using the Point tool as shown.

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Next, we want to create a copy of . To do this, we can use the Compass tool. Select vertices A and C ﬁrst then select point D. This creates a circle centered at D with a radius of length AC.

We can then use the Point tool to create a new point E on the arc.

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Connect D and E using the Segment tool. This is a copy of congruent to .

. Because

is the radius of circle D it must be

To create a copy of ∠A and ∠C, we need to create an arc centered at A that intersects both and use the Circle with Center through tool. Select point A and any other point on which we will call F.

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. To do this,

Find the point of intersection of the arc and

which we will call G and add it using the Point tool.

Use the Compass tool to create a copy of circle A centered at point D. To do this, select A and F, and then select D.

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Locate the intersection of the arc of the smaller circle D and

using the Point tool. This point H is a copy of G.

Create an arc using the Compass tool. Select F and G, then select H.

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Use the point tool to add the point I at the intersection of the arcs of circles D and H above point I to be the same distance from H as point F is from G. This means .

. We have created the

Use the Line tool. Select I and then select D. ∠IDH is a copy of ∠A.

We need to repeat the previous steps to copy ∠C. Use the Circle with Center through tool. Select C and then another point J on .

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Locate K, the intersection of the arc and

, using the Point tool.

Use the Compass tool to create a copy of the arc on E. Select C and J, and then select E.

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Locate the intersection of the arc and

using the Point tool. This point L is a copy of K.

Create an arc using the Compass tool. Select J and K, then select L.

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Use the Line tool. Select M, the intersection of the arc centered at E and the arc centered at L, and then select E. ∠MEL is a copy of ∠C.

Locate N, the intersection of

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and

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, using the Point tool.

Finally, we can create △DNE using the Polygon tool. This △DNE is copy of △ABC. We know the triangles are congruent by ASA because we constructed copies of two angles and their included side.

Example 6 Justify that the following steps construct the perpendicular bisector of

1. Using a compass, construct an arc centered at A with a radius with length that is more than half the length of

2. Using the same compass setting, construct an arc centered at B. The two arcs must intersect at two points C and D. 3. Draw

. This is a perpendicular bisector of

. C

Create a strategy We can use the fact that we used the same setting of the compass when drawing the arcs to justify that some segments are congruent. Using SSS, ASA and CPCTC, we will prove that is a bisector of and that the segments are perpendicular, and therefore that is a perpendicular bisector if .

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Apply the idea Start by marking point E, the intersection of

and

E A

C and D lie on the arc centered at A, and so . C and D also lie on the arc centered at B, and so . Since by construction the two arcs have the same radius, we know that . This means △ACD ≅ △BCD by SSS congruence. Using CPCTC, ∠ACD ≅ ∠BCD. This also means that ∠ACE ≅ ∠BCE.

△ACB is isosceles, so ∠CAE ≅ ∠CBE. By ASA congruence, we know that △ACE ≅ △BCE. By CPCTC, we have . This means is a bisector of . Using the same congruence statement and CPCTC, we know that ∠AEC ≅ ∠BEC. Since ∠AEC and ∠BEC are also linear pair, then m∠AEC = m∠BEC = 90°. This means that , and by extension , is perpendicular to .

E A

D C

E A

This proves that

is a perpendicular bisector of

Reﬂect and check It is important that the compass width is greater than half of AB to ensure that there will be two points of intersection. Setting the width to exactly half AB will only result in one intersection point and setting the width to less than half of AB will result in no intersection points.

Idea summary To show that two triangles are congruent, it is suﬃcient to demonstrate the following: • •

318

Angle-side-angle, or ASA: If two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent. Angle-Angle-Side, or AAS: If two angles and the non-included side of one triangle are congruent to the corresponding parts of another triangle, the triangles are congruent.

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Practice What do you remember? 1

I G

63°

28°

63°

28°

E D

4 A

U 4

Identify a pair of additional congruent parts needed to prove the triangles congruent by: i

Angle-Angle-Side congruence (AAS)

Angle-Side-Angle congruence (ASA)

△ADC ≅ △ADB

△ABC ≅ △MKJ

B J

B M

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Find the value of the variable(s) that make the triangles congruent: a

D 40°

40°

53° A

54°

13.53 ft

21 ft E

F X

Q 7 ft

87° 40°

B 75°

112°

25°

(12x − 3) ft

54°

9 ft

40°

(x + 17)°

65°

91 40°

C E

(2y − 5)°

2x + y

112° G

F F

Let’s practice 4

Consider the given diagram: a

State which two triangles are congruent.

State the congruence postulate the triangles satisfy.

For each pair of triangles, identify a congruence test that justiﬁes the congruence. If a test does not exist, determine whether the triangles are not congruent or if there is not enough information to justify the congruence. a

K Q S H

R L

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I G

E F

63°

28°

A D

N 63°

28°

B C

Fill in the blanks to complete the proof:

To prove: △PQR ≅ △RSP Statements Reasons ∠PSR ≅ ∠RQP Given

4. 5. SOL

Given

2. 3.

⬚

△PQR ≅ △RSP

⬚

Select the reasons for the last three statements of this proof.

Given: ∠MON ≅ ∠PNO; Prove: △MON ≅ △PNO

To prove: △MON ≅ △PNO Statements Reasons 1. 2.

∠MON ≅ ∠PNO ∠PON ≅ ∠MNO

3. 4.

△MON ≅ △PNO

Given ⬚

⬚ ⬚

Base angles of an isosceles triangle are congruent

Angle-Side-Angle (ASA) Postulate

Corresponding parts of of congruent triangles are congruent

Side-Angle-Side (SAS) Postulate

Reflexive property 8

A teacher provides his class of Geometry students with two angles and one side of an unknown triangle. Determine if the students have enough information to construct a congruent triangle. Explain.

Consider △TRS.

Construct a triangle that is congruent to △TRS using the ASA Congruence Theorem. Use a compass and straightedge.

Construct a triangle that is congruent to △RST using the AAS Congruence Theorem. Use a compass and straightedge.

How were the processes similar?

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How were they different?

Prove each congruence statement: a

△PQR ≅ △STR

△ABD ≅ △CBD

A P B

Q R D T C

△DEC ≅ △FEG

△WBY ≅ △ZAX

Given XZ = WY

X F

Y B

Given:

• •

Prove: △ABX ≅ △DCX B

Given:

• • ∠S ≅ ∠Q Prove: △PRS ≅ △RPQ 13

Given: • •

R A

bisects ∠ADC

Prove: △ABD ≅ △CBD

Given: • •

bisects

Prove: △AXD ≅ △CXB

X C

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For each diagram: i

Name the congruence test (AAS, SAS, or SSS) that the two triangles satisfy.

Solve for the variable.

6.5 + 5a

(22 + 8m)°

(16m)° 6a + 3

d (2.4x + 17.12)°

4( y − 29)°

SOL

(5x + 12)°

What values for x and y make △MNO ≅ △PRT ? N 25

R 37.5°

3x − 5

1.5y° 3y° 75°

A 17

x = 10, y = 25

T O

x = 10, y = 27

x = 15, y = 13

Then ﬁnd the values of x and y that prove △ABC ≅ △DEC.

x = 15, y = 25

A 4y − 6 C B

3y + 1

2x + 6

4x E

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Let’s extend our thinking 18

Explain why Angle-Angle-Angle (AAA) is not a valid triangle congruence theorem.

Explain how the triangles in an hourglass that sits on a ﬂat surface satisﬁes Angle-SideAngle (ASA) or Angle-Angle-Side (AAS) conditions for triangle congruence.

The Angle-Side-Angle congruency theorem states: If two angles and the included side of one triangle are congruent to the corresponding parts of another triangle, the triangles are congruent. Write a proof plan for how to justify the Angle-Side-Angle congruency theorem using rigid transformations.

Explain how proving that Angle-Side-Angle (ASA) is a valid criterion for triangle congruence, tells us that AngleAngle-Side (AAS) must also be a criterion for triangle congruence.

Explain why Side-Side-Angle (SSA) is not a valid triangle congruence test.

Draw a pair of triangles that are not congruent but would satisy the Side-Side-Angle (SSA) congruence test.

In the triangle △ABC we are given that ∠ABC and ∠ACB are congruent. By constructing to be the angle bisector of ∠BAC, give a proof for the converse of base angles theorem.

△STP ≅ △QTP. If the size of ∠SPQ is 78°, use triangle congruence criteria to find the measure of the highlighted angle ∠SPT.

D R

T Q

Prove that the following steps construct a line perpendicular to external point C. 1.

from an

Using a compass, construct an arc centered at C and intersecting at two points D and E.

2. Using the same compass setting, construct an arc centered at P and another arc centered at Q. Label their intersection F. 3. Draw

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6.05 Right triangle congruence After this lesson, you will be able to... • prove triangles are congruent using HL triangle congruence. • solve problems using HL triangle congruence. • use constructions to create a triangle congruent to a given triangle with the HL congruence criteria.

Right triangle congruence Exploration Tawana says that the congruence theorem: Side-Side-Angle (SSA) is a valid congruence theorem and works for all triangles. 1.

Explain why SSA is not valid by providing a counterexample for Tawana’s claim.

Tawana changes her claim and states that SSA is valid for right triangles. Explain why this statement made by Tawana is valid.

In general, side-side-angle is not a valid congruence theorem. When the congruent angle pair are right angles, however, we introduce a new triangle congruence theorem speciﬁc to right triangles: Hypotenuse-Leg (HL) congruency theorem If the hypotenuse and leg of one right triangle are congruent to the hypotenuse and leg of another right triangle, then the two triangles are congruent. Remember, the hypotenuse is the side opposite the right angle. D

A F

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Example 2 Given:

• • Prove: △ILK ≅ △JLK K

Create a strategy When proving triangles congruent, start by labeling and identifying the given information as well as anything that we can conclude directly from the diagram (such as vertical angles or reﬂexively congruent parts). Since we are trying to prove two triangles congruent, we need to decide which triangle congruence theorem best ﬁts the given information. Then, when writing the proof we always start with the given information, build any conclusions based on the given information, and end with the conclusion we are trying to prove.

Apply the idea We are given that which means that ∠IKL and ∠JKL are both right angles by the definition of perpendicular lines. That means △ILK and △JLK are both right triangles by the definition of right triangles. We are also given that which means we have two congruent hypotenuses, and we know that by the reflexive property so we have a pair of congruent legs as well. This is enough information to conclude that △ILK ≅ △JLK by the HL theorem.

Reﬂect and check Triangle proofs can be written in any style - just make sure that each congruent part needed to justify the triangle congruence theorem has its own statement and reason included.

Example 3 Consider the triangles shown:

X G

a Identify the additional information needed to prove these triangles congruent by HL congruence.

Create a strategy From the diagram we know that ∠G and ∠N are both right angles making both triangles right triangles. We also know that which is a pair of corresponding legs. We are missing congruent hypotenuses to complete the theorem.

Apply the idea

Reﬂect and check Having a right angle alone is not enough to justify HL congruency. Make sure that the hypotenuses are congruent and one pair of legs are also congruent.

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b What other information could be given about the triangles to be able to prove congruence?

Create a strategy By taking the original diagram into consideration, we can seek to use congruence between unknown corresponding angles to show that the triangles are congruent.

Apply the idea If we are given that ∠M ≅ ∠X, we can prove the two triangles congruent by ASA congruence.

Example 4 Find the value for each of the variables that makes △ABC ≅ △CDE. A

B 8 F

Create a strategy Since m∠ABC and m∠CDE = 90°, we know that △ABC and △CDE are both right triangles. We know that if the hypotenuse and leg of one triangle are congruent to the hypotenuse and leg of another triangle, then the triangles are congruent. We need to choose values for x and y that will make the hypotenuses and the legs congruent. We can use that and to solve for the variables. By the deﬁnition of congruence, since and , the two pairs of segments will be equal in length.

Apply the idea

AC = CE

Subtract

Add 4 to both sides

AB = CD

Multiply both sides by 4

Subtract 2 from both sides

Divide both sides by 3

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x from both sides

Example 5 Construct a triangle congruent to △ABC using one leg and hypotenuse. C

Create a strategy To construct a triangle congruent to △ABC, we will use the Hypotenuse-Leg congruence criteria and construct a congruent hypotenuse and leg.

Apply the idea We will ﬁrst copy one of the legs of the right triangle. Let’s choose Start by creating a horizontal line

using the Line tool.

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Next, we create a point F somewhere on the line by using the Point tool.

Then create an arc with a radius of length (one of the legs). Once again, we can use the Compass tool. Select vertices A and B ﬁrst then select point F. This ensures that the radius of the circle is the same as the length of .

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Next, we mark the points of intersection of the line and the arcs using the Point tool. to .

and

are both congruent

We now create two arcs with a radius of length (the hypotenuse) using the Compass tool. For the first arc, select vertices B and C first then select point G. For the second arc, select vertices B and C first then select point H.

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Next, we mark the point of intersection of the two arcs we have created using the Point tool.

Finally, we can create △FHI using the Polygon tool. This △FHI is congruent to △ABC.

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Reﬂect and check If we want to conﬁrm that the constructed triangle is a copy of the original triangle, we can look at the side lengths of both triangles in the Algebra view.

Idea summary To show that two right triangles are congruent, it is suﬃcient to demonstrate Hypotenuse-leg, (or HL): If the hypotenuse and leg of one right triangle are congruent to the hypotenuse and leg of another right triangle, then the two triangles are congruent.

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Practice What do you remember? 1

For each of the following right triangles, label each side as a leg or hypotenuse. a

M N

B C

G K

A F

J C

D 7

4 A

7 M B

5 4

Q Q

R S

U R

8 T

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Identify a pair of additional congruent parts needed to prove the triangle pairs congruent by the HypotenuseLeg (HL) congruency theorem: a

△UTV ≅ △WYX

△CED ≅ △FDE

T U

Y F

Find the value of x that would make the triangles congruent: a

E 15

N x+6

11.6 G

F 9.5

X H

P 2x + 5

Let’s practice 5

Triangle A

Triangle B

Triangle C 5

2 5

Triangle D

Which two triangles are congruent?

Which congruence test is satisﬁed?

Use the given diagram to: a

State which two triangles are congruent.

State the congruence theorem that justiﬁes the congruence.

U Z

W X

Fleur and Jeung have each drawn a triangle. The longest sides of their triangles have the same length, as do the shortest sides of their triangles. They claim that their triangles must be congruent by the Hypotenuse-Leg (HL) congruency theorem, since the longest side of a triangle is always the hypotenuse. Are they correct? Why or why not?

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Use the information given in the diagram to complete the paragraph proof that △MQP ≅ △MNP: From the diagram we know that ≅ by deﬁnition of congruence. We also know that ⬚ and ⬚ are right angles so △MNP and △MQR are both ⬚ triangles by the ⬚. Next, we know that ⬚ ≅ ⬚ by the reﬂexive property. Therefore △MQP ≅ △MNP by the ⬚ congruence theorem.

For each of the following diagrams: i

State the theorem and at least two congruence tests that can be used to justify the two triangles are congruent.

Find x.

35 − 6x

6x°

4x + 5

(5x − 7)°

(3x + 15)° (3x + 5)°

Given: • •

(8x − 18)°

(7x − 23)°

∥ ≅

Prove: △HJK ≅ △LKJ L

Given:

• m∠DCE = 90° • bisects

Prove: △ABC ≅ △EDC

In the ﬁgure shown, △ABC is equilateral.

E A

Prove △ACD ≅ △ABD

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Construct a right angle given one leg and hypotenuse. Show all steps of your construction.

Let’s extend our thinking 14

You notice two right triangles in the tile ﬂoor of a hotel lobby. You want to determine whether the triangles are congruent, but you only have a piece of string. Can you determine whether the triangles are congruent? Explain.

A rectangle can be divided into two triangles by either of its diagonals. Explain how these two triangles satisfy the Hypotenuse-Leg (HL) conditions for triangle congruence.

Justify the converse of angle bisector theorem.

Consider the following information: • SP = SR • ⊥ • ⊥ Show that: m∠PQS = m∠RQS R

P S

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7 Similarity Big ideas • A dilation preserves angle measure but not distance resulting in figures that are similar but not congruent (assuming the scale factor is not 1).

Chapter outline 7.01 7.02 7.03 7.04

Dilations Similarity transformations Proving triangles similar Applications of similarity

340 352 364 377

Reﬂect and check We could use a ruler and protractor to verify: 1. corresponding segments are an equal distance apart (thus parallel) 2. the length of each segment on the image is

the size of the corresponding segment of the pre-image

3. corresponding angles are congruent These steps would verify that we have dilated the ﬁgure correctly.

Idea summary When a ﬁgure is dilated with a center of dilation at point P by a factor of k: • • • •

The distance between P and any point on the image will be k times the size of the distance between P and the corresponding point on A. The length of the image segments will be k times the size of the corresponding segment lengths of A. When P is not a point on A, the image will be parallel to A. When P lies on a line segment of A, its image will lie on the same line.

In coordinate notation, a point dilated with respect to the origin and a scale factor of k is written as (x, y) → (kx, ky)

Practice What do you remember? 1

Find the scale factor for each pair of ﬁgures: a

C B 4 A

B′

25 D

C′

12 B′

12 A′

O 21

C′

A′ 5

D′

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7.02 Similarity transformations After this lesson, you will be able to... • identify a dilation or a combination of transformations that maps one figure onto another. • use a dilation or a combination of transformations to identify similar figures.

Similarity transformations When two ﬁgures are similar, we express this using a similarity statement and can identify the similarity ratio of each pair of corresponding sides. Similarity ratio

Similarity statement

The ratio of two corresponding side lengths in a pair of similar ﬁgures.

A statement that indicates two polygons are similar by listing the vertices in the order of the correspondence.

The similarity statement for the triangles in the diagram shown is △ABC ∼ △DEF.

B C D

Similarity statements can be written in any order which keeps corresponding vertices in order. For this example we could have written the similarity statement in several different ways, though it is most common to arrange the vertices in alphabetical order (when possible).

If two ﬁgures are similar, then their corresponding angles are congruent and corresponding sides are in equal proportion.

Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 7.02 to answer these questions. 1.

What sequence of transformations could map the pre-image to the image?

Does the sequence of transformations preserve the angles of the pre-image, the side lengths of the pre-image, or both the angles and the side lengths? How do you know?

When a sequence of transformations is made up of rigid transformations, we can say that the image is congruent and therefore similar to its pre-image. The side lengths and angle measures of the pre-image are preserved. Two ﬁgures are said to be similar if there exists a similarity transformation which maps the pre-image to the image.

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Similar

Similarity transformation

Two ﬁgures are similar if their corresponding angles are congruent and their corresponding sides are proportional.

A series of one or more transformations which results in the image being similar to the pre-image.

Rotations, reﬂections, and translations all result in an image congruent to the pre-image. Since all congruent ﬁgures can be considered similar with a ratio of 1 : 1 (that is, with a scale factor of k = 1), these are all similarity transformations as well. As a dilation enlarges or reduces a shape, the image and pre-image’s corresponding angles will be congruent, and the corresponding sides will be proportional. This means that dilations are also similarity transformations. Any combination of these four transformations will maintain similarity.

Example 1 The following sequences of transformations are applied to figure ABCD. Without performing the transformations, determine whether each final image will be similar, congruent, or both. Explain your reasoning. a A reflection across the x-axis followed by a rotation 90° clockwise about the origin.

Create a strategy A reﬂection across the x-axis has a transformation mapping: (x, y) → (x, −y) A rotation 90° clockwise about the origin has a coordinate mapping: (x, y) → ( y, −x)

Apply the idea Since a reﬂection across the x-axis has a transformation mapping (x, y) → (x, −y), we know that the coordinates were not multiplied by a scale factor, meaning no dilation occurred. After this rigid transformation, a rotation 90° clockwise about the origin with a coordinate mapping (x, y) → ( y, −x) also does not change the segments of the ﬁgure, since there is no scale factor other than 1. The sequence of rigid transformations will change the orientation of ABCD, while preserving its angles and side lengths. Since the sequence of transformations will lead to a congruent image to the pre-image, the sequence is both a congruency transformation and a similarity transformation.

b A dilation by a scale factor of

with the center of dilation at the origin followed by a rotation 180°.

Apply the idea Since the scale factor k < 1, any segment on the image will be k times the size of the line segments of the pre-image, so the side lengths will be proportional, but not congruent, while the angles will be preserved. A rotation will preserve both the side lengths and angle measures of the ﬁgure, after it has been dilated. The sequence of transformations will be a similarity transformation.

Reﬂect and check The coordinate mapping for the sequence of transformations is (x, y) → Multiplying the x- and y-coordinates by the same scale factor indicates a dilation, which creates a similar, but not congruent image.

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Side-Side-Side similarity (SSS ∼) theorem

If the lengths of the corresponding sides of two triangles are proportional, then the triangles are similar.

z az

Example 1 Explain why similarity theorems work using transformations. a Why can triangles be proven similar with only two corresponding congruent angles, and not all 3?

Create a strategy We need to consider why the AA similarity theorem works using transformations. We know that the triangle sum theorem states the sum of the angles in a triangle is 180°. Draw a simple diagram as an example and use it to show why AA similarity works.

Apply the idea Given: ∠L ≅ ∠X and ∠M ≅ ∠Y, we can prove: △LMN ∼ △XY Z Z

M Y L

To prove: △LMN ∼ △XY Z Statements

4. 5. 6.

∠L ≅ ∠X ∠M ≅ ∠Y m∠L = m∠X m∠M = m∠Y m∠L + m∠M + m∠N = 180 m∠X + m∠Y + m∠Z = 180 m∠L + m∠M + m∠N = m∠X + m∠Y + m∠Z m∠X + m∠Y + m∠N = m∠X + m∠Y + m∠Z m∠N = m∠Z

△LMN ∼ △XY Z

1. 2. 3.

Reasons Given Deﬁnition of congruent angles Triangle sum theorem Transitive property of equality Congruent angles Subtraction property of equality Since the triangles have three corresponding congruent angles, a similarity transformation exists that maps △LMN to △XY Z

Since we know transformations lead to similar ﬁgures, we can conclude that triangles can be proven similar with only two corresponding congruent angles.

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Example 3 Determine whether the given pairs of triangles are similar. If so, state the theorem which proves their similarity. If not, explain how you know. a

1.5 85° 65°

85°

65°

Create a strategy

Apply the idea

We can use the given information about the angles in the two triangles to determine whether we can show that they are similar.

Since there are two pairs of congruent angles, we can state that the triangles are similar using the AA similarity theorem.

We are given two angle measures and a side length in each triangle. The side lengths given are not corresponding sides.

12 47°

29°

R B 104° Q

4 47° 2 P

Create a strategy There are two angles in the triangle that are 47° which, if the triangles are similar, indicates that these are corresponding angles. We will use this information to determine two pairs of corresponding sides, then use these three pairs of corresponding parts to determine similarity.

Apply the idea First, we must determine if the corresponding side lengths are proportional. Since ∠CAB and ∠RPQ have the same measure, we will assume these are corresponding angles. From this, we can say corresponds to and corresponds to . To check if the sides are proportional, we need to set up our corresponding ratios. This shows the corresponding side lengths are proportional with a scale factor of 3, and the angle measure between the side lengths is congruent. Therefore, the triangles are similar using the SAS similarity theorem.

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C 16

B 24

A E

Create a strategy We will need to use the side-side-side similarity theorem to show that the pair of triangles are similar, since the only parts of the triangles given are the side lengths. If a common scale factor exists that will map one triangle to the other, we can state that the corresponding side lengths are proportional and that the triangles are similar.

Apply the idea We can show that the triangles are similar by showing that their corresponding side lengths are proportional. and because they have the shortest side lengths in the triangles: First, we’ll test

We will test the next set of side lengths,

and

because they are the next shortest:

Finally, if the last pair of sides have a proportional scale factor to the first two sets, we can confirm similarity. We have:

Since there is not a common factor between the three sides of the triangles, the triangles are not similar.

7.8 D

3.4

8.84 5.2

Create a strategy We will need to show that the pair of triangles satisﬁes the side-side-side similarity theorem since there is no information given about the angle measures of the triangles.

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Apply the idea We can test each side that could be corresponding to ﬁnd a common factor. Since the two shortest sides are 2 from △PQR and 5.2 from △DEF,

Next, we will test the next-shortest set of corresponding sides

Finally, we’ll test the ﬁnal pair of corresponding sides,

and

Since the corresponding side lengths of the triangles have a common scale factor and are thus proportional, we can conclude that the triangles are similar by the SSS similarity theorem.

Reﬂect and check We cannot rely on triangles to have corresponding parts in alphabetical order, so looking at the side lengths in ascending or descending order and pairing them up helps us look for a common factor. For this pair of triangles, △PQR ∼ △FED.

Example 4 Determine whether the pair of triangles are similar. If so, write a similarity statement and justify with a similarity theorem. If not, explain why not.

J 2 K 4

3 L 6 N

Create a strategy First, we should consider whether the given information is enough to establish similarity between the two triangles. In this case, we are given some side lengths, but also note that there is an angle common to both triangles. ∠J is shared between the two triangles. We also see that JK is part of the total length of JM, and JL is part of the total length of JN. So, we will test △JKL ∼ △JMN. We also know some side lengths on two pairs of sides, so it makes sense to test the SAS similarity theorem.

Apply the idea Based on the given lengths, we can see that

The included angle, ∠J belongs to both △JMN and △JKL. Therefore, △JMN ∼ △JKL by the SAS similarity theorem.

Reﬂect and check A triangle like this that is split proportionally has special properties, which we will learn about in the next lesson.

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Example 5 Find x. Show your work and justify your steps.

C 4

54°

(5x − 6)° 2.5

D 2

Create a strategy In order to ﬁnd x, we’re going to need to know the measure of ∠FDE. The only angle measure we’re given is ∠CGD. It would be helpful if those two angles happen to be congruent. So, let’s see if we can ﬁnd similar triangles and match those corresponding angles.

Apply the idea Notice that

, and

. That gives us two corresponding sides with the same proportions, so now

we need to test the third sides. From the diagram, we know DF = FG, so DG = 2DF and

= 2.

, we can use SSS similarity theorem to say △GCD ∼ △DEF.

Since

Since the triangles are similar, we know that their corresponding angle measures are congruent. ∠CGD ≅ ∠EDF, so we can ﬁnd x by solving the equation 5x − 6 = 54. 5x − 6 = 54

m∠CGD = m∠EDF by deﬁnition of congruent angles

5x = 60

Add 6 to both sides

x = 12

Divide by 5 on both sides

Idea summary Use the theorems about the angles and side lengths of triangles to prove their similarity: • • •

370

Angle-angle similarity, or AA ∼: The two triangles have two pairs of congruent angles. Side-side-side similarity, or SSS ∼: The two triangles have three pairs of sides whose lengths are in the same proportion. Side-angle-side similarity, or SAS ∼: The two triangles have two pairs of sides whose lengths are in the same proportion, and the angles between these sides are also congruent.

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Practice What do you remember? 1

What is the difference between congruent shapes and similar shapes?

Given: △ECD ∼ △LMN a

Match the corresponding angles by ﬁlling in the blanks: ∠D corresponds to ∠⬚ ∠E corresponds to ∠⬚

∠C corresponds to ∠⬚

State what must be true about these pairs of angles.

State which side in △LMN corresponds to

Write a proportion that relates two pairs of sides.

C N

Determine whether each statement is true or false: a

If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.

If two sides of one triangle are congruent to two corresponding sides of another triangle, then the triangles are similar.

If two triangles are similar then their corresponding sides are congruent.

If two triangles are similar then their corresponding angles are congruent.

If two triangles are congruent, then they must be similar.

If two triangles are not congruent, then they can not be similar.

Name the triangle similarity theorem described by each statement. a

Two pairs of consecutive angles are equal and a pair of corresponding side lengths are proportional.

All pairs of corresponding side lengths are equal.

Two pairs of corresponding angles are equal and the pair of included sides are proportional.

Two pairs of corresponding sides are proportional and the pair of included angles is equal.

Let’s practice 5

For each set of triangles: i

Identify which two triangles are similar.

Name the similarity test they satisfy.

72° 27

31°

C 10

31° 73°

12 27

32°

B 18

73°

32°

73°

4 8 9

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C 27°

21 27°

27°

27° 19

41°

C 12

41°

D 41°

41°

81° 56°

57° 81°

43° 80° 8

81°

42°

For each pair of triangles: i

Identify the similarity test they satisfy.

List the sides and angles of the triangles that satisfy this similarity test.

49 M

9 76°

14 L

H 42

76° F

143

85° C

85°

121

S 56

Determine whether each pair of triangles is similar. If so, write a similarity statement and justify with a similarity postulate or theorem. If not, explain why not. a

36.9° 8 Y

10 P

25 53.1°

78 ft

54°

63 ft

48 ft

57° D Mathspace Virginia SOL Geometry mathspace.co

X 39 ft

372

38°

33 ft 85° E 24 ft

A B

40° Z

C R

21 B SOL

Given: Q lies on

40°

and S lies on P

Q T

Which condition proves △PRT ≅ △QRS? A 9

∠PQS ≅ ∠TSQ

∠PTR ≅ ∠TPR

For each of the similar triangles, ﬁnd the value of the missing side lengths. a

f E

30 g

30 24 N

I b

2 58° 33

7 32° y

32° 10

N a

L 12

14 P

58° x

Using the given diagram: a

Identify two similar triangles.

Which similarity test do the triangles satisfy?

If KL = 20, what is the length of MN ?

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Prove that each pair of triangles is similar: a

△CDE and △LMN

△ZY C and △ZXB

D 4 E

7 5

C L 42

30 D

24 N

△JKL and △MNJ

△ADE and △ABC

J K

A 6

N 10 L B

Let’s extend our thinking 17

Determine whether each of the following is enough to prove △ABD ∼ △CBE. Justify your answer. a

b c

∠BCE ≅ ∠BEC C

d 18

Given:

and ∠B ≅ ∠E

Prove: △ABC ∼ △DEF A

C E

Given: Prove: △AEB ∼ △DEC

B D

C E

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For image 1:

Image 1:

Find AC in terms of a and c.

Find XZ in terms of a, c and k.

For image 2:

Find AB in terms of a and b.

Find DE in terms of a, b and k.

D A

What do part (a) and part (b) tell us about testing for similarity in right triangles?

Image 2:

iii Show that the two triangles are similar using the SSS similarity theorem. c

iii Show that the two triangles are similar using the SSS similarity theorem.

To show △ABC ∼ △DEF, Tadashi creates △A′B′C′ which is a dilation of △ABC by a factor of Describe how Tadashi can use their diagram and transformations to show that △ABC ∼ △DEF. C B

C′ B′

A′

Explain how transformations could be used to justify the SSS similarity criteria for triangles.

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Example 1 Formulate a proof for the side-splitter theorem.

Create a strategy First, we draw a ﬁgure with the characteristics from the theorem, and state what we know and what we’re trying to prove. We know

. We need to prove the two sides are divided proportionally, or

E C

We know similar triangles have congruent angles and proportional side lengths, so the triangle similarity theorems (AA, SAS, SSS) may be helpful. We also know some things about congruent angles when parallel lines are cut by transversals. Extend to be a line with auxiliary points: a point F such that D lies between F and E, and a point G such that E lies between D and G. and are transversals of parallel lines and . A

F D B

E C

Apply the idea To prove: Statements Given

1. 2. 3. 4. 5.

m∠BDF = m∠CBD m∠CEG = m∠BCE m∠BDF = m∠EDA m∠CEG = m∠DEA m∠BDF = m∠CBD = m∠EDA m∠CEG = m∠BCE = m∠DEA △ADE ∼ △ABC

Alternate interior angles are congruent Vertical angles are congruent Transitive property of equality AA similarity theorem Deﬁnition of similar triangles

6. 7.

Reasons

AB = AD + BD AC = AE + CE

Deﬁnition of collinear line segments

Substitution property of equality

Common denominators

10.

Deﬁnition of one

11.

Subtraction property of equality

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Example 2 Use theorems to solve the problems that follow. . Justify your answer.

a Determine whether

L 1.6 K 1 J

8 M 5 N

Create a strategy We can use the converse of the side-splitter theorem to determine whether and are parallel.

. If

, then the sides

Apply the idea We have that Therefore,

and

based on the converse of the side-splitter theorem.

b The Eastern Garbage Patch in the Paciﬁc Ocean is a collection of marine debris that is diﬃcult to measure directly. Using the diagram, how can we indirectly determine the length of the Eastern Garbage Patch?

C E

Create a strategy We can use the side-splitter theorem to indirectly measure the length of the Eastern Garbage Patch.

Apply the idea Using the side-splitter theorem, we see that

. We can calculate the distances of BD, AC, and CE and use

those to ﬁnd the length of the Eastern Garbage Patch.

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c Xiker is training for a triathlon. He wants to use a lake nearby to train for the swimming portion. To determine how far the length of the lake is, he paces out a triangle, counting his paces, as shown in the diagram: If Xiker’s strides are 2.75 feet, determine the distance he must swim across the lake for his training.

40 125

Create a strategy Use the triangle midsegment theorem to determine the length of the lake in paces, then convert the paces to feet.

Apply the idea The triangle midsegment theorem states that the line connecting the midpoints of two sides of a triangle is parallel to the third side of the triangle which is the midsegment, and the length of this midsegment is half the length of the third side. The third side of the triangle is 125 paces, so the length of the lake is 62.5 paces. If each of Xiker’s paces is 2.75 feet, then the length of the lake is approximately 172 feet.

Example 3 Consider the diagram shown where

x+3

D 5 A 4 E 2x C

Find x. What else can we say about

Create a strategy The side-splitter theorem states that if a line intersects two sides of a triangle and is parallel to the third side of the triangle, then it divides those two sides proportionally. Solve a proportion with the side lengths.

Apply the idea We have: Since

, the sides are split proportionally (side-splitter theorem)

Substitute the lengths of each segment

Multiply both sides by 4

Multiply both sides by 5

Subtract 4x from both sides

Divide both sides by 6

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By substituting 2 for x in the triangle, we know that BD = 5 and CE = 4. This means that of the large triangle.

is also the midsegment

Reﬂect and check We couldn’t assume that was the midsegment with the information given initially and assume that we could solve the equations BD = AD → x + 3 = 5 and CE = AE → 2x = 4.

Idea summary We can use these theorems to help us solve problems with triangles: • •

Side-splitter theorem: If a line intersects two sides of a triangle and is parallel to the third side of the triangle, then it divides those two sides proportionally. Triangle midsegment theorem: The midsegment connecting the midpoints of two sides of a triangle is parallel to the third side of the triangle, and the length of this midsegment is half the length of the third side.

Right triangle similarity Exploration These four triangles are similar:

45° 5

Triangle A

Triangle B

45°

Triangle C

Triangle D

How could we show that Triangle A and Triangle B are similar?

How could we show that Triangle A and Triangle D are similar?

How could we show that Triangle B and Triangle C are similar?

If we know two triangles are right triangles, what additional information do we need to prove they are similar?

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When we are dealing with right triangles, we only need to know that the hypotenuse and leg are proportional to determine similarity. This is because the Side-Side-Side similarity theorem is indirectly satisﬁed by the Pythagorean theorem. Consider a right triangle ABC with leg a and hypotenuse c, and a second right triangle XY Z with a proportional side ka and proportional hypotenuse kc. The scale factor between the known sides is k. Length of unknown leg in △ABC

Length of unknown leg in △XY Z

Factor out k2

Product of radicals property

This shows the corresponding unknown sides are also proportional with a scale factor of k. Hypotenuse-Leg similarity (HL ∼) theorem If the ratio of the hypotenuse and leg of one right triangle is equal to the ratio of the hypotenuse and leg of another right triangle, then the two triangles are similar.

An altitude of a triangle is a perpendicular line segment drawn from one vertex to the opposite side of the triangle. Right triangle similarity theorem The altitude to the hypotenuse of a right triangle divides the triangle into two triangles that are similar to the original triangle and to each other. The altitude in a right triangle creates three similar right triangles: C

△ABC ∼ △CBD A

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△ABC ∼ △ACD

D C

△CBD ∼ △ACD

D D

Example 4 Use theorems to solve the problems that follow. a A ﬂagpole that stands 4.9 meters high casts a shadow of 4.5 meters. At the same time, the shadow of a nearby building falls at the same point S. The shadow cast by the building measures 13.5 meters. Find h, the height of the building.

4.9 m

4.5 m

13.5 m

Create a strategy The building and ﬂagpole are both perpendicular to the ground, creating a set of congruent angles. Both triangles also share ∠S. So, the AA similarity theorem would be used to show the two triangles are similar. Equate the ratios of the corresponding sides of the similar triangles.

Apply the idea The corresponding sides are h and 4.9, and 13.5 and 4.5. Corresponding side lengths are proportional in similar triangles

Multiply both sides by 4.9

The height of the building is 14.7 meters.

Reﬂect and check We can check the similarity ratio by conﬁrming that the triangles are proportional.

Since the corresponding side lengths of the triangles are proportional, we can conﬁrm that the height of the building is 14.7 meters.

b Determine whether △ABC ∼ △ADE. Justify your reasoning. B 19 ft D

28 ft

19 ft

14 ft C

Create a strategy The hypotenuse-leg similarity theorem states that if the ratio of the hypotenuse and leg of one right triangle is proportional to the ratio of another hypotenuse and leg of another triangle, the triangles are similar.

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Apply the idea If the ratio of the hypotenuse and leg of △ABC is proportional to the ratio of the hypotenuse and leg of △ADE, then we can state that the triangles are similar.

Since the ratios of the hypotenuse and leg of each triangle are proportional, the triangles are similar by the HL similarity theorem.

Example 5 Solve for the length of

Create a strategy We know the triangles are similar by the right triangle similarity theorem. We can separate the similar triangles to help us visualize which ones to use to create the similarity ratios. • △ABC ∼ △CBD • △ABC ∼ △ACD • △ACD ∼ △CBD

Notice that △ABC does not have has a side, so we want to use triangles ACD and CBD to set up the ratios.

D D

Apply the idea Since △ACD ∼ △CBD, we know that so their proportions will be equal.

and

are corresponding sides and

Ratios of corresponding sides are equal

Substitute AD = 8 and BD = 4

Cross multiply

Square root both sides

Simplify the radical

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and

are corresponding sides,

Reﬂect and check Recall there are two methods we can use to simplify radicals. Prime factorization method: Find prime factorization of radicand

Group factors according to the index

Multiplication property of radicals

Simplify perfect squares

Multiply the coeﬃcients

Perfect square method: Find largest perfect square factor of radicand

Multiplication property of radicals

Simplify perfect square

Idea summary We can use these theorems to help us solve problems with triangles: • •

Hypotenuse-Leg similarity theorem: If the ratio of the hypotenuse and leg of one right triangle is equal to the ratio of the hypotenuse and leg of another right triangle, then the two triangles are similar. Right triangle similarity theorem: The altitude to the hypotenuse of a right triangle divides the triangle into two triangles that are similar to the orginal triangle and to each other.

Practice What do you remember? 1

Find the value of the variables in each pair of similar, right triangles: a

s°

8 55°

r°

b 1.6

m 3 n

h 30

10 28

24 d

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Decide whether the two triangles are similar: a

Two sides of one triangle are proportional to two corresponding sides of another triangle.

Three pairs of corresponding sides are proportional.

Three corresponding angles are equal.

Two pairs of corresponding sides are proportional and a pair of corresponding, non-included angles are equal.

Two pairs of corresponding legs of two right triangles are proportional.

Given △ABC and △XY Z that are both acute triangles, complete each statement using the given similarity test. a

∠A = ∠⬚, ∠⬚ = ∠Z; AAA similarity test

= ⬚ = ⬚; SSS similarity test

c d

∠C = ⬚, = 5,

= ⬚, ⬚ = 5; SSS similarity test

= ⬚, ∠⬚ = ∠Z; SAS similarity test

e 4

; SAS similarity test

For two right triangles △ABC and △XY Z, is each statement true or false? a

Since they are both right triangles, they are similar.

If they have one acute angle in common, they are similar.

If they have the same hypotenuse length, they are similar.

If △ABC ∼ △XY Z, then

Draw 2 similar triangles from the following ﬁgures: a

4.24 cm 20 3 cm

K 20

10 24

N M

A 10

9m D E

12.025 m

9.23

3.7 m

386

16 cm

Mathspace Virginia SOL Geometry mathspace.co

Let’s practice 6

The right triangles in the diagram are similar:

△ABC ∼ ⬚

Find the values of x and y.

y cm D

16 cm E

5 cm

x cm B

8 cm

Given: △ABE ∼ △ACD a

Solve for x.

Solve for AC.

A 6.25

x−1

x+2

Given that ∠S = ∠V

Find the value of x that makes △V UW ∼ △SUT.

5x + 11 U

18 U

Given: △ABE ∼ △DCE a

Solve for AE.

Solve for ED.

C A

x−1 2

x+5

Given: △DCE ∼ △BCA

Solve for x. 6 x+7 A

12 − x C 4 B

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Solve for the variable(s) in each diagram: a

y+3

8y − 2 B x

S C

2y + 15

d 5y − 2

7y + 5

3x + 2 x

f 5

x 12

8 y

A large tree casts a shadow of 30 m. At the same time, a 2 m high stick, standing vertically, casts a shadow of 3 m. B

h Sun Ray C 2m A

30 m

a 13

Explain why △ABE and △DCE similar.

Find the height of the tree, h.

A horizontal brace lies between the slanted sides of an A-shaped ladder. In this position, the top of the ladder is 220 cm from the ground. The brace is 44 cm in length and sits 71 cm from the top of the ladder. How far apart are the feet of the ladder? Round your answer to the nearest whole centimeter.

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The Super Slide at Big Kahuna’s Water Park is shown. Assume the two triangles created are similar. Determine how tall the Super Slide is. Give your answer to two decimal places. h

6.7 4

ft 85

Ms. Shin is 5.5 ft tall. She stands near a ﬂagpole that casts a 22 ft long shadow. If Ms. Shin’s shadow is 7 ft long, determine how tall the ﬂagpole is. Round your answer to two decimal places.

Peter constructs a ramp in the shape of a right-angled triangle. To give the ramp more strength, he wants to place a brace from the base point across to the hypotenuse. The brace should meet the hypotenuse at a right angle, as shown in the diagram:

Identify a triangle similar to △ABC.

How long does the brace, BD, need to be?

Find the length x.

Consider the given ﬁgure: Solve for the height of the roof. Give your answer to two decimal places.

8.2 m

5.1 m

9.66 m

A developer is selling beachfront lots in Destin. The developer will determine the price of each lot based on how much beachfront access it provides. The greater the beachfront, the higher the price of the lot. Of the lots shown, determine which one will have the highest and lowest list prices.

Gulf of Mexico

180 Lot 031

yds

Lot 032 Lot 033

Explain your answer. 48 yds

55 yds

61 yds

Gulf Shore Drive

Consider the given triangle: a

Solve for CD.

Solve for CB.

Solve for AC.

D 3

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In the given diagram, ﬁnd the following segments in exact values: a

9 R S

7.2

9 P

X 2.1 W

Let’s extend our thinking 21

. Justify your answer.

Determine whether there is enough information to conclude a

L 8

N 23

18 M

K 5

20 L

7.5 N

Prove the converse of the side-splitter theorem:

Given: Prove:

Using the diagram:

Prove that △AOB and △DOC are similar.

Prove that AB ∥ CD.

B 3

5 O

Using the diagram, prove that

D A

B E

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Sybil wants to find the height of her school’s flagpole. During recess, she measures the length of the flagpole’s shadow to be 132 in. Her friend then measures Sybil’s shadow, which turns out to be 37 in. a

Determine a reasonable range of heights for the ﬂagpole based on Sybil’s unknown height. Use a model to justify your solution.

At a later point in the day, Sybil’s friend measures her shadow again. This time, Sybil’s shadow is measuring 45 in and the ﬂagpole’s shadow is 161 in. Revise the model drawn in part (a) to construct a new model and determine the possible height of the ﬂagpole. Round your answer to the nearest inch.

Explain another way Sybil and her friend could calculate a reasonable range of heights for the ﬂagpole.

Esther visits Tokyo Skytree Tower which is 2080 ft tall. If Esther is 5 ft 3 in tall, determine how far away she must stand in order for the length of her shadow to be 1 foot less than

of her distance to the tower. Use a diagram

or model to justify your solution, and make a reasonable choice about what to round your solution to.

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8 Right Triangles & Trigonometry Big ideas • Right triangles have special properties that allow their side lengths to be determined using the Pythagorean theorem or ratios. • Trigonometry connects the angle measures and side lengths of a triangle and provides a set of functions that can be used to find unknown angle measures and side lengths.

Chapter outline 8.01 8.02 8.03 8.04

Right triangles and the Pythagorean theorem Special right triangles Trigonometric ratios Solving right triangles

394 406 419 433

Pythagorean triple Whole number side length measures of right triangles. Example: (3, 4, 5) and (5, 12, 13) Given a Pythagorean triple, we can ﬁnd another set by multiplying by a whole number to make a similar triangle with proportional sides. H E B 3

5 A

15 F

12 G

The Pythagorean theorem and its converse describe how the side lengths of right triangles are related. Pythagorean theorem

A 2

If △ABC is a right triangle, then a + b = c . b C

a2 + b2 = c2 a

is the length of one of the legs (shorter sides) of the right triangle

is the length of the other leg of the right triangle

is the length of the hypotenuse of the right triangle

Converse of Pythagorean theorem If a2 + b2 = c2, then △ABC is a right triangle. If a triangle is not a right triangle, then the Pythagorean inequality theorem can be used to determine the type of triangle based on angles. Where c is the longest side, we have that: • If c2 < a2 + b2, then the triangle is acute • If c2 > a2 + b2, then the triangle is obtuse

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Example 1 Find the value of c for each triangle. a

Create a strategy The triangle is a right triangle so the hypotenuse, c, can be found using the Pythagorean theorem.

Apply the idea a2 + b2 = c2

Pythagorean theorem

142 + 482 = c2

Substitution

196 + 2304 = c

Simplify the exponents

2500 = c

Combine like terms

50 = c

Evaluate the square root of both sides of the equation

Reﬂect and check Notice that in this example, 2500 is a perfect square so our answer for c is an integer. This means {14, 48, 50} is a Pythagorean triple. We can use the Pythagorean theorem to solve for missing leg lengths as well.

c 4

2 5

Create a strategy The value of c is the hypotenuse of the top triangle, and we need the leg lengths of the triangle to solve for it. We can find the other missing leg length, which is one of the legs of the bottom triangle, using the Pythagorean theorem first and then use it to find the value of c.

Apply the idea Let’s call common, unlabeled side b. Pythagorean theorem

Substitution

Evaluate the exponents

Combine like terms

Evaluate the square root of both sides of the equation

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Example 3 For the given triangle: 16

a Is the triangle a right triangle? Explain.

Create a strategy We want to use the converse of the Pythagorean theorem to determine if the triangle is a right triangle. Let a and b represent the two shorter side lengths. The hypotenuse will be c. Once we have labeled the sides, we want to ﬁnd the value of a2 + b2 and the value of c2. If a2 + b2 = c2 then the triangle is a right triangle by the converse of the Pythagorean theorem. If a2 + b2 ≠ c2 then the triangle is not a right triangle for the same reason.

Apply the idea Let a = 9, b = 16, and c = 18. Now we can calculate the following: a2 + b2 = 92 + 162 = 337 And c2 = 182 = 324 2

This is not a right triangle because 337 ≠ 324 so a + b ≠ c2

b Classify the triangle in terms of its sides and angles.

Create a strategy From part (a), we know that a2 + b2 = 337 and c2 = 324, we can compare these to classify the angles in the triangle.

Apply the idea We can see that all the sides are different lengths, so this is a scalene triangle. Since c2 < a2 + b2, using the Pythagorean inequality theorem, we can say that it is an acute triangle. This triangle is an acute, scalene triangle.

Reﬂect and check If the triangle were a right triangle, we could also classify it this way. We could have a scalene right triangle or an isosceles right triangle, but we can never have an equilateral right triangle because the hypotenuse must always be longer than each leg.

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Example 4 Archeologists have uncovered an ancient pillar which, after extensive digging, remains embedded in the ground. The lead researcher wants to record all of the dimensions of the pillar, including its height above the ground. However, the team can only take certain measurements accurately without risking damage to the artifact. These measurements are shown in the diagram.

h cm 232.1 cm

x cm 178.3 cm

304.6 cm

Find the value of the variables.

Create a strategy The right triangle at the bottom of the pillar shows the length of a leg and its hypotenuse. Use the Pythagorean theorem to ﬁnd the length of the unknown leg and its variable. Then, the length of that unknown side is a leg of the triangle at the top of the pillar. Use the Pythagorean theorem to ﬁnd the length of the unknown hypotenuse and its variable.

Apply the idea Start with the triangle with hypotenuse length of 304.6 cm, one leg with length 232.1 cm, and the other leg of length x cm. a2 + b2 = c2 2

Pythagorean theorem

x + 232.1 = 304.6

Substitution

x + 53 870.41 = 92 781.16

Evaluate the exponents

x2 = 38 910.75

Subtract 53 870.41 from both sides of the equation

x = 197.26

Evaluate the square root of both sides of the equation

x = 197.26. Next, look at the triangle with hypotenuse length of h cm, one leg with length x = 197.26 cm, and the other leg of length 178.3 cm. a2 + b2 = c2 2

197.26 + 178.3 = h

70 702.4 = h2 265.9 = h

Pythagorean theorem Substitution Evaluate the exponents and addition Evaluate the square root of both sides of the equation

h = 265.9.

Reﬂect and check When there are multiple triangles involved, we need to carefully determine if a length is a leg or a hypotenuse. A length may be a leg in one triangle and a hypotenuse in another.

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Idea summary Given two sides in a right triangle, we can ﬁnd the third side length using the Pythagorean theorem. Pythagorean theorem: If a triangle is a right triangle, then the square of the length of the hypotenuse is equal to the sum of the squares of its leg lengths. a2 + b2 = c2

b C

Where a and b represent the lengths of the legs, and c represents the length of the hypotenuse.

c a

The Pythagorean inequality theorem can be used to determine if a non-right triangle is acute or obtuse. Converse of the Pythagorean theorem: If the lengths a, b and c of the three sides of a triangle satisfy the relationship a2 + b2 = c2, then the triangle is a right triangle. If a, b, and c satisfy a2 + b2 = c2, and are positive integers, then we say that they form a Pythagorean triple. Any whole number scalar multiple of a Pythagorean triple is also a Pythagorean triple.

Practice What do you remember? 1

Determine if each statement is true or false for the given triangle. a

a2 + b2 = c2

a2 = b2 + c2

c represents the length of the hypotenuse.

a and b represent the lengths of the legs.

To ﬁnd a we could use a =

a < b < c is always true.

We know the hypotenuse is the longest side because it is opposite the largest angle.

The triangle has two acute angles.

If c2 < a2 + b2, then the triangle is obtuse.

(12, 16, 26)

(9, 12, 15)

(12, 16, 20)

Calculate the value of x, as simpliﬁed radicals. All values of x are positive. a

400

Are these Pythagorean triples? a

Mathspace Virginia SOL Geometry mathspace.co

x2 = 22 + 32

82 = 52 + x2

Find the value of the variable in each triangle. Give your answer as a simpliﬁed radical. a

17 cm

c cm

9 cm

c cm

3 cm

24 b

13 m

Let’s practice 5

The screen on a handheld device has dimensions 29 cm by 21 cm, and a diagonal of length x cm. Find the value of x. Round your answer to two decimal places. 29 cm

x cm

21 cm

Fred and Carlo are playing football together. An opposition player who is 11 ft away is about to tackle Fred, and Carlo is 17 ft away from the opposition player and perpendicular to Fred. Fred passes the ball to Carlo. Find the distance the ball travels, rounding your answer to two decimal places.

11 ft 17 ft

Find the length of the unknown side, b, in each right triangle. Round your answer to two decimal places. a

A triangle whose hypotenuse is 3 ft in length and whose other side is 2 ft in length.

A triangle whose hypotenuse is 13 in in length and whose other side is 8 in in length.

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For each triangle: i

Determine whether the triangles are right triangles or not.

If the triangle is not a right triangle, determine if it is an acute or obtuse triangle.

b 35

10 23 37

25 1

SOL

Which of the following sets of lengths can represent the measures of the sides of a right triangle? A

SOL

(4, 5, 7)

(6, 12, 16)

(7, 11, 18)

(9, 40, 41)

Select the measures that could be the three side lengths of a right triangle. 5 cm

12 cm

35 cm

37 cm

Use the Pythagorean triple (22, 120, 122) to create two more Pythagorean triples. Explain your reasoning.

Andriana knows the two largest numbers in a Pythagorean triple are 41 and 40. Find the number that Andriana needs to complete the triad.

Each set of numbers represents the side lengths of a right triangle. Use Pythagorean triples to ﬁnd the missing side length of each triangle. a

402

⬚, 15, 17

7, ⬚, 25

For each set of three side lengths:

28, ⬚, 53

14, 48, ⬚

Determine if they could form a valid triangle.

If a valid triangle can be formed, determine if it is a right triangle. If they can’t form a valid triangle, explain why not.

1, 2, 4

Mathspace Virginia SOL Geometry mathspace.co

10, 14, 15

14, 48, 50

5, 6, 15

A city council plans to build a seawall and boardwalk along a local coastline. According to safety regulations, the seawall needs to be 5.25 m high and 7.66 m deep and will be built at the bottom of a 13.59 m long sloped section of shoreline. This means that the boardwalk will need to be built 2.43 m above the seawall, so that it is level with the public area near the beach. Find the width of the boardwalk, x m, rounding your answer to two decimal places. xm

2.43 m 7.66 m 5.25 m

13.59 m

The size of a computer monitor is often given as the length of its diagonal. Rowena has just bought a 19 inch monitor, meaning that the display has a diagonal length of 19 inches. The screen has a 4 : 3 width-to-height ratio (aspect ratio). Determine the width and height of Rowena’s new monitor.

Let’s extend our thinking 23

The set {6, 8, 10} forms a Pythagorean triple. a

Draw a triangle ABC with side lengths of {6 in, 8 in, 10 in}. Classify △ABC. Justify your classiﬁcation.

Draw a triangle DEF, such that △DEF ∼ △ABC. Classify △DEF. Justify your classiﬁcation.

For any triangle that is similar to △ABC, determine if its side lengths will form a Pythagorean triple.

Tricia is concerned that her neighborhood community center has some structural issues. She measures the length of one side of the roof, the height of one of the walls, and the length from the midpoint of the front edge to the vertex of the roof.

18 ft 21 ft 13 ft

The given diagram shows her measurements. Determine whether or not the community center may have a structural issue. Explain your reasoning. 25

404

Angelia hikes south of her starting position for 834 m and then 691 m east, before stopping for a lunch break. She then travels south again for 426 m before arriving at her ﬁnal destination. a

Find the shortest distance between where Angelia started and where she stopped for lunch. Round your answer to two decimal places.

Find the shortest distance between where Angelia started and where she ﬁnished her journey. Round your answer to two decimal places.

Write a problem that uses the Pythagorean theorem and builds off of the diagram. Solve the problem.

Mathspace Virginia SOL Geometry mathspace.co

30 ft Start

834 m

691 m 426 m

A farmer wants to build a fence around the entire perimeter of a chicken pen, as shown in the diagram. The fencing costs $37 per meter. Assume fencing is sold by meter. a

Determine a plan to ﬁnd the cost for building a fence along the entire perimeter of the land.

Find the cost for the farmer to build the fence along the entire perimeter of the land.

A 1m F

5m 7m

E 2m

y C

A soda can has a height of 11 cm and a radius of 4 cm. Find L, the length of the longest straw that can ﬁt into the can.

4 cm

Round your answer down to the nearest cm to ensure it ﬁts inside the can. 11 cm

L cm

By constructing a second triangle, △DEF, prove the converse of the Pythagorean theorem.

Given:

• AC2 = AB2 + BC2 Prove that: • 29

Given:

a E

• ABCD is a rectangle •

• Prove: w2 + x2 = y2 + z2

V G

x F

A right triangle has a hypotenuse with a length of 65 units. If the lengths of the legs are integers, ﬁnd two possible pairs of lengths that can complete the triangle. Explain your reasoning.

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8.02 Special right triangles After this lesson, you will be able to... • apply the properties of special right triangles to solve problems, including contextual problems.

Special right triangles Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 8.02 to answer these questions. 1.

Drag one of the sliders. How would you classify the triangle that forms?

What are the angle measures of the triangle?

If the length of the base is 1 unit, what are the lengths of the hypotenuse and the other leg?

Write out the measure of each side length and the opposite angle’s measure. What do you notice?

Repeat questions 1-3 with the other triangle.

While the Pythagorean theorem can apply to any kind of right triangle, there are particular types of right triangles whose side lengths and angles have helpful properties. We will look at two special right triangles. 45°- 45°- 90° triangle theorem

In a 45°- 45°- 90° triangle, given leg length = x, we have: x 45°

hypotenuse = x ⋅ This means that the ratio of sides is 1 : 1 :

45° x

The 45°- 45°- 90° triangle is an isosceles right triangle. 30°- 60°- 90° triangle theorem

In a 30°- 60°- 90° triangle, given short leg = x, we have: hypotenuse = x ⋅ 2

30°

longer leg = x ⋅ This means that the ratio of sides 1 :

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60° x B

The 30°- 60°- 90° triangle can be created by starting with an equilateral triangle and constructing an altitude. When solving problems involving special right triangles, radicals are often involved. When dividing by a radical, answers should be given with a rationalized denominator. A rationalized denominator is a denominator that does not contain a radical. To rationalize the denominator that contains a radical, we multiply the numerator and denominator by the radical term from the denominator. = 1 to make an equivalent fraction

Use that

= 2 to make the denominator rational

Use that

Divide out the common factor of 2

Write in simplest form

We can check on the calculator that

Example 1 Consider the triangle below:

45° 1

a Find the length of the hypotenuse.

Create a strategy Call the hypotenuse c. Use the Pythagorean theorem to ﬁnd the missing side of the triangle.

Apply the idea Pythagorean theorem

Substitution

Evaluate the exponents and addition

Evaluate the square root of both sides of the equation

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b Find and justify the ratio of proportionality between the side lengths of any 45°- 45°- 90° triangle.

Create a strategy Draw a triangle with the given angle measures. 45°

45°

Apply the idea All 45°- 45°- 90° triangles are similar because they have three corresponding congruent angles. We also know a 45°- 45°- 90° triangle is an isosceles triangle by the converse of the base angles theorem. Therefore, its legs are congruent.

45°

Let the legs of the triangle be x. Call the hypotenuse of the triangle y.

We can use the Pythagorean theorem to show that: Pythagorean theorem

Substitution

Combine like terms

Evaluate the square root of both sides of the equation

45° x

Using this, we can say that the ratio of the side lengths of any 45°- 45°- 90° triangle is x : x :

, or 1 : 1 :

Reﬂect and check Notice that this means if the legs were length 5, then the hypotenuse would have length

Example 2 Consider the triangle below: 30° 30° 2

60°

a Find the height of the triangle.

Create a strategy Equiangular triangles are also equilateral, so the base is also 2 units. The altitude of an equilateral triangle is also the median, so half of the base must be 1 unit. We can use the Pythagorean Theorem to ﬁnd the missing side of the triangle, which we will call b.

30° 2

60° 1

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Apply the idea Pythagorean theorem

Substitution

Evaluate the exponents

Subtract 1 from both sides of the equation

Evaluate the square root of both sides of the equation

Reﬂect and check Notice that the smallest angle is opposite the shortest side and the largest angle is opposite the longest side. • Side with length 1 is opposite the 30° angle ≈ 1.732 is opposite the 60° angle • Side with length • Side with length 2 is opposite the 90° angle

30° 2

60° 1

b Find and justify the ratio of proportionality between the side lengths of any 30°- 60°- 90° triangle.

Create a strategy Draw and label an equilateral triangle with a 30°- 60°- 90° triangle drawn. 30° 30°

60°

Apply the idea All 30°- 60°- 90° triangles are similar because they have three corresponding congruent angles. We also know a 30°- 60°- 90° triangle is half of an equilateral triangle. Let the short leg of the triangle be x. Then the hypotenuse of the triangle must be 2x. Call the missing leg of the triangle y.

30° 30° 2x y

We can use the Pythagorean theorem to show that: Pythagorean theorem

Substitution and evaluate the exponent 2

60°

Subtract x from both sides of the equation

Evaluate the square root of both sides of the equation and apply the commutative property of multiplication Using this, we can say that the ratio of the side lengths of any 30°- 60°- 90° triangle is x :

: 2x, or 1 :

: 2.

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Example 3 Consider the triangle shown:

15 45° c

a Find the exact value of a.

Create a strategy

Apply the idea

The triangle has a right angle and a 45° angle shown. This means that this triangle is a 45°- 45°- 90° triangle because the sum of interior angles in a triangle is equal to 180°.

Legs in 45°- 45°- 90° special right triangles are congruent, and the ratio of the sides is 1 : 1 : . We are given that 15 is the length of one leg from the diagram. a is the other leg of the triangle as a is opposite a 45° angle. So, a = 15.

b Find the exact value of c.

Create a strategy Using the 45°- 45°- 90° triangle theorem, the hypotenuse is property to ﬁnd c.

times the length of the leg. We want to use this

Apply the idea We know the length of the legs is 15 and c is the hypotenuse as it is opposite the right angle. So, c =

Reﬂect and check We could have also used the Pythagorean theorem to solve for c. a2 + b2 = c2

Pythagorean theorem

Substitute the side lengths a = 15 and b = 15

450 = c

Evaluate the exponents and addition

15 + 15 = c 21.21 = c

Evaluate the square root of both sides of the equation

≈ 21.21.

Example 4 Find the value of θ. 15

Create a strategy We can see that this is a right triangle where the ratio between the legs is 15 : we know it cannot be a 45°- 45°- 90° triangle.

. Since the legs are not congruent,

To conﬁrm it is a 30°- 60°- 90° triangle, we should simplify the ratio between the sides. If it is a 30°- 60°- 90° triangle, then we can use the side-angle relationship to determine the angle measure.

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Reﬂect and check We can check our solution using that the 45°- 45°- 90° triangle theorem says: If leg = x, hypotenuse = So since leg =

, hypotenuse =

This can be simpliﬁed to hypotenuse =

= 3 ⋅ 2 = 6, which is what we were given.

Example 6 Find the exact value of each variable in the diagram shown. p q 6 30° 30° s

Create a strategy Both triangles are special 30°- 60°- 90° triangles. We can use that if short leg = x, then hypotenuse = 2 ⋅ x and longer leg = x ⋅ . In other words, that the side lengths of a 30°- 60°- 90° triangle have a ratio of 1 :

Apply the idea For the bottom triangle, we have that hypotenuse = 6, so we get: hypotenuse = 2 ⋅ short leg Using 30°- 60°- 90° triangle theorem 6 = 2 ⋅ r Side opposite 30° angle is the shorter leg 3 = r Divide both sides by 2 And then: Using 30°- 60°- 90° triangle theorem

Using that r = 3 is the shorter leg

For the top triangle, we have that longer leg = 6, so we get: Using 30°- 60°- 90° triangle theorem

Side opposite 30° angle is the shorter leg

Divide both sides by

Rationalize the denominator

Evaluate products

Write fraction in simplest form

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: 2.

And then: Using 30°- 60°- 90° triangle theorem Using p =

is the shorter leg

Evaluate the product

So we have that: • p= • q= • r=3 • s=

Reﬂect and check We could also have found one length in each triangle using the 30°- 60°- 90° triangle theorem, and then found the third side using the Pythagorean theorem.

Idea summary We can use special right triangles to ﬁnd missing side lengths or angles in particular right triangles. The 45°- 45°- 90° triangle can be drawn by creating an isosceles right triangle with leg lengths of 1 or x. The 45°- 45°- 90° triangle theorem says that, given leg length = x, we have:

A x 45°

hypotenuse = x ⋅ 45° x

So the ratio of sides is 1 : 1 :

The 30°- 60°- 90° triangle can be drawn by creating an equilateral triangle with side lengths of 2 or 2x and then bisecting the top vertex with an altitude and looking only at one half. The 30°- 60°- 90° triangle theorem says that, given short leg = x, we have:

A 2x

hypotenuse = x ⋅ 2

60° x B

longer leg = x ⋅

30° C

So the ratio of sides 1 :

Practice What do you remember? 1

Use the Pythagorean Theorem to ﬁnd the exact length of the missing side, c. c

45° 3

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413

For the given triangles: i

Find the exact value of a.

Find the value of θ.

30°

Determine if each statement is true or false. a

All isosceles right triangles are congruent.

For a right triangle with side lengths 2, 4, and

For a triangle with side lengths of 1, 2, and angle.

, the angle opposite the side with length , has angles measuring 30°- 60°- 90°

, the side with length

A triangle with side lengths of

The legs of an isosceles triangle are always congruent.

The base angles of an isosceles right triangle will always be 45°.

, 3, and

is the hypotenuse. will be the largest

Label the triangles using the given measurements. a

Sides 1 Angles

30°

60°

Sides

Angles

45°

Let’s practice 5

For each triangle, identify if it is a 45°- 45°- 90° or 30°- 60°- 90° triangle. If it is not a special triangle, explain why not. a

A 414

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The equilateral triangle has side lengths of 2 cm. The perpendicular height of the triangle is shown in the diagram: a

Find θ.

Find the exact value of the height equilateral triangle.

Find the exact value of the perimeter of one of the smaller triangles.

60° 2 cm

Find the value of x in simplest radical form. A

x= 30° x

Fill in the blanks:

For the given triangle, h = ⬚, because the side opposite the 60° angle is ⬚ units, so this is the basic 30°- 60°- 90° triangle, where the side opposite the 30° angle has length ⬚.

30°

60° B

Find the value of the variable (s) in the following triangles. Express your answer in simplest radical form. a

c 45°

60° x

30° v y

30°

e a

f 14

b 60°

60°

45°

9 y

45°

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415

SOL

An equilateral triangle with height of 1 cm is folded in half.

1 cm 60°

60° x

What is x, the base of the equilateral triangle? B

A 11

Find θ in the following triangles. a

b θ 12

△ABC is an equilateral triangle with side lengths of 2x units. An altitude is drawn from A to meet side This information is shown in the diagram below: A

Find the m∠B.

Given that

bisects ∠BAC, ﬁnd the measure of ∠BAD.

Given that

bisects

, ﬁnd the length of

in terms of x.

Determine the length of the altitude

State whether the following statements are true or false. If false, correct the statement so it is true. i

The longer leg of a 30°- 60°- 90° triangle is twice the length of the shorter leg.

The shorter leg of a 30°- 60°- 90° triangle is opposite the angle measuring 30°.

iii The hypotenuse of a 30°- 60°- 90° triangle is

times the length of the longer leg.

iv The ratio of sides in a 30°- 60°- 90° triangle is 1 :

SOL

: 2.

Find the exact side length of an equilateral triangle with a perpendicular height of

The diagonals of a square measure 24 cm. Which is the length of a side of the square? A

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in.

at D.

Each half of a drawbridge open to let boats through. When it is down, the drawbridge is 30 ft above the water. When open, the highest point of the drawbridge is 40 ft above the water. Find x, the length of one half of the drawbridge.

x 30°

30 ft

40 ft

The bed of a dump truck is approximately 18 feet long. How high, to the nearest foot, is the top of the truck bed from the frame when the angle of elevation, x, is: a

30°?

45°?

60°?

18 ft

Let’s extend our thinking 17

Given: △ABC is a 45°- 45°- 90° triangle

Prove: c = 45° c

45° a

Zeki claims that all isosceles right triangles are similar to each other. Determine if Zeki is correct. Explain your reasoning.

Prove that △BCD ∼ △RPQ.

R 20 30° Q

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Vinod is trying to build this bookcase, but does not know all of the interior angles of the triangle. Vinod knows that the hypotenuse is 8 ft and the base is 4 ft. Explain why a special right triangle can be used to solve the problem.

A farmer has two ﬁelds. On the given diagram, the cows are in a rectangular section and the goats are in a triangular section. a

Find m.

Find t.

Determine which animals have the larger section. Justify your answer.

6 10

30° t

Using a compass and straightedge, construct a 30°- 60°- 90° triangle using the given equilateral triangle.

60°

Using a compass and straightedge, construct a 45°- 45°- 90° triangle in two different ways using the given rectangle.

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60°

With the given notation of ratios in mind, we then deﬁne the following three trigonometric ratios: Sine (sin)

Tangent (tan)

The ratio between the length of the side opposite to a given angle and the hypotenuse of the right triangle

The ratio between the sides opposite and adjacent to a given angle of a right triangle

Cosine (cos) The ratio between the length of the side adjacent to a given angle and the hypotenuse of the right triangle

That is, for a given reference angle θ, we have: sin θ =

cos θ =

tan θ =

Example 1 Write the following ratios for the given triangle: A α

θ B

a sin θ

Create a strategy We want to ﬁrst label the side lengths of the right triangle as shown in the diagram.

as it is directly across from the The reference angle is θ, so the opposite side is angle of reference, θ. The hypotenuse is as it is opposite the right angle. The Opposite adjacent side is as the side is connected to the angle of reference and is not the hypotenuse. Now that we have the sides of the right triangle labeled, we want to write the correct trigonometric ratio for sine. Sine is the ratio between the side opposite to the given angle and the hypotenuse of the right triangle.

Adjacent

Hypotenuse

θ B

Apply the idea sin θ =

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b cos θ

Create a strategy We want to reference the sides of the right triangle that we labeled in part (a) to write the correct trigonometric ratio for cosine. Cosine is the ratio between the side adjacent to the given angle and the hypotenuse of a right triangle.

Apply the idea cos θ =

c tan θ

Create a strategy We want to reference the sides of the right triangle that we labeled in part (a) to write the correct trigonometric ratio for tangent. Tangent is the ratio between the sides opposite and adjacent to the given angle of a right triangle.

Apply the idea tan θ =

Example 2 Consider the triangle in the ﬁgure. If sin θ = : B

a Which angle is represented by θ ?

Create a strategy In a right triangle, we know that the sine of an angle is equal to the ratio of the side length opposite that angle and the length of the hypotenuse. Since 4 is in the numerator, the side length with length 4 is the opposite side and the hypotenuse has a length of 5. We need to look at the diagram and ﬁnd the angle that is across from the opposite side with length of 4.

Apply the idea ∠BCA

Reﬂect and check We could have also named this angle ∠ACB, or even just ∠C since there is no ambiguity as to what that represents.

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b Find the value of cos θ.

Create a strategy From part (a), we labeled the opposite and hypotenuse sides. The adjacent side is labeled x in the diagram. In order to find cos θ, we need to find the value of x and then write the trigonometric ratio. To find the value of x, we can use the Pythagorean theorem.

Apply the idea a2 + b2 = c2 2

x +4 =5 2

x + 16 = 25

Pythagorean Theorem Substitution Simplify

x =9

Subtract 16 to both sides

x=3

Square root both sides

Now we want to write the trigonometric ratio for cosine, cos θ =

Reﬂect and check We could also realize that this a Pythagorean triple of 3, 4, and 5 at the start and then write the trigonometric ratio for cos θ. c Find the value of tan θ.

Create a strategy From part (a), we labeled the opposite and hypotenuse sides. From part (b) we labeled and found the value of the adjacent side. We can use these values to write the tangent ratio, which is the ratio between the sides opposite and adjacent to the given angle of the right triangle.

Apply the idea tan θ =

Example 3 Explain why sin (x) is the same for any of the triangles in the ﬁgure. F J H D A

x° B

Create a strategy Notice that each triangle shares ∠x, and each triangle also has a right angle. Translate each triangle to the right, so that we can see the four triangles from the ﬁgure separately.

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J H x°

D x°

x° A

B A

C A

x° G

E A

We can use similar triangles and trigonometric ratios to explain why sin (x) is the same for any triangle in the ﬁgure.

Apply the idea Since each triangle has a right angle and shares ∠x, we can say that each triangle is similar by the AA similarity theorem. sin (x) =

for the ﬁrst, small triangle. Since the triangles are similar, there is a constant of proportionality,

k, between the lengths of opposite sides and between the lengths of the hypotenuses. So, for any other triangle, sin (x) =

Since both the lengths of the opposite and hypotenuse sides change by the same factor, the ratio between them remains the same.

Example 4 Find the height, HC, of the tree. T 62.29 m

63 m

31.14 m

46 m

23 m

Create a strategy Notice that the skyscraper and the tree create two right triangles with a common ∠A. So, by the AA similarity theorem, △ATB ∼ △AHC. Choose one of the acute angles as the reference angle and set up trigonometric ratios. We can set ratios from the small and large triangles equal to one another, and solve for the height of the tree.

Apply the idea For ∠A, we have sin ∠A = that we have sin ∠A =

for the small triangle. We can use the same reference angle for the large triangle so .

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Since the triangles are similar, we know that their corresponding side lengths are proportional so

Multiply both sides of the equation by 31.14

Evaluate the division and multiplication

The height of the tree is 21 m.

Reﬂect and check We can conﬁrm that the height of the tree will lead △HAC to be a right triangle by evaluating its side lengths in the Pythagorean theorem: a2 + b2 = c2 2

Pythagorean theorem 2

21 + 23 = 31.14

970 = 969.7

Substitution Evaluate the exponents and addition

Since the sum of the squares of the legs is approximately equal to the square of the hypotenuse, the height of the tree creates a valid right triangle.

Idea summary Use the following notation for writing the three trigonometric ratios given the acute reference angle θ in a right triangle: sin θ =

cos θ =

tan θ =

Relationships between ratios Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 8.03 to answer these questions. 1.

What is the relationship between m∠ABC and m∠BAC?

What do you notice about the trigonometric ratios of the angles?

For the acute angles in a right triangle which are complementary, we can state that: • The sine of any acute angle is equal to the cosine of its complement: sin (θ ) = cos (90 − θ ) • The cosine of any acute angle is equal to the sine of its complement: cos (θ ) = sin (90 − θ ) From reference angle A

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From reference angle B

Example 5 Use the diagram to show that tan θ =

. C

1 unit

θ A

Create a strategy Write the side lengths of the given triangle for their respective trigonometric ratios: sin θ =

cos θ =

tan θ =

Apply the idea Using the given triangle, we have: sin θ =

= y cos θ =

= x tan θ =

If sin θ = y, cos θ = x, and tan θ = , we have that:

Example 6 Consider the following triangle: A β

θ B

a Write a rule to describe the relationship between θ and β.

Create a strategy

Apply the idea

We don’t know the angle measures of θ and β, but we Since m∠C = 90°, and the triangle sum theorem can use what we know about the triangle sum theorem to states that the sum of the angles in a triangle is 180°, write a rule for their relationship. we know that the sum of θ and β must be equal to 90°. So θ = 90 − β.

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b Use the rule you found in part (a) to determine the relationship between sin θ and cos β.

Create a strategy Evaluate sin θ and cos β, then compare their values.

Apply the idea sin θ =

Reﬂect and check We could also verify that cos θ = sin β :

and cos β = .

If cos θ = and sin β = , we have shown that cos θ = sin β. This shows that the cosine of the acute

sin θ is equal to cos β. This shows that the sine of the acute angle θ is equal to the cosine of its complement.

angle θ is equal to the sine of its complement.

Example 7 Complete the statement: If cos (30°) =

, then sin (?) =

Apply the idea The cosine of any acute angle is equal to the sine of its complement, so since cos (30°) = the complement of 30° must be 60°.

= sin (?), we know that

Reﬂect and check We could also draw a diagram with the given information from the trigonometric ratio cos (30°) =

30°

We know that the trigonometric ratio sin θ =

, so using the diagram we have the side opposite of a missing

angle and the hypotenuse of a right triangle. This means that the reference angle for sine must be the missing angle, which is 60° by the triangle sum theorem.

Idea summary For the acute angles in a right triangle that are complementary, we can state that: •

The sine of any acute angle is equal to the cosine of its complement: sin (θ ) = cos (90 − θ )

•

The cosine of any acute angle is equal to the sine of its complement: cos (θ ) = sin (90 − θ )

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Practice What do you remember? 1

For the following triangles, with reference angle θ, identify: i

The adjacent side

iii

The hypotenuse

The opposite side

b Z

Y C

θ B

θ b

θ X

c a

Identify the reference angle in each of the following triangles given the labeled side: B ITE POS

AD JA CE

A C C

CE JA D A A

OPPO

SITE

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Consider the following triangle:

State the opposite side to angle θ.

State the adjacent side to angle θ.

State the opposite side to angle α.

State the adjacent side to angle α.

Identify the angle that is opposite the hypotenuse.

θ B

Identify the measure of ∠F. 15

37° 9

12 G

Find the length of the adjacent side using the Pythagorean theorem. C

67° B

For the given triangle: a

m∠C = ⬚

⬚60° =

iii

tan 60° =

60° C

Determine if each statement is true or false i

Fill in the blanks.

sin 30° =

sin 30° = cos 30°

iii

sin 60° =

cos 60° =

Write down the following ratios for the given triangle: a

sin θ

sin α

cos θ

cos α

tan θ

tan α

A α

θ B

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Let’s practice 8

Evaluate the following, rounding your answer to two decimal places: a

sin 82°

cos 28°

tan 84°

tan 10°

cos 77°

tan 80°

sin 30°

cos 89°

For each of the given triangles, the angle measures have been rounded to the nearest whole number. i

Identify which trigonometric ratio we can ﬁnd without any additional information.

Use the triangle side lengths to write the value of the identiﬁed trigonometric ratio as a fraction. Convert the fraction to a decimal and round to two places.

iii

Use a calculator to ﬁnd the value of the identiﬁed trigonometric ratio of the given angle, rounded to two decimal places.

b 37

17 35

71°

28° 15

13°

9 53°

40 12

For the given triangle where all values have been rounded to the nearest whole number. a

Evaluate sin 37° i

Using a calculator

Using a ratio from the triangle

8 4

iii Explain why they are not exactly the same. b

Evaluate tan 53° i

Using a calculator

Using a ratio from the smaller triangle

37° 53° 5

D 10

53° B

iii Using a ratio from the larger triangle iv Explain why the ratios using the triangles are the same as one another, but are different from the calculator. 11

Consider the triangle below: A 13

θ B

Select the ratio that represents cos θ : A

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Consider the triangle below: A α

θ B

Select the ratio that represents sin θ : B

A 13

For each triangle, ﬁnd each of the following ratios. i

sin θ

cos θ

ii 21

iii

tan θ

B 17

A 15

8 C

θ B

13 12

For the given triangles: 60

30°

60° 30

430

sin 30°

cos 30°

iii

sin 60°

cos 60°

cos 20°

iii

sin 70°

cos 70°

Using △DEF, ﬁnd: i

Using △ABC, ﬁnd: i

20°

70°

sin 20°

What do you notice? Which ratios are equal?

Mathspace Virginia SOL Geometry mathspace.co

Consider the isosceles triangle below: a

Find the value of c.

Find the following trigonometric ratios.

Give your answer with a rationalized denominator. i

sin 45°

cos 45°

iii

tan 45° 45° 1

In the given ﬁgure, △ABC is an equilateral triangle: a

i b

Find the following trigonometric ratios: sin 60°

cos 60°

iii

tan 60°

30° 2x

Find the following trigonometric ratios: i

sin 30°

cos 30°

iii

tan 30° 60° B

In the following triangle, sin θ = :

Identify which angle is represented by θ.

Find the value of cos θ.

Find the value of tan θ.

In the following triangle, tan θ =

Identify which angle is represented by θ.

Find cos θ.

Find sin θ.

Consider the following triangle:

Write down the value of sin (90° − θ ).

Write down the value of cos θ.

Determine the relationship between sin (90° − θ ) and cos θ.

90° − θ 5

θ B

Complete the statement: If cos (72°) =

, then sin (⬚) =

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Let’s extend our thinking 21

Beth is building a house, as shown.

Roof height

She wants the roof height to be between 3.5 ft and 5 ft. She must decide the angle measure to use for the slant of the roof when the slant height is d ft.

d x°

State the inequality that Beth can use in order to ensure that her roof height will be within the desired range. 50 feet

= tan θ.

With reference to the diagram, show that

A α

B′

State whether the two right triangles are similar. Justify your answer. and

Find expressions for the length of

Describe the relationship of the trigonometric ratios from △ABC and △A′B′C′. Justify your answer.

Kaleigh is trying to prove that the SSA B′ theorem is enough to show that two triangles are similar. To do this, she draws the following ﬁgures: She said that using △ABC, the cosine ratio for α is cos α = and using △A′B′C′, the cosine ratio for α is cos α =

c α

A′

. Because

a α

C′

With reference to the diagram, determine if sin θ = cos α. Explain why or why not.

C′

the cosine ratio are the same, the triangles must be similar. Explain why Kaleigh’s argument is not valid. 25

Consider the following ﬁgure: a

A′

B 5

4 3 C

A right triangle is enlarged by a factor of 4. Explain why the trigonometric ratios are the same for both triangles.

Explain why tan 30° =

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. Include a diagram with your explanation.

Example 1 Find the missing side length for each triangle. Round your answer to two decimal places. a

f in 47°

8 in

Create a strategy We want to start by looking at the ﬁgure and determine how the labeled side lengths and angles are related. The triangle is a right triangle. The hypotenuse side length is 8 in. With respect to the 47° angle, f is the adjacent side. We can use this information to write the appropriate trigonometric ratio and then solve for f.

Apply the idea

Substitution

Multiply both sides of the equation by 8

Evaluate the multiplication and round to two decimal places

Reﬂect and check The answer for f should be less than the hypotenuse, since the hypotenuse is the longest side of the triangle. This is a good check when solving for missing side lengths.

11 ft

g ft

42°

Create a strategy We want to look at the diagram and determine how the given information is related. The hypotenuse is g since the side is opposite the right angle. The side with length of 11 ft is the opposite side with respect to the labeled angle of 42°. We can use this information to write the appropriate trigonometric ratio and then solve for g.

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32 in 52°

x in 52° 21 in

11 in

The side length adjacent to the reference angle in the small right triangle of the parallelogram is given, and we will need to solve for hypotenuse x, so we can use cosine of the reference angle to solve.

Apply the idea

Substitution

Multiply both sides of the equation by x

Divide both sides of the equation by cos (52°)

Evaluate the division and round to the nearest whole number

Reﬂect and check Note that the diagram is not drawn to scale, so the length of the top of the parallelogram, while shorter than the side of the parallelogram, may look longer.

Idea summary If we know an acute angle measure in the right triangle and a side length, we can solve for another side length of the triangle using sine, cosine, or tangent: 1. Highlight the reference angle. 2. Identify which sides are the hypotenuse, opposite, and adjacent - label if desired. 3. Determine which trigonometric ratio to use. Choose the ratio for which we have one of the required sides given and one is the missing side: • Sine: opposite and hypotenuse • Cosine: adjacent and hypotenuse • Tangent: opposite and adjacent 4. Set up the chosen ratio: sin θ =

cos θ =

tan θ =

5. Solve for the variable using inverse operations.

Solving for angles in a right triangle The trigonometric ratios take angles and give side ratios, but we are going to need functions that take side ratios and give angles. We are already familiar with certain operations having inverse operations that “undo” them - such as addition and subtraction, or multiplication and division. In a similar manner, we can apply inverse trigonometric functions to reverse, or “undo”, the existing trigonometric functions. 8.04 Solving right triangles mathspace.co

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The functions sine, cosine, and tangent each take an angle measure as input and return a side ratio as output. Inverse trigonometric functions Functions that determine the measure of an acute angle of a right triangle, given a ratio between two of its sides For a given angle θ we have: sin θ =

cos θ =

Hypotenuse

tan θ =

Hypotenuse

Opposite

Adjacent

Opposite

Adjacent

The inverse functions work in reverse, taking in the ratio of two known sides as the input and returning a missing angle measure as the output. So, the three inverse trigonometric functions are:

We can use these inverse trigonometric functions to determine an unknown angle measure when we know a pair of side lengths of a right-triangle. As these inverse functions “undo” the original functions, we have sin − 1 (sin (x)) = x cos − 1 (cos (x)) = x

tan − 1 (tan (x)) = x

and sin (sin − 1 (x)) = x cos (cos − 1 (x)) = x tan (tan − 1 (x)) = x Input

Input

sinθ

sin−1

sin Output

Output

sinθ

It is important to note that while the notation used to represent these inverse trigonometric functions looks like a power of −1, these functions are not the same as the reciprocals of the trigonometric functions. That is:

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Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 8.04 to answer these questions. 1.

Change the hypotenuse to match the desired ratio. Calculate the inverse trigonometric ratio, then check the box to show the angle. Does your calculated measurement match the actual angle measure?

Repeat by clicking New Goal.

Example 4 If cos θ = 0.256, ﬁnd θ. Round your answer to two decimal places.

Create a strategy We want to undo the cosine function, that is take the inverse trigonometric functions of both sides, to solve for θ.

Apply the idea cos θ = 0.256 −1

cos (cos (θ )) = cos− 1 (0.256) −1

Apply the inverse function to both sides

θ = cos (0.256)

Evaluate the inverse function on the left side of the equation

θ = 75.17°

Evaluate the inverse trigonometric function

Reﬂect and check We need to use a calculator here to evaluate cosine inverse. Make sure your calculator is in degree mode when working with inverse trigonometric ratios.

Example 5 Solve for the value of x. Round your answer to two decimal places.

25 cm

x°

20 cm

Create a strategy With respect to x, we are given the opposite side and adjacent side of the right triangle. The opposite side has a length of 20 cm and the adjacent side has a length of 25 cm. Use this information to set up the appropriate trigonometric ratio and use the correct inverse trigonometric ratio to solve for x.

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Apply the idea

Apply the inverse function to both sides

Evaluate the inverse function on the left side of the equation

Evaluate the inverse trigonometric function

Example 6 A ladder of length 3.46 m needs to make an angle with the wall of between 10° and 20° for it to be safe to stand on. At steeper angles, the ladder is at risk of toppling backwards when the climber leans away from it and at shallower angles, the ladder may lose its grip on the ground.

The base of the ladder is placed at 190 cm, 42 cm, or 86 cm from the wall. At which of these base lengths is the ladder safest to climb? Explain how you found your answer.

Create a strategy We know the ladder is safest to climb when the ladder makes an angle with the wall between 10° and 20°. So, let’s call the angle the ladder makes with the wall x°. We can also see that the ladder, wall, and base length from wall to ladder makes a right triangle, so we can use trigonometric ratios to help find x. We can label the diagram with the given information for each scenario, then decide which trigonometric ratio will be most useful in helping us find the missing angle measure since the ladder creates a right triangle when leaning against the wall. Since the distance from the base of the ladder to the wall is given in centimeters and the ladder is given in meters, we will first convert the distances to meters:

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Apply the idea First, draw the diagram with a 1.9 m distance from the base of the ladder to the wall. Since the opposite side of the reference angle and the hypotenuse of the triangle are given, we will use the sine function.

x°

3.46 m

1.9 m

Apply the inverse function to both sides

Evaluate the inverse function on the left side of the equation

Evaluate the inverse trigonometric function

Next, draw the diagram with a 0.42 m distance from the base of the ladder to the wall. Set up the trigonometric ratio and solve for x.

x°

3.46 m

0.42 m

Apply the inverse function to both sides

Evaluate the inverse function on the left side of the equation

Evaluate the inverse trigonometric function

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Finally, draw the diagram with a 0.86 m distance from the base of the ladder to the wall. Set up the trigonometric ratio and solve for x.

x°

3.46 m

0.86 m

Apply the inverse function to both sides

Evaluate the inverse function on the left side of the equation

Evaluate the inverse trigonometric function

The distance that leads to an angle with the wall of between 10° and 20° is when the ladder is placed 86 cm from the wall. This forms a safe 14.4° angle between the ladder and the wall.

Example 7 An isosceles triangle has equal side lengths of 10 in and a base of 8 in as shown.

10 in

8 in

Construct an altitude from C to . Since the altitude of an isosceles triangle is also its median, the segment constructed is also a perpendicular bisector of , so the length of is cut in half and we have two right triangles.

Find m∠A, to the nearest tenth of a degree.

Create a strategy

10 in

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4 in

10 in

4 in

The reference angle ∠A in the newly construct right triangle is shown with its hypotenuse and adjacent side. We can use the cosine function to ﬁnd the unknown angle.

Apply the idea

Apply the inverse function to both sides

Evaluate the inverse function on the left side of the equation

Evaluate the inverse trigonometric function

Idea summary We can use these inverse trigonometric functions to determine an unknown angle measure when we know a pair of side lengths of a right-triangle:

Practice What do you remember? 1

Evaluate the following, rounding your answer to two decimal places: a

9 sin 63°

8 tan 28°

Find the value of the following, rounding your answer to the nearest degree: a

sin − 1(0.7035)

cos − 1(0.4447)

tan − 1(0.5085)

sin − 1(0.2734)

cos − 1(0.6484)

tan − 1(8.1519)

sin − 1(0.8176)

cos − 1(0.7035)

For each of the following, ﬁnd θ to the nearest degree: a

sin θ = 0.5

tan θ = 1

cos θ = 0.5

sin θ = 0.906

tan θ = 2.2

sin θ =

cos θ =

tan θ =

Determine if each statement is true or false. a

m∠A = 28°

For ∠B, the side with length 15 is the opposite side.

For ∠B, the side with length 17 is the adjacent side.

For ∠A, the side with length 8 is the adjacent side.

The hypotenuse is opposite the right angle.

The shortest side is opposite the smallest angle.

C 8 15

62°

17 A

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443

Consider the following triangle: 11 in

h in

42°

Select the trigonometric ratio that could be used to solve for the side h. A 6

cos 42° =

sin 42° =

sin 30° =

cos θ =

Consider the following triangle:

30° x

Select the trigonometric ratio that could be used to solve for x. A 7

cos 30° =

sin 30° =

Consider the following triangle:

θ 12

Select the trigonometric ratio that could be used to solve for angle θ. A 8

sin θ =

cos θ =

Rose needs to ﬁnd tan 20° without using a calculator. Using the given triangle: a

Find the length of the adjacent side using the Pythagorean theorem.

Find tan 20° without using a calculator.

B 20° 4

11.7

If sin 30° = 0.5, which statement is true? A

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sin 60° = 0.5

Mathspace Virginia SOL Geometry mathspace.co

cos 60° = 0.5

sin 60° = 0.25

cos 30° = 0.5

Let’s practice 10

For each triangle: i

Identify which ratio can be used to ﬁnd the missing side

Find the missing side, rounding your answer to two decimal places.

b 29° f ft

7 in

27°

f in

8 ft

y in

25°

12 in 4 in

58° f in

Find the value of the variable in the following triangles, rounding your answer to two decimal places: a

a yd 65°

43 ft

15 yd f ft 32°

d 60°

35°

x b yd

15 yd

30°

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445

Find the value of the variable in the following triangles, rounding your answer to two decimal places: a

f mm

x ft 56°

25°

11 ft

60°

11 mm

43 in

b yd

33° 6 cm

x in

y cm 63°

32°

3.75 yd

Solve for all missing sides and angles. a

2313 ft

29° 14 yd

55°

c ft a yd

a ft

b yd

A tree has a height of h. The tip of the shadow of the tree has an angle of 29° and its distance from the top of tree is 23 m.

23 m

29°

Select the trigonometric ratio that could be used to solve for height of the tree. A sin 29° =

446

sin 29° =

Find h, correct to two decimal places.

Mathspace Virginia SOL Geometry mathspace.co

cos 29° =

Find the value of the variable (s) in the following diagrams, rounding your answer to two decimal places: a

b 54°

b 60°

26 in

x ft 43° h ft

h in

70 ft

45°

30 ft

Find the value of each variable, rounding your answer to the nearest degree: a

x°

23 in 25

x°

x° C

18 in

x°

17 x°

x° A

h 63.6 in

68.5 in 26

y°

4 x° B

j θ 12

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For each triangle, ﬁnd the value of: i

θ A

θ B

Consider the following ﬁgure: a

Find y, rounding your answer to one decimal place.

Find x, rounding your answer to one decimal place.

x°

y° 7

12 SOL

SOL

Jessica is stringing lights between two lamp posts in her garden on a slope. If the string is 100 in long and forms a 30° angle with the top post, what is the distance x, between the posts? A

50 in

57 in

87 in

115 in

100 in 30°

A skier is observing a snow fence marking a ski course as they descend a snowy slope. If the snow fence stretches from the top of the slope directly down to the skier as shown:

35°

Which is closest to the total length of the snow fence? A

25.8 ft

55.9 ft

64.3 ft

78.5 ft 45 ft

Consider the following ﬁgure. If l is the length of the ladder in meters, ﬁnd l, to an appropriate level of precision. 1.36 m

44°

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Let’s extend our thinking 27

Antima needs to ﬁnd all missing side and angle measures in △ABC.

Knowing that she has to ﬁnd the missing sides and angles of the triangle, in what order should she ﬁnd them and how?

Is this the only possible way she could solve the triangle? Explain. 27°

Find the value of the variable(s) in the given diagram, rounding your answer to two decimal places: y in

x in

8 in 64°

Consider the following ﬁgure: a

Find x, rounding your answer to two decimal places.

Find y, rounding your answer to two decimal places.

19 in

E y°

8 in x° G

33 in

Consider the given ﬁgure: y°

Find the value of each variable, rounding your answers to two decimal places: a

z° x° 30

Consider the shape given: a

Find the value of d, rounding your answer to one decimal place.

Find the value of f, rounding your answer to one decimal place. 6.4 in f in

d in

3.9 in

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51°

is perpendicular to the radius of the circle with center O.

is 24 in long and cuts the circle at C.

Find the length of the radius of the circle, rounding your answer to one decimal place.

Find the length of

, rounding your answer to one decimal place.

Find the length of

, rounding your answer to one decimal place.

62°

C 24 in

Calculate the size of angle DEC.

6 x+7 A

12 − x C 4 B

Consider the following ﬁgure: a

Determine which angle has the greater cosine value. Explain your reasoning.

Determine which angle has the greater sine value. Explain your reasoning.

Determine which angle has the greater tangent value. Explain your reasoning.

A ramp of length 311 cm is being built. The Construction Code says that “ramps shall have a running slope not steeper than 1 unit vertical in 12 units horizontal.” a

Determine the greatest possible angle that the ramp can make with the ground, rounded to two decimal places.

311 cm

Determine the maximum height the ramp could have, rounded to two decimal places.

If the ramp covers a horizontal distance of 300 cm, determine if it would be allowable under the Construction Code.

Make a recommendation for the height and horizontal distance covered by the ramp. Explain your reasoning.

height

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9 Polygons Big ideas • The relationships between the sides, angles, and diagonals of a polygon can be used to classify the polygon and solve problems.

Chapter outline 9.01 9.02 9.03 9.04

Angles of polygons Parallelograms Special parallelograms Trapezoids

454 466 483 496

Exploration Consider the quadrilateral shown:

What ways could you break the quadrilateral into the least number of triangles? How many triangles does this create?

Determine the sum of the interior angles of the quadrilateral.

Draw a hexagon and determine the least number of triangles you could break the polygon into, and determine the sum of its interior angles.

What can you say about the relationship between the number of sides on any polygon, the number of triangles it can be divided into, and the sum of its interior angles?

The sum of the interior angle measures of a polygon depends on the number of sides of the polygon. A polygon with n sides (or an n-gon) can always be divided into (n − 2) non-overlapping triangles. This fact and the triangle angle sum theorem helps us calculate interior angle sums and individual angle measures of regular polygons. Polygon interior angle sum theorem

Corollary to the polygon interior angle sum theorem

The sum of the measures of the interior angles of a convex n-gon is equal to (n − 2) 180°.

The measure of each interior angle of a regular n-gon is

Example 1 Consider the polygon angle sum theorem. a Prove the interior polygon angle sum theorem works for a pentagon.

Create a strategy First, draw a pentagon with diagonals drawn from a vertex. D

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Apply the idea

Reﬂect and check

Given a pentagon with the diagonals drawn from a vertex, we can identify the number of triangles in the pentagon. The pentagon is broken into 3 triangles, and the sum of the interior angles of each triangle is 180°. Using this information and the angle sum addition postulate, we know the sum of the interior angles of the pentagon is (5 − 2) ⋅ 180° = 3 ⋅ 180° = 540°.

We can use another polygon to prove the polygon angle sum theorem, such as an octagon.

Draw an octagon with the diagonals drawn from one vertex. The total number of triangles drawn is 6. We know that the sum of the interior angles of each triangle is 180°, so the sum of the angles must be 1080°. This also follows from (8 − 2) ⋅ 180° = 6 ⋅ 180° = 1080°.

b Explain why the polygon angle sum theorem will work for any convex polygon.

Create a strategy We will use a convex polygon and draw triangles to conceptualize the theorem.

Apply the idea Consider a convex polygon with n sides and n vertices. To divide any n-gon into non-overlapping triangles, we will start with a random vertex, A1, and draw a diagonal to every other non-adjacent vertex. A1 is adjacent to A2 and An, so those diagonals are already part of the polygon. The diagonal between A1 and A3 creates the ﬁrst triangle, the diagonal from A1 to A4 creates the second triangle, and this pattern continues until we connect A1 to An − 1, creating the last triangle. Thus, we will have created (n − 2) non-overlapping triangles. By the Triangle Angles Sum Theorem, we know that the sum of the interior angles of a triangle is 180°, so the sum of the interior angles of (n − 2) triangles is (n − 2) ⋅ 180°.

A1 An

n−2

An − 1

2 A3

Example 2 For a regular 24-gon:

a Find the sum of the interior angles.

Create a strategy We use the polygon angle sum theorem to ﬁnd the sum of the interior angles. The polygon angle sum theorem states that the sum of the interior angles of a convex polygon is equal to (n − 2) 180° for n sides. A 24-gon is a convex polygon and has 24 sides. So we want to substitute 24 for n in the expression and evaluate.

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Apply the idea (24 − 2) 180° = 3960° The sum of interior angles is 3960°

b Find the measure of a single interior angle.

Create a strategy Since we want to ﬁnd the measure of a speciﬁc interior angle of a regular polygon, we want to use the corollary to the polygon angle sum theorem. This theorem states that the interior angle is equal to the sum of the interior angles divided by the number of sides, Now we need to substitute 24 for n and evaluate.

Apply the idea

Reﬂect and check

The measure of an interior angle is 165°

Note that we can only ﬁnd the measure of the interior angle because this is a regular polygon and all angles are the same. If it was a non-regular 24-gon, the interior angle sum would still be the same, but each interior angle could be a different measure.

Example 3 Find the value of y. 57° y 55°

Create a strategy We know the measures of 3 angles in the given quadrilateral, so we can use the polygon angle sum theorem to determine the sum of the interior angles and write and solve an equation to ﬁnd y.

Apply the idea (n − 2)180 = (4 − 2)180 = 360

Polygon angle sum theorem for a quadrilateral Evaluate the parentheses and multiplication

Since the sum of the interior angles of a quadrilateral are 360°, we have 57 + 90 + 55 + y = 360 202 + y = 360 y = 158

Sum of the interior angles of the quadrilateral Combine like terms Subtract 202 from both sides

y = 158°.

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Example 4 Determine the number of sides of a regular polygon when each interior angle has a measure of 150°.

Create a strategy To ﬁnd the number of sides of the regular polygon, we can use the corollary to the polygon angle sum theorem:

where I is the measure of each interior angle and n is the number of sides. We can substitute the given interior angle measure and solve for n.

Apply the idea Given that each interior angle has a measure of 150°, we can substitute this value into the equation and solve for n: Substitute I = 150°

Multiply both sides by n

Distribute 180°

Subtract 180°n from both sides

Divide by −30° on both sides

This means that the regular polygon has 12 sides.

Reﬂect and check We can check our answer by substituting n = 12 back into the formula for the measure of each interior angle of a regular polygon: Substitute n = 12

Evaluate the subtraction

Evaluate the multiplication and division

Since the calculated interior angle measure matches the given value of 150°, our answer of 12 sides for the regular polygon is correct.

Idea summary We can use the polygon angle sum theorem and its corollary to ﬁnd unknown angles of convex and regular n-gons:

458

•

The sum of the interior angles of a convex n-gon is equal to (n − 2) 180°

•

The measure of each interior angle of a regular n-gon is

Mathspace Virginia SOL Geometry mathspace.co

Exterior angles in polygons Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 9.01 to answer these questions. 1.

How would you describe the angles measured in the applet?

What do you notice about the angle measurements?

We can take any side of a polygon and extend it to create an exterior angle.

The exterior angle and the corresponding interior angle form a linear pair (add to 180°). The sum of all of the exterior angles of a polygon is 360°. Linear pair

Exterior angle Sum

Polygon exterior angle sum theorem The sum of the exterior angles of a convex polygon is 360°.

Corollary to the polygon exterior angle sum theorem The measure of each exterior angle of a regular n-gon is

⋅ 360°.

Example 5 Prove the polygon exterior angle sum theorem.

Create a strategy The theorem states that the sum of the exterior angles of any polygon is 360°. We already know the sum of the interior angles is 180 (n − 2) for an n-sided polygon.

Apply the idea Let N be the sum of the exterior angles of an n-sided polygon. Let’s focus on one exterior angle ﬁrst for an n-sided polygon.

x°1

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For this exterior angle with measure x1°, we can see that it forms a linear pair with the related interior angle. This means the interior angle would measure (180 − x1)°. This will be true for each exterior-interior angle pair, so for any exterior angle measuring xn, the interior angle will be (180 − xn). So we have: • Sum of the exterior angles: N = x1 + x2 + … + xn • Sum of the interior angles that they form linear pairs with exterior angles: S = (180 − x1) + (180 − x2) + … + (180 − xn) Commutative, associative, and distributive properties S = 180n − (x1 + x2 + … + xn) of multiplication S = 180n – N • Sum of interior angles using polygon angle sum theorem:

Substitution

S = 180 (n − 2) Putting together the two ways to express the interior angle sum we get: S = 180(n − 2) 180 (n − 2) = 180n – N

Transitive property of equality

180n − 360 = 180n – N

Distributive property

−360 = − N

Subtract 180n from both sides

360 = N

Multiply both sides by −1

So the sum of the exterior angles of a polygon is 360°.

Example 6 Determine the value of y: 60° 112° 115° (3y + 18)° 42.5°

Create a strategy We want to solve for y which is part of an expression of an exterior angle. To do so, we can use the polygon exterior angle theorem to write an equation relating the exterior angles of the polygon after calculating the measures of the exterior angles.

Apply the idea The exterior angle that is supplementary to 115° must be 65°, and the exterior angle that is supplementary to 112° must be 68°. 60 + 65 + 42.5 + (3y + 18) + 68 = 360 3y + 253.5 = 360

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Exterior angle formula Combine like terms

3y = 106.5

Subtract 253.5 from both sides

y = 35.5

Divide both sides by 3

Example 7 Determine the number of sides of a regular polygon when each exterior angle has a measure of 20°.

Create a strategy To ﬁnd the number of sides of the regular polygon, we can use the corollary to the polygon exterior angle sum theorem:

where E is the measure of each exterior angle and n is the number of sides. We can substitute the given exterior angle measure and solve for n.

Apply the idea

Reﬂect and check

Given that each exterior angle has a measure of 20°, we can substitute this value into the equation and solve for n:

We can check our answer by substituting n = 18 back into the equation for the measure of each exterior angle of a regular polygon: Substitute n = 18

Substitute E = 20°

Multiply both sides by n

Divide both sides by 20°

Since the calculated exterior angle measure matches the given value of 20°, our answer of 18 sides for the regular polygon is correct.

This means that the regular polygon has 18 sides.

Evaluate

Idea summary We can use the polygon exterior angle sum theorem and its corollary to ﬁnd unknown angles of convex and regular n-gons: •

The sum of the exterior angles of any polygon is 360°.

•

The measure of each exterior angle of a regular n-gon is

⋅ 360°.

Practice What do you remember? 1

Name the following regular polygons. a

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461

For each of the following regular polygons: i

Find the sum of the interior angles.

Find the measure of an interior angle.

Quadrilateral

Dodecagon

SOL

Heptagon

22-gon

The measure of each exterior angle of a regular polygon is 24°. a

What is the sum of the exterior angles of this regular polygon?

Determine the number of sides in this regular polygon. Explain how you used the exterior angle to ﬁnd this answer.

State the sum of the exterior angles of a 13-gon.

Let’s practice 5

Find the value of x in the following quadrilaterals: a

b 60°

80°

120°

95° x°

155° 70°

x°

74°

x°

107°

83°

131° x°

55°

x°

(3x + 2)° 114°

97°

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128°

SOL

Solve for x in the following polygons: a

(2x − 50)°

(x + 40)°

(2x + 5)° L x°

(x + 30)° P

(x + 20)°

x° (2x)° O

A landscape architect is designing a ﬁve-sided garden plot. If four of the angles in the plot measure 80°, 100°, 95°, and 85°, what is the measure of the ﬁfth angle? A

80°

150°

61° N

50°

80°

100°

120°

Solve for x. a

b (x + 40)°

60° 73°

(2x + 20)°

x° 3x°

46°

6x°

A polygon has n sides. Which of the following represents the sum of its interior angles? A

SOL

180(n − 2)°

180n°

360°

The ﬁgure shown is a regular hexagon. i

What is the length of diagonal AC?

Explain your answer. in

A C

18 in

B D

12 in

360(n − 2)° A

A house is built with a pitched roof. The engineer states that the angle of the pitch is 14°. Solve for x.

6 in

D 14°

x°

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SOL

Determine the number of sides of a regular polygon when: a

The interior angles sum to 360°

Each interior angle has a measure of 135°

Each exterior angle has a measure of 24°

Each exterior angle has a measure of 15°

The interior angles sum to 1980°

Each interior angle has a measure of 160°

This figure is composed of a regular octagon and a rectangle.

What is the value of each of the angles labeled as x? A

45°

90°

135°

180°

Let’s extend our thinking 14

If possible, draw an example of each of the following quadrilaterals. If not possible, explain why not. a

A regular quadrilateral.

A quadrilateral with four congruent sides with no interior angles with measure of 90°.

A convex quadrilateral with an exterior angle of 300°.

A quadrilateral with two pairs of congruent angles.

A quadrilateral with exactly one pair of congruent angles.

In a regular polygon, each interior angle measures x degrees and each exterior angle measures y degrees. Write an equation that shows the relationship between x and y. Explain how you know.

ABCDE is a regular pentagon. A x G 72°

B I

F J D

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What is the measure of ∠AED? Explain how you know.

Explain why x = 36°.

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The pattern on a soccer ball is a tessellation, or a repeated pattern with no overlaps or gaps. On a threedimensional soccer ball, the tessellation consists of regular pentagons and hexagons:

When laid ﬂat, a piece of the pattern looks like this: K

B C D

J A

H F

Explain why the pattern on a soccer ball is impossible to tesselate on a two-dimensional surface. 18

For each of the following conjectures: i

Explain their error.

Use their diagram to show how we could prove that the sum of the interior angles of a hexagon is 720°.

Ignacio noticed that a hexagon can be divided into 5 triangles by connecting all of the vertices a point on one of the sides. He made the conjecture: “A hexagon can be divided into 5 triangles, so the sum of interior angles is (5) ⋅ 180° = 900°.”

Kelley noticed that around a point inside a hexagon, we can draw 6 triangles. She made the conjecture: “A hexagon can be divided into 6 triangles, so the sum of interior angles is (6) ⋅ 180° = 1080°.”

Which of these could be the measure of an interior angle in a regular polygon? Choose every option that is correct. Justify your choice(s). A

95°

108°

120°

190°

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9.02 Parallelograms After this lesson, you will be able to... • use the definition of a parallelogram to identify and prove the properties of a parallelogram. • solve problems using the properties of parallelograms. • determine if a polygon is a parallelogram using its properties. • use constructions to verify properties of parallelograms.

Explore properties of parallelograms Quadrilateral

Parallelogram

A polygon with exactly four sides and four vertices

A quadrilateral with both pairs of opposite sides parallel.

Consecutive angles are angles of a polygon that share a side.

Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 9.02 to answer these questions. 1. 2. 3. 4.

A quadrilateral is a parallelogram if and only if its opposite sides are ⬚.

A quadrilateral is a parallelogram if and only if its opposite angles are ⬚. In a parallelogram, consecutive angles will be ⬚.

A quadrilateral is a parallelogram if and only if its diagonals ⬚ each other.

We have many tools in our mathematical tool box to help with proofs now, for example:

The diagonal of a parallelogram is a transversal between a pair of parallel lines

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The other diagonal of a parallelogram is also a transversal between a pair of parallel lines

If we extend the sides of a parallelogram, one pair of sides can be seen as transversals of the other pair of parallel sides.

We can utilize congruent triangle theorems since the diagonals of a parallelogram break it into triangles: • Side-side-side, or SSS: The two triangles have three pairs of congruent sides • Side-angle-side, or SAS: The two triangles have two pairs of congruent sides, and the angles between these sides are also congruent • Angle-side-angle, or ASA: If two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent • Angle-Angle-Side, or AAS: If two angles and the non-included side of one triangle are congruent to the corresponding parts of another triangle, the triangles are congruent

Example 1 Consider the quadrilateral shown:

a If ABCD is a parallelogram, prove the opposite sides are congruent.

Create a strategy Start by stating what is given and the deﬁnition of a parallelogram to build the proof.

Apply the idea

1. 2. 3.

To prove: Opposite sides of a parallelogram are congruent Statements Reasons ABCD is a parallelogram Given ∥ and ∥ ∠BAC ≅ ∠DCA and ∠ACB ≅ ∠CAD

Alternate interior angles theorem Reﬂexive property of congruence

4. 5.

Deﬁnition of a parallelogram

△ABC ≅ △CDA and

ASA congruence theorem CPCTC

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b Prove that if opposite sides of a quadrilateral are congruent, then it is a parallelogram.

Create a strategy Construct a diagonal from B to D. Include congruence markings based on what is given. A

Apply the idea To prove: If opposite sides of a quadrilateral are congruent, then it is a parallelogram Statements Reasons 1.

Given

and

Reﬂexive property of congruence

2. △ABD ≅ △CDB ∠ABD ≅ ∠CDB and ∠ADB ≅ ∠CBD

3. 4. 5.

∥

SSS congruence theorem CPCTC Since ∠ABD ≅ ∠CDB by converse of alternate interior angles theorem Since ∠ADB ≅ ∠CBD by converse of alternate interior angles theorem Deﬁnition of a parallelogram

ABCD is a parallelogram

c Verify the opposite sides of a parallelogram are congruent using constructions.

Create a strategy Construct a copy of the opposite sides to two consecutive sides to show that the parallelogram’s opposite sides are congruent.

Apply the idea

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Step 2.

Step 1. Identify the side we want to copy. Let’s start with

Place the compass point on vertex A.

A B

D D

Step 3.

Step 4.

Adjust the compass width to vertex B.

Without changing the compass width, move the compass to vertex D.

If the other end of the compass lines up with C, then

is congruent to

Now let’s check the other pair of sides. B

Step 2.

Step 1. Identify the side we want to copy. Let’s start with

Place the compass point on vertex D.

Step 3.

Step 4.

Adjust the compass width to vertex A.

Without changing the compass width, move the compass to vertex B.

If the other end of the compass lines up with C, then

is congruent to

Example 2 Consider the quadrilateral shown: A

B P

a If ABCD is a parallelogram, prove that the diagonals bisect each other.

Create a strategy Use the alternate interior angles theorem, congruence theorems, and the deﬁnition of a bisector to help prove that the diagonals bisect each other.

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Apply the idea ABCD is a parallelogram (Given)

AB || DC and AD || BC (Definition of a parallelogram) ∠ ABD ≅ ∠ CDB ∠ ADB ≅ ∠ CBD (Alternate interior angles theorem)

∠ ABD ≅ ∠ CDB ∠ BAC ≅ ∠ DCA (Alternate interior angles theorem)

BD ≅ BD (Reflexive property of congruence)

△ABP ≅ △CDP (ASA Congruence theorem)

△ABD ≅ △CDB (ASA Congruence theorem)

AB ≅ CD (CPCTC)

BP ≅ DP and AP ≅ CP (CPCTC) AC bisects BD and BD bisects AC (Definition of a bisector)

b Prove that if the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Create a strategy Draw congruence markings using the given diagram to support starting a proof. B

A P

Apply the idea To prove: If the diagonals of a quadrilateral bisect each other, then it is a parallelogram Statements Reasons 1. 2. 3. 4. 5. 6. 7.

470

and ∠APB ≅ ∠CPD and ∠APD ≅ ∠BPC △ABP ≅ △CDP and △APD ≅ △CPB

Given

and ∠ABP ≅ ∠CDP and ∠ADP ≅ ∠CBP

CPCTC

∥

and

∥

ABCD is a parallelogram

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Vertical angles theorem SAS congruence theorem CPCTC Since ∠ABP ≅ ∠CDP and ∠ADP ≅ ∠CBP by converse of alternate interior angles theorem Deﬁnition of a parallelogram

c Verify that both halves of each diagonal are congruent using constructions.

Create a strategy Construct a copy of one half of the diagonals for each diagonal to show that both halves of each diagonal in a parallelogram are congruent.

Apply the idea

B P

Step 2.

Step 1. Identify the side we want to copy. Let’s start with A

Place the compass point on vertex A. A

B P

P D

Step 3.

Step 4.

Adjust the compass width to vertex P.

Without changing the compass width, move the compass to vertex C.

If the other end of the compass lines up with P, then

is congruent to

Now let’s check the other diagonal.

B P

P D

Step 2.

Step 1. Identify the side we want to copy. Let’s start with A

Place the compass point on vertex B. A

B P

Step 3.

Step 4.

Adjust the compass width to vertex P.

Without changing the compass width, move the compass to vertex D.

If the other end of the compass lines up with P, then

is congruent to

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Example 3 Use geometric constructions to show that ABCD is a parallelogram if its opposite angles are congruent.

Create a strategy Construct a copy of the opposite angles to two consecutive angles to show that if the quadrilateral’s opposite angles are congruent, it must be a parallelogram.

Apply the idea A

Step 1.

Step 2.

Identify the angle we want to copy. Let’s start with ∠A.

Draw a ray from vertex C that will form one of the legs of the copied angle.

Step 3.

Step 4.

With the compass point on vertex A, use the compass to draw an arc that intersects both legs.

Copy the arc in Step 3 by placing the point end of the compass onto vertex C.

Step 5.

Step 6.

On ∠A, use the compass to measure the distance between the points where the legs of ∠A meet the arc drawn in Step 3.

Without changing the compass width, copy the distance by placing the compass point where the ray meets the copied arc and draw an intersecting arc.

Step 7. Draw a ray that shares its end point with the ray from Step 2, and goes through the intersection found in Step 6.

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If the constructed copy of ∠A lines up with ∠C, then ∠A is congruent to ∠C. A

Step 1.

Step 2.

Identify the angle we want to copy. Let’s start with ∠B.

Draw a ray from vertex D that will form one of the legs of the copied angle.

Step 3.

Step 4.

With the compass point on vertex B, use the compass to draw an arc that intersects both legs.

Copy the arc in Step 3 by placing the point end of the compass onto vertex D.

Step 5.

Step 6.

On ∠B, use the compass to measure the distance between the points where the legs of ∠B meet the arc drawn in Step 3.

Without changing the compass width, copy the distance by placing the compass point where the ray meets the copied arc and draw an intersecting arc.

Step 7. Draw a ray that shares its end point with the ray from Step 2, and goes through the intersection found in Step 6. If the constructed copy of ∠B lines up with ∠D, then ∠B is congruent to ∠D.

Idea summary We can use the deﬁnition of a parallelogram, theorems about congruency, and transversals to prove properties of parallelograms.

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Apply properties of parallelograms Parallelograms have special properties regarding side lengths, angles, and diagonals. We can use these properties to ﬁnd unknown angles or sides of parallelograms, or to prove that a quadrilateral is a parallelogram. Parallelogram opposite sides theorem

Parallelogram consecutive angles theorem

A quadrilateral is a parallelogram if and only if its opposite sides are congruent

If a quadrilateral is a parallelogram, then its consecutive angles are supplementary Example: ∠ADC and ∠DAB are supplementary A

Parallelogram opposite angles theorem A quadrilateral is a parallelogram if and only if its opposite angles are congruent

Parallelogram diagonals theorem A quadrilateral is a parallelogram if and only if its diagonals bisect each other A

B E C

We may use these properties to solve problems when we are told that a diagram is a parallelogram.

Example 4 Find the missing parts of the parallelograms. a Given parallelogram PQRS, ﬁnd RS. Q

5.08

2.41 P

Create a strategy Since we know PQRS is a parallelogram, we want to use the theorems about parallelograms to determine RS.

Apply the idea Opposite sides of a parallelogram are congruent so

RS = PQ = 2.41

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b Given parallelogram DEFG, ﬁnd m∠DGF. E

F (5x)°

(2x + 5)° D

Create a strategy The two labeled angles, ∠DEF and ∠EDG, are consecutive angles. Since DEFG is a parallelogram, the consecutive angles are supplementary. We want to write an equation relating the two labeled angles and then solve for x. Once we solve for x, we then want to use the theorem that states that opposite angles of a parallelogram are congruent. Using this theorem, we know that ∠DEF ≅ ∠DGF. We want to substitute the value we solved for x and into 5x and evaluate m∠DEF as this will be the same as m∠DGF.

Apply the idea (5x) + (2x + 5) = 180 7x + 5 = 180

Consecutive angles are supplementary Combine like terms

7x = 175

Subtract 5 from both sides of equation

x = 25

Divide both sides of equation by 7

Since we know that ∠DEF ≅ ∠DGF, we know that ∠DGF = 5x Substituting 25 for x and evaluating, we get 5 (25) = 125. m∠DGF = 125°

Example 5 Determine whether or not each of the given quadrilaterals is a parallelogram. a

101°

79°

101°

Create a strategy We know that if a quadrilateral is a parallelogram, its opposite angles are congruent and its consecutive angles are supplementary. Use the polygon angle sum theorem to ﬁnd the missing angle and determine if the quadrilateral satisﬁes the conditions of a parallelogram.

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Apply the idea By the polygon angle sum theorem, we know that the sum of the angles of an quadrilateral must be (4 − 2) ⋅ 180° = 2 ⋅ 180° = 360°. Let the missing angle be x. For the given quadrilateral, we have 79 + 101 + 101 + x = 360 281 + x = 360 x = 79

Polygon angle sum theorem Combine like terms Subtract 281 from both sides

Since the unknown angle in the quadrilateral is 79°, we know that the opposite angles are congruent and therefore the quadrilateral is a parallelogram.

Reﬂect and check We can also use that consecutive angles are supplementary in a parallelogram, so 101° + x = 180° and the unknown angle must be 79°, so we have that opposite angles are congruent and therefore the quadrilateral is a parallelogram.

42 m 42 m

30 m 30 m

Apply the idea Since 42 m ≠ 30 m, the diagonals of the quadrilateral do not bisect each other and the quadrilateral is not a parallelogram.

Reﬂect and check If the segments were instead the same length along each diagonal, we could use the fact that if a quadrilateral is a parallelogram, then its diagonals bisect each other.

15 ft

Create a strategy We aren’t given information about the angles or diagonals of the quadrilateral, so we rely on determining if the quadrilateral meets the criteria: If a quadrilateral is a parallelogram, then its opposite sides are congruent.

Apply the idea Since the opposite sides of the quadrilateral are congruent, the quadrilateral is a parallelogram.

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Example 6 Solve for the unknown variables in the diagram that make the quadrilateral a parallelogram. 4x 74°

32° 5y − 5 4y + 6

4z°

Create a strategy We can use the fact that the diagonals of the parallelogram form transversals, so we can use the alternate interior angles theorem to state that 32 = 4z. We know that diagonals of a parallelogram bisect each other, so 4y + 6 = 5y − 5. Since opposite sides of a parallelogram are congruent, 4x = 16.

Apply the idea We have 32 = 4z

Alternate interior angles theorem

8=z

Divide both sides by 4

4y + 6 = 5y – 5

Diagonals of a parallelogram bisect each other

4y + 11 = 5y

Add 5 to both sides

11 = y

Subtract 4y from both sides

4x = 16

Opposite sides of a parallelogram are congruent

x=4

Divide both sides by 4

Idea summary Use the following about quadrilaterals to solve problems involving parallelograms: • • • •

A quadrilateral is a parallelogram if and only if its opposite sides are congruent A quadrilateral is a parallelogram if and only if its opposite angles are congruent In a parallelogram, consecutive angles will be supplementary A quadrilateral is a parallelogram if and only if its diagonals bisect each other

Practice What do you remember? 1

Decide whether each statement below is true or false. If false, give a counterexample. a

Opposite angles of a parallelogram are congruent.

In a parallelogram, all angles are right angles.

In any parallelogram, diagonals bisect each other.

In a parallelogram, adjacent angles are supplementary.

All parallelograms must have all sides of equal length. 9.02 Parallelograms mathspace.co

477

Based on the given markings, determine whether or not we can conclude the ﬁgure is a parallelogram. a

Consider parallelogram ABCD. Decide where each pair of angles is supplementary, congruent, or neither. a

∠A and ∠B

∠A and ∠C

∠C and ∠B

Let’s practice 4

Solve for each variable in the following parallelograms: a

a° b°

k+4

8 11

m 101°

2z + 1

w+3 5 q−3

4w 4z − 5

Given: Parallelogram ABCD a

Solve for x.

Solve for y.

Solve for the perimeter of ABCD.

Given: Parallelogram FGHJ a

If HF = 32, ﬁnd KF.

If ∠F = 55°, ﬁnd ∠H.

If ∠G = 127°, ﬁnd ∠H.

2y − 6 3x − 3

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C G

F K

478

y+2

x+7

Find the value of x if the perimeter of the parallelogram given is 48 m. 4x 8x

Consider the following proof to show that the opposite sides of parallelogram ABCD are congruent:

To prove: ? 1. 2. 3.

Statements ABCD is a parallelogram ∥ and ∥ ∠BAC ≅ ∠DCA and ∠BCA ≅ ∠DAC

Deﬁnition of a parallelogram Alternate Interior Angles Theorem Reﬂexive property of congruence

4. 5.

Reasons Given

△ABC ≅ △CDA and

Angle-Side-Angle Congruence Corresponding parts of congruent triangles are congruent (CPCTC)

Determine whether each of the following statements is shown to be true from the given proof.

If a quadrilateral is a parallelogram, then its opposite angles are congruent.

If a quadrilateral has opposite sides congruent, then it is a parallelogram.

If a quadrilateral is a parallelogram, then its opposite sides are congruent.

If a quadrilateral has consecutive interior angles supplementary, then its opposite sides are congruent.

etermine what additional piece of information about the angles in PQRS D would allow you to prove it is a parallelogram.

Determine what additional piece of information about the sides of PQRS would allow you to prove it is a parallelogram.

P T

Find the value of a that makes PQRS a parallelogram.

(4a − 12)° 12a° P

Find the value of y that makes WXY Z a parallelogram.

3a°

(4x + 6)°

(6x − 8)°

(2y + 16)°

Solve for LN in parallelogram KLMN.

Z L

x y

+ 10

−8

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Given the parallelogram ABCD, write a statement that can be deduced from the proof.

To prove: ? Statements 1. 2. 3. 14

Reasons

ABCD is a parallelogram

Given

∥ and ∥ ∠A and ∠B, ∠C and ∠D, and ∠D and ∠A are supplementary.

Deﬁnition of a parallelogram Consecutive interior angles postulate

Consider the quadrilateral PQRS. a

Complete the following two-column proof to prove that the consecutive angles are supplementary given that PQRS is a parallelogram: To prove: Consecutive angles of a parallelogram are supplementary Statements Reasons 1. PQRS is a parallelogram Given 2. 3. -

Complete the following two-column proof to prove the converse of part (a):

1. 2. 3. 15

To prove: PQRS is a parallelogram Statements ∠P and ∠S, ∠S and ∠R, and ∠R and ∠Q are supplementary. -

Consider that x = 11 and

Reasons Given -

Find the measure of ∠ABE, ∠EBC and ∠ECB.

Find the measure of ∠BEC and ∠ABC.

Classify ABCD as precisely as possible. Prove your classiﬁcation.

C E A

(4x + 1)°

(56 − x)°

(6x − 21)° B

Use geometric constructions to show that ABCD is a parallelogram if its diagonals bisect each other.

480

Consider parallelogram ABCD. What property of parallelograms could this construction be used to prove? A

Opposite sides are parallel.

The two diagonals bisect each other.

Opposite angles are congruent.

Adjacent angles are congruent.

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C A

Let’s extend our thinking 18

Given the parallelogram PQRS, prove that the opposite angles are congruent.

To prove: Opposite angles of a parallelogram are congruent. Statements Reasons PQRS is a parallelogram Given ∥

2. 3. 4. SOL

and

Deﬁnition of a parallelogram

∥

The diagonals of rectangle WXY Z intersect at the point (−4, 5). One of the vertices of rectangle WXY Z is located at (2, 1). Rectangle Diagonals Intersection

8 7 6 5 4 3 2 1

−8−7−6−5−4−3−2−1−1

Intersection

1 2 3 4 5 6 7 8

−2 −3 −4 −5 −6 −7 −8

Which of the following could be the location of another vertex of this rectangle? A 20

(10, 0)

(8, −3)

(0, 6)

(−8, 12)

Given: Parallelogram ABCD

B 2

Prove: E is the midpoint of

3 C

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Complete the given proof using the ﬁgure of quadrilateral ABCD:

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

∥ ∠ABE ≅ ∠CDE △ABE ≅ △CDE ⬚ ABCD is a parallelogram ∠AED ≅ ∠CEB △AED ≅ △CEB ∠EAD ≅ ∠ECB

12.

bisects opposite angles ∠EDA ≅ ∠EBC

13.

bisects opposite angles

14.

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To prove: ABCD is a rhombus Statements Reasons ⬚ ∠BAE ≅ ∠DCE ≅ ∠EAD

⬚

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⬚ ⬚

⬚ ⬚ CPCTC Parallelogram diagonals converse ⬚ ⬚ ⬚ ⬚ ⬚

⬚ ⬚

These are two examples of rhombi:

Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 9.03 to answer these questions. 1.

Which polygon(s) are always rectangles? Can you create a rectangle with any of the polygons?

How do you know which polygon(s) form a rhombus versus a square?

Which polygon(s) are always parallelograms? Can you create a parallelogram with any of the polygons?

The following theorems relate to the special parallelograms: Rectangle diagonals theorem

Rhombus opposite angles theorem

A parallelogram is a rectangle if and only if its diagonals are congruent.

A parallelogram is a rhombus if and only if each diagonal bisects a pair of opposite angles.

Q P

R S

Rhombus diagonals theorem A parallelogram is a rhombus if and only if its diagonals are perpendicular. Q P

R S

Squares have the same properties as both a rectangle and rhombus.

Note that these theorems are for parallelograms, so if we are only told that a polygon is a quadrilateral, then they may not meet the conditions stated. 484

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Example 1 List all classiﬁcations of quadrilaterals that apply to the ﬁgures. Explain your reasoning. a

Create a strategy Recall what we know about parallelograms: • A quadrilateral is a parallelogram if and only if its opposite sides are congruent. • A quadrilateral is a parallelogram if and only if its opposite angles are congruent. • In a parallelogram, consecutive angles will be supplementary. • A quadrilateral is a parallelogram if and only if its diagonals bisect each other. Recall the special types of parallelograms: • A rectangle is a quadrilateral with four right angles. • A square is a quadrilateral with four right angles and four congruent sides. • A rhombus is a quadrilateral containing four congruent sides.

Apply the idea The quadrilateral is a parallelogram and rectangle. A parallelogram’s opposite angles are congruent, and a rectangle is a quadrilateral with four right angles by deﬁnition.

Create a strategy We are given information about the side lengths of the ﬁgure. Use this information to classify the ﬁgure.

Apply the idea The ﬁgure is a parallelogram and rhombus. A parallelogram’s opposite sides are congruent, and a quadrilateral with four congruent sides is a rhombus by deﬁnition.

Apply the idea The ﬁgure can be classiﬁed as a parallelogram, square, rhombus, and rectangle because it has four right angles and consecutive congruent sides.

Reflect and check Most specifically, since the angle measures are all right angles and the sides are all congruent, the figure is a square.

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Create a strategy This ﬁgure cannot be classiﬁed as a parallelogram because its opposite sides are not congruent.

Apply the idea The ﬁgure is a quadrilateral because it has four sides.

Example 2 Consider the diagram that illustrates the rhombus diagonals theorem: If a parallelogram is a rhombus, then its diagonals are perpendicular bisectors of one another. Step 1

Step 2

Step 3 B

Step 4 B

B E

a Add reasoning to each step of the diagram.

Create a strategy Start with the given statement, and use the initial step to identify what is given. From there, the diagram provides more information to use when attempting to prove that the diagonals are perpendicular bisectors of one another.

Apply the idea B

We are given a parallelogram that is a rhombus, where all sides are congruent and opposite sides are parallel to one another.

The diagonals of the rhombus bisect one another since the diagonals of a parallelogram bisect one another.

Step 1 A

C Step 2 A E

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The triangles formed by the diagonals are congruent by SSS congruence, so we know that their corresponding angles are also congruent by CPCTC.

Since the triangles are congruent, we know that ∠AED ≅ ∠CED and they are linear pairs along . Since they are congruent and linear pairs, they must each be 90°. This means that , so the diagonals are perpendicular bisectors of one another.

Step 3 A E

C Step 4 A E

b Write a formal proof of the theorem.

Create a strategy Use the diagram from part (a), deﬁnitions, and theorems to write a formal proof.

Apply the idea

Consider the rhombus ABCD:

B E

To prove: Statements 1.

ABCD is a rhombus

is a bisector of

and

is a perpendicular bisector of

△AED ≅ △AEB ≅ △CEB ≅ △CED ∠AED ≅ ∠AEB ≅ ∠CEB ≅ ∠CED m∠AED = m∠DEC m∠AED + m∠DEC = 180° 2 ⋅ m∠AED = 180° m∠AED = 90° m∠AED = m∠AEB = m∠CEB = m∠CED = 90°

Definition of bisecting Side-side-side congruency theorem Corresponding parts of congruent triangles theorem Definition of congruence ∠AED and ∠DEC are a linear pair Substitution Divide by 2 Definition of congruence Definition of perpendicular

12. 13.

Reasons Given If a quadrilateral is a parallelogram, then its diagonals bisect each other. Sides of rhombus are congruent

4. 5. 6. 7. 8. 9. 10. 11.

and another

are perpendicular bisectors of one

Perpendicular and bisect one another

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Example 3 Prove that in the given rectangle ABCD,

Create a strategy It may be useful to mark up a diagram with information that we are given and that we already know using the deﬁnition of a rectangle, then use that to help build the proof.

Apply the idea ABCD is a rectangle (Given)

AD ≅ BC (If a quadrilateral is a parallelogram, then its opposite sides are congruent)

ABCD is a parallelogram ∠ ADC and ∠ BCD are right angles (Definition of a rectangle)

DC ≅ DC

∠ ADC ≅ ∠ BCD

(Reflexive property of congruence)

(All right angles are congruent)

△ ACD ≅ △ BDC (Side-angle-side congruency theorem)

AC ≅ BD (CPCTC)

Example 4 Given: • ABCD is a rhombus • m∠ACD = (5x + 8)° • m∠BCD = (12x + 2)°

Complete the following: a Solve for x.

Create a strategy Since parallelogram ABCD is a rhombus, the diagonals in a rhombus bisect a pair of opposite angles. In other words, m∠ACD = m∠ACB and m∠ACD + m∠ACB = m∠BCD

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We want to use this information to write an equation relating the two given angles. (5x + 8) + (5x + 8) = (12x + 2) We want to use the above equation to solve for x.

Apply the idea (5x + 8) + (5x + 8) = (12x + 2)

m∠ACD + m∠ACB = m∠BCD

10x + 16 = 12x + 2

Combine like terms

10x + 14 = 12x

Subtract 2 from both sides of equation

14 = 2x

Subtract 10x from both sides of equation

7=x

Divide both sides of equation by 2

b Solve for m∠ABC.

Create a strategy ∠BCD and ∠ABC are consecutive angles. Since ABCD is a rhombus and a rhombus is a parallelogram, we know that consecutive angles are supplementary. So m∠BCD + m∠ABC = 180°. Use the given information and the value for x we solved for in part (a) to ﬁnd m∠BCD. Then use that information and the above equation to solve for m∠ABC.

Apply the idea Substituting 7 for x, we get m∠BCD = (12 (7) + 2)°. So m∠BCD = 86°. Now we want to use this angle measure and the fact that consecutive angles are supplementary to solve for m∠ABC. m∠BCD + m∠ABC = 180° 86° + m∠ABC = 180° m∠ABC = 94°

Consecutive angles in a parallelogram are supplementary Substitution Subtract 86° from both sides of the equation

Example 5 If

, show that ABCD is not a rhombus.

2x + 5

3x − 2

4x − 17

Create a strategy We know that a rhombus has four congruent sides by definition, so to show that the figure is not a rhombus, we need to show that the side lengths are not congruent. If , we know that AB = CD by the definition of congruence. Start by solving the equation 2x + 5 = 4x − 17 for x, then use the value of x to determine the side lengths of the quadrilateral.

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Apply the idea 2x + 5 = 4x – 17 2x + 22 = 4x

Add 17 to both sides of the equation

22 = 2x

Subtract 2x from both sides of the equation

x = 11

Divide both sides of the equation by 2

Since x = 11, we know that BC = 3x − 2 = 3 (11) − 2 = 22 − 2 = 31. Since AD ≠ BC, ABCD is not a rhombus because we have shown that at least two of the sides are not congruent.

Example 6 W

Use geometric constructions and properties of rhombuses to verify that parallelogram WXY Z is a rhombus. Z

Create a strategy If a parallelogram has perpendicular diagonals, it’s a rhombus. So, we may construct perpendicular bisectors for the diagonals of the parallelogram to verify that it is a rhombus.

Apply the idea W

Step 1: Draw the diagonals point E.

and

, intersecting at

Step 2: Construct an arc centered at point W so that the arc is longer than half of .

Z Z′

X X′

Z Z′

E E′

X X′

Step 3: Construct an arc centered at point Y using the Step 4: Draw . Label the point E′ where same compass setting, such that this arc intersects the . E′ is the midpoint of . arc constructed in Step 2 at two distinct points. Label the points X′ and Z′.

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intersects

at the midpoint of

, so

is the perpendicular bisector of

and are perpendicular bisectors of each other and pass through the intersection point E′. The diagonals Since they coincide with the diagonals and and the intersection point E, parallelogram WXY Z is a rhombus.

Idea summary Use the theorems relating to the special parallelograms to solve problems: • • •

Rectangle diagonals theorem: A parallelogram is a rectangle if and only if its diagonals are congruent. Rhombus diagonals theorem: A parallelogram is a rhombus if and only if its diagonals are perpendicular. Rhombus opposite angles theorem: A parallelogram is a rhombus if and only if each diagonal bisects a pair of opposite angles.

Practice What do you remember? 1

Determine whether each statement is true or false: a

All quadrilaterals are parallelograms.

All rectangles are parallelograms.

All parallelograms are rhombi.

A parallelogram with four congruent sides is a square.

A rectangle with consecutive sides congruent is a square.

A rhombus is a parallelogram.

Classify each quadrilateral. Explain your reasoning. a

91°

83°

97° 8

8 83°

97° 15

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Complete each statement with “always”, “sometimes”, or “never”, and then explain your reasoning. a b c d e f

A square is ⬚ a rhombus.

A rectangle is ⬚ a square.

A rectangle ⬚ has congruent diagonals.

The diagonals of a square ⬚ bisect its angles. A rhombus ⬚ has four congruent angles.

A rectangle ⬚ has perpendicular diagonals.

What conclusions, if any, can be drawn from the following statements? Statement A: Figure X is a rectangle. Statement B: If a ﬁgure is a rectangle, then it has four right angles.

Given: ABCD is a rectangle.

B 5

Solve for each missing angle in the diagram.

3 7

24°

Given: RSTV is a rhombus.

C R

Solve for each missing angle in the diagram.

S 1 3

36°

Solve for x using the given ﬁgure:

(13x − 1)°

(6x − 10)°

Let’s practice 8

Sheldon draws a quadrilateral and covers it up. He tells Wanda that the quadrilateral has two pairs of opposite sides congruent. From this information, determine the most specific classification that Wanda can determine for the figure. Explain your answer.

Given:

• ABCD is a rhombus. • m∠DAC = 4x + 9 • m∠DAB = 11x − 3 a

Solve for x.

Solve for m∠BAC.

Solve for m∠ABC.

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Given that ABCD is a rhombus, solve for x. Justify your solution.

(8x − 3)°

(5x + 2)° B

Given that ABCD is a square: a

Solve for x. Justify your solution.

Solve for y. Justify your solution.

(9y + 9)° x°

Let AT = 4x − 10 and BD = 5x + 20. a

Classify the quadrilateral.

Solve for x.

T C

Consider the given parallelogram ABCD, where a

Classify the parallelogram.

If m∠ABD = (3y + 12) °. Justify your solution.

Solve for x. Justify your solution.

Find CD. Justify your solution.

B x° 7 ft E

Use the following statements to prove that ﬁgure X is not a rhombus. Statement A: For a quadrilateral X, no two sides are parallel. Statement B: If a quadrilateral is a rhombus, then it has two pairs of parallel sides.

Complete the following proof for: If a parallelogram is a rectangle, then its diagonals are congruent. Consider that a parallelogram ABCD is a rectangle, then we want to show that ⬚ ≅ A

If ABCD is a rectangle, then ∠ABC, ∠BCD, ∠CDA, and ∠DAB are all ⬚ angles. It also tells us that ≅⬚ and ⬚ ≅ . This means that we have two congruent right triangles △ABC and △DCB, using the ⬚ theorem. By ⬚ we get that .

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You are asked to build a frame for a window. The window will be installed in the opening shown in the diagram. a

The opening must be a rectangle. Given the measurements in the diagram, can you assume that it is? Explain.

You measure the diagonals of the opening. The diagonals are 60.8 inches and 61.3 inches, respectively. What can you conclude about the shape of the opening?

37 in

50 in

37 in

Which construction can be used to prove that a given quadrilateral is a rectangle? Explain your reasoning. A

Constructing congruent diagonals

Constructing a pair of congruent and parallel opposite sides

Constructing an angle bisector that also bisects the opposite side

Constructing perpendicular bisectors of all four sides

You want to mark off a square region for a garden at school. You use a tape measure to mark off a quadrilateral on the ground. Each side of the quadrilateral is 3.5 meters long. Explain how you can use the tape measure to make sure that the quadrilateral is a square.

Consider parallelogram WXY Z. The diagram gives the ﬁrst half of the geometric constructions that illustrates the rhombus opposite angles theorem: a parallelogram is a rhombus if and only if each diagonal bisects a pair of opposite angles. W

Y W Z

Y Step 1

494

X Z

Y Step 2

X Z

X Z Y Step 3

WW′

Y Step 4

X Y Y′ Step 5

Add reasoning to each step of the diagram.

Complete the geometric constructions for the other pair of opposite angles to show that the parallelogram is a rhombus using the rhombus opposite angle theorem. Include reasoning in each step of your diagram.

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Let’s extend our thinking 20

Order the terms in a diagram so that each term build off the previous term(s). Explain why each ﬁgure is in the location you choose. Quadrilateral

Square

Rectangle

Rhombus

Parallelogram

Your friend claims a rhombus will never have congruent diagonals because it would have to be a rectangle. Is your friend correct? Explain your reasoning.

Will a diagonal of a rhombus ever divide the rhombus into two equilateral triangles? Explain your reasoning.

Sketch each of the following, labeling all sides and angles accordingly:

A parallelogram that is neither a rectangle or a rhombus.

A parallelogram with a pair of consecutive congruent sides and a pair of acute angles.

A parallelogram with a right angle.

Prove that if a parallelogram is a rhombus, then its diagonals are angle bisectors of a pair of opposite angles.

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9.04 Trapezoids After this lesson, you will be able to... • use the definition of a parallelogram to identify and prove the properties of a trapezoid. • solve problems using properties of trapezoids and the relationship between the bases and midsegment of a trapezoid. • use constructions to verify properties of trapezoids.

Trapezoids A trapezoid is a quadrilateral with exactly one pair of parallel sides. The parallel sides are called bases and the nonparallel sides are called legs.

Base 2

Leg 3

Leg

Consecutive angles in a trapezoid whose common side is a base are base angles. A trapezoid has two pairs of base angles.

4 Base

In this case, ∠1 and ∠2 are base angles to the top base and ∠3 and ∠4 are base angles to the bottom base. An isosceles trapezoid is a quadrilateral with one set of opposite sides parallel and the other set of opposite sides congruent. The following theorems are related to isosceles trapezoids: Isosceles trapezoid diagonals theorem

Isosceles trapezoid base angles theorem

If a quadrilateral is an isosceles trapezoid, then its diagonals are congruent.

If a quadrilateral is an isosceles trapezoid, then each pair of base angles is congruent.

A midsegment of a trapezoid is a line segment that bisects both legs. Trapezoid midsegment theorem The midsegment of a trapezoid is parallel to each base and the length of the midsegment is equal to the sum of the length of the bases divided by two. ∥

∥

Length of midsegment = or

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(b1 + b2)

M P

and

Length of midsegment =

R N

Example 1 ABCD is a trapezoid. 3x + 2

2x − 2

Solve for x.

Create a strategy Use the trapezoid midsegment theorem: Length of midsegment = midsegment length = 15.

, where b1 = 3x + 2, b2 = 2x − 2 and

Apply the idea Substitute the values of base and midsegment lengths

Combine like terms in the numerator

Multiply both sides of equation by 2

Divide both sides of equation by 5

Symmetric property of equality

Example 2 Given: Trapezoid PQRS Q

63°

(12x + 9) S

Solve for x.

Create a strategy Combine the measurements of all the angles, and remember that all quadrilaterals’ interior angles sum to 360°.

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Apply the idea 2 (m∠PQR) + 2 (m∠PSR) = 360 2 (m∠PQR + m∠PSR) = 360

Factor out the common factor of 2

2 (12x + 9 + 63) = 360

Substitute m∠PQR = 12x + 9 and m∠PSR = 63

(12x + 9 + 63) = 180

Divide both sides of equation by 2

12x + 72 = 180

Combine like terms

12x = 108

Subtract 72 from both sides of equation

x=9

Divide both sides of the equation by 12

Reﬂect and check Bases of a trapezoid are parallel so ∠QPS and ∠PSR are supplementary using the consecutive interior angles theorem. We could have also used theorems of parallel lines to solve for x.

Example 3 ABFE is a trapezoid. 10x − 10 A

B 8x + 2

E 140

a Explain whether or not m∠A = 73° and m∠F = 117° would make this diagram valid or not.

Create a strategy Notice that , so this is an isosceles trapezoid. This means that the base angles should be congruent, so we need m∠A = m∠B = 73° and m∠F = m∠E = 117° for it to be valid.

Apply the idea Since ABFE is a trapezoid ∥ , so using as a transversal we get that ∠A and ∠E are supplementary. as a transversal we get that ∠B and ∠F are supplementary. Similarly, using If ∠A and ∠E are supplementary, then m∠E = 107°. If ∠B and ∠F are supplementary, then m∠B = 63°. Since both of these violate the requirement for base angles to be congruent, m∠A = 73° and m∠F = 117° makes this diagram invalid.

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b Solve for x.

Create a strategy Use the trapezoid midsegment theorem: Length of midsegment = midsegment length = 8x + 2.

, where b1 = 140, b2 = 10x − 10 and

Apply the idea

Substitution

Multiply both sides by 2

Combine like terms

Subtract 4 and 10x from both sides

Divide both sides by 6

Example 4 Given: Trapezoid PQRS • PR = 4x − 1 • QS = 2x + 8 Q

Solve for x.

Create a strategy The isosceles trapezoid diagonals theorem tells us that if a quadrilateral is an isosceles trapezoid, then its diagonals are congruent.

Apply the idea The markings tell us that this is an isosceles trapezoid, so PR = QS. We can use this to solve for x. Isosceles trapezoid diagonals theorem

Substitute

Add 1 to both sides

Subtract 2x from both sides

Divide both sides by 2

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Example 5 Prove the trapezoid midsegment theorem: the midsegment of a trapezoid is parallel to each base and the length of the midsegment is equal to the sum of the length of the bases divided by two.

Create a strategy Given: Trapezoid ABCD • E is the midpoint of • F is the midpoint of We will draw an auxiliary point G to help with this proof. A E

B F

Apply the idea To prove:

Statements Trapezoid ABCD

Given

E is the midpoint of

Given

F is the midpoint of

Given Deﬁnition of trapezoid

4. 5. 6.

∠BAF ≅ ∠CGF ∠AED ≅ ∠CEF △ABF ≅ △GCF

Angle-Side-Angle congruency theorem Corresponding parts of congruent triangles

9. 10.

Alternate interior angles theorem Vertical angles congruence theorem Deﬁnition of midpoint

7. 8.

Reasons

is a midsegment

Deﬁnition of midsegment

11.

Triangle midsegment theorem

12.

Triangle midsegment theorem

13. 14.

DG = DC + CG CG = AB

Segment addition postulate Corresponding parts of congruent triangles

15.

Substitution property

16.

Substitution property

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Example 6 Use geometric constructions and properties of trapezoids to verify that quadrilateral PQRS is a trapezoid given that its non-parallel sides are and . Q

Create a strategy If a quadrilateral has at least one pair of parallel sides, it is a trapezoid. So, we may use a parallel line construction to verify that one pair of opposite sides is indeed parallel.

Apply the idea 1. Choose a point T on

through which to construct a line parallel to Q

2. Set the compass width to the distance ST. Q

3. Construct an arc centered at P with the radius ST. Q

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4. Set the compass width to the distance PT. T

5. Construct an arc centered at S with the radius PT. Q

6. Construct a line through T and the intersection of the two arcs, labeling it U. This line is parallel to Q

Since is parallel to and coincides with , we have shown that one pair of opposite sides of the quadrilateral are parallel and thus the quadrilateral is a trapezoid.

Idea summary A trapezoid is a quadrilateral with exactly one pair of parallel sides. An isosceles trapezoid has congruent legs, base angles, and diagonals. Length of midsegment =

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or Length of midsegment =

(b1 + b2)

Practice What do you remember? 1

Determine whether each statement is true or false. a

A trapezoid has exactly one pair of parallel sides.

A trapezoid must be a parallelogram.

The diagonals of an isosceles trapezoid are congruent.

The base angles of any trapezoid are congruent.

Determine whether each of the following statements about isosceles trapezoid JKLM is true or false and explain how you know. a

SOL

∣∣

∠J ≅ ∠L

∠K and ∠M are supplementary.

SOL

Which shape must have opposite sides that are parallel and congruent, and diagonals that are perpendicular bisectors of each other? A

Square

Kite

Trapezoid

Rhombus

Which of the following polygons are trapezoids? A

A quadrilateral has a set of parallel sides and a set of congruent sides. What is the most speciﬁc classiﬁcation you can make for this shape?

Select each property that is valid about the diagonals of an isosceles trapezoid. The diagonals of an isosceles trapezoid — A

bisect each other

are congruent

do not bisect each other

are not congruent

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Consider the diagram shown, given that ABCD is a trapezoid. a

Identify which sides are parallel.

Identify which sides are the bases.

Identify which sides are the legs.

Find m∠ABC.

D 120° 49°

Find m∠ADC.

Let’s practice 8

Consider the following trapezoid. a

Find the value of m∠B.

Find the value of m∠C.

Find the value of m∠D.

52° D

Solve for each of the missing angle measures in the following trapezoids: a

R Q 3

1 K 111°

77° P

3 J 2 M

C B 1

2 Y

60°

D X

105° 3 W

Consider the following trapezoid JKLM, where KM = 20.

Find the value of JL.

Given:

M G

• Trapezoid DEFG • EG = 3.5 Solve for FD.

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Given:

• Trapezoid JKLM • KM = 1.26

Solve for JL.

Given:

5.5

• Trapezoid PQRS • is a midsegment. Solve for WV.

Solve for the value of x that proves PQRS is a trapezoid.

12.9

51° 28x − 11 R

Find the value of x that makes TUVW a trapezoid.

Given:

−x + 21

• Trapezoid FEDC • EC = 4x + 11 • FD = 3x + 17 Solve for x.

Solve for x.

Solve for BC.

Solve for TU.

Solve for KH.

8x + 3

4x + 7.5 T

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Given:

B 2

• ∠BAD = (5x + 32)° • ∠CDA = (7x2)°

Find the value of x that proves trapezoid ABCD is isosceles.

A D

Given:

∥

Prove: ∠BCE ≅ ∠ADE and ∠BAD ≅ ∠ABC Note that D, E, and C are collinear. If you were to complete this proof, which of the following would you prove?

If a trapezoid is isosceles, then base angles are congruent.

If a trapezoid has one pair of congruent base angles, then it is isosceles.

If a quadrilateral is a trapezoid, then base angles are supplementary.

If a quadrilateral has one pair of supplementary base angles, then it is a trapezoid.

Given: • ABCD is a trapezoid with bases • ∠BCD ≅ ∠ADC •

and

and A

intersect at E

Prove: ABCD is an isosceles trapezoid D

If you were to complete this proof, which of the following would you prove?

If a trapezoid has congruent legs, then it is isosceles.

If a trapezoid is isosceles, then it has congruent legs.

If a trapezoid has one pair of congruent base angles, then it is isosceles.

If a trapezoid is isosceles, then it has one pair of congruent base angles.

Given that ABCD is an isosceles trapezoid with altitudes

and

Complete the following two-column proof that proves that each pair of base angles are congruent.

To prove: Each pair of base angles of trapezoid ABCD are congruent. Statements Reasons ABCD is an isosceles trapezoid Given

Deﬁnition of an isosceles trapezoid

Altitudes of a trapezoid are congruent

4. 5. 6. 7. 8. 9. 10.

506

∠DEA and ∠CFB are right angles ∠DEA ≅ ∠CFB ⬚ ⬚ ∠BCD and ∠CBA are supplementary ∠ADC and ∠BAD are supplementary ∠BAD ≅ ∠ABC

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Deﬁnition of an altitude All right angles are congruent ⬚ ⬚ Consecutive interior angles postulate Consecutive interior angles postulate Congruent supplements theorem

Given that ABCD is an isosceles trapezoid with bases

and

Complete the following two-column proof that proves that

To prove: 1.

Statements ABCD is an isosceles trapezoid

Reasons Given Deﬁnition of an isosceles trapezoid

2. 3.

⬚

∠ADC ≅ ∠BCD

△ADC ≅ △BCD

⬚ If a trapezoid is isosceles, then each pair of base angles is congruent. Side-angle-side congruence theorem Corresponding parts of congruent triangles are congruent (CPCTC)

6. 23

Consider quadrilateral ABCD. Shown are the steps to use geometric constructions to begin verifying the isosceles trapezoid base angles theorem. A pair of base angles are shown to be congruent by constructing the copy of one of its angles. A

Step 1: Identify the angle we want to copy. We will start with ∠C. A

Step 2: Draw a ray with its endpoint at D that will form one of the legs of the copied angle. A

Step 3: With the compass point on C, use the compass to draw an arc that intersects both legs. A

Step 4: Without changing the compass width, copy the arc in Step 3 by placing the point end of the compass onto D. A

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Step 5: On ∠C, use the compass to measure the distance between the points where the legs of the angle meets the arc drawn in Step 3. A

Step 6: Without changing the compass width, copy the distance by placing the compass point where the ray meets the copied arc and draw an intersecting arc. A

Step 7: Draw a ray that shares its end point with the ray from D in Step 2, and goes through the intersection found in Step 6. A

Since the copy of ∠C with its vertex at D coincides with ∠D, we have veriﬁed that one pair of base angles are congruent. Show that the other pair of base angles are congruent in order to verify the isosceles trapezoid base angles theorem.

Let’s extend our thinking 24

Given: ∠BAD = (5x2 + 32)°

∠CDA = (7x )°

Find the value of x that makes Trapezoid ABCD isosceles.

A D

The midsegment of a trapezoid has a length of 3z + 8. Give possible lengths for the two bases of the trapezoid.

Given: • Trapezoid ABCD with altitudes •

and

To prove: ∠A ≅ ∠D A

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Given:

• Trapezoid ABCD • • •

To prove: Trapezoid ABCD is isosceles.

Given:

• Trapezoid ABCD is isosceles. • ∠BAK ≅ ∠BKA To prove: BKDC is a parallelogram. B

Given the coordinates of a quadrilateral A (−3, 2), B (1, 5), C (4, 3), and D (0, 0). a

Calculate the slopes of all four sides.

Prove that

Calculate the midpoints of diagonals

What can you conclude about the diagonals

and

are parallel. and

. of quadrilateral ABCD?

and

A They bisect each other.

They are equal in length.

C They are perpendicular to each other.

None of the above.

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10 Modeling in 2 & 3 Dimensions Big ideas • Physical objects can be modeled with 2D and 3D geometric figures whose properties can be applied to solve real-world problems.

Chapter outline 10.01 10.02 10.03 10.04 10.05 10.06

Area and perimeter Cross sections of solids Prisms and cylinders Cones and pyramids Spheres Similarity in 2D and 3D ﬁgures

512 524 531 548 567 577

Example 1 Simy is building a fence around her vegetable patch to keep her dog from digging up the potatoes. The vegetable patch is a 24 ft by 9 ft rectangle. She will create the fence with wire netting supported by posts placed every 3 ft, and with a single gate placed between two of the posts. Estimate the total cost to fence off the vegetable patch if wire netting costs $0.60 per foot, the posts cost $17.50 each, and the gate costs $75.

Create a strategy Both the amount of wire netting and number of posts depend on the perimeter, P, of the vegetable patch. So, we can ﬁrst ﬁnd the perimeter of the patch using P = 2l + 2w, where l and w are the length and width of the rectangle respectively. The amount of wire netting required is the perimeter minus 3 ft for the gap with the gate. The number of posts required will be the perimeter divided by three, for the 3 ft gaps between each post.

Apply the idea Finding the perimeter: P = 2l + 2w

Perimeter formula

= 2 ⋅ 24 + 2 ⋅ 9

Substitute l = 24 and w = 9

= 66 ft

Evaluate the multiplication and addition

The vegetable patch has a perimeter of 66 ft, so we require P − 3 = 63 ft of wire netting, and Total cost: C = $0.60 ⋅ feet of wire + $17.50 ⋅ number of posts + cost of gate

= 22 posts.

Set up equation

= $0.60 ⋅ 63 + $17.50 ⋅ 22 + $75

ubstitute 63 feet of wire, 22 posts, S and $75 gate

= $497.80

Evaluate multiplication and addition

The total cost of the fence should be approximately $497.80.

Example 2 The wheel of Kirara’s bicycle has a diameter of 26 in. Determine the number of feet she would travel if the wheels made 240 complete revolutions.

Create a strategy On each revolution Kirara will travel approximately the circumference of the wheel. We can multiply the circumference of the wheel by the number of revolutions to ﬁnd the distance traveled in inches and then convert to feet.

Apply the idea First, we need to ﬁnd the radius which is half the diameter. So r =

⋅ 26 = 13

Circumference multiplied by the number of revolutions

Substitute d = 13

Evaluate the multiplication

Convert inches to feet

Evaluate the division

Kirara traveled 520π ft, which is approximately 1634 ft.

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Reﬂect and check What might affect the accuracy of the estimated distance? For example, if the tire is ﬂat and reduced the diameter of the wheel by half an inch, by how much would the distance calculated above overestimate the actual distance traveled?

Example 3 The side length of a square Q is doubled to create square R. How does this affect the perimeter of square R relative to square Q?

Create a strategy Let the side length of square Q be s. Since the side length of square R is double that of square Q, the side length of square R is 2s. Calculate the perimeters of both squares and compare them to determine the relationship between the perimeters.

Apply the idea We know that the formula for the perimeter of a square is P = 4l where l is the side length. Perimeter of square Q with side s: PQ = 4(s) = 4s

Substitute the side length s Simplify

Perimeter of square R with side 2s: PR = 4(2s) = 8s

Substitute the side length 2s Simplify

The perimeter of square R is twice the perimeter of square Q.

This means doubling the side length of square Q to create square R also doubles the perimeter of the square.

Reflect and check Doubling all sides of square Q is the same as dilating square Q by a scale factor of 2. As we saw previously, dilating all sides of a figure creates a similar figure. So, square Q is similar to square R.

Idea summary Perimeter is a term for the distance around the boundary of a two-dimensional shape. To calculate the perimeter of any polygon we simply add up all the lengths of the sides. The perimeter of a circle is called the circumference and can be found using the formula:

C = 2π r C r

circumference radius

A change in one dimension of a figure results in a predictable change in perimeter. The resulting figure may or may not be similar to the original figure.

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Area Area is the measure of the space enclosed by the boundary of a two-dimensional shape. We have previously encountered several formulas for simple shapes that can be used in a wide variety of real-world problems and can also be used to build or approximate the area of more complex ﬁgures. The area of a rectangle is given by Area = length ⋅ width, or

width

A=l⋅w

length

The area of a square is given by Area = side ⋅ side, or A = s ⋅ s = s2

side

The area of a triangle is given by Area = height

⋅ base ⋅ height, or bh

base

The area of a parallelogram is given by Area = base ⋅ height, or

height

A = bh base

The area of a circle is given by radius

base a

Area = π ⋅ radius ⋅ radius, or A = π r2

The area of a trapezoid is given by: Area =

height

⋅ (base a + base b) ⋅ height, or (a + b)h

base b

A change in one dimension of a figure results in a predictable change in area. The resulting figure may or may not be similar to the original figure.

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Example 4 Find the area of the following ﬁgure to two decimal places: 4 cm

Create a strategy We can use the area formula for a circle, A = π r2, and multiply by

to ﬁnd area of the given ﬁgure.

Apply the idea The proportion of the circle times the area of the circle

Substitute r = 4

Evaluate the multiplication

Approximate to 2 decimal places

Example 5 An area of ﬂoor measuring 2280 cm2 is to be paved with identical tiles in the shape of parallelograms. Each tile measures 12 cm along the base, and has a perpendicular height of 5 cm.

How many tiles are needed to cover the whole area?

5 cm

12 cm

Create a strategy Find the area of a single tile and divide the ﬂoor area by this value.

Apply the idea Area single tile: Area = bh = 5 ⋅ 12 2

= 60 cm

Formula for area of a parallelogram Substitute b = 5 and h = 12 Evaluate the multiplication

Number of tiles required: Number of tiles = Floor area ÷ Area of tile

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Set up equation

= 2280 ÷ 60

Substitute Area of tile = 60

= 38

Evaluate the division

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Reﬂect and check The formula for the area of a parallelogram is equal to the area of a rectangle with the same base and perpendicular height, thus A = bh. This can be observed by cutting off a triangle and rearranging to form a rectangle as shown below. h b

h b

This can also be seen as a 2-Dimensional application of Cavalieri’s principle. Where slices taken parallel to the base, through the rectangle and parallelogram, would produce lines of the same length at each height.

Example 6 By considering a trapezoid as a composite ﬁgure made up of two triangles, ﬁnd a general formula for the area of a trapezoid in terms of its perpendicular height, h, and parallel side lengths, a and b.

a h b

Create a strategy We can split the trapezoid into two triangles using one of the diagonals. Then use the formula for the area of a triangle A =

base ⋅ height to ﬁnd a general formula for a trapezoid.

Apply the idea Splitting the trapezoid as follows: a 1 2

Triangle 1 has base length a and height h and Triangle 2 has base length b and height h. Area of trapezoid: Breaking the area into the two triangles

Using formula for area of a triangle

Factoring

and h

Thus, the area of a trapezoid is one half of the product of the height and the sum of the lengths of the bases, where h is the height and a and b are the bases.

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Reﬂect and check There are several ways to justify the formula for the area of a trapezoid. For example, we could create a parallelogram out of two identical trapezoids as follows: a

Then, area of trapezoid:

Using formula for area of a parallelogram Substitution

Can you think of any other ways to justify the area of a trapezoid?

Example 7 The area of a square is initially 25 cm2 before its dimensions are altered. a If one side of the square is increased by a factor of 4, what is the new side length of the square?

Create a strategy

Apply the idea 2

Since the area of a square is given by s , where s is the length of a side, we can ﬁnd the original side length by taking the square root of the area. We can then multiply this original side length by the given factor to ﬁnd the new side length. To distinguish between the original side length and the new side length, we can use the notation s′ for the new side length.

Find the original side length

Evaluate the square root

Multiply the original side length by 4

Evaluate the multiplication

Therefore, the new side length of the square is 20 cm.

b If one side of the square is decreased by a factor of 2, what is the new side length of the square?

Create a strategy In the previous part, we found the original side length to be s = 5 cm. Since the side is now being decreased, we can divide this length by the given factor to ﬁnd the new side length. Recall s represents the original side, s′ represents the side after the enlargement, so we will use s″ to represent the side after the reduction.

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Apply the idea Find the original side length

Evaluate the square root

Divide the side length by the factor

Evaluate the division

Therefore, the new side length of the square is 2.5 cm.

c Describe how these changes affect the area and perimeter of the square.

Create a strategy The area of a square is given by s2, and the perimeter is given by 4s, where s is the side length of the square. In the ﬁrst case, we multiplied the side length by 4, which is represented generally as 4s. In the second case, we divided the side length by 2, which is represented generally by . By substituting these general side lengths into the formulas, we can compare the new area and perimeter of the square in both cases to the original area and perimeter of the square.

Apply the idea The area of a square is given by s2. If the side length is multiplied by 4, the new area becomes (4s)2 = 16s2. This means the new area is 16 times more than the original area. The perimeter of a square is given by 4s. Multiplying the side by 4 gives a new perimeter of 4(4s) = 16s. This means the new perimeter is 4 times the original perimeter. If the side length is halved, the new area becomes The new perimeter becomes 4

. This means the new area is

the original area.

= 2s, which means the new perimeter is half the original perimeter.

In general, if the side of a square is dilated by a factor of k, the new area will be k2 times the original area and the new perimeter will be k times the original perimeter.

Reﬂect and check We can check our answers by ﬁnding the new areas and perimeters and comparing them to the original area and perimeter of the square. The original area of the square is 25 cm2, and the original perimeter is 4 (5) = 20 cm. The area of the square after the enlargement: Substitute s′ = 20

Evaluate the square

Divide new area by original area

This shows that the new area is 16 times (or 42 times) larger than the original area. The perimeter of the square after the enlargement: Substitute s′ = 20

Evaluate the multiplication

Divide new perimeter by original perimeter

This shows the new perimeter is 4 times the original perimeter. The area of the square after the reduction: Substitute s″ = 2.5

Evaluate the square

Divide new area by original area

This shows the new area is

times the original area.

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The perimeter of the square after the reduction: Substitute s″ = 2.5

Evaluate the multiplication

Divide new perimeter by original perimeter

This shows the new perimeter is half the original perimeter.

Idea summary The area of a trapezoid is given by:

The area of a rectangle is given by:

A = lw

l length w width

a b h

The area of a triangle is given by:

A= b h

(a + b)h

base a base b height

The area of a circle is given by:

A = π r2

base height

radius

The area of a parallelogram is given by:

A = bh b h

base height

General formulas for shapes with special properties such as parallelograms, trapezoids, kites, rhombuses, and regular polygons can be often be derived by breaking the shape down into components of simpler shapes. A change in one dimension of a figure results in a predictable change in area. The resulting figure may or may not be similar to the original figure.

Practice What do you remember? 1

Find the perimeter of each polygon: a

b 18 cm

80 mm

19 cm

26 cm

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9 mm

7 cm

Two identical trapezia are put together to make a parallelogram: a

Find an expression for the area of the entire parallelogram in terms of a, b and h.

Find an expression for the area of one trapezia in terms of a, b and h.

Explain how the image helps to ﬁnd the area of the rhombus. Consider the transformation of the rhombus into a rectangle and use the given diagonal lengths of x and y.

Width Length

Find the area of each polygon: a

11 cm

18 m 10 m

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f 6 mm

8 cm 16 mm

9 mm

9 cm

Determine whether each of the formulas represents an area or a perimeter. Explain how you know. a

2π r

l⋅w

2l + 2w

⋅b⋅h

a+b+c

b⋅h

π r2

⋅h

A 1-inch square has an area of 1 square inch. Determine whether a 3-inch square has an area of 3 square inches. Explain your answer.

Let’s practice 7

Sharon has purchased a rectangular piece of fabric measuring 12 yd by 7 yd. Determine the area of the largest triangular piece she can cut out from the fabric.

Humberto has a custom dartboard, where each ring has a width of 1.25 in. The bullseye, or the center circle on the dartboard has a total width of 2.5 in. a

Determine the area of the entire dart board.

Determine the area of the dart board, excluding the center bullseye.

2.5 in

1.25 in

A farmer is fertilizing his pasture: a

Find the area of the pasture.

If 150 kg of fertilizer covers 100 square meters, determine how many kilograms of fertilizer the farmer needs to buy.

52 m

201 m

47 m

A rectangle has been enlarged with a scale factor of k. Describe the relationship between the perimeters of the original and the enlarged rectangles.

A circle has an area of approximately 78.5 cm2. Explain why a circle with a radius twice as large has an area of approximately 314 cm2.

Consider a square mosaic with a side length of 3 m. If a border of square tiles, each with a side length of 10 cm, is added around the mosaic, explain how this modiﬁcation alters the area of the original square mosaic.

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Provide an explanation of the increase in area, taking into account the new overall dimensions of the mosaic with the border.

State the increase in area from adding the border to the mosaic.

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The area of a rectangle is initially 48 u2 before its dimensions are altered. a

If the length of the rectangle is increased by a factor of 3, what is the new length of the rectangle?

If the width of the rectangle is decreased by a factor of 2, what is the new width of the rectangle?

Describe how these changes affect the area and perimeter of the rectangle. Then, calculate the new area and perimeter. The new area of the rectangle is ⬚ square units.

SOL

The new perimeter of the rectangle is ⬚ units.

The height of a trapezoid is decreased by a factor of . The other dimensions, including the lengths of the parallel sides, are unchanged. How is the area of the trapezoid affected by this change?

The area is decreased by a factor of .

The area is decreased by a factor of

The side lengths of triangle P are doubled to create triangle Q. a

How does this transformation affect the area of triangle Q relative to triangle P ?

Given the same transformation, how does this affect the perimeter of triangle Q relative to triangle P ?

Let’s extend our thinking 16

A rectangular prism has its length and width doubled, but its height remains the same. Explain how this change will affect the perimeter of each face of the prism. h

w l

A square garden has its side lengths increased by 50%. The new area is 576 square feet. what was the original area?

Prove that doubling the radius of a circle doubles its circumference.

Given a square with side length s, draw a graph showing the relationship between the side length and the area of the square as the side length increases. Describe the pattern you observe. 90

Area, in units2

80 70 60 50 40 30 20 10

Side length, in units 1 2 3 4 5 6 7 8 9 10

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10.02 Cross sections of solids After this lesson, you will be able to... • identify the shape of a two-dimension cross section of a three-dimensional solid. • identify possible three-dimensional solids given a two-dimensional cross section.

Cross sections of solids A solid is a term used when talking about a three-dimensional object. By slicing through a solid, we produce a two-dimensional shape called a cross section. The ﬁgure created from a cross section depends on the orientation or angle of the intersecting plane. Cross section

Apex

The two-dimensional shape made by slicing through a three-dimensional ﬁgure

The point (vertex) furthest from the base of an object

Exploration 1.

Draw at least two solid ﬁgures that would produce circular cross sections.

Draw at least two solid ﬁgures that would produce triangular cross sections.

Draw a solid ﬁgure that would produce a non-rectangular polygon cross section.

What additional information would be useful in deciding what three-dimensional solid to draw from the cross section shape?

A solid may form many different shapes by taking different cross sections. In particular, knowing a cross section of a solid isn’t enough information to uniquely determine the original solid. Many different solids can produce identical cross sections. Cross sections are formed when taking a slice of a ﬁgure horizontally, vertically, or diagonally. The shapes that form from cross sections may be more or less obvious, like when we take a vertical slice of a cube versus a diagonal slice. By taking a slice of a cube that is parallel to one of its sides, we form a cross section that is a square which is congruent to the sides of the cube.

By taking a slice of a cube that passes through the midpoint of three edges, as if we were cutting off a corner of the cube, we form a cross section that is an equilateral triangle in shape.

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Example 1 Find a cross section parallel to the base and identify the shape formed by the cross section.

Create a strategy

Apply the idea

Slice the shape parallel to the base.

The cross section is the intersection of the shape and the plane. The cross section is a pentagon.

Example 2 Consider the following cross section sliced from a solid ﬁgure:

Draw two different ﬁgures that the cross section could have come from.

Create a strategy We can slice a solid horizontally, vertically, or diagonally in order to get a cross section. We can think of a few triangular-shaped solids that could produce the cross section shown.

Apply the idea

A rectangular pyramid cut vertically from its apex could produce the triangular cross section.

A cone cut vertically could produce the triangular cross section.

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Reﬂect and check A cube, like the image shown for the equilateral triangle in the lesson could be cut at a steeper angle to produce an isosceles triangle.

Example 3 Consider the cylinder shown:

Identify three differently shaped cross sections, including at least one that comes from a diagonal slice.

Create a strategy Imagine slicing a plane through the cylinder vertically, horizontally, and diagonally at different points to produce differently-shaped cross sections.

Apply the idea Three shapes that could be formed are a rectangle, an ellipse (oval), and a circle.

Reﬂect and check An ellipse is a special shape that is formed by taking diagonal cross sections of cylinders or cones. An ellipse is a special type of oval that has two perpendicular lines of symmetry which divide it into four congruent regions.

It is also a diagonal cross section of a cone.

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Example 4 A 3D printer uses computer assistance to stack layers of material that make a three-dimensional shape. The printer creates an object out of several layers to create a physical model of a computer image. Shown are the layers of a model that a 3D printer has created. What solid ﬁgure is created by the printer?

Create a strategy

Apply the idea

If a 3D printer is stacking the layers, we can imagine the cross sections being cut horizontally to the printing surface.

The printer is printing a cone.

Idea summary A single three-dimensional solid may have cross sections of different shapes, depending on how the solid is sliced (vertical, horizontal, or diagonal).

Practice What do you remember? 1

Match each two-dimensional shape and three-dimensional solid with its definition: a

A three-dimensional surface of which all points are equidistant from a ﬁxed point.

A solid ﬁgure with two congruent circular bases that lie in parallel planes.

iii

A polyhedron with a polygonal base and triangular faces meeting in a common vertex.

A solid that has one circular base, an apex, and a lateral surface.

A solid that is half of a sphere with one ﬂat, circular side.

A four-sided ﬁgure with at least one pair of parallel sides.

vii A shape resembling a stretched circle or an ellipse. viii A three-dimensional solid object which has six faces, all of which are rectangles. ix

A three-sided shape with three corners, or vertices, and three edges, or sides.

A shape consisting of all the points in a plane that are the same distance from a ﬁxed point, known as the center.

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Identify the shape formed by the intersection of a horizontal plane with the given solid: a

Describe the cross-section formed by cutting straight down from the dotted line:

Let’s practice 4

A solid is cut parallel to its base and the resulting uniform cross-section is shown to the right. State whether the following could be the solid: a

List at least two solids that can have triangular cross-sections. Describe the plane that forms the triangular cross-section.

A solid is cut parallel to one of its faces to give the cross-section shown to the right. State the possible solid with this cross-section.

Specify what solids could have the cross-section shown:

Given the following cube, state whether the following cross-sections could be formed by slicing it: a

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Rectangle

Triangle

Identify the solid for each of the following set of parallel cross-sectional blocks of the solid: a

Describe the cross-section formed by the intersection of the stated plane with the following solids: a

A plane perpendicular to the prism

A plane parallel to the base

A plane perpendicular to the base

A diagonal plane passing through its center

diagonal plane passing through its center, not A parallel to any of the faces

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The volume of any prism or cylinder can be calculated using the formula, by simply substituting the area of the base for B:

V = Bh B

area of the base

perpendicular height between bases

Here are some speciﬁc volume formulas: Vrectangular prism = length ⋅ width ⋅ height V = lwh Height

Width

Length

Vcube = side3 V = s3

s s

Vtriangular prism = Area of base ⋅ prism height V = Area of triangle ⋅ height Height

bhH

Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 10.03 to answer these questions. 1.

Set the dimensions of the triangular prism: b = 5, h = 8, Height = 5. What is the volume?

Set the dimensions of the rectangular prism: width = 2, length = 10, Height = 5. What do you notice about the Height and the volume of the triangular prism and rectangular prism from part (1)?

Find a set of dimensions for the two prisms, so that they have the same volume, but different heights.

If you double the Height of both shapes, what do you notice about the volumes of the triangular prism and the rectangular prism?

Example 1 Find the volume of a cylinder rounded to one decimal place if its radius is 5 cm and its height is 13 cm.

Create a strategy To solve this, we will use the formula for the volume of a cylinder with the given radius and height lengths.

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Apply the idea

Reﬂect and check

We have been given the values for the radius, r = 5, and the height, h = 13, so we can substitute these values into the volume formula.

As an exact value, this would be 325π cm3.

V = π r2h

Formula for volume of a cylinder 2

= (π ⋅ 5 ) ⋅ 13

Substitute the values

= π ⋅ 25 ⋅ 13

Evaluate the squares

= 1021.0 cm3

Evaluate

Example 2 A 13 cm concrete cylindrical pipe has an outer radius of 6 cm and an inner radius of 4 cm as shown. Find the volume of concrete required to make the pipe, rounded to two decimal places.

13 cm 4 cm 6 cm

Create a strategy To create the pipe, we must ﬁnd the difference in the area of the outer circle and the inner circle. Once we have found the difference in the area of the circles, we can use that area to ﬁnd the volume using the formula V = Bh.

Apply the idea To ﬁnd the base area, subtract the area of the inner circle from the outer circle. For this problem, we will let the base area B = Abase. Abase = Aouter circle − Ainner circle 2

Find the area of the pipe face

=π⋅6 −π⋅4

Area of circle is A = π r2

= 36π − 16π

Evaluate the squares

= 20π cm

Evaluate as an exact value

V = Abase h

Volume formula

= 20π ⋅ 13

Substitute the values 3

= 816.81 cm

Evaluate 3

The volume of concrete required is 816.81 cm .

Reﬂect and check We can identify the volume in one set of workings: V = Bh

Use the formula

= ( Aouter circle − Ainner circle) h 2

= (π ⋅ (6) − π ⋅ (4) ) ⋅ 13 cm 3

= (π ⋅ 36 − π ⋅ 16) ⋅ 13 cm 3

= 816.81 cm

Substitute B with desired base area Substitute the values in the areas Evaluate the squares Evaluate

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This can help with accuracy, because if we had rounded the area of the base to 20π = 62.83 cm2, then our answer would be V = 62.83 ⋅ 13 = 816.79 cm3. Being off by 0.02 sometimes does not make a difference, but if we make that rounding error for 1 000 000 pipes, then it can add up to be signiﬁcant.

Example 3 Find the volume of the triangular prism shown.

4 mm 11 mm 4 mm

Create a strategy To find the volume, we must first find the area of the triangle. Once we have the area of the triangle, we can use the volume of a right prism formula, V = Bh.

Apply the idea Use the area of a triangle formula

Substitute b = 4 and h = 4

Evaluate

V = Bh

Use the volume of a right prism formula

= 8 ⋅ 11 = 88 mm

Substitute B = 8 and h = 11 3

Evaluate

Example 4 Find the volume of this ﬁgure, rounded to two decimal places.

10 m 4m

Create a strategy This ﬁgure is composed of a rectangular prism and a half cylinder. First, we will ﬁnd the volume of each shape and then add the two volumes together. The half cylinder has a diameter of 4 m which means its radius is 2 m. The rectangular prism has a side length of 4 m and a height of 10 m.

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Apply the idea Cylinder volume = π r2 h

Formula for volume of a cylinder

= π ⋅ 2 ⋅ 10

Substitute r = 2, h = 10

= 125.66

Evaluate

Since this the volume of a full cylinder, we need to divide this value by 2 to ﬁnd the volume of half of it. 125.66 ÷ 2 = 62.83 m3 Next, we will ﬁnd the volume of the rectangular prism: Rectangular prism volume = lwh

Formula for rectangular prism

= 4 ⋅ 10

Substitute l = w = 4, h = 10

= 160 m

Evaluate

Total volume = 62.83 + 160 = 222.83 m

Add the volumes Evaluate

Reﬂect and check An alternative solution would be to ﬁnd the area of the base of this solid and then multiply it by the height of the solid.

Start with the formulas Substitute s = 4 and r = 2 Evaluate powers V = Bh

Simplify multiplication using exact values Start with the formula

= (16 + 4π ) ⋅ 10 3

= 222.83 m

Substitute h = 10 and area of base Evaluate on the calculator

Example 5 A prism has a volume of 990 cm3. If it has a base area of 110 cm2, ﬁnd the height of the prism.

Create a strategy We can use the volume of a right prism formula, V = Bh, to substitute our known values for the volume, 990 cm3, and the base area, 110 cm2. Then, we will solve for the missing value of h.

Apply the idea Use the volume of a right prism formula Substitute B = 110 and V = 990 Divide both sides by 110 Evaluate

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Example 6 A manufacturer is designing a new storage box to replace its current cube-shaped model. • The volume of the new storage box will be 27 times greater than that of the original. • The new storage box will have the same length and width as the original box. What will be the height of the new storage box compared to the original box?

Create a strategy We know that only the height is going to change, which means the base area will stay the same. We can use the volume of a prism formula, V = Bh, where B is the base area and h is the height.

Apply the idea If the new volume is 27 times greater, then Vnew = 27 ⋅ V = 27 ⋅ Bh As the new cube has the same length and width, the only dimension that can change to increase the volume is the height. That means the height of the new box must be 27 times the height of the original box to achieve the new volume.

Idea summary The volume, V, of a prism or cylinder is calculated using the formula V = Bh, where B represents the area of the base and h represents the height of the prism.

Surface area of prisms and cylinders Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 10.03 to answer these questions. 1.

Create a triangular prism by changing N to 3 and dragging the sliders to close the net. How can you use the net to find the surface area of the prism?

What happens as you drag the N slider to its largest value?

A net is a diagram of the faces of a three-dimensional figure arranged in such a way that the diagram can be folded to form the three-dimensional figure. We can find the surface area from a net by calculating the sum of the areas of the lateral faces and the bases. Lateral faces The faces in a prism or pyramid that are not bases

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A cube and its net

A right cylinder and its net

Note that even some solids with curved faces, such as cylinders, have nets consisting of ﬂat, two-dimensional shapes. Additionally, a solid can be represented with multiple different, equivalent nets. The surface area of a cylinder can be calculated identically; by adding the area of two circular bases to the product of the circumference and the perpendicular height between bases. We can ﬁnd the lateral area of a prism or a cylinder by ignoring the bases. Lateral area The sum of the areas of the lateral faces of a prism or pyramid, or the area of the lateral surface of a cylinder or cone LA = Ph where P is the perimeter of the base and h is the height between bases

We can write a general formula for the surface area of prisms and cylinders:

SA = LA + 2B LA

total lateral area

area of one base

Example 7 Consider the can of tuna shown:

a Draw the net of the can of tuna and label its dimensions.

Create a strategy

Apply the idea

The net of a cylinder will have two circular bases and a rectangle for its lateral face.

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b Find the area of each part of the net and ﬁnd the total surface area of tin that a company must produce per can of tuna.

Create a strategy We have enough information from the can of tuna to find the area of each of the circular bases. The diameter of each base, is 2.75 in so we use half the diameter for the radius, 1.375 in. However, we will need the length of the can in order to find the area of the lateral face. We can calculate the length of the can by finding the circumference of a circular base.

Apply the idea First, calculate the area of each base: A = π r2

Formula for area of a circle 2

= π (1.375) = 5.940

Substitute the length of the radius of the base Evaluate the exponent and multiplication 2

The area of each base is 5.940 in . Then, calculate the circumference of a base to ﬁnd the length of the can’s lateral side: C = 2π r

Formula for circumference

= 2π (1.375)

Substitute the length of the radius of the base

= 8.639

Evaluate the multiplication

The length of the can is 8.639 in. Now, the area of the lateral side, which is a rectangle, is found by multiplying the length and width of the can. So, the area of the lateral side is 8.639 in ⋅ 1.125 in = 9.719 in2. Finally, we can ﬁnd the surface area of the tuna can by adding the area of each base to its lateral side: 5.94 in2 + 5.94 in2 + 9.719 in2 = 21.599 in2.

Reﬂect and check While we could have used fractions throughout the problem, converting the given dimensions to decimals may be easier to work with. We rounded the decimals to 3 places, since the fractions we are using are exact to 3 decimal places. We may also want to keep that level of precision because the company will want to be as precise as possible when producing each can.

c The formula for ﬁnding the surface area of a prism or cylinder is SA = 2B + Ph, where SA represents surface area, B represents the area of the base, P represents the perimeter of the base, and h represents the height of the prism or cylinder. Explain what each part of the formula for surface area represents, and relate it to ﬁnding the surface area for the can of tuna.

Create a strategy We explain why variables are multiplied and added a certain way, then relate it back to the can of tuna.

Apply the idea The surface area formula is the sum of two products. The ﬁrst product is twice the area of the base. This makes sense because each prism or cylinder will have two identical bases, so their areas should be added together, or in this formula multiplied by 2. The can of tuna had circular bases, so we used the Area formula for a circle to ﬁnd the Base area.

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The second product is the result of multiplying the perimeter of the base and the height of the prism or cylinder. This is the area of the rectangular lateral side, since the perimeter of the base is also the width of the rectangle and the height of the prism is the height of the rectangle. For the can of tuna, the base was circular, so we used the formula for the Circumference of a circle as the perimeter. By calculating the sum of these products, we have found the combined areas of each part of the net, which is the surface area of the entire can of tuna.

Example 8 4 cm

Find the surface area of the triangular prism shown.

3 cm 5 cm

19 cm

Create a strategy Add the areas of the ﬁve faces of the prism, where the top and bottom face are identical:

Apply the idea Because the top and bottom face are congruent, we can ﬁnd twice the area of one face. Notice that the top and bottom are right triangles, so the base and height will be the lengths of each leg.

Area of left rectangle = 19 ⋅ 3 = 57 cm Area of front rectangle = 19 ⋅ 5 2

Area of back rectangle = 19 ⋅ 4 = 76 cm

Evaluate Multiply the length and width

= 95 cm

Evaluate Multiply the length and width

Surface area = 12 + 57 + 95 + 76 = 240 cm

Evaluate Multiply the length and width

Substitute b = 4 and h = 3

Evaluate Add the areas Evaluate

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Example 9 Yuki is decorating a wedding cake with some icing. He wants to cover the outer facing surface of the cake with a layer of icing that is an eighth of an inch thick. 2 in 3 in 4 in

12 in 8 in 4 in

Determine how much icing Yuki will need to decorate the cake.

Create a strategy We need to determine whether the entire top of each layer needs to be iced, and then the next layer added, or if the layers are placed ﬁrst, and then only the exposed portions are iced. After ﬁnding the surface area, we will want to multiply that value by

inches. Depending on the approach to the

problem, we could end up with different solutions to how much icing Yuki needs for the cake. Let’s assume that the layers are placed, and then only exposed cake will be iced. That means we should calculate the lateral area of each layer, then calculate the tops of the tiers shown in the diagram using the areas of the circular tops.

Apply the idea For the lateral area of each tier, we need the circumference of each. Starting with the first tier we have C = 2π r → C = 2π (2 in) → C = 12.57 in So, the lateral area of the first tier is 12.57 in ⋅ 2 in = 25.14 in2. For the middle tier we have C = 2π r → C = 2π (4 in) → C = 25.13 in So, the lateral area of the middle tier is 25.13 in ⋅ 3 in = 75.39 in2. For the bottom tier we have C = 2π r → C = 2π (6 in) → C = 37.7 in So, the lateral area of the bottom tier is 37.7 in ⋅ 4 in = 150.8 in2. The top of the first tier will be iced, so we can find the area of the circular top. A = π r2 → A = π (2 in)2 = 12.57 in2 The exposed portion of the middle layer can be found by subtracting the area of the first tier’s base from the area of the middle tier’s base. The area of the top of the middle tier can be calculated as A = π r2 → A = π (4 in)2 = 50.27 in2 Then, we can subtract the area of the first tier’s base 50.27 in2 − 12.57 in2 = 37.7 in2

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The exposed portion of the bottom layer can be found by subtracting the area of the middle tier’s base from the area of the bottom tier’s base. The area of the top of the bottom tier can be calculated as A = π r2 → A = π (6 in)2 = 113.1 in2 Then, we can subtract the area of the middle tier’s base 113.1 in2 − 50.27 in2 = 62.83 in2 Finally, we will add the total surface area of icing needed for the cake and multiply the total by of icing needed for Yuki’s cake:

in to ﬁnd the amount

25.14 in2 + 75.39 in2 + 150.8 in2 + 12.57 in2 + 37.7 in2 + 62.83 in2 = 364.43 in2 364.43 in2 ⋅

in = 45.55 in3

Example 10 A rectangular prism has a length 16 cm, a width of 10 cm, and a height of 9 cm. If the height of the rectangular prism is increased by 4 cm while the other dimensions remain unchanged, calculate and compare the new volume and surface area to the original. a Compare the new volume of the rectangular prism to the original.

Create a strategy First, we need to calculate the volume of both prisms. To calculate the volume, we use the volume of rectangular prism formula, V = l ⋅ w ⋅ h, and substitute the values given for length, width and height. Once we have calculated both volumes, we can ﬁnd the difference between the two volumes to compare.

Apply the idea Original volume = 16 ⋅ 10 ⋅ 9 3

= 1440 cm

New volume = 16 ⋅ 10 ⋅ (9 + 4) = 16 ⋅ 10 ⋅ 13 3

= 2080 cm

Substitute l = 16, w = 10, h = 9 Evaluate Increased height by 4 cm Evaluate the addition Evaluate

Now, we can ﬁnd the difference between the volumes: 2080 − 1440 = 640 The volume has increased by 640 cm3.

Reﬂect and check Because only one dimension changed, the rectangular prisms will not be similar. Recall that objects are only similar when all linear dimensions are changed proportionally.

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b Compare the new surface area of the rectangular prism to the original.

Create a strategy First, we need to calculate the surface area of both prisms. To calculate the surface area, we use the surface area of a rectangular prism formula, 2(wl + hl + hw), and substitute the values given for length, width and height. Once we have calculated both surface areas, we can ﬁnd the difference between them to compare.

Apply the idea Original SA = 2 ⋅ (10 ⋅ 16 + 9 ⋅ 16 + 9 ⋅ 10)

Substitute l = 16, w = 10, h = 9

= 2 ⋅ 394 = 788 cm

Evaluate the operations inside the parentheses 2

Evaluate

New SA = 2 ⋅ (10 ⋅ 16 + 13 ⋅ 16 + 13 ⋅ 10)

Substitute l = 16, w = 10, h = 13

= 2 ⋅ 498

Evaluate the operations inside the parentheses

= 996 cm2

Evaluate

Now, we can ﬁnd the difference between the surface areas: 996 − 788 = 208 The surface area has increased by 208 cm2.

Reﬂect and check Changing the height impacts not only the volume, but also the surface area, though in a different way. The increase in height alters the areas of the sides connected to the height, leading to an overall increase in surface area.

Idea summary We can calculate the surface area of any prism or cylinder using the formula

SA = LA + 2B LA B

total lateral area area of one base

Practice What do you remember? 1

542

Match each solid to its name. a

Right rectangular prism

Right cylinder

Right triangular prism

Right prism

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Identify the solids used to make up each of these composite solids. a

15 m

4 cm

11 cm

7 cm

7m 8 cm

8 cm

9 mm 16 mm

11 cm

10 mm

6 cm 8 cm

4 cm

36 mm

53 mm

Nataraja is building a storage chest in the shape of a rectangular prism. The chest will be 93 cm long, 80 cm deep, and 18 cm high. Find the lateral surface area of the chest.

Fill in the blanks with the appropriate terminology for the cylinder shown: Base

Volume, lateral surface area, base area, surface area.

Height (h) Base Radius (r)

Determine the base area of the triangular prism shown:

4.08 cm

15 cm 6 cm

Let’s practice SOL

Consider the dimensions of three ﬁgures: • Rectangular prism with a length of 10 cm, a width of 2 cm, and a height of 5 cm • Triangular prism with a base of 5 cm, a height of 4 cm, and a length of 10 cm • Cylinder with a radius of 2.39 cm and a height of 10 cm a

Which ﬁgures have the same volume? Approximate π to 3.14 for your calculations.

Which ﬁgures have the same lateral surface area? Approximate π to 3.14 for your calculations.

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The lid of this treasure chest is found to be exactly one half of a cylindrical barrel. Find the surface area of the chest.

30 cm

42 cm

59 cm

The given diagram shows the design for a marquee (tent). The roof of the marquee has a height of 3 m. The material for the marquee costs $44/m2. a

Find the total surface area of the marquee. Consider that the only measurements needed from the rectangular prism is its lateral area.

Find the total cost of the marquee material.

3m 7m 10 m 8m

The larger the surface area of an ice cube, the more heat it absorbs and thus the quicker it melts. Draw a model for each scenario a

A cylindrical ice cube with a radius of 1.5 cm and a height of 4 cm

Three rectangular ice cubes, each with length 3 cm, width 3 cm, and height 1 cm

1 cm

1 cm 3 cm

3 cm

1 cm 3 cm

c 10

3 cm

Which ice will melt faster?

A triangular tunnel is made through a rectangular prism. a

Find the volume of the solid formed using two different methods: i

Find the volume of the rectangular prism and subtract the volume of the triangular prism.

Find the area of the shaded face and use Volume of Prism = A × h. 3 cm

Explain the process of finding the surface area of the solid formed, including the inside of the tunnel.

3 cm

6 cm 8 cm

7 cm

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The larger prism has two identical holes carved out of it, each of which is a rectangular prism. All measurements are in meters. a

Explain how to find, and then calculate, the volume of the remaining solid. Round your answer to two decimal places.

Explain how to find, and then calculate, the surface area of the remaining solid.

Mathspace Virginia SOL Geometry mathspace.co

6.5 5 6.5

Find the surface area of the following solids: a

8 cm

8 cm 7 cm

8 cm

16 cm

3 cm 16 cm

9 mm

2 cm

16 mm 3 cm

10 mm

3 cm 53 mm

36 mm

10 cm

Find the volume of the following solids: a

b 11

3 5 cm

14 cm 6 cm

16 cm 6 cm

9 cm 9 cm

72 mm 36 mm

15 cm

Queso Fresco is sold in wheels. The volume of a wheel of Queso Fresco is 2002 cm3. If the height of a wheel of Queso Fresco is 6 cm, determine its radius.

Parmigiano-Reggiano is formed in wheels, and then cut into twenty equal slices before being packaged and sold. The volume of a wheel of Parmigiano-Reggiano is 8000π cm3:

Find the volume of a slice of Parmigiano-Reggiano that has been packaged for sale.

Find the height of the wheel of Parmigiano-Reggiano.

20 cm

The volume of a cylinder is initially 500 cubic inches. If the height of the cylinder is doubled, with the radius remaining unchanged, calculate the new volume of the cylinder. Explain your reasoning. 10.03 Prisms and cylinders mathspace.co

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Imagine a rectangular prism with a length of 12 cm, a width of 7 cm, and a height of 6 cm. If the width of the rectangular prism is decreased by 2 cm while the other dimensions remain unchanged, calculate and compare the new volume and surface area to the original. a

Compare the new volume of the rectangular prism to the original.

Compare the new surface area of the rectangular prism to the original.

A company is creating a new rectangular prism container to replace its original rectangular prism container. • The new container will have 8 times the volume of the original container. • The height of the new container will remain the same as the height of the original container. What will be the length of the base area of the new container compared to the original container? A

2 times the base area of the original container

4 times the base area of the original container

8 times the base area of the original container

16 times the base area of the original container

Let’s extend our thinking 19

Brenton wants to know the volume of the basketball hoop stand at school, including the base, pole, backboard, and hoop. He’s confident that he can measure the dimensions of the different parts of the stand, but isn’t sure how to break up the volume into simple solids.

Using the image of the basketball hoop stand, identify the simple solids that make up the stand and describe the dimensions that Brenton would need to measure to find their volume. 20

Quicker Oatmeal company is choosing between two new packages for their Chocolate Chip Oats: • Option 1 is a canister with a diameter of 6 in and a height of 10 in. • Option 2 is a box with dimensions 3 in × 6.5 in × 9.5 in. a

Which package should Quicker Oatmeal use in order to minimize the amount of materials used? Explain.

No matter which package Quicker Oats chooses, they will sell it for $4.99 per container. Which package should Quicker Oats use to maximize their proﬁts? Explain.

As part of their moonshots program, Quicker Oats offers limited edition packaging for one month of the year that is neither a cylinder or a rectangular prism. The specifications for the moonshots packaging is that it will sell for $6.99 and contain roughly the same amount of Chocolate Chip Oats as their other packages. Design a package prototype that fits these specifications.

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U-Pack Moving Company sells two sizes of moving boxes. U-Pack claims to keep their prices lower than competitors by selling their boxes ﬂat, and having customers assemble the boxes themselves. Box 1 sells for $2.29:

Box 2 sells for $0.99: 7.25 in

17.5 in 12 in 10 in 33.25 in 44 in

If you have 60 ft3 of stuff to pack into boxes for your move, determine which of the following combinations of boxes you would buy. Explain. A 22

15 × Box 1

36 × Box 2

Nabila wants to determine the thickness of her window blinds but her measuring stick is not precise enough to measure such a short distance. Nabila observes that when her blinds can be tightly rolled up into a cylindrical shape when they are up, and is like a rectangular sheet when they are fully rolled out. Describe a method for Nabila to approximate the thickness of her blinds.

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10.04 Cones and pyramids After this lesson, you will be able to... • create models and solve problems, including those in context, involving surface area of pyramids, cones, and composite solids consisting of pyramids and cones. • solve multistep problems, including those in context, involving volume of pyramids, cones, and composite solids consisting of pyramids and cones. • determine unknown measurements of pyramids and cones using information such as length of a side, area of a face, or volume. • describe how changes in one or more dimensions of a pyramid or cone affect the total surface area and volume of the figure. • describe how changes in surface area and/or volume of a pyramid or cone affect the measures of one or more dimensions of the figure. • solve problems involving changing the dimensions or derived measures of a pyramid or cone.

Volume of cones While a cylinder is formed by a pair of congruent circles joined by a curved surface, a cone is formed from a single circle with a curved surface that meets itself at the apex (or vertex). Right cone A cone in which the apex lies directly above the center of the base

Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 10.04 to answer these questions. 1.

How many times does the liquid in the cone fill the cylinder?

How do you think the formula for the volume of a cone will relate to the formula for the volume of a cylinder?

The volume of a cone is exactly one-third the volume of a cylinder formed from the same base with the same perpendicular height. That is, the volume of a cone is given by the general formula:

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area of the base

perpendicular height

We can rewrite this formula more speciﬁcally for cones as: Volume = V= h

Area of Base ⋅ Height

π r2 h

where r is the radius of the base and h is the perpendicular height of the cone. r

A = π r2

Example 1 Find the volume of this cone and round your answer to two decimal places.

6 cm

2 cm

Create a strategy

Apply the idea

We can ﬁnd the volume of the cone using the

We are given the radius, r = 2, and height, h = 6. We can substitute these into the formula to ﬁnd the volume of the cone.

formula V =

π r h.

V= =

π r2 h

Write the formula

⋅ π ⋅ 22 ⋅ 6

Substitute the values

= 25.13 cm2

Evaluate and round

Example 2 A cone is sawed in half to create the following solid: 11 cm

5 cm

What is the volume of the solid?

Create a strategy Calculate the volume of the original cone, then take half the volume. We need the height of the cone, so use the Pythagorean theorem with the height of the slanted lateral face as the hypotenuse and the radius of the base as the length of the triangle’s short leg.

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Apply the idea First, we’ll ﬁnd the height of the cone: a2 + b2 = c2 2

Pythagorean theorem

5 + b = 11

Substitute a = 5 and c = 11

Evaluate the exponents

b = 96

Subtract 25 from both sides

Evaluate the square root of both sides

25 + b = 121

The height of the cone is

cm.

Now, we need the area of the base of the original cone. We know that the radius of the base is 5 cm, so the area of the base is π (52) = 25π cm2. We can calculate the volume of the original cone as follows: Volume of a cone

Substitute B = 25π and h =

Evaluate the multiplication

Finally, since the cone was sawed in half, we can take half of the total cone of the original cone to be approximately 128.25 cm3.

Example 3 A cone has a volume of 160 mm3. If the perpendicular height and radius of the cone are equal in length, ﬁnd the radius r in mm. Round your answer to two decimal places.

Create a strategy The volume of a cone is found by V =

π r2 h.

Since the perpendicular height and radius are the same, the volume is equal to: Volume of a cone

Substitute h = r

Apply the idea We are given the volume is 160 mm3, so we can substitute this into our equation and solve for r. Solving for r, we have Substitute V = 160

Multiply 3 on both sides

Divide π on both sides

Evaluate the division

Take the cube root of both sides of the equation

Evaluate

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Reﬂect and check Let’s check our answer by ﬁnding the volume of the cone using 5.35 mm for the radius and height: Substitute 5.35 for r and h Evaluate Notice that this volume is slightly differently from the given volume of 160 mm3. This is because we rounded r to two decimal places. Rounding can introduce slight differences in values, especially those involving π and cube roots. It’s important to consider the effects of rounding when stating or checking answers.

Example 4 An ice cream shop currently has one size ice cream cone. They are adding a new, larger size option. • The new cone will have 2 times the volume of the original cone. • The height of the new cone will remain the same as the height of the original cone. How does the new radius compare to the radius of the original cone?

Create a strategy We can set up an equation given the following: • The volume of a cone is given by V =

π r2h.

• The height of both cones remains the same. • The new cone has 2 times the volume of the original cone. Let’s denote the radius of the original cone as r and the new cone as rnew.

Apply the idea Let the volume of the original cone be V =

π r2h.

Let the volume of the new cone be Vnew =

Since the new cone has twice the volume of the original cone, we can say Vnew = 2 ⋅

We can use the expressions for Vnew to write and solve an equation.

Equate the expressions for the volume of the new cone

Divide both sides by

Square root both sides

So, the length of the radius of the new cone is

πh

times the length of the radius of the original cone.

Reﬂect and check To verify that rnew = ⋅ r, we can calculate the volumes of both cones and see if the new cone’s volume is twice of the original cone’s volume. As an example, let’s set r = 1 in, h = 2 in, and π = 3.14. Let’s ﬁrst calculate the volume of the original cone: Volume of the original cone =

(3.14) (1)2 (2)

= 2.09 in3

Substitute r = 1, h = 2, and π = 3.14 Evaluate and round to two decimal places

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Let’s calculate the volume of the new cone. Note that the radius of the new cone is of the original cone: Substitute r =

times the length of the radius

, h = 2, and π = 3.14

Evaluate and round to two decimal places

This veriﬁes that the new cone’s radius is indeed 2 times the original. The slight discrepancy is due to rounding.

Idea summary The volume of a cone can be found by taking one-third the volume of a cylinder with the same base area and height. The formula for the volume of a cone is given by:

B h

area of the base perpendicular height

This can be rewritten speciﬁcally for cones as:

r h

radius of the base perpendicular height

Volume of pyramids A pyramid is a ﬁgure formed from a polygonal base and a set of triangular faces. The triangular faces connect to one side of the base and all join together at the apex. Rectangular pyramid

Triangular pyramid

A pyramid with a rectangular base

A pyramid with a triangular base

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Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 10.04 to answer these questions. 1.

How many pyramids fit into the prism?

If the volume of a prism is found using the formula V = Bh, how do you think we can find the volume of a pyramid?

Similarly to the volume of a cylinder and a cone, the volume of a pyramid is also calculated by taking one-third the volume of a prism that has the same height and base area.

area of the base

perpendicular height

A more speciﬁc formula is dependent on the shape of the base.

Example 5 The Pyramid of Giza is a square pyramid, that is 280 Egyptian royal cubits high and has a base length of 440 Egyptian royal cubits. What is the volume of the Pyramid of Giza?

Create a strategy First we ﬁnd the area of the base, then we can use that to ﬁnd the volume. Since this solid is a square pyramid, the base is a square.

Apply the idea Finding the area of the base, we have: B = side length2 = 440

= 193 600

Area of a square Substitute the side length Simplify

We can now use this to calculate the volume: Volume of a pyramid

Substitute known values

Simplify

So the volume of the Pyramid of Giza is 18 069 333

cubic Egyptian royal cubits.

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Example 6 A small square pyramid of height 4 cm was removed from the top of a large square pyramid of height 8 cm forming the solid shown. Find the exact volume of the solid.

4 cm

4 cm 4 cm

8 cm

Create a strategy Subtract the volume of the smaller pyramid from the volume of the larger pyramid.

Apply the idea Start by calculating the volume of the larger pyramid. We have Volume of a pyramid Substitute B = 64 for the area of the base and h = 8 for the height of the larger pyramid

Evaluate the multiplication

The volume of the larger pyramid is exactly

cm3.

Now we can calculate the volume of the smaller pyramid. We have Volume of a pyramid Substitute B = 16 for the area of the base and h = 4 for the height of the smaller pyramid

Evaluate the multiplication

The volume of the smaller pyramid is exactly

cm3.

The volume of the solid after removing the smaller pyramid is

cm3 –

cm3 =

cm3.

Reﬂect and check We could choose to leave the solution in exact form, since the directions do not specify rounding requirements, or we could simplify the solution and round to a measure we ﬁnd appropriate.

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Example 7 The height and base side of a triangular pyramid are tripled. The triangular base is equilateral. Determine the effect this has on the volume of the pyramid.

H = 24 in h = 17.3 in

b = 20 in

Create a strategy To determine the effect on the volume, recall that the volume of a triangular pyramid is given by V = is the area of the triangular base and H is the perpendicular height. We can calculate B using the triangle area formula A =

BH, where B

bh, where b is the base and h is the height of the triangle.

We can determine the effect on the volume by tripling the perpendicular height and the base side, calculating the new volume, then dividing the new volume by the original volume.

Apply the idea We are given the following: • Perpendicular height of the pyramid (H) = 24 in • Base side (b) = 20 in • Height of the triangular base (h) = 17.3 in Let’s calculate the original volume:

Volume of a right pyramid

Substitute B =

Substitute b = 20, h = 17.3, and H = 24

Evaluate

Let’s calculate the new volume using the tripled dimensions: • Perpendicular height of the pyramid (H) = 72 in • Base side (b) = 60 in • Height of the triangular base (h) = 51.9 in

Volume of a right pyramid

Substitute B =

Substitute b = 60, h = 51.9, and H = 72

Evaluate

The effect of on the volume is:

Scaling the pyramid’s height and base side length by a factor of 3 causes the volume to increase by a factor of 27.

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Reﬂect and check Tripling the height and base of the original pyramid, a, results in a new pyramid, b, that maintains proportional dimensions with the original. These two pyramids are similar ﬁgures, and the ratio between their volumes is a3 : b3. So, increasing the dimensions of the original triangle by a factor of 3 results in the volume being multiplied by 33 or 27.

Idea summary The volume of a pyramid can be found by taking one-third the volume of a prism with the same base area and height. The formula for the volume of a pyramid is given by:

B area of the base h perpendicular height

Surface area of cones and pyramids Similar to prisms, we can consider the nets of pyramids and cones to determine their surface area and lateral surface area.

Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 10.04 to answer these questions. 1.

The formula for the lateral surface area of a cone is given by LA = π rhs, where r is the radius of the base of the cone and hsis the slant height of the cone. How does this formula relate to the area of the rearranged lateral face on the cone?

The surface area of a pyramid or cone is the sum of the area of the base and the area of the lateral face. The formula can be written generally as:

SA = LA + B LA

lateral area

area of base

For a pyramid, the lateral area can be found by calculating the area of each of the triangular faces.

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For a cone we must ﬁnd the area of the circular base and add that to the area of the sector that creates the lateral face. r l

Area of the sector = π l2 ⋅

= π rl

We can then combine this area with the area of the circular base to get a formula for the surface area of a cone given by: Surface area of right cone = Area of base + Area of sector SA = π r2 + π rl where r is the cone’s base radius and l is the slant height.

Example 8 Consider the diagram of the right square pyramid and its net shown:

l b b

Let b = the base of each triangle and l = the height of each triangle. a Explain a process for ﬁnding the surface area of the square pyramid.

Apply the idea To find the surface area of the square pyramid, we will need to calculate the sum of the area of the pyramid’s base and the area of the pyramid’s lateral faces. First, we find the area of the base of the pyramid. Then, we find the area of the lateral faces of the pyramid. Since the base of this pyramid is a square, which is a regular polygon, each of the lateral face triangles are congruent. So, we can calculate the area of one of the faces and multiply that by the number of faces on the pyramid. Together, these areas make up the entire surface area of the square pyramid, which is the area of its net.

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Reﬂect and check We could use the given dimensions to write an expression for the surface area of the given pyramid. For this pyramid, the area of the base, B, is B = b ⋅ b or B = b2. Then the area of each triangle, A, is A =

bl. Since there are 4 faces, we multiply the area of the triangular face by 4.

Therefore, the total surface area of the square pyramid can be expressed as SA = b2 + 4

or SA = B + 4A.

b Now, consider the regular right octagonal pyramid shown below, where a = the distance from the midpoint of a side of the base to the center of the base:

l a b

Draw the net of the right octagonal pyramid and describe how to ﬁnd the surface area of the pyramid.

Create a strategy We can draw the net of the octagonal pyramid the same way as the net in part (a) is drawn, or we can imagine unraveling the lateral faces of the pyramid and drawing them side by side.

Apply the idea

Reﬂect and check We could use the given dimensions to write an expression for the surface area of the given pyramid. This would summarize our description algebraically. For this pyramid, the area of the base, B, can be calculated by ﬁnding the total area of the 8 congruent

l b

b b b

b b

a b

triangles formed by joining the corners of the base to its

For this pyramid, we will again need to calculate the area of the base and the lateral surface area. We can ﬁnd the area of the base by cutting the octagon into eight congruent triangles, and then ﬁnding the sum of the area of each triangle. For the lateral surface area, we can calculate the area of one triangle and multiply by eight triangles. Together, the base area and lateral surface area make up the surface area of the pyramid.

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center B = 8

or B = 4ab.

Then the area of each triangular face, A, is A = bl. Since there are 8 faces, we multiply the area of the triangular face by 8. Therefore, the total surface area of the octagonal pyramid can be expressed as: SA = 4ab + 4bl or SA = B + 8A.

c Write a formula for ﬁnding the surface area of a regular, right n-gon pyramid. Explain your reasoning.

Create a strategy For any pyramid, we know that the net we need the area from will have a base and lateral faces that we need the sum of.

Apply the idea If we let B = the area of the base and A = the area of one of the lateral face triangles, we can say that the surface area, SA, is equivalent to the sum of the area of the base and the area of the lateral faces. For any n-gon, the number of lateral faces is equal to the number of sides of the polygon. A formula for the surface area of any pyramid is SA = B + nA.

Reﬂect and check We can express the formula above in different forms depending on the dimensions given in a problem. For example, if we are given the following dimensions: • b - the base of each triangular face • l - the height of each triangular face • a - the distance from the midpoint of a side of the base to the center of the base Then the formula can also be expressed in terms of these parameters. We have the area of the base, B, will be n triangles of base b and height a. Hence, B = n

. And the area of each face is A =

So, the total surface area is SA = n

, which could also be expressed as SA =

bl. (a + l).

Example 9 What is the surface area of the following cone?

9 cm

3 cm

Create a strategy Calculate the slant height of the cone using the Pythagorean theorem, then use the formula for the surface area of a cone to calculate the surface area.

Apply the idea The slant height of the cone can be found using the Pythagorean theorem as follows: Pythagorean theorem a2 + b2 = c2 Substitute a = 3 and b = 9 32 + 92 = c2 Evaluate the exponents and addition 90 = c2 = c Evaluate the square root of both sides The slant height of the cone is

cm.

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Now, we can calculate the surface area as follows: Surface area of a cone SA = π r2 + π rl = (π ⋅ 32 ) +

Substitute r = 3 and l =

= 37.46π Evaluate the exponent, square root, multiplication, and addition The surface area of the cone is 37.46π cm2.

Example 10 The Pyramid of Giza is a square pyramid, that is 280 Egyptian royal cubits high and has a base length of 440 Egyptian royal cubits. What is the surface area of the exposed walls of Pyramid of Giza?

Create a strategy We will need the height of a lateral face of the pyramid to calculate the lateral surface area. First, ﬁnd the height of one lateral face using the Pythagorean theorem, then calculate the surface area of the pyramid. The diagram that follows shows the dimensions of the Pyramid of Giza:

280 Egyptian royal cubits l

220 Egyptian royal cubits

Apply the idea First, we’ll use the height and half the length of the base as the legs in the Pythagorean theorem: a2 + b2 = c2 2

Pythagorean theorem

Substitute a = 220 and b = 280

126 800 = c

Evaluate the exponents and addition

Evaluate the square root of both sides

220 + 280 = c

The height of one lateral face of the pyramid is exactly

Egyptian royal cubits.

We can use this to calculate the area of one of the faces of the Pyramid of Giza as follows: Area of a triangle Substitute b = 440 and h =

Evaluate the multiplication

The area of one of the faces of the Pyramid of Giza is (220) lateral faces is (880) Egyptian royal cubits2.

Egyptian royal cubits2, so the area of all four

The surface area of the exposed walls of Pyramid of Giza to the nearest square royal cubit is 313 359 Egyptian royal cubits2.

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Reflect and check The actual length of an Egyptian royal cubit is about 20 in, consider how the surface area compares to area of a football field. Note that we did not simplify the square root until the final calculation in order to preserve the precision of the length.

Example 11 A decorative lampshade is designed in the shape of a right circular cone with an original height of 8 in and a radius of 4 in. To ﬁt a larger lamp base, the height of the lampshade is increased by 6 in. Determine the new total surface area of the cone. Provide your answer rounded to the nearest whole number.

Create a strategy To ﬁnd the new total surface area of the cone, we ﬁrst calculate the slant height using the Pythagorean theorem, given the new height and the radius. Remember that the decorative lampshade shaped as a right circular cone has no base. We only need to calculate the lateral surface area Alateral = π rl, where r is the radius and l is the slant height.

Apply the idea The new height of the cone is: 8 in + 6 in = 14 in The slant height l of the cone is

Using π ≈ 3.14, calculate the lateral surface area: Alateral = π ⋅ r ⋅ l

Formula for lateral surface area

=π⋅4⋅

Substitute known values

≈ 183 in2

Evaluate

Idea summary The surface area of a cone or pyramid can be calculated using the formula:

SA = LA + B LA B

lateral area area of base

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Practice What do you remember? 1

Write the formula for the surface area and the volume of each ﬁgure: a

vertex

b lateral surface (curved surface of cone)

slant height (l)

height (h)

slant height (l)

height (h)

area of base (B) base

apex

radius (r) base

perimeter of base ( p)

Calculate the volume of each right pyramid, rounding to the nearest hundredth if necessary: a

b 5 in

8 cm 11 cm 15 cm

7 in 4 in

Calculate the exact volume of each cone: a

9 km

18 km

9 cm

3 cm

Calculate the total surface area of each right solid, rounding to the nearest tenth if necessary: a

b 14.3 cm 10.5 cm

10 in

2 in

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14 m 10 ft 10 m

6 ft 8.7 m

For each solid, calculate the lateral surface area. Round the answer to one decimal place if necessary: a

12 ft

20 mm

10 ft 8.7 ft 10 ft

10 ft

16 mm

d 18 in

9 in

24 ft

14 ft

Let’s practice 6

The Sioux Nation is a group of Native American tribes from the Great Plains region of North America. The Sioux historically built tepees for shelter, intentionally designing their tepees to be slightly slanted to allow for proper ventilation through a smoke hole. A typical Sioux tepees could be 14 feet tall.

If the slant height of a tepee was 16 feet, ﬁnd the approximate lateral surface area, rounded to the nearest ten.

To build a tepee, the Sioux set up wooden poles and covered them in animal hide before painting images of earthly creatures and various forms of symbolism. One animal hide could cover about 26 ft2. Determine how many hides would be required to cover the tepee from part (a). 10.04 Cones and pyramids mathspace.co

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An ice cream cone is made by folding together a sector of pastry, with a small overlap. The dimensions of the cone are shown in the diagram: a

Find the external surface area of the cone in square centimeters. Round your answer to one decimal place.

If the overlap adds an extra 5% to the area, determine how much pastry is required to produce the cone. Round your answer to the nearest square centimeter.

4 cm

10 cm

The Louvre pyramid is a large glass and metal pyramid which serves as the entrance to the Louvre museum in Paris. It is a square pyramid with a perpendicular height of 22 m and a base length of 35 m. a

Draw a model of the Louvre pyramid with its dimensions labeled.

If the lateral surface of the pyramid is entirely covered in glass, determine how many square meters of glass make up the structure. Round your answer to the nearest square meter.

Find the capacity of the ice-cream cone in milliliters, given that 1 cm3 = 1 mL.

9 cm

Round your answer to one decimal place. 13 cm

A paperweight is in the shape of a square-based pyramid with dimensions as shown. The paperweight is ﬁlled with solid glass. Find the volume of glass needed to make 3000 paperweights.

5 cm

4 cm 4 cm

Find the surface area of the following solids, rounding your answers to two decimal places where necessary: a

11 cm

10 m

5 cm 6m

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d 5 cm

5 cm 3 cm 10 cm 10 cm 12 cm

e 8 cm

2 cm

8 cm

4 cm

The radius and the height of a cone are equal, and the volume of the cone is 72π cm3. Find the radius of the cone.

The lateral surface area of a cone with slant height of 17 cm is 255π cm3. a

SOL

Find the radius of the cone.

Find the volume of the cone.

A confectionery company packages its premier chocolate in an elegant package shaped like a square pyramid. The base of the pyramid is a square that measures 8 inches on each side, and the slant height of the pyramid is 10 inches. The packaging for a 16 oz piece of chocolate has a height of 8 inches. a

Draw and label the chocolate package with the given measurements.

Calculate the amount of decorative paper required to cover the packaging of the chocolate, and explain your calculations.

A silo used for storing grain is designed with a cylindrical body and a conical roof. The cylinder is 20 meters tall and has a diameter of 6 meters. The cone on top has a slant height of 5 meters. a

Draw and label the silo with these measurements.

What is the volume of grain the silo can hold?

If the grain is sold for $30.00 per cubic meter, calculate the value of the grain that can be stored in the silo to capacity.

If the farmer plans to paint the outside of the silo, excluding the base, and one can of paint covers 200 square meters, determine the number of cans needed to paint the exterior surface of the silo.

A cosmetics company is designing a new conical package to replace its original conical package for a product. • The new package will have 3 times the volume of the original package. • The height of the new package will remain the same as the height of the original package. The length of the radius of the new package will be — A

SOL

the same length as the radius of the original package

times the length of the radius of the original package

3 times the length of the radius of the original package

times the length of the radius of the original package

The height and radius of a cone are doubled. What effect does this have on the volume of the cone? The volume of the cone is multiplied by ⬚. 10.04 Cones and pyramids mathspace.co

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SOL

The height and base side of a square pyramid are tripled. What effect does this have on the volume of the pyramid? The volume of the pyramid is multiplied by ⬚.

Consider a party hat in the shape of a right circular cone with an original height of 10 in and a radius of 3 in. The height of the original hat will be increased by 5 in to accommodate more fun hat sizes. If the radius of the base is 3 in, what is closest to the new total surface area of the cone? A

113 in2

141 in2

176 in2

201 in2

Let’s extend our thinking 20

An incense manufacturer sells different forms of incense, including sticks, wheels, and cones. The density of . incense is 0.4 a

If the mass of a single incense cone is 10 g, ﬁnd the volume the cone.

The height of an incense cone is 3 cm, ﬁnd the approximate radius of the cone, rounded to one decimal place.

A spire is an architectural feature which originated in the 12th century in Germany and can be seen throughout Europe in Gothic and Baroque architecture. a

A building has two spires with square bases. The width of the base is 4 ft and the slant height of each side of the spire is 35 ft. Determine how much paint is required to cover both spires.

Another building has two spires with the same slant height as in part (a), but the base of the spires is a hexagon with side length 2 ft. Explain how will this building require more or less paint than the previous building.

Assuming they have equal slant heights, determine the side length of the spires with a hexagonal base need to be in order to require the same amount of paint as the spires with a square base in part (a).

The Virginia Agricultural Museum plans to bring in a replica grain silo for a new exhibit. The volume of the conical portion of the silo is

π ft3.

The museum’s doorway has a height clearance of 12 ft. Explain whether the model grain silo will ﬁt through the door. 8 ft

6.5 ft

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So can calculate the surface area of a sphere using the formula:

SA = 4π r2 r

the radius of the sphere

We can calculate the volume of a sphere using the formula:

the radius of the sphere

Example 1 Find the surface area of a sphere with radius 3 cm.

Create a strategy Substitute the given radius into the formula for the surface area of a sphere.

Apply the idea

Reﬂect and check

SA = 4π r

Surface area of a sphere 2

= 4π (3)

Substitute the radius

= 36π

Simplify

The surface area of the sphere is 36π cm2.

We found the exact surface area of the sphere in terms of pi. If we multiply 36 by π and round the result, we would have an approximation of the surface area which is less exact. The approximate surface of the sphere is about 113.1 cm2.

Example 2 The ice cream cones at an ice creamery have the dimensions indicated in the diagram:

a Given that 1 cubic centimeter is equivalent to 1 milliliter, how many milliliters of ice cream can ﬁt in each cone, including the hemisphere scoop on top? Round your answer to the nearest milliliter.

8 cm

15 cm

Create a strategy Calculate the volume of the hemisphere and the cone, then ﬁnd their sum. The radius of the hemisphere and cone is 4 cm.

Apply the idea A hemisphere is half a sphere, so we can calculate half the volume of a sphere to ﬁnd the amount of ice cream on top of the cone as follows: Volume of a hemisphere

Substitute the radius

Evaluate the exponent and multiplication

The volume of the ice cream scoop is 134.04 cm3, or 134.04 mL.

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Now, we can calculate the volume of the ice cream that ﬁlls the cone as follows: Volume of a cone

Substitute the area of the base and height

Evaluate the exponent and multiplication

Since the volume of the ice cream in the cone is 251.33 cm3 or 251.33 mL and the ice cream on top is 134.04 mL, the amount of ice cream that can ﬁt in each cone is 251.33 mL + 134.04 mL ≈ 385 mL.

b The ice cream is bought in 10 L tubs. How many whole cones can be made with a single tub of ice cream?

Create a strategy Convert 10 L to milliliters by multiplying by

, then divide the amount by 385 mL.

Apply the idea

Reﬂect and check

Since 10 L is equivalent to 10 000 mL, we can ﬁnd the number of ice cream cones that a tub of ice cream makes as follows:

While 25.97 could be rounded up in a non-contextual problem, the tub of ice cream will not actually have enough to make the next cone with exactly 385 mL, so we state that there is enough ice cream to make 25 cones.

A tub of ice cream can make 25 cones. c Double cones are served with a second hemispherical scoop of the same dimensions as the ﬁrst scoop. How many double cones can be made with from a 10 L tub?

Create a strategy We will need to calculate the amount of ice cream needed for a double cone, then divide the amount of ice cream in a tub by the amount of ice cream needed per double scoop cone.

Apply the idea Since the amount of ice cream in an ice cream cone is approximately 385.37 mL, by adding another scoop, the amount of ice cream needed for a double cone is 385.37 mL + 134.04 mL ≈ 519 mL We can ﬁnd the number of double scoop ice cream cones that a tub of ice cream makes as follows:

A tub of ice cream can make 19 double scoop ice cream cones.

Reﬂect and check Note that as we calculated the math further, we used the milliliters of ice cream that were rounded to two decimal places. The most accurate way to add the volume of the next scoop would be to keep the calculations in their original forms for as long as possible before rounding, like this:

Our calculations yielded the same amount of ice cream in a double cone, because we only needed accuracy to the nearest integer. While performing calculations, we need to keep in mind the degree of accuracy needed for a speciﬁc context.

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Example 3 A sphere has a radius that is r cm long and a volume of

cm3.

Find the radius of the sphere. Round your answer to two decimal places.

Create a strategy Substitute the given volume into the volume formula V =

π r3 and solve for the radius.

Apply the idea Start with the formula for the volume of a sphere

Substitute the volume

Multiply both sides by 3

Divide both sides by 4π

Simplify

Take the cube root of both sides

Evaluate and round

Example 4 A spherical lamp base, initially designed with a diameter of 10 inches, needs to be resized to ﬁt a new space. Its diameter needs to be reduced by 20 percent. a What is the diameter of the lamp’s base after the redesign?

Create a strategy To ﬁnd the new diameter, we will calculate 20 percent of the original diameter and subtract it from the original diameter.

Apply the idea

Reﬂect and check

The original diameter is 10 inches. Reducing this by 20 percent, we calculate the reduction as

An alternative way to ﬁnd the 20% decrease is by multiplying the original diameter by 80%.

10 ⋅ 20% = 2 Thus, the new diameter is 10 − 2 = 8 inches.

100% ⋅ 10 − 20% ⋅ 10 = 80% ⋅ 10 This allows us to ﬁnd the new diameter in a single step: 0.8 (10) = 8 in

b By what percentage does the surface area decrease due to the design change?

Create a strategy We need to calculate the surface area of the spherical lamp base before and after the redesign using the formula SA = 4π r2, where r is the radius of the spherical lamp. Since we are given the diameter, we need to ﬁnd the radius by dividing the diameter by two.

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Dividing the difference between the surface areas by the surface area before the redesign, and then multiplying by 100% would give the percentage decrease.

Apply the idea Let’s calculate the surface area before the redesign: Given the diameter of 10 in, the radius is 10 ⋅ SAbefore = 4π (5)2

= 5.

Substitute r = 5

= 314.16 in

Evaluate

Let’s calculate the surface area after the redesign: Given the diameter after the redesign in part (a), the radius is 8 ⋅ SAafter = 4π (4)2 = 201.06 in

= 4.

Substitute r = 4 2

Evaluate

Now, let’s calculate the percentage decrease:

This shows that decreasing the diameter by 20% will lead to a 36% decrease in the surface area.

c By what percentage does the volume decrease due to the design change?

Create a strategy We need to calculate the volume of the spherical lamp base before and after the redesign using the formula V=

π r3, where r is the radius of the spherical lamp.

Dividing the difference between the volumes by the volume before the redesign, and then multiplying by 100% would give the percentage decrease.

Apply the idea Let’s calculate the volume before the redesign: The radius is 5 in, as calculated in part (b). V Abefore = π (5)3 = 523.6 in3

Substitute r = 5 Evaluate

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Let’s calculate the volume after the redesign: The radius of the new design is 4 in, as calculated in part (b). Vafter = π (4)3 = 268.08 in

Substitute r = 4 3

Evaluate

Now, let’s calculate the percentage decrease:

This means a 20% decrease in the diameter leads to a 48.8% decrease in the volume.

Example 5 Soledad has two spheres. The volume of the larger sphere is 4 times the volume of the smaller sphere. How much greater is the radius of the larger sphere than the smaller sphere?

Create a strategy We can set up an equation given the following: • The volume of a sphere is given by V = π r3. • The new sphere has 4 times the volume of the original sphere. We will let the volume and radius of the larger sphere be VL and rL respectively. For the smaller sphere, the volume and radius will be VS and rS. Using this information, we can use VL = 4 ⋅ VS or

Then, we can simplify it to ﬁnd the relationship between their radii.

Apply the idea Set up the equation

Substitute the volumes

Divide both sides by

Take the cube root of both sides

Simplify

This means that the radius of the larger sphere is

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times radius of the smaller sphere.

Reﬂect and check To verify our answer, we can choose speciﬁc values for the radii of each sphere and see if the volume of the larger sphere is 4 times the volume of the smaller sphere. Let’s set the radius of the smaller sphere to 2 in, which would make the radius of the larger sphere be in. The volume of the smaller sphere is: VS =

π (2)3

= 33.51 in

Substitute r = 2 3

Simplify and round to two decimal places

The volume of the larger sphere is: Substitute r =

Simplify and round to two decimal places

Then divide the volume of the larger sphere by volume of the smaller sphere:

As shown, the volume of the larger sphere is 4 times the volume of the smaller sphere. This veriﬁes our answer.

Idea summary We can calculate the volume of a sphere using the formula:

the radius of the sphere

We can calculate the surface area of a sphere using the formula:

SA = 4π r2 r

the radius of the sphere

Practice What do you remember? 1

Match the surface area or volume calculations with their correct values. 4π (3)2 = 36π cm2

The surface area of a sphere with radius 3 cm.

The volume of a sphere with radius 3 cm.

π (6)3 = 288π cm3

The surface area of a sphere with radius 6 cm.

The volume of a sphere with radius 6 cm.

iii

4π (6)2 = 144π cm2

π (3)3 = 36π cm3

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For each sphere, calculate the volume and surface area: a

6 cm 9 in

The planet Earth has a radius of 6370 km. Find the surface area of Earth.

The diameter of a bowling ball is 8.6 in. Find the approximate volume of the ball, correct to two decimal places.

Calculate the exact volume of each sphere, given its radius, r, or diameter d. a

r = 8 ft

r = 9.24 cm

r = 6 in

d = 2.4 ft

d = 30 cm

d = 48.6 ft

Let’s practice 6

Find the volume of the solids correct to two decimal places: a

b 4 cm

12 cm

3 cm 9 cm

3 cm

11 cm

4 cm

8 cm

9 cm 15 cm

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Find the surface area of the following compound ﬁgures, correct to two decimal places: a

b 4 cm

11 cm

3 cm 5 cm

9 cm

A student was calculating the volume of the sphere shown and has these workings. Step 1

π ⋅ 93

Step 2

Step 3

≈ 3053.63 cm3

9 cm

Explain what the student’s mistake was, then ﬁnd the actual volume of the sphere.

The volume of a sphere is 972π yd3. Find the diameter of the sphere.

The surface area of a sphere is 81π ft2. Find the volume of the sphere.

The surface area of a sphere is 484π and the diameter is 2(x2 − 14). Solve for x.

The circumference of a golf ball is 4.2π cm.

Draw and label a model of the golf ball.

Find the surface area of the ball.

A spherical chocolate truﬄe has a volume of 0.5625π cm3 a

Draw and label a model of the spherical chocolate truﬄe.

Find the radius of a chocolate truﬄe.

If the truffle is dipped in sprinkles, find the surface area of the truffle that will be covered in sprinkles.

The solid shown is constructed by cutting out a quarter of a sphere from a cube. Find its surface area if the side length is 14.2 cm and the radius of the sphere is half the side length. 14.2 cm

Consider the following diagram constructed by removing a hemisphere from a cylinder. Report each calculated measurement to an appropriate degree of precision. a

Find the surface area of the hemispherical part.

Find the surface area of the cylindrical part, including the circular base.

Find the total surface area of the shape.

22 cm

15 cm

The planet Jupiter has a radius of 69 911 km, and planet Mars has a radius of 3390 km. State approximately how many times bigger the surface area of Jupiter is than Mars. 10.05 Spheres mathspace.co

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If the radius of a sphere tripled, what happens to its volume?

Consider a sphere with a radius of 4 cm and another sphere with a radius of 12 cm. Compare their surface areas and explain the difference.

A spherical water tank is being designed to ﬁt within a new building structure. The original design has a diameter of 8 meters. Due to space constraints, the ﬁnal design requires the diameter to be reduced by 25 percent. Calculate the following:

What is the diameter of the ﬁnal design of the tank?

By what percentage does the surface area decrease due to the design change?

By what percentage does the volume decrease due to the design change?

By what factor will the surface area increase if the radius of a sphere is doubled?

Let’s extend our thinking 21

Are these statements true or false? a

If two spheres have the same surface area, they also have the same volume.

If the volume of a sphere is 972π cubic units, the radius of the sphere is 9 units.

A sphere with radius r = 5 has a smaller volume than a sphere with r = 8.

For these two composite shapes, explain the similarities and differences in calculating the volumes of these shapes. Shape A Shape B

Tennis balls are packaged and sold in cylindrical containers, with three balls per container. The height of a standard tennis ball container is 21 cm. a

Find the volume of a tennis ball.

Find the volume of the container.

Find the percentage of the container that is occupied by the tennis balls.

Consider a hollow metal sphere used as a buoy, with an initial outer radius of 5 meters and a metal thickness of 0.1 meters. If the outer radius of the buoy is increased to 10 meters while keeping the metal thickness constant, calculate the following: a

The factor by which the outer surface area of the buoy increases.

The factor by which the volume of metal used in the buoy increases.

Note: The volume of metal is the volume of the outer sphere minus the volume of the inner sphere, which is the outer sphere minus the thickness.

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Dilating a solid to produce a similar solid will scale each length dimension, affecting both the surface area and the volume of the solid. ×2

Scaling a solid by a linear scale factor of d causes each side length to scale by a factor of d. This means that • the surface area will scale by a factor of d2 • the volume will scale by a factor of d3

×2

The cube in the image has been scaled by a factor of 2. Each face now has 4 times the area of the initial faces, and the overall volume could ﬁt 8 of the original cubes inside.

Keep in mind that scaling all of the length dimensions of a figure by a common scale factor results in a similar figure, but only scaling one dimension does not. Adding or subtracting the same number from all dimensions will also not create similar figures.

Example 1 Consider the square pyramid shown:

a Dilate each dimension of the square pyramid by a scale factor of d. What happens to the total surface area of the ﬁgure?

Create a strategy Use the given dimensions to write an equation for the surface area of the square pyramid. Recall that the surface area of a pyramid is calculated with the equation SA = B + nA, where B is the area of the base polygon, n is the number of sides of the base polygon, and A is the area of one triangular face. Then, dilate each dimension of the square pyramid by a scale factor of d, and write an equation to represent the surface area of the new solid. Finally, compare the two surface area equations.

Apply the idea For the surface area of the square pyramid, we have SA = a2 + 4 ⋅

as = a2 + 2as

By dilating the dimensions of the pyramid, we know that each dimension will be multiplied by a scale factor of d. For the new surface area, we have SAd = (ad)2 + 4 ⋅

ad ⋅ sd = a2 d2 + 2asd2 = d2 (a2 + 2as)

Comparing the equations shows us that the surface area of the dilated ﬁgure is d2 times bigger than the original ﬁgure. When each dimension of the pyramid is dilated by a scale factor of d, its surface area is dilated by a scale factor of d2.

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Reﬂect and check This ratio of area makes sense because we know that area of a 2D ﬁgure will increase by a scale factor of d2 when the dimensions increase by a scale factor of d.

b Dilate each dimension of the square pyramid by a scale factor of d. What happens to the total volume of the dilated ﬁgure?

Create a strategy Recall that the formula for calculating the volume of a pyramid is V =

Bh where B is the area of the base polygon

and h is the height of the pyramid. Write the equation to represent the volume of the given pyramid. Then, apply a scale factor of d to each dimension and write the equation to represent the volume again and compare the equations.

Apply the idea For the volume of the square pyramid, we have V=

a2 h

By dilating the dimensions of the pyramid, we know that each dimension will be multiplied by a scale factor of d. For the new volume, we have

When each dimension of the pyramid is dilated by a scale factor of d, its volume is dilated by a scale factor of d3.

Example 2 Lakendra likes baking miniature cakes. First, she makes a full size cake with volume 250 in3. Then she makes a miniature cake with volume 2 in3. What is the scale factor?

Create a strategy To calculate the scale factor, we need to evaluate the cube root of the ratio of the miniature cake to the full size cake.

Apply the idea

Simplify the fraction The miniature cake is

Evaluate the cube root

the size of the full-sized cake.

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Example 3 Consider these two cylinders:

10 ft 8 ft

30 ft

24 ft

a Are the cylinders similar? Explain.

Create a strategy To determine if the cylinders are similar, we will compare the ratios of corresponding dimensions (height and radius) between the two cylinders. If these ratios are equal, the cylinders are similar.

Apply the idea

Reﬂect and check

For the ﬁrst cylinder, the height is 30 ft and the radius is 10 ft. For the second cylinder, the height is 24 ft and the radius is 8 ft. The ratio of the heights is

This comparison reveals a fundamental property of similarity in geometry: objects are similar if their corresponding dimensions are in proportion. In this case, the constant ratio of the dimensions indicates that one of the cylinders was dilated, thus creating a similar cylinder.

30 : 24 or 5 : 4 The ratio of the radii is 10 : 8 or 5 : 4 Since both ratios are equal, the cylinders are similar.

b Find the ratio of the surface area of the smaller cylinder to the larger cylinder.

Create a strategy The surface area of a cylinder is given by 2π r (h + r), where r is the radius and h is the height. To ﬁnd the ratio of the surface areas, we calculate the surface area for each cylinder, then form a ratio of the smaller to the larger.

Apply the idea

Reﬂect and check

Surface area of the ﬁrst cylinder:

Comparing the height and radius from the smaller cylinder to the bigger cylinder, we found that they each have a ratio of 4 : 5. Squaring this ratio, we would get 42 : 52 or 16 : 25, which is the ratio of the surface areas. This shows that if the ratios between dimensions of similar ﬁgures is a : b, the ratio of their areas is a2 : b2.

2π (10) (30 + 10) = 800π Surface area of the second cylinder: 2π (8) (24 + 8) = 512π The ratio of the smaller to the larger surface area is 512π : 800π = 16 : 25

c Find the ratio of the volume of the smaller cylinder to the larger cylinder.

Create a strategy The volume of a cylinder is given by π r2 h. To ﬁnd the ratio of the volumes, we calculate the volume for each cylinder and then form a ratio of the smaller to the larger.

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Apply the idea

Reﬂect and check

Volume of the ﬁrst cylinder:

The cylinders’ height and radius each had a ratio of 4 : 5. Cubing this ratio, we get 43 : 53 or 64 : 125. This shows that if the ratio between the dimensions of similar ﬁgures is a : b, the ratio of their volumes is a3 : b3.

π (10) (30) = 3000π Volume of the second cylinder:

π (8)2 (24) = 1536π The ratio of the smaller to the larger volume is 1536π : 3000π = 64 : 125

d If the measurements of the smaller cylinder are doubled, calculate the new ratios for the surface area and volume of the cylinders.

Create a strategy Doubling the measurements of the smaller cylinder affects its surface area and volume. The new dimensions will be used to calculate the updated surface area and volume, and then compare them to the original larger cylinder.

Apply the idea After doubling, the smaller cylinder’s height becomes 48 ft and the radius 16 ft. New surface area: 2π (16) (48 + 16) = 2048π New volume: π (16)2 (48) = 12 288π New surface area ratio: 2048π : 800π = 64 : 25 New volume ratio: 12 288π : 3000π = 512 : 125

Reflect and check Let’s take a look at how the ratios of surface areas and volumes compare to the scale factor. We can use corresponding parts of the cylinders to determine if they are similar and find scale factor. • After doubling the smaller cylinder, the new radius is 16 ft and the new height is 48 ft • The original large cylinder has a radius of 10 ft and a height of 30 ft The ratio of the heights is 48 : 30 or 8 : 5 The ratio of the raii is 16 : 10 or 8 : 5 So, the cylinders are similar with a ratio between the dimension of 8 : 5. With similar figures, scaling the dimensions affects the area by the square of the scale factor. 82 : 52 or 64 : 25 We can confirm this was our new surface area ratio. Scaling the dimensions affects the volume by the cube of the scale factor. 83 : 53 or 512 : 125 We can confirm this was our new surface area ratio.

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Idea summary For two-dimensional ﬁgures, the scale factor can be summarized in the following table: Ratio (scale factor) of sides d

Ratio of perimeters d

Ratio of area d2

For three-dimensional ﬁgures, the scale factor can be summarized in the following table: Ratio (scale factor) of sides d

Ratio of surface areas d2

Ratio of volume d3

Practice What do you remember? 1

Match the following ﬁgures with their correct deﬁnition based on the concept of similarity. a

25 C B 38°

S 14 8 52° 38° U

52°

12 B C

5 3 10

6 582

iii Figures that have proportional side lengths, surface areas, and volumes

ii Figures where all corresponding sides are proportional and all corresponding angles are congruent

i Figures that have the same shape and their radii are proportional.

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iv Two ﬁgures that have exactly the same size and shape

Let’s practice

SOL

Two given ﬁgures are similar pentagons with a scale factor of 3.25. Describe the relationship between their perimeters.

The areas of two circles are in a ratio of 1 : 49. What is the ratio of their diameters? A

1:7

1 : 14

1 : 49

7:1

Describe the relationship between the volume of the two cylinders shown:

r Cylinder A

SOL

The surface areas of two cubes are in a ratio of 1 : 64. What is the ratio of the lengths of their edges? A

Cylinder B

1:4

1 : 16

1:8

1 : 64

The Sanford Seminole Aquatic Center has a swimming pool with dimensions proportionally similar to the Marcom Aquatic Center’s pool, which has a volume of 27 500 cubic yards. The Sanford pool measures 25 yards by 55 yards by 2.5 yards deep. Sanford Seminole Aquatic Center

Marcom Aquatic Center

2.5 yards deep 55 yards 25 yards

Compare the volumes of the two pools and state which has the greater volume.

Label the missing dimensions of the Marcom Aquatic Center’s pool. Round the dimensions into two decimal places.

Determine the exact scale factor between the dimensions of the Sanford pool and the Marcom pool, assuming the pools are similar ﬁgures.

The volume of a regulation adult beach volleyball is 106π in3. The volume of a regulation youth volleyball is

π in3. Find the approximate ratio of the radius of the youth volleyball to the adult beach volleyball. 15

584

Two cones are geometrically similar. The larger cone has a volume of 2400 cubic centimeters. If the diameter of the smaller cone is one-third the diameter of the larger cone, determine: a

The volume of the smaller cone.

The ratio of the volume of the larger cone to the volume of the smaller cone.

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SOL

Aristea crafted a geometrically similar model of the Eiffel Tower. The actual Eiffel Tower is approximately 300 meters tall. Eva’s model is 250 millimeters tall. If the base width of the actual Eiffel Tower is 125 meters, which of the following is closest to the width of the base of the model?

25 mm

104 mm

125 mm

250 mm

For each pair of right solids: i

Write ratios comparing each of the dimensions of the smaller solid to the larger solid.

Determine whether the solids are similar.

4 cm

8 cm

28 in

2 cm

35 in 6 cm

12 cm

8 in

12 in

4 cm 10 in

4 mm

d 14.4 ft

6.4 mm

25 mm

35 mm

15 in

24 ft

6 ft

10 ft

Consider these two rectangular prisms:

5.74 m

Are the prisms similar? Explain.

Find the ratio of the surface area of the smaller prism to the larger prism.

Find the ratio of the volume of the smaller prism to the larger prism.

If the measurements of the smaller prism are doubled, calculate new ratios for the surface area and volume of the prisms.

2.38 m 10.78 m

7.7 m 4.1 m 1.7 m

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Sophia was making a practice birthday cake for her son. The dimensions are 8 cm for the length, 3 cm for the width and 3 cm for the height. The actual cake will have dimensions 24 cm for the length, 9 cm for the width and 9 cm for the height. a

Find the ratio of the dimensions of the actual cake to that of the practice cake.

Find the ratio of the volume of the actual cake to that of the practice cake.

To make the actual cake, determine the number of times the quantities of the ingredients of the practice cake Sophia should use.

Would the practice cake and the actual cake be considered similar solids? Explain.

Consider three square pyramids, Pyramid P, Pyramid Q, and Pyramid R. Pyramid P

Pyramid Q

Pyramid R

4.5

18 3 6

Analyze the statements and determine whether they are true or false.

Pyramids P and Q are similar with a scale factor of 2.

The ratio of the volumes of Pyramid P to Pyramid Q is 8 : 1.

The ratio of the base perimeters of Pyramid P to Pyramid R is 1 : 2.

The ratio of the surface areas of Pyramid Q to Pyramid R is 1 : 16.

An independent artist makes hexagonal artwork. The most popular piece has a side length of 11 in and depth of 3.8 in. She ships them in protective containers. To ensure the artwork is not damaged the containers must be similar to the shape of the artwork but large enough to ﬁt packing material. She builds a container with a depth of 5.51 in and a side length of 16.5 in. Is the container safe to use for shipping? Explain.

Let’s extend our thinking 22

Consider the following cylinders: a

Show that dilating the dimensions of a cylinder by a scale factor of s will dilate the volume of the cylinder by s3.

Show that dilating the dimensions of a cylinder by a scale factor of s will dilate the surface area of the cylinder by s2.

r sh h

Cylinder A

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Cylinder B

Cylinder Y has radius r and height h. Cylinder Z has radius 5r and height

Write an equation that describes the relationship between the volume of Cylinder Y and the volume of Cylinder Z. 24

The Super Cereal Company sells their Frosty O’s cereal in a rectangular box with dimensions 12 in × 10 in × 2.5 in. In order to reduce their environmental impact, Super Cereal wants to reduce the surface area of their boxes while maintaining the proportions of the current packaging. a

List a scale factor that Super Cereal could use to reduce the total surface area of Frosty O’s boxes to somewhere between 75 in2 and 100 in2.

Give the dimensions for a Frosty O’s box that satisﬁes the required reduction in surface area.

A game company makes six-sided dice two sizes: a standard size with side lengths of s cm and a jumbo size with side lengths of j cm. a

A six-sided die is a cube. Explain why a cube is always similar to another cube.

Give an expression for the scale factor for the two dice in terms of s and j.

Give an expression for the volume scale factor for the two dice in terms of s and j.

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Connecting Algebra & Geometry 11 Through Coordinates Big ideas • The position in space of a geometric figure can be represented in the coordinate plane. Using coordinate algebra, the properties of that figure can be uncovered and applied to solve problems.

Chapter outline 11.01 11.02 11.03 11.04 11.05

Distance in the coordinate plane Midpoints Parallel and perpendicular lines Classify polygons in the coordinate plane Similarity and congruence in the coordinate plane

590 602 611 626 639

We can use the Pythagorean theorem to help us ﬁnd distances and lengths on the coordinate plane. y

The distance between two points can be calculated by creating a right triangle, where the line segment is the hypotenuse, then using the Pythagorean theorem.

(x2, y2)

Length of a: a = ∣x2 − x1∣ b

Length of b: b = ∣y2 − y1∣ Length of d using the Pythagorean theorem: d2 = a2 + b2

(x1, y1)

d2 = (x2 − x1)2 + ( y2 − y1)2

d= This equation is known as the distance formula.

distance between the two points

(x1, y1)

coordinates of the ﬁrst point

(x2, y2)

coordinates of the second point

Example 1 Find the distance between A (−1, 9) and B (−4, 1). Leave your answer in exact form.

Create a strategy We can either plot on the coordinate plane and use the Pythagorean theorem, or we can use the distance formula. Using the distance formula will likely be more eﬃcient.

Apply the idea We will use A (−1, 9) as (x1, y1) and B (−4, 1) as (x2, y2).

Distance formula

Substitute (x1, y1) = (−1, 9) and (x2, y2) = (−4, 1)

Evaluate the parentheses

Evaluate the squares

Evaluate the addition

There are no perfect squares as factors of 73, so there is no further simpliﬁcation to do. The distance between A and B is

units.

Reﬂect and check A good ﬁnal check is that our answer is positive as it is a length, so it must be a positive value. We could also have used A (−1, 9) as (x2, y2) and B (−4, 1) as (x1, y1) and gotten the same answer.

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Practice What do you remember? 1

Find the length of each vertical segment on the coordinate planes: a

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

y 9

A (−7, −1) and B (−7, −6) x y −9 −8 −7 −6 −5 −4 −3 −2 −1 −1 A −2

−3

−4 −5

−6

−7

2 1 1

−8

x 2

−9

A (4, −5) and B (4, 5)

A (−2, 2) and B (−2, −3)

y B 5 4 3 2 1

−4 −3 −2 −1 −1 −2 −3 −4 −5

A x 1

5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 B −4 −5

x 1 2 3 4 5

Find the length of each horizontal segment on the coordinate planes: a

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

A (−4, 5) and B (−7, 5)

9 8

7 3

B A

6 5 4 3 2

1 x

x 1

596

2 3 4 5 6 7 8 9

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−9 −8 −7 −6 −5 −4 −3 −2 −1

Let’s extend our thinking 21

A triangle is formed by three points: A (4, −3), B (1, 0) and C (7, 0). Is this triangle equilateral, isosceles or scalene?

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

The exact length of diagonals: i

The exact area of the rhombus.

The hexagon ABCDEF has all sides congruent and vertices at A (−1, 3), B (4, 3), C (7, −1), D (4, −5), E (−1, −5) and F (−4, −1). Determine:

The perimeter of ABCDEF.

The area of ABCDEF rounded to the nearest whole number.

Shantelle is putting up a fence around her chicken coop. She has posts at poles at A (0, 3), B (−1, −1), C (1, −3) and D (6, −2). Each unit on the grid represents 1 foot. She purchased 18 feet of fencing. a

Zenaida tells Shantelle that there is not going to be enough fencing. Explain how Zenaida has drawn this conclusion.

The posts at B (−1, −1), C (1, −3) and D (6, −2) have all be put into the ground with concrete, so can not be moved, but the post at point A has not been put in the ground yet. Determine if she can move the post at point A so that she has enough fencing. If she can, give an example of a new location for point A. If not, explain why not.

The area of the triangle with vertices A (3, 0), B (8, 2) and C (8, −6) has been calculated. a

If the triangle is translated up 3 units, determine if the area will change. Explain your answer.

If only the vertex A (3, 0) is translated up 3 units, determine if the area will change. Explain your answer.

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11.02 Midpoints After this lesson, you will be able to... • calculate the coordinates of the midpoint of a line segment. • solve for one endpoint of line segment given the midpoint and the other endpoint.

Midpoints We can divide a line segment into two congruent pieces by ﬁnding the midpoint. A

Midpoint

Exploration Three line segments are graphed. 11

Find the midpoint of each of the segments.

Describe your strategy for ﬁnding the midpoint of (2, 3) and (6, 11).

Think of and describe a second strategy for ﬁnding the midpoint of (2, 3) and (6, 11).

How are the two strategies related?

7 5 4 3

x 1

We can ﬁnd the midpoint of a line segment using the following formula, which shows that the coordinates of the midpoint are the average of the coordinates of the endpoints.

(xM, yM)

coordinates of the midpoint

(x1, y1)

coordinates of the ﬁrst endpoint

(x2, y2)

coordinates of the second endpoint

Another strategy for ﬁnding the midpoint is to use similar triangles. We can count squares or use absolute values to ﬁnd the lengths of the legs, the rise and run, of the right triangle with hypotenuse . Then, we can create a similar triangle that is

602

the size of the original triangle.

Mathspace Virginia SOL Geometry mathspace.co

12 10

The legs of the new triangle need to be half the size of the legs of the original triangle:

y Leg 1 = 7.5 Leg 2 = 8

8 6 4 2 −6 −4 −2 −2

M x 2

8 10

−4

New run =

⋅ 15 = 7.5

New rise =

⋅ 16 = 8

From point A (−5, 12) moving right 7.5 units and down 8 units, we get the midpoint: M = (−5 + 7.5, 12 − 8) = (2.5, 4)

Reﬂect and check We could have used the midpoint formula to ﬁnd the answer instead, but we can also use it to check our answer.

Formula for midpoint

Substitute coordinates of the endpoints

Evaluate the numerators

Evaluate the division

Example 2 The midpoint of A (1, 4) and B (a, b) is M (9, 7). a Find the value of a.

Create a strategy We can solve for the missing endpoint by substituting the midpoint and known endpoint into the formula for midpoint. To ﬁnd a, we only need to look at the x-coordinate.

Apply the idea Formula for midpoint Only the x-coordinate Substitute the x-coordinates Multiply both sides by 2 Subtract 1 from both sides Symmetric property of equality

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Create a strategy We can ﬁnd the midpoint algebraically and then check the map to see which city is closest and ensure we get a reasonable answer.

Apply the idea Finding the midpoint between: Dalhart, Texas(36.06°N, −102.52°W) and Liberal, Kansas (37.05°N, −100.92°W) Formula for midpoint

Substitute the coordinates

Evaluate the addition and division

The two cities that are close to this midpoint are: • Texhoma, (36.51°N, −101.78°W) • Goodwell, (36.60°N, −101.64°W) The x-coordinates are both 0.045° away, but for Texhoma, the y-coordinate is 0.6° away, while Goodwell is 0.8° away. This means that Texhoma is closer to the halfway mark between Dalhart and Liberal.

Reﬂect and check Using the map, we could estimate that it would have been one of Texhoma and Goodwell, but we need to work algebraically to conﬁrm which is closer.

Idea summary We can ﬁnd the midpoint of a line segment using the midpoint formula:

M (x1, y1) (x2, y2)

coordinates of the midpoint coordinates of the ﬁrst endpoint coordinates of the second endpoint

or using similar triangles. To create a triangle that is half the size of the original, we ﬁnd: ⋅ rise ⋅ run and add those values to the coordinates of the leftmost point.

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Practice What do you remember? 1

Indicate whether each statement is true or false. a

The midpoint of a segment is always equidistant from each endpoint of the segment.

The midpoint of a horizontal line segment has the same x-coordinate as its endpoints.

The y-coordinate of the midpoint of a horizontal line segment is the same as the y-coordinates of its endpoints.

The midpoint of a diameter is the center of a circle.

For the segment with endpoints, A (x1, y1 ) and B (x2, y2 ) state the formula for the midpoint, M, of

Verify if point M on the given coordinate plane is the midpoint of

y A (1, 3)

3 2 1 −1

−1

x 1

3 M

−2 −3 −4 −5

B (4, −5)

Consider the two points A and B with their midpoint M plotted on the coordinate plane. For each pair of coordinates, ﬁnd their midpoint M: a

A (5, −6) and B (5, 2)

y 4 3 2 1

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

1 2 3 4 5 6 7 8 9 10 M

A (6, 5) and B (14, 7) 10 9 8 7 6 5 4 3 2 1

B M A

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

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For each of the following, ﬁnd the coordinates of M, the midpoint of a

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

y B

10 9 8 7 6 5 4 3 2 1

A (−9, −8) and B (−2, −8) x −10−9 −8 −7 −6 −5 −4 −3 −2 −1 y −1 −2 −3 −4 −5 −6 −7 A B −8 −9 −10

1 2 3 4 5 6 7 8 9 10

A (0, 0) and B (0, 16)

A (0, 0) and B (−5, 0)

A (0, 0) and B (10, 8)

A (0, 0) and B (−12, 4)

Let’s practice 6

For each of the following, ﬁnd the coordinates of M, the midpoint of a

A (−8, −4) and B (2, 8)

A (−7, 3) and B (1, −7)

y B 8 7 6 5 4 3 2 1

−8−7−6−5−4−3−2 −1 −1

3 2 1

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

x 1 2 3 4

−2 −3 −4

: y x 1 2 3

For each pair of points A and B given: i

Find the coordinates of the midpoint M.

Plot

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

A (−9, 1) and B (5, −7)

A (−1, −4) and B (8, −7)

A (−7, 3) and B (−4, −7)

and B

and the point M on a coordinate plane.

and B

Find the midpoint of A (2m, 5n) and B (6m, n).

Given that M is the midpoint of A and B, ﬁnd the coordinates of A in the following given points:

608

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

B (16, 7) and M (10, 2)

B (8, 8) is M (−2, 0)

B (2, −9) and M (−1, −7)

Mathspace Virginia SOL Geometry mathspace.co

The graph shows the annual net profit (in millions) of a company over the last few years. It shows that its profit has been growing approximately linearly from $19 million in 2008 to $39 million in 2014. By finding the midpoint of the line segment, determine the company’s net profit in 2011. Profit (millions) 40 30 20 10 Year 2009

2011

2013

2015

Lines of latitude and longitude measure position on the Earth’s surface and work like coordinates. The first coordinate represents how far above or below the equator you are, and the second coordinate measures how far from Greenwich Mean Time you are. A plane starts its flight at (20°N, 62°E). It is bound for its destination at (52°N, 146°E). Assuming the plane flies directly to its destination, find its position half way through the flight.

Let’s extend our thinking 17

The points P, Q, R, S and T are collinear, and PQ = QR = RS = ST. Find the points Q, R and S given P (−4, −2) and T (−12, 10).

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

Find the value of p.

Find the value of q.

Point C (1, 1) is one quarter of the way from the point A (−3, −2) to point B. Find the coordinates of point B. Explain your process.

M is the midpoint of . The coordinates of point A are (xA, 12), the coordinates of point M are (xM, 4), and the coordinates of B are (6, −4). Given that AB = 20, ﬁnd the x-coordinates of A and M.

Tom, a treasure hunter, embarks on a journey to ﬁnd a treasure hidden on Mystic Island. Tom arrives on the island at (15 km N, 40 km W), and the treasure is rumored to be located at (30 km S, 20 km E). Deciding to take a shortcut in a straight line, determine Tom’s current position when he is halfway to the treasure, considering that South and West coordinates would be considered negative on the coordinate plane.

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

Refer to the given image. Using coordinate geometry: a

Prove that the midpoint of the hypotenuse of a right triangle is equidistant from the three vertices.

Prove that the length of the median to the hypotenuse of a right triangle is half the length of the hypotenuse.

y B (0, 2b)

M x O (0, 0)

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A (2a, 0)

Example 1 Prove that the lines are parallel.

3 2 1 −5 −4 −3 −2 −1 −1

x 1 2 3 4 5

−2 −3 −4

Create a strategy We will use the slopes of parallel lines theorem to prove the lines are parallel by showing the slopes are equal. If we ﬁnd nice points, we can count the rise and run to determine the slope.

Apply the idea y

Counting the rise and run, we see the top line has a slope of

4 3

Doing the same for the bottom line, the slope is also

Therefore, m1 = m2 = parallel lines theorem.

1 −5 −4 −3 −2 −1 −1

so these lines are parallel by the slopes of

1 2 3 4 5

−2 −3 −4

Example 2 The line AB passes through the points (−2, 9) and (3, −21). a Write the equation of the line.

Create a strategy To find the equation, we need to know the slope and the y-intercept. We will find the slope using m = will find the y-intercept using y = mx + b.

Apply the idea Slope formula

Substitute (x1, y1) and (x2, y2)

Evaluate the subtraction

Evaluate the division

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, and we

The slope of the line is m = −6. Now, we will use y = mx + b with the slope we found and one of the points. We can use either point because either will result in the same answer. y = mx + b

Slope-intercept form of a linear equation

9 = −6(−2) + b

Substitute m = −6 and (x1, y1)

9 = 12 + b

Evaluate the multiplication

−3 = b

Subtraction property of equality

b = −3

Reﬂexive property of equality

This means the y-intercept is at (0, −3). Substituting m = −6 and b = −3 into slope-intercept form of a linear equation, we ﬁnd the equation of the line to be y = −6x − 3.

Reﬂect and check The equation of the line in standard form is 6x + y = −3.

b Find the equation of the line that passes through (1, 5) and is parallel to the line AB.

Create a strategy Since this line is parallel to the line AB, we know that it will have the same slope as to ﬁnd the y-intercept of the parallel line.

which was −6. We only need

Apply the idea Just like we did in the previous part, we will substitute the slope and the x- and y-values of the point into y = mx + b. y = mx + b

Slope-intercept form of a linear equation

5 = −6(1) + b

Substitute m = −6 and the point (1, 5)

5 = −6 + b

Evaluate the multiplication

11 = b

Addition property of equality

b = 11

Reﬂexive property of equality

The equation of the parallel line is y = −6x + 11.

Reﬂect and check Using technology to graph the lines, we can see that they are parallel, and they pass through the speciﬁed points from parts (a) and (b). y 10 5 −3 −2 −1 −5

x 1

−10 −15 −20

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Example 3 Prove that two non-vertical lines are parallel if their slopes are the same.

Create a strategy Begin with the deﬁnition of parallel lines: “Parallel lines are lines in the same plane that do not intersect.”

Apply the idea We can prove this informally using deﬁnitions and diagrams. We will start with one line on the coordinate plane and try to ﬁnd a line which never intersects it. y

b a

For a line to never intersect the one that we started with, it must be a translation of our starting line without any reflection or rotation. A translation is defined as “a transformation in which every point in a figure is moved in the same direction and by the same distance.” This diagram shows the possible vertical translations.

Whether we translate the line up or down, the slope triangle is translated along with the line. Translations preserve distance, so the distances a and b are preserved.

b a

This means the rise and run remain the same, so the slope of the line remains the same.

This diagram shows the possible horizontal translations. Whether we translate the line left or right, the slope triangle is translated with the line.

Using the same reasoning as we did above, we know the slope of the line remains the same. a

Since translating a line does not change its slope, we know that two lines are parallel if and only if their slopes are the same.

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Reﬂect and check Another strategy for proving this is to make a system of two non-vertical linear equations which has no solution. We can start with:

If b1 = b2, then system will have a solution when x = 0, so we must have b1 ≠ b2. Now, we can consider what a solution to the equation will look like. If we solve the equation by letting the y-values be equal, we get: Equate y-values Add b1 and subtract m2 x from both sides

Factor out x

Divide both sides by m1 − m2 We can see that there will be no viable solutions to the system only when m1 = m2 because that would force us to divide by zero which is not possible. So, two lines are parallel if and only if their slopes are the same.

Idea summary If two lines are parallel, they have the same slope. If two lines have the same slope, they are parallel.

Perpendicular lines Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 11.03 to answer these questions. 1.

What do you notice about the slopes of the two perpendicular lines?

Do you think this is true for all non-vertical perpendicular lines?

How could we prove two lines are perpendicular?

Perpendicular lines have slopes with opposite signs and they are reciprocals of one another. A vertical and horizontal line are perpendicular. Slopes of perpendicular lines theorem

Two non-vertical lines are perpendicular if and only if the product of their slopes is −1.

1 x m1 m2 = −1

We may also call the slopes of perpendicular lines opposite reciprocals which refers to them being reciprocals with opposite signs. We use this theorem to prove lines are perpendicular and to ﬁnd the equations of perpendicular lines. 11.03 Parallel and perpendicular lines mathspace.co

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Example 4 Consider the lines on the given coordinate plane.

y 5 c 4 3 2 b 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 f −5 a

d x 2 3 4 5

a Identify all pairs of parallel lines.

Create a strategy We are looking for pairs of lines with the same slope. We can count the rise and run to determine the slope.

Apply the idea Line a and line d both have a slope of ma = 3 = md, so they are parallel.

Reﬂect and check Lines b and c have similar slopes, but they are not equal. mb = − mc = −

b Identify all pairs of perpendicular lines.

Create a strategy We are looking for pairs of lines with slopes that are opposite reciprocals, so their slopes should have a product of −1. Looking at the coordinate plane, the pairs that look like they might be perpendicular are a and b, a and c, and e and f.

Apply the idea From the coordinate plane, we can ﬁnd that:

Line a and line b have slopes with a product of 3

= −1, so they are perpendicular.

In part (a), we determined that lines a and d are parallel, so any line that is perpendicular to a is also perpendicular to d by the perpendicular transversal theorem. In particular, that means that line b is perpendicular to line d. Line e and line f have slopes with a product of

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(−2) = −1, so they are perpendicular.

Example 6 A mirror is placed along the x-axis. A laser beam is projected along the line y = −x + 4 which reﬂects off the mirror.

y 7 6 5 4 3 2 1 −1

a A normal is a line which is perpendicular to the surface of the mirror at the point of reﬂection. Find the equation of the normal.

x 1

−1

Create a strategy

Apply the idea

We can do a quick sketch of the normal to help:

Since the mirror is a horizontal line, the normal must be a vertical line if it is to be perpendicular. This means it will be of the form x = a.

Since it goes through the point where the laser hits the mirror, (4, 0), the equation of the normal will be x = 4.

6 Normal

5 4 3

Laser

2 1 −1

−1

x 1

2 3 4 5 6 7 Surface of the mirror

b The angles that the laser and its reflection make with the normal will be congruent. If the angle between the laser beam and the normal is 45°, find the equation of the path of the reflection.

Create a strategy Since the angles are congruent, know that the angle formed between the normal and the reﬂection will also be 45°.

We can label this on our diagram:

Normal

4 3 2

45°

1 −1

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−1

45° x

2 3

To ﬁnd a line that is at a right angle to our starting line, we can just rotate it by 90°. y b a

Notice that rotating the line by 90° also rotated the right triangle by 90°. This shows us that the values of the rise and run have switched, so the slope of the new line will be the reciprocal of the starting line. We can also see that the slope has changed sign, since rotating any line by 90° will change its slope from positive to negative or vice versa. The sign of the slope of the new line will be the negative of the starting line. Therefore, two lines are perpendicular if their slopes are opposite reciprocals, which is the same as their product being −1.

Reﬂect and check To show that two non-vertical lines whose slopes have a product of −1 are perpendicular, we would need to begin with a line that has a slope of

and a line that has a slope of

and show they meet at a 90° angle.

Idea summary The slopes of perpendicular lines are reciprocals with opposite signs. When multiplied together, they have a product of −1.

Practice What do you remember? 1

Fill in the blanks for the following two theorems: a b

Two non-vertical lines are parallel if and only if their slopes are ⬚. Any two ⬚ lines are parallel.

Two non-vertical lines are perpendicular if and only if the product of their slopes is ⬚. ⬚ and ⬚ lines are perpendicular.

Give the graphs of the lines: • y=x+5

• y=x+2

• y=x−3

Determine whether the following statements are true or false:

They all have the same slope.

They all have positive y-intercepts.

y=x+5

They all have positive x-intercepts.

−5

They are all parallel to the line y = x.

They all have slope equal to 1.

y=x+2 −5 y=x−3

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x 5

Given the slope of a line is 3, which of the following is the slope of a line that is parallel to this line? A

−3

4 Given the slope of a line is −2, which of the following is the slope of a line that is perpendicular to this line? A 5

–

−2

Identify the slope of each line. y 8

4 2

−8 −6 −4 −2 −2

6 8

−4 C −6

−8

Let’s practice 6

What is the slope of the line that is parallel to x = −2. Is this line a vertical or horizontal line?

What is the slope of the line that is parallel to y = −4. Is this line a vertical or horizontal line?

The segment joining point A (1, 10) and B (4, 9) is parallel to the segment joining point C (−2, 5) and D (−11, y). a

Find the slope of

Find the value of y.

Consider the points A (5, 6) and B (−13, −22). .

Find the slope of

Find the midpoint of

Find the slope of the perpendicular bisector of

. .

Using the given graph, determine if:

a and b are perpendicular

b and e are perpendicular

a and d are parallel

b and f are parallel

e a

−4 −3 −2 −1

−1

−2 −3 −4

d x 4

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Using the given graph, determine if: a

and

are perpendicular.

and

are perpendicular.

and

are parallel.

and

are parallel.

−8

−7

G −6 −5

C −4

−3

6 5 4 3 2 1

−2

−1 −1 N −2 E −3 D −4 −5 −6

B L

I 1

x 4

5 F

In the town of Gridland, the streets form a perfect grid pattern. There are four important landmarks identiﬁed by their coordinates on the map: the Diner at (2, 3), the Library at (6, 3), the Pet Store at (2, 7), and the Gym at (6, 7). a

Which pair of routes are parallel? Select all that apply. A The Diner to the Library and the Pet Store to the Gym. B The Diner to the Pet Store and the Library to the Diner. C The Diner to the Gym and the Library to the Pet Store. D None of the routes are parallel to each other.

Which pair of routes are perpendicular? Select all that apply. A The Library to the Gym and the Diner to the Library. B The Diner to the Gym and the Pet Store to the Library. C The Diner to the Library and the Diner to the Pet Store. D The Library to the Gym and the Diner to the Pet Store.

In a rectangular garden, four types of plants are planted at speciﬁc coordinate points: Roses at (3, 4), Tulips at (7, 7), Daisies at (5, 10), and Sunﬂowers at (1, 7). a

Which pair of routes are parallel? Select all that apply. A Roses to Tulips and Daisies to Sunflowers. B Roses to Daisies and Tulips to Sunflowers. C Roses to Sunflowers and Tulips to Daisies. D None of the paths are parallel to each other.

Which pair of routes are perpendicular? Select all that apply. A Roses to Tulips and Daisies to Sunflowers. B Roses to Daisies and Tulips to Sunflowers. C Roses to Sunflowers and Tulips to Daisies. D None of the paths are perpendicular to each other.

The vertices of a right triangle are at A (−8, 1), B (−1, 2), C (−2, −1). Find the sides that are perpendicular.

Trapezoid QRST has one pair of parallel sides and a right angle. The vertices are at Q (−1, −1), R (2, 0), S (3, −3), and T (−3, −5).

622

Find the sides that are parallel.

Find the sides that are perpendicular.

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For each polygon:

Determine if the two speciﬁed line segments are perpendicular.

Justify your answer.

and

Lines

and

y K

−4 −3 −2 −1

−4 −3 −2 −1 −1 J −2

Lines

and

Lines

−3

and y 4

2 3 4 5 6 7 8 9 F

3 2

−2

1 x

−3

−9 −8 −7 −6 −5 −4 −3 −2 −1 −1

−4

−2

−5

−1

x 1

−4

y 1

−2

and

Lines

−3

Lines

−4

and

y x

2 3 4 5 6 7 8 9

−1

−3

6 5

P I

−4

8 M

−2

−5

4 3 2

−6

−9 −8 −7 −6 −5 −4 −3 −2 −1

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For each plotted quadrilaterals:

Determine if the two speciﬁed line segments are parallel.

Justify your answer.

Lines

and

Lines

y 6 A

−1

−4 −3 −2 −1

Lines

and

2 3 4 5 6 7 8 9

H x

and

x 1

−2

Lines

and

y x

−9 −8 −7 −6 −5 −4 −3 −2 −1

1 2 3 4

−2 I

2 L

−3

−5 −6

−3 −2 −1 −1

−4

−1

−2 −3

−7

Given the plotted shape: 9 A H G

7 6 5 4

3 2 E F −9 −8 −7 −6 −5 −4 −3 −2 −1

C 1

x 9

Determine whether each pair of segments are parallel, perpendicular, or neither. Justify your answer.

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and

Let’s extend our thinking 18

A builder is constructing a new house in a region with heavy snowfall and has planned the side view of the house on a coordinate grid. The design has two sloped roofs that meet in the middle that must be perpendicular to each other. Does the side view shown match the builder’s speciﬁcation? Justify your answer. 18 y 16 D 14 12 10 8 6 4 2

B C −18 −16 −14 −12 −10 −8 −6 −4 −2−2 −4 −6 −8 −10 −12 −14 −16 A −18

F 2

x 16

Given the lines: • p passing through the points A (6, 10) and B (−9, 5) • q passing through the points C (−3, −2) and D (−6, 7) Show that

is perpendicular to

Write the equations of three lines that will form a rectangle with the y = 3x + 5.

Consider the line x = 0.

Determine the equation of a line that is perpendicular to x = 0.

Determine if there is more than one possible answer for part (a). Explain.

The points A (xa, ya ) and B (xb, yb ) form Determine if

and

. The points C (xa + 3, ya − 4) and D (xb + 3, yb − 4) form

are parallel or not. Justify your answer.

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Apply the idea Find AB: Distance formula

Substitute the coordinates

Evaluate the parentheses

Evaluate the squares

Evaluate the addition

Find BC:

Distance formula

Substitute the coordinates

Evaluate the parentheses

Evaluate the squares

Evaluate the addition

Find AC: Distance formula

Substitute the coordinates

Evaluate the parentheses

Evaluate the squares

Evaluate the square root

None of the sides are congruent, so it is a scalene triangle. Find the slope of

Slope formula

Substitute the coordinates

Evaluate the subtraction

Evaluate the division

Find the slope of

Slope formula

Substitute the coordinates

Evaluate the subtraction

Simplify the fraction

These two sides are perpendicular because their slopes are opposite reciprocals, so they form a right angle. △ABC is a right-scalene triangle.

Reﬂect and check If and were not perpendicular, then we would also need to ﬁnd the slope of two sides are perpendicular.

to check whether or not any

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11.05 Similarity and congruence in the coordinate plane After this lesson, you will be able to... • use coordinates to prove triangles are congruent or similar.

Similarity and congruence in the coordinate plane Similarity and congruency theorems are useful in proving whether two triangles are similar or similar and congruent. Remember the similarity theorems for triangles: • Side-Side-Side (SSS∼) • Side-Angle-Side (SAS∼) • Angle-Angle (AA) Remember the congruency theorems for triangles: • SSS • SAS • Angle-Side-Angle (ASA) • Angle-Angle-Side (AAS) • Hypotenuse-Leg (HL) We may now see triangles on a coordinate grid without given side lengths and angle measures, where tools such as the distance formula and the slope formula help prove similarity or congruence between them.

d the distance between two points x1, x2 the x-coordinates of (x1, y1) and (x2, y2) y1, y2 the y-coordinates of (x1, y1) and (x2, y2)

m x 1, x 2 y1, y2

the slope between two points the x-coordinates of (x1, y1) and (x2, y2) the y-coordinates of (x1, y1) and (x2, y2)

To write a direct proof of a theorem using the coordinate plane, we can follow the steps below: 1. Represent the given information with a labeled diagram on the coordinate plane. 2. Use the coordinates of the key points to determine other properties of the diagram. 3. Use formulas like the distance and slope formulas on the coordinate grid to prove the theorem, depending on what’s required (e.g., side lengths, right angles). An indirect proof for triangle congruence might involve assuming two triangles are not congruent despite having three pairs of congruent sides, and then showing this assumption leads to a contradiction.

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Apply the idea Calculate the lengths of the sides for triangle ABC and triangle PQR, then compare the ratios. For △ABC: AB =

BC =

CA =

For △PQR: PQ =

= 10

QR =

RP =

= 10

The ratios of the corresponding sides are equal: =

= ,

△ABC and △PQR are similar by the SSS similarity criterion.

Reﬂect and check When showing two non-right triangles are or are not similar, the Side-Side-Side (SSS) similarity criterion is usually easiest. However, the AA similarity theorem works as well. This theorem requires two pairs of corresponding angles to be congruent, which we can do by identifying transformations and calculating slopes. For this example, triangle PQR has been dilated by a factor of 2 where the center of dilation is the origin. This means the orientation of the triangles are the same, so the slopes of corresponding sides will be parallel. Slope of Slope of Slope of Slope of Slope of Slope of Because the corresponding sides of each triangle are parallel, the angles between them are congruent.

Idea summary To write a direct proof of a theorem using the coordinate plane, we can follow the steps below: 1. Represent the given information with a labeled diagram on the coordinate plane. 2. Use the coordinates of the key points to determine other properties of the diagram. 3. Use formulas like the distance and slope formulas on the coordinate grid to prove the theorem, depending on what’s required (e.g., side lengths, right angles).

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12 Circles Big ideas • All circles are similar. Congruence and similarity criteria can be applied to prove relationships in circles and solve problems. • The relationships between the parts of a circle can be used to find the measures of other parts. • The position in space of a geometric figure can be represented in the coordinate plane. Using coordinate algebra, the properties of that figure can be uncovered and applied to solve problems.

Chapter outline 12.01 12.02 12.03 12.04

Arc length and sector area Measures of arcs Circles in the coordinate plane Equations of circles

650 664 675 686

Apply the idea

Reﬂect and check

The measure of the central angle is 160°.

The ratio of the central angle to the total number of degrees in a circle is proportional to the ratios of other parts of a circle. For example, the ratio of the arc length to circumference and the sector area to the area of the circle are both proportional to the ratio of the central angle to 360°.

Reduce by dividing by 40 The simplified ratio is . b Find the length of the solid arc.

Create a strategy

Apply the idea

The full circle is 360° and has a length of 144 cm. If we can determine what proportion 160° is of the circle, then we can use that to determine the length of the arc of that sector as a proportion of the circumference.

The angle in the middle is

of the entire circle,

so the arc length of the sector is going to be circumference.

of the

The arc length of the sector is 64 cm.

Example 2 The sector shown has a radius of 10 cm and an arc length of 30 cm. Find the measure of the central angle θ in degrees. Round your answer to two decimal places.

30 cm

θ 10 cm

Create a strategy Use the formula for the arc length of a sector: s =

⋅ 2π r.

Apply the idea We were given s = 30 and r = 10. Formula for arc length

Substitute known values

Evaluate the multiplication and division

Multiply by 18

Divide by π

The central angle of the sector measures approximately 171.89°.

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Example 3 For the sector shown, AB = 5 inches:

A 68°

Find the area of the sector.

Create a strategy We can ﬁnd the area of the sector using a process similar to ﬁnding the arc length of the sector. First, we determine the central angle’s proportion of the full circle. The sector’s area is the same proportion of the full circle’s area.

Apply the idea Recall the area of a circle is A = π r2. Let’s begin by ﬁnding the area of the full circle with radius 15 in: Acircle = π (5)2 = 25π in2 Next, we need to determine the central angle’s proportion of the full circle:

This means the area of the sector will be

The area of the sector is

ths of the area of the circle.

in2.

Reﬂect and check Since there was no instruction to approximate the solution by rounding the answer, we should keep the answer as an exact value.

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Degrees and radians We have established that sectors with the same central angle will have an arc length that is proportional to the radius. We deﬁne this constant of proportionality as the radian measure of the central angle. Radian measure (of a central angle) The ratio of the arc length divided by the radius, θ = , where θ is the central angle in radians, s is the arc length, and r is the radius of the circle. From the time of the ancient Babylonians, it has been the practice to divide circles into 360 small arcs. The central angle of any one of those arcs is called one degree. In effect, an arc of the circle is used as a measure of its central angle. We have deﬁned the central angle in radians as the ratio of the arc length divided by the radius. So, the central angle of an arc whose length is equal to the radius is 1 radian.

r 1 rad r

Radians are an alternate way to describe angles and are the international standard unit for measuring angles. Because angles in radian measure are in essence just fractions of the circle, they do not require a unit.

Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 12.01 to answer these questions. 1.

What do you notice about a sector with a radian measure of 1?

What is the measure of 1 radian in degrees?

What is the measure of the central angle of a semicircle in degrees and radians?

What is the measure of the central angle of a full circle in degrees and radians?

How can we use this information to convert from radians to degrees and from degrees to radians?

As we previously explored in this lesson, the arc length of a sector is part of the circumference of a full circle. We just learned that a radian is deﬁned as the ratio of the arc length and the radius. This means the central angle in radians of a full circle is

θ=

= 2π

When we compare this to degrees, we see that: 2π rad = 360°

π rad = 180°

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Example 5 Convert the following degrees to radians. a 90°

Create a strategy We can use proportions to ﬁnd this result since we know 180° = π radians. First, we will need to ﬁnd what proportion 90° is of 180°.

Apply the idea 90° is

Reﬂect and check

of 180°, so in radians, the measure will be

of π.

We could also have written the proportion as

⋅π= This shows 90° =

radians.

Solving for x, we get

Notice that this is the same as multiplying the angle by . =

90 ⋅ b 216°

Create a strategy As we found in the reﬂection of the previous part, we can multiply an angle in degrees by

to convert it to radians.

Apply the idea

Reﬂect and check

216 ⋅

Using proportions to check our answer:

This shows 216° =

radians. 216° is of the whole circle. In radians, the full rotation of the circle is 2π. ⋅ 2π = of the circle in radians is

Example 6 Convert the following radians to degrees. a 1.8 rad Round to one decimal place.

Create a strategy Since we know π = 180°, we will set up an equivalent proportion between radians and degrees and solve for the angle in degrees.

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Apply the idea

Reﬂect and check When we set up an equivalent proportion, we converted from radians to degrees by multiplying the radian measure by

Using a calculator to evaluate, then rounding to one decimal place, we ﬁnd 1.8 rad ≈ 103.1°.

rad

Create a strategy As we found in the reﬂection of the previous part, we can multiply an angle in radians by

Apply the idea ⋅

Reﬂect and check We can check this answer by ﬁnding what part of the

= 120°

This shows

to convert it to degrees.

rad = 120°.

circle radians is, then determine if 120° is the same fraction of the circle in degrees.

Both are

of the full circle, so this shows

rad = 120°.

Example 7 The sector of a circle with radius 7 is formed from an angle of size

a Find the exact length of the arc.

Create a strategy We deﬁned the measure of a radian as the ratio of the arc length to the radius, θ = . We can use this formula to solve for the arc length.

Apply the idea We were given θ =

Reﬂect and check and r = 7. Deﬁnition of a radian

Substitute known values

Multiply both sides by 7

The arc length is

By the deﬁnition of a radian, we can ﬁnd the length of any arc when given the radius and central angle in radians by s = r ⋅ θ.

units.

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b Find the area of the sector.

Create a strategy We derived the sector area formula in degrees by taking the amount of the circle covered by the sector and multiplying by the total area of the circle. We can do the same thing in radians since 360° = 2π rad. Sector area formula in degrees

Sector area formula in radians

Simplify since

Apply the idea Using the formula for the area of a sector in radians with θ =

and r = 7:

Sector area formula in radians

Substitute known values

Evaluate the numerator

Evaluate the division

Evaluate the multiplication

The area of the sector is

units2.

Example 8 The sector has an area of 3.6 m2 and a radius of 2 m. Find the value of θ in the sector shown.

Create a strategy We are given the area and the radius of the sector, so we can use the formula A =

Apply the idea Sector area formula in radians

Substitute A = 3.6 and r = 2

Multiply by 2 and evaluate the exponent

Divide by 4

Therefore, the central angle of the sector is 1.8 radians.

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to solve for θ.

Find the area of each circle, rounding to the nearest hundredth: a

5 cm

Consider a circle with a central angle of measure m°. Fill in each statement: times the ⬚ of the circle.

The arc that the angle subtends has a length of

The sector that subtends the angle has an area of

The ratio of the arc length to circumference is the same as the ratio of ⬚ area to ⬚ area.

d 5

0.4 in

times the ⬚ of the circle.

m°

Without any units, the arc length is ⬚ times the sector area.

Consider each of the following arcs. What fraction of the entire circumference is each arc?

A B

105°

30° 20° 70°

The area of this sector is what fraction of the full circle’s area?

220° A C

660

Convert the following to radians: a

360°

180°

112°

29°

161.17°

330°

120°

140°

Convert the following to degrees: a

Mathspace Virginia SOL Geometry mathspace.co

d 3.4 radians

4.2 radians

C D

Let’s practice 9

The arc length is the distance from one point to another on a circle. Suppose that ∠B subtends

as ∠B changes.

. Describe what happens to the length of

Find the length of the given arc in terms of π :

π C

N M

G 12 cm

6.5 mi

300°

55° K

H L

Describe how the area of a sector changes relative to its central angle while keeping the radius constant.

For each sector, ﬁnd the area in terms of π : a

b 270°

4 in

10 mm 90°

d 135°

180° 5m

12 yd

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For each sector, ﬁnd the area and arc length. Round your answers to the nearest hundredth. a

9 cm 45°

70.6 mm

d 240°

30.5 yd 17.1 ft

SOL

Given: Circle Q Which is closest to the area of the shaded sector of circle Q? A

20 cm2

58 cm

79 cm2

154 cm2

If a circular plate has a radius of 10 inches, ﬁnd the length of an arc corresponding to a central angle of radians.

An engineer is designing a fan blade that is part of a circular section with a radius of 15 cm. If the blade covers a central angle of

radians, what is the area of the blade?

The length of an arc that is formed by a 29° angle is 10 cm. Find the approximate radius of the circle.

Consider a circle with arc length of 36 cm and a corresponding central angle of 105°. Which of the following is closest to the area of the sector created by this arc and central angle? A

SOL

135°

7 cm

123 cm2

39 cm2

1212 cm2

353 cm2

An arc on the surface of Saturn is formed by a 25° angle at the center of the planet. If the radius of Saturn is 58 232 km, ﬁnd the approximate length of the arc correct to one decimal place.

An architect used this diagram to design a curved balcony. She drew a circle with a radius of 45 feet and a central angle of 80 degrees to determine the length of railing needed for the balcony.

Railing

Which is closest to the length of railing needed for the curved section of the balcony?

Balcony

662

31 ft

63 ft

141 ft

251 ft

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80° 45 ft

A beam of light from a lighthouse can be seen from as far as 16 nautical miles away. If the light spreads at an angle of 20°, what is the area of the region where the light is visible? Give your answer to the nearest nautical mile.

20° 16 NM

The length of

is 10.1 cm.

Find the measure of the central angle, ∠BAC.

Find the area of the shaded sector in circle A.

B 5 cm C

The sector shaded in circle M has an area of 376.3 cm. a

Find the measure of central angle that forms major arc

Find the length of the arc

. M

14 cm

Consider the circle with the given sector area. a

Find the diameter of the circle.

Find the area of the circle.

A = 15 cm2 M L

110°

Let’s extend our thinking 25

Consider circles A and B. The sector area on both circles is the same. Circle A has a radius with twice the length of circle B. What must be true about the central angles? 2r

B r

Find the perimeter of each semicircle given their diameters: a

D = 8 in

D = 17.23 cm

Mikey is planning for the city park and wants to make sure the area is well-lit. A single lamp projects a beam of light with a central angle of 110° arc. a

How many lamps facing outward from the center of a circle would be needed to form a full circle of light at the center of a park?

If the light from the lamp can be seen from 35 yards away, what would be the area of the overlap of these beams?

Stormie says that in a sector with radius r, if you double the angle θ and keep the radius the same, you will double the area. Is she correct? Explain how you know. 12.01 Arc length and sector area mathspace.co

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12.02 Measures of arcs After this lesson, you will be able to... • solve for arc measures and angles in a circle formed by central angles. • solve for arc measures and angles in a circle involving inscribed angles. • use the arc length or area of a sector to solve for unknown measurements of the circle.

Arcs of central angles Central angle An angle that has its vertex at the center of a circle with radii as its sides.

Arcs of a circle can be further classiﬁed as follows: Semicircle An arc of a circle whose endpoints lie on a diameter. O

Minor arc An arc smaller than a semicircle.

Major arc An arc larger than a semicircle.

The notation we use to denote a minor arc with endpoints at A and B is

To distinguish between a major arc and a minor arc, we use a third point that lies between the endpoints. If the . endpoints of an arc are A and B and point P lies between them on the major arc, we use the notation Measure of an arc The measure of an arc is equal to the measure of its central angle. The measure of an arc is different from the length of the arc. While arc length refers to the distance from one endpoint of the arc to the next, the measure of an arc refers to the measure of its central angle. We always use the notation when talking about arc measure and when talking about arc length.

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Adjacent arc measures can be combined by the following postulate: Arc addition postulate

The measure of an arc formed by two adjacent arcs is the sum of the measures of the two arcs.

C D

This theorem is helpful to connect the central angle with the minor arc. Congruent central angles theorem

In the same circle, or in congruent circles, two minor arcs are congruent if and only if their corresponding central angles are congruent.

B C

Example 1 Consider the given diagram: J Q M

115° 9°

a Find m∠JQM given

= 166°.

Create a strategy

Apply the idea

The measure of ∠JQM is equivalent to

b Find

Since ∠JQM is a central angle, we know that the measure of corresponding arc, , is equivalent to m∠JQM. So, m∠JQM = 166°.

Create a strategy The measure of arc JL is made up of arc JK and arc KL, so we will need to use the arc addition postulate to add the arc measures. The measure of arc JK is the measure of ∠JQK and the measure of arc KL is the measure of ∠KQL.

Apply the idea =

Reﬂect and check +

Arc addition postulate

= 115° + 9°

Substitute known values

= 124°

Evaluate the addition

The corresponding major arc of arc JL is arc JML. A minor arc and its corresponding major arc add to 360°, so we know that = 360 − 124 = 236°.

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Example 2 Let m∠EDH = (6x − 5)° and

= (5x + 20)°.

D G H

a Solve for x.

Create a strategy Since ∠EDH is a central angle, we know its measure is equivalent to the measure of its corresponding arc, We can also see from image that is congruent to .

Apply the idea First, set up the equation, knowing m∠EDH will be equal to

. Then solve for x.

6x − 5 = 5x + 20 Set up equation x − 5 = 20 Subtract 5x from both sides x = 25 Add 5 to both sides

Reflect and check We can confirm our solution by substituting x = 25 back into the original equation. 6x − 5 = 5x + 20 Write original equation 6 (25) − 5 = 5 (25) + 20 Substitute x = 25 150 − 5 = 125 + 20 Evaluate the multiplication 145 = 145 Simplify Substituting our x-value has resulted in a true statement, we have confirmed our solution. b Find m∠EDH.

Create a strategy Since we have an expression representing m∠EDH, we can substitute the value of x we found in part (a) to ﬁnd the measure of the central angle.

Apply the idea m∠EDH = 6x – 5

Given

= 6 (25) – 5

Substitute x = 25

= 150 – 5

Evaluate the multiplication

= 145 Evaluate the subtraction The m∠EDH = 145°

Reﬂect and check Since the measure of the central angle m∠EDH = 145°, this means that its corresponding arc = 145°. From the image we can also tell that the central angle m∠FDG = 145° and its corresponding arc = 145°.

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Example 3 The area of a sector of a circle with radius 8 cm is

π cm2.

a Find the measure of the central angle.

Create a strategy We know the formula for the arc length of a sector to be A =

⋅ π r2. We can substitute the known values of sector

area and radius to solve for the central angle that we will call θ.

Apply the idea Formula for sector area Substitute r = 8 cm and A =

Evaluate the exponent

Divide both sides by 64π

Divide out common factor of π

Simplify the fraction

Multiply both sides by 360

We have just found that the central angle, θ = 30°.

b Find the measure of the corresponding arc.

Create a strategy The measure of the corresponding arc is equal to the measure of its central angle.

Apply the idea We found in part (a) that the measure of the central angle is 30°, which means its corresponding arc also measures 30°.

Idea summary Measure of arc = Measure of its central angle By the congruent central angles theorem, two minor arcs are congruent if and only if their corresponding central angles are congruent. We can ﬁnd the sum of adjacent arcs using the arc addition postulate. By this, we know the minor arc of a central angle and its corresponding major arc will sum to 360° or 2π radians.

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Arcs of inscribed angles Angles formed by chords of a circle are known as inscribed angles. Inscribed angle in a circle An angle whose vertex is a point on the circle and whose sides contain chords of the circle.

Interactive exploration Explore online to answer the questions

mathspace.co Use the interactive exploration in 12.02 to answer these questions. 1.

What relationship do you notice between the central angle and the inscribed angle?

How does this relationship translate to the inscribed angle and the arc it intercepts?

The following theorems relate to angles inscribed in circles: Inscribed angle theorem

Congruent inscribed angle theorem

If an angle is inscribed in a circle, then its measure is half the measure of its intercepted arc.

If two inscribed angles of a circle intercept the same arc, then the angles are congruent. C

θ°

2θ ° B

Example 4 Solve for x.

(−10 + x)°

(4 + x)°

Create a strategy We can use the inscribed angle theorem to write an equation and then solve for x.

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Apply the idea 2(m∠ATB) = 2 (−10 + x) = 4 + x

Substitution

−20 + 2x = 4 + x

Distribute the 2

2x = 24 + x

Add 20 to both sides

x = 24

Subtract x from both sides

x = 24

Reﬂect and check The value of x is 24, which means the inscribed angle m∠ATB = 14° and the intercepted arc

= 28°.

Example 5 Given m∠CEB = 4x + 11 and m∠CDB = 12x − 5. Find m∠CDB.

B D

Create a strategy By the congruent inscribed angle theorem, we know m∠CEB = m∠CDB. We want to write an equation relating the two angles and then solve for x.

Apply the idea 4x + 11 = 12x – 5

Congruent inscribed angle theorem

4x + 16 = 12x

Add 5 to both sides

16 = 8x

Subtract 4x from both sides

2=x

Divide both sides by 8

x=2

Symmetric property of equality

We have established x = 2, so we can substitute that into the equation for ∠CDB. Substituting x = 2 into 12x − 5 we get 12 (2) − 5 = 19. Therefore, m∠CDB = 19°.

Reﬂect and check Note that since m∠CDB = m∠CEB, we could have used the expression for m∠CEB to calculate the size of the angle instead: m∠CEB = 4x + 11

Given

= 4 (2) + 11

Substitute x = 2

= 19

Simplify the expression

which is the same result.

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Idea summary Measure of inscribed angle =

⋅ Measure of the intercepted arc

If two inscribed angles intercept the same arc, the angles are congruent.

Practice What do you remember? 1

Given that circle A is drawn to scale, identify the following: a

An inscribed angle.

A central angle.

A radius.

A diameter.

F C

A E B D

Consider the given circle where and are diameters. Identify each arc as a major arc, minor arc, or semicircle. Then ﬁnd its measure. a

K J

L 18° 47°

P M

Describe the difference between arc measure and arc length.

Consider ∠ABC inscribed in circle O. If the measure of ∠ABC is x°, what do we know about the measure of its intercepted arc ?

O C x° B

Consider the given ﬁgure: a

Use the Arc Addition Postulate to write an expression that represents

Find

. 105° L E

15°

50° 65°

B C

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Let’s practice 6

Consider the given ﬁgure. a

Solve for m∠UPT.

Solve for

Q 154°

86°

R T

60°

Consider each ﬁgure: a

Solve for m∠CDB.

Solve for m∠SQT.

70°

64° S Q

T D

Solve for m∠PSR.

Solve for ∠MPN.

S Q

70°

57°

56°

R N

Solve for m∠HKJ.

Solve for m∠XFD.

H 122°

X 114°

F D

L E

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Consider each ﬁgure: a

Solve for

B Q

29°

110°

Solve for

M 53°

70°

56° N

Solve for

122°

. X

F 53° D E

Let m∠RPQ = (8x + 16)° and m∠SPT = (12x − 32)°. a

Solve for x.

Solve for m∠SPT.

P T Q

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Solve for x.

G (31x + 3)°

192°

Consider the diagram, assuming ∠ACD ≅ ∠ECB: a

Solve for

A 40° E (18x − 40)°

(12x + 10)°

5 C

Given:

is a diameter of the circle • • = (9x + 18)° • m∠E = 34° a

Solve for x.

Solve for

. E

Consider the right triangle △ABC inscribed in the circle.

What do we know about the hypotenuse of △ABC? B

A sector of a circle has an arc length of 10 cm and a central angle of 72°. Find the radius of the circle.

A circle has a radius of 15 cm. The length of an arc is 12 cm. Find the measure of the arc in degrees.

A sector of a circle has an area of 28 cm2. The radius of the circle is 7 cm. Find the measure of the central angle of the sector.

The area of a sector is 50 cm2 and the central angle is 60°. Find the radius of the circle.

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Let’s extend our thinking 18

Determine the relationship between the inscribed angles in the circle. Justify your answer.

1 2 3 4 5

674

The graph shows the results of a survey in which students were asked what their favorite snack is. a

What would be the arc measures associated with the burger and nachos categories?

Describe the kinds of arcs associated with the ﬁrst category and the last category.

Are there any congruent arcs in this graph? Explain.

The table shows the results of a survey in which students were asked where they went on vacation last year. a

If you were to construct a circle graph of this information, what would be the arc measures associated with the beach and roadtrip categories?

Describe the kind of arcs associated with the beach and roadtrip categories.

Are there any congruent arcs in this graph? Explain.

Type of vacation Beach Camping Roadtrip Snow

String art can be made by weaving string through various geometric patterns. This particular design consists of 32 equally spaced points, where the string has been woven to form angles that reach the points 12 positions away from its vertex on either side. a

Find the measure of an angle formed by two segments sharing a common endpoint on the circle.

If the angles were modiﬁed so that each endpoint is 8 positions from the vertex of its angle, ﬁnd the measure of each.

If the angles were modiﬁed so that each endpoint is n positions from the vertex of its angle, ﬁnd the measure of each.

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14% 17% 36%

25%

Burger

Ice Cream

Sandwich

Pizza

Nachos

Tacos

Number of friends 20% 20% 50% 10%

The Pythagorean theorem, or distance formula, can help us ﬁnd distances in the coordinate plane.

distance between the two points

(x1, y1)

coordinates of the ﬁrst point

(x2, y2)

coordinates of the second point

The center is the midpoint of the diameter, so we can use the midpoint formula to ﬁnd the center when we know the endpoints of the diameter.

(xM, yM) coordinates of the midpoint (x1, y1)

coordinates of the ﬁrst endpoint

(x2, y2)

coordinates of the second endpoint

Given center (h, k), we can determine whether points in the coordinate plane lie on a circle. • If (x1 − h)2 + ( y1 − k)2 < r2 then (x1, y1) is inside the circle. • If (x1 − h)2 + ( y1 − k)2 = r2 then (x1, y1) is on the circle. • If (x1 − h)2 + ( y1 − k)2 > r2 then (x1, y1) is outside the circle.

Example 1 The given circle has a diameter with endpoints (−1, 3.65) and (5, −1.65):

5 (−1, 3.65)

4 3 2 1

−3 −2 −1 −1

x 1

2 3 4 5 6

−2 −3

(5, −1.65)

a Find the center of the circle.

Create a strategy A diameter has endpoints on the circle and passes through center. We know that the given line segment is a diameter, which means it passes through the center. Notice that the points (−2, 1), (2, 5), (6, 1) and (2, −3) also lie on the circle. The vertical and horizontal lines connecting these points appear to be diameters, which means they too will pass through the center. To ﬁnd the center, we can ﬁnd the point where all 3 lines intersect.

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Apply the idea The point of intersection of the three diameters, and the center of the circle, is (2, 1).

5 4 3 2

(2, 1)

1 −3 −2 −1 −1

x 1

2 3 4 5 6

−2 −3

Reﬂect and check We could also use the midpoint formula to calculate the center. Start with the midpoint formula

Substitute values from the points

Evaluate each numerator

Simplify

b Find the length of the diameter of the circle.

Create a strategy We can ﬁnd the length of the diameter by ﬁnding the distance between any two points that lie on the circle, such that the line between the two points passes through the center. 6

y (2, 5)

5 4

(−1, 3.65)

3 2 (−2, 1)

(6, 1)

−3 −2 −1 −1

x 1

2 3 4 5 6

−2 −3

The most efficient method for finding the length of the diameter of this circle is to find the distance of either the vertical or the horizontal diameter.

(5, −1.65) (2, −3)

Apply the idea Because the vertical and horizontal distances are both whole numbers, we can use either one to determine the length of the diameter. By counting the distances, we ﬁnd that the diameter is 8 units long.

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Reﬂect and check We could also use the distance formula using the given points to calculate the diameter. Let (x1, y1) be (−1, 3.65) and let (x2, y2) be (5, −1.65). Start with distance formula

Substitute values from the points

Evaluate the subtraction

Evaluate the exponents

Evaluate the addition

Simplify

Because the points on the circle are rounded values, there is a slight difference between the exact length of the diameter and the length we ﬁnd when using the distance formula.

Example 2 Find the center of the circle whose endpoints of a diameter are (−1.5, 4) and (4.5, −2).

Create a strategy The diameter of the circle passes through the center and is twice the radius. From this, we also know that the center is the midpoint of the diameter.

Apply the idea

Reﬂect and check

Recall the midpoint formula:

We can plot the given endpoints and use technology to graph the circle. We can see that the center appears to be at (1.5, 1), which conﬁrms our answer.

Substituting the values from the endpoints of the diameter, we ﬁnd the center of the circle to be:

5 4 3 2 1 −3 −2 −1 −1

x 1

2 3 4 5 6

−2 −3

Example 3 Determine if the point (4, 3) lies on the circle with a center at (1, 1) and radius of 3.

Create a strategy We want to ﬁnd the distance between the point (4, 3) and the center of the circle and compare it to the radius. We can do this by ﬁnding the distance between the point and the center and seeing if it is equal to the radius.

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Apply the idea

Start with the distance formula

Substitute values from the points

Evaluate the subtraction

Evaluate the exponents

Evaluate the addition

Because the distance between the center and the point is not equal to the length of the radius, this point does not lie on the circle.

Reﬂect and check The distance between the point (1, 1) and (4, 3) is approximately 3.6 units. Since this is greater than 3, the point lies outside the circle. We can use the radius to ﬁnd other points on the circle, such as: • The point directly right of the center: (1 + 3, 1) = (4, 1) • The point directly left of the center: (1 − 3, 1) = (−2, 1) • The point directly above the center: (1, 1 + 3) = (1, 4) • The point directly below the center: (1, 1 − 3) = (1, −2)

y 5 4 3 2

We can use technology to graph the circle and check that these points lie on the circle, and the given point lies outside the circle.

1 −3 −2 −1 −1

x 1

−2 −3

Example 4 A circle has a center of (−1, −4) and contains the point (4, 8). a Find the length of the radius.

Create a strategy The distance from the center to any point on the circle is the radius. So, to ﬁnd the radius, we can calculate the distance from the center to the given point using the distance formula.

Apply the idea We can use the coordinates of the two points with the distance formula: (x1, y1) = (−1, −4) and (x2, y2) = (4, 8)

Start with the distance formula

Substitute values from the points

Evaluate the subtraction

Evaluate the exponents

Evaluate the addition

Evaluate the square root

The length of the radius is 13 units.

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Reflect and check We could have also used the Pythagorean theorem to find the length of d since the distance formula comes from the Pythagorean theorem. To do this, we can plot the points and connect them to visualize the radius. Then, we will find the vertical and horizontal distances between the center and the given point.

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

The horizontal distance is 5 and the vertical distance is 12. We can use these values in the Pythagorean theorem to ﬁnd the length of the radius. x 1 2 3 4 5 6 7 8

c2 = a2 + b2

Begin with the Pythagorean Theorem

c = 5 + 12

Substitute vertical and horizontal lengths

c = 15 + 144

Evaluate the exponents

c = 169

Evaluate the addition

Square root both sides

c = 13

Evaluate

The radius is 13.

b Find the diameter of the circle.

Create a strategy Since the radius is half of the diameter, we can double the length of the radius.

Apply the idea d = 2(r)

Double the radius

= 2(13)

Substitute the length of the radius

= 26

Evaluate the multiplication

The length of the diameter is 26.

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Idea summary We can use the distance formula and midpoint formula to help determine the length and location of points on the circle and the center, radius, and diameter of a circle. Distance formula

d (x1, y1) (x2, y2)

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

(xM, yM) (x1, y1) (x2, y2)

coordinates of the midpoint coordinates of the ﬁrst endpoint coordinates of the second endpoint

Midpoint formula

Practice What do you remember? 1

Find the radius of the circles given the following information: a

Center: (0, 0)

Point on circle: (−5, 0) 2

What is the diameter of the circle? A

Center: (−3, 0) Point on circle: (−3, −1) 9 8 7 6 5 4 3 2 1 −5 −4 −3 −2 −1 −1

x 1 2 3 4 5

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Find the center and radius of each circle: a

1 −6 −5 −4 −3 −2 −1

y 5

−1

3 2

−2

−3

−4

−4 −3 −2 −1 −1

−5

−2

−6

−3

x 1

d 10

x −1

−1

−2

−3

−10 −8 −6 −4 −2

2 −2

−4

−3 −2

Using the graph of the circle: a

What are the coordinates of the center?

What is the length of the diameter?

y 4 2 −4 −2 −2

x 2

−4 −6 −8

Let’s practice 5

Match the given coordinates of the endpoints of diameters with the correct center coordinates of the circles. a

682

Endpoints (3, 6) and (9, 6)

Center (−2, 2)

Endpoints (−2, 4) and (4, 4)

Center (3, −1)

Endpoints (0, −1) and (6, −1)

iii

Center (5, −3)

Endpoints (−5, 2) and (1, 2)

Center (6, 6)

Endpoints (2, −3) and (8, −3)

Center (1, 4)

Find the centers of the circles with the following diameter endpoints: a

(0, 0) and (0, 4)

(−4, 2) and (6, 4)

and

(1.7, −2.8) and (−0.3, 3)

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Find the center of the circle. a

(1.5, 2.8)

2 1 −2 −1

−1

x 1

−4 −3 −2 −1

(4.7, −1.3)

−2

Center: (8, −2)

−1

Center: (−2, 3) Point on the circle: (−6, 6)

Find the center of the circle.

Find the exact length of the diameter of the circle.

(−1, 3) and (4, −2)

(1, 1) and (−8, 4)

(2, −5) and (−3, 4)

(0, 0) and (6, 8)

(1.1, 2.2) and (3.4, 5.6)

and

A circle has a radius of

units. Which of the following could represent the center and a point on the circle?

Center: (2, −2)

Point on Circle: (5, 2)

Point on Circle: (−3, 4)

Center: (−6, −1)

Center: (5, 6)

B D

Point on Circle: (−3, −3)

x 1

Consider each of the following set of endpoints of the diameters of a circle:

(1.3, 1.5)

−2

Point on the circle: (2, 6)

Find the diameter of the circle using the center and a point on the circle. a

(−2.1, 3.7)

Center: (0, −2)

Point on Circle: (−1, 2)

Find the length of the radius of a circle when given the endpoints of its diameter. Round your answer to two decimal places. a

Endpoints (0, 0) and (6, 8)

Endpoints (3, 4) and (7, 9)

Endpoints (5, 2) and (−3, −4)

Endpoints (−2, 3) and (4, −1)

Endpoints

Endpoints (1.1, 2.2) and (3.4, 5.6)

Endpoints (0.4, 1.2) and (2.8, −0.8)

Endpoints

and

Determine the coordinates of the center of the circle when given the endpoints of its diameter. a

Endpoints (2, 4) and (−2, −4)

Endpoints (1, 2) and (3, 4)

Endpoints (−3, −5) and (1, 7)

Endpoints

Endpoints (0, 5) and (6, 0)

Endpoints

Endpoints (1.1, 2.2) and (3.4, 5.6)

Endpoints (−1.5, 2.5) and (4, −3.5)

and and

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Zoe claims the center of this circle is (4, 3). Do you agree or disagree? How do you know?

y 7 6

(5.9, 4.9)

5 4 3

2 1 −1

A circle has a center at (−3, 6) and a radius of A

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

−9−8−7−6−5−4−3−2 −1−1

1 2 3 4

−9−8−7−6−5−4−3−2 −1−1

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

−9−8−7−6−5−4−3−2 −1−1

1 2 3 4

−9−8−7−6−5−4−3−2 −1−1

x 1 2 3 4

. Select the graph of the circle.

y 18 16 14 12 10 8 6 4 2

−12 −10−8 −6 −4 −2 −2 −4 −6

684

(1.7, 0.3) 2 3 4

. Select the graph of the circle.

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

A circle has a radius of A

−1

x 2 4 6 8 10 12

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16 14 12 10 8 6 4 2 −8 −6 −4 −2 −2 −4 −6 −8

2 4 6 8 10 12 14 16

x 5

8 6 4 2

16 12 8 4 −12 −8 −4 −4

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10 −12

x 4

−8 −12

x 2 4 6 8 10

Let’s extend our thinking 16

The points (5, 3) and (5, 7) lie on the circumference of a circle. Which of the following could be the coordinates of the center of the circle? A

SOL

(0, 4)

(−1, 6)

(4, 3)

(4, 5)

(3, 7)

(−1, 11)

(−4, 3)

Given: Circle G Center G: (−1, 3) Radius: 8 units Which point lies on circle G? A

(0, 4)

Determine whether the point (5, 4) is inside, outside, or on the circle with center (1, 2) and diameter of 6.

Find two potential points on the circle with center (4, 3) and radius 5.

Could the points (−4, 2) and (4, 2) both lie on a circle with center (1, 1)? Explain why or why not.

Consider the circle with center at (3, 4). A sector is created from point A to the center to another point on the circle. The area of this sector is approximately 18.4 cm2. Which of the following could be the other point on the sector?

(3, −2)

(8.3, 1.1)

(3, 10)

(−2, 7.2)

D (−2, 7.2)

11 10 9 8 7 6 5 4 3 2 1

−3−2−1 −1 −2 −3

C (3, 10)

(3, 4) (8.3, 1.1)

1 2 3 4 5 6 7 8 9 10 11 A (3, −2)

Telecommunication towers can be used to transmit cellular phone calls. A graph with units measured in kilometers shows towers at points (0, 0), (0, 4), and (−2, 2). These towers have a range of about 3 kilometers. a

Sketch a graph and locate the towers. Are there any locations that may receive calls from more than one tower? Explain your reasoning.

The center of City A is located at (−3, 2.5), and the center of City B is located at (4, 7). Each city has a radius of 1.5 kilometers. Which city seems to have better cell phone coverage? Explain your reasoning.

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12.04 Equations of circles After this lesson, you will be able to... • derive the equation of a circle of given the center and radius using the Pythagorean Theorem. • identify the coordinates of the center of the circle when given a graph or the equation of a circle in standard form. • identify the length of the radius or diameter of the circle when given the equation of a circle in standard form. • identify the coordinates of a point or multiple points on the circle when given the coordinates of the center and length of the radius of a circle. • determine the equation of a circle given a graph of a circle with a center with integer coordinates. • determine the equation of a circle given coordinates of the center and a point on the circle. • determine the equation of a circle given coordinates of the center and the length of the radius or diameter. • determine the equation of a circle given coordinates of the endpoints of the diameter.

Equations of circles All points on a circle are the same distance from the center. The radius tells us the distance from the center to any point on the circle.

Exploration Consider the circle with a radius of 13 units shown below: y 10 5 −20 −15 −10 −5 (−6, −8)

x 5

Verify the point (−11, 4) lies on the circle.

Write an equation that would allow you to ﬁnd any point, (x, y), on the circle.

Rewrite your equation to represent any circle with a center at (h, k) and radius, r.

−5 −10 −15 −20

The standard form of the equation of a circle is

(x − h)2 + ( y − k)2 = r2 r (h, k) (x, y)

radius of the circle center of the circle coordinates of any point on the circle

To check whether a point (x1, y1) is inside, on or outside a circle, we can compare the distance between that point and the center of the circle to the value of the radius. Using the Pythagorean theorem, we can write these conditions as: • If (x1 − h)2 + ( y1 − k)2 < r2 then (x1, y1) is inside the circle. • If (x1 − h)2 + ( y1 − k)2 = r2 then (x1, y1) is on the circle. • If (x1 − h)2 + ( y1 − k)2 > r2 then (x1, y1) is outside the circle. Notice that these conditions are the same as substituting the point into the equation of the circle and comparing the values on each side. 686

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Example 1 Derive the equation of a circle with center (h, k) and radius r.

Create a strategy A circle is comprised of an inﬁnite number of points that are equidistant from the center, (h, k). The distance from the center to any of these points, (x, y), is the length of the radius, r. y (x, y) r (h, k)

We can derive the equation of a circle by ﬁnding the length of r.

Apply the idea We can begin by drawing a right triangle with a radius as the hypotenuse. Then, we need to ﬁnd the lengths of each leg of the right triangle.

y (x, y)

The lengths of the lengths are simply the horizontal and vertical distances between the center, (h, k), and a point on the circle, (x, y).

r (h, k)

The horizontal distance between the two points is the absolute value of the difference between the x-values.

∣x − h∣

(x, y)

(h, k) x

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The vertical distance between the two points is the absolute value of the difference between the y-values.

y (x, y)

∣y − k∣

y (h, k)

Now that the lengths of the legs of the triangle are known, we can use the Pythagorean theorem to ﬁnd the distance between the two points.

y (x, y)

∣x − h∣2 + ∣y − k∣2 = r2

r (h, k)

Since squaring a expression will always result in a positive value, the absolute value bars are not necessary. Therefore, the equation of a circle with center (h, k) and radius r is (x − h)2 + ( y − k)2 = r2

Reﬂect and check We could have also used the distance formula to ﬁnd the length of r since the distance formula comes from the Pythagorean theorem.

Example 2 Consider the circle shown.

6 5 4 3 2 1 −4−3−2 −1 −1

a State the coordinates of the center.

x 1 2 3 4 5 6 7 8

−2 −3 −4 −5 −6

Create a strategy Notice that this circle has several points with integer coordinates. By connecting the points that lie directly across the circle, we can assume these lines will be close to the diameter. Since we know a diameter passes through the center, the point where these lines intersect will be the center.

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Apply the idea 6 5 4 3 2 1 −4−3−2 −1 −1

x 1 2 3 4 5 6 7 8

−2 −3 −4 −5 −6

The center of the circle is at the point (2, 0).

b State the radius of the circle.

Create a strategy The length of a circle’s radius is the distance from the center point to any point on the circumference of the circle.

Apply the idea In the previous part, we identiﬁed the center of the circle, (2, 0), and several other points on the circle. To ﬁnd the radius, we can count the number of units the center is from one of the points on the circle. 6 5 4 3 2 1 −4−3−2 −1 −1

x 1 2 3 4 5 6 7 8

−2 −3 −4 −5 −6

The distance from (2, 5) to (2, 0) is ∣5 − 0∣ = 5, so the length of the radius is 5 units.

c State the diameter of the circle.

Create a strategy The diameter of a circle is any straight line segment that passes through the center of the circle and whose endpoints lie on the circumference of the circle. The diameter is deﬁned as twice the length of the radius of the circle.

Apply the idea We found in the previous part that the radius of the circle is 5 units. Multiplying the radius by 2 will give us the diameter. This circle has a diameter length of 10 units.

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d State the equation of the circle.

Create a strategy The standard form of the equation of a circle is (x − h)2 + ( y − k)2 = r2, where r is the radius and (h, k) is the center.

Apply the idea

Reﬂect and check

The center coordinates are (h, k) = (0, 2) and the length of the radius it r = 5.

We can use technology to graph the circle represented by our equation and conﬁrm that it matches the original graph.

Substituting these values into the standard form of the equation of a circle, we get: (x − 2)2 + ( y − 0)2 = 52 Simplifying the equation, we get: (x − 2)2 + y2 = 25

Example 3 Write the equation of the circle with the given conditions. a Center at (−1, 2) and a radius of 4.

Create a strategy To write the equation of a circle, we use the standard form (x − h)2 + ( y − k)2 = r2, where (h, k) represents the center of the circle and r is the radius. Given the center (−1, 2) and radius 4, we can directly substitute these values into the equation.

Apply the idea

Reﬂect and check

Substituting the values of h = −1, k = 2 and r = 4 into standard form of the circle equation gives us:

Recall the standard form of the equation of a circle is (x − h)2 + ( y − k)2 = r2. The values of h and k can be directly related to translations of circles from the parent circle, x2 + y2 = r2, centered at the origin.

(x − (−1))2 + ( y − 2)2 = 42 Simplifying further, we get: (x + 1)2 + ( y − 2)2 = 16

The value of h represents a horizontal translation. In this case, h shifts the circle left 1 unit. The value of k represents a vertical translation. In this case, k shifts the circle up 2 units. Combining the movements, our circle was translated left 1 unit and up 2 units from the origin. 6 5 4 3 2 1 −6−5−4−3−2 −1 −1

−2 −3 −4 −5 −6

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x 1 2 3 4 5 6

b Center at (2, 4) and a point on the circle at (−2, 1).

Create a strategy To ﬁnd the equation of a circle, we need to use the formula (x − h)2 + ( y − k)2 = r2, where (h, k) is the center of the circle and r is the radius. The radius can be calculated using the distance formula between two points, which in this case are the center and the given point on the circle.

Apply the idea First, we calculate the radius r by ﬁnding the distance between the center, (h, k) = (2, 4), and the given point on the circle, (x1, y1) = (−2, 1).

Distance formula

Substitute the values of the points

Evaluate the subtraction

Evaluate the squares

Evaluate the addition

Square root of a perfect square

With r = 5, we can now write the equation of the circle: (x − 2)2 + ( y − 4)2 = 52 (x − 2)2 + ( y − 4)2 = 25

Reﬂect and check We have the equation of the circle to be (x − 2)2 + ( y − 4)2 = 25. To check the accuracy of the equation, we can substitute the coordinates of the given point (−2, 1) to make sure they satisfy this equation: (x − 2)2 + ( y − 4)2 = 25 2

(−2 − 2) + (1 − 4) = 25 16 + 9 = 25 25 = 25

Equation of the circle Substitute (x, y) = (−2, 1) Simplify 

This point does satisfy the equation, showing the equation we found for the circle is correct.

c Endpoints of a diameter are (−1.5, 4) and (4.5, −2).

Create a strategy To ﬁnd the equation of the circle, we need to know the coordinates of the center of the circle and the length of the radius. The diameter of the circle passes through the center and is twice the radius. From this, we also know that the center is the midpoint of the diameter.

Apply the idea Recall the midpoint formula:

Substituting the values from the endpoints of the diameter, we ﬁnd the center of the circle to be:

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Next, we need to find the length of the radius which we can do by finding the distance from the center to either of the endpoints of the diameter. Using the first endpoint and the center, we find the radius to be:

State the distance formula

Substitute values from the points

Evaluate the subtraction

Evaluate the exponents

Evaluate the addition

Now, we can substitute the center and radius into the standard form of the equation of a circle. The equation of the circle will be:

which can be simpliﬁed to: (x − 1.5)2 + ( y − 1)2 = 18

Reﬂect and check Using technology to graph the circle, we can see that the equation of the circle is correct because the given endpoints do lie on a diameter of the circle. 6

5 4 3 2 1 −3 −2 −1 −1

x 1

2 3 4 5 6

−2 −3

Example 4 Darnell shines a torch at a wall which lights up a circular region with a diameter of 4 meters. The center of the light is positioned 3 meters above the ground, and 5 meters horizontally from the left side of the wall. a Let the bottom left corner of the wall be the origin. Determine the equation of the circle which describes the edge of lighted area.

Create a strategy We are told that the center of the light is 5 meters from the left side of the wall, which is 5 meters to the right of the origin. It is also 3 meters above the ground, which is 3 meters up from the origin. We are also given that the diameter of the lighted area is 4 meters. We can use this information to determine the equation of the circle of the lighted area.

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Apply the idea The center of the lighted area will have the coordinates (5, 3), and the radius will be 2 meters. The equation of the circle which describes the edge of the lighted area is (x − 5)2 + ( y − 3)2 = 22 which we can simplify to (x − 5)2 + ( y − 3)2 = 4

b Yvonne has a height of 1.66 meters and is standing against the wall, 5 meters from the left side. Determine if any part of Yvonne is in the lighted area.

Create a strategy Notice that the x-coordinate of Yvonne’s position is the same of the x-coordinate of the center of the circle. This means we only need to determine if the bottom edge of the circle will reach Yvonne.

Apply the idea Since the radius of the lighted area is 2 meters, the lowest point that the light reveals will be 2 meters below the y-value of the center. The lighted area will begin from a height of 3 − 2 = 1 meter. Since both Yvonne and the center of the light are positioned 5 meters from the left side of the wall, and Yvonne is taller than 1 meter, part of Yvonne will be in the lighted area.

c Kayoko is standing against the wall, 6 meters from the left side of the wall. Determine the greatest height that Kayoko can be without being in the lighted area. Round your answer to the nearest centimeter.

Create a strategy For Kayoko to not be in the lighted area, her height must be less than or equal to the lowest point that the light reveals at 6 meters from the left side of the wall. In other words, the greatest height Kayoko can have without being seen is equal to the smallest y-value of the edge of the lighted area when x = 6.

Apply the idea We will substitute x = 6 into the standard form of the equation we found in part (a) and solve for the height, y. (x − 5)2 + ( y − 3)2 = 4 2

Standard form of the equation of the circle

(6 − 5) + ( y − 3) = 4

Substitute x = 6

( y − 3) = 4 − (6 − 5)

Subtract (6 − 5)2 from both sides

( y − 3)2 = 3

Evaluate the right-hand side of the equation

y−3= ± y= ±

Square root property +3

y = 4.73, y = 1.26

Add 3 to both sides Evaluate the expression, rounding down to the nearest centimeter

Taking the smallest y-value, we get that Kayoko can be up to 1.26 meters or 126 centimeters tall without being in the lighted area.

Reﬂect and check Recall there are 100 centimeters in 1 meter which is why we rounded to 2 decimal places. In meters, the numbers may seem small for heights. But if we converted 1.26 meters to feet, it would be about 4 feet and 2 inches which is a reasonable height for a student.

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Let’s practice 5

For each of the following circle graphs: i

State the coordinates of the center.

State the radius of the circle.

iii

State the diameter of the circle.

State the equation of the circle.

5 4 3 2 1

x 1

−5 −6

6 4

x 3

−3

−4

−2

1 2 3 4 5

−1 −1

−3 −2 −1 −1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

−2

−8 −6 −4 −2 −2

−3

−4

−6

x 2

For each of the following equations: i

Find the center of the circle.

iii

Graph the circle.

x2 + y2 = 16

Find the radius of the circle.

(x + 4)2 + ( y − 2)2 = 25

For each of the following equations for circles: i

State the center of the circle.

Calculate the diameter of the circle.

(x − 2)2 + ( y + 3)2 = 75

(x − 2.5)2 + y2 = 24.01

x2 + ( y − 4)2 = 13

For each of the following descriptions, write the equation of the circle. a

A circle with center at (3, 3) and a radius of 6 units.

A circle with center at (0, −6) and a diameter of 6 units.

A circle with center at

and a diameter of

A circle with center at

and a radius of

units. units.

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Using the given points, ﬁnd the equations for the circles: a

Center at (−2, 5), passing through the point (3, −7)

Center at (0, 0), passing through the point (5, −10)

Center at (0.5, −2), passing through the point (5, 6.4)

Diameter with endpoints at (4, 4), and (−2, 6)

Diameter with endpoints at (0, −2), and (13.6, 0.4)

Diameter with endpoints at (10, 6) and (−12, −14)

Given: Circle P with center at (−3, 5). Which equation could represent circle P ?

(x − 3)2 + ( y − 5)2 = 41

(x − 3)2 + ( y + 5)2 = 41

(x + 3)2 + ( y − 5)2 = 41

(x + 3)2 + ( y + 5)2 = 41

A circle has a center (4, 0) and goes through the point (3, −2). Which of the following could be the equation of the circle? A

(x − 4)2 + y2 = 5

(x + 4)2 + y2 = 5

(x − 4)2 − y2 = 5

x2 + ( y − 4)2 = 5

Fill in the blanks to complete the equation of the circle. Circle Q with diameter

: G(−5, 3) and H(3, −1).

Create the equation of this circle. The Equation of the Circle (x + 1)2 13

(x − 1)2

( y − 1)2

⬚

( y + 1)2

= 20

⬚

A circle has a center at (3, −5) and a radius of 6 units. Create the equation of this circle. The Equation of the Circle (x − 3) (x − 3)2 (x − 5) (x − 5)2 + 32

696

⬚

(x + 3) (x + 3)2 (x + 5) (x + 5)2 − 62

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⬚

Find the equation of each circle: a

−3 −2 −1 −1 −2 −3 −4 −5

−5 −4 −3 −2 −1 −1

3 2 1

x 1 2 3 4 5 6 7

11 10 9 8 7 6 5 4 3 2 1 −1−1

x 1 2 3 4 5

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

Determine whether the following points are inside, outside, or on the circle with equation x2 + y2 = 25. a

9 8 7 6 5 4 3 2 1

1 2 3 4 5 6 7

−3 −2 −1 −1 −2 −3 −4 −5 −6 −7

5 4 3 2 1

(−3, 2)

(4, 3)

(1, 6)

(−5, 0)

(16, 4)

(8, 8)

(8, 3)

Circle O has a center at (−1, 3) and a diameter of 10 units. Which point lies on circle O? A

(5, 11)

Which point lies on the circle represented by the equation (x − 3)2 + ( y − 5)2 = 52? A

(−6, −5)

(0, 1)

(0, 4)

(−1, 5)

A wind turbine has 20 m long blades that are attached to a tower 30 m high. The distance from the origin to the base of the wind turbine is 25 m. Find the standard form of the equation of the circle represented by the path of the blades of the wind turbine.

20 m

30 m x 25 m

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The circle shown has a radius of 5. Explain how the Pythagorean theorem can be used to derive the equation for this circle.

(x, y) (3, 4)

Let’s extend our thinking 20

A soccer match is being televised. One of the cameras is mounted on a drone which is programmed to zoom in on the ball only when it is inside the center circle. The drone uses a coordinate system to track the position of the ball, where the origin is at the bottom left corner of the ﬁeld and each unit corresponds to 1 m.

Center Circle 75 m

9.15 m

99 m

698

The center of the center circle is in the exact center of the ﬁeld. Find the coordinates of the circle’s center.

State the equation for the circle.

State the domain and range of the circle in interval notation.

A circle has a diameter with endpoints of A (−1, 1) and B (9, −7). a

Determine the equation of the circle.

Prove that the point P (8, 3) is outside of the circle.

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Coralee needs to use her school’s laser cutter to make a cork component for her design project. Using technology to sketch the component, she knows that the intersection of two different sized circles inscribed on a piece of cork gives the shape that she wants. Two such circles have the equations (x − 12)2 + ( y − 8)2 = 49 and (x − 20)2 + ( y − 8)2 = 36, where each unit is 1 cm and (0, 0) is the bottom left corner of the given cork board.

State the domain and range of the leftmost circle in interval notation.

State the domain and range of the rightmost circle in interval notation.

Find the width of the component that Coralee wants to make.

Find the least area of material that could be used to make Coralee’s shape, considering the following constraints: • The laser cutter will only accept rectangular pieces of material. • The height of the intersection is 10.17 cm. • There also needs to be a gap of at least 1 cm between any cut and the edge of the piece.

Three receiving stations are located on a coordinate plane, where each unit is one mile, at the points (−3, −2), (−1, 2) and (3, 1). The epicenter of an earthquake is determined to be 2 miles, 4 miles and 5 miles from these three points, respectively. a

Determine the coordinates of the epicenter of the earthquake.

The earthquake can be felt 10 miles away. Determine if the town located at (8, −7) would feel the earthquake.

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