Category: Envision Math

Go through the enVision Math Common Core Grade 8 Answer Key Topic 8 Solve Problems Involving Surface Area and Volume and finish your homework or assignments.

enVision Math Common Core 8th Grade Answers Key Topic 8 Solve Problems Involving Surface Area And Volume

Topic 8 Essential Question
How can you find volumes and surface areas of three-dimensional figures?
Answer:
The “Surface area” is the sum of the areas of all faces (or surfaces) on a 3D shape.
Ex:
A cuboid has 6 rectangular faces. To find the surface area of a cuboid, add the areas of all 6 faces
We know that,
The volume of a three-dimensional figure = Cross-sectional area × length

3-ACT MATH

Measure Up
Have you ever heard of the terms griffin beaker, Erlenmeyer flask, or graduated cylinder? Maybe you’ve used them in your science class. Each piece of equipment in a chemistry lab has a specific purpose, so containers come in many shapes. It’s sometimes necessary to pour a solution from one container to another. Think about this during the 3-Act Mathematical Modeling lesson.

Topic 8 ënVision STEM Project

Did You Know?
The production of packaging is a huge industry employing over five million people with annual sales of more than 400 billion dollars.
Packaging materials protect and deliver food and products to consumers.

A plastic bottle takes 450–1,000 years to biodegrade.

Seabirds are dying of starvation with stomachs full of plastic and Styrofoam.

Polystyrene foam lasts forever!
Eco-friendly packaging materials are being made from mushrooms and bamboo.

There is even a drink bottle made from recyclable paper.

New technology results in packaging materials that are both affordable and biodegradable.
Environmentally friendly companies are producing sustainable packaging. In addition to using recyclable materials, they reduce the water, natural resources, and energy needed for production. They minimize waste when designing products.

Your Task: Wrap it Up!

Engineers consider several factors when designing product packaging. These factors include cost efficiency and eco-friendly design so that materials are disposable, recyclable, biodegradable, and not wasted. Suppose you are an engineer working for Liquid Assets, an environmentally friendly company that designs, builds, and packages water purifiers. You and your classmates will use your knowledge of volume and surface area to determine an environmentally sound way to package the purifiers.

Topic 8 GET READY!

Review What You Know!

Vocabulary
Choose the best term from the box to complete each definition.
base
diameter
radius
three-dimensional
two-dimensional

Question 1.
The __________ is the distance from the center to the edge of a circle.
Answer:
We know that,
The “Radius” is the distance from the center to the edge of a circle
Hence, from the above,
We can conclude that the best term to complete the given definition is: Radius

Question 2.
A shape that has length, width, and height is ___________.
Answer:
We know that,
A shape that has a length, width, and height is known as “Three-dimensional”
Hence, from the above,
We can conclude that the best term to complete the given definition is: Three dimensional

Question 3.
Any side of a cube can be considered a __________.
Answer:
We know that,
Any side of a cube can be considered a “Base”
Hence, from the above,
We can conclude that the best term to complete the given definition is: Base

Question 4.
A shape that has length and width, but not height, is ___________.
Answer:
We know that,
A shape that has length and width, but not height is known as “Two-dimensional”
Hence, from the above,
We can conclude that the best term to complete the given definition is: Two-dimensional

Question 5.
The _____________ of a circle is a line segment that passes through its center and has endpoints on the circle.
Answer:
We know that,
The “Diameter” of a circle is a line segment that passed through its center and has endpoints on the circle
Hence, from the above,
We can conclude that the best term to complete the given definition is: Diameter

Multiplying with Decimals

Find the product.
Question 6.
14 ∙ 3.5 = _______
Answer:
The given expression is:
14 × 3.5
So,
14 × 3.5 = 49.0
Hence, from the above,
We can conclude that the value for the given expression is: 49

Question 7.
9 ∙ 3.14 = _________
Answer:
The given expression is:
9 × 3.14
So,
9 × 3.14 = 28.26
Hence, from the above,
We can conclude that the value for the given expression is: 28.26

Question 8.
4.2 ∙ 10.5 = _________
Answer:
The given expression is:
4.2 × 10.5
So,
4.2 × 10.5 = 44.1
Hence, from the above,
We can conclude that the value for the given expression is: 44.1

Areas of Circles

Find the area of each circle. Use 3.14 for π.
Question 9.

A = ________
Answer:
The given circle is:

From the given circle,
The radius is: 8 cm
Now,
We know that,
The area of the circle = πr²
So,
The area of the given circle = 3.14 × 8²
= 200.96 cm²
Hence, from the above,
We can conclude that the area for the given circle is: 200.96 cm²

Question 10.

A = _________
Answer:
The given circle is:

From the given circle,
The diameter is: 12 cm
Now,
We know that,
Radius = \(\frac{Diameter}{2}\)
Radius = \(\frac{12}{2}\)
Radius = 6 cm
Now,
We know that,
The area of the circle = πr²
So,
The area of the given circle = 3.14 × 6²
= 113.04 cm²
Hence, from the above,
We can conclude that the area of the given circle is: 113.04 cm²

Use the Pythagorean Theorem

Find the missing side length of the triangle.
Question 11.

x = _________
Answer:
The given triangle is:

Now,
Fro the given figure,
We can observe that the triangle is a right triangle
Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length of the hypotenuse
a and b are the side lengths
So,
13² = 12²+ x²
x²= 169 – 144
x² = 25
x = \(\sqrt{25}\)
x = 5 in.
Hence, from the above,
We can conclude that the missing side length of the given triangle is: 5 in.

Question 12.

x = __________
Answer:
The given triangle is:

Now,
Fro the given figure,
We can observe that the triangle is a right triangle
Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length of the hypotenuse
a and b are the side lengths
So,
30² = 24²+ x²
x²= 900 – 576
x² = 324
x = \(\sqrt{324}\)
x = 18 in.
Hence, from the above,
We can conclude that the missing side length of the given triangle is: 18 in.

Language Development

Complete the word web. Write keywords, ideas, examples, or illustrations that connect to each new vocabulary term.

Topic 8 Pick A Project

PROJECT 8A
What makes a concert rock?
PROJECT: DESIGN PROPS OR STAGE STRUCTURES

PROJECT 8B
What is the most interesting museum you have visited?
PROJECT: MAKE A MODEL OF A MUSEUM

PROJECT 8C
Where around the United States can you find quarries?
PROJECT: POUR AND MEASURE SAND

PROJECT 8D
If you were cast in a play, would it be a comedy or a drama? Why?
PROJECT: WRITE A SKIT

Lesson 8.1 Find Surface Area of Three-Dimensional Figures

Explore It!
Andrea is designing the packaging for a tube-shaped container.

I can… find the surface areas of cylinders, cones, and spheres.

A. Model with Math What two-dimensional shape represents the top and bottom of the container? What two-dimensional shape represents the tube? Draw a net of the tube-shaped container.
Answer:
It is given that Andrea is designing the packaging for a tube-shaped container.
Now,
The given arrangement for tube shaped container is:

Now,
From the given arrangement,
We can observe that
The two-dimensional shape that represents the top and bottom of the container is: Circle
The two-dimensional shape that represents the tube is: Rectangle
Hence,
The representation of the tube-shaped container is:

B. Look for Relationships The circular top and bottom fit perfectly on the ends of the container. How are the measures of the circles and the rectangle related?
Answer:
The representation of the tube-shaped container is:

Now,
From the given figure,
We can observe that the tube-shaped container is made up of 2 circles and 1 rectangle
So,
The total surface area of the tube-shaped container is the sum of the areas of 2 circles and 1 rectangle
Now,
We know that,
The area of a circle = πr²
The area of a rectangle = Height× Base
Where,
The base of the rectangle is a circle
So,
The circumference of the circle = 2πr
Hence,
The surface area of the tube-shaped container = 2πr² + 2πrh

Focus on math practices
Model with Math
How can you check whether the net that you drew accurately represents the tube-shaped container?
Answer:
The representation of the tube-shaped container is:

Now,
When the top and bottom of the container correctly fit the tube,
That is the representation that the net you drew accurately represents the tube-shaped container

Essential Question
How are the areas of polygons used to find the surface area formulas for three-dimensional figures?
Answer:
We know that,
A three-dimensional figure is a combination of some two-dimensional figures
Ex:
We can make a “Cuboid” from the combination of “Rectangles”
We can make a “Sphere” from the combination of ‘Circles”
So,
The total surface area of a three-dimensional figure can be given as the sum of all the areas of the two-dimensional figures that are used to make the three-dimensional figure

Try it

What is the surface area of a cylinder with a height of 9.5 inches and a radius of 2.5 inches?
The surface area of the cylinder is __________ square inches.
Answer:
It is given that
The height of a cylinder is: 9.5 inches
The radius of a cylinder is: 2.5 inches
Now,
We know that,
The surface area of the cylinder (S.A) = 2πr² + 2πrh
= 2 × 3.14 × (2.5)² + 2 × 3.14 × 9.5 × 2.5
= 39.25 + 149.15
= 188.4 square inches
Hence, from the above,
We can conclude that the surface area of the given cylinder is: 188.4 square inches

Convince Me! How can you find the surface area of a cylinder if you only know its height and the circumference of its base?
S.A. = 2(πr²) + (2πr)h
= 2π(________²) + 2π(_______)(________)
= _______π + _______π
= _______π
Answer:
It is given that
We know only the height of the cylinder and the circumference of its base
Now,
We know that,
The surface area of a cylinder (S.A) = 2πr² + 2πrh
So,
S.A = 2π (r²) + 2π (r) (h)
S.A = 2π + π
S.A = 3π
Hence, from the above,
We can conclude that the surface area of a cylinder with only its height and the circumference of its base is: 3π

Try It!

a. What is the surface area of a cone with a radius of 7 feet and a slant height of 9 feet? Use \(\frac{22}{7}\) for π.
Answer:
It is given that
The radius of the cone is (r): 7 feet
The slant height of the cone is (l): 9 feet
Now,
We know that,
The surface area of the cone (S.A) = πr² + πrl
So,
S.A = \(\frac{22}{7}\) × 7² + \(\frac{22}{7}\) × 7 × 9
= 154 + 198
= 352 square feet
Hence, from the above,
We can conclude that the surface area of the given cone is: 352 square feet

b. What is the surface area of a sphere with a diameter of 2.7 inches? Use 3.14 for π.
Answer:
It is given that,
The diameter of a sphere is: 2.7 inches
Now,
We know that,
Radius = \(\frac{Diameter}{2}\)
So,
Radius of the sphere = \(\frac{2.7}{2}\)
So,
The radius of the sphere (r) = 1.35 inches
Now,
We know that,
The surface area of the sphere (S.A) = 4πr²
So,
S.A = 4 × 3.14 × (1.35)²
= 22.89 square inches
Hence, from the above,
We can conclude that the surface area of the given sphere is: 22.89 square inches

KEY CONCEPT

Formulas for finding the area of polygons can be used to find the surface areas of cylinders, cones, and spheres.

Do You Understand?
Question 1.
Essential Question How are the areas of polygons used to find the surface area formulas for three-dimensional figures?
Answer:
We know that,
A three-dimensional figure is a combination of some two-dimensional figures
Ex:
We can make a “Cuboid” from the combination of “Rectangles”
We can make a “Sphere” from the combination of ‘Circles”
So,
The total surface area of a three-dimensional figure can be given as the sum of all the areas of the two-dimensional figures that are used to make the three-dimensional figure

Question 2.
Reasoning Why is the length of the base of the rectangle the same as the circumference of the circles in the net of a cylinder?
Answer:
The representation of the cylinder is:

Now,
If you look at the net, the curved surface of the cylinder is rectangular in shape. The length of the rectangle is the same as the circumference of the circle. Since the length of the rectangle wraps around the circle rim, it is the same length as the circumference of the circle.

Question 3.
Construct Arguments Aaron says that all cones with a base circumference of 8 inches will have the same surface area. Is Aaron correct? Explain.
Answer:
It is given that
Aaron says that all cones with a base circumference of 8 inches will have the same surface area
Now,
We know that,
The surface area of a cone = πr² + πrl
Now,
From the above formula,
We can conclude that the surface area of the cone does not depend only on the circumference of the base but also we need the side length of the cone part as well
So,
All cones with a base circumference of 8 inches will not have the same surface area.
Hence, from the above,
We can conclude that Aaron is not correct

Do You Know How?
Question 4.
What is the surface area of the cylinder? Use 3.14 for π, and round to the nearest tenth.

Answer:
The given figure is:

From the given figure,
The diameter of the cylinder is: 2 mm
The height of the cylinder is: 10 mm
So,
Radius of the cylinder = \(\frac{Diameter of the cylinder}{2}\)
= \(\frac{2}{2}\)
= 1 mm
Now,
We know that,
The surface area of the cylinder (S.A) = 2πr² + 2πrh
So,
S.A = 2 × 3.14 × 1² + 2 × 3.14 × 1 × 10
= 6.28 + 62.8
= 69.08 mm²
Hence, from the above,
We can conclude that the surface area of the given cylinder is: 69.08 mm²

Question 5.
What is the surface area of the cone to the nearest tenth? Use 3.14 for π.

Answer:
The given figure is:

From the given figure,
The slant height of the cone is (l): 4 ft
The radius of the cone is: 3 ft
Now,
We know that,
The surface area of the cone (S.A) = πr² + πrl
So,
S.A = 3.14 × 3² + 3.14 × 3 × 4
= 28.26 + 37.68
= 65.94 ft²
Hence, from the above,
We can conclude that the surface area of the given cone is: 65.94 ft²

Question 6.
What is the surface area of the sphere in terms of π?

Answer:
The given figure is:

From the given figure,
The diameter of the sphere is: 2 cm
Now,
We know that,
The radius of the sphere (r) = \(\frac{Diameter of the sphere}{2}\)
= \(\frac{2}{2}\)
= 1 cm
Now,
We know that,
The surface area of the sphere (S.A) = 4πr²
So,
S.A = 4 × 3.14 × 1²
= 12.56 cm²
Hence, from the above,
We can conclude that the surface area of the given sphere is: 12.56 cm²

Practice & Problem Solving

Leveled Practice In 7-8, find the surface area.
Question 7.
What is the surface area of the cylinder? Use 3.14 for π, and round to the nearest tenth.

S.A. = 2(πr²) + (2πr)h
= 2π(________²) + 2 π(_______)(________)
= 2 π(________) + 2 π(_________)
= _______π + _______π
= _______π
≈ ________ cm²
Answer:
The given figure is:

From the given figure,
The radius of the cylinder is: 3 cm
The height of the cylinder is: 5 cm
Now,
We know that,
The surface area of the cylinder (S.A) = 2πr² + 2πrh
So,
S.A = 2 × 3.14 × 3² + 2 × 3.14 × 3 × 5
= 56.52 + 94.2
= 150.72 cm²
Hence, from the above,
We can conclude that the surface area of the given cylinder is: 150.72 cm²

Question 8.
What is the surface area of the cone? Use 3.14 for π.

S.A. = πr² + πlr
= π(________²) + π(_______)(________)
= ________ π + _________π
= _______π
≈ ________ cm²
Answer:
The given figure is:

From the given figure,
The radius of the cone (r) is: 7 cm
The slant height of the cone (l) is: 13 cm
Now,
We know that,
The surface area of the cone (S.A) = πr² + πrl
So,
S.A = 3.14 × 7² + 3.14 × 7 × 13
= 153.86 + 285.74
= 439.6 cm²
Hence, from the above,
We can conclude that the surface area of the given cone is: 439.6 cm²

Question 9.
Construct Arguments Sasha incorrectly claimed that the surface area of the cylinder is about 76.9 square inches. Explain her likely error and find the correct surface area of the cylinder.

Answer
The given figure is:
:
Now,
From the given figure,
The diameter of the cylinder is: 7 in.
The height of the cylinder is: 19 in.
Now,
We know that,
Radius = \(\frac{Diameter}{2}\)
So,
Radius of the circle = \(\frac{7}{2}\)
= 3.5 in.
Now,
We know that,
The surface area of the cylinder (S.A) = 2πr² + 2πrh
So,
S.A = 2 × 3.14 × (3.5)² + 2 × 3.14 × 3.5 × 19
= 76.93 + 417.62
= 494.55 in.²
Hence, from the above,
We can conclude that
The correct surface area of the given cylinder is: 494.55 in.²
The mistake made by Sasha is:
She adds only the area of the top and bottom but not the area of the rectangle and the area of the top and bottom of the cylinder

Question 10.
A theme park has a ride that is located in half a sphere. The ride goes around the widest part of the sphere, which has a circumference of 514.96 yards. What is the surface area of the sphere? Estimate to the nearest hundredth using 3.14 for π.

Answer:
It is given that
A theme park has a ride that is located in half a sphere. The ride goes around the widest part of the sphere, which has a circumference of 514.96 yards
Now,
The given sphere is:

Now,
We know that,
Circumference = 2πr
So,
Circumference of the theme park = 514.96 yd
2πr = 514.96
r = \(\frac{514.96}{2 × 3.14}\)
= 2.38 yd
Now,
We know that,
The surface area of the sphere (S.A) = 4πr²
So,
S.A = 4 × 3.14 × (2.38)²
= 71.14 yd²
Hence, from the above,
We can conclude that the surface area of the given sphere is: 71.14 yd²

Question 11.
Find the amount of wrapping paper you need to wrap a gift in the cylindrical box shown. You need to cover the top, the bottom, and all the way around the box. Use 3.14 for π, and round to the nearest tenth.

Answer:
The given figure is:

Now,
From the given figure,
The radius of the cylinder is 9 in.
The height of the cylinder is: 8 in
Now,
We know that,
The surface area of the cylinder (S.A) = 2πr² + 2πrh
So,
S.A = 2 × 3.14 × 9² + 2 × 3.14 × 9 × 8
= 508.68 + 452.16
= 960.84 in.²
Hence, from the above,
We can conclude that the amount of wrapping paper you need to wrap a gift in the cylindrical box is: 960.84 in.²

Question 12.
Donna paints ornaments for a school play. Each ornament is made up of two identical cones, as shown. How many bottles of paint does she need to paint 70 ornaments?

Answer:
It is given that
Donna paints ornaments for a school play. Each ornament is made up of two identical cones, as shown
Now,
From the given figure,
We can observe that
The radius of a cone is: 4.1 cm
The slant height of a cone is: 8.9 cm
Now,
We know that,
The surface area of a cone (S.A) = πr² + πrl
So,
S.A = 3.14 × (4.1)² + 3.14 × 4.1 × 8.9
= 52.78 + 114.57
= 167.35 cm²
So,
The surface area of the second cone (S.A) = 167.35 cm²
So,
The S.A of 70 ornaments = 70 × (167.35 × 2)
= 23,429 cm²
So,
The number of bottles of paint she needed to paint 70 ornaments = \(\frac{23,429}{2,000}\)
= 11.7
≅ 12 bottles
Hence, from the above,
We can conclude that she need 12 bottles of paint to paint 70 ornaments

Question 13.
Higher-Order Thinking
a. What is the surface area of the cone? Use 3.14 for π, and round to the nearest whole number.

Answer:
The given figure is:

Now,
From the given figure,
The diameter of the cone is: 6 cm
The slant height of the cone is: 12 cm
Now,
We know that,
Radius = \(\frac{Diameter}{2}\)
Radius = \(\frac{6}{2}\)
Radius = 3 cm
Now,
We know that,
The surface area of the cone (S.A) = πr² + πrl
So,
S.A = 3.14 × 3² + 3.14 × 3 × 12
= 28.26 + 113.04
= 141.3 cm²
Hence, from the above,
We can conclude that the surface area of the given cone is: 141.3 cm²

b. Reasoning Suppose the diameter and the slant height of the cone are cut in half. How does this affect the surface area of the cone? Explain.
Answer:
It is given that the diameter and the slant height of the cone are cut in half
So,
S.A = π(\(\frac{d}{2}\))² + π (\(\frac{d}{2}\)) (\(\frac{l}{2}\))
= 3.14 × \(\frac{6²}{4}\) + 3.14 × 3 × 6
= 28.26 + 56.52
= 84.78 cm²
Now,
From part (a),
The S.A of the cone is: 141.3 cm²
So,
The ratio of S.A of the cones obtained from part (a) and part (b) respectively = \(\frac{141.3}{84.78}\)
= 1.666
Hence, from the above,
We can conclude that the S.A of the cone we obtained in part (b) is 1.666 times of the S.A of the cone we obtained in part (b)

Assessment Practice
Question 14.
What is the surface area of the sphere? Use 3.14 for π, and round to the nearest tenth.

A 254.5 cm²
B. 56.55 cm²
C. 1,017.4 cm²
D. 4,071.5 cm²
Answer:
The given figure is:

From the given figure,
The radius of the sphere is: 9 cm
Now,
We know that,
The surface area of the sphere (S.A) = 4πr²
So,
S.A = 4 × 3.14 × 9²
= 1,017.36 cm²
≈ 1,017.4 cm²
Hence, from the above,
We can conclude that the S.A of the given sphere is: 1,017.4 cm²

Question 15.
What is the approximate surface area of the cone, in square inches? Use 3.14 for π, and round to the nearest whole number.

Answer:
The given figure is:

Now,
From the given figure,
The slant height of the cone (l) is 40 in.
The diameter of the cone (d) is 40 in.
Now,
We know that,
Radius (r) = \(\frac{Diameter}{2}\)
r = \(\frac{40}{2}\)
r = 20 in.
Now,
We know that,
The surface area of the cone (S.A) = πr² + πrl
So,
S.A = 3.14 × 20² + 3.14 × 20 × 40
= 1,256 + 2,512
= 3,768 in.²
Hence, from the above,
We can conclude that the surface area of the given cone is: 3,768 in²

Lesson 8.2 Find Volume of Cylinders

Explain It!
Jenna and Ricardo are buying a new fish tank for the growing population of zebrafish in their science lab. Jenna says the tanks hold the same amount of water because they have the same dimensions. Ricardo says that he can fill the bottom of the rectangular tank with more cubes, so it can hold more water.

I can… use what I know about finding volumes of rectangular prisms to find the volume of a cylinder.

A. Look for Relationships How are the shapes of the two fish tanks alike? How are they different?
Answer:
When we observe the fish tanks of Jenna and Ricardo,
We can observe that
a. The heights of the two fish tanks are the same
b. The number of cubes that can be filled at the bottom is different
c. The number of cubes filled in Jenna’s fish tank is less than that of the number of cubes filled in Ricardo’s fish tank
d. The amount of water that can hold in Jenna’s fish tank is less than the amount of water that can hold in Ricardo’s fish tank

B. Critique Arguments Who do you think is correct, Ricardo or Jenna? Explain.
Answer:
It is given that
Jenna and Ricardo are buying a new fish tank for the growing population of zebrafish in their science lab. Jenna says the tanks hold the same amount of water because they have the same dimensions. Ricardo says that he can fill the bottom of the rectangular tank with more cubes, so it can hold more water.
Now,
To find which fish tank holds more water, find the volume of the two fish tanks
Now,
For Jenna’s fish tank:
The fish tank is in the form of a cylinder
Now,
We know that,
The volume of the cylinder (V) = πr²h
= \(\frac{πd²h}{4}\)
So,
V = \(\frac{3.14 × 24 × 24 × 48}{4}\)
= \(\frac{86,814.72}{4}\)
= 21,073.68 in³
For Ricardo’s fish tank:
The fish tank is in the form of a rectangular prism
Now,
We know that,
The volume of a rectangular prism (V) = Length × Width × Height
So,
V = 24 × 24 × 48
= 27,648 in³
Hence, from the above,
We can conclude that Ricardo is correct on the basis of volumes of their fish tanks

Focus on math practices
Use Structure How can you use what you know about areas of two-dimensional figures and volumes of prisms to compare the volumes of the fish tanks?
Answer:
We know that,
Volume = Area × Length (or) Height (or) Depth
Where,
“Area” is the area of two-dimensional figures like rectangles, circles, etc.

Essential Question
How is the volume of a cylinder related to the volume of a rectangular prism?
Answer:
Rectangular prisms and cylinders are somewhat similar because they both have two bases and a height.
The formula for the volume of a rectangular solid is
V=Bh
can also be used to find the volume of a cylinder
Where,
“B” in the rectangular prism is the area of the rectangle
“B” in the cylinder is the area of the circle

Try It!

The area of the base of the cylinder is 78.5 in.². What is the volume of the cylinder?

V = Bh
= _______ ∙ _______
= _______
The volume of the cylinder is ________ cubic inches.
Answer:
It is given that
The area of the base of the cylinder is 78.5 in.²
Now,
The given figure is:

Now,
We know that,
The volume of a cylinder (V) = Bh
Where,
“B” is defined as the area of the circle
So,
V = 78.5 × 11
= 863.5 in.³
Hence, from the above,
We can conclude that the volume of the given cylinder is: 863.5 cubic inches

Convince Me!
Why can you use the formula V = Bh to find the volume of a cylinder?
Answer:
The representation of the cylinder is:

Now,
From the given figure,
We know that,
The two circles that are in the top and bottom positions are congruent
So,
The area for both the circles is also the same
Now,
We know that,
Volume = Area × Height
Now,
We know that,
Area of the circle = πr²
Hence,
The volume of the cylinder (V) = πr²h

Try It!

Lin is building a cylindrical planter with a base diameter of 15 inches. She has 5,000 cubic inches of soil to fill her planter. What is the height of the largest planter Lin can build? Use 3.14 for π, and round to the nearest inch.
Answer:
It is given that
Lin is building a cylindrical planter with a base diameter of 15 inches. She has 5,000 cubic inches of soil to fill her planter.
So,
From the given information,
The volume of the cylindrical planter = 5,000 cubic inches
The diameter of the cylindrical planter = 15 inches
Now,
We know that,
The volume of a cylinder (V) = πr²h
= \(\frac{πd²h}{4}\)
So,
5,000 = \(\frac{3.14 × 15 × 15 × h}{4}\)
5,000 = 176.625h
h = \(\frac{5,000}{176.625}\)
h = 28.3 inches
Hence, from the above,
We can conclude that the height of the largest planter Lin can build is: 28.3 inches

KEY CONCEPT
The formula for the volume of a cylinder is the same as the formula for the volume of a prism. The formula for the volume of a cylinder is V= Bh, where B is the area of the circular base and h is the height of the cylinder.

Do You Understand?
Question 1.
Essential Question How is the volume of a cylinder related to the volume of a rectangular prism?
Answer:
Rectangular prisms and cylinders are somewhat similar because they both have two bases and a height.
The formula for the volume of a rectangular solid is
V=Bh
can also be used to find the volume of a cylinder
Where,
“B” in the rectangular prism is the area of the rectangle
“B” in the cylinder is the area of the circle

Question 2.
Use Structure What two measurements do you need to know to find the volume of a cylinder?
Answer:
We know that,
The volume of a cylinder (V) = πr²h
Now,
From the given formula,
We can observe that
π is a constant
Hence, from the above,
We can conclude that the two measurements that needed to be known are:
a. Radius of the cylinder
b. The height of the cylinder

Question 3.
Reasoning Cylinder A has a greater radius than Cylinder B. Does Cylinder A necessarily have a greater volume than Cylinder B? Explain.
Answer:
It is given that
Cylinder A has a greater radius than Cylinder B
Now,
We know that,
The volume of a cylinder (V) = πr²h
Now,
Fro the above,
We can observe that
Volume (V) ∝ Radius²
So,
When we increase the value of the radius, the value of the volume will automatically increase
Hence, from the above,
We can conclude that cylinder A has a greater volume than Cylinder B

Do You Know How?
Question 4.
What is the volume of the cylinder? Express your answer in terms of π.

Answer:
The given figure is:

Now,
We know that,
The volume of a cylinder (V) = Bh
Where,
B = πr²
So,
V = 4π × 10
V = 40π mm³
Hence, from the above,
We can conclude that the volume of the given cylinder in terms of π is: 40π mm³

Question 5.
What is the approximate height of the cylinder? Use 3.14 for π, and if necessary, round to the nearest tenth.

Answer:
The given figure is:

Now,
From the given figure,
We can observe that
The volume of a cylinder (V) = 314 ft³
The radius of a cylinder = 10 ft
Now,
We know that,
The volume of a cylinder (V) = πr²h
So,
314 = 3.14 × 10 × 10 × h
h = \(\frac{314}{3.14 × 10 × 10}\)
h = 1 ft
Hence, from the above,
We can conclude that the height of the given cylinder is: 1 ft

Question 6.
What is the volume of the cylinder? Use 3.14 for π, and if necessary, round to the nearest tenth.

Answer:
The given figure is:

Now,
From the given figure,
We can observe that
The height of a cylinder = 4 cm
The circumference of a circle = 22.4 cm
Now,
We know that,
The circumference of a circle = 2πr
So,
2πr = 22.4
r = \(\frac{22.4}{2 × 3.14}\)
r = 3.56 cm
Now,
We know that,
The volume of a cylinder (V) = πr²h
So,
V = 3.14 × 3.56 × 3.56 × 4
= 159.2 cm³
Hence, from the above,
We can conclude that the volume of the given cylinder is: 159.2 cm³

Practice & Problem Solving

Question 7.
Leveled Practice What is the volume of a cylinder with a radius of 5 centimeters and height of 2.5 centimeters? Use 3.14 for π.
V = π ________ ² ∙ ________
= π ______ ∙ ________
= _______ π
The volume of the cylinder is about _______ cubic centimeters.
Answer:
It is given that
The radius of a cylinder (r) = 5 cm
The height of a cylinder (h) = 2.5 cm
Now,
We know that,
The volume of a cylinder (V) = πr²h
So,
V = 3.14 × 5 × 5 × 2.5
= 196.25 cm
Hence, from the above,
We can conclude that the volume of the given cylinder is: 196.25 cm³ (or) 62.5π cm³

Question 8.
Find the volume of each cylinder in terms of. Which cylinder has the greater volume?
Cylinder A: Area of Base = 61 ft², height = 10 ft
Cylinder B: Circumference = 6π ft, height = 6 ft
Answer:
The given data is:
Cylinder A: Area of Base = 61 ft², height = 10 ft
Cylinder B: Circumference = 6π ft, height = 6 ft
Now,
We know that,
The volume of a cylinder (V) = Bh
Where,
B is the area of the circle
Now,
For Cylinder A:
V = 61 × 10
= 610 ft³
For Cylinder B:
We know that,
Circumference = 2πr
So,
2πr = 6π
r = \(\frac{6π}{2π}\)
= 3 ft
So,
V = 3.14 × 3 × 3 × 6
= 169.56 ft³
Hence, from the above,
We can conclude that Cylinder A has the greatest volume when we compare the volumes of Cylinder A and Cylinder B

Question 9.
The volume of a cylinder is 2251 cubic inches, and the height of the cylinder is 1 inch. What is the radius of the cylinder?
Answer:
It is given that
The volume of a cylinder is 2251 cubic inches, and the height of the cylinder is 1 inch
Now,
We know that,
The volume of a cylinder (V) = πr²h
So,
2251 = 3.14 × r² × 1
r² = \(\frac{2251}{3.14}\)
r² = 716.87 in²
r = 26.77 in
Hence, from the above,
We can conclude that the radius of the given cylinder is: 26.77 in.

Question 10.
A company is designing a new cylindrical water bottle. The volume of the bottle is 103 cubic centimeters. What is the radius of the water bottle? Estimate using 3.14 for π, and round to the nearest hundredth.

Answer:
It is given that
A company is designing a new cylindrical water bottle. The volume of the bottle is 103 cubic centimeters
Now,
The given figure is:

Now,
We know that,
The volume of a cylinder (V) = πr²h
So,
103 = 3.14 × r² × 8.1
r² = \(\frac{103}{3.14 × 8.1}\)
r² = 4.05cm²
r = 2.01 cm
Hence, from the above,
We can conclude that the radius of the given cylinder is: 2.01 cm

Question 11.
Use the figure at the right.

a. Find the volume of the cylinder in terms of π.
Answer:
The given figure is:

Now,
From the given figure,
We can observe that
The radius of the cylinder (r) = 4 in.
The height of the cylinder (h) = 3 in
Now,
We know that,
The volume of a cylinder (V) = πr²h
So,
V = π × 4 × 4 × 3
V = 48π in.³
Hence, from the above,
We can conclude that the volume of the given cylinder in terms of π is: 48π in.³

b. Is the volume of a cylinder, which has the same radius but twice the height, greater or less than the original cylinder? Explain.
Answer:
From part (a),
The radius of the cylinder (r) = 4 in.
The height of the cylinder (h) = 3 in
So,
Now,
For part (b),
The radius of the cylinder (r) = 4 in.
The height of the cylinder (h) = 6 in.
So,
V = π × 4 × 4 × 6
= 96π in.³
Hence, from the above,
We can conclude that the volume of the cylinder we obtained in part (b) is greater than the volume of the cylinder we obtained in part (a)

Question 12.
Reasoning A rectangular piece of cardboard with dimensions 6 inches by 8 inches is used to make the curved side of a cylinder-shaped container. Using this cardboard, what is the greatest volume the cylinder can hold? Explain.
Answer:
It is given that
A rectangular piece of cardboard with dimensions 6 inches by 8 inches is used to make the curved side of a cylinder-shaped container.
Now,
Let the height of the cylinder be: 6 inches (or) 8 inches
Now,
We know that,
The volume of a rectangular prism (V) = Length × Width × Height
So,
For h = 6 inches:
V = 6 × 8 × 6
= 288 cubic inches
For h = 8 inches:
V = 6 × 8 × 8
= 384 cubic inches
Hence, from the above,
We can conclude that the greatest volume the given cylinder can hold is: 384 cubic inches

Question 13.
The cylinder shown has a volume of 885 cubic inches.

a. What is the radius of the cylinder? Use 3.14 for π.
Answer:
It is given that
The cylinder shown has a volume of 885 cubic inches.
Now,
The given figure is:

Now,
From the given figure,
We can observe that
The height of a cylinder (h) = 11.7 in.
Now,
We know that,
The volume of a cylinder (V) = πr²h
So,
885 = 3.14 × r² × 11.7
r² = \(\frac{885}{3.14 × 11.7}\)
r² = 24.08
r = 4.9 in.
Hence, from the above,
We can conclude that the radius of the given cylinder is: 4.9 in.

b. Reasoning If the height of the cylinder is changed, but the volume stays the same, then how will the radius change? Explain.
Answer:
We know that,
The volume of a cylinder (V) = πr²h
Now,
It is given that
Volume —–> Constant
Height —– > Changed
Now,
From the given formula,
If the volume is constant, then
h ∝ \(\frac{1}{r²}\)
Hence, from the above relation,
We can conclude that
If the value of height increases, then the value of radius decreases
If the value of height decreases, then the value of radius increases

Question 14.
Toy rubber balls are packaged in a cylinder that holds 3 balls. Find the volume of the cylinder. Use 3.14 for π, and round to the nearest tenth.

Answer:
It is given that
Toy rubber balls are packaged in a cylinder that holds 3 balls
Now,
The given figure is:

Now,
From the given figure,
We can observe that
The height of the cylinder (h) = 20.7 cm
The diameter of the cylinder (d) = 6.9 cm
Now,
We know that,
The volume of a cylinder (V) = πr²h
= \(\frac{πd²h}{4}\)
So,
V = \(\frac{3.14 × 20.7 × 20.7 × 6.9}{4}\)
= 2,321 cm³
Hence, from the above,
We can conclude that the volume of the given cylinder is: 2,321 cm³

Question 15.
Higher-Order Thinking
An insulated collar is made to cover a pipe. Find the volume of the material used to make the collar. Let r = 3 inches, R= 5 inches, and h = 21 inches. Use 3.14 for π, and round to the nearest hundredth.

Answer:
It is given that
An insulated collar is made to cover a pipe.
Now,
The given data is:
r = 3 inches, R= 5 inches, and h = 21 inches
Now,
The radius of insulated collar = R – r
= 5 – 3
= 2 inches
Now,
The volume of a cylinder (V) = πr²h
So,
The volume of the material that is used to make the collar (V) = 3.14 × 2 × 2 × 21
= 263.76 cubic inches
Hence, from the above,
We can conclude that the volume of the material that is used to make the collar is: 263.76 cubic inches

Assessment Practice
Question 16.
The volume of a cylinder is 1,029π cubic centimeters. The height of the cylinder is 21 centimeters. What is the radius, to the nearest centimeter, of the cylinder?
Answer:
It is given that
The volume of a cylinder is 1,0291 cubic centimeters. The height of the cylinder is 21 centimeters.
Now,
We know that,
The volume of a cylinder (V) = πr²h
So,
1,029π = πr² × 21
r² = \(\frac{1,029}{21}\)
r² = 49
r = 7 cm
Hence, from the above,
We can conclude that the radius of the given cylinder is: 7 cm

Question 17.
The diameter of a cylinder is 7 yards. The height is 12 yards. What is the volume, in terms of π and to the nearest cubic yard, of the cylinder?
Answer:
It is given that
The diameter of a cylinder is 7 yards. The height is 12 yards
Now,
We know that,
The volume of a cylinder (V) = πr²h
= \(\frac{πd²h}{4}\)
So,
V = \(\frac{3.14 × 7 × 12 × 7}{4}\)
= 1,846.32 yards³ (or) 588π yards³
Hence, from the above,
We can conclude that the volume of the given cylinder is: 1,846.32 cubic yards (or) 588π cubic yards

Question 18.
A cylinder is shown. What statements about the cylinder are true?

☐ The radius of the cylinder is 2 ft.
☐ The diameter of the cylinder is 4 yd.
☐ The height of the cylinder is 8 in.
☐ The volume of the cylinder is 32 in.².
☐ The volume of the cylinder is 32π in.³.
Answer:
The given figure is:

Now,
From the given figure,
We can observe that
The diameter of the cylinder = 4 in.
The height of the cylinder = 8 in.
So,
The radius of the cylinder = \(\frac{Diameter}{2}\)
= 2 in.
Now,
We know that,
The volume of a cylinder (V) = πr²h
So,
V = 3.14 × 2 × 2 × 8
= 32π in.³
= 100.48 in.³
Hence, from the above,
We can conclude that the statements that are true about the given cylinder are:

Topic 8 MID-TOPIC CHECKPOINT

Question 1.
Vocabulary Select all the statements that describe surface area and volume. Lessons 8-1 and 8-2
☐ Surface area is the sum of the areas of all the surfaces of a figure.
☐ Volume is the distance around a figure.
☐ The surface area is a three-dimensional measure.
☐ Volume is the amount of space a figure occupies.
☐ Volume is a three-dimensional measure.
Answer:
The above statements that describe the surface area and volume are:

In 2-4, use the figure at the right. Sallie packed a cone-shaped cup inside of a cylindrical package.

Question 2.
The cone-shaped cup is made out of paper. How much paper was used to make the cup, excluding the opening at the top of the cup? Use 3.14 for π, and round to the nearest tenth. Lesson 8-1
Answer:
It is given that
Sallie packed a cone-shaped cup inside of a cylindrical package and the cone-shaped cup is made out of paper
Now,
The given figure is:

Now,
From the given figure,
We can observe that
The slant height of cone (l) is: 33 cm
The diameter of the cone (d) is: 20 cm
So,
The radius of cone (r) = \(\frac{Diameter}{2}\)
= 10 cm
Now,
We know that,
The surface area of a cone (S.A) = πr² + πrl
So,
S.A = 3.14 × 10 × 10 + 3.14 × 10 × 33
= 314 + 1,036.2
= 1,350.2 cm²
Hence, from the above,
We can conclude that
We have to use 1,350.2 cm² of paper was used to make the cup, excluding the opening at the top of the cup

Question 3.
The cylindrical package is made out of cardboard. In terms of π, how much cardboard was used to make the package? Lesson 8-1
Answer:
It is given that the cylindrical package is made out of cardboard
Now,
The given figure is:

Now,
From the given figure,
We can observe that
The height of the cylinder (h) is: 33 cm
The diameter of the cylinder (d) is: 20 cm
So,
Radius = \(\frac{Diameter}{2}\)
= \(\frac{20}{2}\)
= 10 cm
Now,
We know that,
The surface area of a cylinder (S.A) = 2πr² + 2πrh
So,
S.A = 2π × 10² + 2π × 10 × 33
= 200π + 660π
= 860π cm²
Hence, from the above,
We can conclude that 860π cm² of cardboard was used to make the package

Question 4.
How much space does the package occupy in terms of π? Lesson 8-1
Answer:
The given figure is:

Now,
From the above,
We can observe that the given figure is a cylinder
Now,
From the given figure,
We can observe that
The height of the cylinder (h) is: 33 cm
The diameter of the cylinder (d) is: 20 cm
So,
Radius (r) = \(\frac{Diameter}{2}\)
r = \(\frac{20}{2}\)
r = 10 cm
Now,
We know that,
The volume of a cylinder (V) = πr²h
So,
V = π × 10² × 33
= 3,300π cm³
Hence, from the above,
We can conclude that 3,300π cm³ of space does the package occupy in terms of π

Question 5.
What is the surface area of the sphere in terms of π? Lesson 8-1

Answer:
The given figure is:

Now,
From the given figure,
We can observe that
The radius of the sphere (r) is: 3 ft
Now,
We know that,
The surface area of a sphere (S.A) = 4πr²
So,
S.A = 4π × 3²
= 36π ft²
Hence, from the above,
We can conclude that the surface area of the given sphere is: 36π ft²

Question 6.
The volume of the cylinder is 400π cm³. What is the height of the cylinder? Lesson 8-2
A. 5 cm
B. 16 cm
C. 25 cm
D. 80 cm
Answer:
It is given that
The volume of the cylinder is 400π cm³
Now,
The given figure is:

Now,
From the given figure,
We can observe that
The diameter of the cylinder (d) is: 10 cm
So,
Radius = \(\frac{Diameter}{2}\)
= \(\frac{10}{2}\)
= 5 cm
Now,
We know that,
The volume of a cylinder (V) = πr²h
So,
400π = π × 5² × h
h = \(\frac{400}{25}\)
h = 16 cm
Hence, from the above,
We can conclude that the volume of the given cylinder is: 16 cm

Topic 8 MID-TOPIC PERFORMANCE TASK

Melissa designed a sculpture in which a cylinder-shaped section was removed from a cube.

PART A
Before painting the surface of the sculpture, Melissa wants to sand the surface where the cylinder section was removed. What is the surface area of the section she will sand? Use 3.14 for π. Explain how you found the surface area.
Answer:
It is given that
Melissa designed a sculpture in which a cylinder-shaped section was removed from a cube.
Now,
The given figure is:
Now,
From the given figure,
The diameter of the cylinder (d) is: 5 cm
The height of the cylinder (d) is: 10 cm
So,
The radius of the cylinder (r) = \(\frac{Diameter of the cylinder}{2}\)
r = 2.5 cm
Now,
We know that,
We know that,
The surface area of a cylinder (S.A) = 2πr² + 2πrh
So,
S.A = 2 × 3.14 × 2.5 × 2.5 + 2 × 3.14 × 2.5 × 10
= 39.25 + 157
= 196.25 cm²
Hence, from the above,
We can conclude that 196.25 cm²  is the surface area of the section Melissa will sand

PART B
Melissa has a can of spray paint that covers about 6,500 square centimeters. Can Melissa apply two coats of paint to the entire sculpture? Explain. Use 3.14 for π.
Answer:
The given figure is:

Now,
From the above,
We can observe that
The scripture is the combination of the cube and the cylinder
Now,
We know that,
The surface area of a cube (S.A) = 6a²
Where,
a is the side of the cube
So,
S.A of the cube = 6 × 10²
= 600 cm²
Now,
From part (a),
The S.A of the cylinder = 196.25 cm²
So,
The S.A of the scripture = 600 + 196.25
= 796.25 cm²
So,
6,500 > 796.25 × 2
Hence, from the above,
We can conclude that Melissa can apply two coats of paint to the entire sculpture

PART C
What is the volume of the sculpture? Use 3.14 for π.
Answer:
The given figure is:

Now,
From the above,
We can observe that
The scripture is the combination of the cube and the cylinder
Now,
The diameter of the cylinder (d) is: 5 cm
The height of the cylinder (h) is: 10 cm
So,
The radius of the cylinder (r) is: 2.5 cm
Now,
We know that,
The volume of the cylinder (V) = πr²h
The volume of the cube (V) = a³
Where,
a is the side of the cube
So,
The volume of the scripture (V) = πr²h + a³
So,
V = 3.14 × 2.5 × 2.5 × 10 + 10³
= 196.25 + 1,000
= 1,196.25 cm³
Hence, from the above,
We can conclude that the volume of the scripture is: 1,196.25 cm³

Lesson 8.3 Find Volume of Cones

Solve & Discuss It!
A landscape architect uses molds for casting rectangular pyramids and rectangular prisms to make garden statues. He plans to place each finished pyramid on top of a prism. If one batch of concrete mix makes one prism or three pyramids, how does the volume of one pyramid compare to the volume of one prism? Explain.

I can… find the volume of cones.
Answer:
It is given that
A landscape architect uses molds for casting rectangular pyramids and rectangular prisms to make garden statues. He plans to place each finished pyramid on top of a prism and one batch of concrete mix makes one prism or three pyramids
Now,
We know that,
The prism is in the shape of the cylinder
The pyramid is in the shape of the cone
Now,
We know that,
The volume of a prism (V) = πr²h
The volume of a pyramid (V) = \(\frac{1}{3}\)πr²h
So,
The volume of a prism = 3 × The volume of a pyramid
Hence, from the above,
We can conclude that
The volume of a prism is 3 times the volume of a pyramid

Look for Relationships
What do you notice about the dimensions of the bases of the pyramid and prism? How are the heights of the two solids related?
Answer:
The given figure is:

Now,
From the given figure,
We can observe that
a. The dimensions of the bases of the prism and the pyramid are the same
b. The heights of the pyramid and the prism are the same

Focus on math practices
Make Sense and Persevere If the architect mixes 10 batches of concrete, how many sculptures combining 1 prism and 1 pyramid could he make? Explain.
Answer:
It is given that
one batch of concrete mix makes one prism or three pyramids
So,
The ratio of the mix of prism and pyramid in 1 batch is: 1 : 3
Hence,
For 10 batches of the mix of concrete,
We can make sculptures combining 1 prism and 1pyramid are: 10 prisms and 30 pyramids

Essential Question
How is the volume of a cone related to the volume of a cylinder?
Answer:
The volumes of a cone and a cylinder are related in the same way as the volumes of a pyramid and a prism is related. If the heights of a cone and a cylinder are equal, then the volume of the cylinder is 3 times as much as the volume of a cone

Try It!
Find the volume of the cone. Use 3.14 for π.

The volume of the cone is about ________ cubic inches.
V = ____ πr²h
≈ _______(3.14) (______)² (4)
= ________ (3.14) (______)(4)
= _________
Answer:
The given figure is:

From the above figure,
We can observe that
The radius of the cone (r) is 1.5 in.
The height of the cone (h) is 4 in.
Now,
We know that,
The volume of the cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × 3.14 × (1.5)² × 4
= 9.42 in.³
Hence, from the above,
We can conclude that the volume of the given cone is: 9.42 in.³

Convince Me!
If you know the volume of a cone, how can you find the volume of a cylinder that has the same height and radius as the cone?
Answer:
We know that,
If the cylinder and the cone has the same height and radius, then
The volume of cone = \(\frac{1}{3}\) × The volume of cylinder

Try It!

Find the volume of each cone.
a. Use \(\frac{22}{7}\) for π. Express the answer as a fraction.

Answer:
The given figure is:

Now,
From the given figure,
We can observe that
The slant height (l) is: 5 mm
The height of the cone (h) is: 3 mm
Now,
We know that,
l² = r²+ h²
So,
5² = r² + 3²
r² = 25 – 9
r² = 16
r = 4 mm
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 4² × 3
= \(\frac{1,056}{21}\) mm³
Hence, from the above,
We can conclude that the volume of the given cone in the fraction form is: \(\frac{1,056}{66}\) mm³

b. Express the volume in terms of π.

Answer:
The given figure is:

From the given figure,
We can observe that
The height of the cone (h) is: 21 ft
The circumference of the circle (C) is: 16π ft
Now,
We know that,
The circumference of the circle (C) = 2πr
So,
2πr = 16π
r = 8 ft
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × π × 8² × 21
= 448π ft³
Hence, from the above,
We can conclude that the volume of the given cone in terms of π is: 448π ft³

KEY CONCEPT

The volume of a cone is \(\frac{1}{3}\) the volume of a cylinder with the same base and height. The formula for the volume of a cone is V = \(\frac{1}{3}\)Bh, where B is the area of the base and his the height of the cone.

Do You Understand?
Question 1.
Essential Question How is the volume of a cone related to the volume of a cylinder?
Answer:
The volumes of a cone and a cylinder are related in the same way as the volumes of a pyramid and a prism is related. If the heights of a cone and a cylinder are equal, then the volume of the cylinder is 3 times as much as the volume of a cone

Question 2.
Use Structure What dimensions do you need to find the volume of a cone?
Answer:
To find the volume of a cone,
The dimensions we need are:
a. The radius of the cone
b. The height of the cone

Question 3.
Look for Relationships If you know a cone’s radius and slant height, what must you do before you can find its volume?
Answer:
If you know a cone’s radius (r) and slant height(l), then
The volume of a cone (V) is given as:
V = \(\frac{1}{3}\)πr² × \(\sqrt{l² – r²}\)

Do You Know How?
Question 4.
Wanda found a cone-shaped seashell on the beach. The shell has a height of 63 millimeters and a base radius of 8 millimeters. What is the volume of the seashell? Estimate 63 mm using \(\frac{22}{7}\) for π.

Answer:
It is given that
Wanda found a cone-shaped seashell on the beach. The shell has a height of 63 millimeters and a base radius of 8 millimeters
Now,
The given figure is:

Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 8² × 63
= 4,224 mm³
Hence, from the above,
We can conclude that the volume of the seashell is: 4,224 mm³

Question 5.
What is the volume of the cone? Estimate using 3.14 for π, and round to the nearest tenth.

Answer:
The given figure is:

Now,
From the given figure,
We can observe that
The height of the cone (h) is: 40mm
The slant height of the cone (l) is: 41 mm
Now,
We know that,
l² = r² + h²
So,
41² = r²+ 40²
r² = 81
r = 9 mm
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × 3.14 × 9² × 40
= 3,391.2 mm³
Hence, from the above,
We can conclude that the volume of the given cone is: 3,391.2 mm³

Question 6.
What is the volume of the cone in terms of π if the circumference of the base is 1.4π feet?

Answer:
The given figure is:

Now,
From the given figure,
We can observe that
The height of the cone (h) is: 2.7 ft
Now,
We know that,
Circumference (C) = 2πr
So,
2πr = 1.4π
r = 0.7 ft
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × π × (0.7)² × 2.7
= 0.441π ft³
Hence, from the above,
We can conclude that the volume of the given cone is: 0.441π ft³

Practice & Problem Solving

Multimedia Leveled Practice In 7 and 8, find the volumes of the cones.
Question 7.
What is the volume of the cone? Write your answer in terms of π.

Answer:
From the given figure,
We can observe that
The radius of the cone (r) is: 3 cm
The height of the cone (h) is: 4 cm
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × π × 3² × 4
= 12π cm³
Hence, from the above,
We can conclude that the volume of the given cone is: 12π cm³

Question 8.
What is the volume of the cone to the nearest hundredth? Use 3.14 for π.

Answer:
From the given figure,
We can observe that
The radius of the cone (r) is: 16 units
The height of the cone (h) is: 36 units
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × 3.14 × 16² × 36
= 9,646.08 units³
Hence, from the above,
We can conclude that the volume of the given cone is: 9,646.08 units³

Question 9.
If a cone-shaped hole is 3 feet deep and the circumference of the base of the hole is 44 feet, what is the volume of the hole? Use \(\frac{22}{7}\) for π.
Answer:
It is given that
A cone-shaped hole is 3 feet deep and the circumference of the base of the hole is 44 feet
Now,
We know that,
Circumference (C) = 2πr
So,
2πr = 44
r = \(\frac{44}{2π}\)
r = 7 feet
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 7² × 3
= 154 feet³
Hence, from the above,
We can conclude that the volume of the given cone-shaped hole is: 154 feet³

Question 10.
The volume of the cone is 462 cubic yards. What is the radius of the cone? Use \(\frac{22}{7}\) for π.

Answer:
The given figure is:

Now,
From the given figure,
We can observe that
The height of the cone (h) is: 9 yd
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
462 = \(\frac{1}{3}\) × \(\frac{22}{7}\) × r² × 9
r² = \(\frac{462}{9.428}\)
r = 7 yd
Hence, from the above,
We can conclude that the radius of the given cone is: 7 yd

Question 11.
A city engineer determines that 5,500 cubic meters of sand will be needed to combat erosion at the city’s beach. Does the city have enough sand to combat erosion? Use \(\frac{22}{7}\) for π. Explain.

Answer:
It is given that
A city engineer determines that 5,500 cubic meters of sand will be needed to combat erosion at the city’s beach.
Now,
The given figure is:

Now,
From the given figure,
We can observe that
The slant height of the cone(l) is: 37 m
The height of the cone (h) is: 35 m
Now,
We know that,
l² = r²+ h²
37² = r² + 35²
r² = 37² – 35²
r = 12 m
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 12² × 35
= 5,280 m³
So,
5,500 > 5,280
Hence, from the above,
We can conclude that the city has enough sand to combat erosion

Question 12.
A water tank is shaped like the cone shown.

a. How much water can the tank hold? Use 3.14 for π, and round to the nearest tenth.
Answer:
The given figure is:

Now,
From the given figure,
We can observe that
The slant height of the cone (l) is: 61 ft
The height of the cone (h) is: 60 ft
Now,
We know that,
l² = r²+ h²
61² = r² + 60²
r² = 61² – 60²
r = 11 ft
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × 3.14 × 11² × 60
= 7,598.8 ft³
Hence, from the above,
We can conclude that the amount of water that the tank can hold is: 7,598.8 ft³

b. If water is drained from the tank to fill smaller tanks that each hold 500 cubic feet of water, how many smaller tanks can be filled?
Answer:
It is given that the water is drained from the tank to fill smaller tanks that each holds 500 cubic feet of water
Now,
From part (a),
We know that,
The amount of water that the tank can hold is: 7,598.8 ft³
Now,
The number of smaller tanks that can be filled = \(\frac{The amount of water that the tank can hold}{The amount of water that the smaller tank can hold}\)
= \(\frac{7,598.8}{500}\)
≅ 15
Hence, from the above,
We can conclude that the number of smaller tanks that can be filled is: 15

Question 13.
An ice cream cone is filled exactly level with the top of a cone. The cone has a 9-centimeter depth and a base with a circumference of 91 centimeters. How much ice cream is in the cone in terms of π?
Answer:
It is given that
An ice cream cone is filled exactly level with the top of a cone. The cone has a 9-centimeter depth and a base with a circumference of 91 centimeters
Now,
We know that,
Circumference (C) = 2πr
So,
2πr = 91
r = \(\frac{91}{2π}\)
r = 14.49 cm
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × π × (14.49)² × 9
= 629.88π cm³
Hence, from the above,
We can conclude that the amount of ice cream that is in the cone in terms of π is: 629.88π cm³

Question 14.
In the scale model of a park, small green cones represent trees. What is the volume of one green cone? Use \(\frac{22}{7}\) for π.

Answer:
The given green cone is:

Now,
From the given green cone,
We can observe that
The slant height of the cone (l) is: 65 mm
The height of the cone (h) is: 63 mm
Now,
We know that,
l² = r²+ h²
65² = r² + 603²
r² = 65² – 63²
r = 16 mm
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 16² × 63
= 16,896 mm³
Hence, from the above,
We can conclude that the volume of one green cone is: 16,896 mm³

Question 15.
Reasoning Compare the volumes of two cones. One has a radius of 5 feet and a slant height of 13 feet. The other one has a height of 5 feet and a slant height of 13 feet.
a. Which cone has the greater volume?
Answer:
The given data is:
Cone 1: Radius: 5 feet       Slant height: 13 feet
Cone 2: Height: 5 feet       Slant height: 13 feet
Now,
We know that,
l² = r²+ h²
So,
For cone 1:
13² = h² + 5²
h² = 13² – 5²
h = 12 feet
For cone 2:
13² = r² + 5²
r² = 13² – 5²
r = 12 feet
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
For Cone 1:
V = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 5² × 12
= 314.28 feet³
For Cone 2:
V = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 12² × 5
= 754.28 feet³
Hence, from the above,
We can conclude that Cone 2 has the greater volume

b. What is the volume of the larger cone in terms of π?
Answer:
From part (a),
We can observe that,
Cone 2 has a greater volume
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × π × 12² × 5
= 240π feet³

Question 16.
An artist makes a cone-shaped sculpture for an art exhibit. If the sculpture is 7 feet tall and has a base with a circumference of 24.492 feet, what is the volume of the sculpture? Use 3.14 for π, and round to the nearest hundredth.
Answer:
It is given that
An artist makes a cone-shaped sculpture for an art exhibit and the sculpture is 7 feet tall and has a base with a circumference of 24.492 feet
Now,
We know that,
Circumference (C) = 2πr
So,
2πr = 24.492
r = \(\frac{24.492}{2π}\)
r = 3.9 feet
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × 3.14 × 3.9² × 7
= 111.43 feet³
Hence, from the above,
We can conclude that the volume of the given cone-shaped sculpture is: 111.43 feet³

Question 17.
Higher-Order Thinking A cone has a radius of 3 and a height of 11.
a. Suppose the radius is increased by 4 times its original measure. How many times greater is the volume of the larger cone than the smaller cone?
Answer:
It is given that
A cone has a radius of 3 and a height of 11
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × 3.14 × 3² × 11
= 103.62 units³
Now,
If the radius is increased by 4 times of its original measure, then
V = \(\frac{1}{3}\) × 3.14 × (3 × 4)² × 11
= 1,657.92 units³
Now,
The number of times the volume of the larger cone than the smaller cone = \(\frac{1,657.92}{103.62}\)
= 16
Hence, from the above,
We can conclude that the volume of the larger cone is 16 times greater than the volume of the smaller cone

b. How would the volume of the cone change if the radius were divided by four?
Answer:
From part (a),
We know that,
The volume of the cone with a radius of 3 and a height of 11 units is: 103.62 units³
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × 3.14 × (3 ÷ 4)² × 11
= 6.476 units³
Now,
The ratio of the volume of the changed cone to the volume of the original cone = \(\frac{6.476}{103.62}\)
= 0.06
Hence, from the above,
We can conclude that the volume of the cone will be 0.06 times of the original cone

Assessment Practice
Question 18.
List the cones described below in order from least volume to greatest volume.
• Cone 1: radius 6 cm and height 12 cm
• Cone 2: radius 12 cm and height 6 cm
• Cone 3: radius 9 cm and height 8 cm
A. Cone 2, Cone 3, Cone 1
B. Cone 1, Cone 3, Cone 2
C. Cone 2, Cone 1, Cone 3
D. Cone 1, Cone 2, Cone 3
Answer:
The given data is:
a. Cone 1: radius 6 cm and height 12 cm
b. Cone 2: radius 12 cm and height 6 cm
c.  Cone 3: radius 9 cm and height 8 cm
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
For Cone 1:
V = \(\frac{1}{3}\) × 3.14 × 6² × 12
= 452.16 cm³
For Cone 2:
V = \(\frac{1}{3}\) × 3.14 × 12² × 6
= 904.32 cm³
For Cone 3:
V = \(\frac{1}{3}\) × 3.14 × 9² × 8
= 678.24 cm³
Hence, from the above,
We can conclude that the order of the volumes from the least to the greatest is:
Cone 1 < Cone 3 < Cone 2

Question 19.
What is the volume, in cubic inches, of a cone that has a radius of 8 inches and a height of 12 inches? Use 3.14 for π, and round to the nearest hundredth.
Answer:
It is given that
A cone has a radius of 8 inches and a height of 12 inches
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × 3.14 × 8² × 12
= 803.84 inches³
Hence, from the above,
We can conclude that the volume of the given cone is: 803.84 inches³

Lesson 8.4 Find Volume of Spheres

Explore It!
Marshall uses the beaker to fill the bowl with water.

I can… find the volume of a sphere and use it to solve problems.

A. Draw and label three-dimensional figures to represent the beaker and the bowl.
Answer:
The representation of the three-dimensional figures that represent the beaker and bowl are:

B. Marshall has to fill the beaker twice to completely fill the bowl with water. How can you use an equation to represent the volume of the bowl?
Answer:
The given figures are:

Now,
It is given that
Marshall has to fill the beaker twice to completely fill the bowl with water
Now,
From part (a),
We can observe that the beaker is in the form of a cone
Now,
We know that,
The volume of a cone = \(\frac{1}{3}\)πr²h
Now,
According to the given situation,
The volume of the bowl(V’) = 2 × The volume of the beaker(V)
V’ = 2 × \(\frac{1}{3}\)πr²h
Now,
From the given figures,
We can observe that
The height of a bowl is 2 times its radius
So,
V’ = \(\frac{2}{3}\)πr² (2r)
V’ = \(\frac{4}{3}\)πr³
Hence, from the above,
We can conclude that the volume of the bowl is: \(\frac{4}{3}\)πr³

Focus on math practices
Reasoning How is the volume of a sphere and the volume of a cone related? What must be true about the radius and height measurements for these relationships to be valid?
Answer:
We know that,
The relationship between the volume of the sphere and the volume of the cone is:
The volume of the sphere = 2 × The volume of the cone
Now,
For the above relationship to be valid,
a. The heights of the sphere and the cone must be the same
b. The radius of the sphere and the cone must be the same

Essential Question
How is the volume of a sphere related to the volume of a cone?
Answer:
The volume of a sphere is twice the volume of a cone that has the same circular base and height. i.e.,
The volume of the sphere = 2 × The volume of the cone

Try It!

What is the volume of a ball with a diameter of 6 centimeters? Use 3.14 for π.
V = \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) π _______³
≈ ________ ∙ ________
= __________
The volume of the ball is about _______ cm³
Answer:
It is given that
The diameter of a ball is: 6 cm
Now,
We know that,
The ball is in the form of a sphere
Now,
We know that,
Radius (r) = \(\frac{Diameter}{2}\)
r = \(\frac{6}{2}\)
r = 3 cm
Now,
We know that,
The volume of a sphere (V) = \(\frac{4}{3}\)πr³
So,
V = \(\frac{4}{3}\) × 3.14 × 3³
= 113.04 cm³
Hence, from the above,
We can conclude that the volume of the given ball is about 113.04 cm³

Convince Me!
How is the volume of a sphere related to the volume of a cone that has the same circular base and height?
Answer:
The relationship between the volume of a cone and the volume of a sphere that has the same circular base and height is:
The volume of the sphere = 2 × The volume of the cone

Try It!

What is the volume of the composite figure shown? Use 3.14 for π.

Answer:
The given composite figure is:

Now,
We know that,
A composite figure is made up of 2 or more two-dimensional figures
Now,
The given composite figure is made up of Hemisphere and Cone
Now,
We know that,
The volume of the given composite figure = The volume of the Hemisphere + The volume of the cone
Now,
From the given composite figure,
We can observe that
Radius (r) = 2 in.
Height (h) = 6 in.
Now,
We know that,
The volume of the Hemisphere (V) = \(\frac{2}{3}\)πr³
So,
V = \(\frac{2}{3}\) × 3.14 × 2³
= 16.74 in.³
Now,
We know that,
The volume of the cone (V’) = \(\frac{1}{3}\)πr²h
So,
V’ = \(\frac{1}{3}\) × 3.14 × 2² × 6
= 25.12 in.³
So,
The volume of the given composite figure = 16.74 + 25.12
= 41.84 in.³
Hence, from the above,
We can conclude that the volume of the given composite figure is: 41.84 in.³

KEY CONCEPT

The volume of a sphere is twice the volume of a cone that has the same circular base and height. The formula for the volume of a sphere with radius r is V = \(\frac{4}{3}\) πr³.

Do You Understand?
Question 1.
Essential Question How is the volume of a sphere related to the volume of a cone?
Answer:
The volume of a sphere is twice the volume of a cone that has the same circular base and height. i.e.,
The volume of the sphere = 2 × The volume of the cone

Question 2.
Critique Reasoning Kristy incorrectly says that the volume of the sphere below is 144π cubic units. What mistake might Kristy have made?

Answer:
The given sphere is:

Now,
It is given that
Kristy incorrectly says that the volume of the sphere below is 144π cubic units
Now,
From the given sphere,
We can observe that
Radius (r) = 6 units
Now,
We know that,
The volume of a spher (V) = \(\frac{4}{3}\)πr³
So,
V = \(\frac{4}{3}\) × π × 6³
= 288π units³
Hence, from the above,
We can conclude that the mistake done by Kristy is:
Considering the sphere as the hemisphere and calculated the volume of the hemisphere instead of the sphere

Question 3.
Generalize Mehnaj has a set of blocks that are all the same height. The cone-shaped block has a volume of 125 cubic inches. The sphere-shaped block has a volume of 250 cubic inches. What do you know about the radius of the base of the cone-shaped block? Explain.
Answer:
It is given that
Mehnaj has a set of blocks that are all the same height. The cone-shaped block has a volume of 125 cubic inches. The sphere-shaped block has a volume of 250 cubic inches
So,
From the given situation,
We can observe that
a. The heights of the two cone-shaped blocks are the same
b. The volumes of the two cone-shaped blocks are different
Now,
We know that,
The volume of a cone = \(\frac{1}{3}\)πr²h
So,
V₁ = \(\frac{1}{3}\)πr₁²h₁
V₂ = \(\frac{1}{3}\)πr₂²h₂
So,
V₁/V₂ = r₁²/ r₂²
r₁²/ r₂² = \(\frac{125}{250}\)
r₁² / r₂² = 0.5
r₁ / r₂ = 0.707
Hence, from the above,
We can conclude that the radius of the first cone-based block is 0.707 times the radius of the second cone-shaped block

Do You Know How?
Question 4.
Clarissa has a decorative bulb in the shape of a sphere. If it has a radius of 3 inches, what is its volume? Use 3.14 for π.
Answer:
It is given that
Clarissa has a decorative bulb in the shape of a sphere and it has a radius of 3 inches
Now,
We know that,
The volume of a sphere (V) = \(\frac{4}{3}\)πr³
So,
V = \(\frac{4}{3}\) × 3.14 × 3³
= 113.04 inches³
Hence, from the above,
We can conclude that the volume of the given decorative bulb is: 113.04 inches³

Question 5.
A sphere has a surface area of about 803.84 square centimeters. What is the volume of the sphere? Use 3.14 for π and round to the nearest whole number.
Answer:
It is given that
A sphere has a surface area of about 803.84 square centimeters
Now,
We know that,
The surface area of a sphere (S.A) = 4πr²
So,
4πr² = 803.84
r² = \(\frac{803.84}{4π}\)
r² = 64
r = 8 cm
Now,
We know that,
The volume of a sphere (V) = \(\frac{4}{3}\)πr³
So,
V = \(\frac{4}{3}\) × 3.14 × 8³
= 2,143.57 cm³
≈2,144 cm³
Hence, from the above,
We can conclude that the volume of the given sphere is about 2,144 cm³

Question 6.
A water pipe is a cylinder 30 inches long, with a radius of 1 inch. At one end of the cylinder, there is a hemisphere. What is the volume of the water pipe? Explain.
Answer:
It is given that
A water pipe is a cylinder 30 inches long, with a radius of 1 inch. At one end of the cylinder, there is a hemisphere.
So,
The volume of the water pipe = The volume of the cylinder + The volume of the hemisphere
Now,
From the given water pipe,
We can observe that
Radius (r) = 1 in.
Height (h) = 30 in.
Now,
We know that,
The volume of the Hemisphere (V) = \(\frac{2}{3}\)πr³
So,
V = \(\frac{2}{3}\) × 3.14 × 1³
= 2.09 in.³
Now,
We know that,
The volume of the cylinder (V’) = πr²h
So,
V’ = 3.14 × 1² × 30
= 94.2 in.³
So,
The volume of the given water pipe = 2.09 + 94.2
= 96.29 in.³
Hence, from the above,
We can conclude that the volume of the given water pipe is: 96.29 in.³

Practice & Problem Solving

Question 7.
Leveled Practice What is the amount of air, in cubic centimeters, needed to fill the stability ball? Use 3.14 for π, and round to the nearest whole number. Use the formula

The volume of the stability ball is approximately __________ cubic centimeters.
Answer:
From the given figure,
We can observe that
The stability ball is in the form of a sphere
Now,
The diameter of the stability ball (d) is: 55 cm
Now,
We know that,
Radius (r) = \(\frac{Diameter}{2}\)
r = \(\frac{55}{2}\)
r = 27.5 cm
Now,
We know that,
The volume of a sphere (V) = \(\frac{4}{3}\)πr³
So,
V = \(\frac{4}{3}\) × 3.14 × (27.5)³
= 87,069.53 cm³
≈87,070 cm³
Hence, from the above,
We can conclude that the volume of the stability ball is approximately 87,070 cm³

Question 8.
The spherical balloon has a 22-inch. diameter when it is fully inflated. Half of the air is let out of the balloon. Assume that the balloon remains a sphere. Keep all answers in terms of π.
a. Find the volume of the fully-inflated balloon.
Answer:
It is given that
The spherical balloon has a 22-inch. diameter when it is fully inflated. Half of the air is let out of the balloon.
Now,
From the given situation,
The diameter of the fully inflated balloon (d) is: 22 inches
Now,
We know that,
Radius (r) = \(\frac{Diameter}{2}\)
r = \(\frac{22}{2}\)
r = 11 cm
Now,
We know that,
The volume of a sphere (V) = \(\frac{4}{3}\)πr³
So,
V = \(\frac{4}{3}\) × π × 11³
= 1,774.6π cm³
Hence, from the above,
We can conclude that the volume of the fully inflated balloon is: 1,774.6π cm³

b. Find the volume of the half-inflated balloon.
Answer:
It is given that
The spherical balloon has a 22-inch. diameter when it is fully inflated. Half of the air is let out of the balloon.
So,
The diameter of half-inflated balloon (d) = \(\frac{The diameter of fully inflated balloon}{2}\)
d = \(\frac{22}{2}\)
d = 11 cm
Now,
We know that,
Radius (r) = \(\frac{Diameter}{2}\)
r = \(\frac{11}{2}\)
r = 5.5 cm
Now,
We know that,
The volume of a sphere (V) = \(\frac{4}{3}\)πr³
So,
V = \(\frac{4}{3}\) × π × (5.5)³
= 221.8π cm³
Hence, from the above,
We can conclude that the volume of the half-inflated balloon is: 221.8π cm³

c. What is the radius of the half-inflated balloon? Round to the nearest tenth.
Answer:
From part (b),
We can observe that
The diameter of half-inflated balloon (d) = \(\frac{The diameter of fully inflated balloon}{2}\)
d = \(\frac{22}{2}\)
d = 11 cm
Now,
We know that,
Radius (r) = \(\frac{Diameter}{2}\)
r = \(\frac{11}{2}\)
r = 5.5 cm
Hence, from the above,
We can conclude that the radius of the half-inflated balloon is: 5.5 cm

Question 9.
Find the volume of the figure. Use 3.14 for π, and round to the nearest whole number.

Answer:
The given figure is:

Now,
From the given figure,
We can observe that
The figure is a combination of a hemisphere and a cone
So,
From the given figure,
Diameter (d) =14 cm
Height (h) = 17 cm
Now,
We know that,
Radius (r) = \(\frac{Diameter}{2}\)
r = \(\frac{14}{2}\)
r = 7 cm
Now,
We know that,
The volume of the figure (V) = The volume of a hemisphere + The volume of a cone
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × 3.14 × 7² × 17
= 871.87 cm³
Now,
We know that,
The volume of a hemisphere (V’) = \(\frac{2}{3}\)πr³
So,
V’ = \(\frac{2}{3}\) × 3.14 × 7³
= 718.01 cm³
So,
The volume of the given figure = 871.87 + 718.01
= 1,589.88 cm³
≈1,590 cm³
Hence, from the above,
We can conclude that the volume of the given figure is: 1,590 cm³

Question 10.
The surface area of a sphere is about 2,826 square millimeters. What is the volume of the sphere? Use 3.14 for π, and round to the nearest whole number.
Answer:
It is given that
The surface area of a sphere is about 2,826 square millimeters
Now,
We know that,
The surface area of a sphere (S.A) = 4πr²
So,
4πr² = 2,826
r² = \(\frac{2,826}{4π}\)
r² = 225
r = 15 mm
Now,
We know that,
The volume of a sphere (V) = \(\frac{4}{3}\)πr³
So,
V = \(\frac{4}{3}\) × 3.14 × 15³
= 14,130 mm³
Hence, from the above,
We can conclude that the volume of the given sphere is: 14,130 mm³

Question 11.
A sphere has a volume of 1,837.35 cubic centimeters. What is the radius of the sphere? Use 3.14 for π, and round to the nearest tenth.
Answer:
It is given that
A sphere has a volume of 1,837.35 cubic centimeters
Now,
We know that,
The volume of a sphere (V) = \(\frac{4}{3}\)πr³
So,
1,837.35 = \(\frac{4}{3}\)πr³
πr³ = \(\frac{1,837.35 × 3}{4}\)
πr³ = 1,378.01
r³ = \(\frac{1,378.01}{π}\)
r³ = 438.85
r = 0.333 cm
Hence, from the above,
We can conclude that the radius of the given sphere is: 0.333 cm

Question 12.
Find the volume of the solid. Use 3.14 for π, and round to the nearest whole number.

Answer:
The given figure is:

From the above,
We can observe that the given figure is a composite figure
Now,
From the given figure,
We can observe that
Radius (r) = 4 m
Height (h) = 17 m
Now,
We know that,
The volume of the given figure = The volume of a hemisphere + The volume of a cylinder
Now,
We know that,
The volume of the Hemisphere (V) = \(\frac{2}{3}\)πr³
So,
V = \(\frac{2}{3}\) × 3.14 × 4³
= 133.97 m³
Now,
We know that,
The volume of the cylinder (V’) = πr²h
So,
V’ = 3.14 × 4² × 17
= 854.08 m³
So,
The volume of the given figure = 133.9 + 854.08
= 987.98 m³
Hence, from the above,
We can conclude that the volume of the given figure is: 987.98 m³

Question 13.
Your friend says that the volume of a sphere with a diameter of 3.4 meters is 164.55 cubic meters. What mistake might your friend have made? Find the correct volume. Use 3.14 for π and round to the nearest hundredth.
Answer:
It is given that
Your friend says that the volume of a sphere with a diameter of 3.4 meters is 164.55 cubic meters
Now,
From the given information,
Diameter (d) = 3.4 m
Now,
We know that,
Radius (r) = \(\frac{Diameter}{2}\)
r = \(\frac{3.4}{2}\)
r = 1.7 m
Now,
We know that,
The volume of a sphere (V) = \(\frac{4}{3}\)πr³
So,
V = \(\frac{4}{3}\) × 3.14 × (1.7)³
= 20.56 m³
Hence, from the above,
We can conclude that
The mistake made by your friend is:
Consideration of diameter as radius and find the volume of the given sphere

Question 14.
A solid figure has a cone and hemisphere hollowed out of it. What is the volume of the remaining part of the solid? Use 3.14 for π, and round to the nearest whole number.

Answer:
It is given that
A solid figure has a cone and hemisphere hollowed out of it
So,
The volume of the remaining part of the solid = | The volume of a hemisphere – The volume of a cone |
Now,
From the given solid,
We can observe that
Radius (r) = 6 in.
Height = 23 in.
Now,
We know that,
The volume of a hemisphere (V) = \(\frac{2}{3}\)πr³
So,
V = \(\frac{2}{3}\)π × 6³
= 144π in.³
Now,
We know that,
The volume of a cone (V’) = \(\frac{1}{3}\)πr²h
So,
V’ = \(\frac{1}{3}\)π × 6² × 23
= 276π in.³
So,
The volume of the remaining part of the solid = |144π – 276π|
= 132π
= 132 × 3.14
= 414.48 in.³
≈415 in.³
Hence, from the above,
We can conclude that the volume of the remaining part of the given solid is about 415 in.³

Question 15.
Higher-Order Thinking A student was asked to find the volume of a solid where the inner cylinder is hollow. She incorrectly said the volume is 2,034.72 cubic inches.

Answer:
It is given that
A student was asked to find the volume of a solid where the inner cylinder is hollow. She incorrectly said the volume is 2,034.72 cubic inches.

a. Find the volume of the solid. Use 3.14 for π. Round to the nearest whole number.
Answer:
The given solid is:

Now,
From the given solid,
We can observe that it is a combination of a cone and a cylinder
So,
The volume of the given solid = The volume of a cylinder + The volume of a cone
Now,
From the given solid,
We can observe that
The radius of the cylinder is 3 in.
The height of the cylinder is 15 in.
The diameter of the cone is 12 in.
The slant height of the cone is 9 in.
Now,
We know that,
The volume of a cylinder (V) = πr²h
So,
V = 3.14 × 3² × 15
= 423.9 in.³
Now,
The volume of a cone:
We know that,
Radius (r) = \(\frac{Diameter}{2}\)
r = \(\frac{12}{2}\)
r = 6 in.
Now,
We know that,
l² = r² + h²
So,
9² = 6² + h²
h² = 9² – 6²
h² = 45
h = 6.70 in.
Now,
We know that,
The volume of a cone (V’) = \(\frac{1}{3}\)πr²h
So,
V’ = \(\frac{1}{3}\) × 3.14 × 6² × 6.70
= 252.45 in.³
So,
The volume of the given solid = 423.9 + 252.45
= 676.3 in.³
≈676 in.³
Hence, from the above,
We can conclude that the volume of the given solid is about 676 in.³

b. What mistake might the student have made?
Answer:

Assessment Practice
Question 16.
The spherical boulder is 20 feet in diameter and weighs almost 8 tons. Find its volume. Use 3.14 for π. Round to the nearest cubic foot.
Answer:
It is given that
The spherical boulder is 20 feet in diameter and weighs almost 8 tons.
Now,
We know that,
The volume of a sphere (V) = \(\frac{4}{3}\)πr³
Now,
We know that,
Radius (r) = \(\frac{Diameter}{2}\)
r = \(\frac{20}{2}\)
r = 10 feet
So,
V = \(\frac{4}{3}\) × 3.14 × 10³
= 4,186.6 feet³
≈ 4,187 feet³
Hence, from the above,
We can conclude that the volume of the given spherical boulder is: 4,187 feet³

Question 17.
A bowl is in the shape of a hemisphere (half a sphere) with a diameter of 13 inches. Find the volume of the bowl. Use 3.14 for π, and round to the nearest cubic inch.
Answer:
It is given that
A bowl is in the shape of a hemisphere (half a sphere) with a diameter of 13 inches. Find the volume of the bowl
Now,
We know that,
Radius (r) = \(\frac{Diameter}{2}\)
r = \(\frac{13}{2}\)
r = 6.5 inches
Now,
We know that,
The volume of a hemisphere (V) = \(\frac{2}{3}\)πr³
So,
V = \(\frac{2}{3}\) × 3.14 × (6.5)³
= 574.88 inches³
≈ 575 inches³
Hence, from the above,
We can conclude that the volume of the given bowl is about 575 inches³

3-ACT MATH

3-Act Mathematical Modeling:
Measure Up

ACT 1
1. After watching the video, what is the first question that comes to mind?

Answer:

Question 2.
Write the Main Question you will answer.
Answer:

Question 3.
Construct Arguments Predict an answer to this Main Question. Explain your prediction.
Answer:

Question 4.
On the number line below, write a number that is too small to be the answer. Write a number that is too large.

Answer:

Question 5.
Plot your prediction on the same number line.
Answer:

ACT 2
Question 6.
What information in this situation would be helpful to know? How would you use that information?

Answer:

Question 7.
Use Appropriate Tools What tools can you use to solve the problem? Explain how you would use them strategically.
Answer:

Question 8.
Model with Math
Represent the situation using mathematics. Use your representation to answer the Main Question.
Answer:

Question 9.
What is your answer to the Main Question? Is it higher or lower than your prediction? Explain why.

Answer:

ACT 3
Question 10.
Write the answer you saw in the video.

Answer:

Question 11.
Reasoning Does your answer match the answer in the video? If not, what are some reasons that would explain the difference?
Answer:

Question 12.
Make Sense and Persevere Would you change your model now that you know the answer? Explain.

Answer:

ACT 3 Extension
Reflect
Question 13.
Model with Math
Explain how you used a mathematical model to represent the situation. How did the model help you answer the Main Question?
Answer:

Question 14.
Make Sense and Persevere When did you struggle most while solving the problem? How did you overcome that obstacle?

Answer:

SEQUEL
Question 15.
Generalize Suppose you have a graduated cylinder half the height of the one in the video. How wide does the cylinder need to be to hold the liquid in the flask?
Answer:

Topic 8 REVIEW

Topic Essential Question
How can you find volumes and surface areas of three-dimensional figures?
Answer:
The “Surface area” is the sum of the areas of all faces (or surfaces) on a 3D shape.
Ex:
A cuboid has 6 rectangular faces. To find the surface area of a cuboid, add the areas of all 6 faces
We know that,
The volume of a three-dimensional figure = Cross-sectional area × length

Vocabulary Review
Complete each definition and then provide an example of each vocabulary word.
Vocabulary
composite figure
cone
cylinder
sphere

Answer:

Use Vocabulary in Writing
Draw a composite figure that includes any two of the following: a cylinder, a cone, a sphere, and a hemisphere. Label each part of your drawing. Then describe each part of your composite figure. Use vocabulary terms in your description.

Concepts and Skills Review

Lesson 8.1 Find Surface Area of Three-Dimensional Figures

Quick Review
Surface area is the total area of the surfaces of a three-dimensional figure. The chart gives formulas for finding the surface area of a cylinder, a cone, and a sphere.

Example
What is the surface area of the cylinder? Use 3.14 for π.
Answer:

Practice
Question 1.
What is the surface area of the cone? Use 3.14 for π.

Answer:
The given figure is:

From the given figure,
We can observe that
The slant height of the cone (l) is: 13 m
The radius of the cone (r) is: 5 m
Now,
We know that,
The surface area of a cone (S.A) = πr² + πrl
So,
S.A = 3.14 × 5² + 3.14 × 5 × 13
= 78.5 + 204.1
= 282.6 m²
Hence, from the above,
We can conclude that the surface area of the given cone is: 282.6 m²

Question 2.
What is the surface area of the sphere in terms of π?

Answer:
The given figure is:

From the given figure,
We can observe that
The diameter of the sphere (d) is: 10 cm
Now,
We know that,
Radius (r) = \(\frac{Diameter}{2}\)
r = \(\frac{10}{2}\)
r = 5 cm
Now,
We know that,
The surface area of a sphere (S.A) = 4πr²
So,
S.A = 4 × 3.14 × 5²
= 314 cm²
Hence, from the above,
We can conclude that the surface area of the given sphere is: 314 cm²

Question 3.
What is the surface area of the cylinder in terms of π?

Answer:
The given figure is:

Now,
From the given figure,
We can observe that,
The diameter of the cylinder (d) is 12 in.
The height of the cylinder (h) is 15 in.
Now,
We know that,
Radius (r) = \(\frac{Diameter}{2}\)
r = \(\frac{12}{2}\)
r = 6 in.
Now,
We know that,
The surface area of a cylinder (S.A) = 2πr² + 2πrh
So,
S.A = 2 × 3.14 × 6² + 2 × 3.14 × 6 × 15
= 226.08 + 565.2
= 791.28 in.³
Hence, from the above,
We can conclude that the surface area of the given cylinder is: 791.28 in.³

Lesson 8.2 Find Volume of Cylinders

Quick Review
The volume of a cylinder is equal to the area of its base times its height.
V = area of base · height, or V = πr²h

Example
What is the volume of the cylinder? Use 3.14 for π.
Answer:

Practice
Question 1.
What is the volume of the cylinder in terms of π?

Answer:
The given figure is:

From the given figure,
We can observe that
The diameter of the cylinder (d) is: 2 m
The height of the cylinder (h) is: 6 m
Now,
We know that,
Radius (r) = \(\frac{Diameter}{2}\)
r = \(\frac{2}{2}\)
r = 1 m
Now,
We know that,
The volume of a cylinder (V) = πr²h
So,
V = 3.14 × 1² × 6
= 18.84 m³
Hence, from the above,
We can conclude that the volume of the given cylinder is: 18.84 m³

Question 2.
The volume of the cylinder is 141.3 cubic centimeters. What is the radius of the cylinder? Use 3.14 for π.

Answer:
It is given that
The volume of the cylinder is 141.3 cubic centimeters
Now,
The given figure is:

Now,
From the given figure,
We can observe that
The height of the cylinder (h) is: 5 cm
Now,
We know that,
The volume of a cylinder (V) = πr²h
So,
141.3 = 3.14 × r² × 5
r² = \(\frac{141.3}{3.14 × 5}\)
r²= 9
r = 3 cm
Hence, from the above,
We can conclude that the radius of the given cylinder is: 3 cm

Lesson 8.3 Find Volume of Cones

Quick Review
To find the volume of a cone, use the formula V = \(\frac{1}{3}\)πr²h.

Example
What is the volume of the cone? Use 3.14 for π.
Answer:

Practice
Question 1.
What is the volume of the cone in terms of π?

Answer:
The given figure is:

Now,
From the given figure,
We can observe that
The radius of the cone (r) is 3 in.
The height of the cone (h) is 8 in.
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × 3.14 × 3² × 8
= 75.36 in.³
Hence, from the above,
We can conclude that the volume of the given cone is: 75.36 in.³

Question 2.
What is the volume of the cone? Use 3.14 for π.

Answer:
The given figure is:

Now,
From the given figure,
We can observe that
The radius of the cone (r) is: 4 cm
The slant height of the cone (h) is: 5 cm
Now,
We know that,
l² = r² + h²
5² = 4² + h²
h² = 5² – 4²
h² = 9
h = 3 cm
Now,
We know that,
The volume of a cone (V) = \(\frac{1}{3}\)πr²h
So,
V = \(\frac{1}{3}\) × 3.14 × 4² × 3
= 50.24 cm³
Hence, from the above,
We can conclude that the volume of the given cone is: 50.24 cm³

Lesson 8.4 Find Volume of Spheres

Quick Review
To find the volume of a sphere, use the formula V = \(\frac{4}{3}\)πr³

Example
Find the volume of the composite figure. Use 3.14 for π.

Answer:
First, find the volume of the sphere. Use 3.14 for π.
V = \(\frac{4}{3}\)πr³
= \(\frac{4}{3}\)π(3.5)³ → Substitute 3.5 for r.
= 57.17π ≈ 179.5 cm³
Divide by 2 to find the volume of the hemisphere: 179.5 ÷ 2 ≈ 89.75 cubic centimeters.
Then, find the volume of the cone. Use 3.14 for π.
V = \(\frac{1}{3}\)πr² h
= \(\frac{1}{3}\)π(3.5)²(14) → Substitute 3.5 for r and 14 for h.
= 57.171 ≈ 179.5 cm³
The volume of the composite figure is approximately 89.75 + 179.5 ≈ 269.25 cubic centimeters.

Practice
Question 1.
What is the volume of the sphere? Use \(\frac{22}{7}\) for π.

Answer:
The given figure is:

Now,
From the given figure,
We can observe that
The radius of the sphere (r) is: 14 cm
Now,
We know that,
The volume of a sphere (V) = \(\frac{4}{3}\)πr³
So,
V = \(\frac{4}{3}\) × \(\frac{22}{7}\) × 14³
= 11,498.6 cm³
Hence, from the above,
We can conclude that the volume of the given sphere is: 11,498.6 cm³

Question 2.
The surface area of a sphere is 1,017.36 square inches. What is the volume of the sphere? Use \(\frac{22}{7}\) for π.
Answer:
It is given that
The surface area of a sphere is 1,017.36 square inches.
Now,
We know that,
The surface area of a sphere (S.A) = 4πr²
So,
4πr² = 1,017.36
r² = \(\frac{1,017.36}{4π}\)
r² = 80.92
r = 8.99 inches
Now,
We know that,
The volume of a sphere (V) = \(\frac{4}{3}\)πr³
So,
V = \(\frac{4}{3}\) × \(\frac{22}{7}\) × (8.99)³
= 3,044.68 inches³
Hence, from the above,
We can conclude that the volume of the given sphere is: 3,044.68 inches³

Question 3.
What is the volume of the composite figure? Use \(\frac{22}{7}\) for π.

Answer:
The given figure is:

Now,
From the above,
We can observe that the given figure is a combination of a hemisphere and a cone
So,
The volume of the given figure = The volume of a hemisphere + The volume of a cone
Now,
From the given figure,
We can observe that
Diameter (d) = 4 cm
Height (h) = 10 cm
Now,
We know that,
Radius (r) = \(\frac{Diameter}{2}\)
r = \(\frac{4}{2}\)
r = 2 cm
Now,
We know that,
The volume of a hemisphere (V) = \(\frac{2}{3}\)πr³
So,
V = \(\frac{2}{3}\) × \(\frac{22}{7}\) × 2³
= 16.76 cm³
Now,
We know that,
The volume of a cone (V’) = \(\frac{1}{3}\)πr²h
So,
V’ = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 2² × 10
= 41.90 cm³
So,
The volume of the given figure = 41.90 + 16.76
= 58.66 cm³
Hence, from the above,
We can conclude that the volume of the given figure is: 58.66 cm³

Topic 8 Fluency Practice

Hidden Clue
For each ordered pair, solve the equation to find the unknown coordinate. Then locate and label the corresponding point on the graph. Draw line segments to connect the points in alphabetical order. Use the completed picture to help answer the riddle below.

What do squares, triangles, pentagons, and octagons have in common?

Answer:
Step 1:
For each ordered pair, solve the equation to find the unknown coordinate.

Step 2:
Then locate and label the corresponding point on the graph.
Step 3:
Draw line segments to connect the points in alphabetical order.

Step 4:
Use the completed picture to help answer the riddle below.
All the squares, triangles, pentagons, and octagons have in common the sum of all the total angles i.e., 360°

Go through the enVision Math Common Core Grade 8 Answer Key Topic 7 Understand and Apply the Pythagorean Theorem and finish your homework or assignments.

enVision Math Common Core 8th Grade Answers Key Topic 7 Understand And Apply The Pythagorean Theorem

Topic Essential Question
How can you use the Pythagorean Theorem to solve problems?
Answer:
The Pythagorean Theorem is used to calculate the steepness of slopes of hills or mountains. A surveyor looks through a telescope toward a measuring stick a fixed distance away, so that the telescope’s line of sight and the measuring stick form a right angle.

3-ACT MATH OOO

Go with the Flow
You may have noticed that when you double the base and the height of a triangle, the area is more than doubled. The same is true for doubling the sides of a square or the radius of a circle. So what is the relationship? Think about this during the 3-Act Mathematical Modeling lesson.

Topic 7 enVision STEM Project

Did You Know?
Over two billion people will face water shortages by 2050 according to a 2015 United Nations Environment Program report.

Rainwater can be collected and stored for use in irrigation, industrial uses, flushing toilets, washing clothes and cars, or it can be purified for use as everyday drinking water.
This alternative water source reduces the use of fresh water from reservoirs and wells.

Using water wisely saves money on water and energy bills and extends the life of supply and wastewater facilities.

Roofs of buildings or large tarps are used to collect rainwater.
A rainwater collection system for a building roof that measures 28 feet by 40 feet can provide 700 gallons of water-enough water to support two people for a year—from a rainfall of 1.0 inch.

Even a 5 foot by 7-foot tarp can collect 2 gallons of water from a rainfall total of only 0.1 in.

The rainwater harvesting market is expected to grow 5% from 2016 to 2020.

Your Task: Rainy Days

Rainwater collection is an inexpensive way to save water in areas where it is scarce. One inch of rain falling on a square roof with an area of 100 ft² collects 62 gallons of water that weighs over 500 pounds. You and your classmates will research the necessary components of a rainwater collection system. Then you will use what you know about right triangles to design a slanted roof system that will be used to collect rainwater.
Answer:
It is given that
Rainwater collection is an inexpensive way to save water in areas where it is scarce. One inch of rain falling on a square roof with an area of 100 ft² collects 62 gallons of water that weighs over 500 pounds
Now,
The necessary components of a rainwater collection system are:
A) Catchments B) Coarse mesh C) Gutters D) Conduits E) First-flushing F) Filter G) Storage facility H) Recharge Structures

Topic 7 GET READY!

Review What You Know!

Vocabulary
Choose the best term from the box to complete each definition.
cube root
diagonal
isosceles triangle
perimeter
right triangle
square root

Question 1.
The __________ of a number is a factor that when multiplied by itself gives the number.
Answer:
We know that,
The “Square root” of a number is a factor that when multiplied by itself gives the number
Hence, from the above,
We can conclude that the best term to complete the given definition is a “Square root”

Question 2.
A _________ is a line segment that connects two vertices of a polygon and is not the side.
Answer:
We know that,
A “Diagonal” is a line segment that connects two vertices of a polygon and is not the side
Hence, from the above,
We can conclude that the best term to complete the given definition is a “Diagonal”

Question 3.
The _________ of a figure is the distance around it.
Answer:
We know that,
The “Perimeter” of a figure is the distance around it
Hence, from the above,
We can conclude that the best term to complete the given definition is the “Perimeter”

Question 4.
A ___________ is a triangle with one right angle.
Answer:
We know that,
A “Right triangle” is a triangle with one right angle
Hence, from the above,
We can conclude that the best term to complete the given definition is a “Right angle”

Simplify Expressions with Exponents

Simplify the expression.
Question 5.
3² + 4²
Answer:
The given expression is: 3² + 4²
So,
3² + 4²
= (3 × 3) + (4 × 4)
= 9 + 16
= 25

Question 6.
2² + 5²
Answer:
The given expression is: 2² + 5²
So,
2² + 5²
= (2 × 2) + (5 × 5)
= 4 + 25
= 29

Question 7.
10² – 8²
Answer:
The given expression is: 10² – 8²
So,
10² – 8²
= (10 × 10) – (8 × 8)
= 100 – 64
= 36

Square Roots

Determine the square root.
Question 8.
\(\sqrt {81}\)
Answer:
The given expression is: \(\sqrt{81}\)
Hence,
\(\sqrt{81}\) = 9

Question 9.
\(\sqrt {144}\)
Answer:
The given expression is: \(\sqrt{144}\)
Hence,
\(\sqrt{144}\) = 12

Question 10.
\(\sqrt {225}\)
Answer:
The given expression is: \(\sqrt{225}\)
Hence,
\(\sqrt{225}\) = 15

Distance on a Coordinate Plane

Determine the distance between the two points.
Question 11.

Answer:
The given graph is:

From the given graph,
The given points are: (2, 5), (7, 5)
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
We know that,
Distance between 2 points = √(x₂ – x₁)² + (y₂ – y₁)²
= √(7 – 2)² + (5 – 5)²
= \(\sqrt{5²}\)
= 5 units
Hence, from the above,
We can conclude that the distance between the given points is: 5 units

Question 12.

Answer:
The given graph is:

From the given graph,
The given points are: (3, 2), (3, 9)
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
We know that,
Distance between 2 points =√(x₂ – x₁)² + (y₂ – y₁)²
= √(3 – 3)² + (9 – 2)²
= \(\sqrt{7²}\)
= 7 units
Hence, from the above,
We can conclude that the distance between the given points is: 7 units

Language Development

Complete the word map using key terms, examples, or illustrations related to the Pythagorean Theorem and its Converse.


Answer:

Topic 7 PICK A PROJECT

PROJECT 7A
Where would you like to bike ride in your neighborhood?
PROJECT: PLAN A METRIC CENTURY RIDE

PROJECT 7B
What designs have you seen on kites?
PROJECT: BUILD A KITE

PROJECT 7C
What buildings in your community have unusual shapes as part of their structure or design?
PROJECT: MAKE A SCRAPBOOK

PROJECT 7D
What geometric designs have you noticed on your clothes?
PROJECT: DESIGN A FABRIC TEMPLATE

3-ACT MATH

3-Act Mathematical Modeling: Go with the Flow

АСТ 1
Question 1.
After watching the video, what is the first question that comes to mind?
Answer:

Question 2.
Write the Main Question you will answer.
Answer:

Question 3.
Make a prediction to answer this Main Question.
_________ % will fit in the third square.
Answer:

Question 4.
Construct Arguments Explain how you arrived at your prediction.

Answer:

ACT 2
Question 5.
What information in this situation would be helpful to know? How would you use that information?
Answer:

Question 6.
Use Appropriate Tools What tools can you use to solve the problem? Explain how you would use them strategically.
Answer:

Question 7.
Model With Math
Represent the situation using mathematics. Use your representation to answer the Main Question.
Answer:

Question 8.
What is your answer to the Main Question? Does it differ from your prediction? Explain.

Answer:

АСТ 3
Question 9.
Write the answer you saw in the video.
Answer:

Question 10.
Reasoning Does your answer match the answer in the video? If not, what are some reasons that would explain the difference?
Answer:

Question 11.
Make Sense and Persevere Would you change your model now that you know the answer? Explain.

Answer:

Act 3
Extension
Reflect
Question 12.
Model with Math
Explain how you used a mathematical model to represent the situation. How did the model help you answer the Main Question?

Answer:

Question 13.
Reason Abstractly How did you represent the situation using symbols? How did you use those symbols to solve the problem?

Answer:

SEQUEL
Question 14.
Construct Arguments Explain why you can use an area formula when the problem involves comparing volumes.

Answer:

Lesson 7.1 Understand the Pythagorean Theorem

Explain It!
Kelly drew a right triangle on graph paper. Kelly says that the sum of the areas of squares with side lengths a and b is the same as the area of a square with side length c.

I can… use the Pythagorean Theorem to find unknown sides of triangles.

A. Do you agree with Kelly? Explain.
Answer:
It is given that
Kelly drew a right triangle on graph paper. Kelly says that the sum of the areas of squares with side lengths a and b is the same as the area of a square with side length c.
We know that,
According to the Pythagorean theorem,
Hypotenuse² = Side length 1² + Side length 2²
So,
From the given graph,
We can observe that
Side length 1 is: a
Side length 2 is: b
The hypotenuse is: c
So,
c² = a² + b²
Hence, from the above,
We can conclude that we can agree with Kelly

B. Sam drew a different right triangle with side lengths a = 5, b = 12, and c = 13. Is the relationship Kelly described true for Sam’s right triangle? Explain.
Answer:
It is given that
Sam drew a different right triangle with side lengths a = 5, b = 12, and c = 13
Now,
From part (a),
The relation according to Kelly is:
c² = a² + b²
So,
13² = 12² + 5²
169 = 144 + 25
169 = 169
Hence, from the above,
We can conclude that the relationship Kelly described is true for Sam’s right-angled triangle

Focus on math practices
Generalize Kelly draws another right triangle. What would you expect to be the relationship between the areas of the squares drawn on each side of the triangle? Explain.
Answer:
It is given that Kelly draws another right triangle
Hence,
If in a triangle, the square on one of the sides equals the sum of the squares on the remaining two sides of the triangle, then the angle contained by the remaining two sides of the triangle is right.”

Essential Question
How does the Pythagorean Theorem relate to the side lengths of a right triangle?
Answer:
The Pythagorean equation relates the sides of a right triangle in a simple way so that if the lengths of any two sides are known the length of the third side can be found. Another corollary of the theorem is that in any right triangle, the hypotenuse is greater than any one of the other sides but less than their sum.

Try It!

A right triangle has side lengths 15 centimeters, 25 centimeters, and 20 centimeters. How can you use the Pythagorean Theorem to write an equation that describes how the side lengths are related?
a² + b² = c²
_______² + ________² = _________²
_________ + _________ = ________
Answer:
It is given that
A right triangle has side lengths 15 centimeters, 25 centimeters, and 20 centimeters.
We know that,
According to the Pythagorean Theorem,
The hypotenuse has the greatest length in the right triangle
Now,
Let the hypotenuse be c
Let the other two sides be a and b
So,
From the given information,
c = 25 centimeters, a = 15 centimeters,and b = 20 centimeters
So,
According to the Pythagorean Theorem,
25² = 15² + 20²
625 = 225 + 400
625 = 625
Hence, from the above,
We can conclude that we proved how the Pythagorean Theorem relates to the lengths of the right triangle

Convince Me!
How do you know that the geometric proof of the Pythagorean Theorem shown above can be applied to all right triangles?
Answer:
It can be proven using the law of cosines or as follows: Let ABC be a triangle with side lengths a, b, and c, with a² + b² = c².  Therefore, the angle between the side of lengths a and b in the original triangle is a right angle. This proof of the converse makes use of the Pythagorean theorem itself.

Try It!

A right triangle has a hypotenuse length of 32 meters. It has one leg with a length of 18 meters. What is the length of the other leg? Express your answer as a square root.
Answer:
It is given that
A right triangle has a hypotenuse length of 32 meters. It has one leg with a length of 18 meters.
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the hypotenuse
a and b are the legs
Now,
Let the other leg be x
So,
32² = 18² + x²
x² = 32² – 18²
x² = 1024 – 324
x² = 700
x = \(\sqrt{700}\)
Hence, from the above,
We can conclude that the length of the other leg is: \(\sqrt{700}\)

KEY CONCEPT

The Pythagorean Theorem is an equation that relates the side lengths of a right triangle, a² + b² = c², where a and b are the legs of a right triangle and c is the hypotenuse.

Do You Understand?
Question 1.
Essential Question How does the Pythagorean Theorem relate to the side lengths of a right triangle?
Answer:
The Pythagorean equation relates the sides of a right triangle in a simple way so that if the lengths of any two sides are known the length of the third side can be found. Another corollary of the theorem is that in any right triangle, the hypotenuse is greater than any one of the other sides but less than their sum.

Question 2.
Use Structure A side of each of the three squares forms a side of a right triangle.
Would any three squares form the sides of a right triangle? Explain.

Answer:
It is given that
A side of each of the three squares forms a side of a right triangle.
Now,
We know that,
The length of all the sides in a square is equal
Now,
From the given figure,
We can observe that
Each side of a square from the three squares form a right triangle
Hence, from the above,
We can conclude that the three squares form the sides of a right triangle

Question 3.
Construct Arguments Xavier said the missing length is about 18.5 units. Without calculating, how can you tell that Xavier solved incorrectly?

Answer:
It is given that
Xavier said the missing length is about 18.5 units
Now,
We know that,
According to the Pythagorean Theorem,
The length of the hypotenuse is the greatest
Now,
The given right triangle is:

So,
According to the Pythagorean Theorem,
The length of the missing side should be greater than 21 and 26
Hence, from the above,
We can conclude that Xavier calculated incorrectly

Do You Know How?
Question 4.
A right triangle has leg lengths of 4 inches and 5 inches. What is the length of the hypotenuse? Write the answer as a square root and round to the nearest tenth of an inch.
Answer:
It is given that
A right triangle has leg lengths of 4 inches and 5 inches
Now,
We know that,
According to the Pythagorean theorem,
c² = a²+ b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
c² = 4² + 5²
c² = 16 + 25
c² = 41
c = \(\sqrt{41}\)
c = 6.4 inches
Hence, from the above,
We can conclude that the length of the hypotenuse is: 6.4 inches

Question 5.
Find the missing side length to the nearest tenth of afoot.

Answer:
The given right triangle is:

Now,
We know that,
According to the Pythagorean theorem,
c² = a²+ b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
14² = 8² + b²
b²= 14² – 8²
b² = 196 – 64
b² = 132
b = \(\sqrt{132}\)
b = 11.5 feet
Hence, from the above,
We can conclude that the length of the missing side is: 11.5 feet

Question 6.
Find the missing side length to the nearest tenth of a millimeter.

Answer:
The given right triangle is:

Now,
We know that,
According to the Pythagorean theorem,
c² = a²+ b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
c² = (3.7)² + (7.5)²
c² = 13.69 + 56.25
c² = 69.94
c = \(\sqrt{69.94}\)
c = 8.4 mm
Hence, from the above,
We can conclude that the length of the missing side is: 8.4 mm

Practice & Problem Solving

Leveled Practice In 7 and 8, find the missing side length of each triangle.
Question 7.

The length of the hypotenuse is ________ units.
Answer:

Question 8.

The length of leg b is about ________ inches.
Answer:

Question 9.
What is the length of the hypotenuse of the triangle when x = 15? Round your answer to the nearest tenth of a unit.

Answer:
The given right angle is:

Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
c² = (3x)² + (4x + 4)²
Now,
When x = 15,
c² = (3 × 15)² + (4 × 15 + 4)²
c² = 45² + 64²
c² = 2,025 + 4,096
c² = 6,121
c = \(\sqrt{6,121}\)
c = 78.2 units
Hence, from the above,
We can conclude that the length of the hypotenuse when x= 15 is: 78.2 units

Question 10.
What is the length of the missing side rounded to the nearest tenth of a centimeter?

Answer:
The given right triangle is:

Now,
We know that,
According to the Pythagorean Theorem,
c² = a²+ b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
a² = (12.9)² + (15.3)²
a² = 166.41 + 234.09
a² = 400.5
a = \(\sqrt{400.5}\)
a = 20 cm
Hence, from the above,
We can conclude that the length of the side a is: 20 cm

Question 11.
Use the Pythagorean Theorem to find the unknown side length of the right triangle.

Answer:
The given right triangle is:

Now,
We know that,
According to the Pythagorean Theorem,
c² = a²+ b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
c² = (10)² + (24)²
c² = 100 + 576
c² = 676
c = \(\sqrt{676}\)
c = 26 m
Hence, from the above,
We can conclude that the length of the side a is: 26 m

Question 12.
What is the length of the unknown leg of the right triangle rounded to the nearest tenth of afoot?

Answer:
The given right triangle is:

Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
9² = 2² + b²
b² = 9² – 2²
b² = 81 – 4
b² = 77
b = \(\sqrt{77}\)
b = 8.8 foot
Hence, from the above,
We can conclude that the length of the unknown leg is: 8.8 foot

Question 13.
A student is asked to find the length of the hypotenuse of a right triangle. The length of one leg is 32 centimeters, and the length of the other leg is 26 centimeters. The student incorrectly says that the length of the hypotenuse is 7.6 centimeters.
a. Find the length of the hypotenuse of the right triangle to the nearest tenth of a centimeter.
Answer:
It is given that
A student is asked to find the length of the hypotenuse of a right triangle. The length of one leg is 32 centimeters, and the length of the other leg is 26 centimeters. The student incorrectly says that the length of the hypotenuse is 7.6 centimeters.
Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
c² = 32² + 26²
c² = 1,024 + 676
c² = 1,700
c = \(\sqrt{1,700}\)
c = 41.2 centimeters
Hence, from the above,
We can conclude that the length of the hypotenuse of a right triangle is: 41.2 centimeters

b. What mistake might the student have made?
Answer:
It is given that
The student incorrectly says that the length of the hypotenuse is 7.6 centimeters.
But,
From part (a),
The length of the hypotenuse is: 41.2 centimeters
Hence, from the above,
We can conclude that the mistake the student might make is the misinterpretation of the length of the hypotenuse

Question 14.
Find the length of the unknown leg of the right triangle.

Answer:
The given right triangle is:

Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
(37.25)² = (12.25)² + b²
b² = (37.25)² – (12.25)²
b² = 1,387.56 – 150.06
b² = 1,237.5
b = \(\sqrt{1,237.5}\)
b = 35.17 units
Hence, from the above,
We can conclude that the length of the unknown leg is: 35.17 units

Question 15.
Higher-Order Thinking A right triangle has side lengths 12 centimeters and 14 centimeters. Name two possible side lengths for the third side, and explain how you solved for each.
Answer:
It is given that
A right triangle has side lengths of
12 centimeters and 14 centimeters.
Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
Now,
Let the length of the third side be x
So,
The possible lengths of the third side may be:
x < 12 centimeters and x > 14 centimeters
Hence, from the above,
We can conclude that the two possible side lengths for the third side are:
x < 12 centimeters and x > 14 centimeters

Assessment Practice
Question 16.
Which right triangle has a hypotenuse that is about 39 feet long?

Answer:
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
For Options A and D:
A) c² = 30² + 15²                     D) c² = 30² + 14²
= 33.54 ft                                 = 33.10 ft
For Options B and C:
B) c² = 36² + 12²                      C) c² = 36² + 15²
= 37.94 ft                                    = 39 ft
Hence, from the above,
We can conclude that Option C matches the given situation

Question 17.
Which right triangle does NOT have an unknown leg length of about 33 cm?

Answer:
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
For Options A and D:
A) b² = 35² – 11²                     D) b² = 34² – 9²
= 33.22 cm                              = 32.78 cm
For Options B and C:
B) b² = 30² – 10²                      C) b² = 35² – 12²
= 28.28 cm                              = 32.87 cm
Hence, from the above,
We can conclude that Option B matches the given situation

Lesson 7.2 Understand the Converse of the Pythagorean Theorem

Solve & Discuss It!
Kayla has some straws that she will use for an art project. She wants to glue three of the straws onto a sheet of paper, without overlapping, to make the outline of a right triangle. Which three straws could Kayla use to make a right triangle? Explain.

I can… use the Converse of the Pythagorean Theorem to identify right triangles.
Answer:
It is given that
Kayla has some straws that she will use for an art project. She wants to glue three of the straws onto a sheet of paper, without overlapping, to make the outline of a right triangle.
Now,
We know that,
The converse of the Pythagorean Theorem states that if the square of the third side of a triangle is equivalent to the sum of its two shorter sides, then it must be a right triangle i.e.,
If c² = a²+ b², then the given triangle is a right triangle
So,
From the given straws,
we can observe that
The straws numbered 3, 4, and 5 can be glued to make the outline of a right triangle
The straws numbered 5, 12, and 13 can be glued to make the outline of a right triangle
Hence, from the above,
We can conclude that there are 2 pairs of straws i.e., (3, 4, 5) and (12, 5, 13) to make the outline of a right triangle

Look for Relationships
How could you use the Pythagorean Theorem to determine whether the lengths form a right triangle?
Answer:
According to the converse of the Pythagorean Theorem,
If the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.

Focus on math practices
Use Structure Could Kayla use the straws that form a right triangle to make a triangle that is not a right triangle? Explain.
Answer:
From the given straws,
We can observe that the pair (6, 7, 4) can’t form a right triangle but (3, 4, 5) can form a right triangle
Hence, from the above,
We can conclude that Kayla can use the straws that form a right triangle to make a triangle that is not a right triangle

Essential Question
How can you determine whether a triangle is a right triangle?
Answer:
We can determine the triangle is a right triangle by using the converse of the Pythagorean Theorem
Hence,
According to the converse of the Pythagorean Theorem,
If the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.

Try It!

A triangle has side lengths 4 inches, 5 inches, and 7 inches. Is the triangle a right triangle?

Is a² + b² equal to c²? _______
Is the triangle a right triangle? _________
Answer:
It is given that
A triangle has side lengths 4 inches, 5 inches, and 7 inches.
Now,
We know that,
According to the converse of the Pythagorean Theorem,
If the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.
So,
c² = a² + b²
Where,
c is the hypotenuse that has the longest side length
a and b are the lengths of the legs
So,
7² = 4² + 5²
49 = 16 + 25
49 = 41
So,
49 ≠ 41
Hence, from the above,
We can conclude that
c² ≠ a²+ b²
The given triangle is not a right triangle

Convince Me!
Explain the proof of the Converse of the Pythagorean Theorem in your own words.
Answer:
The converse of the Pythagorean Theorem is:
If the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.

Try It!

A triangle has side lengths 10 feet, \(\sqrt {205}\) feet, and \(\sqrt {105}\) feet. Is this a right triangle? Explain.
Answer:
It is given that
A triangle has side lengths 10 feet, \(\sqrt {205}\) feet, and \(\sqrt {105}\) feet.
Now,
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b², then
The given triangle is a right triangle
We know that,
c is the length of the hypotenuse that has the longest side in a right triangle
So,
(\(\sqrt{205}\))² =(\(\sqrt{105}\))² + 10²
205 = 105 + 100
205 = 205
So,
c² = a² + b²
Hence, from the above,
We can conclude that the triangle with the given side lengths is a right triangle

Try It!

A triangle is inside a trapezoid. Is the triangle a right triangle? Explain.

Answer:
It is given that a triangle is inside a trapezoid
Now,
The given trapezoid is:

From the given trapezoid,
The sides of the triangle are: 17 in., 15 in., \(\sqrt{514}\) in.
Now,
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²,
then, the given triangle is a right triangle
So,
(\(\sqrt{514}\))² = 17² + 15²
514 = 289 + 225
514 = 514
So,
c² = a² + b²
Hence, from the above,
We can conclude that the triangle that is in the trapezoid is a right triangle

KEY CONCEPT

The Converse of the Pythagorean Theorem states that if the sum of the squares of the lengths of two sides of a triangle is equal to the square of the length of the third side, the triangle is a right triangle.

Do You Understand?
Question 1.
Essential Question How can you determine whether a triangle is a right triangle?
Answer:
We can determine the triangle is a right triangle by using the converse of the Pythagorean Theorem
Hence,
According to the converse of the Pythagorean Theorem,
If the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.

Question 2.
Construct Arguments A triangle has side lengths of 3 centimeters, 5 centimeters, and 4 centimeters. Abe used the Converse of the Pythagorean Theorem to determine whether it is a right triangle.
3² + 5² \(\underline{\underline{?}}\) 4²
9 + 25 \(\underline{\underline{?}}\) 16
34 ≠ 16
Abe concluded that it is not a right triangle. Is Abe correct? Explain.
Answer:
It is given that
A triangle has side lengths of 3 centimeters, 5 centimeters, and 4 centimeters
Now,
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²,
then, the triangle is  a right triangle
We know that,
The hypotenuse has the length of the longest side
So,
5²= 3² + 4²
25 = 9 + 16
25 = 25
So,
c² = a² + b²
Hence, from the above,
We can conclude that the given triangle is a right triangle and Abe is not correct

Question 3.
Use Structure When you are given three side lengths for a triangle, how do you know which length to substitute for a, b, or c in the Pythagorean Theorem?
Answer:
First, we will find the squares of all lengths, then we will check which two squares of sides are equal to the square of the third side as per Pythagoras theorem. Hence the two sides would according be a and b and the third side will become c

Do You Know How?
Question 4.
Is the triangle a right triangle? Explain.

Answer:
The given triangle is:

We know that,
According to the converse of the Pythagorean Theorem,
If c² = a² + b²,
then, the triangle is a right triangle
So,
10² = 6² + 8²
100 = 36 + 64
100 = 100
So,
c² = a² + b²
Hence, from the above,
We can conclude that the given triangle is a right triangle

Question 5.
Is the triangle a right triangle? Explain.

Answer:
The given triangle is:

We know that,
According to the converse of the Pythagorean Theorem,
If c² = a² + b²,
then, the triangle is a right triangle
So,
8² = (\(\sqrt{26}\))² + (\(\sqrt{28}\))²
64 = 26 + 28
64 = 54
So,
c² ≠ a² + b²
Hence, from the above,
We can conclude that the given triangle is not a right triangle

Question 6.
Is the purple triangle a right triangle? Explain.

Answer:
The given triangle is:

We know that,
According to the converse of the Pythagorean Theorem,
If c² = a² + b²,
then, the triangle is a right triangle
So,
(20.8)² = 14² + (15.5)²
432.64 = 196 + 240.25
432.64 = 436.25
So,
c² ≠ a² + b²
Hence, from the above,
We can conclude that the purple triangle is not a right triangle

Practice & Problem Solving

Leveled Practice In 7 and 8, determine whether each triangle is a right triangle.
Question 7.

Is the triangle a right triangle? ________
Answer:

Hence, from the above,
We can conclude that the given triangle is not a right triangle

Question 8.

Is the triangle a right triangle? _________
Answer:

Hence, from the above,
We can conclude that the given triangle is a right triangle

Question 9.
Can the sides of a right triangle have lengths 5, 15, and \(\sqrt {250}\)? Explain.
Answer:
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²,
then, the triangle is a right triangle
The hypotenuse has the longest side length
So,
(\(\sqrt{250}\))² = 5² + 15²
250 = 25 + 225
250 = 250
Hence, from the above,
We can conclude that the sides of a right triangle have lengths 5, 15, and \(\sqrt {250}\)

Question 10.
Is ∆PQR a right triangle? Explain.

Answer:
The given triangle is:

We know that,
According to the converse of the Pythagorean Theorem,
If c² = a² + b²,
then, the triangle is a right triangle
So,
(6.25)² = (3.75)² + 5²
39.0625 = 14.0625 + 25
39.0625 = 39.0625
So,
c² = a² + b²
Hence, from the above,
We can conclude that ΔPQR is a right triangle

Question 11.
The green triangle is set inside a rectangle. Is the green triangle a right triangle? Explain.

Answer:
It is given that
The green triangle is set inside a rectangle
Now,
The given figure is:

We know that,
According to the converse of the Pythagorean Theorem,
If c² = a² + b²,
then, the triangle is a right triangle
So,
21² = (\(\sqrt{282}\))² + (\(\sqrt{159}\))²
441 = 282 + 159
441 = 441
So,
c² = a² + b²
Hence, from the above,
We can conclude that the given green triangle is a right triangle

Question 12.
The side lengths of three triangles are shown. Which of the triangles are right triangles?

Answer:
It is given that
The side lengths of the three triangles are shown in the table
Now,
The given table is:

Now,
We know that,
According to the converse of the Pythagorean Theorem,
If c² = a² + b²,
then, the triangle is a right triangle
So
For Triangle 1,
(\(\frac{5}{7}\))² = (\(\frac{3}{7}\))² + (\(\frac{4}{7}\))²
25 = 16 + 9
25 = 25
For Triangle 2,
15² = 8² + 8²
225 = 64 + 64
225 = 128
For Triangle 3,
(\(\frac{13}{17}\))² = (\(\frac{12}{17}\))² + (\(\frac{5}{17}\))²
169 = 144 + 25
169 = 169
So,
The condition
c² = a² + b²
is true for the Triangles 1 and 3
Hence, from the above,
We can conclude that Triangle 1 and Triangle 3 is a right triangle

Question 13.
Construct Arguments Three students draw triangles with the side lengths shown. All three say that their triangle is a right triangle. Which students are incorrect? What mistake might they have made?
Student 1: 22, 33, 55
Student 2: 44, 33, 77
Student 3: 33, 44, 55
Answer:
It is given that
Three students draw triangles with the side lengths shown. All three say that their triangle is a right triangle.
Now,
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²,
then, the triangle is a right triangle
Now,
For student 1,
55² = 33² + 22²
3,025 = 1,089 + 484
3,025 = 1,573
For student 2,
77²= 44² + 33²
5,929 = 1,936 + 1,089
5,929 = 3,025
For student 3,
55² = 44² + 33²
3,025 = 1,936 + 1,089
3,025 = 3,025
So,
The condition
c² = a² + b²
is false for the side lengths of the triangles that are drawn by students 1 and 2
Hence, from the above,
We can conclude that student 1 and student 2 are incorrect

Question 14.
Model with Math
∆JKL is an isosceles triangle. Is KM the height of ∆JKL? Explain.

Answer:
It is given that
ΔJKL is an isosceles triangle
Now,
To find whether KM is the height of ΔJKL,
Find out whether ΔKLM is a right triangle or not
Now,
The given triangle is:

Now,
We know that,
According to the converse of the Pythagorean Theorem,
c² = a²+ b²
So,
From ΔKLM,
(\(\sqrt{340}\))² = 13² + 14²
340 = 169 + 196
340 = 365
So,
c² ≠ a² + b²
So,
ΔKLM is not a right triangle
Hence, from the above,
We can conclude that KM is not the height of ΔJKL

Question 15.
Higher-Order Thinking The side lengths of three triangles are given.
Triangle 1: \(\sqrt{229}\) units, \(\sqrt{225}\) units, 22 units
Triangle 2: \(\sqrt{11 \frac{1}{3}}\) units, \(\sqrt{13 \frac{2}{3}}\) units, 5 units
Triangle 3: 16 units, 17 units, \(\sqrt{545}\) units
a. Which lengths represent the side lengths of a right triangle?
Answer:
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²
then, the triangle is a right triangle
Now,
For Triangle 1,
(\(\sqrt{229}\))² = (\(\sqrt{225}\))² + 22²
229 = 225 + 484
229 = 709
For Triangle 2,
(\(\sqrt{\frac{41}{3}}\))² + (\(\sqrt{\frac{34}{3}}\))² = 5²
\(\frac{41}{3}\) + \(\frac{34}{3}\) = 25
25 = 25
For Triangle 3,
(\(\sqrt{545}\))² = 16² + 17²
545 = 256 + 289
545 = 545
So,
The condition
c² = a²+ b²
is true for triangle 2 and triangle 3
Hence, from the above,
We can conclude that Triangle 2 and Triangle 3 represent the side lengths of a triangle

b. For any triangles that are not right triangles, use two of the sides to make a right triangle.
Answer:

Assessment Practice
Question 16.
Which shaded triangle is a right triangle? Explain.

Answer:
The given figure is:

Now,
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a²+ b²,
then the triangle is a right triangle
Now,
For ΔABC,
144² = 63² + (\(\sqrt{9}\))²
20,736 = 3,969 + 9
20,736 = 3,978
For ΔXYZ,
(\(\sqrt{144}\))² = (\(\sqrt{63}\))² + 9²
144 = 63 + 81
144 = 144
So,
The condition c² = a²+ b² is true for the shaded triangle XYZ
Hence, from the above,
We can conclude that the shaded triangle ΔXYZ is a right triangle

Question 17.
Which triangle is a right triangle?

A. Triangle I only
B. Triangle II only
C. Triangle I and Triangle II
D. Neither Triangle I nor Triangle II
Answer:
The given triangles are:

Now,
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²,
then the triangle is a right triangle
Now,
For Triangle I,
52² = 40² + 48²
2,704 = 1,600 + 2,304
2,704 = 3,904
For Triangle II,
65² = 60² + 25²
4,225 = 3,600 + 625
4,225 = 4,225
Hence, from the above,
We can conclude that Triangle II only is a right triangle

Topic 7 MID-TOPIC CHECKPOINT

Question 1.
Vocabulary How are the hypotenuse and the legs of a right triangle related? Lesson 7-1
Answer:
The relation between the sides and angles of a right triangle is the basis for trigonometry. The side opposite the right angle is called the hypotenuse. The sides adjacent to the right angle are called legs

Question 2.
Given that ∆QPR has side lengths of 12.5 centimeters, 30 centimeters, and 32.5 centimeters, proves ∆QPR is a right triangle. Lesson 7-2
Answer:
It is given that
∆QPR has side lengths of 12.5 centimeters, 30 centimeters, and 32.5 centimeters
Now,
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²,
then, the triangle is a right triangle
So,
(32.5)² = (12.5)² + 30²
1,056.25 = 156.25 + 900
1,056.25 = 1,056.25
So,
The condition
c² = a² + b²
is true for the given side lengths of a triangle
Hence, from the above,
We can conclude that ΔQPR is a right triangle

Question 3.
Ella said that if she knows the lengths of just two sides of any triangle, then she can find the length of the third side by using the Pythagorean Theorem. Is Ella correct? Explain. Lesson 7-1
Answer:
Ella said that if she knows the lengths of just two sides of any triangle, then she can find the length of the third side by using the Pythagorean Theorem.
Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the hypotenuse
a and b are the legs
Now,
If we know a and b, then we can find c
If we know b and c, then we can find a
If we know a and c, then we can find b
Hence, from the above,
We can conclude that Ella is correct

Question 4.
Find the unknown side length. Round to the nearest tenth. Lesson 7-1

Answer:
The given triangle is:

From the given triangle,
We can observe that it is a right triangle
So,
Now,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the hypotenuse
a and b are the legs
Now,
8² = 4² + b²
64 = 16 + b²
b² = 64 – 16
b² = 48
b = \(\sqrt{48}\)
b = 6.9 cm
Hence, from the above,
We can conclude that the length of the unknown side is: 6.9 cm

Question 5.
The height of a shed is 6 m. A ladder leans against the shed with its base 4.5 m away, and its top just reaching the roof. What is the length of the ladder? Lesson 7-1

Answer:
It is given that
The height of a shed is 6 m. A ladder leans against the shed with its base 4.5 m away, and its top just reaching the roof.
Now,
From the given figure,
We can observe that the ladder looks like the hypotenuse of a right triangle for the shed
Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the hypotenuse
a and b are the legs
In this situation,
c is the length of the ladder
a is the height of the shed
b is the length of the base
So,
c² = 6² + (4.5)²
c² = 36 + 20.25
c² = 56.25
c = \(\sqrt{56.25}\)
c = 7.5 m
Hence, from the above,
We can conclude that the length of the ladder is: 7.5 m

Question 6.
Select all the sets of lengths that could represent the sides of a right triangle. Lesson 7-2
☐ 5 cm, 10 cm, 15 cm
☐ 7 in., 14 in., 25 in.
☐ 13 m, 84 m, 85 m
☐ 5 ft, 11 ft, 12 ft
☐ 6ft, 9 ft, \(\sqrt {117}\) ft
Answer:
Let the given options be named as A, B, C, D, and E
Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the hypotenuse and has the longest length
a and b are the legs
So,
A)
15² = 5² + 10²
225 = 25 + 100
225 ≠ 125
B)
25² = 7² + 14²
625 = 49 + 196
625 ≠ 245
C)
85² = 84² + 13²
7,225 = 7,056 + 169
7,225 = 7,225
D)
12² = 11² + 5²
144 = 121 + 25
144 ≠ 146
E)
(\(\sqrt{117}\))² = 6² + 9²
117 = 36 + 81
17 = 117
Hence, from the above,
We can conclude that the side lengths present in options C and E represent the side lengths of a right triangle

Topic 7 MID-TOPIC PERFORMANCE TASK

Javier is standing near a palm tree. He holds an electronic tape measure near his eyes and finds the three distances shown.

PART A
Javier says that he can now use the Pythagorean Theorem to find the height of the tree. Explain. Use vocabulary terms in your explanation.
Answer:
It is given that
Javier says that he can now use the Pythagorean Theorem to find the height of the tree
Now,
According to the Pythagorean Theorem,
In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“. The sides of this triangle have been named Perpendicular (The height of the tree), Base, and Hypotenuse.

PART B
Find the height of the tree. Round to the nearest tenth. Show your work.

Answer:
From the given figure,
We can observe that
To find the height of the tree, we have to find the perpendicular distances of the two right triangles
Now,
Let x be the perpendicular height of the first right triangle
Let y be the perpendicular height of the second right triangle
So,
The height of the tree = x + y
Now,
The side lengths of the first right triangle are: 25 ft, 7 ft, x ft
Now,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the hypotenuse
a and b are the side lengths
So,
25² = 7² + x²
x² = 625 – 49
x² = 576
x = \(\sqrt{576}\)
x = 24 ft
Now,
The side lengths of the second right triangle are: 9 ft, 7ft, y ft
So,
9² = 7² + y²
y² = 81 – 49
y² = 32
y = \(\sqrt{32}\)
y = 5.6 ft
So,
The height of the tree = 24 + 5.6
= 29.6 ft
Hence, from the above,
We can conclude that the height of the tree to the nearest tenth is: 29.6 ft

PART C
Javier moves backward so that his horizontal distance from the palm tree is 3 feet greater. Will the distance from his eyes to the top of the tree also be 3 feet greater? Explain.
Answer:
Yes it will be greater, he is moving back 3 feet so what you are doing is taking the leg (a²) and multiplying it by 3. Once you do, you see triangle 1 has double and so did triangle 2. Triangle 2 was originally 5.6 (rounded to 6) then went up to 26.1. Triangle 1 was originally 24 and went up to 74.7 (or 75)
Step-by-step explanation:
For Triangle 1 (when multiplied by 3):
We know that,
a² + b² = c²
7² + b² = 27²
49 + b² = 729
b² = 680
b  = \(\sqrt{680}\)
b = 26.1 ft
For Triangle 2:
We know that,
a² + b² = c²
7² + b² =75²
b² = 5576
b = \(\sqrt{5,576}\)
b = 74.7 ft

PART D
Could Javier change his horizontal distance from the tree so that the distance from his eyes to the top of the tree is only 20 feet? Explain.
Answer:
Yes, Javier can change his horizontal distance from the tree so that the distance from his eyes to the top of the tree is only 20 feet by moving forward 5 ft

Lesson 7.3 Apply the Pythagorean Theorem to Solve Problems

Solve & Discuss It!
Carlos is giving his friend in another state a new umbrella as a gift. He wants to ship the umbrella in a box he already has. Which box can Carlos use to ship the umbrella? Explain.

I can… use the Pythagorean Theorem to solve problems.
Answer:
It is given that
Carlos is giving his friend in another state a new umbrella as a gift. He wants to ship the umbrella in a box he already has.
Now,
From the given figure,
We can observe that the umbrella is 37.5 inches long.
Now,
If you observe, each box doesn’t have 37 inches wide, however, they are tridimensional figures, which means they have a certain height.
So,
The smaller box is too small to fit the umbrella.
But,
The medium box is perfect because it has 27 inches wide and 27 inches in height, which is enough to fit the umbrella. These are the best dimensions to send the umbrella.
Hence, from the above,
We can conclude that the box that can be used by Carlos to ship the umbrella has the dimensions 27 in × 14 in × 27 in

Make Sense and Persevere
How will the umbrella fit inside any of the boxes?
Answer:
We know that,
The box is a 3-d figure
So,
To fit the whole umbrella in the box, we have to put it in a diagonal manner i.e., like the hypotenuse of a right triangle

Focus on math practices
Construct Arguments Tim says that the diagonal of any of the boxes will always be longer than the sides. Is Tim correct? Explain.
Answer:
We know that,
If we consider a square or any 2-d figure or any 3-d figure that consists of 1 right angle,
then, the diagonal will divide a figure into 2 right triangles
We know that,
We can apply the Pythagorean Theorem for any right triangle
We know that,
In a right triangle,
The hypotenuse is the longest side
We know that,
The hypotenuse in a right triangle is considered as a diagonal in a figure that consists of 1 right angle
Hence, from the above,
We can conclude that Tim is correct

Essential Question
What types of problems can be solved using the Pythagorean Theorem?
Answer:
The Pythagorean theorem is a way of relating the leg lengths of a right triangle to the length of the hypotenuse, which is the side opposite the right angle. Even though it is written in these terms, it can be used to find any of the sides as long as you know the lengths of the other two sides

Try It!

What is the length of the diagonal, d, of a rectangle with length 19 feet and width 17 feet?
leg² + leg² = hypotenuse²
______ ² + _______² = d²
______ + _______ = d²
_______ = d²
________ ≈ d
Answer:
It is given that a rectangle has a length of 19 feet and a width of 17 feet
Now,
We know that,
In a rectangle,
If a diagonal is drawn, then it divides the rectangle into 2 right angles
So,
According to the Pythagorean Theorem,
d² = a² + b²
Where,
d is the diagonal or hypotenuse
a and b are the lengths of the legs
So,
d²= 19² + 17²
d² = 361 + 289
d² = 650
d = \(\sqrt{650}\)
d = 25.4 ft
Hence, from the above,
We can conclude that the length of the diagonal is: 25.4 ft

Convince Me!
If the rectangle were a square, would the process of finding the length of the diagonal change? Explain.
Answer:
We know that,
For any figure i.e., either 2-d figure or 3-d figure with 1 right angle, the diagonal will divide that figure into 2 right angles
We know that,
We will use the Pythagorean Theorem to find the length of any unknown side in the right triangle
Hence, from the above,
We can conclude that even if the rectangle were a square, the process of finding the length of the diagonal will not change

Try It!

A company wants to rent a tent that has a height of at least 10 feet for an outdoor show. Should they rent the tent shown at the right? Explain.

Answer:
It is given that
A company wants to rent a tent that has a height of at least 10 feet for an outdoor show.
Now,
The given figure is:

Now,
To find whether the tent should be rented or not,
We have to find the value of h
We can also observe that the triangle that contains the value of h is a right triangle
Now,
The base of the right triangle = \(\frac{24}{2}\)
= 12 ft
Now,
The side lengths of the right triangle are: h, 15 ft, 12 ft
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
So,
15² = 12² + h²
h² = 225 – 144
h² = 81
h = \(\sqrt{81}\)
h = 9 ft
But,
The given height of the tent is: 10 ft
So,
9 ft < 10 ft
Hence, from the above,
We can conclude that the company should not rent the tent

KEY CONCEPT

You can use the Pythagorean Theorem and its converse to solve problems involving right triangles.

Do You Understand?
Question 1.
Essential Question What types of problems can be solved using the Pythagorean Theorem?
Answer:
The Pythagorean theorem is a way of relating the leg lengths of a right triangle to the length of the hypotenuse, which is the side opposite the right angle. Even though it is written in these terms, it can be used to find any of the sides as long as you know the lengths of the other two sides

Question 2.
Look for Structure How is using the Pythagorean Theorem in a rectangular prism similar to using it in a rectangle?
Answer:
We know that,
The rectangular prism and the rectangle have at least 1 right angle
We know that,
If a 3-d or 2-d figure has 1 right angle, then the diagonal of that figure divides the figure into the right triangles
So,
If we have the right triangle, then we can use the Pythagorean Theorem irrespective of the overall shape of the figure

Question 3.
Construct Arguments Glen found the length of the hypotenuse of a right triangle using \(\sqrt{a^{2}+b^{2}}\). Gigi used \(\sqrt{(a+b)^{2}}\). Who is correct? Explain.
Answer:
It is given that
Glen found the length of the hypotenuse of a right triangle using \(\sqrt{a^{2}+b^{2}}\). Gigi used \(\sqrt{(a+b)^{2}}\).
Now,
We know that,
We can use the Pythagorean Theorem only for the right triangles
The condition for a triangle to become the right triangle is:
Hypotenuse² = Side² + Side²
c² = a² + b²
Hence, from the above,
We can conclude that Glen is correct

Do You Know How?
Question 4.
You are painting the roof of a boathouse. You are going to place the base of a ladder 12 feet from the boathouse. How long does the ladder need to be to reach the roof of the boathouse?

Answer:
It is given that
You are painting the roof of a boathouse. You are going to place the base of a ladder 12 feet from the boathouse.
Now,
From the figure,
We can observe that the roof of a boathouse, ladder, and the base form a right triangle
So,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the height of the boathouse
a is the length of the base
b is the length of the ladder
So,
35² = 12² + b²
b² = 1,225 – 144
b² = 1,081
b = \(\sqrt{1,081}\)
b = 32.9 ft
Hence, from the above,
We can conclude that the length of the ladder that is needed to reach the roof of the boathouse is: 32.9 ft

Question 5.
A box-shaped like a right rectangular prism measures 5 centimeters by 3 centimeters by 2 centimeters. What is the length of the interior diagonal of the prism to the nearest hundredth?
Answer:
It is given that
A box-shaped like a right rectangular prism measures 5 centimeters by 3 centimeters by 2 centimeters.
So,
The dimensions of a rectangular prism is: 5 cm × 3 cm × 2 cm
So,
The length of the rectangular prism is: 5 cm
The width of the rectangular prism is: 3 cm
The height of the rectangular prism is: 2 cm
Now,
We know that,
The length of the diagonal of the prism = \(\sqrt{Length^{2} + Width^{2} + Height^{2}}\)
So,
The length of the diagonal of the rectangular prism = \(\sqrt{5^{2} + 3^{2} + 2^{2}}\)
= \(\sqrt{25 + 9 + 4}\)
= \(\sqrt{38}\)
= 6.16 cm
Hence, from the above,
We can conclude that the length of the interior diagonal of the rectangular prism is: 6.16 cm

Question 6.
A wall 12 feet long makes a corner with a wall that is 14 feet long. The other ends of the walls are about 18.44 feet apart. Do the walls form a right angle? Explain.

Answer:
It is given that
A wall 12 feet long makes a corner with a wall that is 14 feet long. The other ends of the walls are about 18.44 feet apart
Now,
The given figure is:

From the given figure,
We can observe that the given situation forms the right triangle
Now,
We know that,
According to the Pythagorean Therem,
c² = a² + b²
So,
c² = 12² + 14²
c² = 144 + 196
c² = 340
c = \(\sqrt{340}\)
c = 18.44 feet
Hence, from the above,
We can conclude that the walls form a right angle

Practice & Problem Solving

Leveled Practice In 7 and 8, use the Pythagorean Theorem to solve.
Question 7.
You are going to use an inclined plane to lift a heavy object to the top of a shelving unit with a height of 6 feet. The base of the inclined plane is 16 feet from the shelving unit. What is the length of the inclined plane? Round to the nearest tenth of a foot.

The length of the inclined plane is about ________ feet.
Answer:

Question 8.
Find the missing lengths in the rectangular prism.

Answer:

Question 9.
A stainless steel patio heater is shaped like a square pyramid. The length of one side of the base is 19.8 inches. The slant height is 92.8 inches. What is the height of the heater? Round to the nearest tenth of an inch.
Answer:
It is given that
A stainless steel patio heater is shaped like a square pyramid. The length of one side of the base is 19.8 inches. The slant height is 92.8 inches
Now,
We know that,
The slant height is nothing but a length of the diagonal
Now,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the slant height
a is the length of the base of the steel patio heater
b is the height of the heater
So,
(92.8)² = (19.8)² + b²
b² = 8,611.84 – 392.04
c² = 8,219.8
c = \(\sqrt{8,219.8}\)
c = 90.6 inches
Hence, from the above,
We can conclude that the height of the heater is: 90.6 inches

Question 10.
Reasoning What is the measurement of the longest line segment in a right rectangular prism that is 16 centimeters long, 9 centimeters wide, and 7 centimeters tall? Round to the nearest tenth of a centimeter.
Answer:
It is given that
A right rectangular prism is 16 centimeters long, 9 centimeters wide, and 7 centimeters tall
So,
The dimensions of a right rectangular prism are: 16 cm × 9 cm × 7 cm
So,
The length of the right rectangular prism is: 16 cm
The width of the right rectangular prism is: 9 cm
The height of the right rectangular prism is: 7 cm
We know that,
The longest line segment in any 2-d or 3-d figure is a “Diagonal”
Now,
We know that,
The length of the diagonal of the right rectangular prism = \(\sqrt{Length^{2} + Width^{2} + Height^{2}}\)
So,
The length of the diagonal of the right rectangular prism = \(\sqrt{16^{2} + 9^{2} + 7^{2}}\)
= \(\sqrt{256 + 49 + 81}\)
= \(\sqrt{386}\)
= 19.64 cm
Hence, from the above,
We can conclude that the length of the longest line segment in the right rectangular prism is: 19.64 cm

Question 11.
Felipe is making triangles for a stained glass window. He made the design shown but wants to change it. Felipe wants to move the purple triangle to the corner. The purple piece has side lengths of 4.5 inches, 6 inches, and 7 inches. Can the purple piece be moved to the corner? Explain.

Answer:
It is given that
Felipe is making triangles for a stained glass window. He made the design shown as above but wants to change it. Felipe wants to move the purple triangle to the corner. The purple piece has side lengths of 4.5 inches, 6 inches, and 7 inches.
Now,
From the given figure,
We can observe that
If the purple figure is moving to the corner, then one side of the purple figure will become a right angle
So,
Now,
We have to find whether the purple figure will be a right triangle or not if it moves to a corner
Now,
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²,
then, the triangle is a right triangle
We know that,
The hypotenuse has the longest side length
So,
7² = 4.5² + 6²
49 = 20.25 + 36
49 = 56.25
So,
The condition
c² = a² + b²
is false
Hence, from the above,
We can conclude that the purple figure should not

Question 12.
a. What is the longest poster you could fit in the box? Express your answer to the nearest tenth of an inch.

Answer:
The given figure is:

From the given figure,
We can observe that there will be two longest sides for two pairs of different side lengths
we know that,
The longest side is nothing but a diagonal
Now,
The two pairs of side lengths are: (20, 12, x), and (8, 12, y)
Where,
x and y are the lengths of the longest sides
Now,
We know that
Since the figure consists of the right triangles,
According to the Pythagorean Theorem,
c² = a² + b²
So,
x² = 20² + 12²
x² = 400 + 144
x² = 544
x = \(\sqrt{544}\)
x = 23.32 in.
So,
y² = 12² + 8²
y² = 144 + 64
y² = 208
y = \(\sqrt{208}\)
y = 14.42 in.
Hence, from the above,
We can conclude that the longest poster that you can fit in the box is: 23.32 in.

b. Explain why you can fit only one maximum-length poster in the box, but you can fit multiple 21.5-inch posters in the same box.

Answer:
It is given that
you can fit multiple 21.5-inch posters in the same box.
Now,
From part (a),
We can observe that the length of the longest poster that can fit into the box is: 23.32 in.
So,
21.5 in. < 23.32 in.
Hence, from the above,
We can conclude that since the given size of the posters is less than the maximum length of the poster,
We can fit multiple 21.5-inch posters in the same box instead of 1 poster that is of the maximum length

Question 13.

The corner of a room where two walls meet the floor should be at a right angle. Jeff makes a mark along each wall. One mark is 3 inches from the corner. The other is 4 inches from the corner. How can Jeff use the Pythagorean Theorem to see if the walls form a right angle?
Answer:
It is given that
The corner of a room where two walls meet the floor should be at a right angle. Jeff makes a mark along each wall. One mark is 3 inches from the corner. The other is 4 inches from the corner.
Now,
To see whether the walls form a right angle or not,
We have to see whether the length along the walls is greater than the lengths of the marks from the corners
Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length along the walls
a is the length of one mark from the corner
b is the length of another mark from the corner
So,
c² = 3² + 4²
c² = 9 + 16
c² = 25
c = \(\sqrt{25}\)
c = 5
So,
From the above value,
We can observe that
c > a and c > b
Hence, from the above,
We can conclude that the walls form a right angle

Question 14.
Higher-Order Thinking It is recommended that a ramp have at least 6 feet of horizontal distance for every 1 foot of vertical rise along an incline. The ramp shown has a vertical rise of 2 feet. Does the ramp show match the recommended specifications? Explain.

Answer:
It is given that
It is recommended that a ramp have at least 6 feet of horizontal distance for every 1 foot of vertical rise along an incline. The ramp shown has a vertical rise of 2 feet
Now,
6 feet of horizontal distance for every 1 foot,
The ramp shown has a vertical rise of 5 feet.
So,
The rate of change (m)  ≤ \(\frac{1 foot}{6 feet}\)
Convert both to a single unit of inches
We know that,
1 foot = 12 inches
So,
m ≤ \(\frac{12 inches}{72 inches}\)
Divide by 4 into both sides
m ≤ \(\frac{4 inches}{18 inches}\)
Now,
For a ramp that has a vertical distance of 2 feet and the same horizontal distance
m = \(\frac{12 inches}{24 inches}\)
m = \(\frac{4 inches}{6 inches}\)
So,
The rate of change when the vertical distance is 6 feet > The rate of change when the vertical distance is 2 feet
Hence, from the above,
We can conclude that the ramp shown above matched the given specifications

Assessment Practice

Question 15.
A machine in a factory cuts out triangular sheets of metal. Which of the triangles are right triangles? Select all that apply.

☐ Triangle 1
☐ Triangle 2
☐ Triangle 3
☐ Triangle 4
Answer:
It is given that
A machine in a factory cuts out triangular sheets of metal.
Now,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²
then, the triangle is a right triangle
Now,
For Triangle 1,
(\(\sqrt{505}\))² = 12² + 19²
505 = 505
For Triangle 2,
(\(\sqrt{467}\))² = 16² + 19²
467 ≠ 617
For Triangle 3,
(\(\sqrt{596}\))² = 14² + 20²
596 = 596
For Triangle 4,
(\(\sqrt{421}\))² = 11² + 23²
421 ≠ 650
Hence, from the above,
We can conclude that Triangle 1 and Triangle 3 are the right triangles

Question 16.
What is the length b, in feet, of the rectangular plot of land shown?

Answer:
It is given that the given figure is a rectangular plot of land
We know that,
In a rectangle,
The opposite sides are equal and the angles are all 90°
So,
A diagonal forms 2 right triangles in a rectangle
Now,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length of the diagonal
a is the width
b is the length
Now,
325² = 300² + b²
b²= 1,05,625 – 90,000
b² = 15,625
b = \(\sqrt{15,625}\)
b = 125 ft
Hence, from the above,
We can conclude that the length of b is: 125 ft

Lesson 7.4 Find Distance in the Coordinate

Explore It!
Thomas and Jim are outside the haunted castle ride and want to get to the clown tent in time for the next show.

I can… use the Pythagorean Theorem to find the distance between two points in the coordinate plane.

A. How can you represent the starred locations on a coordinate plane?

Answer:
It is given that
Thomas and Jim are outside the haunted castle ride and want to get to the clown tent in time for the next show.
Now,
The given route of the castle is:

Now,
From the given figure,
To represent the route in the coordinate plane,
The given scale is:
X-axis: 1 cm = 500 feet
Y-axis: 1 cm = 500 feet
Now,
From the given figure,
The coordinates of the starred locations can be:
The coordinates of the haunted house are: (500, 1,500),
The coordinates of the clown tent are: (2,000, 500)
Hence,
The representation of the starred locations in the coordinate plane is:

B. Jim says that the marked yellow paths show the shortest path to the tent. Write an expression to represent this and find the distance Jim walks from the haunted mansion to the clown tent.
Answer:
From part (a),
We know that,
The coordinates of the haunted house are: (500, 1,500),
The coordinates of the clown tent are: (2,000, 500)
Now,
We know that
The linear equation in the slope-intercept form is:
y = mx + b
Where,
m is the slope
b is the initial value (or) y-intercept
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
Now,
We know that,
Slope (m) = y₂ – y₁ / x₂ – x₁
So,
Slope (m) = \(\frac{500 – 1,500}{2,000 – 500}\)
Slope (m) = –\(\frac{1,000}{1,500}\)
Slope (m) = –\(\frac{2}{3}\)
So,
The equation in the slope-intercept form is:
y = –\(\frac{2}{3}\)x + b
So,
3y = -2x + 3b
Now,
Substitute (500, 1,500) or (2,000, 500) in the above equation
So,
1,500 = -4000 + 3b
3b = 5,500
So,
The equation that represents the shortest path to the tent is:
3y = -2x + 5,500
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the haunted mansion and the clown tent = \(\sqrt{(2,000 – 500)^{2} + (500 – 1,500)^{2}}\)
= \(\sqrt{1,500^{2} + 1,000^{2}}\)
= \(\sqrt{2,250,000 + 1,000,000}\)
= 1,802.77 feet
Hence, from the above,
We can conclude that
The equation that represents the shortest path to the clown tent is:
3y = -2x + 5,500
The distance between the haunted mansion and the clown tent is: 1,802.77 feet

Focus on math practices
Construct Arguments Why is the distance between two nonhorizontal and nonvertical points always greater than the horizontal or vertical distance?
Answer:
Let us consider a coordinate plane
Now,
When we draw either a horizontal line or the vertical line,
We can observe that the length will be constant
But,
When we draw non-vertical and non-horizontal lines,
We can observe that the lengths are unknown and not constant
Hence, from the above,
We can conclude that the distance between two nonhorizontal and nonvertical points always greater than the horizontal or vertical distance

Essential Question
How can you use the Pythagorean Theorem to find the distance between two points?
Answer:
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the hypotenuse
a and b are the side lengths or lengths of the legs
Now,
Graphically,
The terms of the Pythagorean Theorem can be expressed as:
c is the distance between two points
a and b are the points
So,
c = \(\sqrt{a^{2} + b^{2}}\)

Try It!

What is the distance between points A and B?
The distance between point A and point B is about ________ units.

Answer:
The given graph is:

The representation of the graph in the coordinate plane is:

From the given graph,
The coordinates of A are: (2, 3)
The coordinates of B are: (4, 1)
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points A and B = \(\sqrt{(4 – 2)^{2} + (1 – 3)^{2}}\)
= \(\sqrt{2^{2} + 2^{2}}\)
= \(\sqrt{4 + 4}\)
= 2.83 units
Hence, from the above,
We can conclude that the distance between points A and B is about 2.83 units

Convince Me!
Why do you need to use the Pythagorean Theorem to find the distance between points A and B?
Answer:
The representation of points A and B in the coordinate plane are:

Now,
When we observe the graph,
We can see that A and B can form a right triangle
Now,
We know that,
The Pythagorean Theorem is only applicable to the right triangles
So,
According to the Pythagorean Theorem,
c² = a² + b²
c = \(\sqrt{a^{2} + b^{2}}\)
where,
c is the distance between points A and B
A and B are the given points

Try It!
Find the perimeter of ∆ABC with vertices (2, 5), (5, -1), and (2, -1).
Answer:
It is given that
∆ABC with vertices (2, 5), (5, -1), and (2, -1)
Now,
The names of the vertices are:
A (2, 5), B (5, -1), and C (2, -1)
We know that,
The perimeter of a triangle is the sum of all the side lengths of a triangle
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
Now,
We know that,
In ∆ABC,
AB and BC are the side lengths
Ac is the hypotenuse
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points A and B = \(\sqrt{(5 – 2)^{2} + (-1 – 5)^{2}}\)
= \(\sqrt{3^{2} + 6^{2}}\)
= \(\sqrt{9 + 36}\)
= 6.70 units
The distance between the points B and C (BC) = \(\sqrt{(2 – 5)^{2} + (-1 + 1)^{2}}\)
= \(\sqrt{3^{2} + 0^{2}}\)
= \(\sqrt{9 + 0}\)
= 3 units
The distance between the points A and C (AC) = \(\sqrt{(2 – 2)^{2} + (-1 – 5)^{2}}\)
= \(\sqrt{0^{2} + 6^{2}}\)
= \(\sqrt{0 + 36}\)
= 6 units
So,
The perimeter of ∆ABC = AB + BC + AC
= 6 + 3 + 6.70
= 15.7 units
Hence, from the above,
We can conclude that the perimeter of ∆ABC is about 15.7 units

Try It!

What are the coordinates, to the nearest tenth, of the third vertex in an isosceles triangle that has one side length of 2 and two side lengths of 5, with vertices at (1, 0) and (1, 2)? The third vertex is in the first quadrant.
Answer:
It is given that
An isosceles triangle that has one side length of 2 and two side lengths of 5, with vertices at (1, 0) and (1, 2)
Now,
Let the third vertex be (x, y)
Now,
The given vertices are:
A (1, 0), B (1, 2), and C (x, y)
It is also given that
BC = 2 units, and AC = 5 units
We know that,
An isosceles triangle has any 2 equal side lengths
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points A and B = \(\sqrt{(2 – 0)^{2} + (1 – 1)^{2}}\)
= \(\sqrt{2^{2} + 0^{2}}\)
= \(\sqrt{4 + 0}\)
= 4 units
The distance between the points B and C = \(\sqrt{(x – 1)^{2} + (y – 2)^{2}}\)
Squaring on both sides
So,
BC² = (x – 1)² + (y – 2)²
The distance between the points A and C = \(\sqrt{(x – 1)^{2} + (y – 0)^{2}}\)
Squaring on both sides
So,
BC² = (x – 1)² + y²
So,
(x – 1)² + (y – 2)² = 4 —– (1)
(x – 1)² + y² = 25 —— (2)
So,
From eq (1) and eq (2),
25 – y² + (y – 2)² = 4
-y² + y² – 4y + 4 = -21
-4y = -25
y = \(\frac{25}{4}\)
So,
(x – 1)² = |25 – (\(\frac{25}{4}\))²|
x = \(\frac{19}{4}\)
Hence, from the above,
We can conclude that the third vertex is: (\(\frac{19}{4}\), \(\frac{25}{4}\))

KEY CONCEPT

You can use the Pythagorean Theorem to find the distance between any two points, P and Q, on the coordinate plane.

Do You Understand?
Question 1.
Essential Question How can you use the Pythagorean Theorem to find the distance between two points?
Answer:
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the hypotenuse
a and b are the side lengths or lengths of the legs
Now,
Graphically,
The terms of the Pythagorean Theorem can be expressed as:
c is the distance between two points
a and b are the points
So,
c = \(\sqrt{a^{2} + b^{2}}\)

Question 2.
Model with Math
Can you use a right triangle to represent the distance between any two points on the coordinate plane? Explain.
Answer:
Derived from the Pythagorean Theorem, the distance formula is used to find the distance between two points in the plane. We know that,
According to the Pythagorean Theorem,
a²+b²=c²
where,
a and b are the lengths of the legs adjacent to the right angle
c is the length of the hypotenuse.

Question 3.
Generalize How does the fact that the points are on opposite sides of the y-axis affect the process of finding the distance between the two points?
Answer:
The fact that the points are on opposite sides of the y-axis affects the process of finding the distance between the two points because  We need to find the distance between the two points by adding the distances from each of them to the y-axis.

Do You Know How?
In 4-6, use the coordinate plane below.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The given points are:
C (1, 2), D (2, -1), and E (-2, 1)

Question 4.
Find the distance between points C and D. Round to the nearest hundredth.
Answer:
Compare the points C and D with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points C and D = \(\sqrt{(2 – 1)^{2} + (-1 – 2)^{2}}\)
= \(\sqrt{1^{2} + 3^{2}}\)
= \(\sqrt{1 + 9}\)
= 3.16 units
Hence, from the above,
We can conclude that the distance between points C and D is: 3.16 units

Question 5.
Find the perimeter of ∆CDE.
Answer:
We know that,
The “Perimeter” is defined as the sum of all the side lengths
So,
The perimeter of ∆CDE = CD + DE + CE
So,
The distance between the points C and D = \(\sqrt{(2 – 1)^{2} + (-1 – 2)^{2}}\)
= \(\sqrt{1^{2} + 3^{2}}\)
= \(\sqrt{1 + 9}\)
= 3.16 units
The distance between the points D and E = \(\sqrt{(-2 – 2)^{2} + (1 + 1)^{2}}\)
= \(\sqrt{4^{2} + 2^{2}}\)
= \(\sqrt{16 + 4}\)
= 4.47 units
The distance between the points C and E = \(\sqrt{(-2 – 1)^{2} + (1 – 2)^{2}}\)
= \(\sqrt{3^{2} + 1^{2}}\)
= \(\sqrt{1 + 9}\)
= 3.16 units
So,
The perimeter of ∆CDE = 3.16 + 4.47 + 3.16
= 10.79 units
Hence, from the above,
We can conclude that the perimeter of ∆CDE is: 10.79 units

Question 6.
Point B is plotted on the coordinate plane above the x-axis. ∆BDE is equilateral. What are the coordinates of point B to the nearest hundredth?
Answer:
It is given that
Point B is plotted on the coordinate plane above the x-axis. ∆BDE is equilateral.
Now,
Let the unknown vertex be B (x, y)
So,
The given points are:
B (x, y), D (2, -1), and E (-2, 1)
It is given that ΔBDE is equilateral
So,
BD = DE = EB
BD² = DE² = EB²
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points B and D = \(\sqrt{(x – 2)^{2} + (y – 1)^{2}}\)
Squaring on both sides
So,
BD² = (x – 2)² + (y + 1)²
The distance between the points D and E = \(\sqrt{(1 + 1)^{2} + (-2 – 2)^{2}}\)
= \(\sqrt{2^{2} + 4^{2}}\)
= \(\sqrt{4 + 16}\)
= 4.47 units
The distance between the points E and B = \(\sqrt{(x + 2)^{2} + (y – 1)^{2}}\)
Squaring on both sides
So,
EB² = (x + 2)² + (y – 1)²
Now,
(x – 2)² + (y + 1)² = 4.47 —- (1)
(x + 2)² + (y – 1)² = (x – 2)² + (y + 1)²
x² + 2x + 4 + y² + 1 – 2y = x² + 4 – 4x + y² + 1 + 2y
2x + 4 + 1 – 2y = 4 – 4x + 1 + 2y
6x + 5 = 4y + 5
6x = 4y
3x = 2y
x = \(\frac{2}{3}\)y
Now,
From eq (1),
x² + 4 – 4x + y² + 1 + 2y = 4.47
x² + y² -4x + 2y = -0.47
(\(\frac{2}{3}\)y)² + y² – 4 (\(\frac{2}{3}\))y + 2y = -0.47
4y² + 15y + 4.23 = 0
So,
y = -0.30 (or) y = -3.44
So,
x = \(\frac{2}{3}\) (-0.30) (or) x = \(\frac{2}{3}\) (-3.44)
x = -0.2 (or) x = -2.29
Hence, from the above,
We can conclude that the coordinates of point B are: (-0.2, -0.30) or (-2.29, -3.44)

Practice & Problem Solving

Question 7.
Leveled Practice Use the Pythagorean Theorem to find the distance between points P and Q.
Label the length, in units, of each leg of the right triangle.

The distance between point P and point Q is __________ units.
Answer:
From the given coordinate plane,
There are only 2 vertices
Let the third vertex be R(x, y) and the coordinates of R can be found from the coordinate plane
Now,
From the given coordinate plane,
The vertices are:
P (3, 2), Q (9, 10), and R (9, 2)
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points Q and R = \(\sqrt{(9 – 9)^{2} + (10 – 2)^{2}}\)
= \(\sqrt{0^{2} + 8^{2}}\)
= \(\sqrt{0 + 64}\)
= 8 units
The distance between the points P and R = \(\sqrt{(9 – 3)^{2} + (2 – 2)^{2}}\)
= \(\sqrt{0^{2} + 6^{2}}\)
= \(\sqrt{0 + 36}\)
= 6 units
Now,
From the given coordinate plane,
We can observe that P, Q, R form the right triangle
So,
According to the Pythagorean Theorem,
PQ² = QR² + PR²
PQ² = 8² + 6²
PQ² = 64 + 36
PQ² = 100
PQ = \(\sqrt{100}\)
PQ = 10 units
Hence, from the above,
We can conclude that the length of PQ is: 10 units

Question 8.
Find the perimeter of triangle QPR. Round to the nearest hundredth.
Answer:
From the coordinate plane,
The vertices of ΔPQR are:
P (-5, -2), Q (2, -2), and R (-1, 3)
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points P and Q = \(\sqrt{(2 + 5)^{2} + (2 – 2)^{2}}\)
= \(\sqrt{0^{2} + 7^{2}}\)
= \(\sqrt{0 + 49}\)
= 7 units
The distance between the points Q and R = \(\sqrt{(-1 – 2)^{2} + (3 + 2)^{2}}\)
= \(\sqrt{3^{2} + 5^{2}}\)
= \(\sqrt{9 + 25}\)
= 5.83 units
The distance between the points P and R = \(\sqrt{(1 – 5)^{2} + (-3 – 2)^{2}}\)
= \(\sqrt{4^{2} + 5^{2}}\)
= \(\sqrt{16 + 25}\)
= 6.40 units
So,
The perimeter of ΔPQR = PQ + QR + PR
= 7 + 5.83 + 6.40
= 19.23 units
Hence, from the above,
We can conclude that the perimeter of ΔPQR is: 19.23 units

Question 9.
Determine whether the triangle is equilateral, isosceles, or scalene.
Answer:
We know that,
On the basis of the side lengths,
Scalene Triangle – All the side lengths are different
Equilateral Triangle – All the side lengths are the same
Isosceles Triangle – Any two of the side lengths are the same
So,
From Exercise 8,
We can observe that all the side lengths are different
Hence, from the above,
We can conclude that ΔPQR is a scalene Triangle

Question 10.
You walk along the outside of a park starting at point P. Then you take a shortcut represented by \(\overline{P Q}\) on the graph.

a. What is the length of the shortcut in meters? Round to the nearest tenth of a meter.
Answer:
It is given that
You walk along the outside of a park starting at point P. Then you take a shortcut represented by \(\overline{P Q}\) on the graph.
Now,
From the given figure,
We can observe that
The vertices are:
P (0, 0), Q (40, 85), and R (40, 0)
The shortest path is represented as PQ
Now,
Compare the points P and Q with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points P and Q = \(\sqrt{(40 – 0)^{2} + (85 – 0)^{2}}\)
= \(\sqrt{40^{2} + 85^{2}}\)
= \(\sqrt{1,600 + 7,225}\)
= 93.9 m
Hence, from the above,
We can conclude that the length of the shortest path is: 93.9 meters

b. What is the total length of your walk in the park? Round to the nearest tenth of a meter.
Answer:
We know that,
The total length is nothing but the “Perimeter”
So,
The perimeter of the given triangle = PQ + QR + PR
Now,
The distance between the points P and Q = \(\sqrt{(40 – 0)^{2} + (85 – 0)^{2}}\)
= \(\sqrt{40^{2} + 85^{2}}\)
= \(\sqrt{1,600 + 7,225}\)
= 93.9 m
The distance between the points Q and R = \(\sqrt{(40 – 40)^{2} + (85 – 0)^{2}}\)
= \(\sqrt{0^{2} + 85^{2}}\)
= \(\sqrt{0 + 7,225}\)
= 85 m
The distance between the points P and R = \(\sqrt{(40 – 0)^{2} + (0 – 0)^{2}}\)
= \(\sqrt{0^{2} + 40^{2}}\)
= \(\sqrt{0 + 1,600}\)
= 40 m
So,
The total length of your walk in the park = PQ + QR + PR
= 93.9 + 85 + 40
= 218.9 meters
Hence, from the above,
We can conclude that the total length of your walk in the park is: 218.9 meters

Question 11.
Suppose a park is located 3.6 miles east of your home. The library is 4.8 miles north of the park. What is the shortest distance between your home and the library?
Answer:
It is given that
A park is located 3.6 miles east of your home. The library is 4.8 miles north of the park
So,
The representation of the above situation is:

Now,
According to the Pythagorean Theorem,
(The shortest distance between home and library)² = (The distance from home to park)² + (he distance from park to library)²
(The shortest distance between home and library)² = 3.6² + 4.8²
(The shortest distance between home and library)² = 12.96 + 23.04
The shortest distance between home and library = 6
Hence, from the above,
We can conclude that the shortest distance between home and library is: 6 miles

Question 12.
Use Structure Point B has coordinates (2, 1). The x-coordinate of point A is -10. The distance between point A and point B is 15 units. What are the possible coordinates of point A?
Answer:
It is given that
Point B has coordinates (2, 1). The x-coordinate of point A is -10. The distance between point A and point B is 15 units
Now,
Let the coordinates of A be:
(x, y) = (-10, y)
Now,
Compare the points A and B with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points A and B = \(\sqrt{(-10 – 2)^{2} + (y – 1)^{2}}\)
15 = \(\sqrt{12^{2} + (y – 1)^{2}}\)
15 = \(\sqrt{144 + (y – 1)^{2}}\)
Now,
Squaring on both sides
So,
144 + (y – 1)² = 225
(y  1)² = 225 – 144
(y – 1)² = 81
y – 1 = \(\sqrt{81}\)
y – 1 = 9 (or) y – 1 = -9
y = 9 + 1 (or) y = -9 + 1
y = 10 (or) y = -8
Hence, from the above,
We can conclude that the possible coordinates of A are: (-10, 10), and (-10, -8)

Question 13.
Higher-Order Thinking ∆EFG and ∆HIJ have the same perimeter and side lengths. The coordinates are E(6, 2), F(9, 2), G(8, 7), H(0, 0), and I(0, 3). What are the possible coordinates of point J?
Answer:
It is given that
∆EFG and ∆HIJ have the same perimeter and side lengths. The coordinates are E(6, 2), F(9, 2), G(8, 7), H(0, 0), and I(0, 3)
So,
According to the side lengths,
EF = HI, FG = IJ, and GE = JH
So,
EF² = HI², FG²= JI², and GE² = JH²
Now,
Let the unknown vertex be J (x, y)
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points E and F = \(\sqrt{(9 – 6)^{2} + (2 – 2)^{2}}\)
= \(\sqrt{3^{2} + 0^{2}}\)
= \(\sqrt{9 + 0}\)
= 3 units
The distance between the points F and G = \(\sqrt{(8 – 9)^{2} + (7 – 2)^{2}}\)
= \(\sqrt{1^{2} + 5^{2}}\)
= \(\sqrt{1 + 25}\)
= 5.09 units
The distance between the points G and E = \(\sqrt{(8 – 6)^{2} + (7 – 2)^{2}}\)
= \(\sqrt{2^{2} + 5^{2}}\)
= \(\sqrt{4 + 25}\)
= 5.38 units
Now,
The distance between the points H and I = \(\sqrt{(0 – 0)^{2} + (3 – 0)^{2}}\)
= \(\sqrt{3^{2} + 0^{2}}\)
= \(\sqrt{9 + 0}\)
= 3 units
The distance between the points I and J = \(\sqrt{(x – 0)^{2} + (y – 3)^{2}}\)
5.09 = \(\sqrt{x^{2} + (y – 3)^{2}}\)
Squaring on both sides
So,
x² + (y – 3)² = 25.90 units
The distance between the points J and H = \(\sqrt{(x – 0)^{2} + (y – 0)^{2}}\)
5.38 = \(\sqrt{x^{2} + y^{2}}\)
Squaring on both sides
So,
x² + y² = 28.94 units
Now,
28.94 – y² + (y – 3)² = 25.90
y² + 9 – 6y – y² = 25.90 – 28.94
9 – 6y = -3.04
-6y = -3.04 – 9
6y = 12.04
y = 2
Now,
Substitute the value of y in eq 2
x² + 4 = 28.94
x² = 24.94
x = 4.99 (or) x = -4.99
Hence, from the baove,
We can conclude that the missing vertex is: J (4.99, 2) or J (-4.99, 2)

b. Explain why there can be different possibilities for the coordinates for point J.
Answer:

Assessment Practice
Question 14.
Find the distance, in units, between P and R. Round to the nearest tenth.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The points are:
P (5, 10), and R (12, 14)
Now,
Compare the points P and R with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points P and R = \(\sqrt{(14 – 10)^{2} + (12 – 5)^{2}}\)
= \(\sqrt{4^{2} + 7^{2}}\)
= \(\sqrt{16 + 49}\)
= 8.1 units
Hence, from the above,
We can conclude that the distance between P and R is: 8.1 units

Question 15.
Find the distance, in units, between A(1, 5) and B(5.5, 9.25). Round to the nearest tenth.
Answer:
The given points are:
A (1, 5), and B (5.5, 9.25)
Now,
Compare the points A and B with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points A and B = \(\sqrt{(9.25 – 5)^{2} + (5.5 – 1)^{2}}\)
= \(\sqrt{4.25^{2} + 4.5^{2}}\)
= \(\sqrt{18.06 + 20.25}\)
= 6.2 units
Hence, from the above,
We can conclude that the distance between points A and B is: 6.2 units

Topic 7 REVIEW

Topic Essential Question
How can you use the Pythagorean Theorem to solve problems?
Answer:
Step 1:
Draw a right triangle and then read through the problems again to determine the length of the legs and the hypotenuse. Step 2:
Use the Pythagorean Theorem (a² + b² = c²) to write an equation to be solved.
Step 3:
Simplify the equation by distributing and combining like terms as needed.

Vocabulary Review

Complete each definition and then provide an example of each vocabulary word.
Vocabulary
The converse of the Pythagorean Theorem
hypotenuse
leg
proof
Pythagorean Theorem

Answer:

Use Vocabulary in Writing
All faces of the figure are rectangles. Explain how to find the length of d. Use vocabulary terms in your description.

Answer:
The given figure is:

From the given figure,
We can observe that there are 2 right triangles
Now,
For the first right triangle,
The given side lengths are:
a = 3, b = 4, and c = x
Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
So,
x² = 3² + 4²
x² = 9 + 16
x² = 25
x = \(\sqrt{25}\)
x = 5
Now,
For the second right triangle,
a = 5, b = 12, and c = d
So,
d² = 5² + 12²
d² = 25 + 144
d² = 169
d = \(\sqrt{169}\)
d = 13
Hence, from the above,
We can conclude that the length of d is: 13 units

Concepts and Skills Review

Lesson 7.1 Understand the Pythagorean Theorem

Quick Review
The Pythagorean Theorem states that, in a right triangle, the sum of the squares of the lengths of the legs, a and b, is equal to the square of the length of the hypotenuse, c. So, a² + b² = c².

Example
Find the length of the hypotenuse of a triangle with legs of 7 meters and 24 meters.
Answer:
Substitute 7 for a and 24 for b. Then solve for c.
a² + b² = c²
49 + 576 = c²
\(\sqrt {625}\) = C
The length of the hypotenuse is 25 meters.

Practice
Question 1.
Find the length of the hypotenuse.

Answer:
The given right triangle is:

Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the hypotenuse
a and b are the side lengths
So,
c² = 12² + 5²
c² = 144 + 25
c² = 169
c = \(\sqrt{169}\)
c = 13
Hence, from the above,
We can conclude that the length of the hypotenuse is: 13 cm

Question 2.
Find the unknown side length. Round to the nearest tenth.

Answer:
The given right triangle is:

Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the hypotenuse
a and b are the side lengths
So,
9² = 7² + b²
b² = 81 – 49
b² = 32
b = \(\sqrt{32}\)
b = 5.65
Hence, from the above,
We can conclude that the length of the hypotenuse is: 5.65 in.

Lesson 7.2 Understand the Converse of the Pythagorean Theorem

Quick Review
For a triangle with side lengths a, b, and c, if a² + b² = c², then the triangle is a right triangle by the Converse of the Pythagorean Theorem.

Example
Is a triangle with side lengths of 8 m, 15 m, and 17 m a right triangle? Explain.
Answer:
Substitute 8 for a, 15 for b, and 17 for c.
a² + b² \(\underline{\underline{?}}\) c²
8² + 15² \(\underline{\underline{?}}\) 17²
289 = 289 ✓
Because a² + b² = c², the triangle is a right triangle.

Practice
Question 1.
Is the triangle a right triangle? Explain.

Answer:
The given triangle is:

Now,
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²
then, the given triangle is a right triangle
So,
Now,
37² = 35² + 12²
1,369 = 1,225 + 144
1,369 = 1,369
So,
The condition c² = a² + b² is true
Hence, from the above,
We can conclude that the given triangle is a right triangle

Question 2.
A triangle has side lengths of 1.5 inches, 2 inches, and 3 inches. Is the triangle a right triangle? Explain.
Answer:
It is given that
A triangle has side lengths of 1.5 inches, 2 inches, and 3 inches
Now,
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²
then, the given triangle is a right triangle
We know that,
The longest side is the hypotenuse
So,
Now,
3² = (1.5)² + 2²
9 = 2.25 + 4
9 = 6.25
So,
The condition c² = a² + b² is false
Hence, from the above,
We can conclude that the given triangle is not a right triangle

Question 3.
A triangle has side lengths of 9 feet, 40 feet, and 41 feet. Is the triangle a right triangle? Explain
Answer:
It is given that
A triangle has side lengths of 9 feet, 40 feet, and 41 feet
Now,
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²
then, the given triangle is a right triangle
We know that,
The longest side is the hypotenuse
So,
Now,
41² = 40² + 9²
1,681 = 1,600 + 81
1,681 = 1,681
So,
The condition c² = a² + b² is true
Hence, from the above,
We can conclude that the given triangle is a right triangle

Lesson 7.3 Apply the Pythagorean Theorem to Solve Problems

Quick Review
The Pythagorean Theorem can be used to find unknown side lengths of an object that is shaped like a right triangle. It also can be used to find diagonal measures in certain two-dimensional and three-dimensional objects.

Example
A shipping box is 20 inches long along the diagonal of its base. Each diagonal of the box is 29 inches long. How tall is the box?

Answer:
Substitute 20 for a and 29 for c. Then solve for b.
a² + b² = c²
20² + b² = 29²
400 + b² = 841
b = \(\sqrt {441}\)
The height of the shipping box is 21 inches.

Practice
Question 1.
A basketball court is in the shape of a rectangle that is 94 feet long and 50 feet wide. What is the length of a diagonal of the court? Round to the nearest tenth.
Answer:
It is given that
A basketball court is in the shape of a rectangle that is 94 feet long and 50 feet wide
We know that,
By drawing a diagonal in the rectangle, it will become 2 right triangles
The diagonal will be the hypotenuse of the right triangle
Now,
We know that,
According to the Pythagorean Theorem,
c² = a²+ b²
Where,
c is the length of the diagonal
a and b are the side lengths
So,
c² = 94²+ 50²
c²= 8,836 + 2,500
c² = 11,336
c = \(\sqrt{11,336}\)
c = 106.4 feet
hence, from the above,
We can conclude that the length of the diagonal is: 106.4 feet

Question 2.
A packaging box for a metal rod is 7.5 inches along a diagonal of the base. The height of the box is 18 inches. What is the length of a diagonal of the box?

Answer:
It is given that
A packaging box for a metal rod is 7.5 inches along a diagonal of the base. The height of the box is 18 inches.
Now,
From the given figure,
We can observe that it looks like a right triangle
Now,
We know that,
According to the Pythagorean Theorem,
c² = a²+ b²
Where,
c is the length of the diagonal
a and b are the side lengths
So,
c² = 18² + (7.5)²
c²= 324 + 56.25
c² = 380.25
c = \(\sqrt{380.25}\)
c = 19.5 inches
Hence, from the above,
We can conclude that the length of the diagonal of the box is: 19.5 inches

Lesson 7.4 Find Distance in the Coordinate Plane

Quick Review
The Pythagorean Theorem can be used to find the distance between any two points on the coordinate plane.

Example
Find the distance between the two points on the coordinate plane. Round to the nearest tenth.

Answer:
Draw a right triangle. Determine the lengths of its legs.

The length of the horizontal leg is 5 units.
The length of the vertical leg is 5 units.
Use the relationship a² + b² = c². Substitute 5 for a and 5 for b. Then solve for C.
a² + b² = -2
5² + 5² = c²
25 + 25 = c²
50 = c²
\(\sqrt {50}\) = C
7.1 ≈ c
The distance between the two points is about 7.1 units.

Practice
Question 1.
Points C and D represent the location of two parks on a map. Find the distance between the parks if the length of each unit on the grid is equal to 25 miles. Round to the nearest mile.

Answer:
It is given that
Points C and D represent the location of two parks on a map
Now,
The given coordinate plane is:

From the given coordinate plane,
The points are:
C (-2, 2), and D (4, -1)
Now,
Compare the points C and D with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points C and D = \(\sqrt{(-1 – 2)^{2} + (4 + 2)^{2}}\)
= \(\sqrt{3^{2} + 6^{2}}\)
= \(\sqrt{9 + 36}\)
= 6.70 units
It is given that the length of each unit in the grid is: 25 miles
So,
The distance between the points C and D = 6.25 × 25
= 156.25 miles
Hence, from the above,
We can conclude that the distance between points C and D is: 156.25 miles

Question 2.
Find the perimeter of ∆ABC. Round to the nearest tenth.

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of ΔABC are:
A (1, 2), B (7, 9), and C (7, 2)
We know that,
The perimeter of a triangle is the sum of all the side lengths of a triangle
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
Now,
From the given coordinate plane,
We know that,
In ∆ABC,
AC and BC are the side lengths
AB is the hypotenuse
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points A and C = \(\sqrt{(2 – 2)^{2} + (7 – 1)^{2}}\)
= \(\sqrt{0^{2} + 6^{2}}\)
= \(\sqrt{36 + 0}\)
= 6 units
The distance between the points B and C (BC) = \(\sqrt{(7 – 7)^{2} + (-2 + 9)^{2}}\)
= \(\sqrt{0^{2} + 7^{2}}\)
= \(\sqrt{0 + 49}\)
= 7 units
The distance between the points A and B (AB) = \(\sqrt{(9 – 2)^{2} + (7 – 1)^{2}}\)
= \(\sqrt{7^{2} + 6^{2}}\)
= \(\sqrt{49 + 36}\)
= 9.21 units
So,
The perimeter of ∆ABC = AB + BC + AC
= 6 + 7 + 9.21
= 22.2 units
Hence, from the above,
We can conclude that the perimeter of ∆ABC is about 22.2 units

Question 3.
Triangle JKL is an equilateral triangle with two of its vertices at points J and K. What are the coordinates of point L? Round to the nearest tenth as needed.

Answer:
It is given that
Triangle JKL is an equilateral triangle with two of its vertices at points J and K
Now,
The given coordinate plane is:

Now,
From the given coordinate plane,
The two vertices of ΔJKL are:
J (3, 2), K (9, 2)
Now,
Let the third vertex be L (x, y)
We know that,
In an equilateral triangle, all the side lengths are the same
So,
JK = KL = JL
JK² = KL² = JL²
So,
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points J and K = \(\sqrt{(9 – 3)^{2} + (2 – 2)^{2}}\)
= \(\sqrt{0^{2} + 6^{2}}\)
= \(\sqrt{36 + 0}\)
= 6 units
The distance between the points K and L = \(\sqrt{(x – 9)^{2} + (y – 2)^{2}}\)
6= \(\sqrt{(x – 9)^{2} + (y – 2)^{2}}\)
Squaring on both sides
So,
(x – 9)² + (y – 2)² = 36
The distance between the points J and L = \(\sqrt{(x – 3)^{2} + (y – 2)^{2}}\)
6= \(\sqrt{(x – 3)^{2} + (y – 2)^{2}}\)
Squaring on both sides
So,
(x – 3)² + (y – 2)² = 36
Now,
(x – 9)² + 36 – (x – 3)² = 36
(x – 9)² = (x – 3)²
x² + 81 – 18x = x² + 9 – 6x
18x – 6x = 81 – 9
12x = 72
x = \(\frac{72}{12}\)
x = 6
So,
Substitute the value of x in eq (2)
(6 – 3)² + (y – 2)² = 36
(y – 2)² = 36 – 9
y – 2 = \(\sqrt{27}\)
y = 7.2
Hence, from the above,
We can conclude that the third vertex is: L (6, 7.2)

Topic 7 Fluency Practice

Riddle Rearranging
Solve each equation. Then arrange the answers in order from least to greatest. The letters will spell out the answer to the riddle below.

I can… solve multistep equations.

Why did the coffee shop server love the job? Because there were so

Answer:
The Ascending order of the solutions of the equations is:
M < N < Y < P < A < E < R < K < S

Go through the enVision Math Common Core Grade 8 Answer Key Topic 6 Congruence and Similarity and finish your homework or assignments.

enVision Math Common Core 8th Grade Answers Key Topic 6 Congruence And Similarity

Topic Essential Question
How can you show that two figures are either congruent or similar to one another?
Answer:
When two line segments have the same length, we can say that they are congruent. When two figures have the same shape and size, we can say that the two figures are congruent. These two triangles are congruent. We can also say that their side lengths are the same and that their angle measures are the same

3-ACT MATH

Tricks of the Trade
All kinds of objects in nature have symmetry: beehives, pine cones, butterflies, and snowflakes, to name a few. If you look closely, you will start to see patterns left and right. Think about this during the 3-Act Mathematical Modeling lesson.

Topic 6 ënVision STEM Project

Did You Know?
Trees provide wood for cooking and heating for half of the world’s population.

As trees grow, carbon dioxide is removed from the atmosphere for photosynthesis. Forests are called “carbon sinks” because one acre of forest absorbs six tons of carbon dioxide and puts out four tons of oxygen.

Trees provide lumber for buildings, tools, and furniture. Other products include rubber, sponges, cork, paper, chocolate, nuts, and fruit.
About 30% of the land is covered by forests.

Forests are now being managed to preserve wildlife and old-growth forests, protect biodiversity, safeguard watersheds, and develop recreation, as well as extract timber.

Forests also need to be managed to prevent raging wildfires, invasive species, overgrazing, and disease.

Your Task: Forest Health
The proper management of forests is a growing science. You and your classmates will learn about forest health indicators and use what you know about similar triangles and ratios to gather and interpret data in order to assess the health of a forest.

Topic 6 GET READY!

Review What You Know!

Vocabulary
Choose the best term from the box to complete each definition.
adjacent angles
complementary angles
supplementary angles
vertical angles

Question 1.
_________ have a sum of 90°.
Answer:
We know that,
The “Complementary angles” have a sum of 90°
Hence, from the above,
We can conclude that the best term to complete the given definition is: Complementary angles

Question 2.
_________ share the same ray.
Answer:
We know that,
The “Adjacent Angles” share the same ray
Hence, from the above,
We can conclude that the best term to complete the given definition is: Adjacent angles

Question 3.
_________ are pairs of opposite angles made by intersecting lines.
Answer:
We know that,
The “Vertical Angles” are pairs of opposite angles made by intersecting lines
Hence, from the above,
We can conclude that the best term to complete the given definition is: Vertical angles

Question 4.
__________ have a sum of 180°
Answer:
We know that,
The “Supplementary Angles” have a sum of 180°
Hence, from the above,
We can conclude that the best term to complete the given definition is: Supplementary angles

Multiplying Real Numbers

Simplify the expression.
Question 5.
5 × 2 = ________
Answer:
The given expression is:
5 × 2
Hence, from the above,
We can conclude that
5 × 2 = 10

Question 6.
6 × \(\frac{1}{2}\) = ________
Answer:
The given expression is:
6 × \(\frac{1}{2}\)
So,
6 × \(\frac{1}{2}\)
= \(\frac{6}{2}\)
= 3
Hence, from the above,
We can conclude that
6 × \(\frac{1}{2}\) = 3

Question 7.
12 × \(\frac{1}{3}\) = ________
Answer:
The given expression is:
12 × \(\frac{1}{3}\)
So,
12 × \(\frac{1}{3}\)
= \(\frac{12}{3}\)
= 4
Hence, from the above,
We can conclude that
12 × \(\frac{1}{3}\) = 4

Identifying Points on a Coordinate Plane

Name the location of the point.

Answer:
The given coordinate plane is:

From the given coordinate plane,
We can observe that

Question 8.
point W
Answer:
From the given coordinate plane,
We can observe that point W lies where both the x-axis and y-axis are negative
Hence, from the above,
We can conclude that point W lies in the third quadrant

Question 9.
point X
Answer:
From the given coordinate plane,
We can observe that point X lies where the x-axis is negative and the y-axis is positive
Hence, from the above,
We can conclude that point X lies in the second quadrant

Question 10.
point Y
Answer:
From the given coordinate plane,
We can observe that point Y lies where both the x-axis and y-axis are positive
Hence, from the above,
We can conclude that point Y lies in the first quadrant

Question 11.
point Z
Answer:
From the given coordinate plane,
We can observe that point Z lies where the x-axis is positive and the y-axis is negative
Hence, from the above,
We can conclude that point Z lies in the fourth quadrant

Supplementary Angles

The angles are supplementary. Find the missing angle measure.
Question 12.

Answer:
The given figure is:

We know that,
The sum of the supplementary angles is: 180°
Now,
Let the missing angle measure be: x°
So,
x° + 130° = 180°
x° = 180° – 130°
x° = 50°
Hence, from the above,
We can conclude that the missing angle measure is: 50°

Question 13.

Answer:
The given figure is:

We know that,
The sum of the supplementary angles is: 180°
Now,
Let the missing angle measure be: x°
So,
x° + 139° = 180°
x° = 180° – 139°
x° = 41°
Hence, from the above,
We can conclude that the missing angle measure is: 41°

Language Development
Complete the graphic organizer with an illustration for each transformation. Write either congruent or similar to make the given statement true.


Answer:

Topic 6 PICK A PROJECT

PROJECT 6A
How might an artist use mathematics?
PROJECT: WRITE A BIOGRAPHY

PROJECT 6B
What geometric shapes do you see around you?
PROJECT: RECORD A VIDEO ABOUT SIMILAR FIGURES

PROJECT 6C
What different types of bridges have you crossed?
PROJECT: BUILD A MODEL OF A TRUSS BRIDGE

PROJECT 6D
What shapes tessellate?
PROJECT: DESIGN A TESSELLATION

Lesson 6.1 Analyze Translations

Solve & Discuss It!
Ashanti draws a trapezoid on the coordinate plane and labels it in Figure 1. Then she draws Figure 2. How can she determine whether the figures have the same side lengths and the same angle measures?

I can… translate two-dimensional figures.
Answer:
It is given that
Ashanti draws a trapezoid on the coordinate plane and labels it in Figure 1. Then she draws Figure 2.
The given figure is:

From the figure,
We can observe that the first figure was translated or moved to the right by some units and became a second figure
We know that,
A “Translation” is a transformation that moves every point in a figure the same distance in the same direction
We know that,
The pre-image and image in the “Translation” are similar
We know that,
Two figures are said to be similar if they are the same shape. In more mathematical language, two figures are similar if their corresponding angles are congruent, and the ratios of the lengths of their corresponding sides are equal.
Hence, from the above,
We can conclude that by using the “Translation” property, she can determine whether the figures have the same side lengths and the same angle measures

Focus on math practices
Be Precise How do you know that the method you described shows whether the side lengths and angle measures are equal? Explain.
Answer:
We know that,
A “Translation” is a transformation that moves every point in a figure the same distance in the same direction
We know that,
The pre-image and image in the “Translation” are similar
We know that,
Two figures are said to be similar if they are the same shape. In more mathematical language, two figures are similar if their corresponding angles are congruent, and the ratios of the lengths of their corresponding sides are equal.

Essential Question
How does a translation affect the properties of a two-dimensional figure?
Answer:
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way. If you translate a segment, it remains a segment, and its length doesn’t change. Similarly, if you translate an angle, the measure of the angle doesn’t change.

Try It!

The clients also want the small table below the window moved 5 feet to the right. Where should the architect place the small table? Draw the new location of the table on the plan.
Answer:
It is given that
The clients also want the small table below the window moved 5 feet to the right
Now,
The original plan is:

From the above plan,
We can observe that the window is at most right
So,
When we move the table,
The new location of the table will be below the window
Hence,
The location of the table on the new plan is:

Convince Me!
An equilateral triangle with side lengths 5 inches is translated 3 units down and 2 units right. Describe the shape and dimensions of the translated figure.
Answer:
It is given that
An equilateral triangle with side lengths 5 inches is translated 3 units down and 2 units right.
We know that,
In an equilateral triangle, all the side lengths are equal
Now,
Let the dimensions of an equilateral triangle be (x₁, y₁), (x₂, y₂), and (x₃, y₃)
So,
The dimensions of the translated figure are: (x₁ + 2, y₁ – 3), (x₂ + 2, y₂ – 3), and (x₃ + 2, y₃ – 3)
Now,
Let us suppose the dimensions of an equilateral to be:
(1, 1), (1, 6), and (5, 3)
So,
The dimensions of the translated equilateral triangle are:
(3, -2), (3, 3), and (7, 0)
We know that,
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way
So,
The shape of the translated figure remains the same as the original figure
Hence,
The representation and shape of the original and translated equilateral triangles are:

Try It!

Triangle ABC is translated 5 units right and 1 unit down. Graph and label the image A’ B’C’. If m∠A = 30° in ΔABC, then what is m∠A in ΔA’B’C’?

Answer:
The given figure is:

From the given figure,
The dimensions of ΔABC are:
A (-2, 3), B (-3, -4), and C (-5, -1)
Now,
Let the dimensions of an equilateral triangle be (x₁, y₁), (x₂, y₂), and (x₃, y₃)
So,
The dimensions of the translated figure are: (x₁ + 5, y₁ – 1), (x₂ + 5, y₂ – 1), and (x₃ + 5, y₃ – 1)
So,
The dimensions of the translated triangle (A’B’C’) are:
A (3, 2), B (2, -5), and C (0, -2)
We know that,
A translated figure has the same shape as the original shape i.e, the lengths and the angles of the translated figure are also the same as the original figure
So,
m∠A in Triangle ABC = m∠A in Triangle A’B’C’ = 30°
Hence,
The representation of the original and translated triangles are:

KEY CONCEPT

A translation, or slide, is a transformation that moves Preimage to every point of a figure the same distance and the same direction.

Do You Understand?
Question 1.
Essential Question How does a translation affect the properties of a two-dimensional figure?
Answer:
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way. If you translate a segment, it remains a segment, and its length doesn’t change. Similarly, if you translate an angle, the measure of the angle doesn’t change.

Question 2.
Construct Arguments Triangle L’M’N’ is the image of triangle LMN after a translation. How are the side lengths and angle measures of the triangles related? Explain.
Answer:
We know that,
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way. If you translate a segment, it remains a segment, and its length doesn’t change. Similarly, if you translate an angle, the measure of the angle doesn’t change.
Hence, from the above,
We can conclude that
The side lengths and the angle measures of Triangle L’M’N’ is the same as The side lengths and the angle measures of Triangle LMN

Question 3.
Generalize Sanjay determined that one vertex of a figure was mapped to its image by translating the point 2 units left and 7 units down. What is the rule that maps the other vertices of the figure to their images?
Answer:
It is given that
Sanjay determined that one vertex of a figure was mapped to its image by translating the point 2 units left and 7 units down
Now,
Now,
Let the dimensions of any vertex of the given be (x, y)
So,
The dimensions of any vertex of the translated figure are: (x – 2, y + 7)
Hence, from the above,
We can conclude that
The rule that maps the other vertices of the figure to their images is: (x – 2, y – 7)

Do You Know How?
In 4-6, use the coordinate plane.

Question 4.
Which figure is a translation of Figure A? Explain
Answer:
We know that,
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way. If you translate a segment, it remains a segment, and its length doesn’t change. Similarly, if you translate an angle, the measure of the angle doesn’t change.
So,
When we observe the coordinate plane,
We can see that A and C have the same shape i.e., have the same length and the same angle
Hence, from the above,
We can conclude that
Figure C is a translation of Figure A

Question 5.
Graph the translation of Figure A 3 units right and 4 units up.
Answer:
From the given coordinate plane,
The dimensions of Figure A are:
(-2, -1), (-4, -1), (-4, -2), (-3, -2), (-3, -3), (-4, -3), (-4, -4), and (-2, -4)
So,
The rule that maps the vertices of the given figure to their images is: (x + 3, y + 4)
So,
The vertices of the image are:
(1, 3), (-1, 3), (-1, 2), (0, 2), (0, 1), (-1, 1), (-1, 0), and (1, 0)
Hence,
The representation of Figure A and its translated figure is:

Question 6.
Describe the translation needed to move Figure B to the same position as the image from Item 5.
Answer:
From the given coordinate plane,
The dimensions of figure B are:
(-2, 4), (-4, 4), (-4, 3), (-3, 3), (-3, 2), (-4, 2), (-4, 1), and (-2, 1)
Hence,
The translation needed to move figure B to the same position as the image from Item 5 is:
The X-axis: Translate 3 units right
The Y-axis: Translate 1 unit down

Practice & Problem Solving

Question 7.
Graph G’R’A’M’, the image of GRAM after a translation 11 units right and 2 units up.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The dimensions of GRAM are:
G (-8, 2), R (-8, 6), A (-4, 6), and M (-5, 2)
So,
After a translation of 11 units right and 2 units up,
The dimensions of G’R’A’M’ are:
G’ (-8 + 11, 2 + 2), R’ (-8 + 11, 6 + 2), A’ (-4 + 11, 6 + 2), and M’ (-5 + 11, 2 + 2)
So,
G’ (3, 4), R’ (3, 8), A’ (7, 8), and M’ (6, 4)
Hence,
The representation of GRAM and its image G’R’A’M’ is:

Question 8.
∆A’ B’ C’ is a translation of ∆ABC. Describe the translation.

Answer:
The given coordinate plane is:

From the coordinate plane,
The dimensions of ΔABC are:
A (1, -5), B (-2, -2), and C (0, 0)
Now,
The dimensions of ΔA’B’C’ are:
A’ (0, -1), B’ (-3, 2), and C’ (-1, 4)
So,
By observing the dimensions of ΔABC and ΔA’B’C’,
The translation between ΔABC and ΔA’B’C is:
The x-axis: Translate 1 unit left
The y-axis: Translate 4 units up

Question 9.
Which triangle is the image of ∆DEF after a translation? Describe the translation.

Answer:
The given figure is:

We know that,
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way. If you translate a segment, it remains a segment, and its length doesn’t change. Similarly, if you translate an angle, the measure of the angle doesn’t change.
So,
From the given figure,
We can observe that
ΔMNO has the same shape as ΔDEF
So,
The image of ΔDEF is: ΔMNO
Now,
The dimensions of ΔDEF are:
D (-8, -10), F (-6, -10), and E (-8, -4)
Now,
The dimensions of ΔMNO are:
M (2, -10), O (4, -10), and N (2, -4)
So,
By observing the dimensions of ΔDEF and ΔMNo,
The translation between ΔDEF and ΔMNO is:
The x-axis: Translate 10 units right
The y-axis: No Translation required

Question 10.
The vertices of figure QRST are translated 3 units left and 11 units down to form figure Q’R’S’T’. Explain the similarities and differences between the two figures.
Answer:
It is given that
The vertices of figure QRST are translated 3 units left and 11 units down to form figure Q’R’S’T’.
Hence,
The similarities between figure QRST and figure Q’R’S’T’ are:
a. The two figures have the same side length
b. The two figures have the same angle measure
c. The two figures have the same shape
The differences between figure QRST and figure Q’R’S’T’ are:
a. Different dimensions of the vertices of QRST and Q’R’S’T’
b. Different positions of figure QRST and figure Q’R’S’T’

Question 11.
Graph the image of the given triangle after a translation of 3 units right and 2 units up.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The dimensions of the given triangle are:
(-3, 4), (-5, -1), and (-8, 3)
Now,
After a translation 3 units right and 2 units up,
The dimensions of the given triangle are:
(-3 + 3, 4 + 2), (-5 + 3, -1 + 2), and (-8 + 3, 3 + 2)
(0, 6), (-2, 1), and (-5, 5)
Hence,
The representation of the given triangle and its image is:

Question 12.
Quadrilateral P’Q’R’ S’ is the image of quadrilateral PQRS after a translation.

Answer:
It is given that
Quadrilateral P’Q’R’ S’ is the image of quadrilateral PQRS after a translation.
We know that,
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way. If you translate a segment, it remains a segment, and its length doesn’t change. Similarly, if you translate an angle, the measure of the angle doesn’t change.

a. If the length of side PQ is about 2.8 units, what is the length of side P’ Q’?
Answer:
It is given that
The length of the side PQ is about 2.8 units
Hence, from the above,
We can conclude that the length of side P’Q’ is also about 2.8 units

b. If m∠R = 75°, what is m∠R’?
Answer:
It is given that
m∠R = 75°
Hence, from the above,
We can conclude that
m∠R’ = 75°

Question 13.
Higher-Order Thinking A farmer has a plot of land shaped like the figure in the graph. There is another identical plot of land 120 yards east and 100 yards north of the original plot.

a. Draw the image after the given translation.
Answer:
It is given that
A farmer has a plot of land shaped like the figure in the graph. There is another identical plot of land 120 yards east and 100 yards north of the original plot.
Now,
The given plot of land is:

Now,
From the given plot of land,
The dimensions of land are:
(0, 0), (0, 300), (300, 0), and (300, 300)
Now,
After the translation of 120 units right and 100 units up,
The dimensions of the identical plot are:
(0 + 120, 0 + 100), (0 + 120, 300 + 100), (300 + 120, 0 + 100), and (300 + 120, 300 + 100)
(120, 100), (120, 400), (420, 100), and (420, 400)
Hence,
The representation of the plot and its image after the translation is:

b. Find the combined area of the 2 plots in square yards.
Answer:
From part (a),
We can observe that the shape of the plot of land and its identical is like a square
We know that,
The square has the equal side lengths
Now,
From part (a),
We can observe that the side length of the plot of land and its identical is: 300 yards
Now,
We know that,
The area of square = Side²
So,
The area of the plot of land = 300²
= 90,000 yard²
The area of the identical plot of land = 300²
= 90,000 yard²
So,
The combined area of the 2 plots = 90,000 + 90,000
= 1,80,000 yard²
Hence, from the above,
We can conclude that the combined area of the 2 plots is: 1,80,000 yard²

Assessment Practice
Question 14.
What is true about the preimage of a figure and its image created by a translation? Select all that apply.
☐ Each point in the image moves the same distance and direction from the preimage.
☐ Each point in the image has the same x-coordinate as the corresponding point in the preimage.
☐ Each point in the image has the same y-coordinate as the corresponding point in the preimage.
☐ The preimage and the image are the same size.
☐ The preimage and the image are the same shape.
Answer:
We know that,
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way. If you translate a segment, it remains a segment, and its length doesn’t change. Similarly, if you translate an angle, the measure of the angle doesn’t change.
Hence,
The statements that are true about the preimage of a figure and its image created by a translation are:

Question 15.
The vertices of parallelogram QUAD are Q(-7, -7), U(-6, -4), A(-2,-4), and D(-3, -7).

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The dimensions of the parallelogram QUAD are:
Q (-7, -7), U (-6, -4), A (-2, -4), and D (-3, -7)
Now,
After the translation of 11 units right and 9 units up,
The dimensions of the parallelogram QUAD are:
Q’ (-7 + 11, -7 + 9), U’ (-6 + 11, -4 + 9), A’ (-2 + 11, -4 + 9), and D’ (-3 + 11, -7 + 9)
Q’ (4, 2), U’ (5, 5), A’ (9, 5), and D’ (8, 2)

PART A
Graph and label the image of QUAD after a translation 11 units right and 9 units up.
Answer:
The representation of the parallelogram QUAd and its image is:

PART B
If m∠U = 110°, what is m∠ U’?
Answer:
We know that,
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way. If you translate a segment, it remains a segment, and its length doesn’t change. Similarly, if you translate an angle, the measure of the angle doesn’t change.
Hence, from the above,
We can conclude that
m∠U’ = 110°

PART C
If the length of side UA is 4 units, what is the length of side U’ A’?
Answer:
We know that,
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way. If you translate a segment, it remains a segment, and its length doesn’t change. Similarly, if you translate an angle, the measure of the angle doesn’t change.
Hence, from the above,
We can conclude that
The length of the side U’ A’ is: 4 units

Lesson 6.2 Analyze Reflections

Solve & Discuss It!
Dale draws a triangle on grid paper and labels it in Figure 1. Then using his pencil as a guide, he draws another triangle directly on the opposite side of the pencil so that the vertical side is now one square to the right of the pencil instead of one square to the left of the pencil. He labels this triangle in Figure 2. How are the figures the same? How are they different?

I can… reflect two-dimensional figures.
Answer:
Dale draws a triangle on grid paper and labels it in Figure 1. Then using his pencil as a guide, he draws another triangle directly on the opposite side of the pencil so that the vertical side is now one square to the right of the pencil instead of one square to the left of the pencil. He labels this triangle in Figure 2.
Hence,
The representation of Figure 1 and Figure 2 is:

So,
From the above representation,
The similarities between Figure 1 and Figure 2 are:
a. Both the figures have the same side lengths
b. Both the figures have the same angle measures
c. Both the figures have the same size
d. Both the figures have the same distance
The differences between Figure 1 and Figure 2 are:
a. The positions of both figures are different
b. The orientations of both figures are different
c. The directions of both the figures are different

Look for Relationships
What do you notice about the size, shape, and direction of the two figures?
Answer:
The similarities between Figure 1 and Figure 2 are:
a. Both the figures have the same side lengths
b. Both the figures have the same angle measures
c. Both the figures have the same size
d. Both the figures have the same distance
The differences between Figure 1 and Figure 2 are:
a. The positions of both figures are different
b. The orientations of both figures are different
c. The directions of both the figures are different

Focus on math practices
Reasoning Dale draws a line in place of his pencil and folds the grid paper along the line. How do the triangles align when the grid paper is folded? Explain.
Answer:
It is given that
Dale draws a line in place of his pencil and folds the grid paper along the line
Now,
The representation of Figure 1 and Figure 2 are:

Now,
From the given figure,
We can observe that the two triangles will stack on each other when the grid paper is folded i.e.,
a. The vertical side of figure 2 is on top of the vertical side of figure 1
b. The base of figure 2 is on top of the base of figure 1
c. The hypotenuse of figure 2 is on top of the hypotenuse of figure 1

Essential Question
How does a reflection affect the properties of a two-dimensional figure?
Answer:
When the reflection takes place along the x-axis,
The values of x will remain constant and the values of y will have a sign change
Ex:
When (x, y) and (-x, y) reflects along the x-axis,
The reflection of (x, y) will become (x, -y)
The reflection of (-x, y) will become (-x, -y)
When the reflection takes place along the y-axis,
The values of y will remain constant and the values of x will have a sign change
Ex:
When (x, y) and (x, -y) reflects along the y-axis,
The reflection of (x, y) will become (-x, y)
The reflection of (x, -y) will become (-x, -y)

Try It!

While updating the design, the architect accidentally clicked on the chair and reflected it across the centerline. Draw the new location of the chair on the plan.

Answer:
It is given that
While updating the design, the architect accidentally clicked on the chair and reflected it across the centerline.
Hence,
The representation of the new location of the chair is:

Convince Me!
How do the preimage and image compare after a reflection?
Answer:
A reflection is a transformation that turns a figure into its mirror image by flipping it over a line. The line of reflection is the line that a figure is reflected over. If a point is on the line of reflection then the image is the same as the preimage. Otherwise,
the image is not the same as the preimage. Images are always congruent to preimages

Try It!

Quadrilateral KLMN has vertices at K(2, 6), L(3, 8), M(5, 4), and N(3, 2). It is reflected across the y-axis, resulting in quadrilateral K’L’M’N’. What are the coordinates of point N’?
Answer:
It is given that
Quadrilateral KLMN has vertices at K(2, 6), L(3, 8), M(5, 4), and N(3, 2). It is reflected across the y-axis, resulting in quadrilateral K’L’M’N’
Now,
We know that,
When (x, y) and (x, -y) reflects along the y-axis,
The reflection of (x, y) will become (-x, y)
The reflection of (x, -y) will become (-x, -y)
So,
The reflection of N (3, 2) i.e., the coordinates of N’ is: (-3, 2)
Hence, from the above,
We can conclude that the coordinates of N’ are: (-3, 2)

Try It!

Polygon ABCDE is reflected across the line x = -2. Graph and label the image A’B’C’D’E’. Is m∠A= M∠A? Explain.

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of polygon ABCDE are:
A (-4, 4), B (-3, 3), C (-3, 2), D (-5, 1), and E (-5, 3)
It is given that polygon ABCDE is reflected along x = -2 i.e., alone the x-axis
We know that,
When (x, y) and (-x, y) reflects along the y-axis,
The reflection of (x, y) will become (x, -y)
The reflection of (-x, y) will become (-x, -y)
So,
The vertices of the reflection of polygon ABCDE i..e, A’B’C’D’E’ are:
A’ (-4, -4), B’ (-3, -3), C’ (-3, -2), D’ (-5, -1), and E’ (-5, -3)
We know that,
In reflection,
The side lengths and the angle measures in the image and the preimage are the same
Hence,
The representation of polygon ABCDE and its reflection polygon A’B’C’D’E’ is:

Hence,
m ∠A = M ∠A

KEY CONCEPT

A reflection, or flip, is a transformation that flips a figure across a line of reflection. The preimage and image are the same distance from the line of reflection but on opposite sides. They have the same size and shape but different orientations.

Do You Understand?
Question 1.
Essential Question How does a reflection affect the properties of a two-dimensional figure?
Answer:
When the reflection takes place along the x-axis,
The values of x will remain constant and the values of y will have a sign change
Ex:
When (x, y) and (-x, y) reflects along the x-axis,
The reflection of (x, y) will become (x, -y)
The reflection of (-x, y) will become (-x, -y)
When the reflection takes place along the y-axis,
The values of y will remain constant and the values of x will have a sign change
Ex:
When (x, y) and (x, -y) reflects along the y-axis,
The reflection of (x, y) will become (-x, y)
The reflection of (x, -y) will become (-x, -y)

Question 2.
Generalize What do you notice about the corresponding coordinates of the preimage and image after a reflection across the x-axis?
Answer:
When the reflection takes place along the x-axis,
The values of x will remain constant and the values of y will have a sign change
Ex:
When (x, y) and (-x, y) reflects along the x-axis,
The reflection of (x, y) will become (x, -y)
The reflection of (-x, y) will become (-x, -y)

Question 3.
Construct Arguments Jorge said the y-values would stay the same when you reflect a preimage across the line y = 5 since the y-values stay the same when you reflect a preimage across the y-axis. Is Jorge correct? Explain.
Answer:
It is given that
Jorge said the y-values would stay the same when you reflect a preimage across the line y = 5 since the y-values stay the same when you reflect a preimage across the y-axis.
Now,
We know that,
When the reflection takes place along the y-axis,
The values of y will remain constant and the values of x will have a sign change
Ex:
When (x, y) and (x, -y) reflects along the y-axis,
The reflection of (x, y) will become (-x, y)
The reflection of (x, -y) will become (-x, -y)
Hence, from the above,
We can conclude that Jorge is correct

Do You Know How?
Question 4.
Is AX’ Y’ Z’ a reflection of AXYZ across line g?

Answer:
The given figure is:

Now,
From the given figure,
We can observe that ΔXYZ is reflected across the line g i.e., y-axis
We know that,
When the reflection takes place along the y-axis,
The values of y will remain constant and the values of x will have a sign change
Ex:
When (x, y) and (x, -y) reflects along the y-axis,
The reflection of (x, y) will become (-x, y)
The reflection of (x, -y) will become (-x, -y)
So,
From the figure,
We can observe that
The negative x-coordinates of the vertices of ΔXYZ became the positive x -coordinates for Δ X’Y’Z’
Hence, from the above,
We can conclude that ΔX’Y’Z’ is the reflection of ΔXYZ across the line g

Use the coordinate grid below for 5 and 6.

Question 5.
Describe the reflection of figure EFGH.
Answer:
The given coordinate plane is:

From the given coordinate plane,
We can observe that
The reflection of figure EFGH is: Figure E’F’G’H’
Now,
We can observe that
The reflection of the figure EFGH takes place across the x-axis
So,
The x-coordinates of the vertices of the reflection of the figure EFGH will be constant and only y-coordinates will have a change in value
Hence,
The figure EFGH will flip i.e., top becomes down and vice-versa to form a reflection i.e., figure E’F’G’H’

Question 6.
Draw the image that would result from a reflection of figure E’F’G’H across the line x = -1.
Answer:
The given coordinate plane is:

Now,
From the reflection of EFGH i.e., figure E’F’G’H’,
We can observe that the vertices of the figure E’F’G’H’ are:
E’ (-8, 7), F’ (-5, 5), G’ (-4, 6), and H’ (-2, 6)
Now,
To form the image of the figure E’F’G’H’,
We need to reflect the figure E’F’G’H’ across the y-axis
When the reflection takes place along the x-axis,
The values of x will remain constant and the values of y will have a sign change
Ex:
When (x, y) and (-x, y) reflects along the x-axis,
The reflection of (x, y) will become (x, -y)
The reflection of (-x, y) will become (-x, -y)
So,
The vertices for the image of the figure E’F’G’H’ are:
E’ (-8, -7), F’ (-5, -5), G’ (-4, -6), and H’ (-2, -6)
Hence,
The representation of the figure E’F’G’H’ and its image is:

Practice & Problem Solving

Question 7.
Leveled Practice Trapezoid ABCD is shown. Draw the reflection of trapezoid ABCD across the y-axis.

Plot the points and draw trapezoid A’ B’C’D’.
Answer:
From the given trapezoid ABCD,
The vertices are:
A (2, 8), B (6, 8), C (8, 3), and D (1, 3)
It is given that
Draw the reflection of trapezoid ABCD across the y-axis i.e., y is constant
Now,
We know that,
When the reflection takes place along the y-axis,
The values of y will remain constant and the values of x will have a sign change
Ex:
When (x, y) and (x, -y) reflects along the y-axis,
The reflection of (x, y) will become (-x, y)
The reflection of (x, -y) will become (-x, -y)
So,
The vertices for the reflection of trapezoid ABCD are:
A’ (-2, 8), B (-6, 8), C’ (-8, 3), and D (-1, 3)
So,
The points of the preimage and image are:

Hence,
The representation of trapezoid A’B’C’D’ is:

Question 8.
Reasoning is triangle A’ B’C’a reflection of triangle ABC across the line? Explain.

Answer:
The given figure is:

Now,
From the given figure,
We can observe that the reflection of ΔABC takes place across the y-axis
Now,
When the reflection takes place along the y-axis,
The values of y will remain constant and the values of x will have a sign change
Ex:
When (x, y) and (x, -y) reflects along the y-axis,
The reflection of (x, y) will become (-x, y)
The reflection of (x, -y) will become (-x, -y)
So,
From the vertices of ΔABC,
The x-coordinates are negative and the y-coordinates are positive
So,
For the reflection of ΔABC along the y-axis,
The x-coordinates will have to become positive and the y-coordinates will be positive as in the vertices of ΔABC
Hence, from the above,
We can conclude that ΔA’B’C’ is the reflection of ΔABC across the given line

Question 9.
Your friend gives you the graph of quadrilateral ABCD and its image, quadrilateral A’B’C’D’. What reflection produces this image?

Answer:
It is given that
Your friend gives you the graph of quadrilateral ABCD and its image, quadrilateral A’B’C’D’
Now,
The quadrilateral ABCD and its image quadrilateral A’B’C’D’ is:

Now,
From the given figure,
We can observe that
The quadrilateral ABCD and its image will have the same y-coordinates but the x-coordinates are different
We know that,
When the reflection takes place across the y-axis,
The y-coordinates are the same for the image and the preimage
The x-coordiantes have sign changes with the same values for the image and the preimage
Hence, from the above,
We can conclude that the reflection across the y-axis produces the image quadrilateral A’B’C’D’

Question 10.
Construct Arguments Your friend incorrectly says that the reflection of ∆EFG to its image ∆E’ F’G’ is a reflection across the y-axis.

a. What is your friend’s mistake?
Answer:
The given coordinate plane is:

From the above figure,
We can observe that
For the vertices of the given figure,
The x-coordinates remain the same whereas the y-coordinates are different
We know that,
When the reflection takes place across the x-axis,
The x-coordinates of the preimage and the image are the same
The y-coordinates of the preimage and the image will have sign change with the same values
Hence, from the above,
We can conclude that the reflection of ΔEFG takes pace across the x-axis instead of across the y-axis

b. What is the correct description of the reflection?
Answer:
From the given figure,
We can observe that
For the vertices of the given figure,
The x-coordinates remain the same whereas the y-coordinates are different
We know that,
When the reflection takes place across the x-axis,
The x-coordinates of the preimage and the image are the same
The y-coordinates of the preimage and the image will have sign change with the same values
Hence, from the above,
We can conclude that the reflection of ΔEFG to its image ΔE’F’G’ takes pace across the x-axis

Question 11.
Make Sense and Persevere The vertices of ∆ABC are A(-5, 5), B(-2,5), and C(-2, 3). If ∆ ABC is reflected across the line y = -1, find the coordinates of the vertex C’
Answer:
It is given that
The vertices of ∆ABC are A(-5, 5), B(-2,5), and C(-2, 3) and ∆ ABC is reflected across the line y = -1
We know that,
When the reflection takes place across the y-axis,
The y-coordinates of the preimage and the image are the same
The x-coordinates of the preimage and the image will have sign change with the same values
So,
The reflection of the vertex C (-2, 3) is: (2, 3)
Hence, from the above,
We can conclude that the coordinates of the vertex C’ i.e., the reflection of the vertex C is: (2, 3)

Question 12.
Higher Order Thinking What reflection of the parallelogram ABCD results in image A’B’C’D?

Answer:
The given figure is:

From the given figure,
We can observe that the parallelogram ABCD and its reflection A’B’C’D’ are parallel to the x-axis
Now,
We can observe that,
For both the parallelogram ABCD and its reflection A’B’C’D’,
The x-coordinates are changing but the y-coordinates remain the same
Now,
We know that,
When the reflection takes place across the y-axis,
The y-coordinates of the preimage and the image are the same
The x-coordinates of the preimage and the image will have sign change with the same values
Hence, from the above,
We can conclude that the reflection across the y-axis of the parallelogram ABCD results in image A’B’C’D

Assessment Practice
Question 13.
∆JAR has vertices J(4, 5), A6, 4), and R(5,2). What graph shows ∆JAR and its image after a reflection across the line y = 1?
PART A

Answer:
It is given that
∆JAR has vertices J(4, 5), A6, 4), and R(5,2)
Now,
We know that,
The reflection across the y-axis is:
The y -coordinates are the same
The x-coordinates will have sign changes with the same value
So,
The reflection of the vertices of ΔJAR is:
J’ (-4, 5), A’ (-6, 4), and R (-5, 2)
Hence,f rom the above,
We can conclude that option D matches with the given vertices of the reflection i.e., ΔJ’A’R’

PART B
The measure of ∠A = 90°. What is m∠A’?
Answer:
We know that,
In Reflection,
The side lengths and the angle measures of the preimage and the image are the same
Hence, from the above,
We can conclude that
∠A = ∠A’ = 90°

Lesson 6.3 Analyze Rotations

Explain It!
Maria boards a car at the bottom of the Ferris wheel. She rides to the top, where the car stops. Maria tells her friend that she completed \(\frac{1}{4}\) of the turn before the car stopped.

I can… rotate a two-dimensional figure.

A. Do you agree with Maria? Explain.
Answer:
It is given that
Maria boards a car at the bottom of the Ferris wheel. She rides to the top, where the car stops. Maria tells her friend that she completed \(\frac{1}{4}\) of the turn before the car stopped.
Now,
We know that,
From the given figure,
The Ferris wheel looks like a circle
We know that
The total angle measure of a circle (Ferris wheel) is: 360°
But,
It is given that
Maria tells her friend that she completed \(\frac{1}{4}\) of the turn before the car stopped
But, from the figure,
We can observe that she completed the \(\frac{1}{2}\) of the ride
Hence, from the above,
We can conclude that we can’t agree with Maria

B. How could you use angle measures to describe the change in position of the car?
Answer:
We know that,
From the given figure,
The Ferris wheel looks like a circle
We know that
The total angle measure of a circle (Ferris wheel) is: 360°
So,
The starting position of the car in terms of angle measure is given as 0°
The \(\frac{1}{2}\) of the position of the car in terms of angle measure = \(\frac{360°}{2}\)
= 180°
Hence, from the above,
We can conclude that
In terms of angle measures,
The starting position of the car is: 0°
The \(\frac{1}{2}\) of the car is: 180°

Focus on math practices
Construct Arguments How can you describe Maria’s change in position when her car returns to the position at which she began the ride?
Answer:
We know that,
The starting position of the car is: 0°
The \(\frac{1}{2}\) of the car is: 180°
Now,
If Maria’s car returns to the position at which she began the ride, then
Maria has completed the Ferris wheel (or) Maria returned back from her previous \(\frac{1}{2}\) of the position of the car
Hence,
In terms of angle measures,
The change in Maria’s position when her car returns to the position at which she began the ride is: 360° (or) -180°
Here,
-180° represents that Maria returned back from her previous \(\frac{1}{2}\) of the position of the car

Essential Question
How does rotation affect the properties of a two-dimensional figure?
Answer:
When you rotate a two-dimensional figure, you are just moving it.
Ex:
If you rotate a rectangle, then it will remain a rectangle, just moved wherever you move it. This is similarly the same with an angle and aside length, the measure of the angle and the side length won’t change.

Try It!

The architect continues to rotate the umbrella in a counterclockwise direction until it is in its original position. What is the angle of this rotation?

Answer:
It is given that
The architect continues to rotate the umbrella in a counterclockwise direction until it is in its original position.
Now,
We know that,
If any figure rotates until it comes to its original position again, then that means the figure completed a full cycle (or) revolution and the angle of the full cycle is: 360°
Hence, from the above,
We can conclude that the architect continues to rotate the umbrella for 360° in a counterclockwise direction until it is in its original position

Convince Me!
How does an image compare to its preimage after a -45° rotation?
Answer:

Try It!

The coordinates of the vertices of quadrilateral HIJK are H(1,4), I(3, 2), J(-1,-4), and K(-3, -2). If quadrilateral HIJK is rotated 270° about the origin, what are the vertices of the resulting image, quadrilateral H’I’I’K?
Answer:
It is given that
The coordinates of the vertices of quadrilateral HIJK are H(1,4), I(3, 2), J(-1,-4), and K(-3, -2)
Now,.
We know that,
The change in the x and y-coordinates for the given angle of rotation is:

Now,
It is given that to rotate the quadrilateral HIJK 270° about the origin to form the quadrilateral H’I’J’K’
So,
The vertices of the quadrilateral H’I’J’K’ are:
H’ (4, -1), I’ (2, -3), J’ (-4, 1), K’ (-2, 3)
Hence,
The representation of the quadrilateral HIJK and its image H’I’J’K’ is:

Try It!

Describe the rotation that maps ∆FGH to ∆FG’H’.

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of ΔFGH are:
F (-5, -2), G (-1, -2), and H (-1, 1)
The vertices of ΔF’G’H’ are:
F’ (5, 2), G’ (1, 2), and H’ (1, -1)
Now,
To find the rotation that maps ΔFGH through ΔF’G’H’, the following steps are:
Step 1:
Draw rays from the origin through point G and point G’
Step 2:
Measure the angle formed by the rays
So,
The representation of ΔFGH and its image ΔF’G’H’ with its angle of rotation is:

Hence, from the above,
We can conclude that the angle of rotation that maps ΔFGH through ΔF’G’H’ is: 180°

KEY CONCEPT

A rotation is a transformation that turns a figure about a fixed point called the center of rotation. The angle of rotation is the number of degrees the figure is rotated. The x- and y-coordinates change in predictable ways when rotated.

Do You Understand?
Question 1.
Essential Question How does rotation affect the properties of a two-dimensional figure?
Answer:
When you rotate a two-dimensional figure, you are just moving it.
Ex:
If you rotate a rectangle, then it will remain a rectangle, just moved wherever you move it. This is similarly the same with an angle and aside length, the measure of the angle and the side length won’t change.

Question 2.
Reasoning If a preimage is rotated 360 degrees about the origin how can you describe its image?
Answer:
We know that,
To complete the full cycle i.e., to a preimage to return to its original position, the angle measure is: 360°
Hence, from the above,
We can conclude that
If a preimage is rotated 360° about the origin, then the image and the preimage are the same

Question 3.
Construct Arguments In Example 3, side AB is parallel to side DC. How are side A’ B’ and side D’ C’ related? Explain.

Answer:
In Example 3,
The given two-dimensional figure is a parallelogram
We know that,
In a parallelogram, the opposite sides have the same side lengths
So,
Now,
It is given that
In parallelogram ABCD,
AB is parallel to CD
So,
In parallelogram A’B’C’D’,
A’B’ is parallel to C’D’ since these two sides are just the images of the sides AB and CD
Hence, from the above,
We can conclude that A’B’ is parallel to C’D’

Do You Know How?
Question 4.
The coordinates of the vertices of rectangle ABCD are A(3,-2), B(3, 2), C(-3, 2), and D(-3,-2).
a. Rectangle ABCD is rotated 90° about the origin. What are the coordinates of the vertices of rectangle A’B’C’D’?
Answer:
It is given that
The coordinates of the vertices of rectangle ABCD are A(3,-2), B(3, 2), C(-3, 2), and D(-3,-2).
Now,
We know that,
The change in the x and y-coordinates for the given angle of rotation is:

Now,
It is given that to rotate the rectangle ABCD 90° about the origin to form the rectangle A’B’C’D’
So,
The vertices of the rectangle A’B’C’D’ are:
A’ (2, 3), B’ (-2, 3), C’ (-2, -3), and D’ (2, -3)
Hence,
The representation of the rectangle ABCD and its image rectangle A’B’C’D’ is:

b. What are the measures of the angles of A’B’C’D’?
Answer:
From part (a),
We can observe that all the angles of the rectangle A’B’C’D’ are the same
Hence, from the above,
We can conclude that the angle measures of the rectangle A’B’C’D’ are:
∠A = 90°, ∠B = 90°, ∠C = 90°,and ∠D = 90°

Question 5.
Describe the counterclockwise rotation that maps ∆QRS to ∆Q’R’S.

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of ΔQRS are:
Q (6, 2), R (4, 9), and S (2, 7)
The vertices of ΔQ’R’S’ are:
Q’ (2, -6), R’ (9, -4), and S’ (7, -2)
Now,
To find the rotation that maps ΔQRS through ΔQ’R’S’, the following steps are:
Step 1:
Draw rays from the origin through point Q and point Q’
Step 2:
Measure the angle formed by the rays
So,
The representation of ΔQRS and its image ΔQ’R’S’ with its angle of rotation is:

Hence, from the above,
We can conclude that the angle of rotation that maps ΔQRS through ΔQ’R’S’ is: 90°

Practice & Problem Solving

Question 6.
What is the angle of rotation about the origin that maps △ΡQR to △Ρ’ Ο’ R’?

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of ΔPQR are:
P (3, 3), Q (5, 3), and R (5, 7)
The vertices of ΔP’Q’R’ are:
P’ (-3, 3), Q’ (-3, 5), and S’ (-7, 5)
Now,
To find the rotation that maps ΔPQR through ΔP’Q’R’, the following steps are:
Step 1:
Draw rays from the origin through point P and point P’
Step 2:
Measure the angle formed by the rays
So,
The representation of ΔPQR and its image ΔP’Q’R” with its angle of rotation is:

Hence, from the above,
We can conclude that the angle of rotation that maps ΔPQR through ΔP’Q’R’ is: 90°

Question 7.
Is △X’ Y’Z’ a rotation of △XYZ? Explain.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔXYZ are:
X (-2, 3), Y (-5, 4), and Z (-2, 7)
The vertices of ΔX’Y’Z’ are:
X’ (3, 2), Y’ (4, 5), and Z’ (7, 2)
Now,
To find the angle of rotation,
Compare the x and y-coordinates of ΔXYZ and ΔX’Y’Z’
So,
(x, y) before rotation ——-> (y, -x) after rotation
We know that,

Hence, from the above,
We can conclude that ΔX’Y’Z’ is a rotation of ΔXYZ

Question 8.
△PQR is rotated 270° about the origin. Graph and label the coordinates of P’, Q’, and R’.

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of ΔPQR are:
P (2, 3), Q (4, 6), and R (2, 7)
It is given that
Rotate ΔPQR 270° at the origin
We know that,

So,
The vertices of ΔP’Q’R’ when the angle of rotation is 270° are:
P’ (3, -2), Q’ (6, -4), and R’ (7, -2)
Hence,
The representation of ΔPQR and its image ΔP’Q’R’ is:

Question 9.
Is △P’ Q’R’a 270° rotation of △PQR about the origin? Explain.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔPQR are:
P (3, 3), Q (7, 4), and R (3, 5)
The vertices of ΔP’Q’R’ are:
P’ (-3, 3), Q’ (-4, 7), and R’ (-5, 3)
Now,
To find the angle of rotation,
Compare the x and y-coordinates of ΔPQR and ΔP’Q’R’
So,
(x, y) before rotation ——-> (-y, x) after rotation
We know that,

Hence, from the above,
We can conclude that ΔP’Q’R’ is a rotation of ΔPQR

Question 10.
Reasoning Explain why any rotation can be described by an angle between 0° and 360°.
Answer:
If you rotate an object 360°, it’s like the object never moved because the object would still be in the same spot as if you didn’t move it.
Hence,
Any rotation can be described by an angle of 0° to 360°

Question 11.
Rotate rectangle KLMN 270° about the origin.

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of rectangle KLMN are:
K (-3, 2), L (-5, 2), M (-5, 4), and N (-3, 4)
It is given that
Rotate rectangle KLMN 270° at the origin
We know that,

So,
The vertices of rectangle K’L’M’N’ when the angle of rotation is 270° are:
K’ (2, 3), L’ (2, 5), M’ (4, 5), and N’ (4, 3)
Hence,
The representation of rectangle KLMN and its image rectangle K’L’M’N’ is:

 

Question 12.
Higher-Order Thinking An architect is designing a new windmill with four sails. In her sketch, the sails’ center of rotation is the origin, (0, 0), and the tip of one of the sails, point Q, has coordinates (2, -3). She wants to make another sketch that shows the windmill after the sails have rotated 270° about their center of rotation. What would be the coordinates of?
Answer:
It is given that
An architect is designing a new windmill with four sails. In her sketch, the sails’ center of rotation is the origin, (0, 0), and the tip of one of the sails, point Q, has coordinates (2, -3). She wants to make another sketch that shows the windmill after the sails have rotated 270° about their center of rotation
So,
We have to rotate point Q 270° about the origin
We know that,

So,
When we rotate any point 270° about the origin,
(x, y) before rotation ——–> (y, -x)
So,
The representation of point Q after representation is: (-3, -2)
Hence,
The representation of point Q and its image Q’ is:

Assessment Practice
Question 13.
A rotation about the origin maps △TRI to △T’ R’I’.
PART A
Which graph shows an angle you could measure to find the angle of rotation about the origin?

Answer:
It is given that
A rotation about the origin maps △TRI to △T’ R’I’
Now,
We know that,
We can find the angle of rotation only when the vertex maps with its image but not with any other images
Hence, from the above,
We can conclude that option A matches the given situation

PART B
What is the angle of rotation about the origin?
A. 90°
B. 180°
C. 270°
D. 360°
Answer:
From part (a),
When we observe option A,
The vertices of ΔTRI are:
T (-3, 3), R (-5, 3), and I (-4, 5)
The vertices of ΔT’R’I’ are:
T’ (3, -3), R’ (5, -3), and I’ (4, -5)
We know that,

So,
When we compare the vertices of ΔTRI and ΔT’R’I’,
We can observe that
(x, y) before rotation ———–> (-x, -y) after rotation
Hence, from the above,
We can conclude that the angle of rotation about the origin is: 180°

Lesson 6.4 Compose Transformations

Solve & Discuss It!
How can you map Figure A onto Figure B?

I can… describe and perform a sequence of transformations.
Answer:
The given figure is:

Now,
The steps to obtain Figure B from Figure A are:
Step 1:
Reflect Figure A across the x-axis
Step 2:
Reflect the figure that we obtained in Step 1 across the y-axis
Hence,
The representation of the mapping of Figure A and Figure B using the above steps is:

Focus on math practices
Look for Relationships Is there another transformation or sequence of transformations that will map Figure A to Figure B?
Answer:
Yes, there are another sequence of transformations that will map Figure A to figure B
Now,
The steps to obtain Figure B from Figure A are:
Step 1:
Reflect Figure A across the y-axis
Step 2:
Reflect the figure that we obtained in Step 1 across the x-axis
Hence,
The representation of the mapping of Figure A and Figure B using the above steps is:

Essential Question
How can you use a sequence of transformations to map a preimage to its image?
Answer:
Mathematical transformations involve changing an image in some prescribed manner. There are four main types of transformations They are:
A) Translation B) Rotation C) Reflection D) Dilation

Try It!

Ava decided to move the cabinet to the opposite wall. What sequence of transformations moves the cabinet to its new position?

Answer:
It is given that
Ava decided to move the cabinet to the opposite wall
So,
From the figure,
We can observe that
To move the cabinet wall to the opposite wall, the following sequences of transformations have to be followed:
Step 1:
Translate the cabinet 8 units down
Step 2:
Rotate the cabinet 360° counterclockwise
Hence, from the above,
We can conclude that
The new position of the cabinet is:

Convince Me!
Ava decides that she would like the chairs to be placed directly across from the couch. What is a sequence of transformations that she can use to move the chairs to their new positions?

Try It!

What is another sequence of transformations that maps △ABC onto △A” B” C”?

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔABC are:
A (-5, 5), B (-3, 3), and C (-6, 1)
Now,
Another sequence of transformations that maps ΔABC onto ΔA”B” C” is:
Step 1:
Draw ΔABC and make its reflection across the y-axis and name it ΔA’B’C’
Step 2:
Translate ΔA’B’C’ 2 units to the right and 6 units down
Step 3:
Reflect the figure we obtained in step 2 across the y-axis
Hence, from the above,
We can conclude that
The representation of another sequence of transformations that maps ΔABC onto ΔA”B” C” is:

KEY CONCEPT
You can use a sequence of two or more transformations to map a preimage to its image.
You can map △ABC onto △Α” Β” C” by translation of 3 units right followed by a 90° clockwise rotation about the origin.

Do You Understand?
Question 1.
Essential Question How can you use a sequence of transformations to map a preimage to its image?
Answer:
Mathematical transformations involve changing an image in some prescribed manner. There are four main types of transformations They are:
A) Translation B) Rotation C) Reflection D) Dilation

Question 2.
Make Sense and Persevere A preimage is rotated 180° about the origin and then rotated 180° about the origin again. Compare the preimage and image.
Answer:
It is given that
A preimage is rotated 180° about the origin and then rotated 180° about the origin again
Now,
When the preimage is rotated 180° and again 180°
The image will be the same as the preimage
Hence, from the above,
We can conclude that
When a preimage is rotated 180° about the origin and then rotated 180° about the origin again, the preimage and the image will be the same

Question 3.
Reasoning A figure ABC, with vertices A(2, 1), B(7, 4), and C(2, 7), is rotated 90° clockwise about the origin, and then reflected across the x-axis. Describe another sequence that would result in the same image.
Answer:
It is given that
A figure ABC, with vertices A(2, 1), B(7, 4), and C(2, 7), is rotated 90° clockwise about the origin, and then reflected across the x-axis
So,
The steps for another sequence of transformations that would result in the same image as the given situation is:
Step 1:
Draw the given vertices of Triangle ABC
Step 2:
Rotate Triangle ABC 90° counterclockwise
Step 3:
Reflect the image that we obtained in step 2 across the x-axis so that we will get the same image as in the given situation
Hence,
The representation of another sequence of transformations is:

Do You Know How?
In 4-6, use the diagram below.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of Figure WXYZ are:
W (2, 4), X (5, 4), Y (5, 2), and Z (2, 2)
The vertices of Figure W’X’Y’Z’ are:
W’ (-4, -4), X’ (-4, -1), Y’ (-2, -1), and Z’ (-2, -4)

Question 4.
Describe a sequence of transformations that maps rectangle WXYZ onto rectangle W’X’Y’Z’.
Answer:
The steps that we have to follow to obtain the given sequence of transformations that maps rectangle WXYZ onto rectangle W’X’Y’Z’ are:
Step 1:
Draw rectangle WXYZ
Step 2:
Reflect rectangle WXYZ across the x-axis
Step 3:
Rotate the image we obtained in step 2 90° counterclockwise
Step 4:
Translate the image we obtained in step-3 6 units left so that we can obtain rectangle W’X’Y’Z’
Hence,
The representation of the sequence of transformations for the given situation is:

Question 5.
Describe another way that you could map rectangle WXYZ onto W’X’Y’Z’.
Answer:
The steps for another way of transformations that maps rectangle WXYZ onto W’X’Y’Z’ are:
Step 1:
Draw Rectangle WXYZ
Step 2:
Rotate rectangle WXYZ 90° counterclockwise
Step 3:
Rotate the image we obtained in step-2 180° counterclockwise
Step 4:
Translate the image we obtained in step 3 6 units down so that we can obtain rectangle W’X’Y’Z’
Hence,
The representation of another sequence of transformations is:

Question 6.
Draw the image of rectangle WXYZ after a reflection across the line y = 1 and a translation 1 unit right. Label the image W” X” Y” Z”.
Answer:
The vertices of rectangle WXYZ are:
W (2, 4), X (5, 4), Y (5, 2), and Z (2, 2)
After the reflection across the line y = 1,
The vertices of rectangle WXYZ are:
W’ (2, -5), X’ (5, -5), Y’ (5, -3), and Z’ (2, -3)
After the translation of 1 unit right,
The vertices of rectangle WXYZ are:
W” (3, -5), X” (6, -5), Y” (6, -3), and Z” (3, -3)
Hence,
The representation of rectangle WXYZ and its image W”X” Y”Z” after the above sequence of transformations is:

Practice & Problem Solving

Question 7.
Leveled Practice Describe a sequence of transformations that maps △QRS onto △TUV.

A translation ________ units left and ________ units down, followed by a _________ across the ________.
Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔQRS are:
Q (3, 4), R (7, 4), and S (6, 9)
The vertices of ΔTUV are:
T (0, 0), U (-4, 0), and V (-3, 5)
Now,
In order to obtain the vertices of ΔTUV, the following transformations we have to follow are:
Step 1:
Translate the vertices of ΔQRS 3 units left and 4 units down
Step 2:
Reflect the image we obtained in step 1 across the x-axis in order to get the vertices of ΔTUV
Hence, from the above,
We can conclude that
A translation of 3 units left and 4 units down, followed by a reflection across the x-axis

Question 8.
Model with Math
A family moves a table, shown as rectangle EFGH, by translating it 3 units left and 3 units down followed by a 90° rotation about the origin. Graph E’ F’G’H’ to show the new location of the table.

Answer:
It is given that
A family moves a table, shown as rectangle EFGH, by translating it 3 units left and 3 units down followed by a 90° rotation about the origin
Now,
The given coordinate plane is:

From the given coordinate plane,
The vertices of rectangle EFGH are:
E (3, 3), F (8, 3), G (8, 7), and H (3, 7)
Now,
To obtain rectangle E’F’G’H’,
The following series of transformations are:
Step 1:
By translating 3 units left and 3 units down,
The vertices of rectangle EFGH will become:
E (3 – 3, 3 – 3), F (8 – 3, 3 – 3), G (8 – 3, 7 – 3), and H (3 – 3, 7 – 3)
E (0, 0), F (5, 0), G (5, 4), and H (0, 4)
Step 2:
Rotate the vertices we obtain in step 1 90° counterclockwise about the origin
We know that,
When we rotate a point 90° counterclockwise about the origin,
(x, y) before rotation ——> (-y, x) after rotation
So,
The vertices of rectangle E’F’G’H’ are:
E’ (0, 0), F’ (0, 5), G’ (-4, 5), and H’ (-4, 0)
Hence,
The representation of the new location of the table is:

Question 9.
Describe a sequence of transformations that maps quadrilateral ABCD to quadrilateral HIJK.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of quadrilateral ABCD are:
A (3, 1), B (4, 1), C (4, 3), and D (3, 3)
The vertices of quadrilateral HIJK are:
H (-3, 0), I (-2, 0), J (-2, -2), and K (-3, -2)
Now,
The series of transformations that maps quadrilateral ABCD onto quadrilateral HIJK are:
Step 1:
Draw quadrilateral ABCD
Step 2:
Reflect quadrilateral ABCD across the x-axis
So,
The vertices of quadrilateral ABCD are:
A (3, -1), B (4, -1), C (4, -3), and D (3, -3)
Step 3:
Translate 6 units left and 1 unit up
So,
The vertices that we obtained in step 2 will become (The vertices of quadrilateral HIJK):
H (-3, 0), I (-2, 0), J (-2, -2), and K (-3, -2)
Hence,
The representation of the series of transformations that map quadrilateral ABCD onto quadrilateral HIJK are:

Question 10.
Map △QRS to △Q’R’ S’with a reflection across the y-axis followed by a translation 6 units down.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔQRS are:
Q (-3, 5), R (-2, 4), and S (-5, 3)
Now,
The series of transformations that maps ΔQRS to ΔQ’R’S’ as given above are:
Step 1:
Reflect the vertices of ΔQRS along the y-axis
So,
Q (3, 5), R (2, 4), and S (5, 3)
Step 2:
Translate the vertices that we obtain in step-1 6 units down so that we can obtain the vertices of ΔQ’R’S’
So,
Q’ (3, 5 – 6), R’ (2, 4 – 6), and S’ (5, 3 – 6)
Q’ (3, -1), R’ (2, -2), and S’ (5, -3)
Hence,
The representation of the series of transformations for the given situation is:

 

Question 11.
Higher-Order Thinking A student says that he was rearranging furniture at home and he used a glide reflection to move a table with legs from one side of the room to the other. Will a glide reflection result in a functioning table? Explain.
Answer:
It is given that
A student says that he was rearranging furniture at home and he used a glide reflection to move a table with legs from one side of the room to the other.
We know that,
A glide reflection is a  sequence of translation and reflection
Now,
From the given situation,
We can observe that the table is moving from one room to the other
So,
The “Translation” occurs
Now,
After moving he will rearrange the table in the room
So,
A “Reflection” may take place
Hence, from the above,
We can conclude that a glide reflection result in a functioning table

Assessment Practice
Question 12.
PART A
Which sequence of transformations maps rectangle ABCD onto rectangle A’ B’C’D?

A. translation 6 units down, reflection across the x-axis
B. reflection across the x-axis, translation 6 units right
C. reflection across the x-axis, translation 6 units left
D. translation 6 units left, reflection across the y-axis
Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of rectangle ABCD are:
A (2, -2), B (4, -2), C (4, -3), and D (2, -3)
The vertices of rectangle A’B’C’D’ are:
A’ (-4, 2), B’ (-2, 2), C’ (-2, 3), and D’ (-4, 3)
So,
The sequence of transformations that maps rectangle ABCD to rectangle A’B’C’D’ is:

Hence, from the above,
We can conclude that option C matches the given situation

PART B
Describe a sequence of transformations that maps A’B’C’ D’ onto ABCD.
Answer:
The sequence of transformations that maps A’B’C’D’ onto ABCD is:
Step 1:
Draw the rectangle ABCD
So,
The vertices of rectangle ABCD are:
A (2, -2), B (4, -2), C (4, -3), and D (2, -3)
Step 2:
Reflect rectangle ABCD across the x-axis
So,
The vertices of rectangle ABCD are:
A (2, 2), B (4, 2), C (4, 3), and D (2, 3)
Step 3:
Translate the vertices that we obtained in step 2 6 units left so that we can obtain rectangle A’B’C’D’
So,
A’ (2 – 6, 2), B’ (4 – 6, 2), C’ (4 – 6, 3), and D’ (2 – 6, 3)
So,
A’ (-4, 2), B’ (-2, 2), C’ (-2, 3), and D’ (-4, 3)

Question 13.
PART A
Which figures are the image of Figure A after a reflection across the x-axis and a translation of 4 units right?

A. Figure B
B. Figure C
C. Figure D
D. Figure E
Answer:
We know that,
A “Reflection” is called a “Flip” but the reflection does not affect the shape and length of the figure
So,
From the given figures,
When we observe Figures A and B,
We can say that Figure B is a reflection of A because the shape and length is the same
But,
when we observe the other figures,
We have different shapes and lengths between the preimage and the image
Hence, from the above,
We can conclude that
Figure B is the image of Figure A after a reflection across the x-axis and a translation of 4 units right

PART B
Which figure can be transformed into Figure G after a rotation 90° about the origin, then a translation 13 units right and 4 units down?
A. Figure B
B. Figure D
C. Figure E
D. Figure F
Answer:
From the given figures,
We have to obtain
The coordinates of Figure G are:
(6, -6), (9, -6), (9, -9), (7, -9), and (6, -5)
We know that,
When a point rotates 90° about the origin,
(x, y) before rotation ——-> (-y, x) after rotation
Now,
For the translation of 13 units right and 4 units down,
The vertices will be like: (x + 13, y – 4)
Hence, from the above,
We can conclude that option A matches the given situation

3-ACT MATH

3-Act Mathematical Modeling: Tricks of the Trade

ACT 1
Question 1.
After watching the video, what is the first question that comes to mind?
Answer:

Question 2.
Write the Main Question you will answer.

Answer:

Question 3.
Make a prediction to answer this Main Question.
Answer:

Question 4.
Construct Arguments Explain how you arrived at your prediction.
Answer:

ACT 2
Question 5.
What information in this situation would be helpful to know? How would you use that information?

Answer:

Question 6.
Use Appropriate Tools What tools can you use to solve the problem? Explain how you would use them strategically.
Answer:

Question 7.
Model with Math Represent the situation using mathematics. Use your representation to answer the Main Question.
Answer:

Question 8.
What is your answer to the Main Question? Does it differ from your prediction? Explain.

Answer:

ACT 3
Question 9.
Write the answer you saw in the video.

Answer:

Question 10.
Reasoning Does your answer match the answer in the video? If not, what are some reasons that would explain the difference?
Answer:

Question 11.
Make Sense and Persevere Would you change your model now that you know the answer? Explain.
Answer:

ACT 3 Extension
Reflect
Question 12.
Model with Math Explain how you used a mathematical model to represent the situation. How did the model help you answer the Main Question?

Answer:

Question 13.
Make Sense and Persevere When did you struggle most while solving the problem? How did you overcome that obstacle?
Answer:

SEQUEL
Question 14.
Be Precise Find another optical illusion online involving shapes that look different but are the same. Explain how you know the shapes are the same.

Answer:

Lesson 6.5 Understand Congruent Figures

Solve & Discuss It!
Simone plays a video game in which she moves shapes into empty spaces. After several rounds, her next move must fit the blue piece into the dashed space. How can Simone move the blue piece to fit in the space?

I can… use a sequence of translations, reflections, and rotations to show that figures are congruent.
Answer:
It is given that
Simone plays a video game in which she moves shapes into empty spaces. After several rounds, her next move must fit the blue piece into the dashed space.
Now,
From the given figure,
We can observe that
The blue piece and the dashed  piece are the reflections of each other
Hence,
Simone move the blue piece to fit in the space by the translation followed by reflection

Reasoning
How can you use what you know about sequences of transformations to move the piece?
Answer:
From the given figure,
We can observe that
The blue piece should be moved to the place of the dashed space and it will be possible only due to the “Translation”
But,
We can observe that it does not fit into the dashed space.
So,
Reflect the blue piece that it can fit into the dashed piece
Hence, from the above,
We can conclude that the sequence of transformations we can use to move the transformation are:
a. Translation b. Reflection

Focus on math practices
Construct Arguments How do you know that the piece that fits into the space is the same as the original blue shape? Explain.
Answer:
We know that,
In the reflection,
a. The shape of the preimage and the image are the same
b. The length of the preimage and the image are the same
c. The orientation of the image and the preimage are different
Hence, from the above,
We can conclude that
Due to the properties of the reflection,
We know that the piece that fits into space is the same as the original shape

Essential Question
How does a sequence of translations, reflections, and rotations result in congruent figures?
Answer:
If we copy one figure on tracing paper and move the paper so the copy covers the other figure exactly, then that suggests they are congruent. We can prove that two figures are congruent by describing a sequence of translations, rotations, and reflections that move one figure onto the other so they match up exactly.

Try It!

How can you determine whether the orange and blue rectangles are congruent?

Answer:
The given coordinate plane is:

From the given coordinate plane,
We can observe that
The blue rectangle has a length of 6 units and a width of 5 units
The orange rectangle has a length of 5 units and a width of 6 units
Now,
We can say that
We can obtain the orange rectangle by rotating the blue rectangle
We know that,
Rotations, reflections, and translations are isometric. That means that these transformations do not change the size of the figure. If the size and shape of the figure is not changed, then the figures are congruent
Hence, from the above,
We can conclude that the orange and blue rectangles are congruent

Convince Me!
Quadrilateral PQRS is congruent to quadrilateral P’ Q’R’S. What do you know about how these figures relate?
Answer:
It is given that
Quadrilateral PQRS is congruent to quadrilateral P’ Q’R’S
We know that,
When the two figures are congruent,
a. The shapes of the two figures are the same
b. The lengths of the two figures are the same
c. The angle measures of the two figures are the same
Hence,
In Quadrilateral PQRS and Quadrilateral P’Q’R’S’, the two figures are said to be congruent when
a. PQ = P’Q’ and RS = R’S’
b. ∠P = ∠P’, ∠Q = ∠Q’, ∠R = ∠R’, and ∠S = ∠S’

Try It!

Are the figures congruent? Explain.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of Figure 1 are:
(1, 5), (3, 7), (3, 5), and (2, 3)
The vertices of Figure 2 are:
(6, 3), (8, 3), (10, 2), and (8, 1)
Now,
Find out whether Translation, Reflection, and Rotation is possible between 2 figures or not
So,
The representation of Figure 1 and Figure 2 are:

Now,
From the given figures,
We can observe that none of the transformations is possible
Hence, from the above,
We can conclude that the two figures are not congruent

KEY CONCEPT

Two-dimensional figures are congruent if there is a sequence of translations, reflections, and rotations that maps one figure onto the other.

Do You Understand?
Question 1.
Essential Question How does a sequence of translations, reflections, and rotations result in congruent figures?
Answer:
If we copy one figure on tracing paper and move the paper so the copy covers the other figure exactly, then that suggests they are congruent. We can prove that two figures are congruent by describing a sequence of translations, rotations, and reflections that move one figure onto the other so they match up exactly.

Question 2.
Reasoning Does a sequence of transformations have to include a translation, a reflection, and a rotation to result in congruent figures? Explain.
Answer:
Rotations, reflections, and translations are isometric. That means that these transformations do not change the size of the figure. If the size and shape of the figure is not changed, then the figures are congruent

Question 3.
Construct Arguments Is there a sequence of reflections, rotations, and translations that makes the preimage and image not only congruent, but identical in orientation? Explain.
Answer:
We know that,
In the reflection,
The orientation of the preimage and the image will differ
Hence, from the above,
We can conclude that
There is a sequence of reflections, rotations, and translations that makes the preimage and image only congruent but not identical in orientation

Do You Know How?
Question 4.
A rectangle with an area of 25 square centimeters is rotated and reflected in the coordinate plane. What will be the area of the resulting image? Explain.
Answer:
It is given that
A rectangle with an area of 25 square centimeters is rotated and reflected in the coordinate plane.
We know that,
In a sequence of transformations like Translation, Rotation, and Reflection,
The shapes and side lengths of the image and the preimage are the same
Since the side lengths are the same, the area will also be the same
Hence, from the above,
We can conclude that the area of the image will also be 25 square centimeters

In 5 and 6, use the coordinate grid below.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔABC are:
A (1, 4), B (2, 2), and C (5, 2)
The vertices of ΔDEF are:
D (9, 9), E (8, 7), and F (5, 7)
The vertices of ΔGHI are:
G (6, 6), H (8, 5), and I (8, 1)

Question 5.
Is △ABC ≅ △DEF? Explain.
Answer:
To find whether ΔABC is congruent to ΔDEF or not,
Step 1:
Reflect Triangle ABC across the x-axis
Step 2:
Translate the image that we obtained in step-1 10 units right and 5 units up
So,
The representation of step 1 and step 2 is:

Hence, from the above,
We can observe that the vertices we obtained in step 2 are the same as ΔDEF
Hence,
ΔABC is congruent to ΔDEF

Question 6.
Is △ABC ≅ △GHI? Explain.
Answer:
The representation of △ABC and △GHI is:

From the above,
We can observe that △ABC and △GHI do not have the same size
We know that,
In order to be 2 figures congruent,
a. The sizes of the figures would be the same
b. The shapes of the figures would be the same
c. The side lengths of the figures should be the same
Hence, from the above,
We can conclude that
△ABC is not congruent to △GHI

Practice & Problem Solving

Question 7.
△Q’R’ S’ is the image of △QRS after a reflection across the y-axis and a translation 6 units down. Is the image the same size and shape as the preimage?

△QRS and △Q’R’S’ _________ the same size and shape.
Answer:
It is given that
△Q’R’ S’ is the image of △QRS after a reflection across the y-axis and a translation 6 units down
Now,
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔQRS are:
Q (-3, 5), R (-2, 4), and S (-5, 3)
The vertices of ΔQ’R’S’ are:
Q’ (3, -1), R’ (2, -2), and S’ (5, -3)
Now,
Step 1:
After a reflection of ΔQRS across the y-axis,
The vertices of ΔQRS will be:
Q (3, 5), R (2, 4), and S (5, 3)
Step 2:
After a translation of 6 units down,
The vertices that we obtained in Step 1 are:
Q’ (3, 5 – 6), R’ (2, 4 – 6), and S’ (5, 3 – 6)
Q’ (3, -1), R’ (2, -2), and S’ (5, -3)
So,
The vertices of ΔQ’R’S’ we obtained from the coordinate plane and the vertices of ΔQ’R’S’ we obtained after the sequence of transformations are the same
Hence, from the above,
We can conclude that △QRS and △Q’R’S’ have the same size and shape and the image (ΔQ’R’S’) is the same size and shape as the preimage (ΔQRS)

Question 8.
Is △DEF ≅ △D’ E’F’? Explain.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔDEF are:
D (5, 5), E (6, 3), and F (2, 4)
The vertices of ΔD’E’F’ are:
D’ (-2, -1), E’ (-3, 1), and F’ (1, 0)
So,
The representtaion of the sequence of transformations of ΔDEF to show it is congruent to ΔD’E’F’ is:

Hence, from the above,
We can conclude that ΔDEF is congruent to ΔD’E’F’

Question 9.
Construct Arguments Describe a way to show that quadrilateral ABCD is congruent to quadrilateral A’B’C’D’.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of quadrilateral ABCD are:
A (3, 5), B (5, 5), C (5, 4), and D (3, 4)
The vertices of quadriateral A’B’C’D’ are:
A’ (-3, 0), B’ (-5, 0), C’ (-5, -1), and D’ (-3, -1)
So,
The representtaion of the transformation of sequences that shows quadrilateral ABCd is congruent to quadrilateral A’B’C’D’ is:

Hence, from the above,
We can conclude that quadrilateral ABCD is congruent to quadrilateral A’B’C’D’

Question 10.
You are making two triangular flags for a project and need the flags to be the same shape and size. △XYZ and △X’Y’Z’ are the flags you have drawn. Are the flags the same shape and size? Explain.

Answer:
It is given that
You are making two triangular flags for a project and need the flags to be the same shape and size. △XYZ and △X’Y’Z’ are the flags you have drawn.
Now,
The given coordinate plane is:

From the given coordiate plane,
The vertices of ΔXYZ are:
X (5, 6), Y (6, 1), and Z (2, 2)
he vertices of ΔX’Y’Z’ are:
X’ (-2, 0), Y’ (-3, -5), and Z’ (1, -4)
So,
The representtaion of the sequence of transformation to find whether two flags have the same size and the same shape or not is:

Question 11.
Which two triangles are congruent? Describe the sequence of transformations that maps one figure onto the other.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔABC are:
A (-3, 2), B (-2, 7), and C (-7, 4)
The vertices of ΔQRS are:
Q (4, 7), R (2, 2), and S (7, 2)
The vertices of ΔXYZ are:
X (-2, -7), Y (-2, -3), and Z (-7, -5)
The vertices of ΔDEF are:
D (7, -5), E (2, -2), and F (2, -7)
Now,
We know that,
The two figures are said to be congruent when
a. The 2 figures have the same size i.e., the same length and the same angle measure
b. The 2 figures have the same shape
So,
The figures that are congruent to each other are represented as:

So,
From the above,
We can say that ΔQRS and ΔDEf are congruent
Hence,
The sequence of Transformations to show ΔQRS and ΔDEF are congruent is:

Question 12.
Is △LMN ≅ △XYZ? Explain.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔLMN are:
L (7, 9), M (9, 5), and N (6, 5)
The vertices of ΔXYZ are:
X (2, 2), Y (5, 4), and Z (5, 1)
Now,
We know that,
The two figures are said to be congruent when
a. The 2 figures have the same size i.e., the same length and the same angle measure
b. The 2 figures have the same shape
So,
The sequence of transformations to find whether ΔLMN is congruent to ΔXYZ is:

Hence, from the above,
We can conclude that ΔLMN is not congruent to ΔXYZ since they don’t have the same side lengths

Question 13.
Higher-Order Thinking A student was asked to describe a sequence of transformations that maps △DEF onto △D’ E’F’, given that △DEF ≅ △D’ E’F’. She incorrectly said the sequence of transformations that maps △DEF onto △D’ E’F’ is a reflection across the x-axis, followed by a translation of 6 units right and 4 units up. What mistake did the student likely make?

Answer:
It is given that
A student was asked to describe a sequence of transformations that maps △DEF onto △D’ E’F’, given that △DEF ≅ △D’ E’F’. She incorrectly said the sequence of transformations that maps △DEF onto △D’ E’F’ is a reflection across the x-axis, followed by a translation of 6 units right and 4 units up
Now,
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔDEF are:
D (4, 5), E (5, 1), and F (1, 2)
The vertices of ΔD’E’F’ are:
D’ (-2, -1), E’ (-1, 3), and F’ (-5, 2)
Now,
The correct sequence of Transformations to show ΔDEF and ΔD’E’F’ are congruent is:

Hence, from the above,
We can conclude that the mistake done by a student is the interchange of the sequence of Transformations
Hence,
The correct sequence of Transformations is:
Translation of 6 units left and 4 units down followed by the reflection across the x-axis

Assessment Practice
Question 14.
PART A
How can you determine whether △DEF ≅ △D’ E’ F?

A. Determine whether a sequence of rotations maps △DEF onto △D’E’F’.
B. Determine whether a sequence of transformations maps △DEF onto △D’ E’F’.
C. Determine whether a sequence of translations maps △DEF onto △D’ E’F’.
D. Determine whether a sequence of reflections maps △DEF onto △D’ E’F’.
Answer:
We know that,
If we want to find whether the given figures are congruent or not, then
We have to determine whether a sequence of transformations maps ΔDEF onto ΔD’E’F’
Hence, from the above,
We can conclude that option A matches the given situation

PART B
Is △DEF ≅ △D’ E’ F? Explain.
Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔDEF are:
D (5, 5), E (6, 3), and F (2, 4)
The vertices of ΔD’E’F’ are:
D’ (-2, -1), E’ (-3, 1), and F’ (1, 0)
So,
The representation of the sequence of transformations to find out whether ΔDEf and ΔD’E’F’ are congruent or not is:

Hence, from the above,
We can conclude that ΔDEF is congruent to ΔD’E’F’

Topic 6 MID-TOPIC CHECKPOINT

Question 1.
Vocabulary Describe three transformations where the image and preimage have the same size and shape. Lesson 6-1, Lesson 6-2, and Lesson 6-3
Answer:
We know that,
There are four types of transformations. They are:
a. Translation
b. Reflection
c. Rotation
d. Translation
Now,
Some transformations keep the pre-image and image congruent. Congruent means that they are the same size and shape or that they have the same measurements. They make not have the same orientation
Hence,
Three of the four transformations that preserve the size and shape of the pre-image are: Translation, Rotation, and Reflections

For 2-6, use the figures below.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of the quadrilateral MNPQ are:
M (1, 2), N (2, 4), P (4, 5), and Q (3, 3)
The vertices of the quadrilateral RSTU are:
R (1, -2), S (2, -4), T (4, -5), and U (3, -3)

Question 2.
What are the coordinates of each point after quadrilateral RSTU is rotated 90° about the origin? Lesson 6-3
Answer:
We know that,
When any point is rotated 90° about the origin,
(x, y) before rotation ——–> (-y, x) after rotation
Hence,
After 90° rotation about the origin,
The coordinates of each point of the quadrilateral RSTU will become:
R (2, 1), S (4, 2), T (5, 4), and U (3, 3)

Question 3.
What are the coordinates of each point after quadrilateral MNPQ is translated 2 units right and 5 units down? Lesson 6-1
Answer:
We know that,
When the translation occurs,
(x, y) before Translation ——-> (x + h, y + k) after Translation
Where,
h is the translation on the x-axis
k is the translation on the y-axis
We will take the positive value of h when the translation occurs on the right side of the x-axis
We will take the negative value of h when the translation occurs on the left side of the x-axis
We will take the positive value of k when the translation occurs on the top side of the y-axis
We will take the negative value of k when the translation occurs on the down side of the y-axis
Hence,
After the Translation of 2 units right and 5 units down,
The coordinates of each point of the quadrilateral MNPQ will become:
M (1 + 2, 2 – 5), N (2 + 2, 4 – 5), P (4 + 2, 5 – 5), and Q (3 + 2, 3 – 5)
M (3, -3), N (4, -1), P (6, 0), and Q (5, -2)

Question 4.
What are the coordinates of each point after quadrilateral MNPQ is reflected across the x-axis and then translated into 3 units left? Lessons 6-2 and 6-4
Answer:
We know that,
The vertices of the quadrilateral MNPQ are:
M (1, 2), N (2, 4), P (4, 5), and Q (3, 3)
So,
After the vertices of the quadrilateral, MNPQ reflected across the x-axis,
The vertices of the quadrilateral MNPQ will become:
M (1, -2), N (2, -4), P (4, -5), and Q (3, -3)
Now,
After the Translation of 3 units left,
The vertices of the quadrilateral MNPQ will become:
M (1 – 3, -2), N (2 – 3, -4), P (4 – 3, -5), and Q (3 – 3, -3)
M (-2, -2), N (-1, -4), P (1, -5), and Q (0, -3)
Hence, from the above,
We can conclude that the coordinates of each point after quadrilateral MNPQ is reflected across the x-axis and then translated into 3 units left are:
M (-2, -2), N (-1, -4), P (1, -5), and Q (0, -3)

Question 5.
Which series of transformations maps quadrilateral MNPQ onto quadrilateral RSTU? Lesson 6-4
A. reflection across the x-axis, translation 4 units down
B. reflection across the y-axis, translation 4 units down
C. rotation 180° about the origin, and then reflection across the x-axis
D. rotation 180° about the origin, and then reflection across the y-axis
Answer:
The representation of the series of transformations that maps the quadrilateral MNPQ onto the quadrilateral RSTU is:

Hence, from the above,
We can conclude that option A matches the given situation

Question 6.
Is quadrilateral MNPQ congruent to quadrilateral RSTU? Explain. Lesson 6-5
Answer:
We know that,
The 2 figures are said to be congruent only when:
a. The shapes of the 2 figures are the same
b. The sizes of the 2 figures are the same (Side lengths, and Angle measures)
Now,

From the above,
We can observe that the side lengths are not the same
Hence, from the above,
We can conclude that the quadrilateral MNPQ is not congruent to the quadrilateral RSTU

Topic 6 MID-TOPIC PERFORMANCE TASK

A tessellation is a design in a plane that uses one or more congruent figures, with no overlaps and no gaps, to cover the entire plane. A tessellation of an equilateral triangle is shown.

PART A
Explain how the tessellation of an equilateral triangle is formed using reflections.
Answer:
We know that,
When you cut a shape out of paper, then flip it over, the flipped shape looks like a mirror image of the original shape. So a tessellation made with this technique is called a “Reflection tessellation”
Now,
See if the figure will fit together with no gaps. The answer is yes, the figures will tessellate because it is made up of two shapes that do tessellate

PART B
Explain how the tessellation of an equilateral triangle is formed using rotations.
Answer:
We know that
A “Rotational tessellation” is a pattern where the repeating shapes fit together by rotating 90 degrees
Now,
See if the figure will fit together with no gaps. The answer is yes, the figures will tessellate because it is made up of two shapes that do tessellate

PART C
Which of the regular polygon(s) below can be tessellated using a series of transformations?

Answer:
We know that,
Equilateral triangles, squares, and regular hexagons are the only regular polygons that will tessellate. Therefore, there are only three regular tessellations.
Hence, from the above,
We can conclude that square and pentagon can be tessellated by using a series of transformations

Lesson 6.6 Describe Dilations

Solve & Discuss It!
A landscape architect designs a small splash pad represented by △ABC. Then she decides to make the splash pad larger as shown by △ADE. How are the splash pad designs alike? How are they different?

I can… dilate two-dimensional figures
Answer:
It is given that
A landscape architect designs a small splash pad represented by △ABC. Then she decides to make the splash pad larger as shown by △ADE.
Now,
The given landscape architect is:

Now,
From the given landscape architect,
We can observe that the difference in the splash designs
Hence, from the above landscape architect,
We can conclude that
The splash designs are alike in:
a. Shape b. Angle measures c. Orientation
The splash designs are different in:
a. Side lengths

Look for Relationships
How can you use what you know about scale drawings to compare and contrast the designs?
Answer:
A dilation is a transformation that results in an image with the same shape, angle measures, and orientation as the preimage, but different side lengths.
We know that,
When the scale factor is greater than 1, the dilation is a reduction.
When the scale factor is between 0 and 1, the dilation is an enlargement.

Focus on math practices
Reasoning Paul wants to make two square picnic tables. One table will have side lengths that are \(\frac{1}{2}\) of the lengths of the second table. How do the tablets compare? Explain.
Answer:
It is given that
Paul wants to make two square picnic tables. One table will have side lengths that are \(\frac{1}{2}\) of the lengths of the second table
So,
The size of the tables are in the ratio of \(\frac{1}{2}\) : 1
So,
The size of the tables are in the ratio of 1:2
Hence, from the above,
We can conclude that the side lengths of one table are 2 times the side lengths of the second table

Essential Question
What is the relationship between a preimage and its image after a dilation?
Answer:
After dilation, the pre-image and image have the same shape but not the same size.
In terms of Sides:
In dilation, the sides of the pre-image and the corresponding sides of the image are proportional.

Try It!

F’G’ H’I’ is the image of FGHI after a dilation with center at the origin. What is the scale factor?

The ratio of a side length in FGHI to a corresponding side length in F’GH’I’is: \(\frac{}{}\)
The scale factor is __________.
Answer:
It is given that
F’G’ H’I’ is the image of FGHI after a dilation with center at the origin.
Now,
The given coordinate plane is:

From the given coordinate plane,
The vertices of FGHI are:
F (1, 1), G (1, 2), H (2, 2), and I (2, 1)
The vertices of F’G’H’I’ are:
F’ (5, 5), G’ (5, 10), H’ (10, 10), and I’ (10, 5)
We know that,
The “Scale factor” is the ratio of a length in the image to the corresponding length in the preimage
So,
The ratio of a side length in FGHI to a corresponding side length in F’GH’I’is:
\(\frac{Side length of F’G’}{Side length of FG}\) (or) \(\frac{Side length of H’I’}{Side length of HI}\)
= \(\frac{0 + 5}{0 + 1}\) (or) \(\frac{0 + 5}{0 + 1}\)
= 5
So,
The scale factor is: 5
Hence, from the above,
We can conclude that
The ratio of a side length in FGHI to a corresponding side length in F’GH’I’is: \(\frac{5}{1}\)
The scale factor is 5.

Convince Me!
Quadrilateral WXYZ is the image of quadrilateral FGHI after a dilation with center at the origin and a scale factor of 3.5. What are the coordinates of the vertices of quadrilateral WXYZ?
Answer:
It is given that
Quadrilateral WXYZ is the image of quadrilateral FGHI after a dilation with center at the origin and a scale factor of 3.5
Now,
The given coordinate plane is:

From the given coordinate plane,
The vertices of FGHI are:
F (1, 1), G (1, 2), H (2, 2), and I (2, 1)
It is given that
The scale factor is: 3.5
So,
The vertices of the quadrilateral WXYZ are:
W ( 1 × 3.5, 1 ×3.5), X (1 × 3.5, 2 × 3.5), Y (2 × 3.5, 2 × 3.5), and Z (2 × 3.5, 1 × 3.5)
So,
W (3.5, 3.5), X (3.5, 7), Y (7, 7), and Z (7, 3.5)
Hence, from the above,
We can conclude that
The coordinates of the vertices of quadrilateral WXYZ are:
W (3.5, 3.5), X (3.5, 7), Y (7, 7), and Z (7, 3.5)

Try It!

A dilation maps point L(3, 6) to its image L’ (2, 4). Complete the dilation of figure LMN and label the image L’M’N’. What is the scale factor? What is the length of side M’N’?

Answer:
It is given that
A dilation maps point L(3, 6) to its image L’ (2, 4). Complete the dilation of figure LMN and label the image L’M’N’
Now,
The given coordinate plane is:

From the coordinate plane,
The vertices of ΔLMN are:
L (3, 6), M (3, 3), and N (6, 3)
It is given that the image of L is: L’ (2, 4)
Now,
We know that,
The “Scale factor” is the ratio of a length in the image to the corresponding length in the preimage
So,
The scale factor = \(\frac{Side length of L’}{Side length of L}\)
= \(\frac{4 – 2}{6 – 3}\)
= \(\frac{2}{3}\)
So,
The scale factor is: \(\frac{2}{3}\)
Now,
The coordinates of M’ = The coordinates of M × \(\frac{2}{3}\)
= (3, 3) × \(\frac{2}{3}\)
= (3 × \(\frac{2}{3}\), 3 × \(\frac{2}{3}\))
= (2, 2)
The coordinates of N’ = The coordinates of N × \(\frac{2}{3}\)
= (6, 3) × \(\frac{2}{3}\)
= (6 × \(\frac{2}{3}\), 3 × \(\frac{2}{3}\))
= (4, 2)
Now,
Compare M’ and N’ with (x ₁, y ₁), (x ₂, y ₂ )
We know that,
The distance between 2 points = √(x ₂ – x ₁ ) ² + (y ₂ – y ₁ ) ²
So,
The side length of M’N’ = √(4 – 2 ) ² + (2 – 2 )² 
= 2 units
Hence, from the above,
We can conclude that
The scale factor is: \(\frac{2}{3}\)
The side length of M’N’ is: 2 units

KEY CONCEPT

A dilation is a transformation that results in an image with the same shape, angle measures, and orientation as the preimage, but different side lengths.
When the scale factor is greater than 1, the dilation is an enlargement.

When the scale factor is between 0 and 1, the dilation is a reduction.

Do You Understand?
Question 1.
Essential Question What is the relationship between a preimage and its image after a dilation?
Answer:
After dilation, the pre-image and image have the same shape but not the same size.
In terms of Sides:
In dilation, the sides of the pre-image and the corresponding sides of the image are proportional.

Question 2.
Generalize When will a dilation be a reduction? When will it be an enlargement?
Answer:
When the scale factor is greater than 1, the dilation is an enlargement.

When the scale factor is between 0 and 1, the dilation is a reduction.

Question 3.
Reasoning Flora draws a rectangle with points at (12, 12), (15, 12), (15,9), and (12, 9). She dilates the figure with center at the origin and a scale factor of \(\frac{3}{4}\). what is the measure of each angle in the image? Explain.
Answer:
It is given that
Flora draws a rectangle with points at (12, 12), (15, 12), (15,9), and (12, 9). She dilates the figure with center at the origin and a scale factor of \(\frac{3}{4}\)
So,
The vertices for the image of the given rectangle = (x × \(\frac{3}{4}\), y × \(\frac{3}{4}\))
So,
The vertices for the image of the given rectangle wil be:
( 12 × \(\frac{3}{4}\), 12 × \(\frac{3}{4}\)), (15 × \(\frac{3}{4}\), 12 × \(\frac{3}{4}\)), (15 × \(\frac{3}{4}\), 9 × \(\frac{3}{4}\)), and (12 × \(\frac{3}{4}\), 9 × \(\frac{3}{4}\))
= (9, 9), (11.25, 9), (11.25, 6.75), and (9, 6.75)
Now,
We know that,
In dilation,
a. The preimage and image are the same in shape, orientation, and angle measures
b. the preimage and the image are different in size and the side lengths
So,
The representation of the angle measures in the preimage and image of the given rectangle is:

Hence, from the above,
We can conclude that the measure of each angle in the image is: 90°

Do You Know How?
In 4-6, use the coordinate grid below.

Answer:
The given coordinate plane is:

From the coordinate plane,
The vertices of Figure 1 are:
(4, 4), (6, 8), and (8, 4)
The vertices of Figure 2 are:
(2, 2), (3, 4), and (4, 2)
The vertices of Figure 3 are:
(1, 1), (1.5, 2), and (2, 1)

Question 4.
Figure 3 is the image of Figure 1 after a dilation with a center at the origin. What is the scale factor? Explain.
Answer:
It is given that
Figure 3 is the image of Figure 1 after a dilation with a center at the origin
Now,
We know that,
The “Scale factor” is the ratio of a length in the image to the corresponding length in the preimage
So,
Scale factor = \(\frac{4}{1}\) (or) \(\frac{6}{1.5}\) (or) \(\frac{8}{2}\)
= 4
Hence, from the above,
We can conclude that the scale factor is “4” so that Figure 3 is the image of Figure 1

Question 5.
What are the coordinates of the image of Figure 2 after a dilation with center at the origin and a scale factor of 3?
Answer:
We know that,
The vertices of Figure 2 are:
(2, 2), (3, 4), and (4, 2)
So,
With a scale factor of 3,
The vertices of Figure 2 will become:
(2 × 3, 2 × 3), (3 × 3, 4 × 3), and (4 × 3, 2 × 3)
= (6, 6), (9, 12), and (12, 6)
Hence, from the above,
We can conclude that the coordinates of the image of Figure 2 after a dilation at the origin and a scale factor of 3 are:
(6, 6), (9, 12), and (12, 6)

Question 6.
Which figures represent a dilation with a scale factor of \(\frac{1}{2}\)?
Answer:
We know that,
The vertices of Figure 1 are:
(4, 4), (6, 8), and (8, 4)
The vertices of Figure 2 are:
(2, 2), (3, 4), and (4, 2)
The vertices of Figure 3 are:
(1, 1), (1.5, 2), and (2, 1)
Now,
If we consider Figure 2 as the preimage and Figure 1 as the image, then
Scale factor = \(\frac{2}{4}\) (or) \(\frac{3}{6}\) (or) \(\frac{4}{8}\)
= \(\frac{1}{2}\)
If we consider Figure 3 as the preimage and Figure 2 as the image, then
Scale factor = \(\frac{1}{2}\) (or) \(\frac{1.5}{3}\) (or) \(\frac{2}{4}\)
= \(\frac{1}{2}\)
Hence, from the above,
We can conclude that (Figure 2, Figure 1) and (Figure 3, Figure 2) represent a dilation with a scale factor of \(\frac{1}{2}\)

Practice & Problem Solving

Question 7.
Leveled Practice Draw the image of △DEF after a dilation with center (0, 0) and scale factor of 2.

Find the coordinates of each point in the original figure.
D(____, (____) E(____, (____) F(____, (____)
Multiply each coordinate by _______.
Find the coordinates of each point in the image:
D'(____, (____) E'(____, (____) F'(____, (____)
Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of ΔDEF are:
D (0, 0), E (2, 0), and F (0, 2)
Now,
With a scale factor of 2,
Multiply each coordinate with 2
The coordinates of ΔDEF will become:
D (0 × 2, 0 × 2), E (2 × 2, 0 × 2), and F (0 × 2, 2 × 2)
D (0, 0), E (4, 0), and F (0, 4)
So,
The coordinates of the points for the image of ΔDEF are:
D’ (0, 0), E’ (4, 0), and F’ (0, 4)
Hence,
The representation of ΔDEF and its image ΔD’E’F’ is:

Question 8.
Find the scale factor for the dilation shown.

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of Figure DEFG are:
D (0, 0), E (5, 0), F (5, 6), and G (0, 6)
The vertices of Figure D’E’F’G’ are:
D’ (0, 0), E’ (15, 0), G’ (15, 18), and H’ (0, 18)
Now,
We know that,
The “Scale factor” is the ratio of a length in the image to the corresponding length in the preimage
So,
Scale factor = \(\frac{15}{5}\) (or) \(\frac{18}{6}\)
= 3
Hence, from the above,
We can conclude that
The scale factor for the given dilation is: 3

Question 9.
Critique Reasoning For the dilation with center (0, 0) shown on the graph, your friend says the scale factor is \(\frac{5}{2}\). What is the correct scale factor? What mistake did your friend likely make?

Answer:
It is given that
For the dilation with center (0, 0) shown on the graph, your friend says the scale factor is \(\frac{5}{2}\)
Now,
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔABC are:
A (0, 5), B (-5, 0), and C (5, -5)
The vertices of ΔA’B’C’ are:
A’ (0, 2), B’ (-2, 0), C’ (2, -2)
Now,
We know that,
The “Scale factor” is the ratio of a length in the image to the corresponding length in the preimage
So,
Scale factor = \(\frac{2}{5}\) (or) \(\frac{-2}{-5}\)
= \(\frac{2}{5}\)
Hence, from the above,
We can conclude that
The correct scale factor is: \(\frac{2}{5}\)
The mistake done by your friend is:
The consideration of the scale factor as \(\frac{Length of the preimage}{length of the image}\) instead of \(\frac{Length of the image}{Length of the preimage}\)

Question 10.
The smaller figure is the image of dilation of the larger figure. The origin is the center of dilation. Tell whether the dilation is an enlargement or a reduction. Then find the scale factor of the dilation.

Answer:
It is given that
The smaller figure is the image of dilation of the larger figure. The origin is the center of dilation
Now,
The given coordinate plane is:

From the given coordinate plane,
The coordinates of each point of the preimage are:
(3, 6), (15, 6), (15, 18), and (3, 18)
The coordinates of each point of the image are:
(1, 2), (5, 2), (5, 6), and (1, 6)
Now,
We know that,
The “Scale factor” is the ratio of a length in the image to the corresponding length in the preimage
So,
Scale factor = \(\frac{1}{3}\) (or) \(\frac{2}{6}\) (or) \(\frac{5}{15}\) (or) \(\frac{6}{18}\)
= \(\frac{1}{3}\)
Now,
Since the scale factor is less than 1, the dilation is a reduction
Hence, from the above,
We can conclude that
The given dilation is a reduction
The scale factor is: \(\frac{1}{3}\)

Question 11.
Higher-Order Thinking Q’R’S’T’ is the image of QRST after a dilation with center at the origin.

Answer:
It is given that
Q’R’S’T’ is the image of QRST after a dilation with center at the origin
Now,
The given coordinate plane is:

From the given coordinate plane,
The vertices of the parallelogram QRST are:
Q (4, 4), R (16, 4), S (20, 16), and T (8, 16)
The vertices of the parallelogram Q’R’S’T’ are:
Q’ (1, 1), R’ (4, 1), S’ (5, 4), and T’ (2, 4)

a. Find the scale factor.
Answer:
We know that,
The “Scale factor” is the ratio of a length in the image to the corresponding length in the preimage
So,
Scale factor = \(\frac{4}{1}\) (or) \(\frac{16}{4}\) (or) \(\frac{20}{5}\) (or) \(\frac{8}{2}\)
= 4
Hence, from the above,
We can conclude that the scale factor is: 4

b. Find the area of each parallelogram. What is the relationship between the areas?
Answer:
We know that,
Area of the parallelogram = Base × Height
Now,
the representation of the side lengths of the parallelogram QRST and its dilation Q’R’S’T’ is:

So,
The area of the parallelogram QRST = 12.6 × 12
= 151.2 square units
The area of the parallelogram Q’R’S’T’ = \(\frac{The area of the parallelogram QRST}{4}\)
= 37.8 square units
Hence, from the above,
We can conclude that
The area of the parallelogram QRST is: 151.2 sq. units
The area of the parallelogram Q’R’S’T’ is: 37.8 sq. units
The relationship between the two areas is:
The area of the parallelogram Q’R’S’T’ = \(\frac{The area of the parallelogram QRST}{4}\)

Assessment Practice
Question 12.
Triangle PQR is the image of △JKL after a dilation. Is the dilation an enlargement or a reduction? Explain.

A. An enlargement, because the image is larger than the original figure
B. An enlargement, because the image is smaller than the original figure
C. A reduction, because the image is smaller than the original figure
D. A reduction, because the image is larger than the original figure
Answer:
It is given that
Triangle PQR is the image of △JKL after a dilation
Now,
From the given figure,
We can observe that ΔPQR is larger than ΔJKL
So,
We can say that the dilation is an enlargement
Hence, from the above,
We can conclude that option A matches with the given situation

Question 13.
Rectangle QUAD has coordinates Q(0, 0), U(0, 3), A(6, 3), and D(6, 0). Q’U’ A’D’ is the image of QUAD after a dilation with center (0, 0) and a scale factor of 6. What are the coordinates of point D’? Explain.
Answer:
It is given that
Rectangle QUAD has coordinates Q(0, 0), 4(0, 3), A6, 3), and D(6, 0). Q’U’ A’D’ is the image of QUAD after a dilation with center (0, 0) and a scale factor of 6.
So,
With a scale factor of 6,
The vertices of the rectangle QUAD will become:
Q (0 × 6, 0 × 6), U (0 × 6, 3 × 6), A (6 × 6, 3 × 6) and D (6 × 6, 0 × 6)
= Q (0, 0), U (0, 18), A (36, 18), and D (6, 0)
So,
After a dilation,
The vertices of the rectangle QUAD will become the vertices of the rectangle Q’U’A’D’
So,
The vertices of the rectangle Q’U’A’D’ are:
Q’ (0, 0), U’ (0, 18), A’ (36, 18), and D’ (36, 0)
Hence, from the above,
We can conclude that the coordinates of point D’ are: (36, 0)

Lesson 6.7 Understand Similar Figures

Solve & Discuss It!
Andrew draws the two figures shown on a coordinate plane. How are the two figures alike? How are they different? How do you know?

I can… use a sequence of transformations, including dilations, to show that figures are similar.
Answer:
It is given that
Andrew draws the two figures shown on a coordinate plane
Now,
The given figures are:

From the above,
We can observe that
ΔABC is the preimage
ΔA’B’C’ is the image
We can also observe that
The image is smaller than the image since the transformation we used when drawing the image is a “Dilation”
Hence,
Since the transformation we used is a “Dilation”,
The two figures are similar in terms of:
a. Shape b. Size c. Orientation d. Angle measures
The two figures are different in terms of:
a. Side lengths

Look for Relationships
Is △ABC a preimage of △A’B’C’? How do you know?
Answer:
The given figures are:

From the given figures,
We can clearly observe that the transformation called “Dilation” takes place
Since the dilation takes place,
We can conclude that ΔABC is a preimage of ΔA’B’C’

Focus on math practices
Reasoning How can you use the coordinates of the vertices of the triangles to identify the transformation that maps △ABC to △A’B’C’? Explain.
Answer:
The given figures are:

Now,
Fro the given figures,
We know that a transformation called a “Dilation” takes place
So,
We know that,
The “Scale factor” is the ratio of a length in the image to the corresponding length in the preimage
So,
Scale factor = [altex]\frac{The x-coordinate (or) y-coordinate that maps A’ or B’ or C’}{The x-coordinate (or) y-coordinate that maps A or B or C}[/latex}
Hence, from the above,
We can conclude that
By using the “Scale factor”, the coordinates of the vertices of the triangles can be used to identify the transformation that maps △ABC to △A’B’C’

Essential Question
How are similar figures related by a sequence of transformations?
Answer:
Two figures are similar if and only if one figure can be obtained from the other by a single transformation, or a sequence of transformations, including translations, reflections, rotations, and/or dilations. Similarity transformations preserve shape, but not necessarily size, making the figures “similar”

Try It!

Is △ΑΒC similar to △A’ B’ C?
The triangles _________ similar.

Answer:
The given figures are:

Now,
If we want to find the 2 triangles are similar or not, find the scale factor and find whether they have the same shape or not
If the scale factor for 2 triangles is the same, then those triangles are similar
Now,
We know that,
The scale factor is defined as the distance from this center point to a point on the preimage and also the distance from the center point to a point on the image
Now,
For AB and A’B’,
Scale factor = \(\frac{27}{12}\)
= \(\frac{9}{4}\)
For BC and B’C’,
Scale factor = \(\frac{27}{12}\)
= \(\frac{9}{4}\)
For AC and A’C’,
Scale factor = \(\frac{18}{8}\)
= \(\frac{9}{4}\)
Since the scale factor is equal for all the corresponding sides
We can conclude that ΔABC is similar to ΔA’B’C’

Convince Me!
What sequence of transformations shows that △ABC is similar to △A’ B’C’?
Answer:
The sequence of Transformations that shows ΔABC is similar to ΔA’B’C’ is:
a. Rotation b. Dilation c. Translation

Try It!

a. Graph the image of JKL after a reflection across the line x = 1 followed by dilation with a scale factor of \(\frac{1}{2}\) and center of dilation point J’.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔJKL are:
J (-2, -4), K (-4, 0), and L (-2, 1)
The vertices of ΔPQR are:
P (-4, 2), Q (4, 6), and R (6, 2)
Now,
After the reflection of ΔJKL across the line x = 1,
J (-1, 4), K (-3, 0), and L (-1, -1)
So,
With a scale factor of \(\frac{1}{2}\),
The vertices of ΔJKL will become:
J (-1, 4) × \(\frac{1}{2}\), K (-3, 0) × \(\frac{1}{2}\), and L (-1, -1) × \(\frac{1}{2}\)
So,
J’ (-0.5, 2), K’ (-1.5, 0), and L’ (-0.5, -0.5)
Hence,
The representation of the image of JKL after a reflection across the line x = 1 followed by dilation with a scale factor of \(\frac{1}{2}\) and center of dilation point J’ is:

b. Is △JKL similar to △PQR?
Answer:
We know that,
The vertices of ΔJKL are:
J (-2, -4), K (-4, 0), and L (-2, 1)
The vertices of ΔPQR are:
P (-4, 2), Q (4, 6), and R (6, 2)
Now,
Step 1:
Rotate the vertices of ΔJKL 270° about the origin
We know that,
(x, y) before 270° rotation —-> (y, -x) after 270° rotation
So,
The vertices of ΔJKL will be:
J (-4, 2), K (0, 4), and L (1, 2)
Step 2:
Dilate the vertices of ΔJKL we obtained in step 1 by a scale factor of 2
So,
The vertices of ΔJKL will be:
J (-8, 4), K (0, 8), and L (2, 4)
Step 3:
Translate the vertices of ΔJKL we obtained in step 2 by 4 units right and 2 units down
So,
The vertices of ΔJKL will become:
P (-4, 2), Q (4, 6), and R (6, 2)
So,
The representation of the sequence of transformations is:

Hence, from the above,
We can conclude that ΔJKL is similar to ΔPQR

KEY CONCEPT

Two-dimensional figures are similar if there is a sequence of rotations, reflections, translations, and dilations that maps one figure onto the other.

Do You Understand?
Question 1.
Essential Question How are similar figures related by a sequence of transformations?
Answer:
Two figures are similar if and only if one figure can be obtained from the other by a single transformation, or a sequence of transformations, including translations, reflections, rotations, and/or dilations. Similarity transformations preserve the shape, but not necessarily size, making the figures “similar”

Question 2.
Be Precise How do the angle measures and side lengths compare in similar figures?
Answer:
Matching sides of two or more polygons are called corresponding sides, and matching angles are called corresponding angles. If two figures are similar, then the measures of the corresponding angles are equal and the ratios of the lengths of the corresponding sides are proportional.

Question 3.
Generalize Does a given translation, reflection, or rotation, followed by a given dilation, always map a figure to the same image as that same dilation followed by that same translation, reflection, or rotation? Explain.
Answer:
We know that,
The images formed by reflection, translation, and rotation have the same shape and size whereas dilation has the same shape but a different size
In the same way,
The image formed by dilation has the same shape but a different size whereas the image formed by reflection, translation, and rotation has the same shape and size as the image formed in dilation
Hence, from the above,
We can conclude that
The translation, reflection, or rotation, followed by a given dilation, always map a figure to the same image as that same dilation followed by that same translation, reflection, or rotation

Do You Know How?
Question 4.
Is trapezoid ABCD – trapezoid EFGH? Explain.

Answer:
The given figures are:

From the given figures,
We can observe that
The corresponding angles i.e., the opposite angles are equal
Now,
The ratio of the side lengths of the corresponding figures = \(\frac{32}{16}\) (or) \(\frac{28}{14}\) (or) \(\frac{20}{10}\)
= 2
So,
The ratio of the side lengths of the corresponding figures are the same
Now,
We know that,
Two figures are said to be similar if they are the same shape. In more mathematical language, two figures are similar if their corresponding angles are congruent, and the ratios of the lengths of their corresponding sides are equal. This common ratio is called the scale factor
Hence, from the above,
We can conclude that
Trapezoid ABCD – Trapezoid EFGH

Use the graph for 5 and 6.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔABC are:
A (-4, -2), B (-1, -2), and C (-3, -3)
The vertices of ΔDEF are:
D (9, -5), E (6, -8), and F (1, -5)

Question 5.
△ABC is dilated by a factor of 2 with a center of dilation at point C, reflected across the x-axis, and translated 3 units up. Graph the resulting similar figure.
Answer:
We know that,
The vertices of ΔABC are:
A (-4, -2), B (-1, -2), and C (-3, -3)
So,
Step 1:
After dilation by a factor of 2:
Multiply all the coordinates of each point by 2
So,
The vertices of ΔABC are:
A (-8, -4), B (-2, -4), and C (-6, -6)
Now,
With a dilation at point C,
Add all the coordinates of each point with -3
So,
The vertices of ΔABC are:
A (-11, -7), B (-5, -7), and C (-9, -9)
Step 2:
When the vertices of ΔABC that we obtained in step 1 reflects across the x-axis,
The vertices of ΔABC will be:
A (-11, 7), B (-5, 7), and C (-9, 9)
Step 3:
When the vertices of ΔABC we obtained in step 2 translated by 3 units up,
Add all the y-coordinates of each point with 3
So,
The vertices of ΔABC will be:
A (-11, 10), B (-5, 10), and C (-9, 12)
Hence,
The representation of all the steps is:

Question 6.
Is △ABC similar to △DEF? Explain.
Answer:
We know that,
The vertices of ΔABC are:
A (-4, -2), B (-1, -2), and C (-3, -3)
The vertices of ΔDEF are:
D (9, -5), E (6, -8), and F (1, -5)
We know that,
Two figures are said to be similar if they are the same shape. In more mathematical language, two figures are similar if their corresponding angles are congruent, and the ratios of the lengths of their corresponding sides are equal
So,
The representation of ΔABC and ΔDEF are:

From the above,
We can observe that the ratio of the corresponding side lengths are not in the same proportion
Hence, from the above,
We can conclude that ΔABC is not similar to ΔDEF

Practice & Problem Solving

Question 7.
Leveled Practice RSTU and VXYZ are quadrilaterals. Given RSTU ~ VXYZ, describe a sequence of transformations that maps RSTU to VXYZ.

• reflection across the ________
• translation ______ unit(s) left and _______ unit(s) up
• dilation with center (0,0) and a scale factor of ________
Answer:
The given coordinate plane is:

From the coordinate plane,
The vertices of the quadrilateral VXYZ are:
V (-2, -1), X (-3, 0), Y (-2, 2), and Z (-1, 0)
The vertices of the quadrilateral RSTU are:
R (3, 0), S (6, 3), T (3, 9), and U (0, 3)
Now,
Step 1:
Reflect the vertices of the quadrilateral VXYZ across the y-axis
So,
The vertices of the quadrilateral VXYZ will be:
V (2, -1), X (3, 0), Y (2, 2), and Z (1, 0)
Step 2:
Dilate the vertices of the quadrilateral VXYZ with a scale factor of 3
So,
Multiply all the coordinates of all the vertices of the quadrilateral VXYZ by 3
So,
The vertices of the quadrilateral VXYZ will be:
V (6, -3), X (9, 0), Y (6, 6), and Z (3, 0)
Step 3:
Translate the vertices of the quadrilateral VXYZ 3 units left and 3 units up
So,
The vertices of the quadrilateral VXYZ will become:
R (3, 0), S (6, 3), T (3, 9), and U (0, 3)
Hence,
The sequence of Transformations that map the quadrilateral VXYZ onto the quadrilateral RSTU is:

Question 8.
Reasoning Is △MNO similar to △PQO? Explain.

Answer:
The given coordinate plane is:

From the coordinate plane,
The vertices of ΔMNO are:
M (6, 6 ), N (0, 6), and O (0, 0)
The vertices of ΔPQO are:
P (-12, -9), Q (0, -9), and O (0, 0)
We know that,
Two figures are said to be similar if they are the same shape. In more mathematical language, two figures are similar if their corresponding angles are congruent, and the ratios of the lengths of their corresponding sides are equal
So,
The representation of ΔMNO and ΔPQO are:

From the above,
We can observe that
a. The ratio of the corresponding side lengths are not in the same proportion
b. The angle measures are not the same
Hence, from the above,
We can conclude that ΔMNO is not similar to ΔPQO

Question 9.
△PQR is dilated by a scale factor of 2 with a center of dilation (0, 0) and rotated 180° about the origin. Graph the resulting similar △XYZ.

Answer:
It is given that
△PQR is dilated by a scale factor of 2 with a center of dilation (0, 0) and rotated 180° about the origin.
Now,
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔPQR are:
P (2, 2), Q (4, 2), and R (3, 4)
Now,
Step 1:
Dilation of ΔPQR by a scale factor of 2:
Multiply all the coordinates of each point of ΔPQR by 2
So,
The vertices of ΔPQR will be:
P (4, 4), Q (8, 4), and R (6, 8)
Step 2:
Rotation of 180° counterclockwise about the origin:
We know that,
(x, y) before rotating 180° ——- > (-x, -y) after rotating 180°
So,
The vertices of ΔPQR that we obtained in step 1 will become:
X (-4, -4), Y (-8, -4), and Z (-6, -8)
Hence,
The graph of the resulting ΔXYZ is:

Question 10.
Describe a sequence of transformations that shows that quadrilateral RSTU is similar to quadrilateral VXYZ.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of the quadrilateral VXYZ are:
V (-5, -5), X (-5, -1), Y (-1, -1), and Z (-1, -5)
The vertices of the quadrilateral RSTU are:
R (-16, -14), S (-16, -6), T (-8, -6), and U (-8, -14)
Now,
Step 1:
Translate the quadrilateral VXYZ 3 units left and 2 units down
So,
The vertices of the quadrilateral VXYZ will be:
V (-8, -7), X (-8, -3), Y (-4, -3), and Z (-4, -7)
Step 2:
Dilate the vertices of the quadrilateral VXYZ we obtained in step 1 by a scale factor of 2
So,
The vertices of the quadrilateral VXYZ will be:
R (-16, -14), S (-16, -6), T (-8, -6), and U (-8, -14)
Hence,
The sequence of Transformations that shows the quadrilateral VXYZ is similar to the quadrilateral RSTU is:

Question 11.
Construct Arguments Is △PQR similar to △XYZ? Explain.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔXYZ are:
X (4, 4), Y (4, 8), and Z (8, 8)
The vertices of ΔPQR are:
P (2, -2), Q (4, -2), and R (4, -4)
Now,
Step 1:
Rotate ΔXYZ 270° about the origin
We know that,
(x, Y) before 270° rotation —— > (y, -x) after 270° rotation
So,
The vertices of ΔXYZ will become:
X ( 4, -4), Y (8, -4), and Z (8, -8)
Step 2:
Dilate ΔXYZ with a scale factor of \(\frac{1}{2}\)
So,
The vertices of ΔXYZ that we obtained in step 1 will become:
P (2, -2), Q (4, -2), and R (4, -4)
So,
After the sequence of Rotation and Dilation of ΔXYZ,
We obtained the vertices of ΔPQR
Hence, from the above,
We can conclude that ΔPQR is similar to ΔXYZ

Question 12.
Higher-Order Thinking Given △JKL ~ △XYZ, find two possible coordinates for missing point Y. For each coordinate chosen, describe a sequence of transformations, including a dilation, that will map △JKL to △XYZ.

Answer:
It is given that
Given △JKL ~ △XYZ
Now,
The given coordinate plane is:

From the coordinate plane,
The vertices of ΔJKL are:
J (2, 8), K (6, 2), and L (2, 2)
The vertices of ΔXYZ are:
X (-2, 5), Y (x, y), and Z (-2, 2)
Now,
Step 1:
Reflect ΔJKL across the y-axis
So,
The vertices of ΔJKL will be:
J (-2, 8), K (-6, 2), and L (-2, 2)
Step 2:
Dilate the vertices of ΔJKL that we obtained in step by a scale factor of \(\frac{1}{2}\)
So,
The vertices of ΔJKL will be:
(-1, 4), K (-3, 1), and L (-1, 1)
Step 3:
Translate the vertices of ΔJKL that we obtained in step 2 by 1 unit left and 1 unit up
So,
The vertices of ΔJKL will become:
X (-2, 5), Y (-4, 2), and Z (-2, 2)
Hence, from the above,
We can conclude that the coordinates for the missing point Y are: (-4, 2)

Assessment Practice
Question 13.
Rajesh is making pennants in preparation for a school soccer game. He wants the pennants to be similar triangles. Which of these triangles could he use for the pennants?

A. △QRS and △TVW
B. △QRS and △XYZ
C. △TVW and △JKL
D. △TVW and △XYZ
Answer:
It is given that
Rajesh is making pennants in preparation for a school soccer game. He wants the pennants to be similar triangles.
Now,
If we want to find which of the triangles are similar, find the scale factor and find whether they have the same shape or not
If the scale factor for 2 triangles is the same, then those triangles are similar
Now,
We know that,
The scale factor is defined as the distance from this center point to a point on the preimage and also the distance from the center point to a point on the image
Now,
For ΔQRS and ΔTVW,
Scale factor = \(\frac{3}{1}\) (or) \(\frac{3}{1.5}\)
= 3 (or) 2
For ΔQRS and ΔXYZ,
Scale factor = \(\frac{3}{1.5}\) (or) \(\frac{3}{1.5}\)
= 2
Hence from the above,
We can conclude that ΔQRS and ΔXYZ are similar triangles and he can use these triangles for the pennants

Question 14.
Determine whether the following pairs of triangles are similar or not similar.

Answer:
If we want to find which of the triangles are similar, find the scale factor and find whether they have the same shape or not
If the scale factor for 2 triangles is the same, then those triangles are similar
Now,
We know that,
The scale factor is defined as the distance from this center point to a point on the preimage and also the distance from the center point to a point on the image
Now,
For ΔABC and ΔDEF,
Scale factor = \(\frac{2}{1}\) (or) \(\frac{2}{1}\)
= 2
For ΔABC and ΔLMN,
Scale factor = \(\frac{2}{1.5}\) (or) \(\frac{2}{1}\)
= 1.33 (or) 2
For ΔDEF and ΔLMN,
Scale factor = \(\frac{1}{1}\) (or) \(\frac{1.5}{1}\)
= 1 (or) 1.5
Hence from the above,
We can conclude that ΔABC and ΔDEF are similar triangles

Lesson 6.8 Angles, Lines, and Transversals

Solve & Discuss It!
Draw two parallel lines. Then draw a line that intersects both lines. Which angles have equal measures?

I can… identify and find the measures of angles formed by parallel lines and a transversal.
Answer:
Let a and b be the two parallel lines and the line that is intersecting the parallel lines is a “Transversal”
So,
The representation of the parallel lines along with the transversal is:

Now,
From the above,
The angles that have equal measures are:
Corresponding angles:
∠1 = ∠5, ∠2 = ∠6, ∠3 = ∠7, and ∠4 = ∠8
Alternate Interior angles:
∠4 = ∠6, ∠3 = ∠5
Alternate Exterior angles:
∠1 = ∠7, ∠2 = ∠8
Vertically Opposite Angles:
∠1 = ∠7, ∠2 = ∠8

Use Appropriate Tools
What tools can you use to determine which angles have equal measures?
Answer:
The tools that can be used to determine which angles have equal measures are:
a. Pencil b. Scale c. Protractor d. Setsquare

Focus on math practices
Reasoning What properties or definitions can you use to describe which angles have equal measures?
Answer:
The definitions that you can use to describe which angles have equal measures are:
Corresponding angles:
Corresponding angles are angles that are in the same position relative to lines intersected by a transversal
Alternate Interior angles:
When two parallel lines are crossed by a transversal, the pair of angles formed on the inner side of the parallel lines, but on the opposite sides of the transversal are called alternate interior angles
Alternate Exterior angles:
Alternate exterior angles are the pair of angles that lie on the outer side of the two parallel lines but on either side of the transversal line

Essential Question
What are the relationships among angles that are created when a line intersects two parallel lines?
Answer:
If two parallel lines are cut by a third line, the third line is called the transversal.
So,
The relationships among angles that are created when a line intersects two parallel lines are :
(1) Corresponding angles
(2) Vertically Opposite angles
(3) Alternate interior angles
(4) Alternate exterior angles

Try It!

Which angles are congruent to ∠8?


Which angles are supplementary to ∠8?

Answer:
The given transversal is:

We know that,
The angle relationships that have equal angle measures are:
a. Corresponding angles b. Vertically opposite angles c. Alternate interior angles d. Alternate exterior angles
We know that,
The angle relationships that have supplementary angle measures are:
a. Adjacent angles
Hence, from the above,
We can conclude that
The angles that are congruent to ∠8 are: ∠4, ∠6 and ∠2
The angles that are supplementary to ∠8 are: ∠7, ∠1, ∠2, and ∠6

Convince Me!
Use what you know about other angle relationships to explain why ∠4 and ∠5 are supplementary angles.
Answer:
The given transversal is:

Now,
From the given Transversal,
We can observe that
∠4 and ∠5 are consecutive interior angles
Now,
We know that,
The sum of the consecutive interior angles is always equal to 180°
The angles that have the sum 180° are also called “Supplementary angles”
Hence, from the above,
We can conclude that ∠4 and ∠5 are supplementary angles

Try It!

What are the measures of ∠7 and ∠2? Explain.

Answer:
The given transversal is:

Now,
From the given transversal,
We can observe that
∠4 and 99° are the corresponding angles
∠4 and ∠2 are alternate interior angles
∠5 and ∠7 are vertical angles
∠4 and ∠5 are supplementary angles
So,
Now,
∠4 = 99°
Now,
∠4 + ∠5 = 180°
∠5 = 180° – 99°
∠5 = 81°
Now,
∠2 = 99°
∠7 = 81°
Hence, from the above,
We can conclude that
The values of ∠7 and ∠2 are: 81° and 99° respectively

Try It!

In the figure, a || b. What is the value of x?

Answer:
It is given that a || b
Now,
The given transversal is:

From the given transversal,
We can observe that
59° and (x + 12)° are consecutive exterior angles
Now,
We know that,
The sum of the angles of consecutive exterior angles is: 180°
So,
59°+ (x + 12)° = 180°
x + 71° = 180°
x = 180° – 71°
x = 109°
Hence, from the above,
We can conclude that the value of x is: 109°

KEY CONCEPT

If parallel lines are intersected by a transversal, then
• Corresponding angles are congruent.
• Alternate interior angles are congruent.
• Same-side interior angles are supplementary.

Do You Understand?
Question 1.
Essential Question What are the relationships among angles that are created when a line intersects two parallel lines?
Answer:
If two parallel lines are cut by a third line, the third line is called the transversal.
So,
The relationships among angles that are created when a line intersects two parallel lines are :
(1) Corresponding angles
(2) Vertically Opposite angles
(3) Alternate interior angles
(4) Alternate exterior angles

Question 2.
When parallel lines are cut by a transversal, how can you use a translation to describe how angles are related?
Answer:
Let a and b be the parallel lines
Now,
The given transversal is:

Now,
When the parallel lines are cut by a transversal,
The relation between angles is:
Corresponding angles:
∠1 = ∠5, ∠2 = ∠6, ∠3 = ∠7, and ∠4 = ∠8
Alternate Interior angles:
∠4 = ∠6, ∠3 = ∠5
Alternate Exterior angles:
∠1 = ∠7, ∠2 = ∠8
Vertically Opposite Angles:
∠1 = ∠7, ∠2 = ∠8
Consecutive Interior angles:
∠4 + ∠5 = 180°, ∠3 + ∠6 = 180°
Consecutive Exterior angles:
∠1 + ∠8 = 180°, ∠7 + ∠2 = 180°

Question 3.
How many angles are created when two parallel lines are cut by a transversal? How many different angle measures are there?
Answer:
When two parallel lines are cut by a transversal, the number of angles created are: 8
Now,
The different angle measures are:
a. Corresponding angles
b. Alternate Interior angles
c. Alternate Exterior angles
d. Vertical angles
e. Consecutive interior angles
f. Consecutive exterior angles
g. Adjacent angles

Question 4.
Use Structure How can you use angle measures to tell whether two lines are parallel?
Answer:
The ways to prove two lines are parallel using angle measures are:
a. If corresponding angles are congruent.
b. If alternate interior angles are congruent.
c. If consecutive, or same side, interior angles are supplementary.
d. If alternate exterior angles are congruent.

Do You Know How?
In 5-7, use the figure below.

Question 5.
Which angles are congruent to ∠8?
Answer:
The given transversal is:

Now,
We know that,
The angle relationships that have equal angle measures are:
a. Corresponding angles b. Vertically opposite angles c. Alternate interior angles d. Alternate exterior angles
Hence, from the above,
We can conclude that
The angles that are congruent to ∠8 are: ∠2, ∠4, and ∠6

Question 6.
If m∠4 = 70°, what is m∠6? Explain.
Answer:
The given transversal is:

Now,
From the given transversal,
We can observe that
∠4 and ∠6 are alternate interior angles
We know that,
The alternate interior angles are congruent
Hence, from the above,
We can conclude that
∠4 = ∠6 = 70°

Question 7.
If m∠1 = 95°, write an equation that could be used to find the measure of ∠8. Find m∠8.
Answer:
The given transversal is:

Now,
From the given transversal,
We can observe that
∠1 and ∠8 are consecutive exterior angles
Now,
We know that,
The sum of the consecutive exterior angle measures is 180°
So,
∠1 + ∠8 = 180°
∠8 = 180° – ∠1
∠8 = 180° – 95°
∠8 = 85°
Hence, from the above,
We can conclude that the measure of ∠8 is: 85°

Question 8.
What must x equal if line a is parallel to line b?

Answer:
It is given that line a is parallel to line b
Now,
The given transversal is:

Now,
From the given transversal,
We can observe that
(2x + 35)° and 103° are the corresponding angles
We know that,
The corresponding angles are congruent
So,
(2x + 35)° = 103°
2x° = 103° – 35°
2x° = 68°
x° = \(\frac{68°}{2}\)
x° = 34°
Hence, from the above,
We can conclude that the value of x is: 34°

Practice & Problem Solving

Question 9.
If p || q, what is the value of u?

Answer:
The given transversal is:

From the given transversal,
We can observe that
u and 148° are vertically opposite angles
Now,
We know that.
The vertically opposite angles are always congruent
Hence, from the above,
We can conclude that the value of u is: 148°

Question 10.
Are ∠K and ∠B corresponding angles? Explain.

Answer:
The given transversal is:

Now,
We know that,
Corresponding angles are angles that are in the same position relative to lines intersected by a transversal
So,
From the given transversal,
We can observe that
∠B is internal and ∠K is external. They are equals if the two intersected lines by the transversal are parallel
Hence, from the above,
We can conclude that
∠K and ∠B are called “Corresponding angles” only when m || n

Question 11.
Streets A and B run parallel to each other. The measure of ∠6 is 155°. What is the measure of ∠4?

Answer:
It is given that
Streets A and B run parallel to each other. The measure of ∠6 is 155°
Now,
The given representation of streets A and B are:

Now,
From the given representation,
We can observe that
∠4 and ∠6 are consecutive interior angles
We know that,
The sum of consecutive interior angles is always equal to 180°
So,
∠4 + ∠6 = 180°
∠4 = 180° – ∠6
∠4 = 180° – 155°
∠4 = 25°
Hence, from the above,
We can conclude that the measure of ∠4 is: 25°

Question 12.
Reasoning The figure shows the design of a rectangular windowpane. The four horizontal lines are parallel. The measure of ∠6 is 53°. What is the measure of ∠12? Write and solve an equation to find the answer.

Answer:
It is given that
The figure shows the design of a rectangular windowpane. The four horizontal lines are parallel. The measure of ∠6 is 53°
Now,
The representation of the rectangular windowpane is:

Now,
From the given representation,
We can observe that
∠6 and ∠12 are consecutive interior angles
We know that,
The sum of consecutive interior angles is: 180°
so,
∠6 + ∠12 = 180°
∠12 = 180° – ∠6
∠12 = 180° – 53°
∠12 = 127°
Hence, from the above,
We can conclude that the angle measure of ∠12 is: 127°

Question 13.
In the figure, m || n. If m∠8 is (4x + 7)° and m∠2 is 107°, what is the value of x? Explain.

Answer:
It is given that
m || n and m∠8 is (4x + 7)° and m∠2 is 107°
Now,
The given transversal is:

From the given transversal,
We can observe that
∠2 and ∠8 are vertically opposite angles
We know that,
The vertically opposite angles are always congruent
So,
∠2 = ∠8
107° = (4x + 7)°
4x° = 107°- 7°
4x° = 100°
x° = \(\frac{100°}{4}\)
x° = 25°
Hence, from the above,
We can conclude that the value of x is: 25°

Question 14.
For the given figure, can you conclude m || n? Explain.

Answer:
The given figure is:

Now,
From the given figure,
We can observe that
74° and 74° are alternate interior angles
Now,
We know that,
The Converse of the Alternate Interior Angles Theorem states that if two lines are cut by a transversal and the alternate interior angles are congruent, then the lines are parallel
Hence, from the above,
We can conclude that m || n since the alternate interior angles are congruent

Question 15.
Line m is parallel to line n. Find the value of x and each missing angle measure.

Answer:
It is given that line m is parallel to n
Now,
The given transversal is:

From the given transversal,
We can observe that
(2x + 25)° and 86° are vertically opposite angles
We know that,
The vertically opposite angles are always congruent
So,
(2x + 25)° = 86°
2x° = 86°- 25°
2x°= 61°
x° = \(\frac{61°}{2}\)
x° = 30.5°
So,
The value of x is: 30.5°
Now,
We can observe that
∠1 and 86° are supplementary angles
∠1 and ∠3 are vertical angles
∠2 and ∠3 are supplementary angles
(2x + 25)° and ∠6 are supplementary angles
∠6 and ∠4 are vertical angles
∠4 and ∠5 are supplementary angles
So,
∠1 + 86° = 180°
∠1 = 180° – 86°
∠1 = 94°
∠1 = ∠3 = 94°
∠2 + ∠3 = 180°
∠2 = 180° – ∠3
∠2 = 180° – 94°
∠2 = 86°
Now,
(2x + 25)° + ∠6 = 180°
2 (30.5)° + ∠6 = 180° – 25°
∠6 = 155° – 61°
∠6 = 94°
∠ 6 = ∠4 = 94°
∠4 + ∠5 = 180°
∠5 = 180° – ∠4
∠5 = 180° – 94°
∠5 = 96°
Hence, from the above,
We can conclude that
The value of x is: 30.5°
The missing angle measures are:
∠1 = ∠3 = 94°; ∠4 = ∠6 = 94°; ∠2 = 86°; ∠5 = 96°

Question 16.
Higher Order Thinking
a. Find the value of x given that r || s.

m∠1 = (63 – x)°
m∠2 = (72 – 2x)°
Answer:
It is given that r || s
Now,
The given transversal is:

From the given transversal,
We can observe that
∠1 and ∠2 are corresponding angles
So,
∠1 = ∠2
So,
(63 – x)° = (72 – 2x)°
2x° – x°= 72° – 63°
x° = 9°
Hence, from the above,
We can conclude that the value of x is: 9°

b. Find m∠1 and m∠2.
Answer:
From part (a),
We can observe that the value of x is: 9°
So,
∠1 = (63 – x)°
∠1 = 63° – 9°
∠1 = 56°
∠2 = (72 – 2x)°
∠2 = 72° – 2 (9)°
∠2 = 72° – 18°
∠2 = 54°
Hence, from the above,
We can conclude that
The values of ∠1 and ∠2 are: 56° and 54° respectively

Question 17.
Find the measures of ∠b and ∠d given that m || n.

Answer:
It is given that m || n
Now,
The given transversal is:

From the given transversal,
We can observe that
∠b and 119.3° are supplementary angles
So,
∠b + 119.3° = 180°
∠b = 180° – 119.3°
∠b = 70.7°
Now,
∠d and 136.9° are supplementary angles
So,
∠d + 136.9° = 180°
∠d = 180° – 136.9°
∠d = 43.1°
Hence, from the above,
We can conclude that
The values of ∠b and ∠d are 70.7° and 43.1° respectively

Assessment Practice
Question 18.
In the figure, g || p. Which angles are alternate interior angles? Select all that apply.

☐ ∠q and ∠r
☐ ∠q and ∠t
☐ ∠q and ∠k
☐ ∠r and ∠t
☐ ∠r and ∠k
☐ ∠u and ∠9
Answer:
It is given that g || p
Now,
The given transversal is:

From the given transversal,
We can observe that
∠r and ∠k are alternate interior angles
∠q and ∠t are alternate interior angles
Hence, from the above,
We can conclude that
The alternate interior angles are:

Question 19.
In the figure, p || q. On a recent math test, Jacob incorrectly listed the value of w as 101.

PART A
Find the value of w.
Answer:
It is given that p || q
Now,
The given transversal is:

From the given transversal,
We can observe that
79° and ∠w are the corresponding angles
Hence, from the above,
We can conclude that the value of ∠w is: 79°

PART B
What mistake did Jacob likely make?
Answer:
We know that,
The corresponding angles are always congruent
Hence,
The mistake that Jacob likely made is:
Jacob considers ∠w  and 101° the corresponding angles

Lesson 6.9 Interior and Exterior Angles of Triangles

Solve & Discuss It!
Nell cuts tile to make a decorative strip for a kitchen backsplash. She must cut the tiles precisely to be congruent triangles. She plans to place the tiles between two pieces of molding, as shown. What is m∠2? Explain.

I can… find the interior and exterior angle measures of a triangle.
Answer:
It is given that
Nell cuts tile to make a decorative strip for a kitchen backsplash. She must cut the tiles precisely to be congruent triangles. She plans to place the tiles between two pieces of molding, as shown
Now,
The representation of the decorative strip for a kitchen backsplash is:

Now,
From the given representation,
We can observe that
∠1 of second congruent triangle = ∠1 of the first congruent triangle
∠3 of second congruent triangle = ∠3 of the third congruent triangle
Now,
We know that,
The sum of all the adjacent angles in a transversal is 180°
So,
∠1 + ∠2 + ∠3 = 180°
From the given representation,
∠1 = ∠3 = 65°
So,
∠2 = 180° – (65° + 65°)
∠2 = 180° – 130°
∠2 = 50°
Hence, from the above,
We can conclude that the value of ∠2 is: 50°

Model with Math
How can you use your knowledge of parallel lines and transversals to solve the problem?
Answer:
The representation of the above problem in the form of parallel lines and transversals is:

Now,
From the above figure,
We can observe that
According to the alternate interior angles theorem,
∠1 of second congruent triangle = ∠1 of the first congruent triangle
∠3 of second congruent triangle = ∠3 of the third congruent triangle
Now,
We know that,
The sum of all the adjacent angles in a transversal is 180°

Focus on math practices
Reasoning What assumption(s) did you need to make to find m∠2? Explain why your assumption(s) is reasonable.
Answer:
The assumptions you need to find ∠2 are:
a. ∠1 of the second congruent triangle = ∠1 of the first congruent triangle
b. ∠3 of the second congruent triangle = ∠3 of the third congruent triangle
c. The sum of all the adjacent angles in a transversal is 180°

Essential Question
How are the interior and exterior angles of a triangle related?
Answer:
An exterior angle of a triangle is equal to the sum of the two opposite interior angles. The sum of exterior angle and interior angle is equal to 180 degrees.

Try It!

Find the unknown angle measure in the triangle at the right.

Answer:
The given triangle is:

Now,
We know that,
The sum of all the interior angles in a triangle is: 180°
Now,
Let x° be the unknown angle measure
So,
68° + 40° + x° = 180°
x° = 180° – 108°
x° = 72°
Hence, from the above,
We can conclude that the value of the unknown angle measure is: 72°

Convince Me!
Could a triangle have interior angle measures of 23°, 71°, and 96°? Explain.
Answer:
The given interior angle measures of a triangle are 23°, 71°, and 96°
Now,
We know that,
The sum of all the interior angles in a triangle is: 180°
So,
23° + 71° + 96° = 180°
94° + 96° = 180°
190° ≠ 180°
Hence, from the above,
We can conclude that a triangle could not have given interior angle measures

Try It!

What is the measure of the exterior angle shown?

Answer:
The given triangle is:

Now,
From the given figure,
We can observe that
(7x – 1)° and (8x + 8)° are interior angle measures
16x° is an exterior angle measure
Now,
We know that,
An exterior angle of a triangle is equal to the sum of the two opposite interior angles. The sum of exterior angle and interior angle is equal to 180 degrees.
So,
16x° = (7x – 1)° + (8x + 8)°
16x° = 15x° + 7°
16x° – 15x° = 7°
x° = 7°
So,
16x° = 16 (7)°
= 112°
Hence, from the above,
We can conclude that the exterior angle measure of the given triangle is: 112°

KEY CONCEPT

The sum of the measures of the interior angles of a triangle is 180°.
m∠1 + m∠2 + m∠3 = 180°

The measure of an exterior angle of a triangle is equal to the sum of the measures of its remote interior angles.
m∠2 + m∠3 = m∠4

Do You Understand?
Question 1.
Essential Question How are the interior and exterior angles of a triangle related?
Answer:
An exterior angle of a triangle is equal to the sum of the two opposite interior angles. The sum of exterior angle and interior angle is equal to 180 degrees.

Question 2.
Reasoning Maggie draws a triangle with a right angle. The other two angles have equal measures. What are the possible values of the exterior angles for Maggie’s triangle? Explain.
Answer:
It is given that
Maggie draws a triangle with a right angle. The other two angles have equal measures
Now,
We know that,
The sum of the interior angles in a triangle is: 180°
Now,
Let the unknown two angle measures be x°
So,
90° + x° + x° = 180°
2x° = 180° – 90°
2x°= 90°
x° = \(\frac{90°}{2}\)
x° = 45°
So,
The two unknown angle measures are 45° and 45°
Now,
We know that,
An exterior angle of a triangle is equal to the sum of the two opposite interior angles. The sum of exterior angle and interior angle is equal to 180 degrees.
Now,
Let the exterior angle measure be y°
So,
y° = 45° + 45°
y° = 90°
Hence, from the above,
We can conclude that the possible value for the exterior angle measure of Maggie’s triangle is: 90°

Question 3.
Brian draws a triangle with interior angles of 32° and 87°, and one exterior angle of 93°. Draw the triangle. Label all of the interior angles and the exterior angle.
Answer:
It is given that
Brian draws a triangle with interior angles of 32° and 87°, and one exterior angle of 93°
Now,
We know that,
The sum of all the interior angles in a triangle is 180°
Now,
Let the unknown interior angle measure be: x°
So,
x°+ 32° + 87° = 180°
x° = 180° – 119°
x° = 61°
Now,
We know that,
An exterior angle of a triangle is equal to the sum of the two opposite interior angles. The sum of exterior angle and interior angle is equal to 180 degrees.
Now,
Let the unknown exterior angle measure be y°
So,
From the given interior angle measures,
y° = 61° + 32°
So,
In a triangle,
The interior angle measures are 61°, 32°, and 87°
The exterior angle measure is: 93°
Hence,
The representation of the interior angles and the exterior angle of a triangle is:

Do You Know How?
Use the diagram below for 4 and 5. Assume that a || b.

Question 4.
What are the measures of ∠1 and ∠2? Explain.
Answer:
It is given that a || b
Now,
The given transversal is:

From the given transversal,
We can observe that
∠1 and 37.3° are vertically opposite angles
So,
∠1 = 37.3°
Now,
We know that,
The sum of all the interior angles in a triangle is 180°
So,
∠1 + 79.4° + ∠2 = 180°
37.3° + 79.4° + ∠2 = 180°
∠2 = 180° – 116.7
∠2 = 63.3°
Hence, from the above,
We can conclude that the measures of ∠1 and ∠2 are 37.3° and 63.3° respectively

Question 5.
What are the measures of ∠3 and ∠4? Explain.
Answer:
It is given that a || b
Now,
The given transversal is:

Now,
From the given transversal,
We can observe that
∠3 and ∠4 are the exterior angles
Now,
∠3 = ∠2 + 79.4°
∠3 = 63.3° + 79.4°
∠3 = 132.7°
Now,
∠4 = ∠1 + 79.4°
∠4 = 37.3° + 79.4°
∠4 = 116.7°
Hence, from the above,
We can conclude that the values of ∠3 and ∠4 are: 132.7° and 116.7°

Question 6.
In △ABC, M∠A = x°, m∠B = (2x)°, and m∠C = (6x + 18)°. What is the measure of each angle?
Answer:
It is given that
In ΔABC,
The angle measures of all the interior angles are:
∠A = x°, ∠B = 2x°, and ∠C = (6x + 18)°
Now,
We know that,
The sum of all the interior angles in a triangle is 180°
So,
x° + 2x° + 6x° + 18° = 180°
9x°= 180° – 18°
9x° = 162°
x° = \(\frac{162°}{9}\)
x° = 18°
So,
∠A = 18°
∠B = 2 (18°) = 36°
∠C = (6x + 18)° = 6 (18°) + 18° = 126°
Hence, from the above,
We can conclude that the measure of each angle is: 18°, 36°, and 126°

Practice & Problem Solving

Question 7.
Leveled Practice For the figure shown, find m∠1.

Angle 1 is an _________ angle of the triangle.
m∠1 is equal to the sum of its ___________.
m∠1 = _______° + _______°
m∠1 = _______°
Answer:
The given triangle is:

Now,
From the given triangle,
We can observe that
∠1 is the exterior angle
59° and 56° are the interior angle measures
Now,
We know that,
An exterior angle of a triangle is equal to the sum of the two opposite interior angles. The sum of exterior angle and interior angle is equal to 180 degrees.
So,
∠1 = 59° + 56°
∠1 = 115°
Hence, from the above,
We can conclude that the value of ∠1 is: 115°

Question 8.
Find m∠1 and m∠2.

Answer:
The given triangle is:

From the given triangle,
We can observe that
There are 2 triangles and ∠1 is the exterior angle for the second triangle
Now,
We know that,
An exterior angle of a triangle is equal to the sum of the two opposite interior angles. The sum of exterior angle and interior angle is equal to 180 degrees.
So,
For the first triangle,
138° = ∠1 + 18°
∠1 = 138° – 18°
∠1 = 120°
For the second triangle,
∠1 = ∠2 + 85°
120° = ∠2 + 85°
∠2 = 120° – 85°
∠2 = 35°
Hence, from the above,
We can conclude that the values of ∠1 and ∠2 are: 120° and 35°

Question 9.
In △ABC, what is m∠c?

Answer:
The given triangle is:

Now,
We know that,
The sum of all the interior angles in a triangle is: 180°
So,
x° + (4x)° + (5x – 13)° = 180°
(10x – 13)° = 180°
10x° = 180° + 13°
10x° = 193°
x° = \(\frac{193°}{10}\)
x° = 19.3°
So,
∠C = (5x – 13)°
∠C = 5 (19.3)° – 13°
∠C = 96.5° – 13°
∠C = 83.5°
Hence, from the above,
We can conclude that the value of ∠C is: 83.5°

Question 10.
In the figure, m∠1 = (8x + 7)°, m∠2 = (4x + 14)°, and m∠4 = (13x + 12)°. Your friend incorrectly says that m∠4 = 51°. What is m∠4? What mistake might your friend have made?

Answer:
It is given that
In the given figure, m∠1 = (8x + 7)°, m∠2 = (4x + 14)°, and m∠4 = (13x + 12)°. Your friend incorrectly says that m∠4 = 51°
Now,
The given figure is:

From the given figure,
We can observe that
∠1, ∠2, and ∠3 are the interior angles
∠4 is an exterior angle
Now,
We know that,
An exterior angle of a triangle is equal to the sum of the two opposite interior angles. The sum of exterior angle and interior angle is equal to 180 degrees.
The sum of all the interior angles in a triangle is 180°
So,
∠1 + ∠2 + ∠3 = 180°
∠4 = ∠1 + ∠2
So,
(13x + 12)° = (8x + 7)° + (4x + 14)°
13x° + 12° = 12x° + 21°
13x° – 12x° = 21°- 12°
x°= 9°
So,
∠4 = (13x + 12)°
∠4 = 13 (9)° + 12°
∠4 = 117° + 12°
∠4 = 129°
Hence, from the above,
We can conclude that
The value of ∠4 is: 120°
The mistake mightyour friend has made is:
He considered ∠4 as an interior angle and he subtracted the value of ∠4 from 180°

Question 11.
What is m∠1?

Answer:
The given triangle is:

Now,
From the given triangle,
We can observe that
∠1 is an exterior angle
26° and 90° are interior angles
Now,
We know that,
An exterior angle of a triangle is equal to the sum of the two opposite interior angles. The sum of exterior angle and interior angle is equal to 180 degrees.
So,
∠1 = 26° + 90°
∠1 = 116°
Hence, from the above,
We can conclude that the value of ∠1 is: 116°

Question 12.
Higher Order Thinking Given that m∠1 = (16x)°, m∠2 = (8x + 21)°, and m∠4 = (25x + 19)°, what is an expression for m∠3? What is m∠3?

Answer:
It is given that
Given that m∠1 = (16x)°, m∠2 = (8x + 21)°, and m∠4 = (25x + 19)°, what is an expression for m∠3
Now,
The given triangle is:

From the given figure,
We can observe that
∠1, ∠2, and ∠3 are the interior angles
∠4 is an exterior angle
Now,
We know that,
An exterior angle of a triangle is equal to the sum of the two opposite interior angles. The sum of exterior angle and interior angle is equal to 180 degrees.
The sum of all the interior angles in a triangle is 180°
So,
∠1 + ∠2 + ∠3 = 180°
∠4 = ∠1 + ∠2
So,
(25x + 19)° = (16x)° + (8x + 21)°
25x° + 19° = 24x° + 21°
25x° – 24x° = 21°- 19°
x°= 2°
So,
∠3 = 180° – (16 (2)° + 8 (2)° + 21°)
∠3 = 180° – 69°
∠3 = 111°
Hence, from the above,
We can conclude that
The value of ∠3 is: 111°

Question 13.
A ramp attached to a building is being built to help with deliveries. The angle that the bottom of the ramp makes with the ground is 37.2°. Find the measure of the other acute angle.

Answer:
It is given that
A ramp attached to a building is being built to help with deliveries. The angle that the bottom of the ramp makes with the ground is 37.2°
Now,
The representation of the ramp is:

From the above,
We can observe that the ramp is in the form of a right triangle
Now,
We know that,
The sum of all the interior angles in a triangle is: 180°
So,
37.2° + 90° + x° = 180°
127.2° + x° = 180°
x°= 180° – 127.2°
x°= 62.8°
Hence, from the above,
We can conclude that the measure of the other acute angle is: 62.8°

Assessment Practice
Question 14.
The measure of ∠F is 110°. The measure of ∠E is 100°. What is the measure of ∠D?

A. 150°
B. 80°
C. 70°
D. 30°
Answer:
It is given that
The measure of ∠F is 110°. The measure of ∠E is 100°
Now,
The given transversal is:

From the given transversal,
We can observe that
∠A and ∠D are the vertical angles
Now,
We know that,
The sum of the adjacent angles is: 180°
So,
∠F + ∠C = 180°
∠B + ∠E = 180°
So,
∠C = 180° – ∠F
∠C = 180°- 110°
∠C = 70°
Now,
∠B = 180° – ∠E
∠B = 180° – 100°
∠B = 80°
Now,
We know that,
The sum of all the interior angles in a triangle is: 180°
So,
∠A + ∠B + ∠C = 180°
70° + 80°+ ∠A = 180°
∠A = 180° – 150°
∠A = 30°
So,
∠D = 30°
Hence, from the above,
We can conclude that the measure of ∠D is: 30°

Question 15.
In the figure, m∠1 = (3x + 12)°, m∠2 = (3x + 18)° and m∠3 = (7x + 10)°. What is m∠3 in degrees?

Answer:
It is given that
m∠1 = (3x + 12)°, m∠2 = (3x + 18)° and m∠3 = (7x + 10)°
Now,
The given triangle is:

From the given triangle,
We can observe that
∠1 and ∠2 are the interior angles
∠3 is an exterior angle
Now,
We know that,
An exterior angle of a triangle is equal to the sum of the two opposite interior angles. The sum of exterior angle and interior angle is equal to 180 degrees.
So,
∠3 = ∠1 + ∠2
(7x + 10)° = (3x + 12)° + (3x + 18)°
(7x + 10)° = (6x + 30)°
7x° – 6x° = 30° – 10°
x° = 20°
So,
∠3 = 7 (20)° + 10°
∠3 = 140° + 10°
∠3 = 150°
Hence, from the above,
We can conclude that the measure of ∠3 is 150°

Lesson 6.10 Angle-Angle Triangle Similarity

Explore It!
Justin made two flags for his model sailboat.

I can… use angle measures to determine whether two triangles are similar.

A. Draw and label triangles to represent each flag.
Answer:
It is given that
Justin made two flags for his model sailboat
Hence,
The representation of triangles that represent each flag is:

B. How are the side lengths of the triangles related?
Answer:
The representation of triangles that represent each flag is:

From the given triangles,
We can observe that,
After dilation,
Flag A and Flag B have different side lengths that are divided (or) multiplied by some value of scale factor and the sizes of Flag A and Flag B are different
Hence, from the above,
We can conclude that
After dilation transformation,
The side lengths of Flag A is greater than the side lengths of Flag B

C. How are the angle measurements of the triangles related?
Answer:
The representation of triangles that represent each flag is:

From the given triangles,
We can observe that the apex vertex of Flag A is 46° and the base vertices of Flag B are 67° and 67°
Now,
We know that,
The sum of all the angles in a triangle is 180°
So,
For Flag B,
Let the unknown angle measure be x°
So,
67° + 67° + x° = 180°
x° = 180° – 134°
x° = 46°
So,
For Flag B,
The apex vertex is: 46°
Now,
When we observe two triangles,
The apex vertex of Flag A and Flag B is the same
Hence, from the above,
We can conclude that the corresponding angle measurements for Flag A and Flag B are the same

Focus on math practices
Reasoning Justin makes a third flag that has sides that are shorter than the sides of the small flag. Two of the angles for each flag measure the same. Are the third angles for each flag the same measure? Explain.
Answer:
It is given that
Justin makes a third flag that has sides that are shorter than the sides of the small flag. Two of the angles for each flag measure the same
So,
Two of the angles for each flag are 67° and 67°
Since two angle measures are the same for all the flags, the remaining third angle measure will also be the same
Hence, from the above,
We can conclude that the third angles for each flag have the same measure

Essential Question
How can you use angle measures to determine whether two triangles are similar?
Answer:
The Angle-Angle (AA) Criterion states that if two angles in one triangle are congruent to two angles in another triangle, the two triangles are similar triangles.

∠A ≅ ∠D and B ≅ ∠E,
So,
△ABC – △DEF.

Try It!

Is △ΧΥΖ ~ △LMN?

m∠X = _______
m∠N = ________
The triangles ________ similar.
Answer:
The given triangles are:

Now,
For ΔXYZ:
We know that,
The sum of all the angles in a triangle is: 180°
So,
∠X + ∠Y + ∠Z = 180°
∠X = 180° – 92° – 42°
∠X = 180° – 134°
∠X = 46°
For ΔLMN:
We know that,
The sum of all the angles in a triangle is: 180°
So,
∠L + ∠M + ∠N = 180°
∠N = 180° – 92° – 53°
∠N = 180° – 148°
∠N = 32°
Now,
To find whether the given triangles are similar or not,
Find out the angle measures of the corresponding sides
So,
For ΔXYZ and ΔLMN,
∠X = ∠L, ∠Y = ∠M,  ∠Z = ∠N
But,
We can observe that
∠X ≠ ∠L
Hence, from the above,
We can conclude that
Since the angle measure of the corresponding sides are not the same,
ΔXYZ is not similar to ΔLMN

Convince Me!
Use what you know about transformations and parallel lines to explain why the Angle-Angle Criterion is true for all triangles.

Try It!

If QR || YZ, is △XYZ ~ △XRQ? Explain.

Answer:
It is given that QR || YZ
Now,
The given figure is:

Now,
We know that,
Since the lines are parallel, the angles that corresponding to the sides will also be the same
Hence,
According to the Side – Side criterion,
ΔXYZ is similar to ΔXRQ

Try It!

Find the value of x if the two triangles are similar. Explain.

Answer:
The given triangles are:

It is given that two triangles are similar
So,
According to Angle – Angle criterion,
(15x)° = 90°
So,
x° = \(\frac{90°}{15}\)
x° = 6°
Hence, from the above,
We can conclude that the value of x is: 6°

KEY CONCEPT

The Angle-Angle (AA) Criterion states that if two angles in one triangle are congruent to two angles in another triangle, the two triangles are similar triangles.

∠A ≅ ∠D and B ≅ ∠E, so △ABC – △DEF.

Do You Understand?
Question 1.
Essential Question How can you use angle measures to determine whether two triangles are similar?
Answer:
The Angle-Angle (AA) Criterion states that if two angles in one triangle are congruent to two angles in another triangle, the two triangles are similar triangles.

∠A ≅ ∠D and B ≅ ∠E,
So,
△ABC – △DEF.

Question 2.
Construct Arguments Claire says that the AA Criterion should be called the AAA Criterion. Explain why Claire might say this. Do you agree? Explain.
Answer:
It is given that
Claire says that the AA Criterion should be called the AAA Criterion.
Now,
We know that
AA criterion tells us that two triangles are similar if two corresponding angles are equal to each other
Now,
AAA criterion may be reformulated as the AAA (angle-angle-angle) similarity theorem
According to the AAA criterion,
Two triangles have their corresponding angles equal if and only if their corresponding sides are proportional.
So,
From the above two criterions’ statements,
We can observe that the definition of both the criteria are the same
Hence, from the above,
We can conclude that you can agree with Clara

Question 3.
Reasoning Which triangle pairs below are always similar? Explain.
Two right triangles
Two isosceles right triangles
Two equilateral triangles
Answer:
The given pairs of triangles are:
a. Two right triangles
b. Two isosceles right triangles
c. Two equilateral triangles
Now,
We know that,
Two triangles are said to be similar if their corresponding angles are congruent and the corresponding sides are in proportion
Hence, from the above,
We can conclude that
The triangle pairs that are similar are:
a. Two isosceles right triangles
b. Two equilateral triangles

Do You Know How?
Question 4.
Are the two triangles similar? Explain.

Answer:
The given triangles are:

From the given triangles,
We can observe that the apex angle measure in the first triangle is 44° and the apex angle measure in the second triangle is 46°
Now,
We know that
AA criterion tells us that two triangles are similar if two corresponding angles are equal to each other
So,
44° ≠ 46°
Hence, from the above,
We can conclude that the two triangles are not similar

Question 5.
Is △QRS ~ △QLM? Explain.

Answer:
The given triangles are:

Now,
From the given triangles,
We can observe that
∠R from ΔQRS = ∠L from ΔQLM
Now,
We know that
AA criterion tells us that two triangles are similar if two corresponding angles are equal to each other
So,
90° = 90°
Hence, from the above,
We can conclude that ΔQRS is similar to ΔQLM

Question 6.
Are the triangles similar? What is the value of x?

Answer:
The given triangles are:

Now,
From the given triangles,
We can observe that
The apex angle of the smaller triangle and the apex angle of the larger triangle is the same
Now,
We know that,
AA criterion tells us that two triangles are similar if two corresponding angles are equal to each other
So,
The given two triangles are similar
Now,
According to AAA criterion,
(4x)° and 76° are alternate angles
Now,
We know that,
The alternate angles are always congruent
So,
(4x)° = 76°
x° = \(\frac{76°}{4}\)
x° = 19°
Hence, from the above,
We can conclude that the value of x is: 19°

Practice & Problem Solving

Question 7.
Is △XYZ ~ △XTU?

Answer:
The given triangles are:

Now,
We know that,
The sum of all the angles in a triangle is: 180°
So,
In ΔXTU,
103° + ∠T + 48° = 180°
∠T = 180° – 151°
∠T = 29°
Now,
∠T in ΔXTU and ∠Y in ΔXYZ are not the same
Now,
We know that,
AA criterion tells us that two triangles are similar if two corresponding angles are equal to each other
Hence, from the above,
We can conclude that
ΔXTU and ΔXYZ are not similar

Question 8.
For what value of x are △RST and △NSP similar? Explain.

Answer:
The given triangles are:

Now,
For the given triangles to be equal,
We know that,
AA criterion tells us that two triangles are similar if two corresponding angles are equal to each other
So,
∠T in ΔRST = ∠P in ΔNSP
So,
(x + 19)° = (2x)°
2x° – x°= 19°
x° = 19°
Hence, from the above,
We can conclude that the value of x so that ΔRSt and ΔNSP are similar is: 19°

Question 9.
Is △FGH ~ △JIH? Explain.

Answer:
The given triangles are:

Now,
We know that,
The sum of all the angles in a triangle is equal to 180°
So,
In ΔJIH,
∠J + ∠I + ∠H = 180°
∠J = 180° – (43° + 35°)
∠J = 180° – 78°
∠J = 102°
So,
From the given triangles,
We can observe that
∠J in ΔJIH and ∠F in ΔFGH are not the same
Now,
We know that,
AA criterion tells us that two triangles are similar if two corresponding angles are equal to each other
Hence, from the above,
We can conclude that ΔRST and ΔJIH are not similar

Question 10.
Are △RST and △NSP similar? Explain.

Answer:
The given triangles are:

Now,
For the given triangles to be equal,
We know that,
AA criterion tells us that two triangles are similar if two corresponding angles are equal to each other
So,
∠T in ΔRST = ∠P in ΔNSP
So,
(x + 15)° = (2x)°
2x° – x°= 15°
x° = 15°
Hence, from the above,
We can conclude that the value of x so that ΔRSt and ΔNSP are similar is: 15°

Question 11.
Contruct Arguments Describe how to use angle relationships to decide whether any two triangles are similar.
Answer:
According to AA criterion,
If two pairs of corresponding angles in a pair of triangles are congruent, then the triangles are similar. We know this because if two angle pairs are the same, then the third pair must also be equal. When the three angle pairs are all equal, the three pairs of sides must also be in proportion.

Question 12.
Higher Order Thinking Are the triangles shown below similar? Explain.

Answer:
The given triangles are:

Now,
From the given triangles,
We can observe that
The angle measures for the smaller triangle can be obtained by dividing the angle measures of the larger triangle by 3
But,
We know that,
The angle measures must be equal for both the triangles with the proportionate side lengths
Now,
We know that,
AA criterion tells us that two triangles are similar if two corresponding angles are equal to each other
Hence, from the above,
We can conclude that the given triangles are not similar

Assessment Practice
Question 13.
Which of the following statements are true? Select all that apply.

☐ △XYZ ~ △SQR
☐ △XYZ ~ △QSR
☐ △XYZ ~ △GHI
☐ △GIH ~ △SRQ
☐ △ZXY ~ △GIH
☐ △GHI ~ △SRQ
Answer:
The given triangles are:

Now,
From the given triangles,
We can observe that
∠I in ΔGIH or ΔHIG is equal to ∠R in ΔQRS or ΔSRQ
∠G in ΔGIH or ΔGHI is equal to ∠X in ΔXYZ or ΔXZY
Now,
We know that,
AA criterion tells us that two triangles are similar if two corresponding angles are equal to each other
Hence, from the above,
We can conclude that
a. △GIH ~ △SRQ
b. △XYZ ~ △GHI

Question 14.
Is △GHI ~ △QRS? Explain your reasoning.

Answer:
The given triangles are:

Now,
From the given triangles,
We can observe that
∠G in ΔGHi and ∠Q in ΔQRS are the same
Now,
We know that,
AA criterion tells us that two triangles are similar if two corresponding angles are equal to each other
Hence, from the above,
We can conclude that ΔGHI and ΔQRS are similar

Topic 6 REVIEW

Topic Essential Question
How can you show that two figures are either congruent or similar to one another?
Answer:
When the two figures are congruent,
a. The shapes and sizes of the two figures are the same
b. The side lengths and the angle measures are the same
c. The orientation is not the same
When the two figures are similar,
a. The shapes and orientations are the same
b. The sizes may are not the same
c. The angle measures are the same but the side lengths are different

Vocabulary Review
Complete each sentence by matching each vocabulary word to its definition. Assume pairs of lines are parallel.

Answer:

Use Vocabulary in Writing
Describe a way to show that △ABC is congruent to △DEF Use vocabulary terms from this Topic in your description.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔABC are:
A (3, 5), B (6, 2), and C (2, 2)
The vertices of ΔDEF are:
D (-2, -1), E (-5, -4), and F (-1, -4)
Now,
The sequence of transformations to show ΔABC is congruent to ΔDEF is:
Step 1:
Reflect the vertices of ΔABC across the y-axis
Step 2:
Translate the image we obtained in step 1 by 1 unit right and 6 units down so that we can obtain the vertices of ΔDEF
Hence,
The representation of the sequence of transformations to show ΔABC is congruent to ΔDEF is:

Concepts and Skills Review

Lesson 6.1 Analyze Translations

Quick Review
A translation is a transformation that maps each point of the preimage the same distance and in the same direction.

Example
Translate △XYZ 5 units right and 3 units up.
Answer:

Practice
Question 1.
Draw the image after a translation 3 units left and 2 units up.

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of the preimage are:
(-1, -3), (0.5, 2), (3.5, 2), and (5, -3)
Now,
After the translation of 3 units left and 2 units up,
The vertices of the image will become:
(-1 – 3, -3 + 2), (0.5 – 3, 2 + 2), (3.5 – 3, 2 + 2), and (5 – 3, -3 + 2)
(-4, -1), (-2.5, 4), (0.5, 4), and (2, -1)
Hence,
The representation of the preimage and the image is:

Lesson 6.2 Analyze Reflections

Quick Review
Reflected figures are the same distance from the line of reflection but on opposite sides.

Example
What are the coordinates of the image of △ABC after a reflection across the y-axis?

Answer:
Use the rule (x, y) → (-x, y).
A (-4, 1) → A'(4, 1)
B (-1, 1) → B'(1, 1)
C (-1, 5) → C'(1, 5)

Practice
Use the figure.

Question 1.
What are the coordinates of the image of rectangle WXYZ after a reflection across the X-axis?
Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of the rectangle WXYZ are:
W (-4, -2), X (-1, -2), Y (-1, -4), and Z (-4, -4)
Now,
We know that,
If a point reflects across the x-axis,
The x-coordinate will be the same but the y-coordinate will change the sign
So,
(x, y) before reflection —– > (x, -y) after reflection
Hence,
The vertices of the rectangle WXYZ after reflection across the x-axis will become:
W’ (-4, 2), X’ (-1, 2), Y’ (-1, 4), and Z’ (-4, 4)
Hence,
The representation of the rectangle WXYZ (Preimage) and the rectangle W’X’Y’Z’ (Image) is:

Question 2.
What are the coordinates of the image of WXYZ after a reflection across the y-axis?
Answer:
We know that,
The vertices of the rectangle WXYZ are:
W (-4, -2), X (-1, -2), Y (-1, -4), and Z (-4, -4)
Now,
We know that,
If a point reflects across the y-axis,
The y-coordinate will be the same but the x-coordinate will change the sign
So,
(x, y) before reflection —– > (-x, y) after reflection
Hence,
The vertices of the rectangle WXYZ after reflection across the y-axis will become:
W’ (4, -2), X’ (1, -2), Y’ (1, -4), and Z’ (4, -4)
Hence,
The representation of the rectangle WXYZ (Preimage) and the rectangle W’X’Y’Z’ (Image) is:

Lesson 6.3 Analyze Rotations

Quick Review
A rotation turns a figure about a fixed point, called the center of rotation. The angle of rotation is the number of degrees the figure is rotated.

Example
What are the coordinates of the image of △ABC after a 90° rotation about the origin?

Answer:
Use the rule (x, y) → (-y, x).
A(1, 4) → A'(-4, 1)
B(4, 4) → B'(-4, 4)
C(4, 1) → C'(-1, 4)

Practice
Use the figure.

Question 1.
What are the coordinates of the image of quadrilateral STUV after a 180° rotation about the origin?
Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of the quadrilateral STUV are:
S (-4, -2), T (-2, -2), U (-2, -4), and V (-4, -4)
Now,
We know that,
When any point rotates 180° counterclockwise about the origin,
(x, y) before rotating 180° —– > (-x, -y) after rotating 180°
So,
The vertices of the quadrilateral STUV after the rotation of 180° are:
S’ (4, 2), T (2, 2), U (2, 4), and V (4, 4)
Hence,
The representation of the quadrilateral STUV after the rotation of 180° is:

Question 2.
What are the coordinates of the image of quadrilateral STUV after a 270° rotation about the origin?
Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of the quadrilateral STUV are:
S (-4, -2), T (-2, -2), U (-2, -4), and V (-4, -4)
Now,
We know that,
When any point rotates 270° counterclockwise about the origin,
(x, y) before rotating 270° —– > (y, -x) after rotating 270°
So,
The vertices of the quadrilateral STUV after the rotation of 270° are:
S’ (-2, 4), T (-2, 2), U (-4, 2), and V (-4, 4)
Hence,
The representation of the quadrilateral STUV after the rotation of 180° is:

Lesson 6.4 Compose Transformations

Quick Review
To compose a sequence of transformations, perform one transformation, and then use the resulting image to perform the next transformation.

Example
How can you use a sequence of transformations to map Figure A onto Figure B?

Answer:
Translate Figure A 3 units up, and then reflect Figure A across the y-axis.

Practice
Question 1.
Translate rectangle ABCD 5 units down, and then reflect it across the y-axis.

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of the rectangle ABCD are:
A (-5, 4), B (-1, 4), C (-1, 2), and D (-5, 2)
Now,
The sequence of transformations to draw the image of the rectangle ABCD is:
Step 1:
Translate the rectangle ABCD 5 units down
So,
The vertices of the rectangle ABCD are:
A (-5, 4 – 5), B (-1, 4 – 5), C (-1, 2 – 5), and D (-5, 2 -5)
A (-5, -1), B (-1, -1), C (-1, -3), and D (-5, -3)
Step 2:
Reflect the vertices that we obtained in step 1 across the y-axis
Now,
We know that,
If a point reflects across the y-axis,
The y-coordinate will be the same but the x-coordinate will change the sign
So,
(x, y) before reflection —– > (-x, y) after reflection
So,
The vertices of the rectangle ABCD are:
A (5, -1), B (1, -1), C (1, -3), and D (5, -3)
Hence,
The representation of the sequence of transformations to represent the rectangle ABCD and its image is:

Lesson 6.5 Understand Congruent Figures

Quick Review
Two figures are congruent if a sequence of transformations maps one figure onto the other.

Example
How can you determine if Figure A is congruent to Figure B?

Answer:
Reflect Figure A across the y-axis, and then translate Figure A 6 units up and 1 unit left.

Practice
Question 1.
Is quadrilateral A congruent to quadrilateral B? Explain.

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of quadrilateral A are:
(1, 1), (2, 4), (4, 4), and (5, 1)
The vertices of quadrilateral B are:
(-5, -4), (-4, -1), (-3, -1), and (-2, -4)
Now,
We know that,
The two figures are said to be congruent when
a. The shapes and sizes of the figures must be the same
b. The side lengths of the 2 figures must be the same
c. The angle measures must be the same
So,
The representation of the quadrilateral A and the quadrilateral B is:

Hence, from the above,
We can conclude that quadrilateral A and quadrilateral B are not congruent side their side lengths are not the same

Lesson 6.6 Describe Dilations

Quick Review
A dilation results in an image that is the same shape but not the same size as the preimage.

Example
What dilation maps WXYZ to W’X’Y’Z?

Answer:
A dilation with center at the origin and a scale factor of 2 maps WXYZ to W’X’Y’Z’.

Practice
Use the figure.

Question 1.
What are the coordinates of the image of parallelogram ABCD after a dilation with center (0, 0) and a scale factor of 3?
Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of the parallelogram ABCD are:
A (-4, 1), B (0, 1), C (1, -1), and D (-3, -1)
Now,
After a dilation with center (0, 0) and a scale factor of 3,
The vertices of the parallelogram must be multiplied with 3
So,
A (-4, 1) × 3, B (0, 1) × 3, C (1, -1) × 3, and D (-3, -1) × 3
A’ (-12, 3), B’ (0, 3), C’ (3, -3), and D’ (-9, -3)
Hence, from the above,
We can conclude that the coordinates for the image of the parallelogram ABCD after a dilation with center (0, 0) and a scale factor of 3 is:
A’ (-12, 3), B’ (0, 3), C’ (3, -3), and D’ (-9, -3)

Question 2.
What are the coordinates of the image of parallelogram ABCD after a dilation with center (0, 0) and a scale factor of \(\frac{1}{2}\)?
Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of the parallelogram ABCD are:
A (-4, 1), B (0, 1), C (1, -1), and D (-3, -1)
Now,
After a dilation with center (0, 0) and a scale factor of \(\frac{1}{2}\),
The vertices of the parallelogram must be multiplied with \(\frac{1}{2}\)
So,
A (-4, 1) × \(\frac{1}{2}\), B (0, 1) × \(\frac{1}{2}\), C (1, -1) × \(\frac{1}{2}\), and D (-3, -1) × \(\frac{1}{2}\)
A’ (-2, 0.5), B’ (0, 0.5), C’ (0.5, -0.5), and D’ (-1.5, -0.5)
Hence, from the above,
We can conclude that the coordinates for the image of the parallelogram ABCD after a dilation with center (0, 0) and a scale factor of \(\frac{1}{2}\) is:
A’ (-2, 0.5), B’ (0, 0.5), C’ (0.5, -0.5), and D’ (-1.5, -0.5)

Lesson 6.7 Understand Similar Figures

Quick Review
Two-dimensional figures are similar if there is a sequence of translations, reflections, rotations, and dilations that maps one figure onto the other figure. Similar figures have the same shape, congruent angles, and proportional side lengths.

Example
Is rectangle ABCD ~ rectangle A’B’C’D?

Answer:
All the angles are right angles.
\(\frac{A B}{A^{\prime} B^{\prime}}=\frac{B C}{B^{\prime} C^{\prime}}=\frac{C D}{C^{\prime} D^{\prime}}=\frac{A D}{A^{\prime} D^{\prime}}=\frac{2}{1}\) = 2
The figures have congruent angle measures and proportional side lengths, so they are similar.

Practice
Use the figure.

Question 1.
Is △ABC similar to △A’B’C’ ? Explain.
Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of ΔABC are:
A (-4, 1) B (-4, 5), and C (-1, 1)
The vertices of ΔA’B’C’ are:
A’ (8, 2), B’ (8, 10), and C’ (2, 2)
Now,
The sequence of transformations to find whether △ABC is similar to △A’B’C’ or not is:
Step 1:
Reflect the vertices of △ABC across the y-axis
Step 2:
Dilate the vertices of △ABC we obtained in step 1 with center (0, 0) and a scale factor of 2
Hence,
The representation of the sequence of transformations to represent that △ABC is similar to △A’B’C’ is:

Question 2.
What sequence of transformations shows that △ABC is similar to △A’B’C’?
Answer:
The sequence of transformations that shows △ABC is similar to △A’B’C’ is:
Step 1:
Reflect the vertices of △ABC across the y-axis
Step 2:
Dilate the vertices of △ABC we obtained in step 1 with center (0, 0) and a scale factor of 2

Lesson 6.8 Angles, Lines, and Transversals

Quick Review
When parallel lines are intersected by a transversal, corresponding angles are congruent, alternate interior angles are congruent, and same-side interior angles are supplementary.

Example
If m || n, what is the value of x?

Answer:
m∠3 = 45° (5x + 25)°
45+ (5x + 25) = 180 – n
x = 22

Practice
In the figure, a || b. What is the value of x?

Answer:
It is given that a || b
Now,
The given transversal is:

Now,
From the given transversal,
We can observe that
(3x + 9)° and 129° are alternate angles
Now,
We know that,
The alternate angles are always congruent
So,
(3x + 9)° = 129°
3x° = 129° – 9°
3x° = 120°
x° = \(\frac{120°}{3}\)
x°= 40°
Hence, from the above,
We can conclude that the value of x is: 40°

Lesson 6.9 Interior and Exterior Angles of Triangles

Quick Review
The sum of the measures of the interior angles of a triangle is 180°. The measure of an exterior angle of a triangle is equal to the sum of the measures of its remote interior angles.

Example
Find the missing angle measure.

Answer:
x + 40 = 100, so x = 60

Practice
Question 1.
Find the missing angle measure.

Answer:
The given triangle is:

Now,
From the given triangle,
The angle measures are 48°, 102°
Let the unknown angle measure be x°
Now,
We know that,
The sum of all the angles in a given triangle is equal to 180°
So,
48°+ 102° + x° = 180°
x° = 180° – 150°
x° = 30°
Hence, from the above,
We can conclude that the missing angle measure is: 30°

Question 2.
Find the value of x.

Answer:
The given triangle is:

Now,
We know that,
The angle measure of an exterior angle is equal to the sum of the remote interior angles that is opposite to the given exterior angle
So,
3x° + 2x° = 115°
5x° = 115°
x° = \(\frac{115°}{3}\)
x° = 23°
Hence, from the above,
We can conclude that the value of x is: 23°

Lesson 6-10 Angle-Angle Triangle Similarity

Quick Review
By the AA Criterion, if two angles in one triangle are congruent to two angles in another triangle, then the triangles are similar.

Example
Is △ABC ~ △DEF? Explain.

Answer:
m∠B = 180° – 90° – 37° = 53°
m∠A = m∠D = 90° and m∠B = m∠E = 53°
Because two angles of the triangles are congruent, the triangles are similar by the AA Criterion.

Practice
Question 1.
AB || XY. Is △ABC ~ △XYC? Explain.

Answer:
It is given that AB || XY
Now,
The given triangles are:

Now,
From the given triangles,
We can observe that
∠A in ΔABC = ∠X in ΔXYC
Now,
We know that,
AA criterion tells us that two triangles are similar if two corresponding angles are equal to each other
Hence, from the above,
We can conclude that ΔABC is similar to ΔXYC

Question 2.
Find the values of x and y given that △ABC is similar to △MNC.

Answer:
It is given that ΔABC is similar to ΔMNC
Now,
The given triangles are:

Now,
We know that,
The sum of adjacent angles is equal to 180°
So,
4x° + 2x° = 180°
6x° = 180°
x° = \(\frac{180°}{6}\)
x° = 30°
Now,
Since ∠N = 60°,
∠A = 60° (By AA criterion)
Now,
We know that,
The sum of all the interior angles in a triangle is equal to 180°
So,
5y° + 60° + 30° = 180°
5y° = 180° – 90°
5y° = 90°
y° = \(\frac{90°}{5}\)
y° = 18°
Hence, from the above,
We can conclude that the values of x and y are: 30° and 18° respectively

Topic 6 Fluency Practice

Crisscrossed
Solve each equation. Write your answers in the cross-number puzzle below. Each digit, negative sign, and decimal point of your answer goes in its own box.

I can… solve multistep equations. © 8.EE.C.7b