12 Surface Area and Volume - Big Ideas Math

Mathematical Thinking: Mathematically proficient students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace.

12.1 Three-Dimensional Figures12.2 Surface Areas of Prisms and Cylinders12.3 Surface Areas of Pyramids and Cones12.4 Volumes of Prisms and Cylinders12.5 Volumes of Pyramids and Cones12.6 Surface Areas and Volumes of Spheres12.7 Spherical Geometry

12 Surface Area and Volume

Earth (p. 692)

Tennis Balls (p. 685)

Khafre's Pyramid (p. 674)

Great Blue Hole (p. 669)

Traffic Cone (p. 656)

Earth (p 692)

Tennis Balls (p. 685)

Kh f ' P id ( 674)

Great Blue Hole (p. 669)

TTrTr fafaffifificc CCoConene ((p(p. 6565656)6)6)

SEE the Big Idea

637

Maintaining Mathematical ProficiencyMaintaining Mathematical ProficiencyFinding the Area of a Circle (7.9.B)

Example 1 Find the area of the circle.

A = πr2 Formula for area of a circle

= π ⋅ 82 Substitute 8 for r.

= 64π Simplify.

≈ 201.06 Use a calculator.

The area is about 201.06 square inches.

Find the area of the circle.

1.

9 ft

2.

6 m

3.

20 cm

Finding the Area of a Composite Figure (7.9.C)

Example 2 Find the area of the composite figure.

32 ft 30 ft

16 ft

17 ft The composite figure is made up of a

rectangle, a triangle, and a semicircle.

Find the area of each figure.

Area of rectangle Area of triangle Area of semicircle

A =ℓw A = 1 —

2 bh A =

πr2

— 2

= 32(16) = 1 —

2 (30)(16) =

π(17)2

— 2

= 512 = 240 ≈ 453.96

So, the area is about 512 + 240 + 453.96 = 1205.96 square feet.

Find the area of the composite figure.

4. 7 m

7 m

7 m

20 m

5. 6 in.

3 in.

5 in.

10 in.

6.

7. ABSTRACT REASONING A circle has a radius of x inches. Write a formula for the area

of the circle when the radius is multiplied by a real number a.

8 in.

9 cm

6 cm

6 cm

6 cm

3 cm3

638 Chapter 12 Surface Area and Volume

Mathematical Mathematical ThinkingThinkingCreating a Coherent Representation

Mathematically profi cient students create and use representations to organize, record, and communicate mathematical ideas. (G.1.E)

Monitoring ProgressMonitoring ProgressDraw a net of the three-dimensional fi gure. Label the dimensions.

1.

2 ft

4 ft

4 ft

2. 5 m

12 m

8 m

3.

15 in.

10 in.10 in.

Drawing a Net for a Pyramid

Draw a net of the pyramid.

19 in.19 in.

20 in.

SOLUTIONThe pyramid has a square base.

Its four lateral faces are congruent

isosceles triangles.

Nets for Three-Dimensional FiguresA net for a three-dimensional fi gure is a two-dimensional pattern that can be folded to

form the three-dimensional fi gure.

h

w

base

base

lateral face

lateral face

lateral face

lateral face

www

h

Core Core ConceptConcept

20 in.19 in.

19 in.

Section 12.1 Three-Dimensional Figures 639

12.1 Three-Dimensional Figures

Essential QuestionEssential Question What is the relationship between the numbers

of vertices V, edges E, and faces F of a polyhedron?

A polyhedron is a solid that is bounded

by polygons, called faces.

• Each vertex is a point.

• Each edge is a segment of a line.

• Each face is a portion of a plane.

Analyzing a Property of Polyhedra

Work with a partner. The fi ve Platonic solids are shown below. Each of these solids

has congruent regular polygons as faces. Complete the table by listing the numbers of

vertices, edges, and faces of each Platonic solid.

tetrahedron cube octahedron

dodecahedron icosahedron

Solid Vertices, V Edges, E Faces, F

tetrahedron

cube

octahedron

dodecahedron

icosahedron

Communicate Your AnswerCommunicate Your Answer 2. What is the relationship between the numbers of vertices V, edges E, and

faces F of a polyhedron? (Note: Swiss mathematician Leonhard Euler

(1707–1783) discovered a formula that relates these quantities.)

3. Draw three polyhedra that are different from the Platonic solids given in

Exploration 1. Count the numbers of vertices, edges, and faces of each

polyhedron. Then verify that the relationship you found in Question 2 is

valid for each polyhedron.

MAKING MATHEMATICAL ARGUMENTS

To be profi cient in math, you need to reason inductively about data.

edge

face

vertex

G.10.A

TEXAS ESSENTIAL KNOWLEDGE AND SKILLS


12.1 Lesson What You Will LearnWhat You Will Learn Classify solids.

Describe cross sections.

Sketch and describe solids of revolution.

Classifying SolidsA three-dimensional fi gure, or solid, is bounded by

fl at or curved surfaces that enclose a single region

of space. A polyhedron is a solid that is bounded by

polygons, called faces. An edge of a polyhedron is a

line segment formed by the intersection of two faces.

A vertex of a polyhedron is a point where three or more

edges meet. The plural of polyhedron is polyhedra

or polyhedrons.

To name a prism or a pyramid, use the shape of the base. The two bases of a prism

are congruent polygons in parallel planes. For example, the bases of a pentagonal

prism are pentagons. The base of a pyramid is a polygon. For example, the base of a

triangular pyramid is a triangle.

Classifying Solids

Tell whether each solid is a polyhedron. If it is, name the polyhedron.

a. b. c.

SOLUTION

a. The solid is formed by polygons, so it is a polyhedron. The two bases are congruent

rectangles, so it is a rectangular prism.

b. The solid is formed by polygons, so it is a polyhedron. The base is a hexagon, so it

is a hexagonal pyramid.

c. The cone has a curved surface, so it is not a polyhedron.

polyhedron, p. 640face, p. 640edge, p. 640vertex, p. 640cross section, p. 641solid of revolution, p. 642axis of revolution, p. 642

Previoussolidprismpyramidcylinderconespherebase

Core VocabularyCore Vocabullarry

Core Core ConceptConceptTypes of Solids

prism

pyramid

Polyhedra

cylinder cone

sphere

Not Polyhedra

face

edgevertex

Pentagonal prism

Bases arepentagons.

Triangular pyramid

Base is atriangle.


Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com

Tell whether the solid is a polyhedron. If it is, name the polyhedron.

1. 2. 3.

Describing Cross SectionsImagine a plane slicing through a solid. The intersection of the plane and the solid

is called a cross section. For example, three different cross sections of a cube are

shown below.

square rectangle triangle

Describing Cross Sections

Describe the shape formed by the intersection of the plane and the solid.

a. b. c.

d. e. f.

SOLUTION

a. The cross section is a hexagon. b. The cross section is a triangle.

c. The cross section is a rectangle. d. The cross section is a circle.

e. The cross section is a circle. f. The cross section is a trapezoid.


Describe the shape formed by the intersection of the plane and the solid.

4. 5. 6.


Sketching and Describing Solids of RevolutionA solid of revolution is a three-dimensional fi gure that is formed by rotating a

two-dimensional shape around an axis. The line around which the shape is rotated is

called the axis of revolution.

For example, when you rotate a rectangle around a line that contains one of its sides,

the solid of revolution that is produced is a cylinder.

Sketching and Describing Solids of Revolution

Sketch the solid produced by rotating the fi gure around the given axis. Then identify

and describe the solid.

a. 9

9

44

b.

2

5

SOLUTION

a. 9

4

b.

2

5

The solid is a cylinder with

a height of 9 and a base

radius of 4. The solid is a cone with

a height of 5 and a base

radius of 2.


Sketch the solid produced by rotating the fi gure around the given axis. Then identify and describe the solid.

7.

3

4

8.

88

6

6 9.

7

7


Exercises12.1 Tutorial Help in English and Spanish at BigIdeasMath.com

In Exercises 3–6, match the polyhedron with its name.

3. 4.

5. 6.

A. triangular prism B. rectangular pyramid

C. hexagonal pyramid D. pentagonal prism

In Exercises 7–10, tell whether the solid is a polyhedron. If it is, name the polyhedron. (See Example 1.)

7. 8.

9. 10.

In Exercises 11−14, describe the cross section formed by the intersection of the plane and the solid. (See Example 2.)

11. 12.

13. 14.

In Exercises 15–18, sketch the solid produced by rotating the fi gure around the given axis. Then identify and describe the solid. (See Example 3.)

15.

8

8

8

8

16.

6

6

17. 3

3

18.

22

5

5

Monitoring Progress and Modeling with MathematicsMonitoring Progress and Modeling with Mathematics

1. VOCABULARY A(n) __________ is a solid that is bounded by polygons.

2. WHICH ONE DOESN’T BELONG? Which solid does not belong with the other three?

Explain your reasoning.

Vocabulary and Core Concept CheckVocabulary and Core Concept Check


19. ERROR ANALYSIS Describe and correct the error in

identifying the solid.

The solid is a

rectangular pyramid.

✗

20. HOW DO YOU SEE IT? Is the swimming pool shown

a polyhedron? If it is, name the polyhedron. If not,

explain why not.

In Exercises 21–26, sketch the polyhedron.

21. triangular prism 22. rectangular prism

23. pentagonal prism 24. hexagonal prism

25. square pyramid 26. pentagonal pyramid

27. MAKING AN ARGUMENT Your friend says that the

polyhedron shown is a triangular prism. Your cousin

says that it is a triangular pyramid. Who is correct?


28. ATTENDING TO PRECISION The fi gure shows a plane

intersecting a cube through four of its vertices. The

edge length of the cube is 6 inches.

a. Describe the shape formed by the cross section.

b. What is the perimeter of the cross section?

c. What is the area of the cross section?

REASONING In Exercises 29–34, tell whether it is possible for a cross section of a cube to have the given shape. If it is, describe or sketch how the plane could intersect the cube.

29. circle 30. pentagon

31. rhombus 32. isosceles triangle

33. hexagon 34. scalene triangle

35. REASONING Sketch the composite solid produced

by rotating the fi gure around the given axis. Then

identify and describe the composite solid.

a. 2

33

b. 8

4 45

11

36. THOUGHT PROVOKING Describe how Plato might

have argued that there are precisely fi ve Platonic Solids (see page 639). (Hint: Consider the angles

that meet at a vertex.)

Maintaining Mathematical ProficiencyMaintaining Mathematical ProficiencyDecide whether enough information is given to prove that the triangles are congruent. If so, state the theorem you would use. (Sections 5.3, 5.5, and 5.6)

37. △ABD, △CDB 38. △JLK, △JLM 39. △RQP, △RTS

A B

CD

Reviewing what you learned in previous grades and lessons

J

K L M

S

TP

Q

R

Section 12.2 Surface Areas of Prisms and Cylinders 645

Essential QuestionEssential Question How can you fi nd the surface area of a prism or

a cylinder?

Recall that the surface area of a polyhedron is the sum of the areas of its faces. The

lateral area of a polyhedron is the sum of the areas of its lateral faces.

Finding a Formula for Surface Area

Work with a partner. Consider the polyhedron shown.

a. Identify the polyhedron. Then sketch its net so that

the lateral faces form a rectangle with the same

height h as the polyhedron. What types of fi gures

make up the net?

b. Write an expression that represents the perimeter P

of the base of the polyhedron. Show how you can

use P to write an expression that represents the

lateral area L of the polyhedron.

c. Let B represent the area of a base of the polyhedron. Write a formula for the

surface area S.


Work with a partner. Consider the solid shown.

a. Identify the solid. Then sketch its net. What types of

fi gures make up the net?

b. Write an expression that represents the perimeter P

of the base of the solid. Show how you can use P to

write an expression that represents the lateral area L

of the solid.

c. Write an expression that represents the area B of a base

of the solid.

d. Write a formula for the surface area S.

Communicate Your AnswerCommunicate Your Answer 3. How can you fi nd the surface area of a prism or a cylinder?

4. Consider the rectangular prism shown.

35

7a. Find the surface area of the rectangular prism by

drawing its net and fi nding the sum of the areas

of its faces.

b. Find the surface area of the rectangular prism by

using the formula you wrote in Exploration 1.

c. Compare your answers to parts (a) and (b).

What do you notice?

APPLYING MATHEMATICS

To be profi cient in math, you need to analyze relationships mathematically to draw conclusions.

G.10.BG.11.C


Surface Areas of Prismsand Cylinders

12.2

height, h

a

b

c

height, h

radius, r


12.2 Lesson What You Will LearnWhat You Will Learn Find lateral areas and surface areas of right prisms.

Find lateral areas and surface areas of right cylinders.

Use surface areas of right prisms and right cylinders.

Finding Lateral Areas and Surface Areas of Right PrismsRecall that a prism is a polyhedron with two congruent faces, called bases, that lie in

parallel planes. The other faces, called lateral faces, are parallelograms formed by

connecting the corresponding vertices of the bases. The segments connecting these

vertices are lateral edges. Prisms are classifi ed by the shapes of their bases.

base

base

lateralfaces

lateraledges

The surface area of a polyhedron is the sum of the areas of its faces. The lateral area

of a polyhedron is the sum of the areas of its lateral faces.

Imagine that you cut some edges of a

polyhedron and unfold it. The two-dimensional

representation of the faces is called a net. The

surface area of a prism is equal to the area of

its net.

The height of a prism is the perpendicular

distance between its bases. In a right prism,

each lateral edge is perpendicular to both

bases. A prism with lateral edges that are not

perpendicular to the bases is an oblique prism.

height

height

Right rectangular prism Oblique triangular prism

lateral faces, p. 646lateral edges, p. 646surface area, p. 646lateral area, p. 646net, p. 646right prism, p. 646oblique prism, p. 646right cylinder, p. 647oblique cylinder, p. 647

Previousprismbases of a prismcylindercomposite solid


Core Core ConceptConceptLateral Area and Surface Area of a Right PrismFor a right prism with base perimeter P, base apothem a,

h

B

P

height h, and base area B, the lateral area L and

surface area S are as follows.

Lateral area L = Ph

Surface area S = 2B + L

= aP + Ph


Finding Lateral Area and Surface Area

Find the lateral area and the surface area of the

9 ft

6 ft7.05 ftright pentagonal prism.

SOLUTION

Find the apothem and perimeter of a base.

a = √—

62 − 3.5252 = √—

23.574375 6 ft

3.525 ft 3.525 ft

6 ft aP = 5(7.05) = 35.25

Find the lateral area and the surface area.

L = Ph Formula for lateral area of a right prism

= (35.25)(9) Substitute.

= 317.25 Multiply.

S = aP + Ph Formula for surface area of a right prism

= ( √—

23.574375 ) (35.25) + 317.25 Substitute.


The lateral area is 317.25 square feet and the surface area is

about 488.40 square feet.


1. Find the lateral area and the surface area of a right rectangular prism with a

height of 7 inches, a length of 3 inches, and a width of 4 inches.

Finding Lateral Areas and Surface Areas of Right CylindersRecall that a cylinder is a solid with congruent circular bases that lie in parallel planes.

The height of a cylinder is the perpendicular distance between its bases. The radius of

a base is the radius of the cylinder. In a right cylinder, the segment joining the centers

of the bases is perpendicular to the bases. In an oblique cylinder, this segment is not perpendicular to the bases.

The lateral area of a cylinder is the area of its curved surface. For a right cylinder, it is

equal to the product of the circumference and the height, or 2πrh. The surface area of

a cylinder is equal to the sum of the lateral area and the areas of the two bases.

Core Core ConceptConceptLateral Area and Surface Area of a Right CylinderFor a right cylinder with radius r,

rr base area

A = r2

h h

2 rππ

lateral areaA = 2 rhπ

base areaA = r2π

2 rπheight h, and base area B, the

lateral area L and surface area S

are as follows.

Lateral area L = 2πrh

Surface area S = 2B + L

= 2πr2 + 2πrh

height

height

right cylinder

oblique cylinder

ATTENDING TO PRECISION

Throughout this chapter, round lateral areas, surface areas, and volumes to the nearest hundredth, if necessary.



Find the lateral area and the surface area of the right cylinder.

8 m

4 m

SOLUTION


L = 2πrh Formula for lateral area of a right cylinder

= 2π(4)(8) Substitute.

= 64π Simplify.


S = 2πr2 + 2πrh Formula for surface area of a right cylinder

= 2π(4)2 + 64π Substitute.

= 96π Simplify.


The lateral area is 64π, or about 201.06 square meters. The surface area is 96π,

or about 301.59 square meters.

Solving a Real-Life Problem

You are designing a label for the cylindrical soup can shown. 9 cm

12 cm

The label will cover the lateral area of the can. Find the

minimum amount of material needed for the label.

SOLUTION

Find the radius of a base.

r = 1 —

2 (9) = 4.5

Find the lateral area.

L = 2πrh Formula for lateral area of a right cylinder

= 2π(4.5)(12) Substitute.

= 108π Simplify.


You need a minimum of about 339.29 square centimeters of material.


2. Find the lateral area and the surface area of the right cylinder.

18 in.

10 in.

3. WHAT IF? In Example 3, you change the design of the can so that the diameter

is 12 centimeters. Find the minimum amount of material needed for the label.


Using Surface Areas of Right Prisms and Right Cylinders

Finding the Surface Area of a Composite Solid

Find the lateral area and the surface area of the composite solid.

SOLUTION

Lateral area

of solid =

Lateral area

of cylinder +

Lateral area

of prism

= 2πrh + Ph

= 2π(6)(12) + 14(12)

= 144π + 168

≈ 620.39

Surface area

of solid =

Lateral area

of solid + 2 ⋅ ( Area of a base

of the cylinder −

Area of a base

of the prism )

= 144π + 168 + 2(πr2 −ℓw)

= 144π + 168 + 2[π(6)2 − 4(3)]

= 216π + 144

≈ 822.58

The lateral area is about 620.39 square meters and the surface area is

about 822.58 square meters.

Changing Dimensions in a Solid

Describe how doubling all the linear dimensions affects 2 ft

8 ft

the surface area of the right cylinder.

SOLUTION

Before change After change

Dimensions r = 2 ft, h = 8 ft r = 4 ft, h = 16 ft

Surface area

S = 2πr2 + 2πrh

= 2π(2)2 + 2π(2)(8)

= 40π ft2

S = 2πr2 + 2πrh

= 2π(4)2 + 2π(4)(16)

= 160π ft2

Doubling all the linear dimensions results in a surface area that

is 160π — 40π

= 4 = 22 times the original surface area.


4. Find the lateral area and the surface area of the composite solid at the left.

5. In Example 5, describe how multiplying all the linear dimensions by 1 —

2 affects the

surface area of the right cylinder.

3 m4 m

12 m

6 m

6 mm

10 mm

2 mm

8 mm


Tutorial Help in English and Spanish at BigIdeasMath.comExercises12.2

Monitoring Progress and Modeling with MathematicsMonitoring Progress and Modeling with MathematicsIn Exercises 3 and 4, fi nd the surface area of the solid formed by the net.

3. 4.

4 in.

10 in.

8 cm

20 cm

In Exercises 5–8, fi nd the lateral area and the surface area of the right prism. (See Example 1.)

5. 6.

8 ft3 ft

2 ft

9.1 m3 m

8 m

7. A regular pentagonal prism has a height of 3.5 inches

and a base edge length of 2 inches.

8. A regular hexagonal prism has a height of 80 feet and

a base edge length of 40 feet.

In Exercises 9–12, fi nd the lateral area and the surface area of the right cylinder. (See Example 2.)

9.

2 in.

0.8 in. 10. 16 cm

8 cm

11. A right cylinder has a diameter of 24 millimeters and

a height of 40 millimeters.

12. A right cylinder has a radius of 2.5 feet and a height

of 7.5 feet.

13. MODELING WITH MATHEMATICS The inside of the

cylindrical swimming pool shown must be covered

with a vinyl liner. The liner must cover the side

and bottom of the swimming pool. What is the

minimum amount of vinyl needed for the liner? (See Example 3.)

24 ft

4 ft

14. MODELING WITH MATHEMATICS The tent shown has fabric

covering all four sides

and the fl oor. What is

the minimum amount

of fabric needed to

construct the tent?

In Exercises 15–18, fi nd the lateral area and the surface area of the composite solid. (See Example 4.)

15.

2 cm

4 cm

8 cm

4 cm

1 cm 16.

7 ft

4 ft

4 ft

6 ft

17.

11 in.

5 in.

2 in. 18.

9 m

7 m

15 m

6 m

5 m

1. VOCABULARY Sketch a right triangular prism. Identify the bases, lateral faces, and lateral edges.

2. WRITING Explain how the formula S = 2B + L applies to fi nding the surface area of both a

right prism and a right cylinder.

Vocabulary and Core Concept Check

4 ft

6 ft8 ft

5 ft



fi nding the surface area of the right cylinder.

S = 2𝛑 (6)2 + 2𝛑(6)(8)

= 168𝛑

≈ 527.79 cm2

✗8 cm

6 cm


fi nding the surface area of the composite solid.

S = 2(20)(7) + 2(18)(7) + 2𝛑 (8)(7)

+ 2[(18)(20) + 𝛑 (8)2]

≈ 2005.98 ft2

✗

20 ft18 ft

7 ft16 ft

In Exercises 21–24, describe how the change affects the surface area of the right prism or right cylinder. (See Example 5.)

21. doubling all the

linear dimensions 22. multiplying all the

linear dimensions

by 1 — 3

5 in.

4 in.17 in.

24 mm

9 mm

23. tripling

the radius 24. multiplying the base

edge lengths by 1 — 4 and

the height by 4

2 m16 m

8 m

In Exercises 25 and 26, fi nd the height of the right prism or right cylinder.

25. S = 1097 m2 26. S = 480 in.2

8.2 m

h

8 in.

15 in.

h

27. MATHEMATICAL CONNECTIONS A cube has a

surface area of 343 square inches. Write and solve an

equation to fi nd the length of each edge of the cube.

28. MATHEMATICAL CONNECTIONS A right cylinder has

a surface area of 108π square meters. The radius of

the cylinder is twice its height. Write and solve an

equation to fi nd the height of the cylinder.

29. MODELING WITH MATHEMATICS A company makes

two types of recycling bins, as shown. Both types of

bins have an open top. Which recycling bin requires

more material to make? Explain.

10 in.12 in.

36 in.36 in.

6 in.

30. MODELING WITH MATHEMATICS You are painting a

rectangular room that is 13 feet long, 9 feet wide, and

8.5 feet high. There is a window that is 2.5 feet wide

and 5 feet high on one wall. On another wall, there

is a door that is 4 feet wide and 7 feet high. A gallon

of paint covers 350 square feet. How many gallons

of paint do you need to cover the four walls with one

coat of paint, not including the window and door?

31. ANALYZING RELATIONSHIPS Which creates a

greater surface area, doubling the radius of a cylinder

or doubling the height of a cylinder? Explain

your reasoning.

32. MAKING AN ARGUMENT You cut a cylindrical piece

of lead, forming two congruent cylindrical pieces

of lead. Your friend claims the surface area of each

smaller piece is exactly half the surface area of

the original piece. Is your friend correct? Explain

your reasoning.

33. USING STRUCTURE The right triangular prisms

shown have the same surface area. Find the height h

of prism B.

20 cm

24 cm

Prism A Prism B

3 cm

8 cmh6 cm20 cm

7 yd

2 yd


Reviewing what you learned in previous grades and lessonsMaintaining Mathematical ProficiencyFind the area of the regular polygon. (Section 11.3)

43.

6 cm

8 cm

44.

9 in.

10.6 in. 45. 7 m

34. USING STRUCTURE The lateral surface area of a

regular pentagonal prism is 360 square feet. The

height of the prism is twice the length of one of the

edges of the base. Find the surface area of the prism.

35. ANALYZING RELATIONSHIPS Describe how

multiplying all the linear dimensions of the right

rectangular prism by each given value affects the

surface area of the prism.

w

h

a. 2 b. 3 c. 1 —

2 d. n

36. HOW DO YOU SEE IT? An open gift box is shown.

a. Why is the area of

the net of the box

larger than the

minimum amount of

wrapping paper needed

to cover the closed box?

b. When wrapping the box, why would you want

to use more than the minimum amount of

paper needed?

37. REASONING Consider a cube that is

built using 27 unit cubes, as shown.

a. Find the surface area of the

solid formed when the red unit

cubes are removed from the

solid shown.

b. Find the surface area of the solid formed when the

blue unit cubes are removed from the solid shown.

c. Explain why your answers are different in parts (a)

and (b).

38. THOUGHT PROVOKING You have 24 cube-shaped

building blocks with edge lengths of 1 unit. What

arrangement of blocks gives you a rectangular prism

with the least surface area? Justify your answer.

39. USING STRUCTURE Sketch the net of the oblique

rectangular prism shown. Then fi nd the surface area.

7 ft

15 ft

8 ft

4 ft

40. WRITING Use the diagram to write a formula that can

be used to fi nd the surface area S of any cylindrical

ring where 0 < r2 < r1.

r2

r1

h

41. USING STRUCTURE The diagonal of a cube is a

segment whose endpoints are vertices that are not on

the same face. Find the surface area of a cube with a

diagonal length of 8 units.

42. USING STRUCTURE A cuboctahedron has 6 square

faces and 8 equilateral triangular faces, as shown. A

cuboctahedron can be made by slicing off the corners

of a cube.

a. Sketch a net for the

cuboctahedron.

b. Each edge of a

cuboctahedron has a

length of 5 millimeters.

Find its surface area.

wn.

d d

Section 12.3 Surface Areas of Pyramids and Cones 653

Essential QuestionEssential Question How can you fi nd the surface area of a pyramid

or a cone?

A regular pyramid has a regular polygon for a base and the segment joining the

vertex and the center of the base is perpendicular to the base. In a right cone, the

segment joining the vertex and the center of the base is perpendicular to the base.


Work with a partner. Consider the polyhedron shown.

a. Identify the polyhedron. Then sketch its net.

What types of fi gures make up the net?

b. Write an expression that represents the

perimeter P of the base of the polyhedron.

Show how you can use P to write an

expression that represents the

lateral area L of the polyhedron.

c. Let B represent the area of a base of the

polyhedron. Write a formula for the surface area S.


Work with a partner. Consider the solid shown.

a. Identify the solid. Then sketch its net. What types

of fi gures make up the net?

b. Write an expression that represents the area B of

the base of the solid.

c. What is the arc measure of the lateral surface of

the solid? What is the circumference and area

of the entire circle that contains the lateral surface

of the solid? Show how you can use these three

measures to fi nd the lateral area L of the solid.

d. Write a formula for the surface area S.

Communicate Your AnswerCommunicate Your Answer 3. How can you fi nd the surface area of a pyramid or cone?

4. Consider the rectangular pyramid shown.

8 ft

12 ft a. Find the surface area of the rectangular pyramid

by drawing its net and fi nding the sum of the

areas of its faces.

b. Find the surface area of the rectangular pyramid

by using the formula you wrote in Exploration 1.

c. Compare your answers to parts (a) and (b).

What do you notice?

APPLYING MATHEMATICS

To be profi cient in math, you need to analyze relationships mathematically to draw conclusions.

G.10.BG.11.C


Surface Areas of Pyramidsand Cones

12.3

slant height,height, h

base edge length, b

radius, r

slant height,


12.3 Lesson What You Will LearnWhat You Will Learn Find lateral areas and surface areas of regular pyramids.

Find lateral areas and surface areas of right cones.

Use surface areas of regular pyramids and right cones.

Finding Lateral Areas and Surface Areas of Regular PyramidsA pyramid is a polyhedron in which the base

is a polygon and the lateral faces are triangles

with a common vertex, called the vertex of the pyramid. The intersection of two lateral faces

is a lateral edge. The intersection of the base

and a lateral face is a base edge. The height

of the pyramid is the perpendicular distance

between the base and the vertex.

A regular pyramid has a regular polygon for a base

and the segment joining the vertex and the center of

the base is perpendicular to the base. The lateral faces

of a regular pyramid are congruent isosceles triangles.

The slant height of a regular pyramid is the height

of a lateral face of the regular pyramid. A nonregular

pyramid does not have a slant height.

vertex of a pyramid, p. 654regular pyramid, p. 654slant hieght of a regular

pyramid, p. 654vertex of a cone, p. 655right cone, p. 655oblique cone, p. 655slant height of a right cone,

p. 655lateral surface of a cone,

p. 655

Previouspyramidconecomposite solid


Core Core ConceptConceptLateral Area and Surface Area of a Regular Pyramid

For a regular pyramid with base perimeter P, slant heightℓ, and

base area B, the lateral area L and surface area S are as follows.

Lateral area L = 1 —

2 Pℓ

Surface area S = B + L = B + 1 —

2 Pℓ


Find the lateral area and the surface area of the regular hexagonal pyramid.

SOLUTION

The perimeter P of the base is 6 ⋅ 10 = 60, feet and the apothem a is 5 √—

3 feet.

The slant heightℓof a face is 14 feet. Find the lateral area and the surface area.

L = 1 —

2 Pℓ Formula for lateral area of a regular pyramid

= 1 —

2 (60)(14) Substitute.

= 420 Simplify.

S = B + 1 —

2 Pℓ Formula for surface area of a regular pyramid

= 1 —

2 ( 5 √

— 3 ) (60) + 420 Substitute.

= 150 √—

3 + 420 Simplify.


The lateral area is 420 square feet and the surface area is

about 679.81 square feet.

lateral edge

base edgelateral faces

Pyramid

heightvertex

base

Regular pyramid

slant heightheight

PB

3 ft

14 ft

10 ft 5


Finding Lateral Areas and Surface Areas of Right ConesA cone has a circular base and a vertex that is not in the same plane as the base. The

radius of the base is the radius of the cone. The height is the perpendicular distance

between the vertex and the base.

In a right cone, the segment joining the vertex and the center of the base is

perpendicular to the base. In an oblique cone, this segment is not perpendicular to the

base. The slant height of a right cone is the distance between the vertex and a point

on the edge of the base. An oblique cone does not have a slant height.

The lateral surface of a cone consists of all segments that connect the vertex with

points on the edge of the base.

Core Core ConceptConceptLateral Area and Surface Area of a Right ConeFor a right cone with radius r, slant heightℓ, and base area B,

the lateral area L and surface area S are as follows.

Lateral area L = πrℓ

Surface area S = B + L = πr2ℓ + πrℓ


Find the lateral area and the surface area of the right cone.

SOLUTION

Use the Pythagorean Theorem (Theorem 9.1) to fi nd the

slant heightℓ.

ℓ = √—

62 + 82 = 10


L = πrℓ Formula for lateral area of a right cone

= π(6)(10) Substitute.

= 60π Simplify.


S = πr2 + πrℓ Formula for surface area of a right cone

= π(6)2 + 60π Substitute.

= 96π Simplify.


The lateral area is 60π, or about 188.50 square meters. The surface area is 96π,

about 301.59 square meters.

4.8 m

8 m

5.5 m

slantheight height

lateralsurface

baser

Right cone

vertex

height

lateralsurface

baser

Oblique cone

vertex

r

6 m

8 m


1. Find the lateral area and the surface area of the regular pentagonal pyramid.


Solving a Real-Life Problem

The traffi c cone can be approximated by a right cone with a radius of 5.7 inches

and a height of 18 inches. Find the lateral area of the traffi c cone.

SOLUTION

Use the Pythagorean Theorem (Theorem 9.1)

to fi nd the slant heightℓ.

ℓ= √——

182 + (5.7)2 = √—

356.49

Find the lateral area.


= π(5.7) ( √—

356.49 ) Substitute.


The lateral area of the traffi c cone is about 338.10 square inches.


2. Find the lateral area and the surface area

of the right cone.

3. WHAT IF? The radius of the cone in Example 3

is 6.3 inches. Find the lateral area.

Using Surface Areas of Regular Pyramids and Right Cones

Finding the Surface Area of a Composite Solid

Find the lateral area and the surface area of the composite solid.

SOLUTION

Lateral area

of solid =

Lateral area

of cone +

Lateral area

of cylinder

= πrℓ + 2πrh

= π(3)(5) + 2π(3)(6)

= 51π

≈ 160.22

Surface area

of solid =

Lateral area

of solid +

Area of a base

of the cylinder

= 51π + πr2

= 51π + π(3)2

= 60π

≈ 188.50

The lateral area is about 160.22 square centimeters and the surface area is

about 188.50 square centimeters.

5.7

18

15 ft

8 ft

5 cm

3 cm

6 cm



Describe how the change affects the surface area

of the right cone.

a. multiplying the radius by 3 —

2

b. multiplying all the linear dimensions by 3 —

2

SOLUTION

a. Before change After change

Dimensions r = 10 m, ℓ= 26 m r = 15 m, ℓ= 26 m

Surface area

S = πr2 + πrℓ = π(10)2 + π(10)(26)

= 360π m2

S = πr2 + πrℓ = π(15)2 + π(15)(26)

= 615π m2

Multiplying the radius by 3 —

2 results in a surface area that is

615π — 360π

= 41

— 24

times

the original surface area.

b. Before change After change

Dimensions r = 10 m, ℓ= 26 m r = 15 m, ℓ= 39 m

Surface area

S = 360π m2 S = πr2 + πrℓ = π(15)2 + π(15)(39)

= 810π m2

Multiplying all the linear dimensions by 3 —

2 results in a surface area that

is 810π — 360π

= 9 —

4 = ( 3 —

2 ) 2 times the original surface area.


4. Find the lateral area and the surface area of the composite solid.

5 ft

6 ft

10 ft

5. Describe how (a) multiplying the base edge

lengths by 1 —

2 and (b) multiplying all the linear

dimensions by 1 —

2 affects the surface area of the

square pyramid.

26 m

10 m

6 m

5 m

ANALYZING MATHEMATICAL RELATIONSHIPS

Notice that while the surface area does not scale

by a factor of 3 — 2 , the lateral

surface area does scale

by a factor or π(15)(26) — π(10)(26)

= 3 — 2 .



Monitoring Progress and Modeling with MathematicsMonitoring Progress and Modeling with MathematicsIn Exercises 3–6, fi nd the lateral area and the surface area of the regular pyramid. (See Example 1.)

3. 8 in.

5 in.

4.

7.2 mm15.4 mm

5. A square pyramid has a height of 21 feet and a base

edge length of 40 feet.

6. A regular hexagonal pyramid has a slant height

of 15 centimeters and a base edge length of

8 centimeters.

In Exercises 7–10, fi nd the lateral area and the surface area of the right cone. (See Example 2.)

7.

8 in.

16 in.

8.

11 cm

7.2 cm

9. A right cone has a radius of 9 inches and a height of

12 inches.

10. A right cone has a diameter of 11.2 feet and a height

of 9.2 feet.


fi nding the surface area of the regular pyramid.

S = B + 1 — 2

Pℓ

= 62 + 1 — 2

(24)(4)

= 84 ft2

✗

6 ft

5 ft4 ft


fi nding the surface area of the right cone.

S = 𝛑r 2 + 𝛑r 2ℓ

= 𝛑(6)2 + 𝛑(6)2(10)

= 396𝛑 cm2

✗ 10 cm

6 cm

8 cm

13. MODELING WITH MATHEMATICS You are

making cardboard party hats like the one

shown. About how much cardboard

do you need for each hat? (See Example 3.)

14. MODELING WITH MATHEMATICS A candle is in

the shape of a regular square pyramid with a

base edge length of 16 centimeters and a height of

15 centimeters. Find the surface area of the candle.

1. WRITING Describe the differences between pyramids and cones. Describe their similarities.

2. DIFFERENT WORDS, SAME QUESTION Which is different? Find “both” answers.

Find the slant height of the regular pyramid.

Find AB.

Find the height of the regular pyramid.

Find the height of a lateral face of the regular pyramid.

Vocabulary and Core Concept Check

6 in.

5 in.A

B

3.5 in.

5.5 in.


In Exercises 15–18, fi nd the lateral area and the surface area of the composite solid. (See Example 4.)

15. 3 yd

4 yd

8 yd

16. 3 in.

7.5 in.

5 in5 in.

5 in.

17. 18. 10 cm

12 cm

7 mm

4 mm

5 mm

In Exercises 19–22, describe how the change affects the surface area of the regular pyramid or right cone. (See Example 5.)

19. doubling the radius 20. multiplying the base

edge lengths by 4 — 5 and

the slant height by 5

3 in.

7.6 in.

10 mm

4 mm

21. tripling all the linear

dimensions 22. multiplying all the

linear dimensions by 4 — 3

2 m

4 m

3.6 ft

2.4 ft

23. PROBLEM SOLVING Refer to the regular pyramid and

right cone.

3

4

4

3

4

a. Which solid has the base with the greater area?

b. Which solid has the greater slant height?

c. Which solid has the greater lateral area?

24. USING STRUCTURE The

sector shown can be rolled

to form the lateral surface

area of a right cone. The

lateral surface area of the

cone is 20 square meters.

a. Use the formula for the area of a sector to fi nd the

slant height of the cone. Explain your reasoning.

b. Find the radius and the height of the cone.

In Exercises 25 and 26, fi nd the missing dimensions of the regular pyramid or right cone.

25. S = 864 in.2 26. S = 628.3 cm2

h

x

15 in.

8 in.

h

27. WRITING Explain why a nonregular pyramid does not

have a slant height.

28. WRITING Explain why an oblique cone does not have

a slant height.

29. ANALYZING RELATIONSHIPS In the fi gure, AC = 4,

AB = 3, and DC = 2.

A B

E

C

D

a. Prove △ABC ∼ △DEC.

b. Find BC, DE, and EC.

c. Find the surface areas of the larger cone and the

smaller cone in terms of π. Compare the surface

areas using a percent.

30. REASONING To make a paper drinking cup, start with

a circular piece of paper that has a 3-inch radius, then

follow the given steps. How does the surface area of

the cup compare to the original paper circle? Find

m∠ABC.

fold fold

3 in.

opencup

A C

B

150°


Reviewing what you learned in previous grades and lessonsMaintaining Mathematical ProficiencyFind the volume of the prism. (Skills Review Handbook)

38.

3 ft

2 ft7 ft

39.

B = 29 mm2

10 mm

31. MAKING AN ARGUMENT Your friend claims that the

lateral area of a regular pyramid is always greater than

the area of the base. Is your friend correct? Explain

your reasoning.

32. HOW DO YOU SEE IT? Name the fi gure that is

represented by each net. Justify your answer.

a.

b.

33. REASONING In the fi gure, a right cone is placed

in the smallest right cylinder that can fi t the cone.

Which solid has a greater surface area? Explain

your reasoning.

34. CRITICAL THINKING A regular hexagonal pyramid

with a base edge of 9 feet and a height of 12 feet is

inscribed in a right cone. Find the lateral area of

the cone.

35. USING STRUCTURE A right cone with a radius of

4 inches and a square pyramid both have a slant

height of 5 inches. Both solids have the same surface

area. Find the length of a base edge of the pyramid.

36. THOUGHT PROVOKING The surface area of a regular

pyramid is given by S = B + 1 —

2 Pℓ. As the number

of lateral faces approaches infi nity, what does the

pyramid approach? What does B approach? What

does 1 —

2 Pℓ approach? What can you conclude from

your three answers? Explain your reasoning.

37. DRAWING CONCLUSIONS The net of the lateral

surface of a cone is a circular sector with radius y,

as shown.

yx°

a. Let y = 2. Copy and complete the table.

Angle measure of lateral surface, x

30 90 120 180 210

Slant height of cone, ℓCircumference of base of cone, C

Height of cone, h

b. What conjectures can you make about the

dimensions of the cone as x increases?

661661

12.1–12.3 What Did You Learn?

Core VocabularyCore Vocabularypolyhedron, p. 640face, p. 640edge, p. 640vertex, p. 640cross section, p. 641solid of revolution, p. 642axis of revolution, p. 642lateral faces, p. 646lateral edges, p. 646

surface area, p. 646lateral area, p. 646net, p. 646right prism, p. 646oblique prism, p. 646right cylinder, p. 647oblique cylinder, p. 647vertex of a pyramid, p. 654regular pyramid, p. 654

slant height of a regular pyramid, p. 654

vertex of a cone, p. 655right cone, p. 655oblique cone, p. 655slant height of a right cone, p. 655lateral surface of a cone, p. 655

Core ConceptsCore ConceptsSection 12.1Types of Solids, p. 640 Cross Section of a Solid, p. 641 Solids of Revolution, p. 642

Section 12.2Lateral Area and Surface Area of a Right Prism, p. 646Lateral Area and Surface Area of a Right Cylinder, p. 647

Section 12.3Lateral Area and Surface Area of a Regular Pyramid, p. 654Lateral Area and Surface Area of a Right Cone, p. 655

Mathematical ThinkingMathematical Thinking1. In Exercises 21–26 on page 644, describe the steps you took to sketch each polyhedron.

2. Sketch and label a diagram to represent the situation described in Exercise 32 on page 651.

3. In Exercise 13 on page 658, you need to make a new party hat using 4 times as much

cardboard as you previously used for one hat. How should you change the given

dimensions to create the new party hat? Explain your reasoning.

Form a study group several weeks before the fi nal exam. The intent of this group is to review what you have already learned while continuing to learn new material.

Study Skills

Form a Final Exam Study Group


12.1–12.3 Quiz

Tell whether the solid is a polyhedron. If it is, name the polyhedron. (Section 12.1)

1. 2. 3.

4. Sketch the composite solid produced by rotating the fi gure around the given

3

7

10

6axis. Then identify and describe the composite solid. (Section 12.1)

Find the lateral area and the surface area of the right prism or right cylinder. (Section 12.2)

5.

7 in.

9 in.5 in.

6. 7 ft

7 ft

7.

10 m

12 m

9 m

8. Find the lateral area and the surface area of the

composite solid. (Section 12.2)

Find the lateral area and the surface area of the regular pyramid or right cone. (Section 12.3)

9. 10 cm

8 cm

10.

16 ft

12 ft 11.

8 m

10 m 3 m4

12. You are replacing the siding and the roofi ng on the house shown. You

have 900 square feet of siding, 500 square feet of roofi ng material, and

2000 square feet of tarp, in case it rains. (Section 12.3)

a. Do you have enough siding to replace the siding on all four sides

of the house? Explain.

b. Do you have enough roofi ng material to replace the entire roof? Explain.

c. Do you have enough tarp to cover the entire house? Explain.18 ft

12 ft

18 ft

12 ft

32 cm

10 cm

10 cm

12 cm

Section 12.4 Volumes of Prisms and Cylinders 663

Volumes of Prisms and Cylinders

Essential QuestionEssential Question How can you fi nd the volume of a prism or

cylinder that is not a right prism or right cylinder?

Recall that the volume V of a

right prism or a right cylinder

is equal to the product of the

area of a base B and the

height h.

V = Bh

Finding Volume

Work with a partner. Consider a

stack of square papers that is in the

form of a right prism.

a. What is the volume of the prism?

b. When you twist the stack of papers,

as shown at the right, do you change

the volume? Explain your reasoning.

c. Write a carefully worded conjecture

that describes the conclusion you

reached in part (b).

d. Use your conjecture to fi nd the

volume of the twisted stack

of papers.

Finding Volume

Work with a partner. Use the conjecture you wrote in Exploration 1 to fi nd the

volume of the cylinder.

a. 2 in.

3 in.

b. 5 cm

15 cm

Communicate Your AnswerCommunicate Your Answer 3. How can you fi nd the volume of a prism or cylinder that is not a right prism

or right cylinder?

4. In Exploration 1, would the conjecture you wrote change if the papers in each

stack were not squares? Explain your reasoning.

USING PRECISE MATHEMATICAL LANGUAGE

To be profi cient in math, you need to communicate precisely to others.

12.4

right prisms right cylinder

8 in.

2 in. 2 in.

G.10.BG.11.D



12.4 Lesson What You Will LearnWhat You Will Learn Find volumes of prisms and cylinders.

Use volumes of prisms and cylinders.

Finding Volumes of Prisms and CylindersThe volume of a solid is the number of cubic units contained in its interior. Volume

is measured in cubic units, such as cubic centimeters (cm3). Cavalieri’s Principle,

named after Bonaventura Cavalieri (1598–1647), states that if two solids have the

same height and the same cross-sectional area at every level, then they have the same

volume. The prisms below have equal heights h and equal cross-sectional areas B at

every level. By Cavalieri’s Principle, the prisms have the same volume.

B B h

Finding Volumes of Prisms

Find the volume of each prism.

a. 4 cm3 cm

2 cm

b. 14 cm3 cm

5 cm

6 cm

SOLUTION

a. The area of a base is B = 1 —

2 (3)(4) = 6 cm2 and the height is h = 2 cm.

V = Bh Formula for volume of a prism

= 6(2) Substitute.

= 12 Simplify.

The volume is 12 cubic centimeters.

b. The area of a base is B = 1 —

2 (3)(6 + 14) = 30 cm2 and the height is h = 5 cm.


= 30(5) Substitute.

= 150 Simplify.

The volume is 150 cubic centimeters.

volume, p. 664Cavalieri’s Principle, p. 664

Previousprismcylindercomposite solid


Core Core ConceptConceptVolume of a PrismThe volume V of a prism is

V = Bh

where B is the area of a base and

h is the height.B

h h

B


Finding Volumes of Cylinders

Find the volume of each cylinder.

a.

6 ft

9 ft b.

7 cm

4 cm

SOLUTION

a. The dimensions of the cylinder are r = 9 ft and h = 6 ft.

V = πr2h = π(9)2(6) = 486π ≈ 1526.81

The volume is 486π, or about 1526.81 cubic feet.

b. The dimensions of the cylinder are r = 4 cm and h = 7 cm.

V = πr2h = π(4)2(7) = 112π ≈ 351.86

The volume is 112π, or about 351.86 cubic centimeters.


Find the volume of the solid.

1.

5 m9 m

8 m

2.

14 ft

8 ft

Consider a cylinder with height h and base radius r and a rectangular prism with the

same height that has a square base with sides of length r √— π .

B hB

r

r πr π

The cylinder and the prism have the same cross-sectional area, πr2, at every level and

the same height. By Cavalieri’s Principle, the prism and the cylinder have the same

volume. The volume of the prism is V = Bh = πr2h, so the volume of the cylinder is

also V = Bh = πr2h.

Core Core ConceptConceptVolume of a CylinderThe volume V of a cylinder is

V = Bh = πr2h

where B is the area of a base, h is the

height, and r is the radius of a base.

h h

B B

r r


Using Volumes of Prisms and Cylinders

Modeling with Mathematics

You are building a rectangular chest.

You want the length to be 6 feet, the

width to be 4 feet, and the volume

to be 72 cubic feet. What should the

height be?

SOLUTION

1. Understand the Problem You know the dimensions

of the base of a rectangular prism and the volume. You

are asked to fi nd the height.

2. Make a Plan Write the formula for the volume of a rectangular prism,

substitute known values, and solve for the height h.

3. Solve the Problem The area of a base is B = 6(4) = 24 ft2 and the volume

is V = 72 ft3.


72 = 24h Substitute.

3 = h Divide each side by 24.

The height of the chest should be 3 feet.

4. Look Back Check your answer.

V = Bh = 24(3) = 72 ✓


3. WHAT IF? In Example 3, you want the length to be 5 meters, the width to be

3 meters, and the volume to be 60 cubic meters. What should the height be?


Describe how doubling all the linear dimensions affects the

4 ft

6 ft

3 ft

volume of the rectangular prism.

SOLUTION


Dimensions ℓ= 4 ft, w = 3 ft, h = 6 ft ℓ= 8 ft, w = 6 ft, h = 12 ft

VolumeV = Bh

= (4)(3)(6)

= 72 ft3

V = Bh

= (8)(6)(12)

= 576 ft3

Doubling all the linear dimensions results in a volume that is 576—72

= 8 = 23 times

the original volume.

V = 72 ft3

6 ft4 ft

h


Notice that when all the linear dimensions are multiplied by k, the volume is multiplied by k3.

Voriginal = Bh =ℓwh

Vnew = (kℓ)(kw)(kh)

= (k3)ℓwh

= (k3)Voriginal


Changing a Dimension in a Solid

Describe how tripling the radius affects the 3 cm

6 cm


SOLUTION


Dimensions r = 3 cm, h = 6 cm r = 9 cm, h = 6 cm

VolumeV = πr2h

= π(3)2(6)

= 54π cm3

V = πr2h

= π(9)2(6)

= 486π cm3

Tripling the radius results in a volume that is 486π — 54π

= 9 = 32 times the

original volume.


4. In Example 4, describe how multiplying all the linear dimensions by 1 —

2 affects the


5. In Example 4, describe how doubling the length and width of the bases affects the


6. In Example 5, describe how multiplying the height by 2 —

3 affects the volume of the

cylinder.

7. In Example 5, describe how multiplying all the linear dimensions by 4 affects the


Finding the Volume of a Composite Solid

Find the volume of the concrete block.

SOLUTION

To fi nd the area of the base, subtract two times the

area of the small rectangle from the large rectangle.

B = Area of large rectangle − 2 ⋅ Area of small rectangle

= 1.31(0.66) − 2(0.33)(0.39)

= 0.6072

Using the formula for the volume of a prism, the volume is

V = Bh = 0.6072(0.66) ≈ 0.40.

The volume is about 0.40 cubic foot.


8. Find the volume of the composite solid.

0.33 ft 0.33 ft0.39 ft

1.31 ft0.66 ft

0.66 ft

6 ft10 ft

3 ft



In Exercises 3–6, fi nd the volume of the prism. (See Example 1.)

3.

1.8 cm

1.2 cm

2 cm2.3 cm

4.

1.5 m

2 m4 m

5. 6. 7 in. 10 in.

5 in.

6 m11 m

14 m

In Exercises 7–10, fi nd the volume of the cylinder. (See Example 2.)

7.

10.2 ft

3 ft 8.

9.8 cm

26.8 cm

9.

8 ft

5 ft 10.

18 m

60°

12 m

In Exercises 11 and 12, make a sketch of the solid and fi nd its volume.

11. A prism has a height of 11.2 centimeters and an

equilateral triangle for a base, where each base edge

is 8 centimeters.

12. A pentagonal prism has a height of 9 feet and each

base edge is 3 feet.

In Exercises 13–18, fi nd the missing dimension of the prism or cylinder. (See Example 3.)

13. Volume = 560 ft3 14. Volume = 2700 yd3

u

8 ft7 ft

v

15 yd12 yd

15. Volume = 80 cm3 16. Volume = 72.66 in.3

w

8 cm

5 cm

x

2 in.

17. Volume = 3000 ft3 18. Volume = 1696.5 m3

y

9.3 ft

15 m

z


fi nding the volume of the cylinder.

4 ft

3 ft V = 2πrh

= 2π(4)(3)

= 24π

So, the volume of the cylinder is

24π cubic feet.

✗


1. VOCABULARY In what type of units is the volume of a solid measured?

2. COMPLETE THE SENTENCE Cavalieri’s Principle states that if two solids have the same ______ and

the same _________ at every level, then they have the same _________.



20. OPEN-ENDED Sketch two rectangular prisms that

have volumes of 100 square inches but different

surface areas. Include dimensions in your sketches.

In Exercises 21–26, describe how the change affects the volume of the prism or cylinder. (See Examples 4 and 5.)

21. tripling all the


linear dimensions

by 3 — 4

3 in.

4 in.

8 in.

12 m

16 m

23. multiplying the

radius by 1 — 2

24. tripling the base and

the height of the

triangular bases

8 cm

7 cm

5 ft

12 ft

12 ft

25. multiplying the

height by 1 — 3

26. multiplying the height

by 4

3 m

5 m

5 m

1 in.

6 in.

In Exercises 27–30, fi nd the volume of the composite solid. (See Example 6.)

27.

2 ft2 ft

5 ft

10 ft6 ft

3 ft 28.

4 in.4 in.

4 in.

29. 3 in.

11 in.

8 in. 30.

4 ft

5 ft

1 ft

2 ft

31. MODELING WITH MATHEMATICS The Great Blue

Hole is a cylindrical trench located off the coast

of Belize. It is approximately 1000 feet wide and

400 feet deep. About how many gallons of water does

the Great Blue Hole contain? (1 ft3 ≈ 7.48 gallons)

32. COMPARING METHODS The Volume Addition Postulate states that the volume of a solid is the sum

of the volumes of all its nonoverlaping parts. Use this

postulate to fi nd the volume of the block of concrete

in Example 6 by subtracting the volume of each

hole from the volume of the large rectangular prism.

Which method do you prefer? Explain your reasoning.

33. WRITING Both of the fi gures shown are made up of

the same number of congruent rectangles. Explain

how Cavalieri’s Principle can be adapted to compare

the areas of these fi gures.

34. HOW DO YOU SEE IT? Each stack of memo papers

contains 500 equally-sized sheets of paper. Compare

their volumes. Explain your reasoning.

35. PROBLEM SOLVING An aquarium shaped like a

rectangular prism has a length of 30 inches, a width

of 10 inches, and a height of 20 inches. You fi ll the

aquarium 3 —

4 full with water. When you submerge a

rock in the aquarium, the water level rises 0.25 inch.

a. Find the volume of the rock.

b. How many rocks of this size can you place in the

aquarium before water spills out?


36. MODELING WITH MATHEMATICS Which box gives

you more cereal for your money? Explain.

10 in.16 in.

4 in. 10 in. 8 in.

2 in.

37. CRITICAL THINKING A 3-inch by 5-inch index card

is rotated around a horizontal line and a vertical line

to produce two different solids. Which solid has a

greater volume? Explain your reasoning.

3 in.3 in.

5 in.

5 in.

38. CRITICAL THINKING The height of cylinder X is twice

the height of cylinder Y. The radius of cylinder X is

half the radius of cylinder Y. Compare the volumes of

cylinder X and cylinder Y. Justify your answer.

39. USING STRUCTURE Find the volume of the solid

shown. The bases of the solid are sectors of circles.

3.5 in.

in.60°23

π

40. ANALYZING RELATIONSHIPS How can you change

the height of a cylinder so that the volume is increased

by 25% but the radius remains the same?

41. MATHEMATICAL CONNECTIONS You drill a circular

hole of radius r through the base of a cylinder of

radius R. Assume the hole is drilled completely

through to the other base. You want the volume of the

hole to be half the volume of the cylinder. Express r

as a function of R.

42. THOUGHT PROVOKING Cavalieri’s Principle states

that the two solids shown below have the same

volume. Do they also have the same surface area?


B B h

43. PROBLEM SOLVING A barn is in the shape of a

pentagonal prism with the dimensions shown. The

volume of the barn is 9072 cubic feet. Find the

dimensions of each half of the roof.

18 ft 36 ft

Not drawn to scale

8 ft 8 ft

x ft

44. PROBLEM SOLVING A wooden box is in the shape of

a regular pentagonal prism. The sides, top, and bottom

of the box are 1 centimeter thick. Approximate the

volume of wood used to construct the box. Round

your answer to the nearest tenth.

4 cm

6 cm

Maintaining Mathematical ProficiencyMaintaining Mathematical ProficiencyFind the surface area of the regular pyramid. (Section 12.3)

45.

2 m

3 m

46.

8 cm

10 cm 47.

20 in.

18 in. 15.6 in.


Section 12.5 Volumes of Pyramids and Cones 671

Volumes of Pyramids and Cones

Essential QuestionEssential Question How can you fi nd the volume of a pyramid

or a cone?

Finding the Volume of a Pyramid

Work with a partner. The pyramid and the prism

have the same height and

the same square base.

When the pyramid is fi lled with sand and poured into the prism, it takes three

pyramids to fi ll the prism.

Use this information to write a formula for the volume V of a pyramid.

Finding the Volume of a Cone

Work with a partner. The cone and

the cylinder have the same height

and the same circular base.

When the cone is fi lled with sand and poured into the cylinder, it takes three

cones to fi ll the cylinder.

Use this information to write a formula for the volume V of a cone.

Communicate Your AnswerCommunicate Your Answer 3. How can you fi nd the volume of a pyramid or a cone?


To be profi cient in math, you need to look closely to discern a pattern or structure.

12.5

G.10.BG.11.D


h

h


12.5 Lesson What You Will LearnWhat You Will Learn Find volumes of pyramids.

Find volumes of cones.

Use volumes of pyramids and cones.

Finding Volumes of PyramidsConsider a triangular prism with parallel, congruent bases △JKL and △MNP. You can

divide this triangular prism into three triangular pyramids.

Triangularprism

Triangularpyramid 1

Triangularpyramid 2

Triangularpyramid 3

K J

ML

N

P

K

ML

N

K J

ML

M

L

N

P

You can combine triangular pyramids 1 and 2 to form a pyramid with a base that is a

parallelogram, as shown at the left. Name this pyramid Q. Similarly, you can combine

triangular pyramids 1 and 3 to form pyramid R with a base that is a parallelogram.

In pyramid Q, diagonal — KM divides ▱JKNM into two congruent triangles, so the

bases of triangular pyramids 1 and 2 are congruent. Similarly, you can divide any cross

section parallel to ▱JKNM into two congruent triangles that are the cross sections of

triangular pyramids 1 and 2.

By Cavalieri’s Principle, triangular pyramids 1 and 2 have the same volume. Similarly,

using pyramid R, you can show that triangular pyramids 1 and 3 have the same

volume. By the Transitive Property of Equality, triangular pyramids 2 and 3 have

the same volume.

The volume of each pyramid must be one-third the volume of the prism, or V = 1 —

3 Bh.

You can generalize this formula to say that the volume of any pyramid with any base is

equal to 1 —

3 the volume of a prism with the same base and height because you can divide

any polygon into triangles and any pyramid into triangular pyramids.

Finding the Volume of a Pyramid

Find the volume of the pyramid.

SOLUTION

V = 1 —

3 Bh Formula for volume of a pyramid

= 1 —

3 ( 1 —

2 ⋅ 4 ⋅ 6 ) (9) Substitute.

= 36 Simplify.

The volume is 36 cubic meters.

Previouspyramidconecomposite solid


Core Core ConceptConceptVolume of a PyramidThe volume V of a pyramid is

V = 1 —

3 Bh

where B is the area of a base

and h is the height.

h

B

h

B

6 m4 m

9 m

K

M

N

Pyramid Q

Pyramid R

J

M

L

N

LP

K


Finding Volumes of ConesConsider a cone with a regular polygon inscribed in the base. The pyramid with the

same vertex as the cone has volume V = 1 —

3 Bh. As you increase the number of sides of

the polygon, it approaches the base of the cone and the pyramid approaches the cone.

The volume approaches 1 —

3 πr2h as the base area B approaches πr2.

Finding the Volume of a Cone

Find the volume of the cone.

SOLUTION

V = 1 — 3 πr2h Formula for volume of a cone

= 1 — 3 π ⋅ (2.2)2 ⋅ 4.5 Substitute.

= 7.26π Simplify.


The volume is 7.26π, or about 22.81 cubic centimeters.



1.

12 cm

20 cm

2.

8 m

5 m

Core Core ConceptConceptVolume of a ConeThe volume V of a cone is

V = 1 —

3 Bh = 1 —

3 πr2h

where B is the area of a base,

h is the height, and r is the

radius of the base.

hh

r rB B

2.2 cm

4.5 cm


Using Volumes of Pyramids and Cones

Using the Volume of a Pyramid

Originally, Khafre’s Pyramid had a height of about 144 meters and a volume of about

2,218,800 cubic meters. Find the side length of the square base.

SOLUTION

V = 1 —


2,218,800 ≈ 1 —

3 x2(144) Substitute.

6,656,400 ≈ 144x2 Multiply each side by 3.

46,225 ≈ x2 Divide each side by 144.

215 ≈ x Find the positive square root.

Originally, the side length of the square base was about 215 meters.


3. The volume of a square pyramid is 75 cubic meters and the height is 9 meters.

Find the side length of the square base.

4. Find the height of the 5. Find the radius of the cone.

triangular pyramid.

6 m

3 m

h

V = 24 m3

V = 351 in.3π

13 in.

r


Describe how multiplying all the linear dimensions by 1 —


rectangular pyramid.

SOLUTION


Dimensions ℓ= 9 m, w = 6 m, h = 18 m ℓ= 3 m, w = 2 m, h = 6 m

Volume

V = 1 —

3 Bh

= 1 —

3 (9)(6)(18)

= 324 m3

V = 1 —

3 Bh

= 1 —

3 (3)(2)(6)

= 12 m3

Multiplying all the linear dimensions by 1 —

3 results in a volume that

is 12

— 324

= 1 —

27 = ( 1 —

3 ) 3 times the original volume.

O

2

Khafre’s Pyramid, Egypt

18 m

6 m9 m


Changing a Dimension in a Solid

Describe how doubling the height affects the volume

of the cone.

SOLUTION


Dimensions r = 3 ft, h = 5 ft r = 3 ft, h = 10 ft

Volume

V = 1 —

3 πr2h

= 1 —

3 π(3)2(5)

= 15π ft3

V = 1 —

3 πr2h

= 1 —

3 π(3)2(10)

= 30π ft3

Doubling the height results in a volume that is 30π — 15π

= 2 times the

original volume.


6. In Example 4, describe how multiplying all the linear dimensions by 4 affects the

volume of the rectangular pyramid.

7. In Example 4, describe how multiplying the height by 1 —


rectangular prism.

8. In Example 5, describe how doubling the radius affects the volume of the cone.

9. In Example 5, describe how tripling all the linear dimensions affects the volume

of the cone.


Find the volume of the composite solid.

SOLUTION

Volume of

solid =

Volume of

cube +

Volume of

pyramid

= s3 + 1 —

3 Bh Write formulas.

= 63 + 1 —

3 (6)2 ⋅ 6 Substitute.

= 216 + 72 Simplify.

= 288 Add.




6 m

6 m6 m

6 m


Notice that when the height is multiplied by k, the volume is also multiplied by k.

Voriginal = 1 — 3 πr2h

Vnew = 1 — 3 πr2(kh)

= (k) 1 — 3 πr2h

= (k)Voriginal

3 ft

5 ft

3 cm

10 cm

5 cm



In Exercises 3 and 4, fi nd the volume of the pyramid. (See Example 1.)

3.

16 m

12 m

7 m 4.

3 in.

4 in.

3 in.

In Exercises 5 and 6, fi nd the volume of the cone. (See Example 2.)

5.

10 mm

13 mm 6.

2 m1 m

In Exercises 7–12, fi nd the missing dimension of the pyramid or cone. (See Example 3.)

7. Volume = 912 ft3 8. Volume = 105 cm3

19 ft

s

7 cmw

15 cm

9. Volume = 24π m3 10. Volume = 216π in.3

3 m

h

r

18 in.

11. Volume = 224 in.3 12. Volume = 198 yd3

12 in.

8 in. 11 yd9 yd


fi nding the volume of the pyramid.

5 ft

6 ft

V = 1 — 3

(6)(5)

= 1 — 3

(30)

= 10 ft3

✗


fi nding the volume of the cone.

V = 1 — 3

𝛑(6)2(10)

= 120𝛑 m3

✗6 m

10 m

15. OPEN-ENDED Give an example of a pyramid and a

prism that have the same base and the same volume.


16. OPEN-ENDED Give an example of a cone and a

cylinder that have the same base and the same

volume. Explain your reasoning.


1. REASONING A square pyramid and a cube have the same base and height. Compare the volume of

the square pyramid to the volume of the cube.

2. COMPLETE THE SENTENCE The volume of a cone with radius r and height h is 1 —

3 the volume

of a(n) __________ with radius r and height h.



In Exercises 17–22, describe how the change affects the volume of the pyramid or cone. (See Examples 4 and 5.)

17. doubling all the


linear dimensions by 1 — 4

4 in.

7 in.

12 cm

16 cm

20 cm

19. multiplying the base

edge lengths by 1 — 3

20. tripling the radius

9 ft

11 ft

5 m

9 m

21. multiplying the

height by 4

22. multiplying the height

by 3 — 2

10 in.

6 in.

18 cm12 cm

14 cm


23.

9 cm12 cm

7 cm

10 cm

24.

8 cm

5 cm

5 cm

25. 3 cm

3 cm

10 cm

26.

9 ft

8 ft

12 ft

27.

12 in.12 in.

12 in.

28.

5.1 m5.1 m

5.1 m

29. REASONING A snack stand serves a small order of

popcorn in a cone-shaped container and a large order

of popcorn in a cylindrical container. Do not perform

any calculations.

3 in. 3 in.

8 in. 8 in.

$1.25 $2.50

a. How many small containers of popcorn do you

have to buy to equal the amount of popcorn in a

large container? Explain.

b. Which container gives you more popcorn for your

money? Explain.

30. HOW DO YOU SEE IT? The cube shown is

formed by three

pyramids, each with

the same square base

and the same height.

How could you use this

to verify the formula

for the volume of

a pyramid?

In Exercises 31 and 32, fi nd the volume of the right cone.

31.

22 ft60°

32.

14 yd32°

33. MODELING WITH MATHEMATICS A cat eats half a cup

of food, twice per day.

Will the automatic

pet feeder hold enough

food for 10 days?


(1 cup ≈ 14.4 in.3)

2.5 in.

7.5 in.

4 in.


34. MODELING WITH MATHEMATICS During a chemistry

lab, you use a funnel to pour a solvent into a fl ask.

The radius of the funnel is 5 centimeters and its height

is 10 centimeters. You pour the solvent into the funnel

at a rate of 80 milliliters per second and the solvent

fl ows out of the funnel at a rate of 65 milliliters

per second. How long will it be before the funnel

overfl ows? (1 mL = 1 cm3)

35. ANALYZING RELATIONSHIPS A cone has height h

and a base with radius r. You want to change the cone

so its volume is doubled. What is the new height if

you change only the height? What is the new radius

if you change only the radius? Explain.

36. REASONING The fi gure shown is a cone that has been

warped but whose cross sections still have the same

area as a right cone with equal radius and height. Find

the volume of this solid. Explain your reasoning.

3 cm

2 cm

37. CRITICAL THINKING Find the volume of the regular

pentagonal pyramid. Round your answer to the

nearest hundredth. In the diagram, m∠ABC = 35°.

A

C

B3 ft

38. THOUGHT PROVOKING A frustum of a cone is the

part of the cone that lies between the base and a plane

parallel to the base, as shown. Write a formula for

the volume of the frustum of a cone in terms of a, b,

and h. (Hint: Consider the “missing” top of the cone

and use similar triangles.)

h

b

a

39. MAKING AN ARGUMENT In the fi gure, the two

cylinders are congruent. The combined height of

the two smaller cones equals the height of the larger

cone. Your friend claims that this means the total

volume of the two smaller cones is equal to the

volume of the larger cone. Is your friend correct?

Justify your answer.

40. MODELING WITH MATHEMATICS Nautical deck

prisms were used as a safe way to illuminate decks

on ships. The deck prism shown here is composed of

the following three solids: a regular hexagonal prism

with an edge length of 3.5 inches and a height of

1.5 inches, a regular hexagonal prism with an edge

length of 3.25 inches and a height of

0.25 inch, and a regular

hexagonal pyramid with

an edge length of 3 inches

and a height of 3 inches.

Find the volume of the

deck prism.

41. CRITICAL THINKING When the given triangle is

rotated around each of its sides, solids of revolution

are formed. Describe the three solids and fi nd their

volumes. Give your answers in terms of π.

25

2015

42. CRITICAL THINKING A square pyramid is inscribed in

a right cylinder so that the base of the pyramid is on

a base of the cylinder, and the vertex of the pyramid

is on the other base of the cylinder. The cylinder has

a radius of 6 feet and a height of 12 feet. Find the

volume of the pyramid.

Maintaining Mathematical ProficiencyMaintaining Mathematical ProficiencyFind the indicated measure. (Section 11.2)

43. area of a circle with a radius of 7 feet 44. area of a circle with a diameter of 22 centimeters

45. diameter of a circle with an area of 256π 46. radius of a circle with an area of 529π


Section 12.6 Surface Areas and Volumes of Spheres 679

Essential QuestionEssential Question How can you fi nd the surface area and the

volume of a sphere?

Finding the Surface Area of a Sphere

Work with a partner. Remove the covering from a baseball or softball.

r

You will end up with two “fi gure 8” pieces of material, as shown above. From the

amount of material it takes to cover the ball, what would you estimate the surface area

S of the ball to be? Express your answer in terms of the radius r of the ball.

S = Surface area of a sphere

Use the Internet or some other resource to confi rm that the formula you wrote for the

surface area of a sphere is correct.

Finding the Volume of a Sphere

Work with a partner. A cylinder is circumscribed about a

sphere, as shown. Write a formula for the volume V of the

cylinder in terms of the radius r.

V = Volume of cylinder

When half of the sphere (a hemisphere) is fi lled with sand and

poured into the cylinder, it takes three hemispheres to fi ll the

cylinder. Use this information to write a formula for the volume V of a sphere in terms of the radius r.

V = Volume of a sphere

Communicate Your AnswerCommunicate Your Answer 3. How can you fi nd the surface area and the volume of a sphere?

4. Use the results of Explorations 1 and 2 to fi nd the surface area and the volume

of a sphere with a radius of (a) 3 inches and (b) 2 centimeters.

SELECTING TOOLSTo be profi cient in math, you need to identify relevant external mathematical resources, such as content located on a website.

Surface Areas and Volumes of Spheres

12.6

r

r 2r

G.10.BG.11.CG.11.D



12.6 Lesson What You Will LearnWhat You Will Learn Find surface areas of spheres.

Find volumes of spheres.

Finding Surface Areas of SpheresA sphere is the set of all points in space equidistant from a given point. This point is

called the center of the sphere. A radius of a sphere is a segment from the center to

a point on the sphere. A chord of a sphere is a segment whose endpoints are on the

sphere. A diameter of a sphere is a chord that contains the center.

C

centerradius

C

chord

diameter

As with circles, the terms radius and diameter also represent distances, and the

diameter is twice the radius.

If a plane intersects a sphere, then the

intersection is either a single point or a circle.

If the plane contains the center of the sphere,

then the intersection is a great circle of the

sphere. The circumference of a great circle

is the circumference of the sphere. Every

great circle of a sphere separates the sphere

into two congruent halves called hemispheres.

To understand the formula for the surface

area of a sphere, think of a baseball. The

surface area of a baseball is sewn from

two congruent shapes, each of which

resembles two joined circles.

So, the entire covering of the baseball

consists of four circles, each with

radius r. The area A of a circle with

radius r is A = πr2. So, the area of the

covering can be approximated by 4πr2.

This is the formula for the surface area

of a sphere.

chord of a sphere, p. 680great circle, p. 680

Previousspherecenter of a sphereradius of a spherediameter of a spherehemisphere


Core Core ConceptConceptSurface Area of a SphereThe surface area S of a sphere is

S = 4πr2

where r is the radius of the sphere.

greatcircle

hemispheres

r

S = 4 r2π

r

leather covering


Finding the Surface Areas of Spheres

Find the surface area of each sphere.

a. 8 in.

b. C = 12 ftπ

SOLUTION

a. S = 4πr2 Formula for surface area of a sphere

= 4π(8)2 Substitute 8 for r.

= 256π Simplify.


The surface area is 256π, or about 804.25 square inches.

b. The circumference of the sphere is 12π, so the radius of the sphere is 12π — 2π

= 6 feet.

S = 4πr2 Formula for surface area of a sphere


= 144π Simplify.


The surface area is 144π, or about 452.39 square feet.

Finding the Diameter of a Sphere

Find the diameter of the sphere.

SOLUTION

S = 4πr2 Formula for surface area of a sphere

20.25π = 4πr2 Substitute 20.25π for S.

5.0625 = r2 Divide each side by 4π.

2.25 = r Find the positive square root.

The diameter is 2r = 2 • 2.25 = 4.5 centimeters.


Find the surface area of the sphere.

1. 40 ft 2.

C = 6 ftπ

3. Find the radius of the sphere.

COMMON ERRORBe sure to multiply the value of r by 2 to fi nd the diameter.

S = 20.25 cm2π

S = 30 m2π


Finding the Volume of a Sphere

Find the volume of the soccer ball.

SOLUTION

V = 4 —

3 πr3 Formula for volume of a sphere

= 4 —

3 π(4.5)3 Substitute 4.5 for r.

= 121.5π Simplify.


The volume of the soccer ball is 121.5π, or about 381.70 cubic inches.

Finding Volumes of SpheresThe fi gure shows a hemisphere and a cylinder with a cone removed. A plane parallel to

their bases intersects the solids z units above their bases.

r2 − z2

r

rr

z

Using the AA Similarity Theorem (Theorem 8.3), you can show that the radius of

the cross section of the cone at height z is z. The area of the cross section formed

by the plane is π(r2 − z2) for both solids. Because the solids have the same height

and the same cross-sectional area at every level, they have the same volume by

Cavalieri’s Principle.

Vhemisphere = Vcylinder − Vcone

= πr2(r) − 1 —

3 πr2(r)

= 2 —

3 πr3

So, the volume of a sphere of radius r is

2 ⋅ Vhemisphere = 2 ⋅ 2 —

3 πr3 =

4 —

3 πr3.

Core Core ConceptConceptVolume of a SphereThe volume V of a sphere is

V = 4 —

3 πr3


r

V = r3π43

4.5 in.



Describe how multiplying the radius by 1 —

4 affects

12 in.the volume of the sphere.

SOLUTION


Dimensions r = 12 in. r = 3 in.

Volume

V = 4 —

3 πr3

= 4 —

3 π(12)3

= 2304π in.3

V = 4 —

3 πr3

= 4 —

3 π(3)3

= 36π in.3

Multiplying the radius by 1 —

4 results in a volume that is

36π — 2304π

= 1 —

64 = ( 1 —

4 ) 3

times the original volume.


Find the volume of the composite solid.

2 in.

2 in.

SOLUTION

Volume

of solid=

Volume of

cylinder −

Volume of

hemisphere

= πr2h − 1 —

2 ( 4 —

3 πr3 ) Write formulas.

= π(2)2(2) − 2 —

3 π(2)3 Substitute.

= 8π − 16

— 3 π Multiply.

= 24

— 3 π −

16 —

3 π Rewrite fractions using least

common denominator. =

8 —

3 π Subtract.


The volume is 8 —

3 π, or about 8.38 cubic inches.


4. The radius of a sphere is 5 yards. Find the volume of the sphere.

5. The diameter of a sphere is 36 inches. Find the volume of the sphere.

6. In Example 4, describe how doubling the radius affects the volume

of the sphere.

7. A sphere has a radius of 18 centimeters. Describe how multiplying the

radius by 1 —

3 affects the volume of the sphere.

8. Find the volume of the composite solid at the left.

1 m

5 m



1. VOCABULARY When a plane intersects a sphere, what must be true for the intersection to be a

great circle?

2. WRITING Explain the difference between a sphere and a hemisphere.


In Exercises 3–6, fi nd the surface area of the sphere. (See Example 1.)

3.

4 ft

4. 7.5 cm

5.

18.3 m

6.

C = 4 ftπ

In Exercises 7–10, fi nd the indicated measure. (See Example 2.)

7. Find the radius of a sphere with a surface area of

4π square feet.

8. Find the radius of a sphere with a surface area of

1024π square inches.

9. Find the diameter of a sphere with a surface area of

900π square meters.

10. Find the diameter of a sphere with a surface area of

196π square centimeters.

In Exercises 11 and 12, fi nd the surface area of the hemisphere.

11.

5 m

12.

12 in.

In Exercises 13–18, fi nd the volume of the sphere. (See Example 3.)

13.

8 m

14.

4 ft

15.

22 yd

16.

14 ft

17. 18. C = 20 cmπ C = 7 in.π

In Exercises 19 and 20, fi nd the volume of the sphere with the given surface area.

19. Surface area = 16π ft2

20. Surface area = 484π cm2


fi nding the volume of the sphere.

6 ft V = 4 —

3 π(6)2

= 48π≈ 150.80 ft3

✗




fi nding the volume of the sphere.

3 in.

V = 4 — 3

π(3)3

= 36π ≈ 113.10 in.3

✗

In Exercises 23 and 24, describe how the change affects the volume of the sphere. (See Example 4.)

23. tripling the radius 24. multiplying the radius

by 2 — 3

1 m

36 cm


25.

9 in.5 in.

26.

12 ft

6 ft

27. 10 cm

18 cm 28. 14 m6 m

In Exercises 29–32, fi nd the surface area and volume of the ball.

29. bowling ball 30. basketball

d = 8.5 in. C = 29.5 in.

31. softball 32. golf ball

C = 12 in. d = 1.7 in.

33. MAKING AN ARGUMENT Your friend claims that if

the radius of a sphere is doubled, then the surface

area of the sphere will also be doubled. Is your friend

correct? Explain your reasoning.

34. REASONING A semicircle with a diameter of

18 inches is rotated about its diameter. Find the

surface area and the volume of the solid formed.

35. MODELING WITH MATHEMATICS A silo has

the dimensions shown. The top of the silo is a

hemispherical shape. Find the volume of the silo.

20 ft20 ft

60 ft

36. MODELING WITH MATHEMATICS Three tennis balls

are stored in a cylindrical container

with a height of 8 inches and a radius

of 1.43 inches. The circumference

of a tennis ball is 8 inches.

a. Find the volume of a tennis ball.

b. Find the amount of space within

the cylinder not taken up by the

tennis balls.

37. ANALYZING RELATIONSHIPS Use the table shown

for a sphere.

Radius Surface area Volume

3 in. 36π in.2 36π in.3

6 in.

9 in.

12 in.

a. Copy and complete the table. Leave your answers

in terms of π.

b. What happens to the surface area of the sphere

when the radius is doubled? tripled? quadrupled?

c. What happens to the volume of the sphere when

the radius is doubled? tripled? quadrupled?

38. MATHEMATICAL CONNECTIONS A sphere has a

diameter of 4(x + 3) centimeters and a surface area

of 784π square centimeters. Find the value of x.


39. MODELING WITH MATHEMATICS The radius of Earth

is about 3960 miles. The radius of the moon is about

1080 miles.

a. Find the surface area of Earth and the moon.

b. Compare the surface areas of Earth and the moon.

c. About 70% of the surface of Earth is water. How

many square miles of water are on Earth’s surface?

40. MODELING WITH MATHEMATICS The Torrid Zone

on Earth is the area between the Tropic of Cancer and

the Tropic of Capricorn. The distance between these

two tropics is about 3250 miles. You can estimate the

distance as the height of a cylindrical belt around the

Earth at the equator.

Tropic of Cancer

TorridZone

Tropic ofCapricorn

3250 mi equator

a. Estimate the surface area of the Torrid Zone.

(The radius of Earth is about 3960 miles.)

b. A meteorite is equally likely to hit anywhere on

Earth. Estimate the probability that a meteorite

will land in the Torrid Zone.

41. ABSTRACT REASONING A sphere is inscribed in a

cube with a volume of 64 cubic inches. What is the

surface area of the sphere? Explain your reasoning.

42. HOW DO YOU SEE IT? The formula for the volume

of a hemisphere and a cone are shown. If each solid

has the same radius and r = h, which solid will have

a greater volume? Explain your reasoning.

r r

h

V = r2hπ13V = r3π2

3

43. CRITICAL THINKING Let V be the volume of a sphere,

S be the surface area of the sphere, and r be the radius

of the sphere. Write an equation for V in terms of r

and S. ( Hint: Start with the ratio V

— S . )

44. THOUGHT PROVOKING A spherical lune is the

region between two great circles of a sphere. Find

the formula for the area of a lune.

rθ

45. CRITICAL THINKING The volume of a right cylinder

is the same as the volume of a sphere. The radius of

the sphere is 1 inch. Give three possibilities for the

dimensions of the cylinder.

46. PROBLEM SOLVING A spherical cap is a portion of a

sphere cut off by a plane. The formula for the volume

of a spherical cap is V = πh

— 6 (3a2 + h2), where a is

the radius of the base of the cap and h is the height

of the cap. Use the diagram and given information to

fi nd the volume of each spherical cap.

r

ah

a. r = 5 ft, a = 4 ft b. r = 34 cm, a = 30 cm

c. r = 13 m, h = 8 m d. r = 75 in., h = 54 in.

47. CRITICAL THINKING A sphere with a radius of

2 inches is inscribed in a right cone with a height

of 6 inches. Find the surface area and the volume

of the cone.

Maintaining Mathematical ProficiencyMaintaining Mathematical ProficiencyUse the diagram. (Section 1.1)

48. Name four points. 49. Name two line segments.

50. Name two lines. 51. Name a plane.


R ST

P

U

Q

Section 12.7 Spherical Geometry 687

Spherical Geometry

Essential QuestionEssential Question How can you represent a line on a sphere?

The endpoints of a diameter of a sphere

are called antipodal points.

Finding the Shortest Distance Between Two Points

Work with a partner. Use a washable marker, a piece of paper, a ball, and a

piece of string.

a. Draw two points on the piece of paper. Label the points A and B. Use the string

to create the shortest path between the points. What geometric term describes the

path of the string? When you extend the path infi nitely in either direction, what

geometric term describes the path?

b. Draw two points that are not antipodal on the ball. Label

the points C and D. Use the string to create the shortest

path between the points. What geometric term describes

the path of the string? Can you extend the path infi nitely

in either direction? What geometric term describes the

path when you extend it all the way around the ball?

c. Compare and contrast the paths on the piece of paper in part (a) and the paths

on the ball in part (b).

Exploring Lines on a Sphere

Work with a partner. Use a ball and three rubber bands.

a. Construct a great circle on the ball using a rubber band.

b. Construct a second distinct great circle. At how many points do the two great

circles intersect? Compare this to the number of points at which two distinct lines

can intersect in the plane.

c. How many great circles can you construct through two points that are not

antipodal? Explain.

d. How many great circles can you construct through two points that are antipodal?

Explain.

e. Is the Two Point Postulate (Postulate 2.1) true for great circles on a sphere? Explain.

Two Point Postulate: Through any two points, there exists exactly one line.

f. A polygon on the surface of a sphere is a shape enclosed by arcs of great circles.

Describe how you can construct a polygon with two sides on the surface of a

sphere. Then construct a two-sided polygon on the ball using rubber bands.

g. Describe how you can construct a triangle on the surface of a sphere. Then

construct a triangle on the ball using rubber bands.

Communicate Your AnswerCommunicate Your Answer 3. How can you represent a line on a sphere?

ANALYZING MATHEMATICALRELATIONSHIPS

To be profi cient in math you need to look closely to discern a pattern or structure.

12.7

G.4.D


CD

great circle

A B

Points A and B are antipodal.


12.7 Lesson What You Will LearnWhat You Will Learn Compare Euclidean and spherical geometry.

Find distances on a sphere.

Find areas of spherical triangles.

Comparing Euclidean and Spherical GeometryIn Euclidean geometry, a plane is a fl at surface that extends without end in all

directions, and a line in the plane is a set of points that extends without end in

two directions. Geometry on a sphere is different.

In spherical geometry, a plane is the surface

of a sphere, a line is a great circle, and the

angle between two lines is the angle between

the planes containing the two corresponding

great circles.

antipodal points, p. 688

Previousgreat circle


Core Core ConceptConceptEuclidean Geometry and Spherical Geometry

Some properties and postulates in Euclidean geometry are true in spherical geometry.

Others are not, or are true only under certain circumstances. For example, in Euclidean

geometry, the Two Point Postulate (Postulate 2.1) states that through any two points,

there exists exactly one line. In spherical geometry, this postulate is true only for

points that are not the endpoints of a diameter of the sphere. The endpoints of a

diameter of a sphere are called antipodal points.

Euclidean Geometry

AP

Plane P contains lineℓand point A

not on the lineℓ.

BA

CP

The vertices of △ABC are points in

plane P and the sides are segments.

The sum of the interior angles of a

triangle is 180°.

m∠A + m∠B + m∠C = 180°

Spherical Geometry

m

center

A

S

Sphere S contains great circle m and

point A not on m. Great circle m is

a line.

S

B

A

C

The vertices of △ABC are points on

sphere S and the sides are arcs of

great circles. The sum of the interior

angles of a spherical triangle is

greater than 180°.

m∠A + m∠B + m∠C > 180°

great circles

center


Finding Distances on a Sphere

The diameter of the sphere is 15 inches, and

m � AB = 60°. Find the distances between

points A and B.

SOLUTION

Find the lengths of the minor arc � AB and the major arc � ACB of the great circle shown.

Let x be the arc length of � AB and let y be the arc length of � ACB .

Arc length of � AB

—— 2πr

= m � AB

— 360°

Arc length of � ACB

—— 2πr

= m � ACB

— 360°

x —

15π =

60° —

360°

y —

15π =

360° − 60° — 360°

x = 2.5π y = 12.5π x ≈ 7.85 y ≈ 39.27

The distances are about 7.85 inches and about 39.27 inches.


2. The diameter of the sphere is 20 centimeters, and m � AB = 120°. Find the distances

between points A and B.

Comparing Euclidean and Spherical Geometry

Tell whether the following postulate in Euclidean geometry is also true in spherical

geometry. Draw a diagram to support your answer.

Parallel Postulate (Postulate 3.1): If there is a line and a point not on the line,

then there is exactly one line through the point parallel to the given line.

SOLUTION

Parallel lines do not intersect. The sphere shows a

line ℓ (a great circle) and a point A not onℓ.

Several lines are drawn through A. Each great

circle containing A intersectsℓ. So, there can

be no line parallel toℓ.

The Parallel Postulate is not true in spherical geometry.


1. Draw sketches to show that the Two Point Postulate (Postulate 2.1), discussed on

the previous page, is not true for antipodal points and is true for two points that

are not antipodal points.

Finding Distances on a Sphere

In Euclidean geometry, there is exactly one distance that can be measured between

any two points. On a sphere, there are two distances that can be measured between any

two points. These distances are the lengths of the major and minor arcs of the great

circle drawn through the points.

A

A C15 in.

BP

A C20 cm

BP


Finding Areas of Spherical Triangles

Find the area of each spherical triangle.

a. △DEF b. △JKL

SOLUTION

a. A = πr2

— 180°

(m∠D + m∠E + m∠F − 180°) Formula for area of a spherical triangle

= π (10)2

— 180°

(95° + 115° + 105° − 180°) Substitute.

= 100π — 180°

(135°) Simplify.

= 75π Simplify.


The area of △DEF is 75π, or about 235.62 square inches.

b. A = πr2

— 180°

(m∠J + m∠K + m∠L − 180°) Formula for area of a spherical triangle

= π(12)2

— 180°

(102° + 90° + 78° − 180°) Substitute.

= 144π — 180°

(90°) Simplify.

= 72π Simplify.


The area of △JKL is 72π, or about 226.19 square meters.


Find the area of the spherical triangle.

3. △MNP 4. △QRS 5. △TUV

P

M

N

90°

90° 90°

8 cm Q

SR

60° 60°120°

9 mm

U

T

V

110°

70° 30°

12 m

Finding Areas of Spherical Triangles

Core Core ConceptConceptArea of a Spherical TriangleThe area of △ABC on a sphere is

A = πr2

— 180°

(m∠A + m∠B + m∠C − 180°)


B

A

Cr

115°E

D

F105°

95°10 in.

90°K

J

L78°

102°12 m



Monitoring Progress and Modeling with MathematicsMonitoring Progress and Modeling with MathematicsIn Exercises 3–8, the statement is true in Euclidean geometry. Rewrite the statement to be true for spherical geometry. Explain your reasoning. (See Example 1.)

3. Given a line and a point not on the line, there is exactly

one line through the point parallel to the given line.

4. If two lines intersect, then their intersection is exactly

one point.

5. The length of a line is infi nite.

6. A line divides a plane into two infi nite regions.

7. A triangle can have no more than one right angle.

8. Two distinct lines in a plane are either parallel or

intersect at exactly one point.

In Exercises 9–14, use the diagram and the given arc measure to fi nd the distances between points A and B.(See Example 2.)

9. m � AB = 90° 10. m � AB = 140°

A C16 cm

B

P

A C

BP

30 mm

11. m � AB = 30° 12. m � AB = 45°

A CB

P12 ft

A C

BP

18 m

13. m � AB = 150° 14. m � AB = 135°

A C

BP

14 yd

A C

BP

40 in.

In Exercises 15–20, fi nd the area of the spherical triangle. (See Example 3.)

15. △ABC 16. △DEF

45°B

A

C53°

127° 2 m D

E F

90°90°

45°6 ft

17. △JKL 18. △MNP

KL

J

75° 132°

108°4 in.

M

PN110°120°

100°5 yd

19. △QRS 20. △TUV

R S

Q

60°

140°

60°16 mm

T

VU

119°111°

85°30 cm

1. WRITING How is a line in Euclidean geometry different from a line in spherical geometry?

2. WHICH ONE DOESN’T BELONG? Which statement does not belong with the other three?


There are no parallel lines.

Two distances can be measured between any two points.

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

A line extends without end.




fi nding the distances between points A and B.


—— 2𝛑r

= m � AB

— 360°

x —

2𝛑(24) =

90° — 360°

x —

48𝛑 = 90° —

360°

x = 12𝛑

≈ 37.70


—— 2𝛑r

= m � ACB

— 360°

y —

2𝛑(24) =

360° − 90° —— 360°

y —

48𝛑 = 270° — 360°

y = 36𝛑

≈ 113.10

✗

A C24 cm

B

P

The distances are about

37.70 centimeters and

about 113.10 centimeters.


fi nding the area of △ABC.

A = 𝛑r2

— 180°

(m∠ A + m∠ B + m∠ C )

= 𝛑(5)2

— 180°

(114° + 106° + 104°)

= ( 25𝛑 —

180° ) 324°

= 45𝛑

≈ 141.37

The area of △ABC

is 45𝛑, or about

141.37 square inches.

✗

B

A

C

114°

106° 104°5 in.

23. MAKING AN ARGUMENT A polygon on a sphere has

arcs of great circles for sides. Your friend claims that a

polygon on a sphere can have two sides. Is your friend

correct? Explain your reasoning.

24. REASONING Determine whether each of the theorems

from Euclidean geometry is true in spherical

geometry. Explain your reasoning.

a. Linear Pair Perpendicular Theorem (Theorem 3.10): If two lines intersect to form a linear pair of

congruent angles, then the lines are perpendicular.

b. Lines Perpendicular to a Transversal Theorem (Theorem 3.12): In a plane, if two lines are

perpendicular to the same line, then they are

parallel to each other.

25. MODELING WITH MATHEMATICS The radius of

Earth is about 3960 miles. Find the distance between

Pontianak, Indonesia, and the North Pole.

North Pole

EquatorPontianak, Indonesia

26. HOW DO YOU SEE IT? In Euclidean geometry, when

three points are collinear, exactly one of the points

lies between the other two. Is this true in spherical

geometry? Explain.

C

BA

C

B

mA

27. REASONING In Euclidean geometry, two

perpendicular lines form four right angles. How

many right angles do two perpendicular lines form in

spherical geometry?

28. THOUGHT PROVOKING What are the possible sums

of the interior angles of a spherical triangle? Explain.

Maintaining Mathematical ProficiencyFind the areas of the sectors formed by ∠DFE. (Section 11.2)

29.

5 ft

D

F EG

120°

30.

D E

G

45°8 yd

F

31.

6 m

D

FE

G

150°

Reviewing what you learned in previous grades and lessonsggg

693

12.4–12.7 What Did You Learn?

Core VocabularyCore Vocabularyvolume, p. 664Cavalieri’s Principle, p. 664

chord of a sphere, p. 680great circle, p. 680

antipodal points, p. 688

Core ConceptsCore ConceptsSection 12.4Cavalieri’s Principle, p. 664Volume of a Prism, p. 664

Volume of a Cylinder, p. 665

Section 12.5Volume of a Pyramid, p. 672 Volume of a Cone, p. 673

Section 12.6Surface Area of a Sphere, p. 680 Volume of a Sphere, p. 682

Section 12.7Euclidean Geometry and Spherical Geometry, p. 688Finding Distances on a Sphere, p. 689

Area of a Spherical Triangle, p. 690

Mathematical ThinkingMathematical Thinking1. Search online for advertisements for products that come in different sizes. Then

compare the unit prices, as done in Exercise 36 on page 670. Do you get results

similar to Exercise 36? Explain.

2. In Exercise 30 on page 677, the cube is formed by three oblique pyramids. If these pyramids

were right pyramids, it would not be possible to form a cube. Explain why the formula for the

volume of a right pyramid is the same as the formula for the volume of an oblique pyramid.

3. In Exercise 38 on page 685, explain the steps you used to fi nd the value of x.

The city council will consider reopening the closed water park if your team can come up with a cost analysis for painting some of the structures, fi lling the pool water reservoirs, and resurfacing some of the surfaces. What is your plan to convince the city council to open the water park?

To explore the answer to this question and more, go to BigIdeasMath.com.

Performance Task

Water Park Renovation

696933

the value of x.

park some

urfacing ty

e Taskk


1212 Chapter Review

Three-Dimensional Figures (pp. 639–644)12.1


The solid is a cylinder with a height of 8 and

a radius of 3.


1.

9

5

2.

7

7

3.

6

8

Describe the cross section formed by the intersection of the plane and the solid.

4. 5. 6.

33

8

8

3

8

Surface Areas of Prisms and Cylinders (pp. 645–652)12.2

Find the lateral area and the surface area of the right rectangular prism.

L = Ph Formula for lateral area of a right prism

= [2(8) + 2(14)](9) Substitute.

= 396 Simplify.

S = 2B + L Formula for surface area of a right prism

= 2(8)(14) + 396 Substitute.

= 620 Simplify.

The lateral area is 396 square inches and the surface area is 620 square inches.

14 in.9 in.

8 in.

Chapter 12 Chapter Review 695

Find the lateral area and the surface area of the right prism or right cylinder.

7. 8. 9.

18 cm 4 cm

17 ft

16 ft

20 ft

11 m

5 m

Surface Areas of Pyramids and Cones (pp. 653–660)12.3

Find the lateral area and the surface area of the right cone.

Use the Pythagorean Theorem (Theorem 9.1) to fi nd the slant heightℓ.

ℓ= √—

152 + 362 = 39



= π(15)(39) Substitute.

= 585π Simplify.


S = πr2 + πrℓ Formula for surface area of a right cone

= π(15)2 + 585π Substitute.

= 810π Simplify.


The lateral area is 585π, or about 1837.83 square meters. The surface area is 810π, or about

2544.69 square meters.

Find the lateral area and the surface area of the regular pyramid or right cone.

10.

9 in.

8 in. 11.

10 cm

24 cm

12.

6 ft3 ft

10 ft

13. Find the lateral area and the surface area of the

composite solid.

36 m

15 m

12 cm

3 cm

4 cm


Volumes of Prisms and Cylinders (pp. 663–670)12.4

Find the volume of the cylinder.

The dimensions of the cylinder are r = 7 ft and h = 10 ft.

V = πr2h Formula for volume of a cylinder

= π(7)2(10) Substitute.

= 490π Simplify.


The volume is 490π, or about 1539.38 cubic feet.


14.

1.5 m2.1 m

3.6 m

15.

8 mm

2 mm

16.

2 yd

4 yd

17. Describe how the change affects the volume of the triangular prism.

a. multiplying the height of the prism by 1 —

3

b. multiplying all the linear dimensions by 2

6 in.

6 in.6 in.

10 in.


Find the volume of the pyramid.

V = 1 —


= 1 —

3 ( 1 —

2 ⋅ 5 ⋅ 8 ) (12) Substitute.

= 80 Simplify.


12 m

8 m5 m

7 ft

10 ft

7 cm

24 cm

30 cm

Volumes of Pyramids and Cones (pp. 671–678)12.5

Chapter 12 Chapter Review 697

Surface Areas and Volumes of Spheres (pp. 679–686)12.6

Find the (a) surface area and (b) volume of the sphere.

a. S = 4πr2 Formula for surface area of a sphere


= 1296π Simplify.


The surface area is 1296π, or about 4071.50 square inches.

b. V = 4 —

3 πr3 Formula for volume of a sphere

= 4 —

3 π(18)3 Substitute 18 for r.

= 7776π Simplify.

≈ 24,429.02 Use a calculator.

The volume is 7776π, or about 24,429.02 cubic inches.

Find the surface area and the volume of the sphere.

24.

7 in.

25.

17 ft

26.

27. The shape of Mercury can be approximated by a sphere with a diameter of 4880 kilometers.

Find the surface area and the volume of Mercury.

28. A solid is composed of a cube with a side length of 6 meters and a hemisphere with a diameter

of 6 meters. Find the volume of the composite solid.


19.

30 cm34 cm

16 cm

20.

18 m

10 m

5 m

21.

13 m

7 m

22. The volume of a square pyramid is 1024 cubic inches. The base has a side length of 16 inches.

Find the height of the pyramid.

23. A cone with a diameter of 16 centimeters has a volume of 320π cubic centimeters. Find the

height of the cone.

18 in.

C = 30 ftπ


Spherical Geometry (pp. 687–692)12.7

a. The diameter of the sphere is 24 inches, and m � AB = 30°. Find the distances between points A and B.

Find the lengths of the minor arc � AB and the major arc � ACB of the

great circle shown. Let x be the arc length of � AB and let y be the

arc length of � ACB .


—— 2πr

= m � AB

— 360°


—— 2πr

= m � ACB

— 360°

x —

24π =

30° —

360°

y —

24π =

360° − 30° — 360°

x = 2π y = 22π x ≈ 6.28 y ≈ 69.12

The distances are about 6.28 inches and about 69.12 inches.

b. Find the area of △DEF.

A = πr2

— 180°

(m∠D + m∠E + m∠F − 180°) Formula for area of a spherical triangle

= π (6)2

— 180°

(116° + 82° + 82° − 180°) Substitute.

= 36π — 180°

(100°) Simplify.

= 20π Simplify.


The area of △DEF is 20π, or about 62.83 square feet.

29. The diameter of the sphere is 18 feet, and m � XY = 90°. Find the distances between points X and Y.

Find the area of the spherical triangle.

30. △JKL 31. △RST 32. △XYZ

J

K L

125°

85° 90°3 in.

S

R

T

88°

88°124°

12 m

Y

X

Z102° 108°

110°

18 cm

A C

B

P24 in.

E

D

F82° 82°

116° 6 ft

X Z18 ft

Y

P

Chapter 12 Chapter Test 699

Chapter Test1212Find the volume of the solid.

1.

15.5 m

8 m

2.

3.2 ft

3.

6 m

4 m3 m

3 m4 m

4.

5 ft 2 ft

4 ft

8 ft

Find the lateral area and the surface area of the solid.

5.

12 m

8.3 m8 m 6. 9 ft

20 ft

7. 8.

9. Sketch the composite solid produced by rotating the fi gure around

the given axis. Then identify and describe the composite solid.

10. The shape of Mars can be approximated by a sphere with a diameter

of 6779 kilometers. Find the surface area and the volume of Mars.

11. You have a funnel with the dimensions shown.

a. Find the approximate volume of the funnel.

b. You use the funnel to put oil in a car. Oil fl ows out of the funnel at a rate of

45 milliliters per second. How long will it take to empty the funnel when it is full

of oil? (1 mL = 1 cm3)

c. How long would it take to empty a funnel with a radius of 10 centimeters and a height

of 6 centimeters if oil fl ows out of the funnel at a rate of 45 milliliters per second?

d. Explain why you can claim that the time calculated in part (c) is greater than the

time calculated in part (b) without doing any calculations.

12. Describe how the change affects the surface area and the volume of the right cone.

a. multiplying the height by 3 —

4

b. multiplying all the linear dimensions by 6 —

5

13. A water bottle in the shape of a cylinder has a volume of 500 cubic centimeters.

The diameter of a base is 7.5 centimeters. What is the height of the bottle? Justify

your answer.

14. Is it possible for a right triangular pyramid to have a cross section that is a quadrilateral?

If so, describe or sketch how the plane could intersect the pyramid.

15. The soccer ball shown has a radius of 4.5 inches. Each interior angle of the spherical

triangles has a measure of 70°. Find the area of one spherical triangle.

3 3

9

6

10 cm

6 cm

7 in.

7 in.

6 in.

4 in.

30 cm

12 cm16 cm

20 yd

15 yd


1212 Standards Assessment

1. Which cross section formed by the intersection of the plane and the solid is a

rectangle? (TEKS G.10.A)

○A ○B

○C ○D

2. The coordinates of the vertices of △DEF are D(−8, 5), E(−5, 8), and F(−1, 4).

Which set of coordinates describes a triangle that is similar to △DEF? (TEKS G.7.A)

○F A(−24, 15), B(−15, 24), and C(−3, 8)

○G J(16, −10), K(10, −16), and L(2, −8)

○H P(−8, 10), Q(−5, 16), and R(−1, 8)

○J U(−10, 5), V(−7, −2), and W(−2, −7)

3. The top of the Washington Monument in Washington, D.C., is a square pyramid, called

a pyramidion. What is the volume of the pyramidion? (TEKS G.11.D)

○A 22,019.63 ft3

55.5 ft

34.5 ft

○B 172,006.91 ft3

○C 66,058.88 ft3

○D 207,530.08 ft3

4. Use the diagram. Which proportion is false? (TEKS G.8.B)

○F DB

— DC

= DA

— DB

○G CA

— AB

= AB

— AD

B

CD

A○H CA

— BA

= BA

— CA

○J DC

— BC

= BC

— CA

Chapter 12 Standards Assessment 701

5. The surface area of the right cone is 200π square feet. What is the slant height of

the cone? (TEKS G.11.C)

○A 10.5 ft ○B 17 ft

16 ft○C 23 ft ○D 24 ft

6. In the diagram, ABCD is a parallelogram. Which statement is not true? (TEKS G.6.B)

E

B

CD

A

○F △AED ≅ △CEB ○G △DEC ≅ △BEA

○H △ABD ≅ △CDB ○J △AEB ≅ △DEC

7. Points X, Y, and Z lie on the surface of a sphere. What is a possible value of the sum of

the interior angles of △XYZ? (TEKS G.4.D)

○A 90° ○B 167°

○C 180° ○D 204°

8. GRIDDED ANSWER The diagram shows a square pyramid and a cone. Both solids have

a height of 23 inches, and the base of the cone has a radius of 8 inches. According to

Cavalieri’s Principle, the solids will have the same volume if the square base has sides of

length ___ inches. Round your answer to the nearest tenth of an inch. (TEKS G.11.D)

B hB

r

9. Which statement about the circle is not true? (TEKS G.12.A)

○F — AD is a diameter.

C

H

D

E

F

G

B

A○G

— BH is a chord.

○H — AD is a chord.

○J — CD is a radius.

12 Surface Area and Volume - Big Ideas Math

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