Post on 07-Feb-2023
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
640 Chapter 12 Surface Area and Volume
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.
Section 12.1 Three-Dimensional Figures 641
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.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
Describe the shape formed by the intersection of the plane and the solid.
4. 5. 6.
642 Chapter 12 Surface Area and Volume
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.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
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
Section 12.1 Three-Dimensional Figures 643
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
644 Chapter 12 Surface Area and Volume
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?
Explain your reasoning.
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.
Finding a Formula for Surface Area
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
TEXAS ESSENTIAL KNOWLEDGE AND SKILLS
Surface Areas of Prismsand Cylinders
12.2
height, h
a
b
c
height, h
radius, r
646 Chapter 12 Surface Area and Volume
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 VocabularyCore Vocabullarry
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
Section 12.2 Surface Areas of Prisms and Cylinders 647
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.
≈ 488.40 Use a calculator.
The lateral area is 317.25 square feet and the surface area is
about 488.40 square feet.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
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.
648 Chapter 12 Surface Area and Volume
Finding Lateral Area and Surface Area
Find the lateral area and the surface area of the right cylinder.
8 m
4 m
SOLUTION
Find the lateral area and the surface area.
L = 2πrh Formula for lateral area of a right cylinder
= 2π(4)(8) Substitute.
= 64π Simplify.
≈ 201.06 Use a calculator.
S = 2πr2 + 2πrh Formula for surface area of a right cylinder
= 2π(4)2 + 64π Substitute.
= 96π Simplify.
≈ 301.59 Use a calculator.
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.
≈ 339.29 Use a calculator.
You need a minimum of about 339.29 square centimeters of material.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
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.
Section 12.2 Surface Areas of Prisms and Cylinders 649
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.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
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
650 Chapter 12 Surface Area and Volume
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
Section 12.2 Surface Areas of Prisms and Cylinders 651
19. ERROR ANALYSIS Describe and correct the error in
fi nding the surface area of the right cylinder.
S = 2𝛑 (6)2 + 2𝛑(6)(8)
= 168𝛑
≈ 527.79 cm2
✗8 cm
6 cm
20. ERROR ANALYSIS Describe and correct the error in
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
652 Chapter 12 Surface Area and Volume
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.
Finding a Formula for Surface Area
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.
Finding a Formula for Surface Area
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
TEXAS ESSENTIAL KNOWLEDGE AND SKILLS
Surface Areas of Pyramidsand Cones
12.3
slant height,height, h
base edge length, b
radius, r
slant height,
654 Chapter 12 Surface Area and Volume
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 VocabularyCore Vocabullarry
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
Finding Lateral Area and Surface Area
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.
≈ 679.81 Use a calculator.
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
Section 12.3 Surface Areas of Pyramids and Cones 655
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
Finding Lateral Area and Surface Area
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
Find the lateral area and the surface area.
L = πrℓ Formula for lateral area of a right cone
= π(6)(10) Substitute.
= 60π Simplify.
≈ 188.50 Use a calculator.
S = πr2 + πrℓ Formula for surface area of a right cone
= π(6)2 + 60π Substitute.
= 96π Simplify.
≈ 301.59 Use a calculator.
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
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
1. Find the lateral area and the surface area of the regular pentagonal pyramid.
656 Chapter 12 Surface Area and Volume
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.
L = πrℓ Formula for lateral area of a right cone
= π(5.7) ( √—
356.49 ) Substitute.
≈ 388.10 Use a calculator.
The lateral area of the traffi c cone is about 338.10 square inches.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
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
Section 12.3 Surface Areas of Pyramids and Cones 657
Changing Dimensions in a Solid
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.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
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 .
658 Chapter 12 Surface Area and Volume
Tutorial Help in English and Spanish at BigIdeasMath.comExercises12.3
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.
11. ERROR ANALYSIS Describe and correct the error in
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
20. ERROR ANALYSIS Describe and correct the error in
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.
Section 12.3 Surface Areas of Pyramids and Cones 659
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°
660 Chapter 12 Surface Area and Volume
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
662 Chapter 12 Surface Area and Volume
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
TEXAS ESSENTIAL KNOWLEDGE AND SKILLS
664 Chapter 12 Surface Area and Volume
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.
V = Bh Formula for volume of a prism
= 30(5) Substitute.
= 150 Simplify.
The volume is 150 cubic centimeters.
volume, p. 664Cavalieri’s Principle, p. 664
Previousprismcylindercomposite solid
Core VocabularyCore Vocabullarry
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
Section 12.4 Volumes of Prisms and Cylinders 665
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.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
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
666 Chapter 12 Surface Area and Volume
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.
V = Bh Formula for volume of a prism
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 ✓
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
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?
Changing Dimensions in a Solid
Describe how doubling all the linear dimensions affects the
4 ft
6 ft
3 ft
volume of the rectangular prism.
SOLUTION
Before change After change
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
ANALYZING MATHEMATICAL RELATIONSHIPS
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
Section 12.4 Volumes of Prisms and Cylinders 667
Changing a Dimension in a Solid
Describe how tripling the radius affects the 3 cm
6 cm
volume of the cylinder.
SOLUTION
Before change After change
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.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
4. In Example 4, describe how multiplying all the linear dimensions by 1 —
2 affects the
volume of the rectangular prism.
5. In Example 4, describe how doubling the length and width of the bases affects the
volume of the rectangular prism.
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
volume of the cylinder.
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.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
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
668 Chapter 12 Surface Area and Volume
Tutorial Help in English and Spanish at BigIdeasMath.comExercises12.4
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
19. ERROR ANALYSIS Describe and correct the error in
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.
✗
Monitoring Progress and Modeling with MathematicsMonitoring Progress and Modeling with Mathematics
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 _________.
Vocabulary and Core Concept CheckVocabulary and Core Concept Check
Section 12.4 Volumes of Prisms and Cylinders 669
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 22. multiplying 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?
670 Chapter 12 Surface Area and Volume
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?
Explain your reasoning.
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.
Reviewing what you learned in previous grades and lessons
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?
ANALYZING MATHEMATICAL RELATIONSHIPS
To be profi cient in math, you need to look closely to discern a pattern or structure.
12.5
G.10.BG.11.D
TEXAS ESSENTIAL KNOWLEDGE AND SKILLS
h
h
672 Chapter 12 Surface Area and Volume
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 VocabularyCore Vocabullarry
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
Section 12.5 Volumes of Pyramids and Cones 673
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.
≈ 22.81 Use a calculator.
The volume is 7.26π, or about 22.81 cubic centimeters.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
Find the volume of the solid.
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
674 Chapter 12 Surface Area and Volume
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 —
3 Bh Formula for volume of a pyramid
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.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
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
Changing Dimensions in a Solid
Describe how multiplying all the linear dimensions by 1 —
3 affects the volume of the
rectangular pyramid.
SOLUTION
Before change After change
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
Section 12.5 Volumes of Pyramids and Cones 675
Changing a Dimension in a Solid
Describe how doubling the height affects the volume
of the cone.
SOLUTION
Before change After change
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.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
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 —
2 affects the volume of the
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.
Finding the Volume of a Composite Solid
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.
The volume is 288 cubic meters.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
10. Find the volume of the composite solid.
6 m
6 m6 m
6 m
ANALYZING MATHEMATICAL RELATIONSHIPS
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
676 Chapter 12 Surface Area and Volume
Tutorial Help in English and Spanish at BigIdeasMath.comExercises12.5
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
13. ERROR ANALYSIS Describe and correct the error in
fi nding the volume of the pyramid.
5 ft
6 ft
V = 1 — 3
(6)(5)
= 1 — 3
(30)
= 10 ft3
✗
14. ERROR ANALYSIS Describe and correct the error in
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.
Explain your reasoning.
16. OPEN-ENDED Give an example of a cone and a
cylinder that have the same base and the same
volume. Explain your reasoning.
Monitoring Progress and Modeling with MathematicsMonitoring Progress and Modeling with Mathematics
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.
Vocabulary and Core Concept CheckVocabulary and Core Concept Check
Section 12.5 Volumes of Pyramids and Cones 677
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 18. multiplying 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
In Exercises 23–28, fi nd the volume of the composite solid. (See Example 6.)
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?
Explain your reasoning.
(1 cup ≈ 14.4 in.3)
2.5 in.
7.5 in.
4 in.
678 Chapter 12 Surface Area and Volume
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
Reviewing what you learned in previous grades and lessons
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
TEXAS ESSENTIAL KNOWLEDGE AND SKILLS
680 Chapter 12 Surface Area and Volume
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 VocabularyCore Vocabullarry
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
Section 12.6 Surface Areas and Volumes of Spheres 681
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.
≈ 804.25 Use a calculator.
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
= 4π(6)2 Substitute 6 for r.
= 144π Simplify.
≈ 452.39 Use a calculator.
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.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
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
682 Chapter 12 Surface Area and Volume
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.
≈ 381.70 Use a calculator.
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
where r is the radius of the sphere.
r
V = r3π43
4.5 in.
Section 12.6 Surface Areas and Volumes of Spheres 683
Changing Dimensions in a Solid
Describe how multiplying the radius by 1 —
4 affects
12 in.the volume of the sphere.
SOLUTION
Before change After change
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.
Finding the Volume of a Composite Solid
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.
≈ 8.38 Use a calculator.
The volume is 8 —
3 π, or about 8.38 cubic inches.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
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
684 Chapter 12 Surface Area and Volume
Tutorial Help in English and Spanish at BigIdeasMath.comExercises12.6
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.
Vocabulary and Core Concept CheckVocabulary and Core Concept Check
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
21. ERROR ANALYSIS Describe and correct the error in
fi nding the volume of the sphere.
6 ft V = 4 —
3 π(6)2
= 48π≈ 150.80 ft3
✗
Monitoring Progress and Modeling with MathematicsMonitoring Progress and Modeling with Mathematics
Section 12.6 Surface Areas and Volumes of Spheres 685
22. ERROR ANALYSIS Describe and correct the error in
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
In Exercises 25–28, fi nd the volume of the composite solid. (See Example 5.)
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.
686 Chapter 12 Surface Area and Volume
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.
Reviewing what you learned in previous grades and lessons
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
TEXAS ESSENTIAL KNOWLEDGE AND SKILLS
CD
great circle
A B
Points A and B are antipodal.
688 Chapter 12 Surface Area and Volume
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 VocabularyCore Vocabullarry
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
Section 12.7 Spherical Geometry 689
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.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
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.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
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
690 Chapter 12 Surface Area and Volume
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.
≈ 235.62 Use a calculator.
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.
≈ 226.19 Use a calculator.
The area of △JKL is 72π, or about 226.19 square meters.
Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com
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°)
where r is the radius of the sphere.
B
A
Cr
115°E
D
F105°
95°10 in.
90°K
J
L78°
102°12 m
Section 12.7 Spherical Geometry 691
Tutorial Help in English and Spanish at BigIdeasMath.comExercises12.7
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?
Explain your reasoning.
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.
Vocabulary and Core Concept CheckVocabulary and Core Concept Check
692 Chapter 12 Surface Area and Volume
21. ERROR ANALYSIS Describe and correct the error in
fi nding the distances between points A and B.
Arc length of � AB
—— 2𝛑r
= m � AB
— 360°
x —
2𝛑(24) =
90° — 360°
x —
48𝛑 = 90° —
360°
x = 12𝛑
≈ 37.70
Arc length of � ACB
—— 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.
22. ERROR ANALYSIS Describe and correct the error in
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
694 Chapter 12 Surface Area and Volume
1212 Chapter Review
Three-Dimensional Figures (pp. 639–644)12.1
Sketch the solid produced by rotating the fi gure around the given axis. Then identify and describe the solid.
The solid is a cylinder with a height of 8 and
a radius of 3.
Sketch the solid produced by rotating the fi gure around the given axis. Then identify and describe the solid.
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
Find the lateral area and the surface area.
L = πrℓ Formula for lateral area of a right cone
= π(15)(39) Substitute.
= 585π Simplify.
≈ 1837.83 Use a calculator.
S = πr2 + πrℓ Formula for surface area of a right cone
= π(15)2 + 585π Substitute.
= 810π Simplify.
≈ 2544.69 Use a calculator.
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
696 Chapter 12 Surface Area and Volume
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.
≈ 1539.38 Use a calculator.
The volume is 490π, or about 1539.38 cubic feet.
Find the volume of the solid.
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.
18. Find the volume of the composite solid.
Find the volume of the pyramid.
V = 1 —
3 Bh Formula for volume of a pyramid
= 1 —
3 ( 1 —
2 ⋅ 5 ⋅ 8 ) (12) Substitute.
= 80 Simplify.
The volume is 80 cubic meters.
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
= 4π(18)2 Substitute 18 for r.
= 1296π Simplify.
≈ 4071.50 Use a calculator.
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.
Find the volume of the 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
698 Chapter 12 Surface Area and Volume
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 .
Arc length of � AB
—— 2πr
= m � AB
— 360°
Arc length of � ACB
—— 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.
≈ 62.83 Use a calculator.
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
700 Chapter 12 Surface Area and Volume
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.