Chapter 9
Chapter 9 Opener
Try It Yourself (p. 353)
1. Area = w
= (11)(9)
= 99
The area of the rectangle is 99 square meters.
2. Area = w
= (8.5)( 4.2)
= 35.7
The area of the rectangle is 35.7 square feet.
Section 9.1
9.1 Activity (pp. 354 –355)
1.
w
w
h
h
w
w
w
w
S = wh + h + wh + h + w + w
= 2w + 2h + 2wh
2. a. Area of base:
1
•3•4 = 6
2
2
3. Area = s
 2
=  
 3
4
=
9
w
w
2
Areas of lateral faces:
3•3 = 9
3 • 5 = 15
3 • 4 = 12
4
The area of the square is square inch.
9
1
bh
2
1
= (13)(6)
2
1
= (78)
2
= 39
S = areas of bases + areas of lateral faces
= 6 + 6 + 9 + 15 + 12
= 48
4. Area =
The area of the triangle is 39 square feet.
1
bh
2
1
= ( 20)(14)
2
1
= ( 280)
2
= 140
5. Area =
The surface area is 48 square units. The solid is a
triangular prism.
b. The triangles are bases because they are identical in
shape and when folded they are at the top and bottom
of the prism.
3. a. Surface Area: 52 in.2
Drawing
Surface
Area
Drawing
Surface
Area
b.
98 in.2
c.
76 in.2
d.
70 in.2
e.
68 in.2
f.
56 in.2
g.
52 in.2
The area of the triangle is 140 square meters.
1
bh
2
1
= (30)(15)
2
1
= ( 450)
2
= 225
6. Area =
The area of the triangle is 225 square centimeters.
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Chapter 9
4. a. S = 2w + 2h + 2 wh
6. Bigger block:
= 2( 4)(3) + 2( 4)( 2) + 2(3)( 2)
S = 2w + 2h + 2 wh
= 24 + 16 + 12
= 2(3)(1) + 2(3)(1) + 2(1)(1)
= 52
= 6+ 6+ 2
= 14
So, the surface area is 52 square inches.
b. S = 2w + 2h + 2 wh
The surface area of the bigger ice block is 14 square feet.
= 2(1)(1) + 2(1)( 24) + 2(1)( 24)
Smaller block:
S = 2w + 2h + 2 wh
= 2 + 48 + 48
= 2(1)(1) + 2(1)(1) + 2(1)(1)
= 98
= 2 + 2+ 2
So, the surface area is 98 square inches.
= 6
c. S = 2w + 2h + 2 wh
= 2( 2)(1) + 2( 2)(12) + 2(1)(12)
The surface area of one smaller ice block is 6 square feet.
The surface area of the three smaller ice blocks is
6(3) = 18 square feet. Because 18 > 14, the three
= 4 + 48 + 24
= 76
smaller ice blocks will melt faster.
So, the surface area is 76 square inches.
d. S = 2w + 2h + 2 wh
= 2(3)(1) + 2(3)(8) + 2(1)(8)
9.1 On Your Own (pp. 356 –358)
1. S = 2w + 2h + 2 wh
= 6 + 48 + 16
= 2( 2)(3) + 2( 2)( 4) + 2(3)( 4)
= 70
= 12 + 16 + 24
= 52
So, the surface area is 70 square inches.
e. S = 2w + 2h + 2 wh
= 2( 4)(1) + 2( 4)(6) + 2(1)(6)
= 8 + 48 + 12
The surface area is 52 square feet.
2. S = 2w + 2h + 2 wh
= 2(8)(8) + 2(8)(5) + 2(8)(5)
= 68
= 128 + 80 + 80
So, the surface area is 68 square inches.
= 288
f. S = 2w + 2h + 2 wh
= 2( 2)( 2) + 2( 2)(6) + 2( 2)(6)
= 8 + 24 + 24
= 56
So, the surface area is 56 square inches.
g. S = 2w + 2h + 2 wh
= 2( 4)( 2) + 2( 4)(3) + 2( 2)(3)
= 16 + 24 + 12
= 52
So, the surface area is 52 square inches.
5. The surface area of any prism is the sum of the areas of
the bases and the lateral faces.
The surface area is 288 square meters.
3.
1
5m
5m
2
3
13 m
12 m
Area of a base:
3m
12 m
1
• 5 • 12 = 30
2
Area of face 1: 5 • 3 = 15
Area of face 2: 13 • 3 = 39
Area of face 3: 12 • 3 = 36
S = areas of bases + areas of lateral faces
= 30 + 30 + 15 + 39 + 36
= 150
The surface area is 150 square meters.
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Chapter 9
4.
2. A rectangular prism and a cube both have the same
4 cm
1
number of faces, vertices, and edges. All of the edges of a
cube have the same length and its faces are congruent
squares, which is not true of all rectangular prisms.
5 cm
3. The one that is different is “Find the area of the bases of
2
the prism.” It asks for just the area of the bases, while the
other three ask for the surface area of the prism.
5 cm
4 cm
Area of the bases:
A = 2w = 2(3)( 4) = 24
3 cm
3
3 cm
The area of the bases is 24 square inches.
Area of a base:
Surface area of the prism:
1
•3•4 = 6
2
S = 2w + 2h + 2wh
Area of face 1: 4 • 5 = 20
= 2(3)( 4) + 2(3)(7) + 2( 4)(7)
Area of face 2: 4 • 4 = 16
= 24 + 42 + 56
Area of face 3: 4 • 3 = 12
= 122
The surface area is 122 square inches.
S = areas of bases + areas of lateral faces
= 6 + 6 + 20 + 16 + 12
= 60
The surface area is 60 square centimeters.
Practice and Problem Solving
Drawing
Surface Area
4.
22 in.2
S = 6s 2 = 6(9) = 6(81) = 486
5.
38 in.2
The surface area of the cube is 486 square centimeters.
6.
32 in.2
5. Surface area of 9 cm × 9 cm × 9 cm cube:
2
Surface area of 5 cm × 15 cm × 7 cm rectangular prism:
S = 2w + 2h + 2wh
= 2(5)(15) + 2(5)(7) + 2(15)(7)
= 150 + 70 + 210
= 430
The surface area of the rectangular prism is
430 square centimeters.
Because 486 is greater than 430, the cube has the greater
surface area.
6. S = 2h + 2 wh
= 2(12)( 20) + 2(12)( 20)
= 480 + 480
= 960
You need to coat 960 square inches with glue.
9.1 Exercises (pp. 359 –361)
Vocabulary and Concept Check
1. One way is to use a formula. If the rectangular prism is a
cube with side length s, then the surface area is equal to
6s 2 . If the rectangular prism is not a cube, then the
surface area is equal to 2w + 2h + 2wh, where  is
the length, w is the width, and h is the height. Another
way is to use a net.
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7. S = 2w + 2h + 2 wh
= 2(6)(16) + 2(6)(3) + 2(16)(3)
= 192 + 36 + 96
= 324
The surface area is 324 square meters.
8. S = 2w + 2h + 2 wh
= 2(5)( 4) + 2(5)(7) + 2( 4)(7)
= 40 + 70 + 56
= 166
The surface area is 166 square millimeters.
9. S = 2w + 2h + 2 wh
 1
 1
= 21 (5) + 21 (3) + 2(5)(3)
 5
 5
36
= 12 +
+ 30
5
1
= 42 + 7
5
1
= 49
5
The surface area is 49
1
square yards.
5
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259
Chapter 9
10.
12.
15 ft
17 ft
17 ft
15 ft
9 in.
8 ft
1
2
13.5 in.
10 in.
9 in.
1
20 ft
2
3
9 in.
Area of a base:
15 ft
17 ft
3
10 in.
1
• 9 • 10 = 45
2
Area of face 1: 9 • 9 = 81
Area of face 2: 9 • 13.5 = 121.5
1
Area of a base: • 15 • 8 = 60
2
Area of face 3: 10 • 9 = 90
S = areas of bases + areas of lateral faces
Area of face 1: 20 • 17 = 340
= 45 + 45 + 81 + 121.5 + 90
Area of face 2: 8 • 20 = 160
= 382.5
Area of face 3: 15 • 20 = 300
The surface area is 382.5 square inches.
S = areas of bases + areas of lateral faces
= 60 + 60 + 340 + 160 + 300
= 920
13. S = 6 s 2 = 6(7)
= 6( 49) = 294
2
The surface area of the cube is 294 square yards.
So, the surface area is 920 square feet.
11.
4m
5m
2
= 6(0.25) = 1.5
2
5m
2
 2
 4
= 6  = 2
3
 3
9
15. S = 6s 2 = 6 
6m
1
2
The surface area of the cube is 1.5 square centimeters.
5m
5m
14. S = 6 s 2 = 6(0.5)
3
7m
The surface area of the cube is 2
2
square feet.
3
16. The areas of the two 4 × 5 faces are not included in the
Area of a base:
1
• 6 • 4 = 12
2
Area of face 1: 5 • 7 = 35
Area of face 2: 6 • 7 = 42
surface area, and the area of the 3 × 5 face is added four
times, but it should only be added twice.
S = 2(3)(5) + 2(3)( 4) + 2(5)( 4)
= 30 + 24 + 40
= 94 in.2
Area of face 3: 5 • 7 = 35
S = areas of bases + areas of lateral faces
= 12 + 12 + 35 + 42 + 35
= 136
The surface area is 136 square meters.
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Chapter 9
8.7 in.
17.
10 in.
21.
10 in.
10 in.
1
4m
10 in.
10 in. 2
1
3
3 in.
4m
2.5 m
2
2m
2.5 m
4m
4m
3
2.5 m
4
1
Area of a base: • 10 • 8.7 = 43.5
2
Area of face 1: 10 • 3 = 30
Area of face 2: 10 • 3 = 30
4m
Area of face 3: 10 • 3 = 30
Area of a base: 4 • 2 = 8
S = areas of bases + areas of lateral faces
Area of face 1: 4 • 4 = 16
= 43.5 + 43.5 + 30 + 30 + 30
Area of face 2: 4 • 2.5 = 10
= 177
The surface area of the tin game case is
177 square inches.
18. S = 6 s 2 = 6(11)
2
Area of face 3: 4 • 4 = 16
Area of face 4: 4 • 2.5 = 10
S = areas of bases + areas of lateral faces
= 6(121) = 726
The least amount of wrapping paper you need to wrap
the gift is 726 square centimeters.
19. Find the sum of the areas of the lateral faces and the
top base. Do not include the area of the bottom base
in the formula.
S = w + 2h + 2 wh
= 8 + 8 + 16 + 10 + 16 + 10
= 68
The surface area of the prism is 68 square meters.
22. To find dimensions for a prism with a surface area of
158 square yards, choose two dimensions and solve
for the third dimension.
Sample answer: Choose 8 yards for the length and
5 yards for the width. Solve for the height.
= (13)(9) + 2(13)(3) + 2(9)(3)
= 117 + 78 + 54
S = 2w + 2h + 2 wh
= 249
You need 249 square inches of frosting to frost the cake.
One can of frosting covers about 280 square inches. So,
one can of frosting is enough to frost the cake.
158 = 2(8)(5) + 2(8)h + 2(5)h
158 = 80 + 16h + 10h
158 = 80 + (16 + 10)h
158 =
20.
− 80
1
12 in.
2
5 in.
5 in.
5 in.
4 in.
4
3
3 in.
6 in.
5 in.
6 in.
Area of a base:
1
(6 + 12)(4) = 36
2
80 + 26h
− 80
78 = 26h
78
26h
=
26
26
3 = h
So, a height of 3 yards gives a surface area of 158 square
yards when the length is 8 yards and the width is 5 yards.
Area of face 1: 5 • 3 = 15
3 yd
Area of face 2: 12 • 3 = 36
Area of face 3: 5 • 3 = 15
Area of face 4: 6 • 3 = 18
5 yd
8 yd
S = areas of bases + areas of lateral faces
= 36 + 36 + 15 + 36 + 15 + 18
= 156
The surface area of the prism is 156 square inches.
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Chapter 9
23. Surface area of the label:
25. The dimensions of the red prism are 3 times the
S = 4(area of one lateral face of the label)
= 4( 3 • 2 )
dimensions of the blue prism.
Blue prism:
S = 2w + 2h + 2 wh
= 24
The surface area of the label is 24 square inches.
= 2( 4)( 2) + 2( 4)(3) + 2( 2)(3)
Find the lateral surface area of the box:
= 16 + 24 + 12
Because the label covers 75% of the lateral surface area
of the box, solve the percent equation.
= 52 m 2
24 = 0.75S
Red prism:
S = 2w + 2h + 2 wh
= 2(12)(6) + 2(12)(9) + 2(6)(9)
24
0.75S
=
0.75
0.75
32 = S
= 144 + 216 + 108
The lateral surface area of the box is 32 square inches.
Use the lateral surface area of the box to find x.
S = 4(area of one lateral face of the box )
32 = 4( 2 x)
= 468 m 2
468 m 2
= 9. So, the
52 m 2
surface area of the red prism is 9 times the surface area of
the blue prism.
The ratio of the surface areas is
32 = 8 x
26. The dimensions of the red cube are
32
8x
=
8
8
4 = x
3
times the
2
dimensions of the blue cube.
Blue cube:
The value of x is 4.
S = 6s
24. Surface area of the bread:
= 6( 4)
S = 2w + 2h + 2 wh
= 2(10)(10) + 2(10)( h) + 2(10)( h)
= 200 + 20h + 20h
= 200 + 40h
C = 4(10 • h)
= 40h
of the surface area of the bread.
C = 0.50
S = 6s 2
2
= 6(16)
= 96 ft
2
= 6(6)
2
= 6(36)
= 216 ft 2
The ratio of the surface areas is
Crust surface area:
Crust is 50%
Red cube:
2
×
216 ft 2
96 ft
2
=
9
. So, the
4
9
surface area of the red cube is times the surface area of
4
the blue cube.
S
C = 0.50 S
40h = 0.5( 200 + 40h)
40h = 100 + 20h
− 20h
− 20h
20h = 100
20h
100
=
20
20
h = 5
The height is 5 centimeters.
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Chapter 9
27. a. Green pedestal:
b. Blue pedestal:
 = 3(8 in.) = 24 in.
1
 = (16 in.) = 8 in.
2
1
w = (16 in.) = 8 in.
2
1
h = ( 24 in.) = 12 in.
2
w = 3(8 in.) = 24 in.
h = 3(12 in.) = 36 in.
Blue pedestal surface area:
S = 2w + 2h + 2 wh
= 2( 24)( 24) + 2( 24)(36) + 2( 24)(36)
Red pedestal surface area:
S = 2w + 2h + 2wh
= 1152 + 1728 + 1728
= 2(16)(16) + 2(16)( 24) + 2(16)( 24)
= 4608 in.2
= 512 + 768 + 768
Green pedestal surface area:
S = 2w + 2h + 2 wh
= 2(8)(8) + 2(8)(12) + 2(8)(12)
= 128 + 192 + 192
Paint amounts:
2
= 512 in.
The green pedestal’s surface area is
4608 in.2
= 9
512 in.2
times the green pedestal’s surface area. So, you need
9 • 0.125 = 1.125 pints of paint to paint the blue
pedestal.
The blue pedestal’s surface area is
= 2048 in.2
512 in.2
2048 in.2
of the red pedestal’s surface area. So, you need
1
• 0.5 = 0.125 pint of paint to paint the green
4
pedestal.
=
1
4
Side lengths:
green pedestal
8 in.
12 in.
1
=
=
=
red pedestal
16 in.
24 in.
2
Paint amounts:
Side lengths:
2
green pedestal
0.125 pint
1
=
=
red pedestal
0.5 pint
4
green pedestal
0.125 pint
1
=
=
blue pedestal
1.125 pints
9
green pedestal
8 in.
12 in.
1
=
=
=
blue pedestal
24 in.
36 in.
3
2
1 1
1
1
  = ,  =
2
4
3
9
 
 
The ratio of paint amounts (or surface areas) is equal
to the square of the ratio of side lengths.
28. a. Each small cube has a surface area of 1.5 square inches.
1.5
= 0.25 square inch.
6
Four square faces have a total area of 1 square inch.
Use four square faces to make a large square with an
area of 1 square inch.
The area of one square face is
0.25 in.2
0.25 in.2
s
0.25 in.2
0.25 in.2
s
s
s
Total area = 1 in.2
The large square has an area of 1 square inch. Because
1 is a perfect square, the side length of the large square
must be 1 inch. The side length of one of the square
faces is half the side length of the large square. So, the
1
side length of each small cube is • 1 = 0.5 inch.
2
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Chapter 9
b. The side length of the entire cube is three times the
side length of each small cube. So, the side length of
the entire cube is 3(0.5) = 1.5 inches.
S = 6s 2
= 6(1.5)
Section 9.2
9.2 Activity (pp. 362–363)
1. a. Sample answer:
2
18.6 cm
= 6( 2.25)
= 13.5
23 cm
®
The surface area of the entire Rubik’s Cube is
13.5 square inches.
Fair Game Review
1
bh
2
1
= ( 20)(16)
2
= 160
29. A =
The area of the triangle is 160 square feet.
1
bh
2
1
= (12)(9)
2
= 54
30. A =
The area of the triangle is 54 square meters.
1
bh
2
1
= (7)(8)
2
= 28
Scale = 1 cm : 10 m
Lateral surface area:
S = 4(area of one lateral face)
1

= 4 • 230 • 186  = 4( 21,390) = 85,560
2


The lateral surface area of the pyramid is about
85,560 square meters.
b. Sample answer:
27 cm
26 cm
31. A =
The area of the triangle is 28 square feet.
32. B; C = π d
≈ 3.14(9)
= 28.26
The circumference of the basketball is about
28.26 inches.
Scale = 1 cm : 1 m
Lateral surface area:
S = 4(area of one lateral face)
1

= 4 • 26 • 27  = 4(351) = 1404
2

The lateral surface area of the pyramid is about
1404 square meters.
c. Sample answer:
28 cm
35 cm
Scale = 1 cm : 1 m
Lateral surface area:
S = 4(area of one lateral face)
1

= 4 • 35 • 28  = 4( 490) = 1960
2


The lateral surface area of the pyramid is about
1960 square meters.
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Chapter 9
d. Sample answer:
3. a. The square pyramid has the greater surface area; The
29 cm
square pyramid has a larger base. Each lateral face in
both pyramids has the same area, and there are 4
lateral faces in the square pyramid and only 3 lateral
faces in the triangular pyramid.
b. Surface area of the square pyramid:
22 cm
Area of base: 82 = 64
Area of a lateral face:
1
(8)(14) = 56
2
S = Area of base + 4( Area of a lateral face)
Scale = 1 cm : 1 m
Lateral surface area:
= 64 + 4(56)
S = 4(area of one lateral face)
1

= 4 • 22 • 29  = 4(319) = 1276
2

The lateral surface area of the pyramid is about
1276 square meters.
= 288
The surface area of the square pyramid is
288 square inches.
Surface area of the triangular pyramid:
Area of base:
2. a. An octagonal pyramid has eight lateral faces.
1
(8)(6.9) = 27.6
2
Area of a lateral face:
b.
1
(8)(14) = 56
2
S = Area of base + 3( Area of a lateral face)
2 mm
4 mm
= 27.6 + 3(56)
= 195.6
The surface area of the triangular pyramid is 195.6
square inches. Because 288 > 195.6, the square
pyramid has the greater surface area.
c. Lateral surface area:
4. Sample answer:
S = 8(area of one lateral face)
1

= 8 • 2 • 4  = 8( 4) = 32
2


The lateral surface area of the pyramid is 32 square
millimeters.
h
h
b
h
h
b
h
To calculate the surface area of a pyramid whose base
is a regular polygon with x sides, first find the area of
the base. Then find the area of one lateral face. Each
lateral face is a triangle with base b, height h (slant
1
height), and area bh. The surface area is the sum of the
2
area of the base and x times the area of one lateral face.
In a pyramid with a square base, where x = 4,
1 
S = b 2 + 4 bh .
2 
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Chapter 9
9.2 On Your Own (pp. 364 –365)
Practice and Problem Solving
4. Area of base: 4 • 4 = 16
1. Draw a net:
1
•4•3 = 6
2
S = area of base + areas of lateral faces
Area of a lateral face:
9 cm
7 cm
= 16 + 6 + 6 + 6 + 6 = 40
The surface area of the pyramid is 40 square inches.
5. Area of a lateral face:
Area of base: 9 • 9 = 81
1
Area of a lateral face: • 9 • 7 = 31.5
2
S = area of base + areas of lateral faces
= 81 + 31.5 + 31.5 + 31.5 + 31.5
= 207
The surface area of the pyramid is 207 square
centimeters.
2. Draw a net:
1
• 10 • 9 = 45
2
S = area of base + areas of lateral faces
= 43.3 + 45 + 45 + 45 = 178.3
The surface area of the pyramid is 178.3 square
millimeters.
6. Area of a lateral face:
1
• 6 • 6 = 18
2
S = area of base + areas of lateral faces
= 61.9 + 18 + 18 + 18 + 18 + 18 = 151.9
The surface area of the pyramid is 151.9 square meters.
7. Draw a net:
9 ft
6 ft
6 ft
10 ft
Area of base:
5.2 ft
1
• 6 • 5.2 = 15.6
2
Area of base: 6 • 6 = 36
1
Area of a lateral face: • 6 • 10 = 30
2
Area of a lateral face:
S = area of base + areas of lateral faces
S = area of base + areas of lateral faces
= 15.6 + 30 + 30 + 30 = 105.6
The surface area of the pyramid is 105.6 square feet.
3. Because one bundle of shingles covers 32 square feet,
it will take 540 ÷ 32 = 16.875 bundles to cover the roof.
So, you should buy 17 bundles.
1
• 6 • 9 = 27
2
= 36 + 27 + 27 + 27 + 27 = 144
The surface area of the pyramid is 144 square feet.
8. Draw a net:
6 cm
9.2 Exercises (pp. 366 –367)
4 cm
Vocabulary and Concept Check
1. no; Because the lateral faces of a pyramid always meet at
a point to form triangles, the lateral faces of a pyramid
could never be rectangles.
2. Because the lateral faces of a pyramid are triangles, the
slant height is used to find the area of a lateral face,
which is used to find the surface area of the pyramid.
3. triangular pyramid; The base of the pyramid is not a
triangle. The other three are names for the pyramid.
266 Big Ideas Math Red Accelerated
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Area of base: 4 • 4 = 16
1
• 4 • 6 = 12
2
S = area of base + areas of lateral faces
Area of a lateral face:
= 16 + 12 + 12 + 12 + 12 = 64
The surface area of the pyramid is 64 square centimeters.
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Chapter 9
9. Draw a net:
12. Area of a lateral face:
10 yd
9 yd
1
• 9 • 7.8 = 35.1
2
Area of a lateral face:
There are six identical lateral faces. So, the lateral surface
area is
40 + 40 + 40 + 40 + 40 + 40 = 240 square inches,
which is the amount of glass needed.
13. Area of base: 5 • 5 = 25
7.8 yd
Area of base:
1
• 8 • 10 = 40
2
1
• 9 • 10 = 45
2
S = area of base + areas of lateral faces
= 35.1 + 45 + 45 + 45 = 170.1
The surface area of the pyramid is 170.1 square yards.
Slant height:
S = area of base + areas of lateral faces
1

85 = 25 + 4 • 5 • h 
2


85 = 25 + 10h
60 = 10h
6 = h
The slant height of the pyramid is 6 meters.
10. Draw a net:
14. Surface area of rectangular prism
10 in.
(not including the top base):
S = lw + 2lh + 2 wh
= 5(5) + 2(5)( 4) + 2(5)( 4)
13 in.
15 in.
= 25 + 40 + 40
= 105
Surface area of regular pyramid (not including the base):
1
Area of base: • 15 • 13 = 97.5
2
Area of a lateral face:
1
• 15 • 10 = 75
2
S = area of base + areas of lateral faces
= 97.5 + 75 + 75 + 75 = 322.5
The surface area of the pyramid is 322.5 square inches.
11. Draw a net:
20 mm
Area of a lateral face:
1
(5)(6) = 15
2
S = areas of lateral faces
= 4(15)
= 60
Total surface area: 105 + 60 = 165
So, the surface area of the composite solid is 165 square
feet.
16 mm
Area of a lateral face:
1
• 16 • 20 = 160
2
S = area of base + areas of lateral faces
= 440.4 + 160 + 160 + 160 + 160 + 160
= 1240.4
The surface area of the pyramid is 1240.4 square
millimeters.
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267
Chapter 9
15. Surface area of triangular prism
17. a. Area of a lateral face:
(not including the top base):
1
• 4 • 5 = 10
2
Because there are 8 sides of the umbrella, it will take
8 • 10 = 80 square feet of fabric to make the
umbrella.
6 cm
1
2
10 cm
3
10 cm
10 cm
10 cm
b. The most efficient way to cut the fabric is to lay out
the triangles with the longer sides adjacent. That way,
instead of cutting 8 long sides, you only have to cut 5.
10 cm
8.7 cm
Area of bottom base:
1
(10)(8.7) = 43.5
2
Area of a lateral face: 10 • 6 = 60
S = area of bottom base + areas of lateral faces
8 ft
2 ft
The diagram shows the layout for one color of the
fabric or one-half of the triangles needed.
c. There are two pieces of fabric with a width of 5 feet
= 43.5 + 3(60)
= 223.5
Surface area of triangular pyramid
(not including the base):
Area of a lateral face:
5 ft
1
(10)(4) = 20
2
S = areas of lateral faces
= 3( 20)
= 60
Total surface area: 223.5 + 60 = 283.5
So, the surface area of the composite solid is 283.5 square
centimeters.
16. Surface area of rectangular prism:
S = 2lw + 2lh + 2 wh
= 2(12)(5) + 2(12)( 4) + 2(5)( 4)
= 120 + 96 + 40
= 256
and a length of 10 feet. So, the amount of fabric
available is 2(5)(10) = 100 square feet. From part (a),
80 square feet of fabric is needed.
Amount of wasted fabric:
A = (amount of fabric available)
− (amount of fabric needed)
= 100 − 80
= 20
The amount of fabric that is wasted is 20 square feet.
18. Because the pyramid height is the shortest distance from
the top of the pyramid to the base, the slant height is
greater than the pyramid height.
19. Because there are three lateral faces of a tetrahedron, the
area of one lateral face is 93 ÷ 3 = 31 square centimeters.
Because the base is congruent to each lateral face and a
tetrahedron has four faces, the surface area of the
tetrahedron is 4 • 31 = 124 square centimeters.
Surface area of regular pyramid (not including the base):
Area of a lateral face:
1
(5)(5) = 12.5
2
Because the regular pyramid rests on top of the
rectangular prism, remove the area of the base
(5 • 5 = 25) of the regular pyramid from the total.
Total surface area: 256 + 4(12.5) − 25 = 281
So, the surface area of composite solid is 281 square feet.
268 Big Ideas Math Red Accelerated
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Chapter 9
20. Sample answer: Use a visual approach with a square
pyramid.
i. Areas of lateral faces = Area of base
23. The radius is 27 ÷ 2 = 13.5 units.
A = πr2
≈ 3.14(13.5)
2
= 3.14 • 182.25
= 572.265
The area is about 572.265 square units.
C = 2π r ≈ 2(3.14)(13.5) = 84.78
The circumference is about 84.78 units.
If you fold the lateral faces to try and form a pyramid,
you will see that the lateral faces meet at a vertex
when they are lying flat on the base. This does not
form a pyramid.
ii. Areas of lateral faces < Area of base
24. B;
2
60
=
x
3
2 • x = 3 • 60
2 x = 180
2x
180
=
2
2
x = 90
So, the distance between bases on a professional baseball
field is 90 feet.
Section 9.3
If you fold the lateral faces to try and form a pyramid,
you will see that the lateral faces cannot meet at a
vertex. This does not form a pyramid.
So, the total area of the lateral faces is greater than the
area of the base. A similar visual approach can be used
to show that this is true for other pyramids.
Fair Game Review
21. A = π r 2
≈ 3.14(12)
2
= 3.14 • 144
= 452.16
1. Answer should include, but is not limited to:
• A discussion on how to find the area of the outside of
the roll.
• Results are shown for measuring to estimate the
circumference and find the height of the roll with a
ruler, and work is shown for estimating the area by
multiplying.
• Work is shown for finding the amount of the flattened
cardboard, and the results are compared to the
estimate.
2. Answer should include, but is not limited to: a paper net
The area is about 452.16 square units.
C = 2π r ≈ 2(3.14)(12) = 75.36
The circumference is about 75.36 units.
22. A = π r 2
≈ 3.14(8)
9.3 Activity (pp. 368 –369)
2
= 3.14 • 64
= 200.96
The area is about 200.96 square units.
C = 2π r ≈ 2(3.14)(8) = 50.24
The circumference is about 50.24 units.
for the can with the shapes described as two circles and
one rectangle; an explanation that one dimension of the
rectangle is the circumference of the can, and the other
dimension of the rectangle is the height of the can; an
explanation of how to find the surface area of the can by
finding the areas of the shapes in the net.
3. Sample answers:
a. radius ≈ 1.5 in.
height ≈ 4 in.
S = 2(area of circle) + area of lateral surface
= 2(π r 2 ) + Ch
= 2π r 2 + 2π rh
≈ 2(3.14)(1.5) + 2(3.14)(1.5)( 4)
2
= 51.81
The surface area of the soup can is about
52 square inches.
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269
Chapter 9
b. radius ≈ 1.25 in.
5. c;
π ≈ 3.14159265
height ≈ 4.5 in.
3.14 = 3.14000000
S = 2(area of circle) + area of lateral surface
22
≈ 3.142857143
7
355
≈ 3.14159292
113
= 2(π r 2 ) + Ch
= 2π r 2 + 2π rh
≈ 2(3.14)(1.25) + 2(3.14)(1.25)( 4.5)
2
π is closest to
= 45.1375
The surface area of the iced tea can is about
45 square inches.
c. radius ≈ 2.75 in.
355
.
113
9.3 On Your Own (pp. 370 –371)
1. S = 2π r 2 + 2π rh
= 2π (6) + 2π (6)(9)
2
height ≈ 6 in.
S = 2(area of circle) + area of lateral surface
= 2(π r 2 ) + Ch
= 72π + 108π
= 180π
≈ 565.2
= 2π r 2 + 2π rh
The surface area is about 565.2 square yards.
≈ 2(3.14)( 2.75) + 2(3.14)( 2.75)(6)
2
= 151.1125
= 2π (3) + 2π (3)(18)
2
The surface area of the vegetable shortening can is
about 151 square inches.
= 18π + 108π
= 126π
d. radius ≈ 1.75 in.
height ≈ 1.5 in.
≈ 395.6
S = 2(area of circle) + area of lateral surface
= 2(π r 2 ) + Ch
The surface area is about 395.6 square centimeters.
3. a. yes; The new height of the can of peas is
2 • 2 in. = 4 in.
= 2π r 2 + 2π rh
≈ 2(3.14)(1.75) + 2(3.14)(1.75)(1.5)
2
= 35.7175
The surface area of the tuna fish can is about
36 square inches.
4. Sample answer:
2. S = 2π r 2 + 2π rh
3 in.
Amount of paper used: S = 2π rh = 2π (1)( 4) = 8π
The amount of paper needed for the label is 8π square
inches. So, when the height doubles, the amount of
paper used in the label doubles.
b. no; Surface area of the new can:
S = 2π r 2 + 2π rh
= 2π (1) + 2π (1)( 4)
2
10 in.
= 2π + 8π
= 10π
Recycle value of the new can:
To find the surface area of a cylinder, add twice the area
of a circular base to the product of the circumference of
the base and the height of the cylinder.
S = 2π r 2 + 2π rh
≈ 2(3.14)(3) + 2(3.14)(3)(10)
2
= 244.92
The surface area is about 245 square inches.
270 Big Ideas Math Red Accelerated
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10π in.2
6π in.2
=
x
$0.01
10π • 0.01 = x • 6π
5
• 0.01 = x
3
5
• 0.01 is not 2 • 0.01, the recycle value
3
does not double. Only the lateral surface area doubled.
So, doubling the height of the can does not double the
recycle value.
Because
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Chapter 9
9.3 Exercises (pp. 372–373)
6. S = 2π r 2 + 2π rh
Vocabulary and Concept Check
1. Because the lateral surface area is the product of the
circumference of the base and the height of the cylinder,
it is represented by 2π rh.
2. To find the surface area of the entire cylinder, substitute
the value of C into C = 2π r and solve for r. Then use
the formula for the surface area of a cylinder.
Practice and Problem Solving
3.
= 2π ( 2) + 2π ( 2)(5)
2
= 8π + 20π
= 28π
≈ 87.92
The surface area is about 87.9 square millimeters.
7. S = 2π r 2 + 2π rh
= 2π (6) + 2π (6)(7)
2
= 72π + 84π
3 ft
= 156π
2 ft
≈ 489.84
The surface area is about 489.8 square feet.
8. S = 2π r 2 + 2π rh
3 ft
= 2π (3) + 2π (3)(12)
2
S = 2π r 2 + 2π rh
= 18π + 72π
= 2π (3) + 2π (3)( 2)
2
= 90π
= 18π + 12π
= 30π
≈ 282.6
≈ 94.2
The surface area is about 94.2 square feet.
4.
The surface area is about 282.6 square centimeters.
9. S = 2π rh = 2π (10)(6) = 120π ≈ 376.8
The lateral surface area is about 376.8 square feet.
1m
10. S = 2π rh = 2π ( 4)(9) = 72π ≈ 226.08
4m
The lateral surface area is about 226.1 square inches.
11. S = 2π rh = 2π (7 )( 2) = 28π ≈ 87.92
1m
The lateral surface area is about 87.9 square meters.
S = 2π r 2 + 2π rh
12. The area of the base must be multiplied by 2 because
= 2π (1) + 2π (1)( 4)
2
there are two circular bases.
= 2π + 8π
S = 2π r 2 + 2π rh
= 10π
= 2π (5) + 2π (5)(10.6)
2
≈ 31.4
The surface area is about 31.4 square meters.
5.
= 50π + 106π
= 156π
≈ 489.8 yd 2
7 ft
5 ft
13. S = 2π r 2 + 2π rh
= 2π ( 4) + 2π ( 4)(50)
2
7 ft
S = 2π r 2 + 2π rh
= 2π (7) + 2π (7)(5)
2
= 98π + 70π
= 32π + 400π
= 432π
≈ 1356.48
The surface area of the tank is about 1356.48 square feet.
= 168π
≈ 527.52
The surface area is about 527.5 square feet.
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271
Chapter 9
14. Surface area of the ottoman (not including the bottom):
b. Surface area (except for cut) of cut wedge:
= 256π + 448π
1
(surface area of cheese before cut )
8
1
= ( 24π )
8
= 704π in.2
= 3π in.2
S = π r + 2π rh
2
S =
= π (16) + 2π (16)(14)
2
Surface area that is green:
Surface area of exposed cheese:
S = 2π rh = 2π (16)(8) = 256π in.2
S = 2 •  • w = 2(3)(1) = 6 in.2
Percent of the surface area of the ottoman that is green:
Surface area of remaining the cheese:
256π = p • 704π
S = (surface area of cheese before cut )
p ≈ 0.364 = 36.4%
The percent of the surface area of the ottoman that is
green is about 36.4%.
15. The lateral surface areas of the two cylinders are the
same. Because the cylinder with a height of 8.5 inches
has a larger radius, the areas of its bases are greater. So,
the cylinder with a height of 8.5 inches has the greater
surface area.
16. a. Surface area of the smaller ganza:
2
= 21π + 6
≈ 71.94
The remaining surface area of the cheese is about
71.9 square inches. Because 71.9 square inches is less
than 75.4 square inches. The surface area decreased.
S = 2π ( 2r ) + 2π ( 2r )( 2h)
2
= 6.125π + 35π
= 2π • 4r 2 + 2π • 4rh
= 41.125π
Surface area of the larger ganza:
S = 2π r 2 + 2π rh
= 2π ( 2.75) + 2π ( 2.75)( 24.5)
2
= 15.125π + 134.75π
= 4( 2π r 2 + 2π rh)
Multiplying both dimensions by a factor of 2 results in
a surface area that is 4 times greater.
Factor of 3:
S = 2π (3r ) + 2π (3r )(3h)
2
= 149.875π
The surface areas of the smaller and larger ganzas are
41.125π , or 129.1 square centimeters and
149.875π , or 470.6 square centimeters, respectively.
149.875π cm
41.125π cm
=
x lb
1.1 lb
149.875π • 1.1 = x • 41.125π
= 24π − 3π + 6
Factor of 2:
= 2π (1.75) + 2π (1.75)(10)
b.
+ (surface area of exposed cheese)
18. a. Original surface area: S = 2π r 2 + 2π rh
S = 2π r 2 + 2π rh
2
− (surface area of cut wedge)
2
3.644 • 1.1 ≈ x
4.0 ≈ x
The weight of the larger ganza is about 4.0 pounds.
17. a. Surface area of cheese before cut:
S = 2π r 2 + 2π rh
= 2π (3) + 2π (3)(1)
2
= 2π • 9r 2 + 2π • 9rh
= 9( 2π r 2 + 2π rh)
Multiplying both dimensions by a factor of 3 results in
a surface area that is 9 times greater.
Factor of 5:
S = 2π (5r ) + 2π (5r )(5h)
2
= 2π • 25r 2 + 2π • 25rh
= 25( 2π r 2 + 2π rh)
Multiplying both dimensions by a factor of 5 results in
a surface area that is 25 times greater.
Factor of 10:
S = 2π (10r ) + 2π (10r )(10h)
2
= 18π + 6π
= 2π • 100r 2 + 2π • 100rh
= 24π ≈ 75.36 in.2
= 100( 2π r 2 + 2π rh)
Multiplying both dimensions by a factor of 10 results
in a surface area that is 100 times greater.
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Chapter 9
b. Multiplying both dimensions by a factor of x results in
a surface area that is x 2 times greater than the original
surface area. So, if both dimensions of a cylinder are
multiplied by a factor of 20, the new surface area is
202 = 400 times greater.
Fair Game Review
21. A =
=
=
=
1
1
bh = (8)( 4) = 16 cm 2
2
2
1
h (b + B )
2
1
(5)(7 + 12)
2
1
(5)(19)
2
47.5 in.2
= 28 + 56 + 16
= 100
The surface area of the prism is 100 square millimeters.
1

= 65 + 6 • 5 • 12 
2


= 65 + 180
= 245
The surface area of the pyramid is 245 square meters.
4. S = area of base + areas of lateral faces
1

= 2 • 2 + 4 • 2 • 6 
2

= 4 + 24
= 28
The surface area of the pyramid is 28 square centimeters.
a = p•w
22. C;
= 2(7)( 2) + 2(7)( 4) + 2( 2)( 4)
3. S = area of base + areas of lateral faces
19. A = w = (5)( 2) = 10 ft 2
20. A =
2. S = 2w + 2h + 2 wh
80 = 0.4 • w
5. S = 2π r 2 + 2π rh
200 = w
= 2π (3) + 2π (3)(10)
2
So, 40% of 200 is 80.
= 18π + 60π
Study Help
= 78π
Available at BigIdeasMath.com.
≈ 244.92
The surface area is about 244.9 square feet.
Quiz 9.1–9.3
1.
3 cm
6. S = 2π r 2 + 2π rh
4 cm
= 2π (5) + 2π (5)(6)
2
5 cm
= 50π + 60π
= 110π
1
2
3
10 cm
≈ 345.4
The surface area is about 345.4 square meters.
7. S = 2π rh
3 cm
4 cm
1
Area of a base: (3)( 4) = 6
2
Area of face 1: 3 • 10 = 30
Area of face 2: 5 • 10 = 50
Area of face 3: 4 • 10 = 40
S = areas of bases + areas of lateral faces
= 6 + 6 + 30 + 50 + 40
= 132
The surface area of the prism is 132 square centimeters.
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= 2π (9)(7)
= 126π
≈ 395.6
The lateral surface area is about 395.6 square centimeters.
8. The diameter is 8 millimeters, so the radius is
4 millimeters.
S = 2π rh
= 2π ( 4)(12.2)
= 97.6π
≈ 306.5
The lateral surface area is about 306.5 square millimeters.
Big Ideas Math Red Accelerated
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273
Chapter 9
9. a. Calculate the lateral surface area of the pyramid:
S = 12(area of a lateral face)
1 
1

= 12 bh  = 12 • 1 • 3 = 12(1.5) = 18
2 
2

The lateral surface area is 18 square feet, which is the
amount of glass needed to make the skylight.
b. yes; The amount of glass needed is 18 square feet and
the sheet of glass is 4 ft × 8 ft = 32 square feet. The
12 glass triangles can be cut from the sheet of glass.
8 ft
2. a. Because the area of the base B is found by multiplying
the length and the width, the volume can also be
represented by V = Bh.
b. Volume of Prism 1:
V = wh = 3 • 2 • 1 = 6 units3
V = Bh = 6 • 1 = 6 units3
Volume of Prism 2:
V = wh = 3 • 2 • 2 = 12 units3
V = Bh = 6 • 2 = 12 units3
Volume of Prism 3:
2
4
6
8
10 12
4 ft
3 ft
1
3
5
7
9
11
6.5 ft
10. Length of the tube:
3 ft ×
12 in.
= 36 in.
1 ft
Lateral surface area of the tube:
S = 2π rh = 2π (1.5)(36) = 108π ≈ 339.12
The lateral surface area is about 339.12 square inches,
which is the least amount of material needed to make the
mailing tube.
11. 4 feet •
12 inches
= 48 inches
1 foot
S = 5s 2 = 5( 48) = 11,520
2
So, the area of the chest to be painted is 11,520 square
inches.
Section 9.4
9.4 Activity (pp. 376 –377)
1. a. Because each pearl is about 1 cubic centimeter, the
number of pearls in the chest can be estimated by
multiplying the length, width, and height of the chest.
b. The number of pearls in the chest can be estimated by
V = wh = 3 • 2 • 3 = 18 units3
V = Bh = 6 • 3 = 18 units3
Volume of Prism 4:
V = wh = 3 • 2 • 4 = 24 units3
V = Bh = 6 • 4 = 24 units3
Volume of Prism 5:
V = wh = 3 • 2 • 5 = 30 units3
V = Bh = 6 • 5 = 30 units3
Both formulas give the same volume.
3. To find the volume of any prism, first find the area of the
base and then multiply it by the height. V = Bh
4. a. A single sheet of paper does have a volume because it
has a length, width, and height. The height would be a
very small number.
b. The length and width of a single sheet of paper can be
easily measured, but the height is difficult to measure
because the paper is so thin. To find the height of a
single sheet, find the height of the ream of paper and
then divide by 500. Then find the volume of a single
sheet of paper by multiplying its length, width, and
height.
5. To find the volume of a prism, calculate the area of the
base, and then multiply it by the height.
6. Sample answer:
counting and weighing a small number of pearls and
by weighing the entire chest of pearls. A proportion
can be used to find the number of pearls in the chest.
6 cm
c. Number of pearls in the chest:
5 cm
120 • 60 • 60 = 432,000
There are about 432,000 pearls in the chest. Because
each pearl is worth about $80, the value of the pearls
in the chest is about 80 • 432,000 = $34,560,000.
7 cm
8 cm
Area of a base:
A =
1
1
1
(b1 + b2 )h = (6 + 8)(5) = (14)(5) = 35 cm 2
2
2
2
V = Bh = 35 • 7 = 245 cm3
The volume of the prism is 245 cubic centimeters.
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Chapter 9
9.4 On Your Own (p. 379)
1. V = s
3
3
= 4 = 64
The volume is 64 cubic feet.
2. V = Bh =
1
(12)(5) • 9 = 30 • 9 = 270
2
The volume is 270 cubic meters.
3. Height of Bag C:
V = Bh
96 = ( 4)( 4.8)( h)
96 = 19.2h
5 = h
7. V = Bh
 1  1 
=  8  4  • 6
 3  5 
25 21
=
•
•6
3
5
5
1
= 210
The volume is 210 cubic yards.
The height is 5 inches.
S = w + 2h + 2 wh
= ( 4.8)( 4) + 2( 4.8)(5) + 2( 4)(5)
= 19.2 + 48 + 40
= 107.2 in.2
The surface area of Bag C is less than the surface areas of
both Bag A and Bag B. So, the theater should choose
your bag.
9.4 Exercises (pp. 380 –381)
Vocabulary and Concept Check
1. Because the volume is calculated from three dimensions,
volume is measured in cubic units.
2. The volume V of a prism is the product of the area of the
base B and the height h of the prism. Use the formula
V = Bh.
3. Volume describes the space occupied by a solid and
surface area describes the sum of the areas of the faces of
a solid.
Practice and Problem Solving
4. V = s3 = 93 = 729
The volume is 729 cubic inches.
5. V = Bh =
1
(6)(8) • 12 = 24 • 12 = 288
2
The volume is 288 cubic centimeters.
1
1
6. V = Bh = 4(7) • 8 = 28 • 8 = 238
2
2
The volume is 238 cubic meters.
1
= 5•7•6
8. V = Bh =
Amount of paper needed:
7
25 • 21
=
•6
3 • 5
1
(4.5)(6) • 9 = 13.5 • 9 = 121.5
2
The volume is 121.5 cubic feet.
9. V = Bh =
1
1
(10.5)(10) • 8 = • 105 • 8 = 420
2
2
The volume is 420 cubic millimeters.
10. V = Bh =
1
(7.2)( 4.8) • 10 = 17.28 • 10 = 172.8
2
The volume is 172.8 cubic meters.
11. V = Bh = 43 • 15 = 645
The volume is 645 cubic millimeters.
12. V = Bh = 166 • 20 = 3320
The volume is 3320 cubic feet.
13. The base of the prism is a triangle, so use the formula for
the area of a triangle for B.
V = Bh =
1
(5)(7) • 10 = 17.5 • 10 = 175 cm3
2
14. Volume of School Locker:
V = Bh = (10)(12) • 60 = 120 • 60 = 7200 in.3
Volume of Gym Locker:
V = Bh = (15)(12) • 48 = 180 • 48 = 8640 in.3
Because the volume of the gym locker is greater than the
volume of the school locker, the gym locker has more
storage space.
15. V = Bh = (9)( 2.5) • 10 = 22.5 • 10 = 225
The volume of the box is 225 cubic inches.
16. V = Bh = (12)(10) • 12 = 120 • 12 = 1440
The volume of the prism is 1440 cubic inches.
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Chapter 9
17. V = Bh =
1
(20)(30) • 24 = 300 • 24 = 7200
2
The volume of the prism is 7200 cubic feet.
23. Sample answer:
Convert the volume to cubic inches:
450 gal ×
18. sometimes; Two prisms that have the same volume can
231 in.3
= 103,950 in.3
1 gal
have the same surface area when the areas of the faces
are the same. Two prisms that have the same volume can
have different surface areas if one prism is longer than
the other.
21 in.
99 in.
19.
1 ft
50 in.
V = Bh = (50)(99) • 21 = 4950 • 21 = 103,950 in.3
1 ft
1 ft
Because 1 foot equals 12 inches, 1 cubic foot is equal to
12 • 12 • 12 = 1728 cubic inches.
20. Volume inside the calendar:
V = Bh =
1
(4)(6) • 8 = 12 • 8 = 96 in.3
2
Because each packet has a volume of 2 cubic inches, you
can fit at most 96 ÷ 2 = 48 packets of chocolate inside
the calendar.
21. Amount of water in cubic centimeters:
2 L ×
3
1000 cm
= 2000 cm3
1 L
Height of the water:
V = Bh
2000 = 100 • h
20 = h
The height of the water is 20 centimeters.
22. Sample answer:
Calculate the dimensions of the tank in inches:
12 in.
= 15 in.
length: 1.25 ft ×
1 ft
12 in.
= 21 in.
width: 1.75 ft ×
1 ft
Volume of the tank:
V = Bh = (15)( 21) • 11 = 315 • 11 = 3465 in.3
Because one gallon equals 231 cubic inches, the volume
of the tank is 3465 ÷ 231 = 15 gallons.
The dimensions of the tank shown are
21 inches × 50 inches × 99 inches.
24. V = Bh
1 
=  w  h
2 
1
= wh
2
Double dimension h: 2h
V = Bh
1 
=  w  2 h
2 
= wh
1

2 wh  = wh
2

When one of its dimensions is doubled, a triangular
prism’s volume is 2 times greater.
Double all three dimensions:
V = Bh
1

=  • 2 • 2 w  2 h
2

= ( 2w)2h
= 4wh
1

8 wh  = 4wh
2

When all three of its dimensions are doubled, a triangular
prism’s volume is 8 times greater.
Amount of gas needed to fill the tank:
Gas needed = 80% ( Volume of tank )
= 0.8(15)
= 12 gallons
Because 12 gallons are needed to fill the tank, if the price
of gas is $3.00 per gallon, it will cost about $36 to fill the
tank.
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Chapter 9
Fair Game Review
25. a = p • w
3.
= 0.20 • 75
= 15
selling price = cost to store + markup
=
75
=
90
+
15
So, the selling price is $90.
26. a = p • w
= 0.60 • 90
= 54
selling price = cost to store + markup
=
90
=
144
+
54
Height
1
2
3
4
5
Area of Base
1
4
9
16
25
Volume
1
3
8
3
9
64
3
125
3
The volume is one-third the area of the base multiplied by
the height. Because a pyramid with a base length and a
height of 20 has a base with an area of 400 square units,
1
8000
the volume of the pyramid is • 400 • 20 ≈
3
3
cubic units.
4. Volume of rectangular prism:
V = wh = (5)(3)( 2) = 30 units 3
Volume of Pyramid a:
V =
So, the selling price is $144.
27. a = p • w
1
1
Bh = (5 • 3)( 2) = 10 units3
3
3
Volume of Pyramid b:
= 0.85 • 130
V =
= 110.5
selling price = cost to store + markup
=
130
=
240.5
+ 110.5
So, the selling price is $240.50.
28. D;
S = 2π r 2 + 2π rh
= 2π (3) + 2π (3)(10)
2
= 18π + 60π
= 78π
≈ 244.92
So, the surface area is approximately 245 square inches.
Section 9.5
Volume of Pyramid c:
V =
of a prism is V = Bh, then the formula for the volume of
1
a pyramid is V = Bh.
3
2. Sample answer: The size of the pyramid and the size of
the pieces used would affect how long it took to build the
pyramid. The Cheops Pyramid in Egypt took longer to
build because its volume is nearly twice the volume of
the Sun Pyramid in Mexico.
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1
1
Bh = (5 • 2)(3) = 10 units3
3
3
The sum of the volumes of the pyramids is
10 + 10 + 10 = 30 cubic units, which is the same as the
volume of the rectangular prism.
5. To find the volume of a pyramid, multiply one-third by
the product of the area of the base and the height.
6. The general formula for the volume of a pyramid is
V =
1
Bh.
3
9.5 On Your Own (p. 385)
1. V =
9.5 Activity (pp. 382–383)
1. It takes three pyramids to fill the box. So, if the volume
1
1
Bh = (3 • 2)(5) = 10 units3
3
3
1
1
Bh = ( 21)(6) = 42
3
3
The volume is 42 cubic feet.
2. V =
1
1
2
Bh = (10)(8)(7) = 186
3
3
3
2
The volume is 186 cubic inches.
3
3. V =
1
1 1

Bh =  • 7 • 18(11) = 231
3
3 2

The volume is 231 cubic centimeters.
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277
Chapter 9
4. Volume of Bottle C:
1
1
V = Bh = (3)( 2)(3) = 6 in.3
3
3
10. V =
The volume is 112 cubic feet.
Unit cost of Bottle C:
cost
$13.20
$2.20
=
=
volume
6 in.3
1 in.3
Because the unit cost of Bottle C is less than the unit cost
of Bottle B, Bottle C is the better buy.
9.5 Exercises (pp. 386 –387)
Vocabulary and Concept Check
1. The formula for the volume of a pyramid is one-third
times the area of the base times the height. The formula
for the volume of a prism is the area of the base times the
height.
2. Sample answer: The amount of water in a vase that is
shaped like a pyramid can be found by calculating the
volume.
3. Because the formula for the volume of a pyramid is
one-third the formula for the volume of a prism, the
volume of the prism is three times the volume of the
pyramid.
Practice and Problem Solving
4. V =
1
1
1
Bh = ( 2)(1)( 2) = 1
3
3
3
1
The volume is 1 cubic feet.
3
5. V =
1
1
Bh = (15)( 4) = 20
3
3
The volume is 20 cubic millimeters.
6. V =
1
1 1
2

Bh =  • 5 • 4 (8) = 26
3
3 2
3

2
The volume is 26 cubic yards.
3
7. V =
1
1 1

Bh =  • 10 • 6 (8) = 80
3
3 2

The volume is 80 cubic inches.
8. V =
1
1
Bh = (3)(1)(7) = 7
3
3
The volume is 7 cubic centimeters.
9. V =
1
1
Bh = (63)(12) = 252
3
3
The volume is 252 cubic millimeters.
1
1
Bh = (6)(8)(7) = 112
3
3
11. V =
1
1 1

Bh =  • 20 • 14 (15) = 700
3
3 2

The volume is 700 cubic millimeters.
12. V =
1
1
Bh = (12)(12)(12) = 576
3
3
The volume of air inside the parachute is 576 cubic yards.
13. Volume of prism:
V = lwh
= 6(6)(3)
= 108
Volume of pyramid:
1
V = Bh
3
1
= (6)(6)( 4)
3
= 48
Total volume: 108 + 48 = 156
So, the volume of the composite solid is 156 cubic feet.
14. Volume of prism:
V = lwh
= 6(6)( 4)
= 144
Volume of pyramid:
1
V =
Bh
3
1
= (6)(6)(8)
3
= 96
Total volume: 144 + 96 = 240
So, the volume of the composite solid is 240 cubic
meters.
15. Volume of prism: V = Bh
1
(8)(6.9) • 10
2
= 27.6 • 10
= 276
=
1
Bh
3
1 1 
=  (8)(6.9)(7)
3 2 
= 64.4
Volume of pyramid: V =
Total volume: 276 + 64.4 = 340.4
So, the volume of the composite solid is 340.4 cubic
inches.
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Chapter 9
16. Volume of Spire A:
1
1
A = Bh = (30)(6) = 60 in.3
3
3
Volume of Spire B:
V =
1
1
Bh = ( 24)(8) = 64 in.3
3
3
Spire B has a greater volume than Spire A. You need
64 − 60 = 4 cubic inches more sand to make Spire B.
17. V =
1
1
Bh = (3)(3)( 4) = 12
3
3
The volume of one paperweight is 12 cubic inches. So,
the volume of 1000 paperweights is 1000 • 12 = 12,000
cubic inches, which is the amount of glass needed.
18. Sample answer:
a. The shape of the base is a 12-sided polygon, or a
dodecagon, because there are 12 support sticks coming
out of the tepee.
b. The width of the base of the tepee appears to be about
the same as its height.
Because the base is close to a circle with a diameter of
10 feet, the area of the base is about
π (5) = 25π ≈ 78.5 square feet.
2
1
1
V = Bh = (78.5)(10) ≈ 262
3
3
The volume of the tepee is about 262 cubic feet.
19. Sample answer:
Area of the base:
1
Bh
3
1
40 = ( B)(6)
3
40 = 2 B
V =
Fair Game Review
21. Two angles are supplementary when the sum of their
measures is 180°.
So, 180° − 27° = 153°.
Two angles are complementary when the sum of their
measures is 90°.
So, 90° − 27° = 63°.
22. Two angles are supplementary when the sum of their
measures is 180°.
So, 180° − 82° = 98°.
Two angles are complementary when the sum of their
measures is 90°.
So, 90° − 82° = 8°.
23. Two angles are supplementary when the sum of their
measures is 180°.
So, 180° − 120° = 60°.
Because 120° is greater than 90°, there is no
complementary angle.
24. C; Use the circumference to find the radius of the circle.
Then use the radius to approximate the area of the circle.
C = 2π r
44 ≈ 2(3.14)r
22 ≈ 3.14r
21 ≈ 3r
7 ≈ r
A = π r 2 ≈ 3.14(7) = 153.86
2
The area of the circle is about 154 square inches.
9.5 Extension (pp. 388 –389)
Practice
1. The intersection is a triangle.
20 = B
2. The intersection is a triangle.
The area of the base is 20 square feet. So, one possible set
of dimensions of the base are 4 feet by 5 feet.
3. The intersection is a rectangle.
20. Volume of rectangular prism:
4. The intersection is a rectangle.
V = Bh = ( x )( y )( z ) = xyz
5. The intersection is a triangle.
Volume of rectangular pyramid:
6. The intersection is a rectangle.
V =
1
1
Bh = ( x)( y )(3 z ) = xyz
3
3
Because the volume of the prism is represented by xyz
and the volume of the pyramid is represented by xyz, the
solids have the same volume.
7. The intersection is the shape of the base.
8. The intersection is a rectangle.
9. The intersection is a circle.
10. The intersection is a line segment.
11. The intersection is a circle.
12. The shape is a circle.
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Chapter 9
13. The shape is a rectangle.
2.
14. The shape is a circle.
17 cm
1
15. The intersection occurs at the vertex of the cone.
15 cm
Quiz 9.4–9.5
1. V = Bh = 7(3) • 8 = 21 • 8 = 168
17 cm
2
The volume is 168 cubic inches.
2. V = Bh =
1
(15)(6) • 8 = 45 • 8 = 360
2
7 cm
The volume is 360 cubic feet.
3. V = Bh =
Area of a base:
1
(10)(8) • 12 = 40 • 12 = 480
2
4. V = Bh = (197)( 25) = 4925
= 60 + 60 + 119 + 105 + 56
The volume is 4925 cubic millimeters.
1
1
Bh = (166)(12) = 664
3
3
The volume is 664 cubic feet.
6. V =
= 400
The surface area is 400 square centimeters.
3.
3m
4m
3m
4m
5m
1
1 1

Bh =  • 5 • 2 (3) = 5
3
3 2

The volume is 5 cubic meters.
1
• 8 • 15 = 60
2
Area of face 1: 7 • 17 = 119
Area of face 2: 7 • 15 = 105
Area of face 3: 7 • 8 = 56
S = areas of bases + areas of lateral faces
The volume is 480 cubic yards.
5. V =
8 cm
8 cm
3
1
2
3
8m
7. The intersection is a pentagon.
8. The intersection is a rectangle.
9. V =
1
1
Bh = ( 40)( 40)( 20) ≈ 10,666.7
3
3
The volume of the roof is about 10,666.7 cubic feet.
10.
1 yd = 3 ft
1 yd = 3 ft
1 yd = 3 ft
Because 1 yard equals 3 feet, 1 cubic yard is equal to
3 • 3 • 3 = 27 cubic feet.
Area of a base:
1
•3•4 = 6
2
Area of face 1: 3 • 8 = 24
Area of face 2: 5 • 8 = 40
Area of face 3: 4 • 8 = 32
S = areas of bases + areas of lateral faces
= 6 + 6 + 24 + 40 + 32
= 108
The surface area is 108 square meters.
Chapter 9 Review
1. S = 2w + 2h + 2 wh
= 2(8)(3) + 2(8)(5) + 2(3)(5)
= 48 + 80 + 30
= 158
The surface area is 158 square inches.
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Chapter 9
7. S = 2π r 2 + 2π rh
4.
3 in.
= 2π (3) + 2π (3)(6)
2
= 18π + 36π
2 in.
= 54π
≈ 169.56
The surface area is about 169.6 square yards.
Area of a base: 2 • 2 = 4
1
Area of a lateral face: • 2 • 3 = 3
2
S = area of base + areas of lateral faces
= 4+3+3+3+3
= 16
The surface area is 16 square inches.
5.
8. S = 2π r 2 + 2π rh
= 2π (0.8) + 2π (0.8)(6)
2
= 1.28π + 9.6π
= 10.88π
≈ 34.16
The surface area is about 34.2 square centimeters.
9. S = 2π rh
= 2π ( 4)(11)
8m
= 88π
≈ 276.32
The lateral surface area is about 276.3 square centimeters.
10 m
8m
6.9 m
10. V = Bh = (8)( 2)(6) = 96
The volume is 96 cubic inches.
Area of base:
1
• 8 • 6.9 = 27.6
2
Area of a lateral face:
1
• 8 • 10 = 40
2
S = area of base + areas of lateral faces
= 27.6 + 40 + 40 + 40
= 147.6
The surface area is 147.6 square meters.
11. V = Bh =
1
(4)(7.5)(8) = 120
2
The volume is 120 cubic meters.
12. V = Bh = (15)( 4.5)(9) = 607.5
The volume is 607.5 cubic millimeters.
13. V =
1
1 1 
Bh =  (17)(15)( 20) = 850
3
3 2 
The volume is 850 cubic feet.
6.
9 cm
14. V =
7 cm
84.3 cm2
1
1
Bh = ( 210)(30) = 2100
3
3
The volume is 2100 cubic inches.
15. V =
1
1
Bh = (8)(8)(9) = 192
3
3
The volume is 192 cubic millimeters.
1
Area of a lateral face =
• 7 • 9 = 31.5
2
16. The intersection is a rectangle.
S = area of base + 5(area of a lateral face)
17. The intersection is a triangle.
= 84.3 + 5(31.5)
= 84.3 + 157.5
= 241.8
The surface area is 241.8 square centimeters.
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Chapter 9
Chapter 9 Test
9. Surface area:
S = areas of bases + areas of lateral faces
1. S = 2w + 2h + 2 wh
= 2(5)( 2) + 2(5)(3) + 2( 2)(3)
1
= 2 • 14 •
2
= 84 + 296.4
= 20 + 30 + 12
= 62
= 497.4
The surface area is 62 square feet.
Because you will need two coats of paint, the surface
area of the part of the ramp to be painted is
2 • 497.4 = 994.8 square feet. Since one quart of paint
covers 80 square feet and 994.8 ÷ 80 ≈ 12.4 quarts, you
should buy 13 quarts of paint.
2. S = area of base + areas of lateral faces
1
= 1 • 1 + 4 • 1 •
2
=1+ 4

2

= 5

6  + 19.5 • 15.2 + 19.5 • 6

+ 117
10. a. Doubled height: h = 2(9 in.) = 18 in.
S = 2w + 2h + 2 wh
The surface area is 5 square inches.
= 2(6)( 2) + 2(6)(18) + 2( 2)(18)
3. S = area of base + areas of lateral faces
= 24 + 216 + 72
1
1

• 11 • 9.5 + 3 • 11 • 15
2
2

= 52.25 + 247.5
Doubled width: w = 2( 2 in.) = 4 in.
= 299.75
S = 2w + 2h + 2 wh
=
= 312 in.2
= 2(6)( 4) + 2(6)(9) + 2( 4)(9)
The surface area is about 299.8 square meters.
= 48 + 108 + 72
4. S = 2π r 2 + 2π rh
= 228 in.2
= 2π ( 2) + 2π ( 2)(3)
2
Doubling the width uses less cardboard because it
does not increase the surface area as much.
= 8π + 12π
= 20π
b. V = Bh
≈ 62.8
= 6( 4) • 9
The surface area is about 62.8 square centimeters.
= 24 • 9
5. The diameter is 22 inches, so the radius is 11 inches.
= 216
The volume of the new graham cracker box is
216 cubic inches.
S = 2π r 2 + 2π rh
= 2π (11) + 2π (11)(12.5)
2
= 242π + 275π
= 517π
≈ 1623.38
The surface area is about 1623.4 square inches.
6. V = Bh =
1
(12)(6)(9) = 324
2
The volume is 324 cubic inches.
7. V = Bh
= 4( 2) • 5.2
= 8 • 5.2
= 41.6
The volume is 41.6 cubic yards.
8. V =
1
1
Bh = (8)(3)(6) = 48
3
3
11.
S = 2π rh
354.2 = 2π ( 4.7)( h)
354.2 = 9.4π h
354.2
9.4π h
=
9.4π
9.4π
h ≈ 12
The height of the can is about 12 centimeters.
Chapter 9 Standards Assessment
1. D; S = 2w + 2h + 2 wh
= 2(8)( 4) + 2(8)( 2) + 2( 4)( 2)
= 64 + 32 + 16
= 112
So, the least amount of wrapping paper is 112 square
inches.
The volume is 48 cubic meters.
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Chapter 9
2. G; percent of increase =
new score − original score
original score
660 − 600
600
60
=
600
1
=
10
= 0.1, or 10%
=
9. A; The two angles make up a right angle. So, the angles
are complementary angles and the sum of their measures
is 90°.
46 + ( 2 x + 4) = 90
50 + 2 x = 90
− 50
2 x = 40
2x
40
=
2
2
x = 20
Her score increased 10%.
3. B; Raj should distribute 8 to get 8 x − 24.
3
x −3
=
8
24
3 • 24 = ( x − 3) • 8
72 = 8 x − 24
96 = 8 x
10. H; The sum of the angle measures of a triangle is 180°.
60° + 50° + 20° = 130°
130° < 180°
40° + 80° + 90° = 210°
210° > 180°
30° + 60° + 90° = 180°
12 = x
change in y
6
=
= 3
4. H; slope =
change in x
2
0° + 90° + 90° = 180°, but 0° is not an angle
measure.
11. Part A:
5. 152;
James wants to lose 8 pounds.
new weight = 160 − 8 = 152 pounds
Cost (dollars)
y
a = p • w = 0.05 • 160 = 8
6. C; S = π r 2 + 2π rh
35
30
25
20
15
10
5
1 2 3 4 5 6 7 x
= π ( 4) + 2π ( 4)(9.5)
2
= 16π + 76π
Matinee tickets
Part B:
= 92π
7. I; 8 pints ×
2 cups
= 16 cups
1 pint
So, 8 pints is equal to 16 cups.
10 servings
x servings
So, a proportion is
=
,
4 cups
16 cups
10
x
=
.
4
16
8. 648;
V = Bh
= 4(3) • 2
= 12 • 2
= 24 in.3
The volume of a prism whose dimensions are three times
greater is 33 • 24 = 27 • 24 = 648 cubic inches.
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Cost (dollars)
y
To make the popcorn container, 92π square inches of
material is needed.
or
− 50
35
30
25
20
15
10
5
9
2
1 2 3 4 5 6 7 x
Matinee tickets
slope =
change in y
change in x
9
2
= 4.5
=
The slope of the line is 4.5. So, the cost is
$4.50 per ticket.
Part C:
The cost per ticket is $4.50. So, it costs 4.5(8) = $36 to
buy 8 matinee movie tickets.
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