Appendix A — Answer Key, Chapter 12: Volume and Surface Area: Prisms and Cylinders

SOL 7.MG.1 · Covers textbook Chapter 12 and the companion workbook. Item numbers match the textbook; workbook items are the same problems with the same numbers, so this key serves both. Reasoning answers show an acceptable response, not the only wording.

Conventions used throughout. Answers containing $\pi$ are given exactly in terms of $\pi$ and then approximately with $\pi \approx 3.14$; because every $\pi$ coefficient in this chapter is a whole number, each decimal shown is the exact product of that whole number and $3.14$. Volume answers carry cubic units; surface area answers carry square units.

Lesson 12.1 — Volume of a Right Cylinder

Formula: $V = \pi r^2 h$, or $V = Bh$ when the base area is given.

Guided practice

1. $V = \pi(2)^2(7) = \pi(4)(7) = 28\pi$ cm$^3 \approx 87.92$ cm$^3$
2. $V = \pi(6)^2(4) = \pi(36)(4) = 144\pi$ in$^3 \approx 452.16$ in$^3$
3. $r = \tfrac{10}{2} = 5$ m. $V = \pi(5)^2(3) = \pi(25)(3) = 75\pi$ m$^3 \approx 235.5$ m$^3$
4. $V = \pi(1)^2(12) = 12\pi$ ft$^3 \approx 37.68$ ft$^3$
5. $V = Bh = (20\pi)(6) = 120\pi$ cm$^3 \approx 376.8$ cm$^3$

Independent practice

6. $V = \pi(4)^2(9) = \pi(16)(9) = 144\pi$ cm$^3 \approx 452.16$ cm$^3$
7. $r = \tfrac{14}{2} = 7$ in. $V = \pi(7)^2(10) = \pi(49)(10) = 490\pi$ in$^3 \approx 1{,}538.6$ in$^3$
8. $V = \pi(3)^2(3) = \pi(9)(3) = 27\pi$ m$^3 \approx 84.78$ m$^3$
9. $V = \pi(10)^2(1) = 100\pi$ ft$^3 \approx 314$ ft$^3$
10. $100\pi = \pi(5)^2h = 25\pi h$, so $h = \dfrac{100\pi}{25\pi} = 4$ in
11. $V = \pi(4)^2(11) = \pi(16)(11) = 176\pi$ cm$^3 \approx 552.64$ cm$^3$
12. The base area $\pi r^2$ multiplies a length by a length, so it is measured in square units. Multiplying that area by the height $h$, another length, multiplies by a third length. Three lengths multiplied together give cubic units, so the volume is in cubic units — for example, cm $\times$ cm $\times$ cm $=$ cm$^3$.

Exit ticket 12.1

13. $V = \pi(2)^2(10) = \pi(4)(10) = 40\pi$ in$^3 \approx 125.6$ in$^3$
14. $r = \tfrac{6}{2} = 3$ cm. $V = \pi(3)^2(5) = \pi(9)(5) = 45\pi$ cm$^3 \approx 141.3$ cm$^3$
15. $V = Bh = (30\pi)(4) = 120\pi$ m$^3 \approx 376.8$ m$^3$
16. She used the diameter as the radius. The radius is $\tfrac{12}{2} = 6$ cm, so $V = \pi(6)^2(2) = \pi(36)(2) = 72\pi$ cm$^3 \approx 226.08$ cm$^3$. Her answer is four times too large, because squaring $12$ instead of $6$ multiplies the result by $4$.

Lesson 12.2 — Surface Area of a Rectangular Prism (and Nets)

Formula: $SA = 2lw + 2lh + 2wh = 2(lw + lh + wh)$; for a cube, $SA = 6s^2$.

Guided practice

17. $SA = 2(12 + 8 + 6) = 2(26) = 52$ cm$^2$
18. $SA = 6(5)^2 = 6(25) = 150$ in$^2$
19. $SA = 2(35 + 14 + 10) = 2(59) = 118$ ft$^2$
20. $SA = 2(60 + 30 + 18) = 2(108) = 216$ m$^2$
21. $SA = 2(24) + 2(16) + 2(6) = 48 + 32 + 12 = 92$ in$^2$

Independent practice

22. $SA = 2(36 + 45 + 20) = 2(101) = 202$ cm$^2$
23. $SA = 6(10)^2 = 6(100) = 600$ ft$^2$
24. $SA = 2(60 + 24 + 10) = 2(94) = 188$ in$^2$
25. $SA = 2(36 + 12 + 12) = 2(60) = 120$ m$^2$
26. Three face shapes, two of each: $7 \times 4 = 28$ cm$^2$ (two of them), $7 \times 3 = 21$ cm$^2$ (two of them), $4 \times 3 = 12$ cm$^2$ (two of them). $SA = 2(28 + 21 + 12) = 2(61) = 122$ cm$^2$
27. Bottom $= (8)(4) = 32$; front and back $= 2(8)(3) = 48$; two ends $= 2(4)(3) = 24$. $SA = 32 + 48 + 24 = 104$ ft$^2$
28. Surface area is a total of face areas, and every face area is a length times a length, which gives square units. A net makes this plain because it lays the solid out flat: once unfolded, there is nothing left but rectangles, and rectangles have area, not volume.

Exit ticket 12.2

29. $SA = 2(20 + 15 + 12) = 2(47) = 94$ in$^2$
30. $SA = 6(4)^2 = 6(16) = 96$ cm$^2$
31. $SA = 2(12) + 2(30) + 2(10) = 24 + 60 + 20 = 104$ ft$^2$
32. He doubled only the first face area and then added the other two once each. Every face has a matching opposite face, so all three products must be doubled: $SA = 2(20 + 15 + 12) = 94$ in$^2$.

Lesson 12.3 — Surface Area of a Right Cylinder

Formula: $SA = 2\pi r^2 + 2\pi rh$; lateral area alone is $2\pi rh$.

Guided practice

33. $SA = 2\pi(1)^2 + 2\pi(1)(4) = 2\pi + 8\pi = 10\pi$ m$^2 \approx 31.4$ m$^2$
34. $SA = 2\pi(5)^2 + 2\pi(5)(2) = 50\pi + 20\pi = 70\pi$ cm$^2 \approx 219.8$ cm$^2$
35. $r = \tfrac{8}{2} = 4$ in. $SA = 2\pi(4)^2 + 2\pi(4)(3) = 32\pi + 24\pi = 56\pi$ in$^2 \approx 175.84$ in$^2$
36. Lateral area $= 2\pi(6)(5) = 60\pi$ ft$^2 \approx 188.4$ ft$^2$
37. $SA = 2\pi(2)^2 + 2\pi(2)(10) = 8\pi + 40\pi = 48\pi$ cm$^2 \approx 150.72$ cm$^2$

Independent practice

38. $SA = 2\pi(3)^2 + 2\pi(3)(4) = 18\pi + 24\pi = 42\pi$ m$^2 \approx 131.88$ m$^2$
39. $SA = 2\pi(10)^2 + 2\pi(10)(1) = 200\pi + 20\pi = 220\pi$ in$^2 \approx 690.8$ in$^2$
40. $r = \tfrac{6}{2} = 3$ cm. $SA = 2\pi(3)^2 + 2\pi(3)(9) = 18\pi + 54\pi = 72\pi$ cm$^2 \approx 226.08$ cm$^2$
41. Lateral area $= 2\pi(2)(7) = 28\pi$ ft$^2 \approx 87.92$ ft$^2$
42. One circle plus the side: $\pi(5)^2 + 2\pi(5)(6) = 25\pi + 60\pi = 85\pi$ in$^2 \approx 266.9$ in$^2$
43. Label area $= 2\pi(4)(10) = 80\pi$ cm$^2 \approx 251.2$ cm$^2$. The flat rectangle is $2\pi r = 8\pi \approx 25.12$ cm wide and $10$ cm tall.
44. Unrolling the curved side gives a rectangle whose width is the distance once around the circular base, and that distance is the circumference, $2\pi r$. Using $r$ as the width would make the rectangle about six times too narrow, so the lateral area — and therefore the whole surface area — would be far too small.

Exit ticket 12.3

45. $SA = 2\pi(2)^2 + 2\pi(2)(3) = 8\pi + 12\pi = 20\pi$ cm$^2 \approx 62.8$ cm$^2$
46. $r = \tfrac{12}{2} = 6$ in. $SA = 2\pi(6)^2 + 2\pi(6)(5) = 72\pi + 60\pi = 132\pi$ in$^2 \approx 414.48$ in$^2$
47. Lateral area $= 2\pi(1)(9) = 18\pi$ m$^2 \approx 56.52$ m$^2$
48. He forgot the two circular bases, worth $2\pi(3)^2 = 18\pi$ cm$^2$. His $48\pi$ is only the lateral area. $SA = 18\pi + 48\pi = 66\pi$ cm$^2 \approx 207.24$ cm$^2$

Lesson 12.4 — Volume or Surface Area? Deciding from the Problem

Guided practice

49. Surface area. $SA = 2(40 + 16 + 10) = 2(66) = 132$ in$^2$
50. Volume. $V = \pi(3)^2(10) = 90\pi$ m$^3 \approx 282.6$ m$^3$
51. Volume. $V = (10)(6)(3) = 180$ in$^3$
52. Surface area. $SA = 2\pi(1)^2 + 2\pi(1)(6) = 2\pi + 12\pi = 14\pi$ in$^2 \approx 43.96$ in$^2$
53. Volume. $V = \pi(1)^2(4) = 4\pi$ ft$^3 \approx 12.56$ ft$^3$

Independent practice

54. a) surface area, square feet (ft$^2$) b) volume, cubic meters (m$^3$) c) surface area, square centimeters (cm$^2$) d) volume, cubic feet (ft$^3$)
55. Surface area. $SA = 2(36 + 24 + 6) = 2(66) = 132$ in$^2$
56. Volume. $V = (12)(3)(2) = 72$ in$^3$
57. Volume. $V = \pi(2)^2(9) = 36\pi$ in$^3 \approx 113.04$ in$^3$
58. Surface area. $SA = 2\pi(2)^2 + 2\pi(2)(9) = 8\pi + 36\pi = 44\pi$ in$^2 \approx 138.16$ in$^2$
59. a) Volume. $V = \pi(2)^2(5) = 20\pi$ ft$^3 \approx 62.8$ ft$^3$ b) Surface area. Bottom plus side $= \pi(2)^2 + 2\pi(2)(5) = 4\pi + 20\pi = 24\pi$ ft$^2 \approx 75.36$ ft$^2$
60. It must have been a covering question — surface area — because cm$^2$ is a square unit. Filling questions produce cubic units such as cm$^3$.

Exit ticket 12.4

61. Surface area (wrapping paper covers the outside), in square units.
62. Volume. $V = (8)(2)(12) = 192$ in$^3$
63. Surface area. $SA = 2\pi(3)^2 + 2\pi(3)(4) = 18\pi + 24\pi = 42\pi$ in$^2 \approx 131.88$ in$^2$
64. Paint covers surfaces, so the amount of wall being painted is an area and must be reported in square units such as cm$^2$. The unit cm$^3$ is a volume unit, so it answers a filling question, not a covering question.

Lesson 12.5 — How Changing One Dimension Changes the Volume

In every item exactly one measured attribute changes, and the factor is one of $\tfrac14$, $\tfrac13$, $\tfrac12$, $2$, $3$, $4$. Volume before $\times$ factor $=$ volume after.

Guided practice

The starting prism is $8 \times 3 \times 4$, so $V = (8)(3)(4) = 96$ cm$^3$.

65. New height $= 4 \times 2 = 8$. $V = (8)(3)(8) = 192$ cm$^3$. Volume factor $2$.
66. New length $= 8 \times \tfrac12 = 4$. $V = (4)(3)(4) = 48$ cm$^3$. Volume factor $\tfrac12$.
67. New width $= 3 \times 3 = 9$. $V = (8)(9)(4) = 288$ cm$^3$. Volume factor $3$.
68. New height $= 4 \times \tfrac14 = 1$. $V = (8)(3)(1) = 24$ cm$^3$. Volume factor $\tfrac14$.
69. $4 \times 200 = 800$ in$^3$
70. $\tfrac13 \times 72 = 24$ m$^3$

Independent practice

71. Before: $(5)(4)(3) = 60$ cm$^3$. After: $(5)(4)(6) = 120$ cm$^3$. Factor $2$.
72. Before: $(12)(6)(2) = 144$ in$^3$. After: $(4)(6)(2) = 48$ in$^3$. Factor $\tfrac13$.
73. Before: $(10)(4)(5) = 200$ ft$^3$. After: $(10)(1)(5) = 50$ ft$^3$. Factor $\tfrac14$.
74. $3 \times 90 = 270$ m$^3$
75. $\tfrac12 \times 48 = 24$ in$^3$
76. Before: $(9)(4)(5) = 180$ in$^3$. After: $(9)(4)(10) = 360$ in$^3$. That is $360 - 180 = 180$ in$^3$ more.
77. In $V = lwh$ the width appears exactly once as a factor. Replacing $w$ with $3w$ gives $V_{\text{new}} = l(3w)h = 3(lwh) = 3V$, so the volume is multiplied by $3$. If all three dimensions were tripled, the factor $3$ would appear three times — $(3l)(3w)(3h) = 27(lwh)$ — which is a completely different situation from the one this lesson asks about.

Exit ticket 12.5

78. Before: $(6)(5)(2) = 60$ cm$^3$. After: $(6)(5)(6) = 180$ cm$^3$.
79. After: $(3)(5)(2) = 30$ cm$^3$ (half of $60$ cm$^3$).
80. $4 \times 100 = 400$ ft$^3$
81. She used a factor of $2$ three times, but only one dimension changed, so the factor applies only once. Doubling the length multiplies the volume by $2$, not by $8$.

Lesson 12.6 — How Changing One Dimension Changes the Surface Area

The factor is $\tfrac12$ or $2$ in every item. Surface area must be recomputed; it does not change by the factor.

Guided practice

The starting prism is $5 \times 4 \times 3$, so $SA = 2(20 + 15 + 12) = 94$ cm$^2$.

82. New dimensions $5 \times 4 \times 6$. $SA = 2(20 + 30 + 24) = 2(74) = 148$ cm$^2$. It did not double; twice $94$ would be $188$ cm$^2$.
83. New dimensions $10 \times 4 \times 3$. $SA = 2(40 + 30 + 12) = 2(82) = 164$ cm$^2$. It did not double.
84. New dimensions $5 \times 2 \times 3$. $SA = 2(10 + 15 + 6) = 2(31) = 62$ cm$^2$. It was not halved; half of $94$ would be $47$ cm$^2$.
85. The top and the bottom, the two $5 \times 4$ faces, each $20$ cm$^2$. Neither one involves the height, so doubling the height leaves them unchanged.
86. Before: $SA = 6(4)^2 = 96$ in$^2$. After ($4 \times 4 \times 2$): $SA = 2(16 + 8 + 8) = 2(32) = 64$ in$^2$.

Independent practice

87. Before ($6 \times 3 \times 2$): $SA = 2(18 + 12 + 6) = 72$ cm$^2$. After ($6 \times 3 \times 4$): $SA = 2(18 + 24 + 12) = 108$ cm$^2$.
88. Before ($8 \times 5 \times 4$): $SA = 2(40 + 32 + 20) = 184$ in$^2$. After ($4 \times 5 \times 4$): $SA = 2(20 + 16 + 20) = 112$ in$^2$.
89. Before ($10 \times 2 \times 3$): $SA = 2(20 + 30 + 6) = 112$ ft$^2$. After ($10 \times 4 \times 3$): $SA = 2(40 + 30 + 12) = 164$ ft$^2$.
90. Before: $SA = 6(6)^2 = 216$ m$^2$. After ($6 \times 6 \times 3$): $SA = 2(36 + 18 + 18) = 144$ m$^2$.
91. Before: $SA = 6(4)^2 = 96$ in$^2$. After ($4 \times 4 \times 8$): $SA = 2(16 + 32 + 32) = 160$ in$^2$. It did not double — twice $96$ would be $192$ in$^2$ — it increased by $160 - 96 = 64$ in$^2$.
92. Before ($12 \times 6 \times 4$): $SA = 2(72 + 48 + 24) = 288$ in$^2$. After ($12 \times 6 \times 2$): $SA = 2(72 + 24 + 12) = 216$ in$^2$. Saved: $288 - 216 = 72$ in$^2$ per box.
93. Volume is the single product $lwh$, so the changed attribute is a factor of the whole thing and the factor passes straight through. Surface area is a sum of six face areas, and the changed attribute appears in only four of them. When the height doubles, the four side faces ($l \times h$ and $w \times h$) double, but the top and bottom ($l \times w$) do not change at all. Since only part of the total doubles, the total grows by less than a factor of $2$.

Exit ticket 12.6

94. Before ($6 \times 4 \times 2$): $SA = 2(24 + 12 + 8) = 88$ cm$^2$. After ($6 \times 4 \times 4$): $SA = 2(24 + 24 + 16) = 128$ cm$^2$.
95. Before: $88$ cm$^2$. After ($3 \times 4 \times 2$): $SA = 2(12 + 6 + 8) = 52$ cm$^2$.
96. $V = (6)(4)(2) = 48$ cm$^3$; after the height doubles, $V = (6)(4)(4) = 96$ cm$^3$. The volume factor is $2$. The surface area did not change by that factor: it went from $88$ cm$^2$ to $128$ cm$^2$, and twice $88$ would be $176$ cm$^2$.
97. Doubling one attribute doubles the one product that gives volume, but only four of the six faces that make up surface area, so the volume exactly doubles while the surface area grows by less than double.

Chapter 12 Review

Part A — Volume of right cylinders (7.MG.1a)

98. $V = \pi(4)^2(5) = \pi(16)(5) = 80\pi$ cm$^3 \approx 251.2$ cm$^3$
99. $r = 5$ in. $V = \pi(5)^2(6) = \pi(25)(6) = 150\pi$ in$^3 \approx 471$ in$^3$
100. $V = \pi(2)^2(11) = \pi(4)(11) = 44\pi$ m$^3 \approx 138.16$ m$^3$
101. $63\pi = \pi(3)^2h = 9\pi h$, so $h = \dfrac{63\pi}{9\pi} = 7$ ft
102. $V = \pi(3)^2(4) = \pi(9)(4) = 36\pi$ ft$^3 \approx 113.04$ ft$^3$

Part B — Surface area of rectangular prisms and right cylinders (7.MG.1b)

103. $SA = 2(21 + 14 + 6) = 2(41) = 82$ cm$^2$
104. $SA = 6(8)^2 = 6(64) = 384$ in$^2$
105. $SA = 2\pi(5)^2 + 2\pi(5)(3) = 50\pi + 30\pi = 80\pi$ m$^2 \approx 251.2$ m$^2$
106. $r = 2$ in. $SA = 2\pi(2)^2 + 2\pi(2)(10) = 8\pi + 40\pi = 48\pi$ in$^2 \approx 150.72$ in$^2$
107. $SA = 2(45) + 2(18) + 2(10) = 90 + 36 + 20 = 146$ ft$^2$
108. Lateral area $= 2\pi(6)(2) = 24\pi$ cm$^2 \approx 75.36$ cm$^2$

Part C — Volume or surface area? (7.MG.1c)

109. a) surface area, ft$^2$ b) volume, m$^3$ c) surface area, cm$^2$ d) volume, in$^3$
110. Surface area. $SA = 2(40 + 30 + 12) = 2(82) = 164$ in$^2$
111. Volume. $V = (10)(4)(3) = 120$ in$^3$
112. a) Volume. $V = \pi(2)^2(6) = 24\pi$ ft$^3 \approx 75.36$ ft$^3$ b) Surface area. $\pi(2)^2 + 2\pi(2)(6) = 4\pi + 24\pi = 28\pi$ ft$^2 \approx 87.92$ ft$^2$

Part D — Changing one attribute: volume (7.MG.1d)

113. Before: $(8)(5)(2) = 80$ cm$^3$. After: $(8)(5)(6) = 240$ cm$^3$ (factor $3$).
114. Before: $(12)(3)(4) = 144$ in$^3$. After: $(3)(3)(4) = 36$ in$^3$ (factor $\tfrac14$).
115. $\tfrac12 \times 60 = 30$ m$^3$
116. Before: $(6)(5)(4) = 120$ ft$^3$. After: $(6)(10)(4) = 240$ ft$^3$. The larger bin holds $240 - 120 = 120$ ft$^3$ more.

Part E — Changing one attribute: surface area (7.MG.1e)

117. Before ($5 \times 4 \times 2$): $SA = 2(20 + 10 + 8) = 76$ cm$^2$. After ($5 \times 4 \times 4$): $SA = 2(20 + 20 + 16) = 112$ cm$^2$.
118. Before ($8 \times 6 \times 3$): $SA = 2(48 + 24 + 18) = 180$ in$^2$. After ($4 \times 6 \times 3$): $SA = 2(24 + 12 + 18) = 108$ in$^2$.
119. Before: $SA = 6(10)^2 = 600$ ft$^2$. After ($10 \times 10 \times 5$): $SA = 2(100 + 50 + 50) = 400$ ft$^2$.

Part F — Mixed application and reasoning

120. a) $V = \pi(3)^2(10) = 90\pi$ in$^3 \approx 282.6$ in$^3$ b) $SA = 2\pi(3)^2 + 2\pi(3)(10) = 18\pi + 60\pi = 78\pi$ in$^2 \approx 244.92$ in$^2$ c) The volume, $90\pi$ in$^3$, because juice fills the inside.
121. Volume multiplies three lengths, $l \times w \times h$, so its unit is a length unit multiplied by itself three times — cubic units. Surface area adds products of two lengths at a time, such as $lw$ and $lh$ and $wh$, so every term is a length unit multiplied by itself twice — square units.
122. $V$ before $= (4)(3)(2) = 24$ cm$^3$; $V$ after $= (4)(3)(4) = 48$ cm$^3$, which is exactly double, so the volume claim is right. $SA$ before $= 2(12 + 8 + 6) = 52$ cm$^2$; $SA$ after $= 2(12 + 16 + 12) = 80$ cm$^2$. Double $52$ would be $104$ cm$^2$, so the surface area claim is wrong.
123. Cutting a cylinder open gives two circles and one rectangle. Each circle has area $\pi r^2$, and there are two of them, which is the $2\pi r^2$ term. The rectangle is the curved side unrolled: its height is $h$ and its width is the circumference $2\pi r$, so its area is $2\pi r h$, which is the second term. Adding the three pieces gives $SA = 2\pi r^2 + 2\pi r h$.
124. a) $V = \pi(4)^2(10) = \pi(16)(10) = 160\pi$ cm$^3 \approx 502.4$ cm$^3$ b) Lateral area $= 2\pi(4)(10) = 80\pi$ cm$^2 \approx 251.2$ cm$^2$ c) The answer to part (a) is in cm$^3$, because how much the can holds is a filling question, and filling is measured by multiplying three lengths together.