Appendix A — Answer Key, Chapter 11: Surface Area and Volume: Pyramids and Cones

SOL 8.MG.2 · Covers textbook Chapter 11 and the companion workbook. Item numbers match the textbook; workbook items are the same problems, so this key serves both. Item numbers run continuously from 1 to 150 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Conventions used throughout, matching the chapter:

• Pi. Every answer containing $\pi$ is given exactly and then approximately with $\pi \approx 3.14$. A calculator's stored $\pi$ gives a slightly different decimal; accept it only if the student states which value was used.
• Units. Volume in cubic units, surface area in square units.
• Formulas. $\ell^2 = h^2 + \left(\frac{s}{2}\right)^2$ · $SA = s^2 + 2s\ell$ · $LA = 2s\ell$ · $V = \frac{1}{3}s^2h$ · $V = \frac{1}{3}\pi r^2 h$
• Half the base edge. Every slant-height computation uses $\frac{s}{2}$, never $s$. This is the error to watch for in student work; it shows up as a slant height that is too long.

Lesson 11.1 — Pyramids, Cones, and Two Different Heights

Guided practice

1. The base is a square. Each lateral face is a congruent isosceles triangle.
2. $5$ faces (one square and four triangles), $8$ edges (four around the base, four rising to the apex), and $5$ vertices (four base corners and the apex).
3. The height $h$ runs from the apex to the center of the base; the slant height $\ell$ runs from the apex to the midpoint of a base edge. The height is drawn dashed because it is inside the solid; the slant height lies on a lateral face.
4. One square and four triangles.
5. Half the base edge is $3$ cm, so $\ell^2 = 4^2 + 3^2 = 16 + 9 = 25$, and $\ell = 5$ cm.
6. Half the base edge is $5$ in. $\ell^2 = 12^2 + 5^2 = 144 + 25 = 169$, so $\ell = 13$ in. Note $13 > 12$, as it must be.
7. In the right triangle formed by slicing the pyramid through the apex, $\ell$ is the hypotenuse and $h$ is a leg, so $\ell^2 = h^2 + \left(\tfrac{s}{2}\right)^2$. Since $s > 0$, the term $\left(\tfrac{s}{2}\right)^2$ is positive, so $\ell^2 > h^2$ and therefore $\ell > h$. The two can never be equal.

Independent practice

8. a) $\ell^2 = 3^2 + 4^2 = 25$, $\ell = 5$ cm b) $\ell^2 = 5^2 + 12^2 = 169$, $\ell = 13$ m c) $\ell^2 = 8^2 + 6^2 = 100$, $\ell = 10$ ft
9. Half the base edge is $9$ in. $\ell^2 = 12^2 + 9^2 = 144 + 81 = 225$, so $\ell = 15$ in.
10. Half the base edge is $8$ cm. $h^2 = 17^2 - 8^2 = 289 - 64 = 225$, so $h = 15$ cm.
11. Half the base edge is $6$ m. $h^2 = 10^2 - 6^2 = 100 - 36 = 64$, so $h = 8$ m.
12. Half the base edge is $5$ ft. $h^2 = 13^2 - 5^2 = 169 - 25 = 144$, so $h = 12$ ft.
13. The radius is $\tfrac{14}{2} = 7$ cm. The height is $9$ cm — the height of a cone runs straight up from the center of the base to the apex, so it is the measurement that is not across the base.
14. The net is one square with a triangle attached to each of its four edges. The four triangles are congruent, each with base $s$. The height of each triangle is the slant height $\ell$ of the pyramid, not the pyramid's height, because the net shows only lengths that lie on the surface of the solid.
15. The student used the whole base edge, $6$, as a leg instead of half of it. The center of the square base is only $\tfrac{s}{2} = 3$ cm from the midpoint of an edge. Correctly, $\ell^2 = 4^2 + 3^2 = 25$, so $\ell = 5$ cm. Using the whole edge always produces a slant height that is too long.
16. Half the base edge is $8$ mm. $\ell^2 = 15^2 + 8^2 = 225 + 64 = 289$, so $\ell = 17$ mm.
17. The apex sits directly above the center of the square base, so it is the same distance from all four base corners. Each lateral face therefore has two equal sides — the two edges rising to the apex — which makes it isosceles. And since all four base edges are equal (the base is a square) and all four rising edges are equal, the four triangles have the same three side lengths and are congruent.

Exit ticket 11.1

18. Half the base edge is $7$ cm. $\ell^2 = 24^2 + 7^2 = 576 + 49 = 625$, so $\ell = 25$ cm.
19. Half the base edge is $15$ in. $h^2 = 25^2 - 15^2 = 625 - 225 = 400$, so $h = 20$ in.
20. The slant height.
21. The height is the distance straight up from the center of the base to the apex; the slant height is the distance from the apex down the middle of a lateral face to the midpoint of a base edge. The slant height is longer, because it is the hypotenuse of a right triangle whose legs are the height and half the base edge.

Lesson 11.2 — Surface Area of a Square-Based Pyramid

Guided practice

22. $SA = s^2 + 4\left(\tfrac{1}{2}s\ell\right) = s^2 + 2s\ell$ — the four halves combine into $2$.
23. $SA = 6^2 + 2(6)(5) = 36 + 60 = 96$ cm$^2$
24. $SA = 8^2 + 2(8)(5) = 64 + 80 = 144$ in$^2$
25. $SA = 4^2 + 2(4)(6) = 16 + 48 = 64$ m$^2$
26. $SA = 10^2 + 2(10)(13) = 100 + 260 = 360$ ft$^2$
27. $LA = 2s\ell = 2(6)(8) = 96$ cm$^2$ — the base is not included.
28. Step 1: half the base edge is $3$ m, so $\ell^2 = 4^2 + 3^2 = 25$ and $\ell = 5$ m. Step 2: $SA = 6^2 + 2(6)(5) = 36 + 60 = 96$ m$^2$.
29. $SA = 12^2 + 2(12)(10) = 144 + 240 = 384$ in$^2$

Independent practice

30. a) $5^2 + 2(5)(8) = 25 + 80 = 105$ cm$^2$ b) $9^2 + 2(9)(12) = 81 + 216 = 297$ m$^2$ c) $3^2 + 2(3)(4) = 9 + 24 = 33$ in$^2$
31. $SA = 14^2 + 2(14)(25) = 196 + 700 = 896$ ft$^2$
32. $SA = 20^2 + 2(20)(15) = 400 + 600 = 1000$ cm$^2$
33. $LA = 2(18)(15) = 540$ m$^2$
34. Half the base edge is $8$ in, so $\ell^2 = 15^2 + 8^2 = 289$ and $\ell = 17$ in. Then $SA = 16^2 + 2(16)(17) = 256 + 544 = 800$ in$^2$.
35. Half the base edge is $6$ cm, so $\ell^2 = 8^2 + 6^2 = 100$ and $\ell = 10$ cm. Then $SA = 12^2 + 2(12)(10) = 144 + 240 = 384$ cm$^2$.
36. Half the base edge is $12$ ft, so $\ell^2 = 5^2 + 12^2 = 169$ and $\ell = 13$ ft. Then $SA = 24^2 + 2(24)(13) = 576 + 624 = 1200$ ft$^2$.
37. $189 = 7^2 + 2(7)\ell = 49 + 14\ell$, so $140 = 14\ell$ and $\ell = 10$ m.
38. Four walls and no floor means lateral area only: $LA = 2(6)(5) = 60$ ft$^2$.
39. The box is closed, so the base is included, and the height was given, so find the slant height first. Half the base edge is $5$ cm: $\ell^2 = 12^2 + 5^2 = 169$, $\ell = 13$ cm. Then $SA = 10^2 + 2(10)(13) = 100 + 260 = 360$ cm$^2$.
40. The student substituted the height $4$ where the slant height belongs. The height is not a length on the net, so it cannot appear in the surface area formula. The answer must be too small because $\ell > h$ always, so $2s\ell > 2sh$. Correctly, $\ell = 5$ cm and $SA = 36 + 2(6)(5) = 96$ cm$^2$ — not $84$ cm$^2$.
41. Each lateral face is a triangle of area $\tfrac{1}{2}s\ell$, and there are four of them, giving $4 \cdot \tfrac{1}{2}s\ell$. The $4$ and the $\tfrac{1}{2}$ multiply to $2$, so the four faces total $2s\ell$. The $4$ has not disappeared; it has been combined with the $\tfrac{1}{2}$ from the triangle area formula.

Exit ticket 11.2

42. $SA = 5^2 + 2(5)(6) = 25 + 60 = 85$ cm$^2$
43. $SA = 2^2 + 2(2)(5) = 4 + 20 = 24$ m$^2$
44. Half the base edge is $8$ in, so $\ell^2 = 6^2 + 8^2 = 100$ and $\ell = 10$ in. Then $SA = 16^2 + 2(16)(10) = 256 + 320 = 576$ in$^2$.
45. $LA = 2(12)(13) = 312$ ft$^2$
46. The surface area formula adds up the areas of the pieces of the net, and every length in it — the base edge and the slant height — is a length you can measure on the net. The pyramid's height runs inside the solid and appears nowhere on the net, so it cannot be substituted. When a problem gives the height, first compute the slant height with $\ell^2 = h^2 + \left(\tfrac{s}{2}\right)^2$, then use that $\ell$ in $SA = s^2 + 2s\ell$.

Lesson 11.3 — Why One Third? Cones and Cylinders, Pyramids and Prisms

Guided practice

47. the same base and the same height
48. $\tfrac{1}{3}(60) = 20$ cm$^3$
49. $3(25) = 75$ in$^3$
50. $\tfrac{1}{3}(81) = 27$ ft$^3$
51. $3(14) = 42$ m$^3$
52. Three.
53. It means two things match: the base of the cone and the base of the cylinder are circles with the same radius, and the two solids have the same height. If either one differs, the one-third relationship does not apply.
54. $\tfrac{2}{3}$

Independent practice

55. Cylinder: $V = \pi(6)^2(10) = 360\pi$ cm$^3$. Cone: $\tfrac{1}{3}(360\pi) = 120\pi$ cm$^3$, or about $120 \times 3.14 = 376.8$ cm$^3$.
56. Cylinder: $V = \pi(2)^2(6) = 24\pi$ in$^3$. Cone: $8\pi$ in$^3$, or about $25.12$ in$^3$.
57. Prism: $V = 9^2(4) = 324$ m$^3$. Pyramid: $\tfrac{1}{3}(324) = 108$ m$^3$.
58. Prism: $V = 12^2(5) = 720$ ft$^3$. Pyramid: $\tfrac{1}{3}(720) = 240$ ft$^3$.
59. $3(30\pi) = 90\pi$ cm$^3$, or about $282.6$ cm$^3$.
60. $3(50) = 150$ in$^3$
61. The cube holds $6^3 = 216$ cm$^3$. The three pyramids are congruent and use up the whole cube, so each holds $\tfrac{216}{3} = 72$ cm$^3$. The formula agrees: each pyramid has base a face of the cube, so $B = 6^2 = 36$ cm$^2$, and height $6$ cm, giving $V = \tfrac{1}{3}(36)(6) = 72$ cm$^3$. Since the dissection produces exactly three equal pieces with no waste, the one-third factor is exact rather than experimental.
62. Slice both solids with a horizontal plane. At every level, the cylinder's cross-section is the full base circle. The cone's cross-section is a smaller circle that shrinks as you rise, and it shrinks in two directions at once, so its area falls off faster than its radius does — halfway up, the cone's cross-section has half the radius but only one quarter of the area. Since the cone's cross-sections are smaller than the cylinder's at every level above the base, and average out to one third of the base area, the cone holds less than half.
63. No — it is much more than one third. With the cylinder's radius $R$, the cone's radius is $2R$, so $V_{\text{cone}} = \tfrac{1}{3}\pi(2R)^2h = \tfrac{1}{3}\pi(4R^2)h = \tfrac{4}{3}\pi R^2h$, while $V_{\text{cylinder}} = \pi R^2 h$. The cone's volume is $\tfrac{4}{3}$ of the cylinder's, so the cone is actually the larger solid. The one-third relationship needs the same base as well as the same height.
64. $\tfrac{1}{3}(900) = 300$ mL
65. The triangle-is-half-a-rectangle fact is about two dimensions, where a shape shrinks in only one direction. A pyramid tapers in two directions at once, so its horizontal cross-sections lose area faster than a triangle's horizontal widths lose length: halfway up, the pyramid's cross-section has half the side length but only one quarter of the area. The correct fraction is $\tfrac{1}{3}$, not $\tfrac{1}{2}$.
66. Both a cone and a square-based pyramid taper from a flat base to a single apex, and for any such solid the volume is one third of the base area times the height. Only the shape of the base differs, and $B$ absorbs that difference: $B = \pi r^2$ for the cone's circular base, and $B = s^2$ for the pyramid's square base.

Exit ticket 11.3

67. $\tfrac{1}{3}(48\pi) = 16\pi$ m$^3$, or about $16 \times 3.14 = 50.24$ m$^3$.
68. $3(21) = 63$ cm$^3$
69. Three.
70. Because three cone-fulls fill the cylinder exactly, and they do so only when the two solids have the same base and the same height. With the same base, both solids start with the same amount of area covering the bottom; with the same height, both stop rising at the same level. The cone then differs from the cylinder only in tapering, and that taper removes exactly two thirds of the space, leaving the cone with one third.

Lesson 11.4 — Volume of Square-Based Pyramids and Cones

Guided practice

71. $V = \tfrac{1}{3}(3)^2(5) = \tfrac{1}{3}(9)(5) = 15$ cm$^3$
72. $V = \tfrac{1}{3}(6)^2(10) = \tfrac{1}{3}(36)(10) = 120$ in$^3$
73. $V = \tfrac{1}{3}(25)(6) = 50$ m$^3$
74. $V = \tfrac{1}{3}(100)(9) = 300$ ft$^3$
75. $V = \tfrac{1}{3}\pi(9)(4) = 12\pi$ cm$^3$, or about $37.68$ cm$^3$
76. $V = \tfrac{1}{3}\pi(25)(9) = 75\pi$ in$^3$, or about $235.5$ in$^3$
77. $V = \tfrac{1}{3}\pi(16)(3) = 16\pi$ m$^3$, or about $50.24$ m$^3$
78. $r = \tfrac{10}{2} = 5$ ft, so $V = \tfrac{1}{3}\pi(25)(6) = 50\pi$ ft$^3$, or about $157$ ft$^3$

Independent practice

79. a) $\tfrac{1}{3}(16)(6) = 32$ cm$^3$ b) $\tfrac{1}{3}(64)(3) = 64$ in$^3$ c) $\tfrac{1}{3}(225)(8) = 600$ m$^3$
80. $V = \tfrac{1}{3}(49)(6) = 98$ ft$^3$
81. $V = \tfrac{1}{3}(400)(6) = 800$ cm$^3$
82. $V = \tfrac{1}{3}\pi(36)(4) = 48\pi$ cm$^3$, or about $150.72$ cm$^3$
83. $r = 4$ in, so $V = \tfrac{1}{3}\pi(16)(6) = 32\pi$ in$^3$, or about $100.48$ in$^3$
84. $V = \tfrac{1}{3}\pi(81)(5) = 135\pi$ m$^3$, or about $423.9$ m$^3$
85. $V = \tfrac{1}{3}\pi(1)(9) = 3\pi$ ft$^3$, or about $9.42$ ft$^3$
86. $240 = \tfrac{1}{3}(144)h = 48h$, so $h = 5$ ft.
87. $100\pi = \tfrac{1}{3}\pi(25)h$. Dividing both sides by $\pi$ gives $100 = \tfrac{25}{3}h$, so $h = \tfrac{300}{25} = 12$ in. Because $\pi$ divides out, the answer is exact.
88. The volume formula needs the height, so find it first. Half the base edge is $3$ m: $h^2 = 5^2 - 3^2 = 25 - 9 = 16$, so $h = 4$ m. Then $V = \tfrac{1}{3}(36)(4) = 48$ m$^3$. (Using $\ell = 5$ by mistake gives $60$ m$^3$, which is too large.)
89. $V = \tfrac{1}{3}Bh = \tfrac{1}{3}(27\pi)(5) = 45\pi$ cm$^3$, or about $141.3$ cm$^3$
90. $V = \tfrac{1}{3}\pi(9)(10) = 30\pi$ cm$^3$, or about $94.2$ cm$^3$
91. $V = \tfrac{1}{3}(100)(12) = 400$ cm$^3$
92. The student left out the factor $\tfrac{1}{3}$ and so computed the volume of the cylinder with the same base and height. Correctly, $V = \tfrac{1}{3}\pi(36)(10) = 120\pi$ cm$^3$, or about $376.8$ cm$^3$ — exactly one third of $360\pi$.

Exit ticket 11.4

93. $V = \tfrac{1}{3}(121)(3) = 121$ cm$^3$
94. $V = \tfrac{1}{3}(4)(9) = 12$ m$^3$
95. $V = \tfrac{1}{3}\pi(36)(9) = 108\pi$ in$^3$, or about $339.12$ in$^3$
96. $r = 3$ ft, so $V = \tfrac{1}{3}\pi(9)(8) = 24\pi$ ft$^3$, or about $75.36$ ft$^3$
97. The $\tfrac{1}{3}$ comes from the relationship in Lesson 11.3: a cone holds one third of the cylinder with the same base and the same height, and a square-based pyramid holds one third of the prism with the same base and the same height, which the three-pour experiment shows and the dissection of a cube into three congruent pyramids proves. It is the height and not the slant height because volume measures how much space the solid encloses, and that depends on how far the apex is above the base — the straight-up distance. The slant height is a distance along the surface and does not measure how tall the solid stands.

Lesson 11.5 — Problems in Context

Guided practice

98. a) surface area — and lateral area only, since just the four sloped faces are painted b) volume c) surface area, including the base, since the box is closed d) volume
99. Filling with sand is a volume question: $V = \tfrac{1}{3}(144)(8) = 384$ cm$^3$.
100. Covering is surface area, all five faces, and the height was given, so find the slant height first. Half the base edge is $5$ cm: $\ell^2 = 12^2 + 5^2 = 169$, $\ell = 13$ cm. Then $SA = 10^2 + 2(10)(13) = 360$ cm$^2$.
101. $V = \tfrac{1}{3}\pi(9)(8) = 24\pi$ cm$^3$, or about $75.36$ cm$^3$
102. Four lateral faces only, so lateral area: $LA = 2(4)(6) = 48$ cm$^2$.
103. $V = \tfrac{1}{3}\pi(4)(9) = 12\pi$ in$^3$, or about $37.68$ in$^3$

Independent practice

104. Shingles cover the four sloped faces only, and the slant height was given: $LA = 2(18)(15) = 540$ ft$^2$.
105. $V = \tfrac{1}{3}\pi(100)(6) = 200\pi$ ft$^3$, or about $628$ ft$^3$
106. Glass on all five faces, and the height was given. Half the base edge is $8$ in: $\ell^2 = 15^2 + 8^2 = 289$, $\ell = 17$ in. Then $SA = 16^2 + 2(16)(17) = 256 + 544 = 800$ in$^2$.
107. $r = 6$ cm, so $V = \tfrac{1}{3}\pi(36)(7) = 84\pi$ cm$^3$, or about $263.76$ cm$^3$.
108. $V = \tfrac{1}{3}(36)(4) = 48$ cm$^3$. Mass $= 48 \times 0.9 = 43.2$ g.
109. Cone: $V = \tfrac{1}{3}\pi(36)(6) = 72\pi \approx 226.08$ cm$^3$. Pyramid: $V = \tfrac{1}{3}(100)(6) = 200$ cm$^3$. The cone holds more, by about $226.08 - 200 = 26.08$ cm$^3$. The cone's volume has to be approximated before the two can be compared, since one answer contains $\pi$ and the other does not.
110. $V = \tfrac{1}{3}\pi(25)(12) = 100\pi$ cm$^3$, or about $314$ cm$^3$. Time $= 314 \div 20 = 15.7$, so about $15.7$ seconds.
111. Four walls, no floor, and the height was given. Half the base edge is $4$ ft: $\ell^2 = 3^2 + 4^2 = 25$, $\ell = 5$ ft. Then $LA = 2(8)(5) = 80$ ft$^2$.
112. $V = \tfrac{1}{3}\pi(9)(4) = 12\pi$ ft$^3$, or about $37.68$ ft$^3$. Cost $= 3 \times 37.68 = \$113.04$.
113. You must find the height first, using $h^2 = \ell^2 - \left(\tfrac{s}{2}\right)^2$. The slant height cannot be used directly because volume depends on how far the apex is straight up from the base, and the slant height is a distance along the surface instead. Substituting $\ell$ for $h$ always gives a volume that is too large, since $\ell > h$.
114. The student used the height $12$ where the slant height belongs. Half the base edge is $5$ in, so $\ell^2 = 12^2 + 5^2 = 169$ and $\ell = 13$ in. Correctly, $SA = 10^2 + 2(10)(13) = 360$ in$^2$, not $340$ in$^2$.
115. A volume answer comes from multiplying three lengths, so its unit carries an exponent of $3$ — cubic units. A surface area answer comes from adding areas, each of which is two lengths multiplied, so its unit carries an exponent of $2$ — square units. If your answer is in cm$^3$ you solved a filling problem; if it is in cm$^2$ you solved a covering problem. A mismatch between the unit and the question is a signal to check which formula you used.

Exit ticket 11.5

116. $V = \tfrac{1}{3}\pi(16)(9) = 48\pi$ cm$^3$, or about $150.72$ cm$^3$
117. $V = \tfrac{1}{3}(576)(5) = 960$ in$^3$
118. All five faces, and the slant height was given directly: $SA = 20^2 + 2(20)(26) = 400 + 1040 = 1440$ cm$^2$.
119. A closed box includes the base: $SA = 6^2 + 2(6)(8) = 36 + 96 = 132$ cm$^2$.
120. Read what is being measured. If the problem is about what goes inside — filling, holding, pouring, capacity — it is volume, and the answer is in cubic units. If it is about what goes on the outside — covering, wrapping, painting, fabric, glass — it is surface area, and the answer is in square units. For a surface area problem, also decide whether the square base is part of the covering.

Chapter 11 Review

Part A — Surface area of square-based pyramids (8.MG.2a)

121. The net has one square and four triangles: the square is the base, with side $s$, and each triangle is one lateral face, with base $s$. The four triangles are congruent. The height of each triangle is the slant height $\ell$ of the pyramid, which is why $\ell$ and not $h$ appears in the surface area formula.
122. Half the base edge is $24$ cm. $\ell^2 = 7^2 + 24^2 = 49 + 576 = 625$, so $\ell = 25$ cm.
123. a) $7^2 + 2(7)(10) = 49 + 140 = 189$ cm$^2$ b) $11^2 + 2(11)(9) = 121 + 198 = 319$ m$^2$
124. Half the base edge is $15$ ft, so $\ell^2 = 20^2 + 15^2 = 400 + 225 = 625$ and $\ell = 25$ ft. Then $SA = 30^2 + 2(30)(25) = 900 + 1500 = 2400$ ft$^2$.
125. $LA = 2(14)(25) = 700$ in$^2$
126. $1200 = 24^2 + 2(24)\ell = 576 + 48\ell$, so $624 = 48\ell$ and $\ell = 13$ in.
127. The student added only three lateral faces instead of four: $3\left(\tfrac{1}{2} \cdot 6 \cdot 5\right) = 3(15) = 45$. A square-based pyramid has four lateral faces, one on each edge of the square base, which the net makes plain. Correctly, $SA = 36 + 4(15) = 36 + 60 = 96$ cm$^2$, which is also $s^2 + 2s\ell = 36 + 2(6)(5)$.
128. Unfold the pyramid. The net is one square of side $s$ and four congruent triangles, each with base $s$ and height $\ell$. Since the net folds back into the pyramid with no gaps and no overlaps, the area of the net is the surface area. The square contributes $s^2$. Each triangle contributes $\tfrac{1}{2}s\ell$, and there are four of them, contributing $4 \cdot \tfrac{1}{2}s\ell$. The $4$ and the $\tfrac{1}{2}$ multiply to $2$, giving $2s\ell$. Adding: $SA = s^2 + 2s\ell$.

Part B — Volume of cones and square-based pyramids (8.MG.2b)

129. a) $\tfrac{1}{3}(36)(4) = 48$ cm$^3$ b) $\tfrac{1}{3}(144)(5) = 240$ in$^3$ c) $\tfrac{1}{3}(100)(12) = 400$ m$^3$
130. $V = \tfrac{1}{3}(256)(15) = 1280$ ft$^3$
131. a) $\tfrac{1}{3}\pi(9)(10) = 30\pi$ cm$^3$, or about $94.2$ cm$^3$ b) $\tfrac{1}{3}\pi(64)(3) = 64\pi$ in$^3$, or about $200.96$ in$^3$
132. $r = 5$ m, so $V = \tfrac{1}{3}\pi(25)(12) = 100\pi$ m$^3$, or about $314$ m$^3$
133. $V = \tfrac{1}{3}\pi(144)(5) = 240\pi$ ft$^3$, or about $753.6$ ft$^3$
134. $600 = \tfrac{1}{3}(225)h = 75h$, so $h = 8$ m.
135. $48\pi = \tfrac{1}{3}\pi r^2(4)$. Dividing by $\pi$ gives $48 = \tfrac{4}{3}r^2$, so $r^2 = 36$ and $r = 6$ cm. (Only the positive root makes sense for a length.)
136. Volume needs the height. Half the base edge is $6$ cm: $h^2 = 10^2 - 6^2 = 64$, so $h = 8$ cm. Then $V = \tfrac{1}{3}(144)(8) = 384$ cm$^3$.

Part C — Explaining the volume relationships (8.MG.2c)

137. Cylinder: $V = \pi(25)(9) = 225\pi$ cm$^3$. Cone with the same base and height: $\tfrac{1}{3}(225\pi) = 75\pi$ cm$^3$, or about $235.5$ cm$^3$.
138. Prism: $V = 10^2(9) = 900$ in$^3$. Pyramid: $\tfrac{1}{3}(900) = 300$ in$^3$.
139. $3(16\pi) = 48\pi$ ft$^3$, or about $150.72$ ft$^3$.
140. Take a cube with edge $s$ and choose one vertex. Join that vertex to the four corners of each of the three faces it does not touch. This cuts the cube into three pieces, and each piece is a square-based pyramid whose base is a face of the cube — so the base area is $s^2$ — and whose height is $s$, the distance from that face to the chosen vertex. The three pyramids are congruent and use up the entire cube with nothing left over, so each one is exactly one third of the cube: $V = \tfrac{1}{3}s^3 = \tfrac{1}{3}s^2 \cdot s = \tfrac{1}{3}Bh$. Since the cube is a prism with the same square base and the same height as each pyramid, a square-based pyramid is one third of that prism.
141. Take a cone and a cylinder that have the same circular base and the same height. Fill the cone with water and empty it into the cylinder; repeat. After the third pour the cylinder is exactly full, with nothing left over and no room remaining. So the cylinder's capacity is three cone-fulls, which means the cone holds $\tfrac{1}{3}$ of the cylinder, and therefore $V_{\text{cone}} = \tfrac{1}{3}\pi r^2h$. Both conditions are required: if the bases differ, the two solids do not even start with the same amount of area on the bottom, and if the heights differ, one solid stops rising before the other. Change either one and the ratio is no longer one third — a cone with $r = 4$, $h = 3$ holds $16\pi$, while a cylinder with $r = 2$, $h = 3$ holds only $12\pi$.
142. The statement is missing its conditions. It should read: a cone holds one third of the cylinder that has the same base and the same height. Counterexample: a cone with radius $4$ and height $3$ has volume $\tfrac{1}{3}\pi(16)(3) = 16\pi$, while a cylinder with radius $2$ and height $3$ has volume $\pi(4)(3) = 12\pi$. Here the cone is larger than the cylinder, so it is certainly not one third of it — the heights match but the bases do not.

Part D — Problems in context (8.MG.2d)

143. $V = \tfrac{1}{3}\pi(16)(6) = 32\pi$ cm$^3$, or about $100.48$ cm$^3$
144. Covering the whole outside is surface area including the base, and the height was given. Half the base edge is $9$ cm: $\ell^2 = 12^2 + 9^2 = 225$, $\ell = 15$ cm. Then $SA = 18^2 + 2(18)(15) = 324 + 540 = 864$ cm$^2$.
145. $r = 7$ ft, so $V = \tfrac{1}{3}\pi(49)(3) = 49\pi$ ft$^3$, or about $153.86$ ft$^3$.
146. Four sloped faces only, slant height given: $LA = 2(30)(25) = 1500$ m$^2$.
147. $V = \tfrac{1}{3}\pi(36)(10) = 120\pi$ cm$^3$, or about $376.8$ cm$^3$. Since $376.8 \div 100 = 3.768$, one funnel fills 3 full jars, with about $76.8$ cm$^3$ left over — not enough for a fourth full jar.
148. a) Soil is a volume question: $V = \tfrac{1}{3}(144)(8) = 384$ in$^3$. b) The liner covers the four lateral faces only, so lateral area, and the height was given. Half the base edge is $6$ in: $\ell^2 = 8^2 + 6^2 = 100$, $\ell = 10$ in. Then $LA = 2(12)(10) = 240$ in$^2$.
149. Volume: $V = \tfrac{1}{3}(36)(4) = 48$ cm$^3$. Surface area: half the base edge is $3$ cm, so $\ell^2 = 4^2 + 3^2 = 25$ and $\ell = 5$ cm, giving $SA = 36 + 2(6)(5) = 96$ cm$^2$. The units differ because the two numbers measure different things. Volume counts unit cubes filling the inside, and a cube is three lengths multiplied, so the unit is cm$^3$. Surface area counts unit squares covering the outside, and a square is two lengths multiplied, so the unit is cm$^2$. The two answers are not comparable as numbers, and it is a coincidence of these particular dimensions that one is twice the other.
150. Sample response. Cone, volume. A cone-shaped paper cup has radius $3$ cm and height $8$ cm. How much water does it hold? Holding water is a volume question: $V = \tfrac{1}{3}\pi(9)(8) = 24\pi$ cm$^3$, or about $75.36$ cm$^3$. Square-based pyramid, surface area. A closed gift box is a square-based pyramid with base edge $10$ cm and height $12$ cm. How much paper covers it? Covering is surface area, all five faces, and the height was given, so first $\ell^2 = 12^2 + 5^2 = 169$ and $\ell = 13$ cm; then $SA = 10^2 + 2(10)(13) = 360$ cm$^2$. Accept any problem where the cone question asks about filling, holding, or capacity and is answered in cubic units, and the pyramid question asks about covering, wrapping, or painting and is answered in square units, with correct arithmetic and correct use of $h$ for volume and $\ell$ for surface area.

Workbook-only items

Page 2, fill in the blanks. The point where all four triangles meet is the apex. The four triangular faces are called lateral faces. The height $h$ runs from the apex to the center of the base. The slant height $\ell$ runs from the apex to the midpoint of a base edge. The longer of the two is always the slant height. Volume uses the height. Surface area uses the slant height.

Page 7, build it from the net. The base is a square with side $s$, so its area is $s^2$. Each lateral face is a triangle with base $s$ and height $\ell$ (the slant height), so its area is $\tfrac{1}{2}s\ell$. There are 4 lateral faces.

Page 12, fill in the blanks. A cone with the same base and height holds one third of the cylinder. A square-based pyramid with the same base and height holds one third of the prism. One formula covers both: $V = \tfrac{1}{3}Bh$, where $B$ is the area of the base. It takes 3 cone-fulls to fill the cylinder.