Appendix A — Answer Key, Chapter 16: Surface Area and Volume of Solids

SOL G.DF.1 (a, b, c, d) · Covers textbook Chapter 16 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 124 across the chapter.

Conventions used in every answer below. A slant height $l$ is always found from the radius (or half the base) and the height, using the Pythagorean Theorem — it is never given directly. Answers involving $\pi$ are left as an exact coefficient, never a rounded decimal. A composite solid's surface area adds each piece's own full surface area, then subtracts the shared, internal face TWICE.

Lesson 16.1 — Cross Sections

Guided practice

1. A rectangle, and it stays the same size everywhere along the prism.
2. The cylinder gives a circle; the cone also gives a circle. Difference: the cylinder's circle is always the same size as its base, but the cone's circle shrinks the closer the cut is to the apex.
3. A circle. A sphere is round in every direction, not just around one axis, so there is no special "parallel" direction to name — any cut at all produces a circle.
4. A rectangle.
5. Both give a triangle. In common: both cuts pass through the apex, where the slanted or curved surface comes to a single point on each side of the cut.
6. Whether the plane is parallel to the base (giving a shape matching, or shrinking from, the base) or passes through the apex or axis (giving a rectangle or a triangle instead).

Independent practice

7. Circle
8. Rectangle
9. Circle (smaller than the base)
10. Triangle
11. Triangle
12. Square (smaller than the base)
13. Rectangle
14. Triangle, congruent to the base
15. Rectangle
16. Circle — the largest possible cross section (a "great circle")
17. Circle, smaller than a great circle
18. Circle, the same size as the base — a cylinder's parallel cross section never changes size
19. Triangle, congruent to the doorstop's own triangular ends.
20. The shape is always a circle, but its size shrinks the closer the cut is to the apex — only a cut at the base itself matches the base's size.
21. Every point of a sphere is the same distance from its center, so any flat slice leaves a boundary that is still a circle, no matter the angle or position of the plane. A cylinder or a cone is round around only one axis, so a slice has to be parallel to the circular base to produce a circle; a slice at a different angle — through the axis or the apex — crosses the straight or slanted sides instead, producing a rectangle or a triangle.

Exit ticket 16.1

22. Circle; no, its size does not depend on where the cut is made.
23. Triangle.

Lesson 16.2 — Prisms

Guided practice

24. $l=10$, $w=6$, $h=4$. $V=10(6)(4)=$ $240$.
25. $SA=2(60+40+24)=$ $248$. Each product is doubled because each of the three rectangular faces has a matching, identical face on the opposite side of the prism.
26. base $=6$, height $=4$, depth $=10$. Area $=\frac{1}{2}(6)(4)=$ $12$.
27. leg $=\sqrt{3^2+4^2}=5$, from half the base ($3$) and the height ($4$). Perimeter $=6+5+5=$ $16$.
28. $V=12(10)=$ $120$.
29. $SA=2(12)+16(10)=24+160=$ $184$. The "two matching ends" are the prism's two triangular faces.

Independent practice

30. $V=24$, $SA=52$
31. $V=60$, $SA=94$
32. $V=42$, $SA=82$
33. $V=90$, $SA=146$
34. $V=216$, $SA=216$
35. $V=160$, $SA=184$
36. $V=40$, $SA=88$
37. $V=180$, $SA=222$
38. leg $=10$, area $=48$, perimeter $=32$. $V=240$, $SA=256$
39. leg $=13$, area $=60$, perimeter $=36$. $V=420$, $SA=372$
40. leg $=17$, area $=120$, perimeter $=50$. $V=720$, $SA=540$
41. leg $=15$, area $=108$, perimeter $=48$. $V=432$, $SA=408$
42. $V=4(2)(3)=$ $24$ ft³. $SA=2(8+12+6)=$ $52$ ft².
43. The student doubled the entire product $lwh$ instead of doubling each of the three different face products. Correct: $SA=2(20+15+12)=$ $94$.
44. Both formulas add two matching "end" faces plus the faces running the depth in between — a rectangular prism's ends are one chosen pair of rectangular faces (with the other two pairs making up $2lh+2wh$), and a triangular prism's ends are its two triangular faces (with the perimeter-by-depth strip making up the rest).

Exit ticket 16.2

45. $V=6(3)(2)=$ $36$. $SA=2(18+12+6)=$ $72$.
46. leg $=10$, area $=48$, perimeter $=32$. $V=48(3)=$ $144$. $SA=2(48)+32(3)=96+96=$ $192$.

Lesson 16.3 — Cylinders and Cones

Guided practice

47. $r=3$, $h=7$. $V=\pi(9)(7)=$ $63\pi$.
48. $SA=18\pi+42\pi=$ $60\pi$. $2\pi rh$ is the area of the rectangle the curved side unrolls into — width $2\pi r$ (the circle's circumference) and height $h$.
49. $r=3$, $h=4$. $l=\sqrt{9+16}=$ $5$.
50. $V=\frac{1}{3}\pi(9)(4)=$ $12\pi$.
51. $SA=9\pi+15\pi=$ $24\pi$. Only one $\pi r^2$ term because a cone has only one flat circular base — it comes to a point instead of a second flat end.
52. $\pi r^2$ is the area of the circular base in both formulas — the "floor" each solid is built on.

Independent practice

53. $V=160\pi$, $SA=112\pi$
54. $V=150\pi$, $SA=110\pi$
55. $V=36\pi$, $SA=44\pi$
56. $V=490\pi$, $SA=238\pi$
57. $V=108\pi$, $SA=90\pi$
58. $l=10$. $V=96\pi$, $SA=96\pi$
59. $l=13$. $V=100\pi$, $SA=90\pi$
60. $l=17$. $V=320\pi$, $SA=200\pi$
61. $l=15$. $V=324\pi$, $SA=216\pi$
62. $l=25$. $V=392\pi$, $SA=224\pi$
63. $V=\pi(16)(9)=$ $144\pi$ ft³. $SA=32\pi+72\pi=$ $104\pi$ ft².
64. The student used $h$ in place of the slant height $l$. Correct $l=\sqrt{25+144}=13$; correct $SA=25\pi+65\pi=$ $90\pi$.
65. $V_{cone}=\frac{1}{3}\pi r^2h$ and $V_{cyl}=\pi r^2h$ share the identical $\pi r^2h$ — the cone's formula simply carries an extra factor of $\frac{1}{3}$, so for any matching $r$ and $h$, the cone's volume is always exactly one third of the cylinder's.

Exit ticket 16.3

66. $V=\pi(36)(5)=$ $180\pi$. $SA=72\pi+60\pi=$ $132\pi$.
67. $l=\sqrt{144+256}=$ $20$. $V=\frac{1}{3}\pi(144)(16)=$ $768\pi$. $SA=144\pi+240\pi=$ $384\pi$.

Lesson 16.4 — Pyramids and Spheres

Guided practice

68. $\frac{b}{2}=4$, $h=3$. $l=\sqrt{16+9}=$ $5$.
69. $V=\frac{1}{3}(64)(3)=$ $64$.
70. $SA=64+80=$ $144$. $2bl$ comes from the pyramid's four triangular faces: each has area $\frac{1}{2}bl$, and four of them sum to $4\left(\frac{1}{2}bl\right)=2bl$.
71. $r=6$. $V=\frac{4}{3}\pi(216)=$ $288\pi$.
72. $SA=4\pi(36)=$ $144\pi$. Only one measurement needed because a sphere is completely described by its radius alone — no separate height or depth.
73. The cone's and the square pyramid's surface-area formulas — both need $l$, found from $r$ (or $\frac{b}{2}$) and $h$ with the Pythagorean Theorem.

Independent practice

74. $l=5$. $V=48$, $SA=96$
75. $l=10$. $V=384$, $SA=384$
76. $l=13$. $V=400$, $SA=360$
77. $l=17$. $V=1280$, $SA=800$
78. $l=15$. $V=1296$, $SA=864$
79. $V=36\pi$, $SA=36\pi$
80. $V=972\pi$, $SA=324\pi$
81. $V=2304\pi$, $SA=576\pi$
82. $V=4500\pi$, $SA=900\pi$
83. half $=20$, $l=\sqrt{400+225}=25$. $V=\frac{1}{3}(1600)(15)=$ $8000$ ft³. Four triangular faces' area $=2bl=2(40)(25)=$ $2000$ ft².
84. The student forgot to multiply by $\frac{1}{3}$. Correct $V=\frac{1}{3}(100)(12)=$ $400$.
85. $V=\frac{4}{3}\pi r^3$ and $SA=4\pi r^2$ have equal coefficients exactly when $\frac{4}{3}r^3=4r^2$, which simplifies (dividing both sides by $4r^2$) to $\frac{r}{3}=1$, so $r=3$ — a coincidence of that one number, not a general property. Even at $r=3$, volume is measured in cubic units and surface area in square units, so the two are still different kinds of quantity that merely share a numeral here.

Exit ticket 16.4

86. $l=\sqrt{49+576}=$ $25$. $V=\frac{1}{3}(196)(24)=$ $1568$. $SA=196+700=$ $896$.
87. $V=\frac{4}{3}\pi(216)=$ $288\pi$. $SA=4\pi(36)=$ $144\pi$.

Lesson 16.5 — Composite Solids

Guided practice

88. cylinder $r=3$, $h=8$; hemisphere radius $3$.
89. $V=72\pi+18\pi=$ $90\pi$. The hemisphere's own flat circle sits exactly where the cylinder's top circle used to be — it is now inside the finished solid rather than on its outer surface, so neither one counts toward $SA$.
90. box: $l=10$, $w=6$, $h=4$; roof: base $6$, height $4$, depth $10$. Shared face $=10(6)=$ $60$.
91. $V=240+120=$ $360$. $SA=248+184-2(60)=432-120=$ $312$. The shared rectangle is subtracted twice because it is counted once inside the box's own full $SA$ and once again inside the roof's own full $SA$ — both counts must be removed, since the face sits nowhere on the finished solid's actual exterior.
92. shared radius $3$. Neither piece has a second flat face: the cone's only flat face is its base, which is the shared circle; the hemisphere's only flat face is its own flat circle, which is that same circle.
93. $V=12\pi+18\pi=$ $30\pi$. $SA=15\pi+18\pi=$ $33\pi$.

Independent practice

94. cone $l=5$. $V=160\pi+16\pi=$ $176\pi$. $SA=80\pi+16\pi+20\pi=$ $116\pi$.
95. box $V=180$, $SA=192$; pyramid $l=5$, $V=48$, full $SA=96$; shared $=36$. Total $V=228$. Total $SA=192+96-72=$ $216$.
96. $V=360\pi+144\pi=$ $504\pi$. $SA=120\pi+36\pi+72\pi=$ $228\pi$.
97. cone $l=10$. $V=96\pi+144\pi=$ $240\pi$. $SA=60\pi+72\pi=$ $132\pi$.
98. $V=360\pi+288\pi=$ $648\pi$. $SA=120\pi+144\pi=$ $264\pi$.
99. The shared circle (area $9\pi$, where the hemisphere meets the cylinder's top) was counted twice instead of zero times. $93\pi-2(9\pi)=93\pi-18\pi=$ $75\pi$.
100. Each piece's own "full surface area," computed as if it stood alone, counts the shared face exactly once — so adding both pieces' full areas counts that face twice, when it should appear zero times on the finished, joined solid (it's internal, not exterior). Subtracting it twice removes exactly that excess, landing on the same total a direct count of only the exposed faces would give.

Exit ticket 16.5

101. The circle where the hemisphere meets the cylinder's top. Area $=\pi(3)^2=$ $9\pi$.
102. $V=$ $30\pi$, $SA=$ $33\pi$. Included: the cone's lateral (slanted) surface and the hemisphere's curved surface — no flat circle appears anywhere in the total.

Lesson 16.6 — Working Backward

Guided practice

103. Given: $V=100\pi$, $r=5$. Unknown: $h$.
104. $100\pi=\pi(25)h \rightarrow$ $h=4$.
105. Given: $SA=96$, $b=6$. Unknown: $l$, then $h$.
106. $96=36+12l \rightarrow 60=12l \rightarrow$ $l=5$.
107. $h^2=5^2-3^2=25-9=16 \rightarrow$ $h=4$.
108. Same: substitute everything known, then undo the remaining operations one at a time. Different: the cylinder needs a single division step, while the pyramid needs two stages — first isolate $l$ algebraically, then a separate Pythagorean Theorem step to reach $h$.

Independent practice

109. $150=50h \rightarrow$ $h=3$.
110. $94=40+18h \rightarrow 54=18h \rightarrow$ $h=3$.
111. $98\pi=49\pi h \rightarrow$ $h=2$.
112. $136\pi=32\pi+8\pi h \rightarrow 104\pi=8\pi h \rightarrow$ $h=13$.
113. $48\pi=12\pi h \rightarrow$ $h=4$.
114. $90\pi=25\pi+5\pi l \rightarrow 65\pi=5\pi l \rightarrow l=13$; $h^2=169-25=144 \rightarrow$ $h=12$.
115. $972\pi=\frac{4}{3}\pi r^3 \rightarrow r^3=729 \rightarrow$ $r=9$.
116. $400\pi=4\pi r^2 \rightarrow r^2=100 \rightarrow$ $r=10$.
117. $400=\frac{1}{3}(100)h \rightarrow$ $h=12$; $l^2=5^2+12^2=169 \rightarrow$ $l=13$.
118. $180=12(\text{depth}) \rightarrow$ depth $=15$.
119. $152=24+16(\text{depth}) \rightarrow 128=16(\text{depth}) \rightarrow$ depth $=8$.
120. $250\pi=25\pi h \rightarrow$ $h=10$ ft.
121. The student forgot to square the radius. Correct: $200\pi=25\pi h \rightarrow$ $h=8$.
122. A cone's or a pyramid's $SA$ formula mixes a squared term ($r^2$ or $b^2$) with a separate slant-height term ($rl$ or $bl$); isolating $l$ algebraically is only the first stage, since $l$ still has to feed into a second, separate Pythagorean Theorem equation to reach $h$. A volume formula uses $h$ directly, with no slant height anywhere in it, so it never needs that second stage.

Exit ticket 16.6

123. $48\pi=16\pi h \rightarrow$ $h=3$.
124. $216\pi=81\pi+9\pi l \rightarrow 135\pi=9\pi l \rightarrow l=15$; $h^2=225-81=144 \rightarrow$ $h=12$.

Chapter 16 Review — answers

Review 1 (G.DF.1a).

• Rectangle.
• Rectangle.
• Circle, smaller than the base — a cone's parallel cross section shrinks toward the apex.
• Circle — every point of a sphere is the same distance from its center, so any flat cut leaves a circular boundary no matter where the plane sits.

Review 2 (G.DF.1 b, c).

• Box $V=8(6)(7)=336$; roof: area $=\frac{1}{2}(6)(4)=12$, $V=12(8)=96$. Total $V=432$.
• Box $SA=2(48+56+42)=292$; roof: leg $=5$, perimeter $=16$, full $SA=2(12)+16(8)=152$; shared face $=6(8)=48$. Total $SA=292+152-2(48)=348$.
• The shared face is counted once in the box's own full $SA$ and once again in the roof's own full $SA$ — both copies must be removed, since that face is internal to the finished shed, not part of its actual exterior.

Review 3 (G.DF.1d).

• $25\pi+5\pi l=90\pi \rightarrow$ $l=13$.
• $h^2=13^2-5^2=169-25=144 \rightarrow$ $h=12$.
• $V=\frac{1}{3}\pi(25)(12)=$ $100\pi$.
• Direct check: $\frac{1}{3}\pi(5)^2(12)=\frac{1}{3}\pi(25)(12)=100\pi$ — matches.

Every item in Chapter 16 is answered above: 1 to 124, plus the three chapter reviews.