Appendix A — Answer Key, Chapter 17: Changing Dimensions

SOL G.DF.2 (a, b, c) · Covers textbook Chapter 17 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 98 across the chapter.

Conventions used in every answer below. Scale factor is always $k$. When every dimension of a figure changes by the same $k$: length-like measures (length, perimeter, circumference) scale by $k$; area-like measures (area, surface area) scale by $k^2$; volume scales by $k^3$. When only one dimension changes, volume — a single product — still scales by that dimension's exact power, but surface area — a sum of differently-shaped terms — generally does not, and must be recomputed directly. Recovering $k$ from a ratio uses the matching root: none for a length ratio, a square root for an area or surface-area ratio, a cube root for a volume ratio.

Lesson 17.1 — Scale Factor: Perimeter and Area

Guided practice

1. $P = 20$, $A = 24$, $k = 2$.
2. $P = 40$, $A = 96$. $40 = 20 \times 2$; $96 = 24 \times 2^2$.
3. Original $P = 24$; new $P = 48$.
4. Original $A = 24$; new $A = 96$. Area needed $k^2$ because it is built from two lengths multiplied together, each contributing one factor of $k$.
5. Perimeter adds lengths; each length contributes exactly one factor of $k$, and adding doesn't multiply factors of $k$ together.
6. Area always multiplies exactly two lengths, whatever shape those two lengths belong to — a base and a height, or two legs — so it always picks up $k$ twice.

Independent practice

7. $P = 21$, $A = 90$
8. $P = 40$, $A = 96$
9. $P = 104$, $A = 640$
10. $P = 60$, $A = 225$
11. $P = 48$, $A = 96$
12. $P = 90$, $A = 270$
13. $P = 16$, $A = 15$
14. $P = 30$, $A = 54$
15. $C = 24\pi$, $A = 144\pi$
16. $C = 20\pi$, $A = 100\pi$
17. $C = 8\pi$, $A = 16\pi$
18. $k = 3$
19. $k = \dfrac{2}{5}$
20. $k = 4$
21. $k = \dfrac{7}{2}$
22. $P: 26 \rightarrow 78$ (3 times as much fencing). $A: 40 \rightarrow 360$ (9 times as much sod).
23. Correct new area: $l=12, w=6 \rightarrow A = 72$, not $8 \times 3 = 24$. The student multiplied the AREA by $k$ instead of by $k^2$; area needs two factors of $k$ because it is a product of two scaled lengths.
24. $A_{\text{new}} = (kl)(kw) = k^2(lw) = k^2 A$ — the two factors of $k$, one from each scaled length, multiply together rather than adding, so area picks up $k$ twice.

Exit ticket 17.1

25. $P = 26 \rightarrow 52$, $A = 36 \rightarrow 144$.
26. $C = 14\pi \rightarrow 42\pi$, $A = 49\pi \rightarrow 441\pi$.

Lesson 17.2 — Scale Factor: Surface Area and Volume

Guided practice

27. $l=2,w=3,h=4$; $k=2$; new dimensions $4,6,8$.
28. $V: 24 \rightarrow 192$; ratio $= 8 = 2^3$.
29. $SA: 52 \rightarrow 208$; ratio $= 4 = 2^2$.
30. $r=1,h=4$; $k=3$; new $r=3,h=12$.
31. $V$ ratio $= 27 = 3^3$; $SA$ ratio $= 9 = 3^2$.
32. The slant height is the hypotenuse of a right triangle whose legs are $r$ and $h$. Both legs scale by $k$, and scaling both legs of a right triangle by $k$ scales its hypotenuse by $k$ too.

Independent practice

33. $V = 192$, $SA = 208$
34. $V = 270$, $SA = 306$
35. $V = 216$, $SA = 216$
36. $V = 108\pi$, $SA = 90\pi$
37. $V = 4\pi$, $SA = 12\pi$
38. $V = 4500\pi$, $SA = 1100\pi$
39. $l = 10$; $V = 96\pi$, $SA = 96\pi$
40. $l = 5$; $V = 12\pi$, $SA = 24\pi$
41. $l = 10$; $V = 384$, $SA = 384$
42. $V = 7776\pi$, $SA = 1296\pi$
43. $k = 3$
44. $k = \dfrac{1}{5}$
45. $k = 5$
46. $k = \dfrac{2}{7}$
47. original: $V=288\pi$, $SA=144\pi$; new: $V=2304\pi$, $SA=576\pi$ — 8 times the air, 4 times the material.
48. Actual new $V$: $r=6,h=15 \rightarrow V = \pi(36)(15) = 540\pi$, which is $27 \times 20\pi$, not $3\times$. The student used $k$ itself instead of $k^3$ for volume.
49. Surface area is always built from products of exactly two scaled lengths, so it always picks up $k^2$; volume is always built from products of exactly three scaled lengths, so it always picks up $k^3$ — true for every solid's formula, whatever shape it is.

Exit ticket 17.2

50. $V = 480$, $SA = 376$.
51. $V = 288\pi$, $SA = 144\pi$.

Lesson 17.3 — Changing One Dimension

Guided practice

52. The length, doubled ($3 \rightarrow 6$).
53. $V: 30 \rightarrow 60$. It scaled by exactly $2$ because $V=lwh$ is a single product, and $l$ is one factor in it.
54. $SA: 62 \rightarrow 104$, not $124$. Only the two terms containing $l$ ($lw$ and $lh$) changed; the third term, $wh$, has no $l$ in it at all and stayed fixed.
55. $V: 24\pi \rightarrow 216\pi$ ($\times 9$), because $r$ is squared in $V=\pi r^2h$, so tripling it alone multiplies $V$ by $3^2$.
56. $SA: 48\pi \rightarrow 80\pi$, not $144\pi$. Only $2\pi rh$ contains $h$; the two end circles, $2\pi r^2$, are unaffected by a height change.
57. A sphere has only one dimension; there is no second, independent length to hold fixed while the radius changes, so any change to it changes every "direction" of the sphere at once — exactly what uniform scaling means.

Independent practice

58. $V = 30 \rightarrow 60$; $SA = 62 \rightarrow 104$
59. $V = 48 \rightarrow 144$; $SA = 88 \rightarrow 168$
60. $V = 60 \rightarrow 30$; $SA = 94 \rightarrow 62$
61. $V = 40 \rightarrow 160$; $SA = 88 \rightarrow 328$
62. $V = 72 \rightarrow 144$; $SA = 108 \rightarrow 180$
63. $V = 27 \rightarrow 54$; $SA = 54 \rightarrow 90$
64. $V = 24\pi \rightarrow 216\pi$; $SA = 32\pi \rightarrow 144\pi$
65. $V = 75\pi \rightarrow 300\pi$; $SA = 80\pi \rightarrow 260\pi$
66. $V = 32\pi \rightarrow 96\pi$; $SA = 48\pi \rightarrow 80\pi$
67. $V = 90\pi \rightarrow 45\pi$; $SA = 78\pi \rightarrow 48\pi$
68. $V = 144\pi \rightarrow 36\pi$; $SA = 120\pi \rightarrow 42\pi$
69. $V = 8\pi \rightarrow 32\pi$; $SA = 18\pi \rightarrow 66\pi$
70. $V = 36\pi \rightarrow 288\pi$; $SA = 36\pi \rightarrow 144\pi$
71. $V = 972\pi \rightarrow 26244\pi$; $SA = 324\pi \rightarrow 2916\pi$
72. original $V = 90\pi$; new $V = 270\pi$ — 3 times as much.
73. original $SA = 72\pi$; new (r doubled to $8$) $SA = 208\pi$, not $144\pi$. Doubling $r$ alone squares its own term ($2\pi r^2 \rightarrow 4\times$) but only doubles the other ($2\pi rh \rightarrow 2\times$) — two different factors, so the total cannot double.
74. $V=lwh$ is a single product with the changed dimension as one factor, so the whole product scales by exactly that factor. $SA=2(lw+lh+wh)$ is a sum of three products; only the ones containing the changed dimension move, so the sum as a whole does not follow a single scale factor.

Exit ticket 17.3

75. $V = 24 \rightarrow 72$; $SA = 52 \rightarrow 132$.
76. $V = 100\pi \rightarrow 400\pi$; $SA = 90\pi \rightarrow 280\pi$.

Lesson 17.4 — Working Backward

Guided practice

77. $k = \sqrt[3]{27} = 3$.
78. $SA$ scales by $k^2 = 9$: $24 \rightarrow 216$.
79. $k = \sqrt{4} = 2$.
80. $V$ scales by $k^3 = 8$: $972\pi \rightarrow 7776\pi$.
81. The radius doubled ($4 \rightarrow 8$ ft), height unchanged. $V: 160\pi \rightarrow 640\pi$ ft³ — $\times 4$.
82. A volume ratio needs a cube root; a surface-area ratio needs a square root.

Independent practice

83. $k=2$; $SA \times 4$
84. $k=5$; $SA \times 25$
85. $k=3$; $V \times 27$
86. $k=5$; $V \times 125$
87. $k=\dfrac{1}{2}$; $SA \times \dfrac{1}{4}$
88. $k=\dfrac{1}{4}$; $V \times \dfrac{1}{64}$
89. Every dimension changed (uniform scaling). $SA \times 4$, $V \times 8$.
90. Only one dimension changed. $V \times 4$ (exact, since $r$ is squared in a single product). $SA$ cannot be found the same way because its formula has one term with $r^2$ and one with just $r$ — they scale differently, so the sum must be recomputed.
91. Every dimension changed. $SA \times 9$, $V \times 27$.
92. Only one dimension changed. $V \times \dfrac{1}{2}$ (exact — height is one factor in $V=lwh$).
93. $k = \dfrac{6}{2} = 3$. Cardboard (SA) $\times 9$; volume $\times 27$.
94. ratio $= \dfrac{576}{64} = 9 \rightarrow k = \sqrt{9} = 3$. Volume $\times 3^3 = 27$.
95. The student treated the volume ratio itself as $k$. The correct scale factor is $k = \sqrt[3]{8} = 2$.
96. $V_{\text{new}} = k^3 \cdot V$, so isolating $k$ from a volume ratio undoes a cube — a cube root. $SA_{\text{new}} = k^2 \cdot SA$, so isolating $k$ from a surface-area ratio undoes a square — a square root. The root always matches the exponent it's undoing.

Exit ticket 17.4

97. $k = \sqrt[3]{64} = 4$; $SA \times 16$.
98. $V \times 3$ (exact — height is a single factor in $V=\pi r^2h$).

Chapter 17 Review — answers

Review 1 (G.DF.2a).

• $P = 30 \rightarrow 90$.
• $A = 54 \rightarrow 486$.
• Triangle: $A = \dfrac{1}{2}(9)(12) = 54 \rightarrow 54 \times 3^2 = 486$.
• Every one of these areas is a product of exactly two lengths, and scaling both of those lengths by $k$ multiplies the product by $k \times k = k^2$ — true no matter which two lengths (base and height, or two legs) the shape's area formula uses.

Review 2 (G.DF.2a).

• $V = \pi(9)(8) = 72\pi$; $SA = 2\pi(9) + 2\pi(3)(8) = 18\pi+48\pi=66\pi$.
• Only $h$ doubles ($8 \rightarrow 16$): $V = \pi(9)(16) = 144\pi$ — exactly double, since $V$ is a single product and $h$ is one factor.
• $SA = 18\pi + 2\pi(3)(16) = 18\pi+96\pi = 114\pi$ — not double ($132\pi$). The two end circles ($18\pi$) don't contain $h$ at all, so they stay fixed while only the lateral term grows.
• Both $r$ and $h$ doubled: $V = \pi(36)(16) = 576\pi$ ($\times 8 = k^3$); $SA = 2\pi(36)+2\pi(6)(16) = 72\pi+192\pi=264\pi$ ($\times 4 = k^2$). This candle's $SA$ scaled cleanly because every dimension changed together — every term of the $SA$ formula picked up the same two factors of $k$, unlike the height-only candle, where the term without $h$ was left out of the growth entirely.

Review 3 (G.DF.2 b, c).

• ratio $= \dfrac{784}{196} = 4 \rightarrow k = \sqrt{4} = 2$.
• $4\pi r^2 = 196\pi \rightarrow r^2 = 49 \rightarrow r = 7$; new radius $= 14$.
• $V$ factor $= k^3 = 8$.
• No. Tripling only the rectangular tank's length is a single-dimension change, which scales its volume by exactly that one factor — $\times 3$ — because $V=lwh$ is a single product with length as one factor. The spherical tank's volume grew by $\times 8$ because ALL of its dimensions (its one radius) scaled together by $k=2$. A single-dimension change and a uniform scaling by the same-looking "growth" almost never produce the same volume factor.

Every item in Chapter 17 is answered above: 1 to 98, plus the three chapter reviews.