Appendix A — Answer Key, Chapter 13: Similar Figures and Scale Drawings

SOL 7.MG.2 · Covers textbook Chapter 13 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.

Lesson 13.1 — What Similarity Means

Guided practice

1. $\dfrac{6}{3} = 2$, $\dfrac{8}{4} = 2$, $\dfrac{10}{5} = 2$. All three ratios agree and the corresponding angles are congruent, so the triangles are similar. The scale factor from the smaller triangle to the larger one is $2$.
2. $\tfrac12$. Each small side is half its partner: $3 \div 6 = 4 \div 8 = 5 \div 10 = \tfrac12$.
3. $\dfrac{8}{4} = 2$ and $\dfrac{10}{6} = \tfrac53 \approx 1.7$. The ratios disagree, so the rectangles are not similar.
4. Yes, similar, with scale factor $1$. Because the scale factor is $1$, they are also congruent.
5. congruent; proportional

Independent practice

6. a) $\dfrac{8}{6} = \dfrac{12}{9} = \dfrac{16}{12} = \tfrac43$. Similar, scale factor $\tfrac43$. b) $\dfrac{10}{5} = 2$, $\dfrac{14}{7} = 2$, $\dfrac{20}{9} \approx 2.2$. Not similar; the third ratio disagrees. (For similarity the third side would have to be $18$.) c) $\dfrac{7}{4}$ for all four sides, and every angle in a square is $90°$. Similar, scale factor $\tfrac74$.
7. $21$, $27$, $33$
8. $4$, $5$, $6$
9. $TUVW$ has sides $20$, $10$, $8$, $10$. The scale factor from $TUVW$ back to $PQRS$ is $\tfrac12$.
10. $\dfrac{9}{3} = 3$ and $\dfrac{15}{5} = 3$. The ratios agree and all angles are right angles, so the rectangles are similar with scale factor $3$.
11. Scale factor $= 10 \div 4 = 2.5$. New length $= 6 \times 2.5 = 15$ inches.
12. Counterexample: a $4$ by $6$ rectangle and an $8$ by $10$ rectangle. Their angles are all right angles, but $\tfrac84 = 2$ while $\tfrac{10}{6} = \tfrac53$, so the sides are not proportional. Jamal checked only the angle condition; similarity also requires proportional corresponding sides.

Exit ticket 13.1

13. $\dfrac{6}{4} = \dfrac{9}{6} = \dfrac{12}{8} = 1.5$. Similar, scale factor $1.5$ (that is, $\tfrac32$).
14. $6$, $8$, $10$
15. $\dfrac{4}{2} = 2$ and $\dfrac{8}{5} = 1.6$. Not similar.
16. Because it leaves out the angle condition and does not say what "same shape" means precisely. Similar figures need corresponding angles congruent and corresponding sides proportional. A $4$ by $6$ rectangle and an $8$ by $10$ rectangle are both rectangles at different sizes, yet they are not similar, because one has been stretched more in one direction than the other.

Lesson 13.2 — Corresponding Angles and Sides

Guided practice

17. $\angle E$
18. $\overline{DF}$ (equivalently $\overline{FD}$)
19. $\angle V$; and $\overline{RS}$ corresponds to $\overline{VW}$
20. $m\angle C = 180° - 50° - 60° = 70°$; $m\angle D = 50°$, $m\angle E = 60°$, $m\angle F = 70°$
21. $\angle H \cong \angle L$; $\overline{GJ}$ corresponds to $\overline{KM}$

Independent practice

22. Angles: $\angle R \cong \angle X$, $\angle S \cong \angle Y$, $\angle T \cong \angle Z$. Sides: $\overline{RS} \leftrightarrow \overline{XY}$, $\overline{ST} \leftrightarrow \overline{YZ}$, $\overline{TR} \leftrightarrow \overline{ZX}$.
23. $\angle G$; and $\overline{DA}$ corresponds to $\overline{HE}$
24. $m\angle F = 180° - 95° - 35° = 50°$. Then $m\angle A = 95°$, $m\angle B = 35°$, $m\angle C = 50°$.
25. $m\angle S = 360° - 110° - 70° - 110° = 70°$. For $TUVW$: $m\angle T = 110°$, $m\angle U = 70°$, $m\angle V = 110°$, $m\angle W = 70°$.
26. False. $\overline{AB}$ corresponds to $\overline{DE}$.
27. $\overline{XY}$. Since $m\angle R = 180° - 42° - 108° = 30°$ and $\angle R \cong \angle Y$, $m\angle Y = 30°$.
28. The three angles of a triangle add to $180°$, so two known measures determine the third. Corresponding angles of similar figures are congruent, so all three measures repeat in the matching positions of the second triangle. That accounts for all six.

Exit ticket 13.2

29. $\angle F$
30. $\overline{EF}$
31. $\angle C \cong \angle F$, so $m\angle C = 81°$. Then $m\angle B = 180° - 64° - 81° = 35°$.
32. Matching arcs mean the two angles have equal measure, which identifies them as a corresponding pair. Once the vertices are paired that way, the sides pair off too, because each side is named by two vertices that already have partners.

Lesson 13.3 — Writing Similarity Statements

Guided practice

33. $\triangle JKL \sim \triangle XYZ$
34. $\triangle ABC \sim \triangle FDE$
35. $\triangle CAB \sim \triangle FDE$ (also acceptable: $\triangle CBA \sim \triangle FED$)
36. $PQRS \sim TUVW$
37. $\angle N \cong \angle R$; $\overline{MP}$ corresponds to $\overline{QS}$

Independent practice

38. a) $\triangle DEF \sim \triangle PQR$ b) $\triangle GHJ \sim \triangle MKL$
39. $ABCD \sim WXYZ$
40. $\triangle BCA \sim \triangle EFD$ and $\triangle CAB \sim \triangle FDE$. (Reversed-direction forms such as $\triangle ACB \sim \triangle DFE$ are also correct, since the columns still read $A$–$D$, $B$–$E$, $C$–$F$.)
41. $\triangle RST \sim \triangle VUW$
42. $\dfrac{AB}{DE} = \dfrac{6}{9} = \dfrac{2}{3}$
43. $\triangle LMN \sim \triangle TUS$
44. The correspondence is lost — which angle matches which, and therefore which side matches which. Without it you cannot tell whether $\angle A$ is congruent to $\angle D$ or to $\angle E$, so you cannot set up a correct proportion. For example, you could not answer "if $AB = 6$ and the scale factor is 2, which side of the second triangle is 12?"

Exit ticket 13.3

45. $\triangle ABC \sim \triangle RST$
46. $\overline{XZ}$
47. $\triangle QRP \sim \triangle YZX$
48. The first says $\angle A \cong \angle D$, $\angle B \cong \angle E$, $\angle C \cong \angle F$. The second says $\angle A \cong \angle E$, $\angle B \cong \angle D$, $\angle C \cong \angle F$. They pair the vertices differently, so they also pair the sides differently and lead to different proportions.

Lesson 13.4 — Proportions from Corresponding Sides

Guided practice

49. $\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{CA}{FD}$ (the reciprocal form $\dfrac{DE}{AB} = \dfrac{EF}{BC} = \dfrac{FD}{CA}$ is equally correct)
50. $\dfrac{10}{6} = \tfrac53$, $\dfrac{15}{9} = \tfrac53$, $\dfrac{20}{12} = \tfrac53$. All three agree, so the triangles are similar with scale factor $\tfrac53$.
51. $\dfrac{8}{4} = 2$, $\dfrac{12}{6} = 2$, $\dfrac{16}{8} = 2$, $\dfrac{18}{10} = 1.8$. The last ratio disagrees, so the quadrilaterals are not similar. The fourth side of $EFGH$ would have to be $20$.
52. $\dfrac{8}{4} = 2$ and $\dfrac{10}{6} = \tfrac53$. Not similar. Congruent angles were not enough because similarity also requires every pair of corresponding sides to have the same ratio, and here one pair doubles while the other does not.
53. $\dfrac{AB}{DE} = \dfrac{BC}{EF}$ (or its reciprocal form $\dfrac{DE}{AB} = \dfrac{EF}{BC}$)

Independent practice

54. a) $\dfrac{15}{5} = \dfrac{18}{6} = \dfrac{21}{7} = 3$. Similar, scale factor $3$. b) $\dfrac{12}{8} = 1.5$, $\dfrac{15}{10} = 1.5$, $\dfrac{20}{12} = \tfrac53 \approx 1.7$. Not similar. c) $\dfrac{9}{3} = \dfrac{12}{4} = \dfrac{15}{5} = \dfrac{18}{6} = 3$. The sides are proportional; with congruent corresponding angles the quadrilaterals are similar with scale factor $3$.
55. $\dfrac{PQ}{ST} = \dfrac{QR}{TU} = \dfrac{RP}{US}$
56. $\dfrac{5}{15} = \dfrac{8}{EF}$. The scale factor from $\triangle ABC$ to $\triangle DEF$ is $\dfrac{15}{5} = 3$. (Solving gives $EF = 24$.)
57. The student paired $\overline{AB}$ with $\overline{EF}$ and $\overline{BC}$ with $\overline{DE}$, but neither pair corresponds: $\overline{AB}$ goes with $\overline{DE}$ and $\overline{BC}$ goes with $\overline{EF}$. A correct proportion is $\dfrac{AB}{DE} = \dfrac{BC}{EF}$.
58. $4$, $6$, $8$, $10$
59. $\dfrac{8}{24} = \tfrac13$ and $\dfrac{12}{36} = \tfrac13$. The ratios agree — one inch of model for every three feet of pool — and both figures are rectangles, so the model has the same shape as the pool.
60. Every ratio is $\dfrac{8}{4} = 2$, so the corresponding sides are proportional. The figures are still not similar because the angle condition fails: the square's angles all measure $90°$, while the rhombus has angles of $60°$ and $120°$, and $90° \neq 60°$.

Exit ticket 13.4

61. $\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{CA}{FD}$
62. $\dfrac{12}{4} = \dfrac{15}{5} = \dfrac{18}{6} = 3$. Similar, scale factor $3$.
63. $\dfrac{9}{6} = 1.5$, $\dfrac{12}{8} = 1.5$, $\dfrac{16}{10} = 1.6$. Not similar. (The third side would have to be $15$.)
64. Because a quadrilateral can be pushed out of shape without changing any side length. A square and a non-square rhombus can have all four side ratios equal and still have different angle measures, so the angle condition must be checked separately.

Lesson 13.5 — Finding a Missing Side Length

Guided practice

65. $\dfrac{6}{9} = \dfrac{8}{EF}$, so $6 \cdot EF = 9 \times 8 = 72$ and $EF = 12$. Check: $6 \times 12 = 72$ and $9 \times 8 = 72$.
66. $\dfrac{6}{9} = \dfrac{10}{FD}$, so $6 \cdot FD = 90$ and $FD = 15$.
67. $\dfrac{10}{4} = \dfrac{15}{FD}$, so $10 \cdot FD = 60$ and $FD = 6$.
68. $\dfrac{12}{8} = \dfrac{9}{VW}$, so $12 \cdot VW = 72$ and $VW = 6$.
69. $\dfrac{12.5}{5} = 2.5$

Independent practice

70. a) $\dfrac{4}{10} = \dfrac{6}{EF}$, so $4 \cdot EF = 60$ and $EF = 15$. b) $\dfrac{9}{6} = \dfrac{12}{DE}$, so $9 \cdot DE = 72$ and $DE = 8$. c) $\dfrac{15}{5} = \dfrac{21}{GH}$, so $15 \cdot GH = 105$ and $GH = 7$.
71. $DE = 16 \times \tfrac34 = 12$, $EF = 20 \times \tfrac34 = 15$, $FD = 28 \times \tfrac34 = 21$
72. The scale factor is $\dfrac{12}{8} = 1.5$, so $x = EF = 6 \times 1.5 = 9$ and $y = FD = 7 \times 1.5 = 10.5$.
73. $\dfrac{7}{10} = \dfrac{9}{EF}$, so $7 \cdot EF = 90$ and $EF = 12.857\ldots \approx 12.9$.
74. $\overline{LM}$ corresponds to $\overline{QR}$, so $\dfrac{18}{12} = \dfrac{LM}{10}$, giving $12 \cdot LM = 180$ and $LM = 15$.
75. $\dfrac{6}{4} = \dfrac{h}{22}$, so $4h = 132$ and $h = 33$ feet.
76. It is the same skill because the similarity statement produces exactly the kind of equation Chapter 5 taught you to solve: two equal ratios with one unknown. The geometry only decides which numbers go where. To choose the two ratios, pick the pair of corresponding sides you know completely, and pair it with the corresponding sides that contain the unknown — keeping the first figure's lengths in the same position on both sides of the equation.

Exit ticket 13.5

77. $\dfrac{5}{15} = \dfrac{7}{EF}$, so $5 \cdot EF = 105$ and $EF = 21$.
78. $\dfrac{24}{9} = \dfrac{32}{EF}$, so $24 \cdot EF = 288$ and $EF = 12$.
79. $\dfrac{CA}{FD} = \dfrac{11}{4} = 2.75$
80. $\dfrac{9}{6} = \dfrac{5}{EF}$, so $9 \cdot EF = 30$ and $EF = 3.333\ldots \approx 3.3$.

Lesson 13.6 — Scale Drawings

Guided practice

81. $\dfrac{1}{6} = \dfrac{3.5}{L}$, so $L = 6 \times 3.5 = 21$ feet.
82. $\dfrac{1}{6} = \dfrac{d}{45}$, so $6d = 45$ and $d = 7.5$ inches.
83. $\dfrac{1}{20} = \dfrac{4.5}{m}$, so $m = 90$ miles.
84. $5 \times 3 = 15$ ft by $8 \times 3 = 24$ ft.
85. $\dfrac{1}{12} = \dfrac{h}{48}$, so $12h = 48$ and $h = 4$ inches.

Independent practice

86. a) $1.5 \times 8 = 12$ ft by $2 \times 8 = 16$ ft. b) $0.5 \times 8 = 4$ ft.
87. $7.4 \times 15 = 111$ km
88. $18 \div 4 = 4.5$ inches
89. $180 \div 12 = 15$ inches
90. Living room: $4 \times 6 = 24$ ft by $3 \times 6 = 18$ ft. Kitchen: $2 \times 6 = 12$ ft by $3 \times 6 = 18$ ft.
91. $30 \div 12 = 2.5$ in by $42 \div 12 = 3.5$ in, so draw a $2.5$ in by $3.5$ in rectangle. On a grid of half-inch squares that is 5 squares by 7 squares.
92. The scale does not compare the two numbers as sizes of the map and the land; it states a correspondence. One inch measured on the map stands for fifty miles measured on the ground, so the map is far smaller than the land it shows. Every real distance has been shrunk by the same factor.

Exit ticket 13.6

93. $6.5 \times 5 = 32.5$ feet
94. $3.2 \times 25 = 80$ km
95. $60 \div 12 = 5$ inches
96. Without the scale, the lengths on the drawing mean nothing — the same rectangle could stand for a closet or a warehouse. The scale is what converts a measured drawing length into an actual length.

Chapter 13 Review

Part A — Corresponding congruent angles and markings (7.MG.2a)

97. $\angle A \cong \angle D$ (right-angle squares), $\angle B \cong \angle E$ (single arcs), $\angle C \cong \angle F$ (double arcs)
98. $\angle P \cong \angle T$, $\angle Q \cong \angle U$, $\angle R \cong \angle V$, $\angle S \cong \angle W$
99. The matching double arcs mean the two angles have the same measure and occupy corresponding positions in the two figures. In symbols, $\angle B \cong \angle E$.

Part B — Corresponding sides (7.MG.2b)

100. $\overline{FD}$
101. $\overline{FG}$
102. $\overline{TU}$. Their lengths are $PQ = 10$ and $TU = 15$, a ratio of $1.5$.

Part C — Similarity statements (7.MG.2c)

103. $\triangle MNP \sim \triangle BCA$
104. $JKLM \sim RSTU$
105. $\triangle BCA \sim \triangle EFD$

Part D — Proportions from corresponding sides (7.MG.2d)

106. $\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{CA}{FD}$
107. $\dfrac{PQ}{TU} = \dfrac{QR}{UV}$
108. $\dfrac{6}{15} = \dfrac{8}{EF}$. The scale factor from $\triangle ABC$ to $\triangle DEF$ is $\dfrac{15}{6} = 2.5$. (Solving gives $EF = 20$.)

Part E — Justifying similarity with ratios (7.MG.2e)

109. $\dfrac{12}{9} = \tfrac43$, $\dfrac{16}{12} = \tfrac43$, $\dfrac{20}{15} = \tfrac43$. All three agree, so the triangles are similar with scale factor $\tfrac43$.
110. $\dfrac{10}{5} = 2$, $\dfrac{14}{7} = 2$, $\dfrac{18}{9} = 2$, $\dfrac{20}{11} \approx 1.8$. The last ratio disagrees, so they are not similar. The fourth side would have to be $22$.
111. $\dfrac{10}{6} = \tfrac53$ and $\dfrac{15}{9} = \tfrac53$. The ratios agree and every angle in a rectangle is a right angle, so the rectangles are similar with scale factor $\tfrac53$.

Part F — Missing side lengths (7.MG.2f)

112. $\dfrac{12}{8} = \dfrac{21}{FD}$, so $12 \cdot FD = 168$ and $FD = 14$.
113. $\dfrac{7}{17.5} = \dfrac{6}{FG}$, so $7 \cdot FG = 105$ and $FG = 15$.
114. $\dfrac{5}{9} = \dfrac{7}{EF}$, so $5 \cdot EF = 63$ and $EF = 12.6$.

Part G — Unknown angle measures (7.MG.2g)

115. $m\angle C = 180° - 38° - 97° = 45°$. Then $m\angle D = 38°$, $m\angle E = 97°$, $m\angle F = 45°$.
116. $m\angle D = 360° - 85° - 95° - 120° = 60°$. Then $m\angle E = 85°$, $m\angle F = 95°$, $m\angle G = 120°$, $m\angle H = 60°$.

Part H — Scale drawings (7.MG.2h)

117. $3.5 \times 9 = 31.5$ ft by $2 \times 9 = 18$ ft
118. $6.5 \times 40 = 260$ km

Part I — Mixed application and reasoning

119. $\dfrac{5}{3} = \dfrac{h}{27}$, so $3h = 135$ and $h = 45$ feet.
120. $3 \times 4 = 12$ ft by $2.5 \times 4 = 10$ ft. Area $= 12 \times 10 = 120$ square feet.
121. Similarity requires both conditions. The angle condition holds — all eight angles are right angles — but the side ratios are $\tfrac84 = 2$ and $\tfrac{10}{6} = \tfrac53$, which are not equal. The second rectangle was stretched more in one direction than the other, so it is not a scaled copy of the first. For similarity, the $8$-wide rectangle would need to be $12$ tall.
122. The letter order records which vertices correspond. $\triangle ABC \sim \triangle DEF$ claims $\angle A \cong \angle D$, $\angle B \cong \angle E$, $\angle C \cong \angle F$, and pairs $\overline{AB}$ with $\overline{DE}$. $\triangle ABC \sim \triangle EDF$ claims $\angle A \cong \angle E$ and $\angle B \cong \angle D$, and pairs $\overline{AB}$ with $\overline{ED}$ read in the other order. Those are different claims about the same two triangles, and they produce different proportions, so at most one of them can match a given diagram.

Workbook-only items

Page 2, fill in the blanks. Two figures are similar when corresponding angles are congruent and corresponding sides are proportional. The number is the scale factor. A scale factor greater than 1 makes an enlargement; between 0 and 1 it makes a reduction. Congruent figures are similar with a scale factor of 1.

Page 5, correspondence table. $\angle A \to \angle D$; $\angle B \to \angle E$; $\angle C \to \angle F$; $\overline{AB} \to \overline{DE}$; $\overline{BC} \to \overline{EF}$; $\overline{CA} \to \overline{FD}$

Page 8, stacked names. $\angle A \cong \angle D$, $\angle B \cong \angle E$, $\angle C \cong \angle F$; $\overline{AB} \leftrightarrow \overline{DE}$, $\overline{BC} \leftrightarrow \overline{EF}$, $\overline{CA} \leftrightarrow \overline{FD}$

Page 11, proportion frames. For the triangles: $\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{CA}{FD}$. For the quadrilaterals: $\dfrac{PQ}{TU} = \dfrac{QR}{UV} = \dfrac{RS}{VW} = \dfrac{SP}{WT}$.

Page 11, trapezoid check. $\dfrac{15}{10} = 1.5$, $\dfrac{7.5}{5} = 1.5$, $\dfrac{6}{4} = 1.5$. Scale factor $1.5$.

Page 14, the frame. Students should read it as: the known corresponding pair on the left, the pair containing the unknown on the right, with the first figure's lengths in the numerators on both sides.

Page 18, item 91 grid drawing. The rectangle is $2.5$ in by $3.5$ in. On a grid of half-inch squares, that is 5 squares by 7 squares.