Appendix A — Answer Key, Chapter 7: Proving Triangles Similar

SOL G.TR.3 (a, b, c, d, e) · Covers textbook Chapter 7 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 144 across the chapter.

Conventions used in every answer below. A similarity statement claims three congruent angle pairs and three sides in one constant ratio, and the letter order fixes which part is compared with which. A scale factor has a direction: from $\triangle ABC$ to $\triangle DEF$ it is $\tfrac{DE}{AB}$, and reading the pair the other way gives the reciprocal. Scale factors are positive. A variable is not a length: solve the proportion, substitute back, then check the ratio. Coordinate lengths are given in simplest radical form. Perimeter scales by $k$; area is G.DF.2's subject and belongs to Chapter 17.

The three criteria, and the two that are not:

Lesson 7.1 — Similar Triangles, Correspondence, and Scale Factor

Guided practice

1. $k = \tfrac{3}{2}$.
2. $\overline{DF}$.
3. $k = \tfrac{2}{3}$ — the reciprocal, because the direction is reversed.
4. $\tfrac{DE}{AB} = \tfrac{9}{6} = \tfrac{3}{2}$ and $\tfrac{EF}{BC} = \tfrac{12}{8} = \tfrac{3}{2}$.
5. Six: three ratios ($\tfrac{DE}{AB}$, $\tfrac{EF}{BC}$, $\tfrac{DF}{AC}$, all equal to $k$) and three angle congruences ($\angle A \cong \angle D$, $\angle B \cong \angle E$, $\angle C \cong \angle F$).
6. Because the order names the correspondence, and the correspondence is what says which side is compared with which and which angle with which. Change the order and you have made a different claim.

Independent practice

7. $k = \tfrac{PQ}{AB} = \tfrac{10}{4} = \tfrac{5}{2}$.
8. $DE = 3(7) = 21$.
9. $k$ from $\triangle JKL$ to $\triangle MNP$ is $\tfrac{MN}{JK} = \tfrac{8}{12} = \tfrac{2}{3}$, so $NP = \tfrac{2}{3}(15) = 10$.
10. $k = \tfrac{DE}{AB} = \tfrac{10}{6} = \tfrac{5}{3}$, so $EF = \tfrac{5}{3}(9) = 15$ and $DF = \tfrac{5}{3}(12) = 20$.
11. $m\angle X = 43°$, $m\angle Y = 76°$, $m\angle Z = 180° - 43° - 76° = 61°$.
12. Yes. $\tfrac{9}{3} = \tfrac{12}{4} = \tfrac{15}{5} = 3$, so $k = 3$.
13. No. $\tfrac{6}{4} = \tfrac{3}{2}$ and $\tfrac{9}{6} = \tfrac{3}{2}$, but $\tfrac{11}{8} \neq \tfrac{3}{2}$. The third ratio settles it.
14. $\tfrac{5}{4}(36) = 45$.
15. $\tfrac{2}{3}$ is the factor from $\triangle DEF$ to $\triangle ABC$, not the other way. From $\triangle ABC$ to $\triangle DEF$ the factor is $\tfrac{DE}{AB} = \tfrac{12}{8} = \tfrac{3}{2}$. The new triangle's side goes on top.
16. Congruent triangles have equal corresponding sides, which is the constant ratio $k = 1$, and congruent corresponding angles — so every congruent pair meets the definition of similar. The reverse fails because similarity allows any positive $k$; a triangle and its double are similar and are not congruent.

Exit ticket 7.1

17. $k = \tfrac{15}{5} = 3$.
18. $EF = \tfrac{1}{2}(14) = 7$.
19. $m\angle B = m\angle E = 80°$ by the correspondence, so $m\angle C = 180° - 55° - 80° = 45°$.
20. That the three pairs of corresponding angles are congruent and the three pairs of corresponding sides are in one constant ratio — six claims, with the letter order fixing which parts correspond.

Lesson 7.2 — AA — Two Angles Are Enough

Guided practice

21. $48°$ and $71°$.
22. Because the three angles of a triangle sum to $180°$. Two matching pairs leave the same remainder in each triangle: $180° - 48° - 71° = 61°$.
23. No. AA fixes the shape and says nothing about the size — the two triangles in the figure are drawn at different scales on purpose.
24. $\angle A \cong \angle A$, because the two triangles share that angle (Reflexive Property); and $\angle ADE \cong \angle ABC$, because $\overline{DE} \parallel \overline{BC}$ makes them corresponding angles.
25. $\tfrac{AD}{AB} = \tfrac{2\sqrt{13}}{3\sqrt{13}} = \tfrac{2}{3}$.
26. $\tfrac{2}{3}$. Once AA has proved $\triangle ADE \sim \triangle ABC$, every pair of corresponding sides is in the same ratio, so $\tfrac{DE}{BC}$ must equal $\tfrac{AD}{AB}$ without being measured.

Independent practice

27. Yes. The first triangle's angles are $35°$, $80°$, $65°$; the second's are $80°$, $65°$, $35°$. Two pairs match, so AA applies.
28. No. The first is $40°$, $60°$, $80°$; the second is $40°$, $70°$, $70°$. Only the $40°$ pair matches.
29. $AB = 6 + 4 = 10$, so $\tfrac{AD}{AB} = \tfrac{3}{5}$. Then $\tfrac{9}{AC} = \tfrac{3}{5}$ gives $AC = 15$, and $EC = 15 - 9 = 6$.
30. $k = \tfrac{AD}{AB} = \tfrac{8}{20} = \tfrac{2}{5}$, so $\tfrac{6}{BC} = \tfrac{2}{5}$ and $BC = 15$.
31. $AB = 5 + 7 = 12$, so $\tfrac{5}{12} = \tfrac{10}{AC}$ gives $AC = 24$ and $EC = 24 - 10 = 14$.
32. Yes. The two right angles are one pair and the two $32°$ angles are the other, so AA applies.
33. $m\angle C = m\angle F = 180° - 90° - 37° = 53°$.
34. Both ramp triangles have a right angle where the ramp meets the vertical, and the angle the ramp makes with the ground is the same because the slopes are equal — that is AA. The scale factor is $\tfrac{36}{24} = \tfrac{3}{2}$, so the rise is $\tfrac{3}{2}(2) = 3$ ft.
35. One pair is not enough. A triangle with angles $50°$, $60°$, $70°$ and one with angles $50°$, $40°$, $90°$ share a $50°$ angle, and no other pair matches, so they are not similar.
36. Because the angle sum supplies the third pair free. Two triangles with two matching angles both have $180°$ minus the same two measures left over, so the third pair is congruent automatically and checking it would add nothing.

Exit ticket 7.2

37. Yes. The first triangle is $25°$, $105°$, $50°$; the second is $105°$, $50°$, $25°$. Two pairs match, so AA applies.
38. $\tfrac{4}{10} = \tfrac{6}{AC}$, so $AC = 15$.
39. Because it says nothing about size. Two triangles can have all three angles congruent and be any two different sizes, so AA cannot force the sides to be equal.
40. The shared angle at the apex (Reflexive Property), and the pair of corresponding angles formed by the parallel line and one side acting as a transversal.

Lesson 7.3 — SSS Similarity and SAS Similarity

Guided practice

41. $\tfrac{9}{6}$, $\tfrac{12}{8}$, $\tfrac{15}{10}$ — all equal to $\tfrac{3}{2}$.
42. Nothing. Two matching ratios leave the third free to disagree, and if it does, the triangles are not similar.
43. $\overline{AB}$ and $\overline{AC}$, with $\angle A$ — the angle between them.
44. Because two sides and a non-included angle is the SSA arrangement, and SSA does not determine a triangle. The two sides can be hinged to two different shapes around a non-included angle, so the criterion fails.
45. The AA / AAA row: not a criterion for congruence, a criterion for similarity.
46. SSA.

Independent practice

47. Yes, SSS similarity. $\tfrac{8}{6} = \tfrac{12}{9} = \tfrac{16}{12} = \tfrac{4}{3}$, so $k = \tfrac{4}{3}$.
48. No. $\tfrac{10}{5} = 2$ and $\tfrac{14}{7} = 2$, but $\tfrac{20}{9} \neq 2$.
49. Yes. $\tfrac{10}{4} = \tfrac{15}{6} = \tfrac{20}{8} = \tfrac{5}{2}$, so $k = \tfrac{5}{2}$.
50. SAS similarity. $\tfrac{9}{6} = \tfrac{15}{10} = \tfrac{3}{2}$, and $\angle A$ and $\angle D$ are included between the two named sides in each triangle.
51. Nothing. $\angle B$ is not between $\overline{AB}$ and $\overline{AC}$ — those two meet at $A$ — so the arrangement is SSA, which is not a criterion for similarity any more than for congruence.
52. Yes. $\tfrac{8}{12} = \tfrac{12}{18} = \tfrac{16}{24} = \tfrac{2}{3}$, so $k = \tfrac{2}{3}$ from the first to the second.
53. $k = \tfrac{21}{14} = \tfrac{3}{2}$, so $EF = \tfrac{3}{2}(18) = 27$.
54. SAS similarity, $k = \tfrac{5}{2}$. $\tfrac{25}{10} = \tfrac{10}{4} = \tfrac{5}{2}$, and $\angle A$ and $\angle D$ are the included angles.
55. No. $\tfrac{6}{3} = 2$ and $\tfrac{8}{4} = 2$, but $\tfrac{11}{5} \neq 2$.
56. $\tfrac{15}{9} = \tfrac{20}{12} = \tfrac{25}{15} = \tfrac{5}{3}$, so the brackets are similar by SSS similarity with $k = \tfrac{5}{3}$.
57. The third ratio, $\tfrac{16}{12}$. It matters because two agreeing ratios prove nothing on their own — a triangle with sides $6$, $9$, $20$ would pass the same two checks and fail the third. (Here the third ratio does equal $\tfrac{4}{3}$, so the conclusion happens to be right; the reasoning was not.)
58. Two sides and a non-included angle is SSA. Chapter 5 showed that swinging the non-included side gives two different triangles from the same data; scaling every length in that picture by $k$ gives two different shapes from the same proportional data, so the same failure carries over word for word.

Exit ticket 7.3

59. Yes. $\tfrac{12}{9} = \tfrac{16}{12} = \tfrac{20}{15} = \tfrac{4}{3}$, so $k = \tfrac{4}{3}$.
60. SAS similarity, $k = \tfrac{3}{2}$ — $\tfrac{12}{8} = \tfrac{18}{12} = \tfrac{3}{2}$ and the congruent angles are the included ones.
61. AA.
62. AA, SSS similarity, and SAS similarity.

Lesson 7.4 — Similarity by Algebra

Guided practice

63. $\dfrac{2x}{3x + 4} = \dfrac{8}{20}$.
64. $20 \cdot 2x = 8(3x + 4)$.
65. $40x = 24x + 32$, so $16x = 32$ and $x = 2$.
66. $AB = 2(2) = 4$ and $DE = 3(2) + 4 = 10$.
67. $\tfrac{4}{10} = \tfrac{2}{5}$ and $\tfrac{8}{20} = \tfrac{2}{5}$; the two ratios agree.
68. Because $x$ is the value of the variable, not a length. $AB$ is what the expression $2x$ evaluates to, which is $4$.

Independent practice

69. $12x = 60$, so $x = 5$.
70. $27x = 54$, so $x = 2$.
71. $\tfrac{x}{15} = \tfrac{8}{20} = \tfrac{2}{5}$, so $5x = 30$ and $x = 6$.
72. $\tfrac{3x}{24} = \tfrac{5}{20} = \tfrac{1}{4}$, so $12x = 24$, $x = 2$, and $AB = 6$. Check: $\tfrac{6}{24} = \tfrac{1}{4}$.
73. $\tfrac{x + 3}{2x} = \tfrac{6}{9} = \tfrac{2}{3}$, so $3(x + 3) = 4x$, $3x + 9 = 4x$, $x = 9$. Then $AB = 12$ and $DE = 18$, and $\tfrac{12}{18} = \tfrac{2}{3}$ checks.
74. $\tfrac{2x - 1}{3x + 4} = \tfrac{5}{10} = \tfrac{1}{2}$, so $2(2x - 1) = 3x + 4$, $4x - 2 = 3x + 4$, $x = 6$. Then $AB = 11$ and $DE = 22$, and $\tfrac{11}{22} = \tfrac{1}{2}$ checks.
75. $\tfrac{x}{x + 6} = \tfrac{10}{4} = \tfrac{5}{2}$, so $2x = 5x + 30$ and $x = -10$. That makes $AB = -10$, which is not a length, so the solution is rejected and no such pair of triangles exists. The data was contradictory from the start: $\tfrac{BC}{EF} = \tfrac{10}{4} > 1$ says $AB$ exceeds $DE$, while $DE = AB + 6$ says the reverse.
76. $3(3x - 2) = 21$, so $3x - 2 = 7$, $3x = 9$, and $x = 3$. The print's edge is $3(3) - 2 = 7$ inches.
77. The second ratio was written upside down. $\overline{BC}$ corresponds to $\overline{EF}$, so the proportion is $\tfrac{6}{9} = \tfrac{y}{12}$, giving $9y = 72$ and $y = 8$. (The student's $y = 18$ makes the smaller triangle's side longer than the larger one's.)
78. Because a proportion is two expressions set equal, which is what an equation is. Cross-multiplying is just multiplying both sides by both denominators at once — a legal step that clears the fractions and leaves an ordinary linear equation.

Exit ticket 7.4

79. $8x = 120$, so $x = 15$.
80. $\tfrac{4x}{18} = \tfrac{6}{27} = \tfrac{2}{9}$, so $36x = 36$, $x = 1$, and $AB = 4$. Check: $\tfrac{4}{18} = \tfrac{2}{9}$.
81. That there is no such pair of triangles. A length must be positive, so a solution producing a negative side is rejected rather than reported.
82. Substitute the value back into both expressions and confirm the two lengths; then confirm the two ratios are equal, which is what the similarity actually claimed.

Lesson 7.5 — Similarity by Coordinates

Guided practice

83. $DE = 3\sqrt{5}$ and $AB = 2\sqrt{5}$.
84. $\tfrac{3\sqrt{5}}{2\sqrt{5}} = \tfrac{3\sqrt{17}}{2\sqrt{17}} = \tfrac{3\sqrt{10}}{2\sqrt{10}} = \tfrac{3}{2}$.
85. SSS similarity, with $k = \tfrac{3}{2}$.
86. Both are $\tfrac{1}{2}$.
87. That $\overline{AB} \parallel \overline{DE}$. Two pairs of parallel corresponding sides make the angle between them congruent, which is an angle congruence with no measure attached.
88. AA.

Independent practice

89. $AB = 4$, $BC = \sqrt{16 + 9} = 5$, $AC = 3$; $DE = 8$, $EF = \sqrt{64 + 36} = 10$, $DF = 6$. Every ratio is $2$, so $\triangle ABC \sim \triangle DEF$ by SSS similarity with $k = 2$.
90. $AB = \sqrt{16 + 4} = 2\sqrt{5}$, $BC = \sqrt{9 + 36} = 3\sqrt{5}$, $AC = \sqrt{1 + 16} = \sqrt{17}$; $ST = \sqrt{64 + 16} = 4\sqrt{5}$, $TU = \sqrt{36 + 144} = 6\sqrt{5}$, $SU = \sqrt{4 + 64} = 2\sqrt{17}$. Every ratio is $2$, so $k = 2$.
91. No. $\tfrac{DE}{AB} = \tfrac{6}{6} = 1$ but $\tfrac{DF}{AC} = \tfrac{6}{4} = \tfrac{3}{2}$, and the two ratios disagree.
92. $\overline{AB}$ and $\overline{ST}$ both have slope $\tfrac{1}{2}$; $\overline{BC}$ and $\overline{TU}$ both have slope $2$; $\overline{AC}$ and $\overline{SU}$ both have slope $-4$. Three pairs of parallel sides, so all three angle pairs are congruent — AA twice over.
93. $k = \tfrac{ST}{AB} = \tfrac{4\sqrt{5}}{2\sqrt{5}} = 2$, and checking with a second pair, $\tfrac{SU}{AC} = \tfrac{2\sqrt{17}}{\sqrt{17}} = 2$.
94. The image is $(-8, 4)$, $(0, 8)$, $(-4, -4)$. The lengths are $2\sqrt{5}$, $2\sqrt{10}$, $2\sqrt{5}$ and $4\sqrt{5}$, $4\sqrt{10}$, $4\sqrt{5}$, so all three ratios are $2$.
95. Two more ratios. One pair of proportional sides is not a criterion; SSS similarity needs all three, and finding only one leaves the possibility that another disagrees.
96. Yes — the two loops are similar by SSS similarity with $k = 2$, from item 89. Loop 1's perimeter is $4 + 5 + 3 = 12$ grid units, or $1200$ m; loop 2's is $24$ units, or $2400$ m. Loop 2 is $1200$ m longer — twice as far.
97. The distance formula returns a length, and three lengths compared as ratios is SSS similarity. The slope formula returns a direction, and two segments with the same direction are parallel, which makes the angle between a pair of them congruent to the angle between the matching pair — that is an angle congruence, and two of them is AA. In Chapter 6 slope could only certify $90°$, because congruence needed equal sides and a specific angle measure; similarity needs only congruent angles, so parallel sides are enough.
98. Equal slopes prove the sides are parallel, so the angles are congruent — that is AA, which gives similarity and nothing more. Congruence would need the corresponding sides to be equal in length, and two parallel segments may be any lengths at all.

Exit ticket 7.5

99. $k = \tfrac{5\sqrt{7}}{3\sqrt{7}} = \tfrac{5}{3}$.
100. That the two sides are parallel — which makes the angle between them and a second matching pair congruent, and is the first half of an AA argument.
101. The distance formula reaches SSS similarity; the slope formula reaches AA.
102. Plot both triangles and write the correspondence you intend to prove; choose the tool — distances for SSS, slopes for AA; compute in exact form and compare in correspondence order; name the criterion and state the similarity with its scale factor.

Lesson 7.6 — Similarity as a Sequence of Transformations

Guided practice

103. A dilation by $\tfrac{3}{2}$ centered at the origin, then a translation $8$ units right.
104. The dilation.
105. The translation.
106. $A'(-9, 3)$, $B'(-3, 6)$, $C'(-6, -6)$.
107. No. Dilating first lands on $(-1, 3)$, $(5, 6)$, $(2, -6)$; sliding first lands on $(3, 3)$, $(9, 6)$, $(6, -6)$.
108. Each step, in order; for a dilation, the centre and the scale factor; for a translation, the direction and distance; for a reflection, the line; for a rotation, the centre, the angle, and the direction.

Independent practice

109. $(-9, 3)$, $(-3, 6)$, $(-6, -6)$.
110. $(-1, 3)$, $(5, 6)$, $(2, -6)$.
111. $(8, -12)$.
112. $(-2, 3)$.
113. A dilation by $2$ centered at the origin, and nothing else: it sends $(1, 1) \to (2, 2)$, $(3, 1) \to (6, 2)$, and $(1, 2) \to (2, 4)$, which is $\triangle DEF$ exactly.
114. A dilation by $3$ centered at the origin gives $(6, 0)$, $(12, 0)$, $(6, 9)$; a rotation of $180°$ about the origin then gives $(-6, 0)$, $(-12, 0)$, $(-6, -9)$, which is $\triangle DEF$.
115. Because rigid motions preserve length. Any sequence of them carries a triangle onto a congruent triangle, so unless the scale factor is $1$ they can never reach the second triangle — the dilation is the only step that changes the size.
116. The dilation gives $(4, -3)$; the reflection over the $y$-axis then gives $(-4, -3)$.
117. The scale factor of the dilation, the centre of the dilation, the direction and distance of the translation, and the order of the two steps.
118. A dilation multiplies every length by $k$ and leaves every angle unchanged, so the image has congruent angles and proportional sides — it is similar to the original. Rigid motions then change neither lengths nor angles, so the final figure still has those same congruent angles and the same constant ratio. The composition therefore always lands on a similar figure.

Exit ticket 7.6

119. $(-2, 3)$.
120. Its centre and its scale factor.
121. No. The two orders agree only in special cases; in general, sliding first moves the figure away from the centre of dilation and the dilation then multiplies that larger displacement.
122. Each step in order, with the centre and scale factor of the dilation and the full description of every rigid motion — the line of a reflection, the direction and distance of a translation, the centre, angle, and direction of a rotation.

Lesson 7.7 — Measured Attributes and Indirect Measurement

Guided practice

123. AA. The sun's rays are parallel, so each makes the same angle with the ground and those two angles are congruent; and both the person and the tree stand perpendicular to the ground, so the two right angles are congruent.
124. $\dfrac{5}{4} = \dfrac{h}{36}$.
125. $4h = 180$, so $h = 45$ ft.
126. Because nobody has to climb it. The person's height and both shadows are measurable from the ground, and the similarity transfers the ratio.
127. No — the tree is nine times the person, and drawing them at one scale would leave the person a smear. What is the same is the shape: both triangles are drawn from the same height-to-shadow ratio, which is the only thing the figure is claiming.
128. The pair at the far end of each shadow, where the ray meets the ground.

Independent practice

129. $\tfrac{6}{8} = \tfrac{h}{44}$, so $8h = 264$ and $h = 33$ ft.
130. $\tfrac{4}{3} = \tfrac{h}{51}$, so $3h = 204$ and $h = 68$ ft.
131. $\tfrac{5}{2}(22) = 55$.
132. $k = \tfrac{20}{8} = \tfrac{5}{2}$, so the perimeter of $\triangle DEF$ is $\tfrac{5}{2}(30) = 75$.
133. $k = \tfrac{15}{9} = \tfrac{5}{3}$, so the perimeter of $\triangle ABC$ is $60 \div \tfrac{5}{3} = 36$.
134. $EF = 3(7) = 21$; $AC = 27 \div 3 = 9$.
135. $\tfrac{AD}{AB} = \tfrac{9}{24} = \tfrac{3}{8}$, so $DE = \tfrac{3}{8}(40) = 15$.
136. $\tfrac{5}{4} = \tfrac{h}{30}$, so $4h = 150$ and $h = 37.5$ ft.
137. $5.5 \times 48 = 264$ inches, and $264 \div 12 = 22$ feet.
138. The proportion pairs a height with a shadow on the same side of the equation. Height must go with height and shadow with shadow: $\tfrac{5}{4} = \tfrac{h}{36}$, giving $h = 45$ ft. The student's version would give $h = \tfrac{5(4)}{36}$, about $0.56$ ft, which the figure alone should have flagged as impossible.
139. Perimeter is a sum of the three sides. If every side is multiplied by $k$, the sum is multiplied by $k$ as well, so $P_{DEF} = k \cdot P_{ABC}$. Area is not a sum of lengths, so the same argument does not apply to it — and the effect of scaling on area is what G.DF.2 asks about, which is Chapter 17's subject rather than this one's.
140. Because a proved similarity makes every pair of corresponding sides share one ratio, so a measurement made on the reachable triangle fixes the matching part of the other one. The similarity is used at the moment you write the proportion — that step is the only one that needs it, and it is where the criterion should be named.

Exit ticket 7.7

141. $\tfrac{5}{6} = \tfrac{h}{42}$, so $6h = 210$ and $h = 35$ ft.
142. $4(13) = 52$.
143. AA, and the fact that the sun's rays arrive parallel — which makes the two rays cut the ground at congruent angles. The two right angles at the base supply the second pair.
144. Prove the similarity; write the correspondence in order; set up one proportion matching the part you want to the parts you know; solve, and name the similarity and the criterion that established it.

Chapter 7 Review — answers

Review 1 (G.TR.3 a, b).

• $\angle A \cong \angle A$ by the Reflexive Property, and $\angle ADE \cong \angle ABC$ because $\overline{DE} \parallel \overline{BC}$ makes them corresponding angles. The criterion is AA.
• $AB = AD + DB = 2x + (x + 4) = 3x + 4$, so $\tfrac{2x}{3x + 4} = \tfrac{2}{5}$ gives $10x = 6x + 8$, $4x = 8$, and $x = 2$. Then $AD = 4$, $DB = 6$, and $AB = 10$. Check: $\tfrac{4}{10} = \tfrac{2}{5}$.
• $DE = \tfrac{2}{5}(35) = 14$, by the similarity — corresponding sides of similar triangles are in the ratio $k$.
• $\overline{DB}$ is not a side of either triangle in the correspondence: the triangles are $\triangle ADE$ and $\triangle ABC$, and $\overline{DE}$ corresponds to $\overline{BC}$ while $\overline{AD}$ corresponds to $\overline{AB}$. The right proportion is $\tfrac{AD}{AB} = \tfrac{DE}{BC}$, which is $\tfrac{4}{10} = \tfrac{14}{35}$. (The classmate's version gives $\tfrac{4}{6} = \tfrac{2}{3}$ on the left and $\tfrac{2}{5}$ on the right, so it is not even true.)

Review 2 (G.TR.3 c, d).

• $AB = \sqrt{16 + 4} = 2\sqrt{5}$, $BC = \sqrt{9 + 36} = 3\sqrt{5}$, $AC = \sqrt{1 + 16} = \sqrt{17}$; $ST = \sqrt{64 + 16} = 4\sqrt{5}$, $TU = \sqrt{36 + 144} = 6\sqrt{5}$, $SU = \sqrt{4 + 64} = 2\sqrt{17}$. Each ratio is $2$, so $\triangle ABC \sim \triangle STU$ by SSS similarity with $k = 2$.
• Slopes: $\overline{AB}$ and $\overline{ST}$ are both $\tfrac{1}{2}$; $\overline{BC}$ and $\overline{TU}$ are both $2$; $\overline{AC}$ and $\overline{SU}$ are both $-4$. Parallel corresponding sides make the angles between them congruent, so the criterion is AA.
• A dilation by $2$ centered at the origin sends $\triangle ABC$ to $(-14, 0)$, $(-6, 4)$, $(-12, -8)$; a translation $16$ units right and $2$ units up then gives $(2, 2)$, $(10, 6)$, $(4, -6)$, which is $\triangle STU$.
• All three establish the same fact — that these two triangles are similar with $k = 2$ — by three different routes the standard names in three different bullets. Distances reach it through SSS, slopes through AA, and the sequence through the definition of a similarity transformation. A result that survives three independent methods is the same result, not three of them.

Review 3 (G.TR.3 e).

• The sun's rays arrive parallel, so the ray at the rod and the ray at the cliff make congruent angles with the level ground; and the rod and the cliff face are both perpendicular to that ground, so the two right angles are congruent as well. Two pairs of congruent angles is AA.
• $\tfrac{4}{5} = \tfrac{h}{145}$, so $5h = 580$ and $h = 116$ ft.
• It pairs the rod's height with the cliff's shadow and the cliff's height with the rod's shadow — height with shadow on each side instead of height with height. It would give $h = \tfrac{4(5)}{145} \approx 0.14$ ft, which the situation alone rules out. The correct proportion is $\tfrac{4}{5} = \tfrac{h}{145}$.
• $k = \tfrac{145}{5} = 29$, the same factor as $\tfrac{116}{4}$. To compute either perimeter you would need the third side of each triangle — the sloping ray from the top to the tip of the shadow — and that is a hypotenuse, so it takes the Pythagorean Theorem, which is Chapter 8. (For the record it is $\sqrt{4^2 + 5^2} = \sqrt{41}$ for the rod, and $29\sqrt{41}$ for the cliff.)