Appendix A — Answer Key, Chapter 6: Congruence by Algebra and Coordinates

SOL G.TR.2 (b, c, e) · Covers textbook Chapter 6 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 88 across the chapter.

Conventions used in every answer below. A variable is not a length: solving an equation gives the variable's value, and the length is what the expression evaluates to. Every solved value is substituted back into both expressions. Three solved side lengths are checked against the Triangle Inequality; two solved angle measures are checked against the $180°$ sum. Coordinate lengths are given in simplest radical form first, and rounded only when a context asks for a measurement. Letter order in a congruence statement is part of the claim, exactly as in Chapter 5.

The three coordinate tools and what each one certifies:

Lesson 6.1 — Congruence from Algebra

Guided practice

1. $\overline{AB}$ and $\overline{DE}$.
2. $4y - 5 = 2y + 9$.
3. $2w + 7 = 5w - 11$ gives $18 = 3w$, so $w = 6$ and $AC = 2(6) + 7 = 19$. Check: $5(6) - 11 = 19$.
4. $x = 5$ is the value of the variable. The length is what the expression evaluates to: $AB = 3(5) + 2 = 17$.
5. The Triangle Inequality. The solved sides are $17$, $23$, and $19$, and $17 + 19 = 36 > 23$, so a triangle exists.
6. ASA. The congruent side $\overline{AB}$ lies between the two congruent angles $\angle A$ and $\angle B$. It would be AAS only if the congruent side were not between them.

Independent practice

7. $2x + 9 = 6x - 7$ gives $16 = 4x$, so $x = 4$ and $AB = DE = 17$.
8. $7y - 4 = 3y + 20$ gives $4y = 24$, so $y = 6$ and $BC = EF = 38$.
9. $5a - 12 = 3a + 8$ gives $2a = 20$, so $a = 10$ and $m\angle A = m\angle D = 38°$.
10. $2b + 31 = 7b - 44$ gives $75 = 5b$, so $b = 15$ and $m\angle B = m\angle E = 61°$.
11. $x = 6$, so $DE = AB = 25$. $y = 7$, so $EF = BC = 19$. $z = 6$, so $DF = AC = 22$. Perimeter $= 25 + 19 + 22 = 66$.
12. The longest is $25$, and $19 + 22 = 41 > 25$, so the three lengths form a triangle.
13. $3a + 5 = a + 41$ gives $a = 18$ and $m\angle D = 59°$. $6b - 7 = 2b + 45$ gives $b = 13$ and $m\angle E = 71°$. Then $m\angle F = 180° - 59° - 71° = 50°$.
14. The student reported the variable. $x = 5$ is correct as a value of $x$, but $AB = 3(5) + 2 = 17$.
15. $2x + 3 = 2x + 9$ reduces to $3 = 9$, which is false for every $x$. There is no value of $x$; the figure's marking is impossible, because two expressions that always differ by $6$ cannot describe one length.
16. No. The two shorter lengths are $8$ and $5$, and $8 + 5 = 13 < 14$, so the Triangle Inequality fails.
17. $96° + 88° = 184° > 180°$. Two angles of a triangle cannot already exceed the whole angle sum, so no triangle has both.
18. $4x + 6 = 7x - 9$ gives $15 = 3x$, so $x = 5$ and the long edge is $4(5) + 6 = 26$ inches. Check: $7(5) - 9 = 26$.
19. Because the variable is not what was asked for. Two steps remain: substitute the value back into both expressions to get the length or angle measure and confirm the two agree, and then check that the solved measures can actually belong to a triangle — the Triangle Inequality for sides, or a positive third angle for angles.
20. The two congruent angles are $\angle A$ and $\angle C$, and the side between those two is $\overline{AC}$, not $\overline{AB}$. So $\overline{AB}$ is a non-included side and the arrangement is AAS, not ASA.

Exit ticket 6.1

21. $6x - 5 = 2x + 27$ gives $4x = 32$, so $x = 8$ and the length is $6(8) - 5 = 43$. Check: $2(8) + 27 = 43$.
22. $2a + 18 = 5a - 21$ gives $39 = 3a$, so $a = 13$ and the measure is $2(13) + 18 = 44°$. Check: $5(13) - 21 = 44$.
23. No. $9 + 12 = 21 < 22$, so the Triangle Inequality fails.
24. Substitute the value back into both expressions and confirm they agree; then confirm the solved measures can belong to a triangle — three sides by the Triangle Inequality, or two angles by leaving a positive third angle.

Lesson 6.2 — Distance, Slope, and Midpoint

Guided practice

25. The legs are $8$ and $6$. The horizontal leg is $|-1 - (-9)| = 8$ and the vertical leg is $|7 - 1| = 6$.
26. $AB^2 = 8^2 + 6^2 = 64 + 36 = 100$, so $AB = 10$.
27. Slope of $\overline{BA}$ is $\tfrac{2}{3}$; slope of $\overline{BC}$ is $-\tfrac{3}{2}$; the product is $-1$.
28. That $\overline{BA} \perp \overline{BC}$, and therefore $m\angle B = 90°$. It proves that one measure and no other.
29. $\left(\tfrac{-5 + 3}{2}, \tfrac{-3 + 5}{2}\right) = (-1, 1)$.
30. Because one coordinate sum is odd. For the second segment $-5 + 2 = -3$, and half of $-3$ is $-1.5$, so the midpoint lands halfway between two grid lines.

Independent practice

31. $\sqrt{(7 - 2)^2 + (9 - (-3))^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.
32. $\sqrt{36 + 64} = \sqrt{100} = 10$.
33. $\sqrt{16 + 9} = \sqrt{25} = 5$.
34. $\sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$.
35. $\sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$.
36. $\dfrac{6 - 2}{2 - (-6)} = \dfrac{4}{8} = \dfrac{1}{2}$.
37. Undefined. The run is $5 - 5 = 0$, so the segment is vertical.
38. Yes. The slopes are $\tfrac{3}{4}$ and $\tfrac{-8}{6} = -\tfrac{4}{3}$, and $\left(\tfrac{3}{4}\right)\left(-\tfrac{4}{3}\right) = -1$.
39. $\left(\tfrac{-7 + 3}{2}, \tfrac{2 + (-6)}{2}\right) = (-2, -2)$.
40. $\left(\tfrac{-4 + 3}{2}, \tfrac{1 + 6}{2}\right) = (-0.5, 3.5)$.
41. $B = (2(1) - (-3),\ 2(4) - (-2)) = (5, 10)$.
42. $\sqrt{(7 - (-5))^2 + (3 - (-2))^2} = \sqrt{144 + 25} = 13$ grid units, and $13 \times 100 = 1300$ metres.
43. The student added the legs instead of finding the hypotenuse. The differences $6$ and $8$ are the legs of a right triangle, and the distance is the hypotenuse: $\sqrt{36 + 64} = 10$.

Exit ticket 6.2

44. $\sqrt{36 + 64} = 10$.
45. $\dfrac{2 - 6}{5 - (-1)} = \dfrac{-4}{6} = -\dfrac{2}{3}$.
46. $\left(\tfrac{-9 + 2}{2}, \tfrac{4 + (-3)}{2}\right) = (-3.5, 0.5)$.
47. Slope. It must show that the two slopes multiply to $-1$ — or, in the special case, that one segment is horizontal and the other vertical. That certifies $90°$ and no other measure.

Lesson 6.3 — Coordinate Proofs of Congruence

Guided practice

48. $AB = DE = 2\sqrt{5}$.
49. $BC = EF = \sqrt{26}$.
50. SSS, and $\triangle ABC \cong \triangle DEF$.
51. Slope of $\overline{PQ}$ is $\tfrac{1}{2}$ and slope of $\overline{PR}$ is $-2$; the product is $-1$, so $\overline{PQ} \perp \overline{PR}$ and $m\angle P = 90°$.
52. SAS. It used $\overline{PQ} \cong \overline{ST}$, $\angle P \cong \angle S$ (both right), and $\overline{PR} \cong \overline{SU}$ — two sides and the angle between them.
53. The second comparison, $KL$ against $NP$: $KL = 5$ but $NP = 2\sqrt{5}$. That failure kills the statement as written, not the triangles. Both triangles have sides $5$, $5$, and $2\sqrt{5}$, so they are congruent; only the pairing of vertices was wrong.

Independent practice

54. $AB = \sqrt{16 + 1} = \sqrt{17}$, $BC = \sqrt{4 + 16} = 2\sqrt{5}$, $AC = \sqrt{4 + 9} = \sqrt{13}$; $DE = \sqrt{16 + 1} = \sqrt{17}$, $EF = \sqrt{4 + 16} = 2\sqrt{5}$, $DF = \sqrt{4 + 9} = \sqrt{13}$. All three pairs match in order, so $\triangle ABC \cong \triangle DEF$ by SSS.
55. $PQ = 2\sqrt{5}$, $QR = \sqrt{4 + 25} = \sqrt{29}$, $PR = \sqrt{4 + 9} = \sqrt{13}$; $ST = \sqrt{4 + 16} = 2\sqrt{5}$, $TU = \sqrt{25 + 4} = \sqrt{29}$, $SU = \sqrt{9 + 4} = \sqrt{13}$. So $\triangle PQR \cong \triangle STU$ by SSS.
56. Not congruent. $AC = 4$ but $DF = 5$, and $4$ appears nowhere among $\triangle DEF$'s lengths $6$, $\sqrt{61}$, $5$ — so no reordering can rescue it.
57. $AB = 5$, $BC = \sqrt{9 + 16} = 5$, $AC = \sqrt{4 + 16} = 2\sqrt{5}$; $DE = 5$, $EF = \sqrt{4 + 16} = 2\sqrt{5}$, $DF = \sqrt{9 + 16} = 5$. As written the second comparison fails: $BC = 5$ but $EF = 2\sqrt{5}$, so $\triangle ABC \cong \triangle DEF$ is false. The vertex with two sides of length $5$ is $B$ in the first triangle and $D$ in the second, so $B \leftrightarrow D$, $A \leftrightarrow E$, $C \leftrightarrow F$, and $\triangle ABC \cong \triangle EDF$ is true.
58. $AB = \sqrt{16 + 4} = 2\sqrt{5}$ and $AC = \sqrt{1 + 4} = \sqrt{5}$; $DE = \sqrt{4 + 16} = 2\sqrt{5}$ and $DF = \sqrt{4 + 1} = \sqrt{5}$. At $A$ the slopes are $\tfrac{2}{4} = \tfrac{1}{2}$ and $\tfrac{2}{-1} = -2$, product $-1$; at $D$ they are $\tfrac{-4}{2} = -2$ and $\tfrac{1}{2}$, product $-1$. So $m\angle A = m\angle D = 90°$, and with $\overline{AB} \cong \overline{DE}$ and $\overline{AC} \cong \overline{DF}$ on either side of it, $\triangle ABC \cong \triangle DEF$ by SAS.
59. Because coordinates hand you distances directly and angle measures almost never. ASA and AAS both need two angle measures, and reading a general angle measure off coordinates requires trigonometry, which is Chapter 9. Distance is available for every pair of points, and three distances is a complete criterion, so SSS is the one that always works.
60. Slopes: slope of $\overline{BA}$ is $\tfrac{1 - 3}{-2 - 2} = \tfrac{-2}{-4} = \tfrac{1}{2}$ and slope of $\overline{BC}$ is $\tfrac{-1 - 3}{4 - 2} = -2$; the product is $-1$, so $m\angle B = 90°$. Squared sides: $AB^2 = 20$, $BC^2 = 20$, $AC^2 = 40$, and $20 + 20 = 40$, so the converse of the Pythagorean Theorem gives the same right angle at $B$.
61. $\sqrt{193} \approx 13.8924$ and $\sqrt{194} \approx 13.9284$, and both round to $13.9$ — but $193 \neq 194$, so the two sides are genuinely different lengths and the triangles are not congruent. Rounding destroyed exactly the information the proof depended on. Coordinate lengths belong in simplest radical form.
62. Yes, one kit fits both. The first bed's sides are $2\sqrt{5}$, $\sqrt{29}$, and $\sqrt{13}$ metres, and the second bed's are the same three lengths in the same correspondence, so the beds are congruent by SSS and their perimeters are equal — about $13.46$ m each.
63. Because SAS needs the measure of the angle between the two sides, and the only angle measure coordinates certify is $90°$, by a slope product of $-1$. For any other included angle there is no way to establish congruence of the angles without trigonometry. When the included angle is not right, compute the third distance in each triangle and use SSS instead.
64. The three verified congruences are $\overline{AB} \cong \overline{DE}$, $\overline{BC} \cong \overline{EF}$, $\overline{AC} \cong \overline{DF}$, which is the correspondence $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$. Writing $\triangle ABC \cong \triangle DFE$ claims $B \leftrightarrow F$ and $C \leftrightarrow E$, which the computations did not show. The correct statement is $\triangle ABC \cong \triangle DEF$.

Exit ticket 6.3

65. $AB = \sqrt{16 + 4} = 2\sqrt{5}$ and $DE = \sqrt{16 + 4} = 2\sqrt{5}$. One pair of congruent sides is not a criterion, so it proves nothing on its own; two more comparisons are needed for SSS.
66. Yes. The slopes are $\tfrac{1 - (-1)}{1 - (-3)} = \tfrac{2}{4} = \tfrac{1}{2}$ and $\tfrac{-3 - 1}{3 - 1} = -2$, and $\left(\tfrac{1}{2}\right)(-2) = -1$.
67. SSS.
68. Plot both triangles and write the correspondence you intend to prove; compute the parts the criterion needs, in exact form; compare them in correspondence order; name the criterion and state the congruence.

Lesson 6.4 — Measured Attributes from Algebra and Coordinates

Guided practice

69. $AB = 8$, $BC = 6$, $AC = 10$.
70. $8 + 6 + 10 = 24$. Because $\triangle ABC \cong \triangle DEF$ by SSS, corresponding sides are congruent by CPCTC, so the perimeters are equal — none of $\triangle DEF$'s sides had to be measured.
71. $m\angle E = 90°$, by CPCTC. $\angle B$ is right because $\overline{BA}$ is horizontal and $\overline{BC}$ is vertical, and $\angle E$ corresponds to it.
72. One grid unit is $10$ metres. The near plot's sides are $4$, $3$, and $5$ grid units, which is $40$ m, $30$ m, and $50$ m.
73. $40 + 30 + 50 = 120$ metres.
74. SSS. The far plot's coordinates give the same three distances as the near plot's, which proves the two plots congruent, and CPCTC then transfers each side length.

Independent practice

75. $DE = 2\sqrt{13}$, $EF = 5$, $DF = 7$ by CPCTC, so the perimeter is $12 + 2\sqrt{13}$, which is about $19.21$.
76. $AB = 8$, $BC = 6$, $AC = \sqrt{64 + 36} = 10$; the perimeter is $24$.
77. $3x + 4 = 5x - 10$ gives $14 = 2x$, so $x = 7$ and $ST = 5(7) - 10 = 25$. Check: $PQ = 3(7) + 4 = 25$.
78. $m\angle D = 47°$ and $m\angle E = 68°$ by CPCTC, and $m\angle F = 180° - 47° - 68° = 65°$.
79. The sides are $\sqrt{16 + 9} = 5$, $4$, and $3$, so the perimeter is $12$. The right angle is at $(-1, 0)$: the side to $(-1, -3)$ is vertical and the side to $(3, 0)$ is horizontal.
80. Congruent triangles have equal perimeters, so $DF = 34 - 11 - 9 = 14$. Since $\overline{AC}$ corresponds to $\overline{DF}$, $AC = 14$.
81. The template's sides are $12$, $\sqrt{144 + 25} = 13$, and $5$ centimetres, so every bracket has perimeter $30$ cm. Every bracket is congruent to the template, so its corresponding sides are congruent by CPCTC and its perimeter must be the same.
82. Each sail's perimeter is $6 + 8 + 10 = 24$ m, so the two together need $48$ m of binding, at a cost of $48 \times 14 = \$672$.
83. $\overline{AC}$ corresponds to $\overline{DF}$, not to $\overline{DE}$. So $DF = 15$, and $DE$ is whatever $AB$ is.
84. Because congruent triangles have all six pairs of corresponding parts congruent. Once the congruence itself is proved, any measurement made on the reachable triangle transfers to the matching part of the other one, and the reason written beside that step is CPCTC.

Exit ticket 6.4

85. $DE = 9$, $EF = 12$, $DF = 15$ by CPCTC, so the perimeter is $36$.
86. $AB = 6$, $BC = \sqrt{64 + 36} = 10$, $AC = 8$; the perimeter is $24$.
87. $\angle F$, by CPCTC.
88. Prove the congruence — from a marked figure, from algebra, or from coordinates; write the correspondence in order; match the part you want to the part you know; name CPCTC as the reason.

Chapter 6 Review — answers

Review 1 (G.TR.2b).

• $5x - 6 = 3x + 10$ gives $2x = 16$, so $x = 8$ and $AB = 5(8) - 6 = 34$; the check is $DE = 3(8) + 10 = 34$. $(4y + 3) = (6y - 25)$ gives $28 = 2y$, so $y = 14$ and $m\angle B = 4(14) + 3 = 59°$. $2z + 11 = 4z - 7$ gives $18 = 2z$, so $z = 9$ and $BC = 2(9) + 11 = 29$.
• SAS. $\angle B$ lies between $\overline{AB}$ and $\overline{BC}$, and both of those sides are in the list of congruences.
• The three variable values are not the answer. What is missing is the substitution back into both expressions of each pair — giving $AB = DE = 34$, $m\angle B = m\angle E = 59°$, and $BC = EF = 29$ — and then the check that these measures can belong to a triangle. Here the check passes immediately: two sides and an included angle strictly between $0°$ and $180°$ always determine a triangle. It is the all-sides case that needs the Triangle Inequality.
• There would be no criterion: the arrangement becomes SSA. The congruent sides would be $\overline{AB}$ and $\overline{AC}$, which meet at $A$, so the angle between them is $\angle A$ — but the congruent angle given is $\angle B$, which is not included. Moving one side from $\overline{BC}$ to $\overline{AC}$ moves the angle out from between the two sides, and SSA is not a criterion.

Review 2 (G.TR.2c).

• $AB = \sqrt{16 + 9} = 5$, $BC = \sqrt{1 + 36} = \sqrt{37}$, $AC = \sqrt{25 + 9} = \sqrt{34}$. $DE = \sqrt{9 + 16} = 5$, $EF = \sqrt{36 + 1} = \sqrt{37}$, $DF = \sqrt{9 + 25} = \sqrt{34}$.
• $\triangle ABC \cong \triangle DEF$ by SSS.
• Because SSS asks only for sides. Three pairs of congruent sides is a complete criterion, and the distance formula supplies all six lengths directly — no angle measure enters the argument at all.
• $\triangle ABC \cong \triangle EDF$ claims $\overline{BC} \cong \overline{DF}$, but $BC = \sqrt{37}$ and $DF = \sqrt{34}$, so the statement is false. The triangles are still congruent — the three lengths $5$, $\sqrt{37}$, $\sqrt{34}$ appear in both — but only under the correspondence $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$. The statement was wrong about the pairing, not about the triangles.

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

• Label Bed 1 as $\triangle ABC$ with $A(-8, -1)$, $B(-2, -1)$, $C(-8, 7)$ and Bed 2 as $\triangle DEF$ with $D(2, 6)$, $E(2, 0)$, $F(10, 6)$. Then $AB = 6$, $BC = \sqrt{36 + 64} = 10$, $AC = 8$, and $DE = 6$, $EF = \sqrt{64 + 36} = 10$, $DF = 8$. All three pairs match in order, so $\triangle ABC \cong \triangle DEF$ by SSS.
• One unit is $4$ metres, so Bed 1 measures $24$ m, $40$ m, and $32$ m.
• $24 + 40 + 32 = 96$ metres. The beds are congruent by SSS, so corresponding sides are congruent by CPCTC and Bed 2's three sides are the same three lengths.
• $96 \times 7.50 = \$720$. The quote comes from the coordinates on the map and the congruence they prove; the crew already has every length it needs, so no measurement on the far side of the wall would add anything.