Appendix A — Answer Key, Chapter 8: The Pythagorean Theorem and Special Right Triangles

SOL G.TR.4 (a, b, c, f) · Covers textbook Chapter 8 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 110 across the chapter.

Conventions used in every answer below. The hypotenuse is the side opposite the right angle and is always the longest; in $a^2 + b^2 = c^2$, $c$ is that side. Two legs given: add. Hypotenuse and one leg given: subtract. Answers are in simplest radical form first, and rounded only where a context asks, labelled about. Every fraction with a radical denominator is rationalized, which G.TR.4 b and c require by name. Before classifying three lengths, sort them and check the Triangle Inequality.

The two special triangles:

Lesson 8.1 — The Pythagorean Theorem

Guided practice

1. The side of length $5$. It is the side opposite the right angle, and it is the longest of the three.
2. $3^2 = 9$, $4^2 = 16$, and $5^2 = 25$.
3. $3^2 + 4^2 = 5^2$, or $9 + 16 = 25$.
4. Two smaller right triangles, each similar to the original by AA — each shares one acute angle with the original, and each has a right angle of its own.
5. The altitude is $4.8$; the hypotenuse is cut into segments of $3.6$ and $6.4$.
6. Two legs given means the hypotenuse is unknown, so the squares are added. A hypotenuse and one leg given means a leg is unknown, so one square is subtracted from the other.

Independent practice

7. $c^2 = 81 + 144 = 225$, so $c = 15$.
8. $c^2 = 49 + 576 = 625$, so $c = 25$.
9. $b^2 = 676 - 100 = 576$, so $b = 24$.
10. $b^2 = 1681 - 81 = 1600$, so $b = 40$.
11. $c^2 = 4 + 9 = 13$, so $c = \sqrt{13}$ — already simplest, since $13$ has no square factor.
12. $c^2 = 25 + 25 = 50$, so $c = \sqrt{50} = 5\sqrt{2}$.
13. $c^2 = 36 + 100 = 136 = 4 \cdot 34$, so $c = 2\sqrt{34}$.
14. $b^2 = 144 - 64 = 80 = 16 \cdot 5$, so $b = 4\sqrt{5}$.
15. $c^2 = 2.25 + 4 = 6.25$, so $c = 2.5$.
16. $h^2 = 169 - 25 = 144$, so the ladder reaches $12$ ft up the wall.
17. $d^2 = 36 + 64 = 100$, so the brace is $10$ ft.
18. Both given sides are legs, so their squares are added, not subtracted. $c^2 = 36 + 100 = 136$ and $c = 2\sqrt{34}$. Subtracting treats $10$ as the hypotenuse, which the problem never said it was.

Exit ticket 8.1

19. $c^2 = 64 + 225 = 289$, so $c = 17$.
20. $b^2 = 625 - 49 = 576$, so $b = 24$.
21. $c^2 = 9 + 49 = 58$, so $c = \sqrt{58}$ — already simplest.
22. Because $c$ in $a^2 + b^2 = c^2$ means one specific side, the one opposite the right angle. Deciding which side that is first is what tells you whether the problem adds two squares or subtracts one from the other.

Lesson 8.2 — The Converse, and Classifying a Triangle

Guided practice

23. $6$, $8$, $10$ is right, because $36 + 64 = 100 = 10^2$.
24. $6$, $8$, $9$ is acute, because $36 + 64 = 100 > 81 = 9^2$.
25. $6$, $8$, $11$ is obtuse, because $36 + 64 = 100 < 121 = 11^2$.
26. Sort them, so that $c$ is the longest of the three rather than whichever was written last.
27. The Triangle Inequality. Three lengths that cannot form a triangle have no classification at all.
28. Sort the lengths, then compare $a^2 + b^2$ with $c^2$: equal means right, greater means acute, less means obtuse.

Independent practice

29. $81 + 1600 = 1681 = 41^2$. Right.
30. $25 + 36 = 61 < 64$. Obtuse.
31. $49 + 64 = 113 > 100$. Acute.
32. $144 + 256 = 400 = 20^2$. Right.
33. $4 + 5 = 9 < 10$, so the Triangle Inequality fails. Not a triangle, and therefore not classifiable.
34. $100 + 576 = 676 = 26^2$. Right.
35. $4 + 9 = 13 < 16$. Obtuse.
36. $36 + 49 = 85 > 81$. Acute.
37. $(\sqrt{2})^2 + (\sqrt{3})^2 = 2 + 3 = 5 = (\sqrt{5})^2$. Right.
38. $121 + 3600 = 3721 = 61^2$. Right.
39. No. $6^2 + 8^2 = 100$ and $10.2^2 = 104.04$, and $100 < 104.04$, so the triangle is obtuse and the corner is more than $90°$.
40. The lengths were not sorted, so $c$ was taken as $6$ instead of $10$. Sorted, the comparison is $6^2 + 8^2 = 100$ against $10^2 = 100$, and the triangle is right.

Exit ticket 8.2

41. $25 + 144 = 169 = 13^2$. Right.
42. $25 + 144 = 169 > 144 = 12^2$. Acute.
43. $25 + 144 = 169 < 196 = 14^2$. Obtuse.
44. Sort the three lengths so that $c$ is the longest, and check the Triangle Inequality so that there is a triangle to classify.

Lesson 8.3 — Pythagorean Triples

Guided practice

45. Three whole numbers that satisfy $a^2 + b^2 = c^2$ — so they are the side lengths of a right triangle, and all three are integers.
46. $64 + 225 = 289$, and $17^2 = 289$.
47. Because multiplying every length by $k$ multiplies both sides of $a^2 + b^2 = c^2$ by $k^2$: $(ka)^2 + (kb)^2 = k^2(a^2 + b^2) = k^2c^2 = (kc)^2$. The equation still holds, and the three new numbers are still whole.
48. $c^2 = 81 + 144 = 225$, so $c = 15$.
49. $9, 12$ is $3 \times (3, 4)$, so $c = 3 \times 5 = 15$.
50. Whenever the triangle is not a multiple of a triple you recognize — for example legs of $6$ and $10$, where the hypotenuse is $2\sqrt{34}$ and no whole-number triple applies.

Independent practice

51. Yes. $36 + 64 = 100$, and it is $2 \times (3, 4, 5)$.
52. Yes. $25 + 144 = 169 = 13^2$.
53. No. $16 + 25 = 41$, but $6^2 = 36$.
54. $21, 28$ is $7 \times (3, 4)$, so the hypotenuse is $7 \times 5 = 35$.
55. $15, 36$ is $3 \times (5, 12)$, so the hypotenuse is $3 \times 13 = 39$.
56. $50 = 10 \times 5$ and $30 = 10 \times 3$, so this is $10 \times (3, 4, 5)$ and the other leg is $10 \times 4 = 40$.
57. $85 = 5 \times 17$ and $40 = 5 \times 8$, so this is $5 \times (8, 15, 17)$ and the other leg is $5 \times 15 = 75$.
58. $14, 48$ is $2 \times (7, 24)$, so the hypotenuse is $2 \times 25 = 50$.
59. $27, 36$ is $9 \times (3, 4)$, so the hypotenuse is $9 \times 5 = 45$.
60. $65 = 5 \times 13$ and $25 = 5 \times 5$, so this is $5 \times (5, 12, 13)$ and the other leg is $5 \times 12 = 60$.
61. $48, 20$ is $4 \times (12, 5)$, so the diagonal is $4 \times 13 = 52$ inches. The triple used is $5$, $12$, $13$.
62. Multiplying all three numbers by $k$ turns $a^2 + b^2 = c^2$ into $(ka)^2 + (kb)^2 = k^2a^2 + k^2b^2 = k^2(a^2 + b^2) = k^2c^2 = (kc)^2$. Both sides pick up the same factor of $k^2$, so the equation survives — and if $k$ is a whole number the three new lengths are whole numbers too.

Exit ticket 8.3

63. $18, 24$ is $6 \times (3, 4)$, so the hypotenuse is $6 \times 5 = 30$.
64. $39 = 3 \times 13$ and $15 = 3 \times 5$, so this is $3 \times (5, 12, 13)$ and the other leg is $3 \times 12 = 36$.
65. No. $81 + 144 = 225$, but $16^2 = 256$.
66. Three whole numbers $a$, $b$, $c$ with $a^2 + b^2 = c^2$.

Lesson 8.4 — 45°-45°-90° Triangles

Guided practice

67. From cutting a square along a diagonal. Each half has two equal legs with a right angle between them, so the other two angles are $45°$ each.
68. leg : leg : hypotenuse $= 1 : 1 : \sqrt{2}$.
69. $5^2 + 5^2 = 25 + 25 = 50$, and $\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$.
70. $7\sqrt{2}$.
71. $\dfrac{10}{\sqrt{2}} = \dfrac{10\sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \dfrac{10\sqrt{2}}{2} = 5\sqrt{2}$.
72. Because it still has a radical in the denominator. It is the right value written in a form the standard asks you to clear, and multiplying above and below by $\sqrt{2}$ clears it.

Independent practice

73. $9\sqrt{2}$.
74. $12\sqrt{2}$.
75. $3\sqrt{2} \cdot \sqrt{2} = 3 \cdot 2 = 6$.
76. $\dfrac{8}{\sqrt{2}} = \dfrac{8\sqrt{2}}{2} = 4\sqrt{2}$.
77. $\dfrac{14}{\sqrt{2}} = \dfrac{14\sqrt{2}}{2} = 7\sqrt{2}$.
78. $\dfrac{6\sqrt{2}}{\sqrt{2}} = 6$.
79. $\dfrac{5}{\sqrt{2}} = \dfrac{5\sqrt{2}}{2}$.
80. $11\sqrt{2}$.
81. $\dfrac{16}{\sqrt{2}} = \dfrac{16\sqrt{2}}{2} = 8\sqrt{2}$.
82. The cut is the diagonal, $10\sqrt{2}$ inches, which is about $14.14$ inches.
83. Half the hypotenuse of the $10$-unit example would be $5$, but the leg is $5\sqrt{2} \approx 7.07$. Halving is the $30°$-$60°$-$90°$ rule and does not apply here; a $45°$-$45°$-$90°$ leg is the hypotenuse divided by $\sqrt{2}$.
84. Because the hypotenuse is the leg multiplied by $\sqrt{2}$, and $\sqrt{2} > 1$. Multiplying a positive length by a number greater than $1$ makes it larger, so no computation is needed. (It is also the side opposite the largest angle, which Chapter 4 already settled.)

Exit ticket 8.4

85. $6\sqrt{2}$.
86. $\dfrac{20}{\sqrt{2}} = \dfrac{20\sqrt{2}}{2} = 10\sqrt{2}$.
87. $7\sqrt{2}$.
88. $1 : 1 : \sqrt{2}$ — the two $1$s are the legs, which are equal, and the $\sqrt{2}$ is the hypotenuse, which is a leg times $\sqrt{2}$.

Lesson 8.5 — 30°-60°-90° Triangles

Guided practice

89. From an equilateral triangle cut by an altitude. The altitude bisects the side it meets and the angle it comes from, leaving a right triangle with angles of $30°$, $60°$, and $90°$.
90. short leg : long leg : hypotenuse $= 1 : \sqrt{3} : 2$.
91. The short leg is opposite the $30°$ angle. In the figure it is the half-side of length $4$, and it is half the hypotenuse.
92. Long leg $6\sqrt{3}$, hypotenuse $12$.
93. $s = \dfrac{9}{\sqrt{3}} = \dfrac{9\sqrt{3}}{3} = 3\sqrt{3}$, and the hypotenuse is $2 \cdot 3\sqrt{3} = 6\sqrt{3}$.
94. The square's diagonal is $6\sqrt{2}$ m; the equilateral triangle's altitude is $5\sqrt{3}$ ft.

Independent practice

95. Long leg $5\sqrt{3}$, hypotenuse $10$.
96. Long leg $11\sqrt{3}$, hypotenuse $22$.
97. Short leg $9$, long leg $9\sqrt{3}$.
98. Short leg $13$, long leg $13\sqrt{3}$.
99. $s = \dfrac{12}{\sqrt{3}} = \dfrac{12\sqrt{3}}{3} = 4\sqrt{3}$, and the hypotenuse is $8\sqrt{3}$.
100. $s = \dfrac{5\sqrt{3}}{\sqrt{3}} = 5$, and the hypotenuse is $10$.
101. $s = \dfrac{21}{\sqrt{3}} = \dfrac{21\sqrt{3}}{3} = 7\sqrt{3}$, and the hypotenuse is $14\sqrt{3}$.
102. The short leg is half the side, $7$, so the altitude is $7\sqrt{3}$.
103. The short leg is $3$, so the altitude is $3\sqrt{3}$.
104. The altitude is the long leg, so the short leg is $\dfrac{9\sqrt{3}}{\sqrt{3}} = 9$. That is half the side, so the side is $18$.
105. The short leg is $2$ ft, so the height is $2\sqrt{3}$ ft, which is about $3.46$ ft.
106. Doubling gives the hypotenuse from the short leg, not from the long one. Return to the short leg first: $s = \dfrac{10}{\sqrt{3}} = \dfrac{10\sqrt{3}}{3}$, and the hypotenuse is $2s = \dfrac{20\sqrt{3}}{3}$, about $11.55$ — not $20$.

Exit ticket 8.5

107. Long leg $8\sqrt{3}$, hypotenuse $16$.
108. Short leg $15$, long leg $15\sqrt{3}$.
109. $s = \dfrac{6}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3} = 2\sqrt{3}$, and the hypotenuse is $4\sqrt{3}$.
110. $1 : \sqrt{3} : 2$ — the $1$ is the short leg, opposite the $30°$ angle; the $\sqrt{3}$ is the long leg, opposite the $60°$; the $2$ is the hypotenuse, which is twice the short leg.

Chapter 8 Review — answers

Review 1 (G.TR.4 a, f).

• Yes, the corner is square. By the converse of the Pythagorean Theorem, $9^2 + 12^2 = 81 + 144 = 225$ and $15^2 = 225$; the two agree, so the triangle is right.
• It is $3 \times (3, 4, 5)$, so the triple is $3$, $4$, $5$ and the multiplier is $3$.
• $9^2 + 12^2 = 225$ and $16^2 = 256$. Since $225 < 256$, the corner is obtuse.
• $9 + 12 = 21 < 22$, so the three lengths fail the Triangle Inequality. They do not form a triangle at all, and a set of lengths with no triangle has no classification.

Review 2 (G.TR.4 b, c).

• The diagonal is $18\sqrt{2}$ inches.
• The gable's height is $9\sqrt{3}$ inches — the short leg is half the side, $9$, and the altitude is that times $\sqrt{3}$.
• $\dfrac{18}{\sqrt{2}} = \dfrac{18\sqrt{2}}{2} = 9\sqrt{2}$, and half the diagonal is $\dfrac{18\sqrt{2}}{2} = 9\sqrt{2}$. They agree — and not by accident: dividing a square's side by $\sqrt{2}$ and halving its diagonal are the same operation, because the diagonal is the side times $\sqrt{2}$ and $\dfrac{s\sqrt{2}}{2} = \dfrac{s}{\sqrt{2}}$. The rationalizing step is what makes that visible.
• $18\sqrt{2} \approx 25.46$ inches and $9\sqrt{3} \approx 15.59$ inches. The exact form comes first because it is the answer; the decimal is that answer rounded, and it is only wanted once someone has to cut to a mark.

Review 3 (G.TR.4 b, c, f).

• A right triangle with a $30°$ angle is a $30°$-$60°$-$90°$. The short leg is half the hypotenuse, so it is $10$ cm, and the long leg is $10\sqrt{3}$ cm.
• Two $45°$ angles make it a $45°$-$45°$-$90°$, so both legs are equal: $\dfrac{20}{\sqrt{2}} = \dfrac{20\sqrt{2}}{2} = 10\sqrt{2}$ cm each.
• The $45°$-$45°$-$90°$ triangle does. Its shortest side is $10\sqrt{2}$ and the other triangle's shortest side is $10$, and $10\sqrt{2} > 10$ because $\sqrt{2} > 1$ — no decimal needed.
• $10^2 + (10\sqrt{3})^2 = 100 + 300 = 400 = 20^2$, and $(10\sqrt{2})^2 + (10\sqrt{2})^2 = 200 + 200 = 400 = 20^2$. The ratios tell you the shape; the verification confirms that the three lengths were assigned to the right roles — that what was called the hypotenuse really is the one the squares add up to. A ratio applied to the wrong side still produces three numbers, and only the theorem catches it.