Appendix A — Answer Key, Chapter 10: The Pythagorean Theorem

SOL 8.MG.4 · Covers textbook Chapter 10 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 132 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Conventions used throughout:

• $a$ and $b$ are the legs and $c$ is the hypotenuse, always.
• Only the positive square root is kept, because every unknown here is a length.
• When a side is irrational, the exact radical is given first and a rational approximation to the nearest hundredth second, following Chapter 2. The two hundredths that bracket the value are shown whenever the rounding decision is close.
• For converse items, $c$ is assigned to the longest side before anything is squared. That assignment is shown, because it is the step where the work usually goes wrong.

Lesson 10.1 — The Parts of a Right Triangle

Guided practice

1. The $15$-cm side. It is the longest side, and in a right triangle the longest side is the one opposite the right angle — the hypotenuse.
2. $\overline{QR}$. It is the only side that does not touch $P$, the right-angle vertex.
3. $\overline{PQ}$ and $\overline{PR}$ — the two sides that meet at $P$.
4. $\overline{QR}$. The right-angle mark is at $P$, so the two sides meeting at $P$ are the legs and the remaining side must be the hypotenuse. No measuring is needed, because the marked right angle settles it.
5. Legs $\overline{PQ}$ and $\overline{PR}$; hypotenuse $\overline{QR}$. Rotating the drawing does not change which sides meet at the right angle.
6. False. The four figures show one triangle in four orientations, and in figure (b) the hypotenuse runs up the left, while in figure (d) it runs across the top. Position on the page says nothing; the side opposite the right angle is the hypotenuse.

Independent practice

7. a) $\overline{ED}$ and $\overline{EF}$ b) $\overline{DF}$
8. The $25$-inch side, because the hypotenuse is the longest side and $25 > 24 > 7$. (It also checks out: $7^2 + 24^2 = 49 + 576 = 625 = 25^2$.)
9. Legs $\overline{CA}$ and $\overline{CB}$ (lengths $9$ and $40$); hypotenuse $\overline{AB}$ (length $41$); the longest side is $\overline{AB}$, which is the hypotenuse.
10. a) $\overline{PQ}$ b) $\overline{YZ}$ c) $\overline{LN}$ — in each case the side that does not touch the right-angle vertex.
11. No. The hypotenuse is the longest side of a right triangle, so no leg can be longer than it. A leg of $10$ with a hypotenuse of $8$ is impossible.
12. The three angles of a triangle total $180^\circ$. If one is $90^\circ$, the other two must total $90^\circ$, so neither can be $90^\circ$ or more on its own. The right angle is therefore the largest angle, and the longest side of any triangle is the side opposite its largest angle — the hypotenuse.
13. The ladder is the hypotenuse. The wall and the ground are the legs, since they meet each other at a right angle.
14. Legs are defined by where they meet, not by which way they point: the legs are the two sides that form the right angle. A vertical side may be a leg or the hypotenuse depending on how the figure is turned. The rule that always works: find the right-angle mark, call the two sides meeting there the legs, and call the remaining side the hypotenuse.

Exit ticket 10.1

15. $\overline{KL}$ — the side that does not touch $J$.
16. The $29$-meter side. (Check: $20^2 + 21^2 = 400 + 441 = 841 = 29^2$.)
17. $\overline{PQ}$ and $\overline{PR}$.
18. Find the mark for the right angle; the two sides that meet there are the legs, and the third side — the one opposite the right angle — is the hypotenuse. Because the procedure never mentions horizontal or vertical, it works in every orientation.

Lesson 10.2 — Verifying the Pythagorean Theorem

Guided practice

19. Square on the leg of length $3$: $3^2 = 9$ square units. Square on the leg of length $4$: $4^2 = 16$ square units. Sum: $9 + 16 = 25$ square units, which is exactly the area of the square on the hypotenuse, $5^2 = 25$.
20. Nine unit squares in one small square and sixteen in the other; twenty-five in the largest. Equation: $9 + 16 = 25$.
21. $6^2 + 8^2 = 36 + 64 = 100$ and $10^2 = 100$. Verified.
22. $5^2 + 12^2 = 25 + 144 = 169$ and $13^2 = 169$. Verified.
23. The measured hypotenuse is $10$ cm. Then $6^2 + 8^2 = 100$ and $10^2 = 100$, so the triangle verifies the theorem.
24. Because a square with side length $a$ has area $a \cdot a = a^2$. Reading $a^2$ as an area turns the equation $a^2 + b^2 = c^2$ into a statement you can check by counting or by cutting paper: the two smaller squares together cover the same area as the largest one.

Independent practice

25. a) $81 + 144 = 225$ and $15^2 = 225$ — verified b) $49 + 576 = 625$ and $25^2 = 625$ — verified c) $144 + 256 = 400$ and $20^2 = 400$ — verified d) $100 + 576 = 676$ and $26^2 = 676$ — verified
26. Areas $25$, $144$, and $169$ square units. Equation: $25 + 144 = 169$.
27. Squares on the legs: $3^2 = 9$ and $5^2 = 25$ square units. Square on the hypotenuse: $9 + 25 = 34$ square units — a tilted square on the grid, but still a square. Hypotenuse: $c = \sqrt{34} \approx 5.83$ units, since $5.83^2 = 33.9889$ and $5.84^2 = 34.1056$, and $34$ is nearer to $33.9889$.
28. The hypotenuse should measure $15$ cm, since $9^2 + 12^2 = 81 + 144 = 225 = 15^2$. A measured check is only approximate because a ruler reads to about the nearest millimeter, the pencil line has width, and the right angle was drawn by hand. A measurement of $14.9$ cm gives $222.01$ rather than $225$, and that gap is about the drawing, not about the theorem.
29. $4^2 + 6^2 = 16 + 36 = 52$ while $8^2 = 64$, and $52 \neq 64$. The triangle is therefore not a right triangle. The relationship $a^2 + b^2 = c^2$ belongs to right triangles specifically, which is why checking it is a real test and not a formality.
30. Tie the string into a loop with $12$ equally spaced knots. Hold it at three knots so that the three sides span $3$, $4$, and $5$ knot-spaces, and pull it taut into a triangle. Lay the square corner of a sheet of paper into the largest corner: it fits exactly, so that corner is a right angle. Then check the counts: $3^2 + 4^2 = 9 + 16 = 25 = 5^2$.
31. She should find exactly $5$ feet, since $3^2 + 4^2 = 25 = 5^2$. Finding $5$ feet $2$ inches means $a^2 + b^2 \neq c^2$, so the corner is not square — and because the measured diagonal is longer than $5$ feet, the corner is wider than $90^\circ$ and needs to be closed up.
32. The student added the side lengths instead of their squares. The theorem is about areas of squares, not about sums of lengths. Correct test: $3^2 + 4^2 = 9 + 16 = 25$ and $5^2 = 25$, so $25 = 25$ and the theorem holds.

Exit ticket 10.2

33. $8^2 + 15^2 = 64 + 225 = 289$ and $17^2 = 289$. Verified.
34. Areas $81$, $144$, and $225$ square units. Equation: $81 + 144 = 225$.
35. The square on the hypotenuse has area $16 + 16 = 32$ square units, so $c = \sqrt{32} \approx 5.66$ units. ($5.65^2 = 31.9225$ and $5.66^2 = 32.0356$; $32$ is nearer to $32.0356$.)
36. Verifying means checking the relationship on particular triangles by counting squares or measuring sides. Each check is evidence, and a measured check is only ever approximate. Proving means showing the relationship must hold for every right triangle, including all the ones nobody will ever draw. No number of verifications adds up to a proof, though the area picture is where a proof starts.

Lesson 10.3 — Finding a Missing Side

Guided practice

37. $9^2 + 12^2 = 81 + 144 = 225$, so $c = \sqrt{225} = 15$.
38. $5^2 + 12^2 = 25 + 144 = 169$, so $c = 13$.
39. $7^2 + 24^2 = 49 + 576 = 625$, so $c = 25$.
40. $b^2 = 10^2 - 6^2 = 100 - 36 = 64$, so $b = 8$. ($8 < 10$, as a leg must be.)
41. $b^2 = 26^2 - 10^2 = 676 - 100 = 576$, so $b = 24$.
42. $2^2 + 6^2 = 4 + 36 = 40$, so $c = \sqrt{40} \approx 6.32$. ($6.32^2 = 39.9424$ and $6.33^2 = 40.0689$; $40$ is nearer to $39.9424$.)
43. $b^2 = 9^2 - 4^2 = 81 - 16 = 65$, so $b = \sqrt{65} \approx 8.06$. ($8.06^2 = 64.9636$ and $8.07^2 = 65.1249$.)
44. $b^2 = 20^2 - 12^2 = 400 - 144 = 256$, so $b = 16$.

Independent practice

45. a) $144 + 256 = 400$, $c = 20$ b) $400 + 441 = 841$, $c = 29$ c) $81 + 1600 = 1681$, $c = 41$ d) $324 + 576 = 900$, $c = 30$
46. a) $625 - 225 = 400$, leg $= 20$ b) $2500 - 196 = 2304$, leg $= 48$ c) $1156 - 256 = 900$, leg $= 30$ d) $3721 - 121 = 3600$, leg $= 60$
47. $5^2 + 5^2 = 50$, so $c = \sqrt{50} \approx 7.07$ meters. ($7.07^2 = 49.9849$ and $7.08^2 = 50.1264$.)
48. $6^2 + 9^2 = 36 + 81 = 117$, so $c = \sqrt{117} \approx 10.82$ feet. ($10.81^2 = 116.8561$ and $10.82^2 = 117.0724$; $117$ is nearer to $117.0724$.)
49. $4^2 + 7^2 = 16 + 49 = 65$, so $c = \sqrt{65} \approx 8.06$ inches.
50. $b^2 = 8^2 - 5^2 = 64 - 25 = 39$, so $b = \sqrt{39} \approx 6.24$. ($6.24^2 = 38.9376$ and $6.25^2 = 39.0625$; the gaps are $0.0624$ and $0.0625$, so $39$ is nearer to $38.9376$ — barely, and this is a good reason to compare both gaps rather than eyeball them.)
51. $b^2 = 14^2 - 9^2 = 196 - 81 = 115$, so $b = \sqrt{115} \approx 10.72$. ($10.72^2 = 114.9184$ and $10.73^2 = 115.1329$.)
52. $b^2 = 15^2 - 11^2 = 225 - 121 = 104$, so $b = \sqrt{104} \approx 10.20$. ($10.19^2 = 103.8361$ and $10.20^2 = 104.04$.) Write the trailing zero: $10.20$, not $10.2$, since the approximation is to the nearest hundredth.
53. $4^2 + 6^2 = 16 + 36 = 52$, so $c = \sqrt{52}$. Since $7^2 = 49$ and $8^2 = 64$, the value lies between $7$ and $8$. Then $7.21^2 = 51.9841$ and $7.22^2 = 52.1284$, so $c = \sqrt{52} \approx 7.21$.
54. It depends on which letter is unknown. In $a^2 + b^2 = c^2$, if $c$ is unknown the two known squares are on the left and you add them to get $c^2$. If a leg is unknown, the known leg's square is already on the left with it, so you undo that addition by subtracting: $b^2 = c^2 - a^2$. Adding in the leg case would produce a leg longer than the hypotenuse, which is impossible.
55. The student added when the hypotenuse was already known. The check that catches it: $13.93$ is longer than the hypotenuse $13$, and a leg can never exceed the hypotenuse. Correct work: $b^2 = 169 - 25 = 144$, so $b = 12$.
56. The student added the leg lengths instead of their squares. Correct work: $6^2 + 8^2 = 36 + 64 = 100$, so $c = 10$. The $3$-$4$-$5$ triangle shows why adding lengths cannot work: $3 + 4 = 7$, but the hypotenuse is $5$, not $7$.

Exit ticket 10.3

57. $10^2 + 24^2 = 100 + 576 = 676$, so $c = 26$.
58. $b^2 = 16^2 - 9^2 = 256 - 81 = 175$, so $b = \sqrt{175} \approx 13.23$. ($13.22^2 = 174.7684$ and $13.23^2 = 175.0329$.)
59. $3^2 + 6^2 = 9 + 36 = 45$, so $c = \sqrt{45} \approx 6.71$. ($6.70^2 = 44.89$ and $6.71^2 = 45.0241$.)
60. Ask which side is unknown. Unknown hypotenuse: square both legs and add, then take the square root. Unknown leg: square the hypotenuse and the known leg and subtract the smaller from the larger, then take the square root. The result must be shorter than the hypotenuse; if it is not, you added when you should have subtracted.

Lesson 10.4 — The Converse: Is It a Right Triangle?

Guided practice

61. $c = 15$. $9^2 + 12^2 = 81 + 144 = 225$ and $15^2 = 225$. Right triangle, with the right angle opposite the $15$ side.
62. $c = 8$. $4^2 + 6^2 = 16 + 36 = 52$ and $8^2 = 64$. Since $52 \neq 64$, not a right triangle.
63. $c = 17$. $8^2 + 15^2 = 64 + 225 = 289$ and $17^2 = 289$. Right triangle.
64. $c = 8$. $5^2 + 6^2 = 25 + 36 = 61$ and $8^2 = 64$. Since $61 \neq 64$, not a right triangle — close, but close is not equal.
65. $c = 25$. $7^2 + 24^2 = 49 + 576 = 625$ and $25^2 = 625$. Right triangle.
66. $c = 15$. $10^2 + 10^2 = 200$ and $15^2 = 225$. Since $200 \neq 225$, not a right triangle.

Independent practice

67. a) $c = 20$: $144 + 256 = 400 = 20^2$ — right triangle b) $c = 10$: $36 + 49 = 85 \neq 100$ — not a right triangle c) $c = 29$: $400 + 441 = 841 = 29^2$ — right triangle d) $c = 16$: $81 + 144 = 225 \neq 256$ — not a right triangle
68. $c = 50$: $196 + 2304 = 2500 = 50^2$. Yes, a right triangle.
69. $c = 85$: $169 + 7056 = 7225 = 85^2$. Yes, a right triangle.
70. $c = 61$: $121 + 3600 = 3721 = 61^2$. Yes, a right triangle.
71. $c = 34$, because $34$ is the longest of the three lengths even though it is listed last. $16^2 + 30^2 = 256 + 900 = 1156 = 34^2$. Yes, a right triangle.
72. $c = 25$, because $25$ is the longest length even though it is listed first. $7^2 + 24^2 = 49 + 576 = 625 = 25^2$. Yes, a right triangle.
73. $c = 4$: $2^2 + 3^2 = 4 + 9 = 13$ and $4^2 = 16$. Since $13 \neq 16$, not a right triangle.
74. In a right triangle the hypotenuse is the longest side, and the theorem puts the hypotenuse alone on one side of the equation. Testing with the wrong side as $c$ tests a statement the theorem never made. Using $15$ as $c$ for the sides $8$, $15$, $17$ gives $8^2 + 17^2 = 64 + 289 = 353$ against $15^2 = 225$; since $353 \neq 225$, you would wrongly conclude "not a right triangle," even though $8$, $15$, $17$ is a genuine right triple.
75. $c = 16$: $9^2 + 12^2 = 225$ and $16^2 = 256$, so $225 \neq 256$ and no corner is square. Because $a^2 + b^2 < c^2$, the side opposite the largest corner is longer than a right corner would allow, so the largest corner is wider than $90^\circ$.
76. The student used $24$ as $c$, but the longest side is $26$. With $c = 26$ and legs $10$ and $24$: $100 + 576 = 676 = 26^2$. It is a right triangle, with the right angle opposite the $26$ side.

Exit ticket 10.4

77. $c = 25$: $225 + 400 = 625 = 25^2$. Yes.
78. $c = 9$: $25 + 49 = 74$ and $9^2 = 81$; $74 \neq 81$. No.
79. $c = 30$: $324 + 576 = 900 = 30^2$. Yes.
80. If the two shorter sides of a triangle have squares that add up to the square of the longest side, then the triangle is a right triangle and its right angle is opposite that longest side. The first step is finding the longest side because the longest side is the only candidate for the hypotenuse; assigning $c$ to a shorter side tests a false equation and can turn a genuine right triangle into a wrong "no."

Lesson 10.5 — Applying the Theorem in Context

Guided practice

81. The unknown is a leg (the wall height), since the ladder is the hypotenuse. $h^2 = 13^2 - 5^2 = 169 - 25 = 144$, so $h = 12$ feet. ($12 < 13$, as required.)
82. $d^2 = 10^2 + 24^2 = 100 + 576 = 676$, so $d = 26$ inches.
83. Horizontal leg $7 - 1 = 6$ units, vertical leg $9 - 1 = 8$ units. $6^2 + 8^2 = 36 + 64 = 100$, so $AB = 10$ units.
84. The rise and the run are the legs. $5^2 + 12^2 = 25 + 144 = 169$, so the sloped surface is $13$ feet.
85. $10^2 + 10^2 = 200$, so the diagonal is $\sqrt{200} \approx 14.14$ meters. ($14.14^2 = 199.9396$ and $14.15^2 = 200.2225$.)
86. Yes. $c = 15$ (the diagonal is the longest measurement), and $9^2 + 12^2 = 81 + 144 = 225 = 15^2$. The tool is the converse of the Pythagorean Theorem: the side lengths satisfy the equation, so the corner is a right angle.

Independent practice

87. The pole and the ground are the legs; the wire is the hypotenuse. $24^2 + 7^2 = 576 + 49 = 625$, so the wire is $25$ feet.
88. $12^2 + 16^2 = 144 + 256 = 400$, so the straight-line distance is $20$ miles.
89. $30^2 + 40^2 = 900 + 1600 = 2500$, so the diagonal is $50$ inches.
90. The string is the hypotenuse, so the height is a leg. $h^2 = 26^2 - 10^2 = 676 - 100 = 576$, so the kite is $24$ meters above the person's hands.
91. $8^2 + 5^2 = 64 + 25 = 89$, so the distance is $\sqrt{89} \approx 9.43$ blocks. ($9.43^2 = 88.9249$ and $9.44^2 = 89.1136$; $89$ is nearer to $88.9249$.)
92. $6^2 + 9^2 = 36 + 81 = 117$, so the diagonal is $\sqrt{117} \approx 10.82$ meters.
93. First corner, $c = 50$: $900 + 1600 = 2500 = 50^2$, so that corner is square. Second corner, $c = 40$: $400 + 900 = 1300$ while $40^2 = 1600$, and $1300 \neq 1600$, so that corner is not square.
94. The ladder is the hypotenuse and the wall height is a leg, so the distance from the wall is the other leg. $d^2 = 20^2 - 16^2 = 400 - 256 = 144$, so $d = 12$ feet.
95. The two arms are the legs. $7^2 + 5^2 = 49 + 25 = 74$, so the brace is $\sqrt{74} \approx 8.60$ inches. ($8.60^2 = 73.96$ and $8.61^2 = 74.1321$.) Keep the trailing zero, since the approximation is to the nearest hundredth.
96. The student added instead of subtracting. The ladder is the hypotenuse, so the height is a leg and cannot be longer than the $13$-foot ladder — yet $13.93 > 13$, which would have the ladder reaching higher than its own length. Correct work: $h^2 = 169 - 25 = 144$, so $h = 12$ feet.

Exit ticket 10.5

97. $6^2 + 8^2 = 36 + 64 = 100$, so the brace is $10$ feet.
98. $4^2 + 8^2 = 16 + 64 = 80$, so the surface is $\sqrt{80} \approx 8.94$ feet. ($8.94^2 = 79.9236$ and $8.95^2 = 80.1025$.)
99. $c = 19$: $11^2 + 15^2 = 121 + 225 = 346$ and $19^2 = 361$; $346 \neq 361$, so there is no square corner.
100. If the problem gives you two sides of a right triangle and asks for a length, use the theorem. If it gives you all three side lengths and asks whether something is square, right-angled, plumb, or true, use the converse — the answer there is yes or no, not a number.

Chapter 10 Review

Part A — Verifying the theorem with diagrams, materials, and measurement (8.MG.4a)

101. Areas $9$, $16$, and $25$ square units. Equation: $9 + 16 = 25$, which is $3^2 + 4^2 = 5^2$.
102. Squares on the legs: $9$ and $25$ square units. Square on the hypotenuse: $9 + 25 = 34$ square units. Hypotenuse: $\sqrt{34} \approx 5.83$ units. ($5.83^2 = 33.9889$ and $5.84^2 = 34.1056$.)
103. $20^2 + 21^2 = 400 + 441 = 841$ and $29^2 = 841$. Verified.
104. Draw a right angle, mark $9$ cm along one side and $12$ cm along the other, and join the two marks. Measure that third side: it should come out $15$ cm. Then compare $9^2 + 12^2 = 81 + 144 = 225$ with $15^2 = 225$. The comparison is between the sum of the squares of the measured legs and the square of the measured hypotenuse; the two should agree to within the precision of the ruler.
105. $4^2 + 6^2 = 16 + 36 = 52$ and $8^2 = 64$, so $52 \neq 64$. If the equation held for every triangle it would say nothing about right angles in particular. Because it can fail, holding is informative — and that is exactly what makes the converse a usable test.
106. Verifying checks the relationship on specific triangles, by counting unit squares, cutting and fitting paper squares, or measuring with a ruler. Each case is evidence, and measured cases are approximate. Proving establishes the relationship for all right triangles at once by reasoning about areas rather than by checking examples, so a proof covers triangles nobody has drawn.

Part B — Identifying the hypotenuse and the legs in any orientation (8.MG.4c)

107. Legs $\overline{PQ}$ and $\overline{PR}$; hypotenuse $\overline{QR}$.
108. Legs $\overline{PQ}$ and $\overline{PR}$; hypotenuse $\overline{QR}$. Same answer as item 107, which is the point: the four figures are the same triangle turned.
109. $\overline{XY}$. The right angle is at $Z$, so the two sides meeting at $Z$ are the legs, and $\overline{XY}$ is the only side that does not touch $Z$.
110. The $34$ side, because the hypotenuse is the longest side. (Check: $16^2 + 30^2 = 256 + 900 = 1156 = 34^2$.)
111. Rotating a triangle changes the picture, not the triangle, so any side can end up horizontal — including the hypotenuse. Procedure: locate the right-angle mark, name the two sides meeting there as the legs, and name the third side the hypotenuse. Equivalently, the hypotenuse is the only side that does not touch the right-angle vertex.
112. The student used position in the picture instead of the definition. "At the bottom" is not a property of a triangle. The student should look for the right-angle mark and then take the side opposite it; a check is that the hypotenuse must also be the longest side.

Part C — Finding the measure of a missing side (8.MG.4d)

113. $12^2 + 35^2 = 144 + 1225 = 1369$, so $c = \sqrt{1369} = 37$.
114. $14^2 + 48^2 = 196 + 2304 = 2500$, so $c = 50$.
115. $b^2 = 30^2 - 18^2 = 900 - 324 = 576$, so $b = 24$.
116. $b^2 = 41^2 - 9^2 = 1681 - 81 = 1600$, so $b = 40$.
117. $2^2 + 5^2 = 4 + 25 = 29$, so $c = \sqrt{29} \approx 5.39$. ($5.38^2 = 28.9444$, gap $0.0556$; $5.39^2 = 29.0521$, gap $0.0521$. The nearer hundredth is $5.39$.)
118. $5^2 + 7^2 = 25 + 49 = 74$, so $c = \sqrt{74} \approx 8.60$. ($8.60^2 = 73.96$ and $8.61^2 = 74.1321$.)
119. $b^2 = 11^2 - 6^2 = 121 - 36 = 85$, so $b = \sqrt{85} \approx 9.22$. ($9.22^2 = 85.0084$ and $9.21^2 = 84.8241$.)
120. $b^2 = 13^2 - 6^2 = 169 - 36 = 133$, so $b = \sqrt{133} \approx 11.53$. ($11.53^2 = 132.9409$ and $11.54^2 = 133.1716$.)

Part D — Deciding whether a triangle is right from three side lengths (8.MG.4b)

121. $c = 26$: $100 + 576 = 676 = 26^2$. Yes, a right triangle.
122. $c = 8$: $25 + 36 = 61 \neq 64$. No.
123. $c = 41$: $81 + 1600 = 1681 = 41^2$. Yes, a right triangle.
124. $c = 21$: $144 + 256 = 400 \neq 441$. No. (Since $400 < 441$, the corner opposite the $21$ side is wider than $90^\circ$.)
125. $c = 25$, because $25$ is the longest of the three lengths even though it is listed last. $7^2 + 24^2 = 49 + 576 = 625 = 25^2$, so yes, it is a right triangle.
126. The equation $a^2 + b^2 = c^2$ only ever holds with $c$ as the hypotenuse, which is the longest side, so testing with any other side as $c$ tests a statement the theorem does not make. For $8$, $15$, $17$ with $c = 15$: $8^2 + 17^2 = 353$ while $15^2 = 225$, and $353 \neq 225$ would wrongly say "not a right triangle." Assigning $c = 17$ gives $64 + 225 = 289 = 17^2$, the correct verdict.

Part E — Applying the theorem and its converse in context (8.MG.4e)

127. The height is a leg. $h^2 = 17^2 - 8^2 = 289 - 64 = 225$, so $h = 15$ feet.
128. $d^2 = 20^2 + 21^2 = 400 + 441 = 841$, so $d = 29$ inches.
129. Horizontal leg $10 - 2 = 8$ units, vertical leg $7 - 1 = 6$ units. $8^2 + 6^2 = 64 + 36 = 100$, so $CD = 10$ units. The right triangle has its right angle at $(10, 1)$, directly right of $C$ and directly below $D$.
130. $12^2 + 12^2 = 288$, so the diagonal is $\sqrt{288} \approx 16.97$ feet. ($16.97^2 = 287.9809$ and $16.98^2 = 288.3204$; the gaps are $0.0191$ and $0.3204$, so $288$ is nearer to $287.9809$.)
131. $c = 30$: $18^2 + 24^2 = 324 + 576 = 900 = 30^2$. Yes, the corner is square, by the converse of the Pythagorean Theorem.
132. The diagonal is the hypotenuse of a right triangle with legs $9$ and $12$: $81 + 144 = 225$, so the diagonal is $15$ meters. Walking the two sides is $9 + 12 = 21$ meters. The diagonal is $21 - 15 = 6$ meters shorter.