# Spectrum Math Grade 8 Chapter 5 Lesson 8 Answer Key Using Pythagorean Theorem

Students can use the Spectrum Math Grade 8 Answer Key Chapter 5 Lesson 5.8 Pythagorean Theorem as a quick guide to resolve any of their doubts.

## Spectrum Math Grade 8 Chapter 5 Lesson 5.8 Pythagorean Theorem Answers Key

If a, b, and c are the lengths of the sides of this triangle , a2 + b2 = c2 If a = 3 and b = 4, what is c?
a2 + b2 = c2
32 + 42 = c2
9 + 16 = c2
25 = c2
$$\sqrt{25}$$= c
5 = c

If a = 4 and b = 6, what is c?
a2 + b2 = c2
42 + 62 = c2
16 + 36 = c2
52 = c2
$$\sqrt{52}$$ = c

Use the Pythagorean Theorem to determine the length of c. Assume that each problem describes a right triangle. Sides a and b are the legs and the hypotenuse is c.

Question 1.
If a = 9 and b = 4, c = or about _____
Answer: c = $$\sqrt{97}$$ or about 9.648
Given, a = 9 and b = 4
a2 + b2 = c2
92 + 42 = c2
81 + 16 = c2
97 = c2
$$\sqrt{97}$$ = c
Therefore, c = $$\sqrt{97}$$ or about 9.648

Question 2.
If a = 5 and b = 7, c = or about _____
Answer: c = $$\sqrt{74}$$ or about 8.60
Given, a = 5 and b = 7
a2 + b2 = c2
52 + 72 = c2
25 + 49 = c2
74 = c2
$$\sqrt{74}$$ = c
Therefore, c = $$\sqrt{74}$$ or about 8.60

Question 3.
If a = 3 and b = 6, c = or about _____
Answer: c = $$\sqrt{45}$$ or about 6.708
Given, a = 3 and b = 6
a2 + b2 = c2
32 + 62 = c2
9 + 36 = c2
45 = c2
$$\sqrt{45}$$ = c
therefore, c = $$\sqrt{45}$$ or about 6.708

Question 4.
If a = 2 and b = 9, c = or about _____
Answer: c = $$\sqrt{85}$$ or about 9.219
Given, a = 2 and b = 9
a2 + b2 = c2
22 + 92 = c2
4 + 81 = c2
85 = c2
$$\sqrt{85}$$ = c
Therefore, c = $$\sqrt{85}$$ or about 9.219

Question 5.
If a = 5 and b = 6, c = or about _____
Answer: c = $$\sqrt{61}$$ or  about 7.81
Given, a = 5 and b = 6
a2 + b2 = c2
52 + 62 = c2
25 + 36 = c2
61 = c2
$$\sqrt{61}$$ = c
Therefore, c = $$\sqrt{61}$$ or  about 7.81

Question 6.
If a = 3 and b = 5, c = or about _____
Answer: c = $$\sqrt{34}$$ or about 5.83
Given, a = 3 and b = 5
a2 + b2 = c2
32 + 52 = c2
9 + 25 = c2
34 = c2
$$\sqrt{34}$$ = c
Therefore, c = $$\sqrt{34}$$ or about 5.83

Question 7.
If a = 7 and b = 6, c = or about _____
Answer: c = $$\sqrt{85}$$ or about 9.219
Given, a = 7 and b = 6
a2 + b2 = c2
72 + 62 = c2
49 + 36 = c2
85 = c2
$$\sqrt{85}$$ = c
Therefore, c = $$\sqrt{85}$$ or about 9.219

Question 8.
If a = 8 and b = 6, c = or about _____
Answer: c = $$\sqrt{100}$$ or  10
Given, a = 8 and b = 6
a2 + b2 = c2
82 + 62 = c2
64 + 36 = c2
100 = c2
$$\sqrt{100}$$ = c
c = 10
Therefore, c = $$\sqrt{100}$$ or  10

Question 9.
If a = 7 and b = 2, c = or about _____
Answer:c = $$\sqrt{53}$$ or  about 7.28
Given, a = 7 and b = 2
a2 + b2 = c2
72 + 22 = c2
49 + 4 = c2
53 = c2
$$\sqrt{53}$$ = c
Therefore, c = $$\sqrt{53}$$ or  about 7.28

Question 10.
If a = 8 and b = 5, c = or about _____
Answer: c = $$\sqrt{89}$$ or about 9.43
Given, a = 8 and b = 5
a2 + b2 = c2
82 + 52 = c2
64 + 25 = c2
89 = c2
$$\sqrt{89}$$ = c
Therefore, c = $$\sqrt{89}$$ or about 9.43

You can use the Pythagorean Theorem to find the unknown length of a side of a right triangle as long as the other two lengths are known.

If a = 12 m and c = 13 m, what is b?
a2 + b2 = c2 122 + b2 = 132
144 + b2 = 169
144 + b2 – 144 = 169 – 144
b2 = 25 b = $$\sqrt{25}$$ b = 5 m

If b = 15 ft. and c = 17 ft., what is a?
a2 + b2 = c2 a2 + 152 = 172
a2 + 225 = 289
a2 + 225 – 225 = 289 – 225
a2 = 64 a = $$\sqrt{64}$$ a = 8 ft.

Assume that each problem describes a right triangle. Use the Pythagorean Theorem to find the unknown lengths. Round to the nearest hundredth as needed.

Question 1
If a = 12 and c = 20, b = or ______
Answer: b = $$\sqrt{256}$$  or 16
Given, a = 12 and c = 20
a2 + b2 = c2
122 + b2 = 202
144 + b2 = 400
144 + b2 – 144 = 400 – 144
b2 = 256
b = $$\sqrt{256}$$
b = 16
Therefore, b = $$\sqrt{256}$$  or 16

Question 2.
If a = 24 and c = 26, a = or ______
Answer: b = $$\sqrt{100}$$  or 10
Given, a = 24 and c = 26
a2 + b2 = c2
242 + b2 = 262
576 + b2 = 676
576 + b2 – 576 = 676 – 576
b2 = 100
b = $$\sqrt{100}$$
b = 10
Therefore, b = $$\sqrt{100}$$  or 10

Question 3.
If c = 8 and a = 5, b = or about ____
Answer: b = $$\sqrt{39}$$  or about 6.244
Given, a = 5 and c = 8
a2 + b2 = c2
52 + b2 = 82
25 + b2 = 64
25 + b2 – 25 = 64 – 25
b2 = 39
b = $$\sqrt{39}$$
Therefore, b = $$\sqrt{39}$$  or about 6.244

Question 4.
If b = 13 and c = 17, a = or about ____
Answer: a = $$\sqrt{120}$$  or about 10.95
Given, b = 13 and c = 17
a2 + b2 = c2
a2 + 132 = 172
a2 + 169 = 289
a +169-169= 289 – 169
a2 = 120
a = $$\sqrt{120}$$
Therefore, a = $$\sqrt{120}$$  or about 10.95

Question 5.
If a = 20 and c = 32, b = or about ____
Answer: b = $$\sqrt{624}$$  or about 24.97
Given, a = 20 and c = 32
a2 + b2 = c2
202 + b2 = 322
400 + b2 = 1024
400 + b2 – 400 = 1024 – 400
b2 = 624
b = $$\sqrt{624}$$
Therefore, b = $$\sqrt{624}$$  or about 24.97

Question 6.
If c = 15 and c = 12, a = or ____
Answer: a = $$\sqrt{81}$$  or 9
Given, c = 15 and b = 12
a2 + b2 = c2
a2 + 122 = 152
a2 + 144 = 225
a + 144 – 144= 225 – 144
a2 = 81
a = $$\sqrt{81}$$
a = 9
Therefore, a = $$\sqrt{81}$$  or 9

Question 7.
If c = 41 and c = 40, a = or ____
Answer: a = $$\sqrt{81}$$  or 9
Given, c = 41 and b = 40
a2 + b2 = c2
a2 + 402 = 412
a2 + 1600 = 1681
a + 1600 – 1600= 1681 – 1600
a2 = 81
a = $$\sqrt{81}$$
a = 9
Therefore, a = $$\sqrt{81}$$  or 9

Question 8.
If a = 36 and c = 85, b = or ____
Answer:  b = $$\sqrt{5929}$$  or 77
Given, a = 20 and c = 32
a2 + b2 = c2
362 + b2 = 852
1296 + b2 = 7225
1296 + b2 – 1296= 7225 – 1296
b2 = 5929
b = $$\sqrt{5929}$$
b = 77
Therefore, b = $$\sqrt{5929}$$  or 77

Question 9.
If c = 73 and c = 48, a = or ____
Answer: a = $$\sqrt{3025}$$  or 55
Given, c = 73 and b = 48
a2 + b2 = c2
a2 + 482 = 732
a2 + 2304 = 5329
a + 2304 – 2304= 5329 – 2304
a2 = 3025
a = $$\sqrt{3025}$$
a = 55
Therefore, a = $$\sqrt{3025}$$  or 55

Question 10.
If a = 14 and c = 22, b = or ____
Answer: b = $$\sqrt{288}$$  or about 16.97
Given, a = 14 and c = 22
a2 + b2 = c2
142 + b2 = 222
196 + b2 = 484
196 + b2 – 196= 484 – 196
b2 = 288
b = $$\sqrt{288}$$
Therefore, b = $$\sqrt{288}$$  or about 16.97

Use the Pythagorean Theorem to solve each problem.

Question 1.
A boat has a sail with measures as shown. How tall is the sail? The sail is _____ feet tall.
Answer: The sail is 21 feet tall.
Let a = 20ft. c = 29 ft.
a2 + b2 = c2
202 + b2 = 292
400 + b2 = 841
400 + b2 – 400= 841 – 400
b2 = 441
b = $$\sqrt{441}$$
b = 21
Therefore, b = $$\sqrt{441}$$  or 21
Therefore, The sail is 21 feet tall.

Question 2.
Kelsey drove on a back road for 15 miles from Benton to a lake. Her friend Paul drove 12 miles on the highway from Middleville to the lake. This area is shown at the right. How long is the road from Benton to Middleville?
The road is _____ miles long. Kelsey drove on a back road for 15 miles from Benton to a lake. Her friend Paul drove 12 miles on the highway from Middleville to the lake.
Let a = 12 mi. c = 15mi.
a2 + b2 = c2
122 + b2 = 152
144 + b2 = 225
144 + b2 – 144= 225-144
b2 = 81
b = $$\sqrt{81}$$
b = 9
Therefore, b = $$\sqrt{81}$$  or 9
The road is 9 miles long.

Question 3.
A 14-foot ladder is leaning against a building as shown. It touches a point 11 feet up on the building. How far away from the base of the building does the ladder stand? A 14-foot ladder is leaning against a building as shown. It touches a point 11 feet up on the building.
Let a = 11 ft. c = 14ft.
a2 + b2 = c2
112 + b2 = 142
121 + b2 = 196
121 + b2 – 121= 196-121
b2 = 76
b = $$\sqrt{76}$$
b = 8.66
Therefore, b = $$\sqrt{76}$$  or 8.66

Question 4.
This gangway connects a dock to a ship, as shown.
How long is the gangway? The gangway is ____ feet long.
Answer: The gangway is 65 feet long.
Let, a = 63 ft. and b = 16 ft.
a2 + b2 = c2
632 + 162 = c2
3969 + 256 = c2
4225 = c2
$$\sqrt{4225}$$ = c
c = 65 ft.
Therefore, c = $$\sqrt{4225}$$ or 65 ft.
The gangway is 65 feet long.

Question 5.
About how long is the lake shown at right?
The lake is about ____ m long. $$\sqrt{234}$$ = c
Therefore, c = $$\sqrt{234}$$ or 15.3 km