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 , a^{2} + b^{2} = c^{2}

If a = 3 and b = 4, what is c?

a^{2} + b^{2} = c^{2}

3^{2} + 4^{2} = c^{2}

9 + 16 = c^{2}

25 = c^{2}

\(\sqrt{25}\)= c

5 = c

If a = 4 and b = 6, what is c?

a^{2} + b^{2} = c^{2}

4^{2} + 6^{2} = c^{2}

16 + 36 = c^{2}

52 = c^{2}

\(\sqrt{52}\) = c

c = about 7.21

**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

a^{2} + b^{2} = c^{2}

9^{2} + 4^{2} = c^{2}

81 + 16 = c^{2}

97 = c^{2}

\(\sqrt{97}\) = c

c = about 9.648

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

a^{2} + b^{2} = c^{2}

5^{2} + 7^{2} = c^{2}

25 + 49 = c^{2}

74 = c^{2}

\(\sqrt{74}\) = c

c = about 8.60

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

a^{2} + b^{2} = c^{2}

3^{2} + 6^{2} = c^{2}

9 + 36 = c^{2}

45 = c^{2}

\(\sqrt{45}\) = c

c = about 6.708

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

a^{2} + b^{2} = c^{2}

2^{2} + 9^{2} = c^{2}

4 + 81 = c^{2}

85 = c^{2}

\(\sqrt{85}\) = c

c = about 9.219

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

a^{2} + b^{2} = c^{2}

5^{2} + 6^{2} = c^{2}

25 + 36 = c^{2}

61 = c^{2}

\(\sqrt{61}\) = c

c = about 7.81

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

a^{2} + b^{2} = c^{2}

3^{2} + 5^{2} = c^{2}

9 + 25 = c^{2}

34 = c^{2}

\(\sqrt{34}\) = c

c = about 5.83

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

a^{2} + b^{2} = c^{2}

7^{2} + 6^{2} = c^{2}

49 + 36 = c^{2}

85 = c^{2}

\(\sqrt{85}\) = c

c = about 9.219

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

a^{2} + b^{2} = c^{2}

8^{2} + 6^{2} = c^{2}

64 + 36 = c^{2}

100 = c^{2}

\(\sqrt{100}\) = 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

a^{2} + b^{2} = c^{2}

7^{2} + 2^{2} = c^{2}

49 + 4 = c^{2}

53 = c^{2}

\(\sqrt{53}\) = c

c = about 7.28

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

a^{2} + b^{2} = c^{2}

8^{2} + 5^{2} = c^{2}

64 + 25 = c^{2}

89 = c^{2}

\(\sqrt{89}\) = c

c = about 9.43

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?

a^{2} + b^{2} = c^{2} 12^{2} + b^{2} = 13^{2}

144 + b^{2} = 169

144 + b^{2} – 144 = 169 – 144

b^{2} = 25 b = \(\sqrt{25}\) b = 5 m

If b = 15 ft. and c = 17 ft., what is a?

a^{2} + b^{2} = c^{2} a^{2} + 15^{2} = 17^{2}

a^{2} + 225 = 289

a^{2} + 225 – 225 = 289 – 225

a^{2} = 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

a^{2} + b^{2} = c^{2}

12^{2} + b^{2} = 20^{2}

144 + b^{2} = 400

144 + b^{2} – 144 = 400 – 144

b^{2} = 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

a^{2} + b^{2} = c^{2}

24^{2} + b^{2} = 26^{2}

576 + b^{2} = 676

576 + b^{2} – 576 = 676 – 576

b^{2} = 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

a^{2} + b^{2} = c^{2}

5^{2} + b^{2} = 8^{2}

25 + b^{2} = 64

25 + b^{2} – 25 = 64 – 25

b^{2} = 39

b = \(\sqrt{39}\)

b = about 6.244

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

a^{2} + b^{2} = c^{2}

a^{2} + 13^{2} = 17^{2}

a^{2} + 169 = 289

a^{2 } +169-169= 289 – 169

a^{2} = 120

a = \(\sqrt{120}\)

a = about 10.95

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

a^{2} + b^{2} = c^{2}

20^{2} + b^{2} = 32^{2}

400 + b^{2} = 1024

400 + b^{2} – 400 = 1024 – 400

b^{2} = 624

b = \(\sqrt{624}\)

b = about 24.97

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

a^{2} + b^{2} = c^{2}

a^{2} + 12^{2} = 15^{2}

a^{2} + 144 = 225

a^{2 } + 144 – 144= 225 – 144

a^{2} = 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

a^{2} + b^{2} = c^{2}

a^{2} + 40^{2} = 41^{2}

a^{2} + 1600 = 1681

a^{2 } + 1600 – 1600= 1681 – 1600

a^{2} = 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

a^{2} + b^{2} = c^{2}

36^{2} + b^{2} = 85^{2}

1296 + b^{2} = 7225

1296 + b^{2} – 1296= 7225 – 1296

b^{2} = 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

a^{2} + b^{2} = c^{2}

a^{2} + 48^{2} = 73^{2}

a^{2} + 2304 = 5329

a^{2 } + 2304 – 2304= 5329 – 2304

a^{2} = 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

a^{2} + b^{2} = c^{2}

14^{2} + b^{2} = 22^{2}

196 + b^{2} = 484

196 + b^{2} – 196= 484 – 196

b^{2} = 288

b = \(\sqrt{288}\)

b = about 16.97

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.

a^{2} + b^{2} = c^{2}

20^{2} + b^{2} = 29^{2}

400 + b^{2} = 841

400 + b^{2} – 400= 841 – 400

b^{2} = 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.

Answer: The road is 9 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.

a^{2} + b^{2} = c^{2}

12^{2} + b^{2} = 15^{2}

144 + b^{2} = 225

144 + b^{2} – 144= 225-144

b^{2} = 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?

The ladder stands about ____ feet from

Answer: The ladder stands about 8.66 feet from

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.

a^{2} + b^{2} = c^{2}

11^{2} + b^{2} = 14^{2}

121 + b^{2} = 196

121 + b^{2} – 121= 196-121

b^{2} = 76

b = \(\sqrt{76}\)

b = 8.66

Therefore, b = \(\sqrt{76}\) or 8.66

The ladder stands about 8.66 feet from

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.

a^{2} + b^{2} = c^{2}

63^{2} + 16^{2} = c^{2}

3969 + 256 = c^{2}

4225 = c^{2}

\(\sqrt{4225}\) = 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.

Answer: The lake is about 15.3 m long.

Let, a = 3 km and b = 15 km

a^{2} + b^{2} = c^{2}

3^{2} + 15^{2} = c^{2
}9 + 225 = c^{2}

234 = c^{2}

\(\sqrt{234}\) = c

c = 15.297 = 15.3 km

Therefore, c = \(\sqrt{234}\) or 15.3 km

The lake is about 15.3 m long.