## Engage NY Eureka Math 8th Grade Module 4 Lesson 31 Answer Key

### Eureka Math Grade 8 Module 4 Lesson 31 Exercise Answer Key

Exercises

Exercise 1.

Identify two Pythagorean triples using the known triple 3, 4, 5 (other than 6, 8, 10).

Answer:

Answers will vary. Accept any triple that is a whole number multiple of 3, 4, 5.

Exercise 2.

Identify two Pythagorean triples using the known triple 5, 12, 13.

Answer:

Answers will vary. Accept any triple that is a whole number multiple of 5, 12, 13.

Exercise 3.

Identify two triples using either 3, 4, 5 or 5, 12, 13.

Answer:

Answers will vary.

Use the system to find Pythagorean triples for the given values of s and t. Recall that the solution in the form of (\(\frac{c}{b}\), \(\frac{a}{b}\)) is the triple a, b, c.

Exercise 4.

s = 4, t = 5

Answer:

x + y + x – y = \(\frac{5}{4}\) + \(\frac{4}{5}\)

2x = \(\frac{5}{4}\) + \(\frac{4}{5}\)

2x = \(\frac{41}{20}\)

x = \(\frac{41}{40}\)

\(\frac{41}{40}\) + y = \(\frac{5}{4}\)

y = \(\frac{5}{4}\) – \(\frac{41}{40}\)

y = \(\frac{9}{40}\)

Then the solution is (\(\frac{41}{40}\), \(\frac{9}{40}\)), and the triple is 9, 40, 41.

Exercise 5.

s = 7, t = 10

Answer:

x + y + x – y = \(\frac{10}{7}\) + \(\frac{7}{10}\)

2x = \(\frac{149}{70}\)

x = \(\frac{149}{140}\)

\(\frac{149}{140}\) + y = \(\frac{10}{7}\)

y = \(\frac{10}{7}\) – \(\frac{149}{140}\)

y = \(\frac{51}{140}\)

Then the solution is (\(\frac{149}{140}\), \(\frac{51}{140}\)), and the triple is 51, 140, 149.

Exercise 6.

s = 1, t = 4

Answer:

x + y + x – y = 4 + \(\frac{1}{4}\)

2x = \(\frac{17}{4}\)

x = \(\frac{17}{8}\)

\(\frac{17}{8}\) + y = \(\frac{4}{1}\)

y = 4 – \(\frac{17}{8}\)

y = \(\frac{15}{8}\)

Then the solution is (\(\frac{17}{8}\), \(\frac{15}{8}\)), and the triple is 15, 8, 17.

Exercise 7.

Use a calculator to verify that you found a Pythagorean triple in each of the Exercises 4â€“6. Show your work below.

Answer:

For the triple 9, 40, 41:

9^{2} + 40^{2} = 41^{2}

81 + 1600 = 1681

1681 = 1681

For the triple 51, 140, 149:

51^{2} + 140^{2} = 149^{2}

2601 + 19600 = 22201

22201 = 22201

For the triple 15, 8, 17:

15^{2} + 8^{2} = 17^{2}

225 + 64 = 289

289 = 289

### Eureka Math Grade 8 Module 4 Lesson 31 Problem Set Answer Key

Question 1.

Explain in terms of similar triangles why it is that when you multiply the known Pythagorean triple 3, 4, 5 by 12, it generates a Pythagorean triple.

Answer:

The triangle with lengths 3, 4, 5 is similar to the triangle with lengths 36, 48, 60. They are both right triangles whose corresponding side lengths are equal to the same constant.

\(\frac{36}{3}\) = \(\frac{48}{4}\) = \(\frac{60}{5}\) = 12

Therefore, the triangles are similar, and we can say that there is a dilation from some center with scale factor r = 12 that makes the triangles congruent.

Question 2.

Identify three Pythagorean triples using the known triple 8, 15, 17.

Answer:

Answers will vary. Accept any triple that is a whole number multiple of 8, 15, 17.

Question 3.

Identify three triples (numbers that satisfy a^{2} + b^{2} = c^{2}, but a, b, c are not whole numbers) using the triple 8, 15, 17.

Answer:

Answers will vary. Accept any triple that is not a set of whole numbers.

Use the system to find Pythagorean triples for the given values of s and t. Recall that the solution, in the form of (c/b, a/b), is the triple, a, b, c.

Question 4.

s = 2, t = 9

Answer:

x + y + x – y = \(\frac{9}{2}\) + \(\frac{2}{9}\)

2x = \(\frac{85}{18}\)

x = \(\frac{85}{36}\)

\(\frac{85}{36}\) + y = \(\frac{9}{2}\)

y = \(\frac{9}{2}\) – \(\frac{85}{36}\)

y = \(\frac{77}{36}\)

Then the solution is (\(\frac{85}{36}\), \(\frac{77}{36}\)), and the triple is 77, 36, 85.

Question 5.

s = 6, t = 7

Answer:

x + y + x – y = \(\frac{7}{6}\) + 6/7

2x = \(\frac{85}{42}\)

x = \(\frac{85}{84}\)

\(\frac{85}{84}\) + y = \(\frac{7}{6}\)

y = \(\frac{7}{6}\) – \(\frac{85}{84}\)

y = \(\frac{13}{84}\)

Then the solution is (\(\frac{85}{84}\), \(\frac{13}{84}\)), and the triple is 13, 84, 85.

Question 6.

s = 3, t = 4

Answer:

x + y + x – y = \(\frac{4}{3}\) + \(\frac{3}{4}\)

2x = \(\frac{25}{12}\)

x = \(\frac{25}{24}\)

\(\frac{25}{24}\) + y = \(\frac{4}{3}\)

y = \(\frac{4}{3}\) – \(\frac{25}{24}\)

y = \(\frac{7}{24}\)

Then the solution is (\(\frac{25}{24}\), \(\frac{7}{24}\)), and the triple is 7, 24, 25.

Question 7.

Use a calculator to verify that you found a Pythagorean triple in each of the Problems 4â€“6. Show your work.

Answer:

For the triple 77, 36, 85:

77^{2} + 36^{2} = 85^{2}

5929 + 1296 = 7225

7225 = 7225

For the triple 13, 84, 85:

13^{2} + 84^{2} = 85^{2}

169 + 7056 = 7225

7225 = 7225

For the triple 7, 24, 25:

7^{2} + 24^{2} = 25^{2}

49 + 576 = 625

625 = 625

### Eureka Math Grade 8 Module 4 Lesson 31 Exit Ticket Answer Key

Use a calculator to complete Problems 1â€“3.

Question 1.

Is 7, 20, 21 a Pythagorean triple? Is 1, \(\frac{15}{8}\), \(\frac{17}{8}\) a Pythagorean triple? Explain.

Answer:

The set of numbers 7, 20, 21 is not a Pythagorean triple because 7^{2} + 20^{2} â‰ 21^{2}.

The set of numbers 1, \(\frac{15}{8}\), \(\frac{17}{8}\) is not a Pythagorean triple because the numbers \(\frac{15}{8}\) and \(\frac{17}{8}\) are not whole numbers.

But they are a triple because 1^{2} + (\(\frac{15}{8}\))^{2} = (\(\frac{17}{8}\))^{2}.

Question 2.

Identify two Pythagorean triples using the known triple 9, 40, 41.

Answer:

Answers will vary. Accept any triple that is a whole number multiple of 9, 40, 41.

Question 3.

Use the system to find Pythagorean triples for the given values of s = 2 and t = 3. Recall that the solution in the form of (\(\frac{c}{b}\), \(\frac{a}{b}\)) is the triple a, b, c. Verify your results.

Answer:

x + y + x – y = \(\frac{3}{2}\) + \(\frac{2}{3}\)

2x = \(\frac{13}{6}\)

x = \(\frac{13}{12}\)

\(\frac{13}{12}\) + y = \(\frac{3}{2}\)

y = \(\frac{3}{2}\) – \(\frac{13}{12}\)

y = \(\frac{5}{12}\)

Then the solution is (\(\frac{13}{12}\), \(\frac{5}{12}\)), and the triple is 5, 12, 13.

5^{2} + 12^{2} = 13^{2}

25 + 144 = 169

169 = 169