## 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).
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.
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.

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

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

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

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 4–6. Show your work below.
For the triple 9, 40, 41:
92 + 402 = 412
81 + 1600 = 1681
1681 = 1681

For the triple 51, 140, 149:
512 + 1402 = 1492
2601 + 19600 = 22201
22201 = 22201

For the triple 15, 8, 17:
152 + 82 = 172
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.
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.
Answers will vary. Accept any triple that is a whole number multiple of 8, 15, 17.

Question 3.
Identify three triples (numbers that satisfy a2 + b2 = c2, but a, b, c are not whole numbers) using the triple 8, 15, 17.
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

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

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

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 4–6. Show your work.
For the triple 77, 36, 85:
772 + 362 = 852
5929 + 1296 = 7225
7225 = 7225

For the triple 13, 84, 85:
132 + 842 = 852
169 + 7056 = 7225
7225 = 7225

For the triple 7, 24, 25:
72 + 242 = 252
49 + 576 = 625
625 = 625

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

Use a calculator to complete Problems 1–3.
Question 1.
Is 7, 20, 21 a Pythagorean triple? Is 1, $$\frac{15}{8}$$, $$\frac{17}{8}$$ a Pythagorean triple? Explain.
The set of numbers 7, 20, 21 is not a Pythagorean triple because 72 + 202 ≠ 212.
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 12 + ($$\frac{15}{8}$$)2 = ($$\frac{17}{8}$$)2.

Question 2.
Identify two Pythagorean triples using the known triple 9, 40, 41.
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.
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.