## Engage NY Eureka Math 8th Grade Module 3 Lesson 13 Answer Key

### Eureka Math Grade 8 Module 3 Lesson 13 Exercise Answer Key

Exercises

Use the Pythagorean theorem to determine the unknown length of the right triangle.

Exercise 1.

Determine the length of side c in each of the triangles below.

a.

Answer:

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

25+144=c^{2}

169=c^{2}

13=c

b.

Answer:

0.5^{2}+1.2^{2}=c^{2}

0.25+1.44=c^{2}

1.69=c^{2}

1.3=c

Exercise 2.

Determine the length of side b in each of the triangles below.

a.

Answer:

4^{2}+b^{2}=5^{2}

16+b^{2}=25

16-16+b^{2}=25-16

b^{2}=9

b=3

b.

Answer:

0.4^{2}+b^{2}=0.5^{2}

0.16+b^{2}=0.25

0.16-0.16+b^{2}=0.25-0.16

b^{2}=0.09

b=0.3

Exercise 3.

Determine the length of \(\overline{Q S}\). (Hint: Use the Pythagorean theorem twice.)

Answer:

15^{2}+|QT|^{2}=17^{2}

225+|QT|^{2}=289

225-225+|QT|^{2}=289-225

|QT|^{2}=64

|QT| = 8

15^{2}+|TS|^{2}=25^{2}

225+|TS|^{2}=625

225-225+|TS|^{2}=625-225

|TS|^{2}=400

|TS|=20

Since |QT|+|TS|=|QS|, then the length of side \(\overline{Q S}\) is 8+20, which is 28.

### Eureka Math Grade 8 Module 3 Lesson 13 Exit Ticket Answer Key

Determine the length of side \(\overline{B D}\) in the triangle below.

Answer:

First, determine the length of side \(\overline{B C}\).

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

144+BC^{2}=225

BC^{2}=225-144

BC^{2}=81

BC=9

Then, determine the length of side \(\overline{C D}\).

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

144+CD^{2}=169

CD^{2}=169-144

CD^{2}=25

CD=5

Adding the lengths of sides \(\overline{B C}\) and \(\overline{C D}\) determines the length of side \(\overline{B D}\); therefore, 5+9=14. \(\overline{B D}\) has a length of 14.

### Eureka Math Grade 8 Module 3 Lesson 13 Problem Set Answer Key

Students practice using the Pythagorean theorem to find unknown lengths of right triangles.

Use the Pythagorean theorem to determine the unknown length of the right triangle.

Question 1.

Determine the length of side c in each of the triangles below.

a.

Answer:

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

36+64=c^{2}

100=c^{2}

10=c

b.

Answer:

0.6^{2}+0.8^{2}=c^{2}

0.36+0.64=c^{2}

1=c^{2}

1=c

Question 2.

Determine the length of side a in each of the triangles below.

a.

Answer:

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

a^{2}+64=289

a^{2}+64-64=289-64

a^{2}=225

a=15

b.

Answer:

a^{2}+0.8^{2}=1.7^{2}

a^{2}+0.64=2.89

a^{2}+0.64-0.64=2.89-0.64

a^{2}=2.25

a=1.5

Question 3.

Determine the length of side b in each of the triangles below.

a.

Answer:

20^{2}+b^{2}=25^{2}

400+b^{2}=625

400-400+b^{2}=625-400

b^{2}=225

b=15

b.

Answer:

2^{2}+b^{2}=2.5^{2}

4+b^{2}=6.25

4-4+b^{2}=6.25-4

b^{2}=2.25

b=1.5

Question 4.

Determine the length of side a in each of the triangles below.

a.

Answer:

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

a^{2}+144=400

a^{2}+144-144=400-144

a^{2}=256

a=16

b.

Answer:

a^{2}+1.2^{2}=2^{2}

a^{2}+1.44=4

a^{2}+1.44-1.44=4-1.44

a^{2}=2.56

a=1.6

Question 5.

What did you notice in each of the pairs of Problems 1–4? How might what you noticed be helpful in solving problems like these?

Answer:

In each pair of problems, the problems and solutions were similar. For example, in Problem 1, part (a) showed the sides of the triangle were 6, 8, and 10, and in part (b), they were 0.6, 0.8, and 1. The side lengths in part (b) were a tenth of the value of the lengths in part (a). The same could be said about parts (a) and (b) of Problems 2–4. This might be helpful for solving problems in the future. If I am given side lengths that are decimals, then I could multiply them by a factor of 10 to make whole numbers, which are easier to work with. Also, if I know common numbers that satisfy the Pythagorean theorem, like side lengths of 3, 4, and 5, then I recognize them more easily in their decimal forms, that is, 0.3, 0.4, and 0.5.