## 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. 52+122=c2
25+144=c2
169=c2
13=c

b. 0.52+1.22=c2
0.25+1.44=c2
1.69=c2
1.3=c

Exercise 2.
Determine the length of side b in each of the triangles below.
a. 42+b2=52
16+b2=25
16-16+b2=25-16
b2=9
b=3

b. 0.42+b2=0.52
0.16+b2=0.25
0.16-0.16+b2=0.25-0.16
b2=0.09
b=0.3

Exercise 3.
Determine the length of $$\overline{Q S}$$. (Hint: Use the Pythagorean theorem twice.) 152+|QT|2=172
225+|QT|2=289
225-225+|QT|2=289-225
|QT|2=64
|QT| = 8

152+|TS|2=252
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. First, determine the length of side $$\overline{B C}$$.
122+BC2=152
144+BC2=225
BC2=225-144
BC2=81
BC=9

Then, determine the length of side $$\overline{C D}$$.
122+CD2=132
144+CD2=169
CD2=169-144
CD2=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. 62+82=c2
36+64=c2
100=c2
10=c

b. 0.62+0.82=c2
0.36+0.64=c2
1=c2
1=c

Question 2.
Determine the length of side a in each of the triangles below.
a. a2+82=172
a2+64=289
a2+64-64=289-64
a2=225
a=15

b. a2+0.82=1.72
a2+0.64=2.89
a2+0.64-0.64=2.89-0.64
a2=2.25
a=1.5

Question 3.
Determine the length of side b in each of the triangles below.
a. 202+b2=252
400+b2=625
400-400+b2=625-400
b2=225
b=15

b. 22+b2=2.52
4+b2=6.25
4-4+b2=6.25-4
b2=2.25
b=1.5

Question 4.
Determine the length of side a in each of the triangles below.
a. a2+122=202
a2+144=400
a2+144-144=400-144
a2=256
a=16

b. 