# Eureka Math Grade 8 Module 3 End of Module Assessment Answer Key

## Engage NY Eureka Math 8th Grade Module 3 End of Module Assessment Answer Key

### Eureka Math Grade 8 Module 3 End of Module Assessment Task Answer Key

Question 1.
Use the diagram below to answer the questions that follow.

a. Dilate △OPQ from center O and scale factor r=$$\frac{4}{9}$$. Label the image △OP’Q’.

b. Find the coordinates of points P’ and Q’.
P’ = (6, 2)
Q’ = (6, $$\frac{38}{9}$$)
$$\frac{\left(P^{\prime} Q^{\prime}\right)}{|P Q|}$$ = $$\frac{4}{9}$$
$$\frac{\left|P^{\prime} Q^{\prime}\right|}{5}$$ = $$\frac{4}{9}$$
|P’Q’| = $$\frac{20}{9}$$
$$\frac{20}{9}$$ + 2 = $$\frac{20}{9}$$ + $$\frac{18}{9}$$
= $$\frac{38}{9}$$

c. Are ∠OQP and ∠OQ’P’ equal in measure? Explain.
Yes ∠OQP = ∠OQ’P’ Since D(△OQP) = △OQ’P’ and dilations are degree preserving, then ∠OQP = ∠OQ’P’.
∠OQP & ∠OQ’P’ are corresponding angles of parallel lines PQ & P’Q’, therefore ∠OQR = ∠OQ’P’

d. What is the relationship between segments PQ and P’Q’? Explain in terms of similar triangles.
The lines that contain $$\overline{\text { PQ }}$$ and $$\overline{p^{\prime} Q^{\prime}}$$ are parallel. ∆OPQ ~ ∆OP’Q’ by the AA criterion (∠D = ∠D, ∠OP’Q’ = ∠OPQ), Therefore by the fundamental theorem of similarity $$\overline{P Q}$$ || $$\overline{P^{\prime} Q^{\prime}}$$

e. If the length of segment OQ is 9.8 units, what is the length of segment OQ’? Explain in terms of similar triangles.
Since ∆OPQ ~ ∆OP’Q’, then the ratios of lengths of corresponding sides will be equal to the scale factor then
$$\frac{\left|O P^{\prime}\right|}{|O P|}$$ = $$\frac{\left|O Q^{\prime}\right|}{|O Q|}$$ = $$\frac{4}{9}$$
$$\frac{4}{9}$$ = $$\frac{\left|O Q^{\prime}\right|}{|9.0|}$$
39.2 = 9(|OQ’|)
4.36 = |OQ’|
|OQ’| is approximately 4.4 units.

Question 2.
Use the diagram below to answer the questions that follow. The length of each segment is as follows: segment OX is 5 units, segment OY is 7 units, segment XY is 3 units, and segment X’Y’ is 12.6 units.

a. Suppose segment XY is parallel to segment X’Y’. Is △OXY similar to △OX’Y’? Explain.
Yes, ∆OXY ~ ∆OX’Y’. Since $$\overline{X Y}$$ || $$X^{\prime} Y^{\prime}$$ and ∠OYX = ∠OY’X’. Because corresponding angles of parallel lines are equal in measure, by AA ∆OXY ~ ∆OX’Y’.

b. What is the length of segment OX’? Show your work.
$$\frac{12.6}{3}$$ = $$\frac{\left|O X^{\prime}\right|}{|5|}$$
5(12.6) = 3(|OX’|)
63 = 3(|OX’|)
21 = |OX’|
|OX’| is 21 units.

c. What is the length of segment OY’? Show your work.
$$\frac{12.6}{3}$$ = $$\frac{\left|OY^{\prime}\right|}{|7|}$$
12.6(7) = (3|OY’|)
88.2 = 3(|OY’|)
29.4 = |OY’|
|OY’| is 29.4 units.

Question 3.
Given △ABC ~△A^’ B^’ C’ and △ABC ~△A”B”C” in the diagram below, answer parts (a)–(c).

a. Describe the sequence that shows the similarity for △ABC and △A’ B’ C’.
$$\frac{B^{\prime} C^{\prime}}{B C}$$ = $$\frac{2}{1}$$ = 2 = r