## 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.

Answer:

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’.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

\(\frac{12.6}{3}\) = \(\frac{\left|O X^{\prime}\right|}{|5|}\)

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.

Answer:

\(\frac{12.6}{3}\) = \(\frac{\left|OY^{\prime}\right|}{|7|}\)

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’.

Answer:

\(\frac{B^{\prime} C^{\prime}}{B C}\) = \(\frac{2}{1}\) = 2 = r

Le D be the dilation from center O and scale factor r=2. Let there be a reflection across the Y-axis. Then the dilation followed by the reflection maps △ABC onto △A’B’C’.

b. Describe the sequence that shows the similarity for △ABC and △A”B”C”.

Answer:

Let D be the dilation from center O and scale factor. 0<r<1. Let there be a rotation of 180° around center O. Then the dilation followed by the rotation maps △ABC onto △A”B”C”.

c. Is △A’B’C’ similar to △A”B”C”? How do you know?

Answer:

Yes △A’B’C’ ~ △A”B”C”. Dilations preserve angle measures and since △ABC ~ △A’B’C and △ABC ~ △A”B”C”, we know ∠A= ∠A’ = ∠A”, ∠B = ∠B’=∠B”, by AA Criterion for similarity △A’B’C’ ~ △ A”B”C”. Also, similarity is Transitive.