# Eureka Math Geometry Module 1 Lesson 34 Answer Key

## Engage NY Eureka Math Geometry Module 1 Lesson 34 Answer Key

Review Exercise:

### Eureka Math Geometry Module 1 Lesson 34 Problem Set Answer Key

Use any of the assumptions, facts, and/or properties presented in the tables above to find x and/or y in each figure below. Justify your solutions.

Question 1.
Find the perimeter of parallelogram ABCD. Justify your solution.

100, 15 = x + 4, x = 11

Question 2.
AC = 34
AB = 26
BD = 28
Given parallelogram ABCD, find the perimeter of CED. Justify your solution.

57
CE = $$\frac{1}{2}$$AC; CE = 17
CD = AB; CE = 26
ED = $$\frac{1}{2}$$BD; ED = 14
Perimeter = 17 + 26 + 14 = 57

Question 3.
XY = 12
XZ = 20
ZY = 24
F, G, and H are midpoints of the sides on which they are located. Find the perimeter of FGH. Justify your solution.

28
The midsegment is half the length of the side of the triangle it is parallel to.

Question 4.
ABCD is a parallelogram with AE = CF. Prove that DEBF is a parallelogram.

AE = CF – Given
AD = BC – Property of a parallelogram
m∠DAE = m∠BCF – If parallel lines are cut by a transversal, then alternate interior angles are equal in measure.
DE = BF – Corresponding sides of congruent triangles are congruent.
AB = DC – Property of a parallelogram
m∠BAE = m∠DCF – If parallel lines are cut by a transversal, then alternate interior angles are equal in measure.
∆BAE ≅ ∆DCF – SAS
BE = DF – Corresponding sides of congruent triangles are congruent.
∴ ABCD is a parallelogram. – If both sets of opposite sides of a quadrilateral are equal in length, then the quadrilateral is a parallelogram.

Question 5.
C is the centroid of RST. RC = 16. CL = 10. TJ = 21

SC = __
TC = __
KC = __