# Eureka Math Geometry Module 4 Lesson 13 Answer Key

## Engage NY Eureka Math Geometry Module 4 Lesson 13 Answer Key

### Eureka Math Geometry Module 4 Lesson 13 Opening Exercise Answer Key

Let A(30, 40), B(60, 50), and C(75, 120) be vertices of a triangle.
a. Find the coordinates of the midpoint M of $$\overline{A B}$$ and the point G1 that is the point one-third of the way along $$\overline{M C}$$, closer to M than to C. M = (30 + $$\frac{1}{2}$$(60 – 30), 40 + $$\frac{1}{2}$$(50 – 40)) = (45, 45) or M = ($$\frac{1}{2}$$(30 + 60), $$\frac{1}{2}$$(40 + 50))= (45, 45)
G1: (45 + $$\frac{1}{3}$$(75 – 45), 45 + $$\frac{1}{3}$$(12o – 45)) = (55, 70) or ($$\frac{1}{3}$$(45 + 45 + 75), $$\frac{1}{3}$$(45 + 45 + 120)) = (55, 70)

b. Find the coordinates of the midpoint N of $$\overline{B C}$$? and the point G2 that is the point one-third of the way along $$\overline{N A}$$, closer to N than to A. N = 60 + $$\frac{1}{2}$$(75 – 60),50 + $$\frac{1}{2}$$(120 – 50)) = (67.5,85) or N = ($$\frac{1}{2}$$(60 + 75), $$\frac{1}{2}$$(50 + 120)) = (67.5,85)
G2; (67.5 + $$\frac{1}{3}$$(30 – 67.5), 85 + $$\frac{1}{3}$$ (40 – 85)) = (55,70) or ($$\frac{1}{3}$$(30 + 67.5 + 67. 5), $$\frac{1}{3}$$(40 + 85 + 85)) = (55,70)

c. Find the coordinates of the midpoint R of $$\overline{C A}$$ and the point G3 that is the point one-third of the way along $$\overline{R B}$$, closer to R than to B. R = (30 + $$\frac{1}{2}$$(75 – 30),40 + $$\frac{1}{2}$$(120 – 40)) = (52.5, 80) or R = ($$\frac{1}{2}$$(75 + 30), $$\frac{1}{2}$$(120 + 40)) = (52. 5, 80)
G3: (52.5 + $$\frac{1}{3}$$(60 – 52.5), 80 + $$\frac{1}{3}$$(50 – 80)) = (55.70) or ($$\frac{1}{3}$$(52.5 + 52.5 + 60), $$\frac{1}{3}$$(50 + 80 + 80)) = (55, 70)

### Eureka Math Geometry Module 4 Lesson 13 Exercise Answer Key

Exercise 1.
a. Given triangle ABC with vertices A(a1, a2), B(b1, b2), and C(c1, c2), find the coordinates of the point of concurrency of the medians.   b. Let A(- 23, 12), B(13, 36), and C(23, – 1) be vertices of a triangle. Where will the medians of this triangle intersect?
($$\frac{1}{3}$$(-23) + $$\frac{1}{3}$$(13) + $$\frac{1}{3}$$(23), $$\frac{1}{3}$$(12) + $$\frac{1}{3}$$(36) + $$\frac{1}{3}$$(-1)) or ($$\frac{1}{3}$$(- 23 + 13 + 23), $$\frac{1}{3}$$(12 + 36 + – 1)) = $$\left(\frac{13}{3}, \frac{47}{3}\right)$$

Exercise 2.
Prove that the diagonals of a parallelogram bisect each other. Students will show that the diagonals are concurrent at theIr midpoints. Stated another way, both diagonals have the same midpoint.
Midpoint of $$\overline{P R}$$: ($$\frac{1}{2}$$(b + a), $$\frac{1}{2}$$h)
Midpoint of $$\overline{Q S}$$: ($$\frac{1}{2}$$(b + a), $$\frac{1}{2}$$h)

### Eureka Math Geometry Module 4 Lesson 13 Problem Set Answer Key

Question 1.
Point M is the midpoint of $$\overline{A C}$$?. Find the coordinates of M:
a. A(2, 3), C(6, 10)
(4, 6.5)

b. A(- 7, 5), C(4, – 9)
(-1. 5, -2)

Question 2.
M(-2, 10) is the midpoint of $$\overline{A B}$$. If A has coordinates (4, -5), what are the coordinates of B?
(-8, 25)

Question 3.
Line A is the perpendicular bisector of $$\overline{B C}$$ with B(-2, -1) and C(4, 1).
a. What is the midpoint of $$\overline{B C}$$?
(1, 0)

b. What is the slope of $$\overline{B C}$$?
$$\frac{1}{3}$$

c. What is the slope of line A? (Remember, it is perpendicular to $$\overline{B C}$$.)
– 3

d. Write the equation of line A, the perpendicular bisector of $$\overline{B C}$$.
y = – 3x + 3

Question 4.
Find the coordinates of the intersection of the medians of ∆ ABC given A(-5, 3), 8(6, -4), and C(10, 10).
(($$\frac{1}{3}$$(-5 + 6 + 10), $$\frac{1}{3}$$(3 + (-4) + 10)) = (3$$\frac{2}{3}$$, 3)

Question 5.
Use coordinates to prove that the diagonals of a parallelogram meet at the intersection of the segments that connect the midpoints of its opposite sides. Question 6.
Given a quadrilateral with vertices E(0, 5), F(6, 5), G(4, 0), and H(- 2, 0):
a. Prove quadrilateral EFGH is a parallelogram.
$$\overline{E F}$$ and $$\overline{G H}$$ are horizontal segments, so they are parallel.
$$\overline{H E}$$ and $$\overline{G F}$$ have slopes of $$\frac{5}{2}$$, they are parallel.
Both pairs of opposite sides are parallel, so the quadrilateral is a parallelogram.

b. Prove (2, 2. 5) is a point on both diagonals of the quadrilateral.
Since EFGH is a parallelogram, the diagonals intersect at their midpoints. (2,2. 5)is the midpoint of $$\overline{H F}$$ and $$\overline{G E}$$, so it is a point on both diagonals.

Question 7.
Prove quadrilateral WXYZ with vertices W(1, 3), X(4, 8), Y(10, 11), and Z(4, 1) is a trapezoid.
$$\overline{W X}$$ and $$\overline{Y Z}$$ have slopes of $$\frac{5}{3}$$, so they are parallel.
$$\overline{W Z}$$ has a slope of –$$\frac{2}{3}$$ and $$\overline{X Y}$$ has a slope of $$\frac{1}{2}$$, so they are not parallel.
When one pair of opposite sides is parallel, the quadrilateral is a trapezoid.

Question 8.
Given quadrilateral JKLM with vertices j(-4, 2), K(1, 5), L(4, 0), and M(- 1, – 3):
a. Is it a trapezoid? Explain.
Yes, one pair of opposite sides is parallel $$\overline{J K}$$ and $$\overline{L M}$$ both have slopes of $$\frac{3}{5}$$.
When one pair of opposite sides is parallel, the quadrilateral is a trapezoid.

b. Is it a parallelogram? Explain.
Yes, both pairs of opposite sides are parallel. $$\overline{J M}$$ and $$\overline{K L}$$ both have slopes of –$$\frac{5}{3}$$.
When both pairs of opposite sides are parallel, the quadrilateral is a parallelogram.

c. Is it a rectangle? Explain. d. Is it a rhombus? Explain.
JK = KL = LM = MJ = √34
Yes, because a parallelogram with four congruent sides is a rhombus.

e. Is it a square? Explain.
Yes, because a rectangle with four congruent sides is a square.

f. Name a point on the diagonal of JKLM. Explain how you know.
(0, 1) is the midpoint of $$\overline{K M}$$ and $$\overline{J L}$$ and is on both diagonals.

### Eureka Math Geometry Module 4 Lesson 13 Exit Ticket Answer Key

Prove that the medians of any right triangle form similar right triangle whose area is $$\frac{1}{4}$$ the area of the original triangle. Prove the area of ∆ RMS is $$\frac{1}{4}$$ the area of ∆ CAB. M = (o, $$\frac{1}{2}$$ c), R = ($$\frac{1}{2}$$b, $$\frac{1}{2}$$c), and s = ($$\frac{1}{2}$$b, 0).
$$\frac{R M}{A B}=\frac{\frac{1}{2} b}{b}=\frac{1}{2}$$
$$\overline{A C}$$ and $$\overline{R S}$$ are both vertical as their slopes are undefined.
$$\overline{A B}$$ and $$\overline{R M}$$ are both horizontal as their slopes are zero.
∆ ABC ~ ∆ RMS by SAS similarity (∠CAB and ∠SRM are both right angles, and ratio of the lengths of segments RS to AC is $$\frac{1}{2}$$).
The area 0f ∆ABC is $$\frac{1}{2}$$b c.
The area of ∆ RMS is $$\frac{1}{2}$$($$\frac{1}{2}$$ b) ∙ ($$\frac{1}{2}$$ c) or $$\frac{1}{4}$$($$\frac{1}{2}$$ b ∙ c) or $$\frac{1}{8}$$ of the area of ∆ ABC.