## 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 G_{1} that is the point one-third of the way along \(\overline{M C}\), closer to M than to C.

Answer:

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)

G_{1}: (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 G_{2} that is the point one-third of the way along \(\overline{N A}\), closer to N than to A.

Answer:

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)

G_{2}; (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 G_{3} that is the point one-third of the way along \(\overline{R B}\), closer to R than to B.

Answer:

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)

G_{3}: (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(a_{1}, a_{2}), B(b_{1}, b_{2}), and C(c_{1}, c_{2}), find the coordinates of the point of concurrency of the medians.

Answer:

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?

Answer:

(\(\frac{1}{3}\)(-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.

Answer:

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)

Answer:

(4, 6.5)

b. A(- 7, 5), C(4, – 9)

Answer:

(-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?

Answer:

(-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 \(\overline{B C}\)?

Answer:

(1, 0)

b. What is the slope of \(\overline{B C}\)?

Answer:

\(\frac{1}{3}\)

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

Answer:

– 3

d. Write the equation of line A, the perpendicular bisector of \(\overline{B C}\).

Answer:

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).

Answer:

((\(\frac{1}{3}\)(-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.

Answer:

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.

Answer:

\(\overline{E F}\) 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.

Answer:

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.

Answer:

\(\overline{W X}\) 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.

Answer:

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.

Answer:

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.

Answer:

d. Is it a rhombus? Explain.

Answer:

JK = KL = LM = MJ = √34

Yes, because a parallelogram with four congruent sides is a rhombus.

e. Is it a square? Explain.

Answer:

Yes, because a rectangle with four congruent sides is a square.

f. Name a point on the diagonal of JKLM. Explain how you know.

Answer:

(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.

Answer:

Placing the triangle on the coordinate plane, as shown to the right, ollows for the most efficient algebraic solution yielding midpoints

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.