# Eureka Math Geometry Module 4 Lesson 5 Answer Key

## Engage NY Eureka Math Geometry Module 4 Lesson 5 Answer Key

### Eureka Math Geometry Module 4 Lesson 5 Example Answer Key

Example 2.

→ Refer to triangle ABO. What must be true about ABO if $$\overline{O A}$$ is perpendicular to $$\overline{O B}$$?
ABO would be a right triangle, so it must satisfy the Pythagorean theorem: OA2 + OB2 = AB2.
→ Determine the expressions that define OA, OB, and AB, respectively.
√(a12 + a22 ), √(b12 + b22 ) , and √((b1-a1)2 + (b2-a2)2 )

→ If $$\overline{O B}$$ ⊥ $$\overline{O B}$$, the distances will satisfy the Pythagorean theorem.
OA2 + OB2 = AB2, and by substituting in the expressions that represent the distances, we get:
($$\sqrt{a_{1}^{2}+a_{2}^{2}}$$)2 + ($$\sqrt{b_{1}{ }^{2}+b_{2}{ }^{2}}$$)2 = (√((b1-a1 )2 + (b2-a2 )2 ))2
a12 + a22 + b12 + b22 = (b1-a1 )2 + (b2-a2)2
a12 + a22 + b12 + b22 = b12-2a1 b1 + a12 + b22-2a2 b2 + a22
0 = -2a1 b1-2a2 b2
0 = a1 b1 + a2 b2.

→ We have just demonstrated that if two segments, $$\overline{O A}$$ and $$\overline{O B}$$ that have a common endpoint at the origin O(0, 0) and other endpoints of A(a1, a2) and B(b1, b2), are perpendicular, then a1 b1 + a2 b2 = 0.
→ Let’s revisit Example 1 and verify our formula with a triangle that we have already proven is right. Triangle OAB has vertices O(0, 0), A(6, 4), and B(-2, 3). Verify that it is right ($$\overline{O A}$$ ⊥ $$\overline{O B}$$) using the formula.
6∙(-2) + 4∙3 = 0; the segments are perpendicular.
→ In Exercise 1, we found that triangle OPQ was not right. Let’s see if our formula verifies that conclusion. Triangle OPQ has vertices O(0, 0), P(3, -1), and Q(2, 3).
3∙(2) + (-1)∙3 ≠ 0; the segments are not perpendicular.

### Eureka Math Geometry Module 4 Lesson 5 Exercise Answer Key

Opening Exercise
In right triangle ABC, find the missing side.

a. If AC = 9 and CB = 12, what is AB? Explain how you know.
Because triangle ABC is a right triangle, and we know the length of two of the three sides, we can use the Pythagorean theorem to find the length of the third side, which in this case is the hypotenuse.
AB2 = AC2 + CB2
AB = $$\sqrt{A C^{2} + C B^{2}}$$
AB = $$\sqrt{9^{2} + 12^{2}}$$
AB = 15

b. If AC = 5 and AB = 13, what is CB?
AB2 = AC2 + CB2
132 = 52 + CB2
CB = 12

c. If AC = CB and AB = 2, what is AC (and CB)?
AC2 + CB2 = AB2
AC2 + AC2 = AB2
2AC2 = 4
AC2 = 2
AC = $$\sqrt{2}$$
Since AC = CB, CB = $$\sqrt{2}$$ as well.

Exercise 1.
Use the grid on the right.

a. Plot points O(0, 0), P(3, -1), and Q(2, 3) on the coordinate plane.

b. Determine whether $$\overline{O P}$$ and $$\overline{O Q}$$ are perpendicular. Support your findings.
No, $$\overline{O A}$$ and $$\overline{O B}$$ are not perpendicular.
Using the distance formula, we determined that OP = $$\sqrt{10}$$, OQ = $$\sqrt{13}$$, and PQ = $$\sqrt{17}$$
($$\sqrt{10}$$)2 + ($$\sqrt{13}$$)2 ≠ ($$\sqrt{17}$$)2; therefore, triangle OPQ is not a right triangle, ∠QOP is not a right angle, and $$\overline{O P}$$ and $$\overline{O Q}$$are not perpendicular.

Exercises 2–3
Exercise 2.
Given points O(0, 0), A(6, 4), B(24, -6), C(1, 4), P(2, -3), S(-18, -12), T(-3, -12), U(-8, 2), and W(-6, 9), find all pairs of segments from the list below that are perpendicular. Support your answer.
$$\overline{O A}$$, $$\overline{O B}$$, $$\overline{O C}$$, $$\overline{O P}$$, $$\overline{O S}$$, $$\overline{O T}$$, $$\overline{O U}$$, and $$\overline{O W}$$
$$\overline{O A}$$ ⊥ $$\overline{O P}$$ because 6(2) + 4(-3) = 0.
$$\overline{O A}$$ ⊥ $$\overline{O W}$$ because 6(-6) + 4(9) = 0.
$$\overline{O S}$$ ⊥ $$\overline{O W}$$ because -18(-6) + (-12)(9) = 0.
$$\overline{O S}$$ ⊥ $$\overline{O P}$$ because -18(2) + (-12)(-3) = 0.
$$\overline{O B}$$ ⊥ $$\overline{O C}$$ because 24(1) + (-6)(4) = 0.
$$\overline{O B}$$ ⊥ $$\overline{O T}$$ because 24(-3) + (-6)(-12) = 0.
$$\overline{O T}$$⊥ $$\overline{O U}$$ because -3(-8) + (-12)(2) = 0.
$$\overline{O C}$$ ⊥ $$\overline{O U}$$ because 1(-8) + 4(2) = 0.

Exercise 3.
The points O(0, 0), A(-4, 1), B(-3, 5), and C(1, 4) are the vertices of parallelogram OABC. Is this parallelogram a rectangle? Support your answer.

We are given that the figure is a parallelogram with the properties that the opposite angles are congruent, and the adjacent angles are supplementary. To prove the quadrilateral is a rectangle, we only need to show that one of the four angles is a right angle.
$$\overline{O A}$$ ⊥ $$\overline{O C}$$ because -4(1) + 1(4) = 0. Therefore, parallelogram OABC is a rectangle.

### Eureka Math Geometry Module 4 Lesson 5 Problem Set Answer Key

Question 1.
Prove using the Pythagorean theorem that $$\overline{A C}$$ is perpendicular to $$\overline{A B}$$ given points A(-2, -2), B(5, -2), and C(-2, 22).
AC = 24, BC = 25, and AB = 7. If triangle ABC is right, AC2 + AB2 = BC2, and 576 + 49 = 625; therefore, the segments are perpendicular.

Question 2.
Using the general formula for perpendicularity of segments through the origin and (90, 0), determine if $$\overline{O A}$$ and $$\overline{O B}$$ are perpendicular.
a. A(-3, -4), B(4, 3)
(-3)(4) + (-4)(3) ≠ 0; therefore, the segments are not perpendicular.

b. A(8, 9), B(18, -16)
(8)(18) + (9)(-16) = 0; therefore, the segments are perpendicular.

Question 3.
Given points O(0, 0), S(2, 7), and T(7, -2), where $$\overline{O S}$$ is perpendicular to $$\overline{O T}$$, will the images of the segments be perpendicular if the three points O, S, and T are translated four units to the right and eight units up? Explain your answer.
O'(4, 8), S'(6, 15), T'(11, 6)
Yes, the points are all translated 4 units right and 8 units up; since the original segments were perpendicular, the translated segments are perpendicular.

Question 4.
In Example 1, we saw that $$\overline{O A}$$ was perpendicular to $$\overline{O B}$$ for O(0, 0), A(6, 4), and B(-2, 3). Suppose we are now given the points P(5, 5), Q(11, 9), and R(3, 8). Are segments $$\overline{P Q}$$ and $$\overline{P R}$$ perpendicular? Explain without using triangles or the Pythagorean theorem.
Yes, the segments are perpendicular. We proved in Example 1 that $$\overline{O A}$$ was perpendicular to $$\overline{O B}$$. P, Q, and R are translations of O, A, and B. The original coordinates have been translated 5 units right and 5 units up.

### Eureka Math Geometry Module 4 Lesson 5 Exit Ticket Answer Key

Question 1.
Given points O(0, 0), A(3, 1), and B(-2, 6), prove $$\overline{O A}$$ is perpendicular to $$\overline{O B}$$.

($$\sqrt{10}$$)2 + ($$\sqrt{40}$$)2 = ($$\sqrt{50}$$)2.
Given points P(1, -1), Q(-4, 4), and R(-2, -2), prove $$\overline{P R}$$ is perpendicular to $$\overline{Q R}$$ without the Pythagorean theorem.
The points given are translated points of the point given in Problem 1. Points A, B, and O have all been translated left 2 units and down 2 units. Since $$\overline{O A}$$ and $$\overline{O B}$$ are perpendicular, translating will not change the angle relationships, so perpendicularity is conserved.