Eureka Math Geometry Module 1 Lesson 5 Answer Key

Engage NY Eureka Math Geometry Module 1 Lesson 5 Answer Key

Eureka Math Geometry Module 1 Lesson 5 Exercise Answer Key

Opening Exercise
You need a makeshift compass made from string and a pencil.
Use these materials to construct the perpendicular bisectors of the three sides of the triangle below (like you did with Lesson 4, Problem Set 2).
Eureka Math Geometry Module 1 Lesson 5 Exercise Answer Key 1
How did using this tool differ from using a compass and straightedge? Compare your construction with that of your partner. Did you obtain the same results?

Exploratory Challenge
When three or more lines intersect in a single point, they are ___, and the point of intersection is the ___.
Answer:
When three or more lines intersect in a single point, they are concurrent, and the point of intersection is the point of concurrency.

You saw an example of a point of concurrency in yesterday’s Problem Set (and in the Opening Exercise today) when all three perpendicular bisectors passed through a common point.

The point of concurrency of the three perpendicular bisectors is the ____.
Answer:
The point of concurrency of the three perpendicular bisectors is the circumcenter of the triangle.

Have students mark the right angles and congruent segments (defined by midpoints) on the triangle.

The circumcenter of △ABC is shown below as point P.
Eureka Math Geometry Module 1 Lesson 5 Exercise Answer Key 2

The questions that arise here are WHY are the three perpendicular bisectors concurrent? And WILL these bisectors be concurrent in all triangles? Recall that all points on the perpendicular bisector are equidistant from the endpoints of the segment, which means the following:
a. P is equidistant from A and B since it lies on the ___ of \(\overline{A B}\).
Answer:
P is equidistant from A and B since it lies on the perpendicular bisector of \(\overline{A B}\)

b. P is also __ from B and C since it lies on the perpendicular bisector of \(\overline{B C}\).
Answer:
P is also equidistant from B and C since it lies on the perpendicular bisector of \(\overline{B C}\)

c. Therefore, P must also be equidistant from A and C.
Hence, AP=BP=CP, which suggests that P is the point of ___ of all three perpendicular bisectors.
You have also worked with angle bisectors. The construction of the three angle bisectors of a triangle also results in a point of concurrency, which we call the ___.
Answer:
Hence, AP=BP=CP, which suggests that P is the point of concurrency of all three perpendicular bisectors.
You have also worked with angle bisectors. The construction of the three angle bisectors of a triangle also results in a point of concurrency, which we call the incenter.
Use the triangle below to construct the angle bisectors of each angle in the triangle to locate the triangle’s incenter.
Eureka Math Geometry Module 1 Lesson 5 Exercise Answer Key 3

Have students label the congruent angles formed by the angle bisectors.

d. State precisely the steps in your construction above.
Answer:
Construct the angle bisectors of ∠A, ∠B, and ∠C.
Label the point of intersection Q.

e. Earlier in this lesson, we explained why the perpendicular bisectors of the sides of a triangle are always concurrent. Using similar reasoning, explain clearly why the angle bisectors are always concurrent at the incenter of a triangle.
Answer:
Any point on the angle bisector is equidistant from the rays forming the angle. Therefore, since point Q is on the angle bisector of ∠ABC, it is equidistant from \(\overrightarrow{B A}\) and \(\overrightarrow{B C}\). Similarly, since point Q is on the angle bisector of ∠BCA, it is equidistant from \(\overrightarrow{C B}\) and \(\overrightarrow{C A}\). Therefore, Q must also be equidistant from \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\), since it lies on the angle bisector of ∠BAC. So Q is a point of concurrency of all three angle bisectors.

f. Observe the constructions below. Point A is the ___ of △JKL. (Notice that it can fall outside of the triangle.) Point B is the ___ of △RST. The circumcenter of a triangle is the center of the circle that circumscribes that triangle. The incenter of the triangle is the center of the circle that is inscribed in that triangle.
Answer:
Observe the constructions below. Point A is the circumcenter of △JKL. (Notice that it can fall outside of the triangle.) Point B is the incenter of △RST. The circumcenter of a triangle is the center of the circle that circumscribes that triangle. The incenter of the triangle is the center of the circle that is inscribed in that triangle.
Eureka Math Geometry Module 1 Lesson 5 Exercise Answer Key 4

On a separate piece of paper, draw two triangles of your own below and demonstrate how the circumcenter and incenter have these special relationships.
Answer:
Answers will vary.

g. How can you use what you have learned in Exercise 3 to find the center of a circle if the center is not shown?
Answer:
Inscribe a triangle into the circle and construct the perpendicular bisectors of at least two sides. Where the bisectors intersect is the center of the circle.

Eureka Math Geometry Module 1 Lesson 5 Problem Set Answer Key

Question 1.
Given line segment AB, using a compass and straightedge, construct the set of points that are equidistant from A and B.
Eureka Math Geometry Module 1 Lesson 5 Problem Set Answer Key 4.1
Answer:
Eureka Math Geometry Module 1 Lesson 5 Problem Set Answer Key 5
What figure did you end up constructing? Explain.
Answer:
I ended up drawing the perpendicular bisector of the segment AB. Every point on this line is equidistant from the points A and B.

Question 2.
For each of the following, construct a line perpendicular to segment AB that goes through point P.
Eureka Math Geometry Module 1 Lesson 5 Problem Set Answer Key 5.1
Answer:
Eureka Math Geometry Module 1 Lesson 5 Problem Set Answer Key 6

Question 3.
Using a compass and straightedge, construct the angle bisector of ∠ABC shown below. What is true about every point that lies on the ray you created?
Eureka Math Geometry Module 1 Lesson 5 Problem Set Answer Key 7
Answer:
Eureka Math Geometry Module 1 Lesson 5 Problem Set Answer Key 7.1
Every point on the ray is equidistant from ray BA and ray BC.

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