## Engage NY Eureka Math Geometry Module 5 Lesson 2 Answer Key

### Eureka Math Geometry Module 5 Lesson 2 Exercise Answer Key

Opening Exercise

Construct the perpendicular bisector of \(\overline{A B}\) below (as you did in Module 1).

Draw another line that bisects \(\overline{A B}\) but is not perpendicular to it.

List one similarity and one difference between the two bisectors.

Answer:

Answers will vary. Both bisectors divide the segment into two shorter segments of equal length. All points on the perpendicular bisector are equidistant from points A and B. Points on the other bisector are not equidistant from points A and B. The perpendicular bisector meets \(\overline{A B}\) at right angles. The other bisector meets at angles that are not congruent.

Exercises

Exercise 1.

Prove the theorem: If a diameter of a circle bisects a chord, then it must be perpendicular to the chord.

Answer:

Draw a diagram similar to that shown below.

Given: Circle C with diameter \(\overline{D E}\), chord \(\overline{A B}\), and AF = BF

Prove: \(\overline{D E}\)âŠ¥\(\overline{A B}\)

AF = BFÂ Â Â Â Given

FC = FCÂ Â Â Â Reflexive property

AC = BCÂ Â Â Â Radii of the same circle are equal in measure.

â–³AFC â‰… â–³BFCÂ Â Â SSS

mâˆ AFC = mâˆ BFCÂ Â Â Corresponding angles of congruent triangles are equal in measure.

âˆ AFC and âˆ BFC are right angles Equal angles that form a linear pair each measure 90Â°.

\(\overline{D E}\) âŠ¥\(\overline{A B}\) Definition of perpendicular lines

OR

AF = BFÂ Â Given

AC = BCÂ Â Â Â Radii of the same circle are equal in measure.

mâˆ FAC = mâˆ FBCÂ Â Â Base angles of an isosceles triangle are equal in measure.

â–³AFC â‰… â–³BFCÂ Â Â SAS

mâˆ AFC = mâˆ BFCÂ Â Â Â Corresponding angles of congruent triangles are equal in measure.

âˆ AFC and âˆ BFC are right angles Equal angles that form a linear pair each measure 90Â°.

\(\overline{D E}\) âŠ¥ \(\overline{A B}\) Definition of perpendicular lines

Exercise 2.

Prove the theorem: If a diameter of a circle is perpendicular to a chord, then it bisects the chord.

Answer:

Use a diagram similar to that in Exercise 1.

Given: Circle C with diameter \(\overline{D E}\), chord \(\overline{A B}\), and \(\overline{D E}\)âŠ¥\(\overline{A B}\)

Prove: \(\overline{D E}\) bisects \(\overline{A B}\)

\(\overline{D E}\)âŠ¥\(\overline{A B}\) Given

âˆ AFC and âˆ BFC are right angles Definition of perpendicular lines

â–³AFC and â–³BFC are right triangles Definition of right triangle

âˆ AFC â‰…âˆ BFC All right angles are congruent.

FC = FC Reflexive property

AC = BC Radii of the same circle are equal in measure.

â–³AFC â‰… â–³BFC HL

AF = BF Corresponding sides of congruent triangles are equal in length.

\(\overline{D E}\) bisects \(\overline{A B}\) Definition of segment bisector

OR

\(\overline{D E}\) âŠ¥\(\overline{A B}\)Given

âˆ AFC and âˆ BFC are right angles Definition of perpendicular lines

âˆ AFC â‰…âˆ BFC All right angles are congruent.

AC = BC Radii of the same circle are equal in measure.

mâˆ FAC = mâˆ FBC Base angles of an isosceles triangle are congruent.

mâˆ ACF = mâˆ BCF Two angles of triangle are equal in measure, so third angles are equal.

â–³AFC â‰… â–³BFC ASA

AF = BF Corresponding sides of congruent triangles are equal in length.

\(\overline{D E}\) bisects \(\overline{A B}\) Definition of segment bisector

Exercise 3.

The distance from the center of a circle to a chord is defined as the length of the perpendicular segment from the center to the chord. Note that since this perpendicular segment may be extended to create a diameter of the circle, the segment also bisects the chord, as proved in Exercise 2.

Prove the theorem: In a circle, if two chords are congruent, then the center is equidistant from the two chords.

Use the diagram below.

Answer:

Given: Circle O with chords \(\overline{A B}\) and \(\overline{C D}\); AB = CD; F is the midpoint of \(\overline{A B}\)and E is the midpoint of \(\overline{C D}\).

Prove: OF = OE

AB = CD Given

\(\overline{O F}\)and \(\overline{O E}\) are portions of diameters Definition of diameter

\(\overline{O F}\)âŠ¥\(\overline{A B}\); \(\overline{O E}\)âŠ¥\(\overline{C D}\) If a diameter of a circle bisects a chord, then the diameter must be perpendicular to the chord.

âˆ AFO and âˆ DEO are right angles Definition of perpendicular lines

â–³AFO and â–³DEO are right triangles Definition of right triangle

E and F are midpoints of \(\overline{C D}\) and \(\overline{A B}\) Given

AF = DE AB = CD and F and E are midpoints of \(\overline{A B}\)and \(\overline{C D}\).

AO = DO All radii of a circle are equal in measure.

â–³AFO â‰… â–³DEO HL

OE = OF Corresponding sides of congruent triangles are equal in length.

Exercise 4.

Prove the theorem: In a circle, if the center is equidistant from two chords, then the two chords are congruent.

Use the diagram below.

Answer:

Given: Circle O with chords \(\overline{A B}\)and \(\overline{C D}\); OF = OE; F is the midpoint of \(\overline{A B}\)and E is the midpoint of \(\overline{C D}\).

Prove: AB = CD

OF = OE Given

\(\overline{O F}\) and \(\overline{O E}\) are portions of diameters Definition of diameter

\(\overline{O F}\) âŠ¥\(\overline{A B}\), \(\overline{O E}\)âŠ¥\(\overline{C D}\) If a diameter of a circle bisects a chord, then it must be perpendicular to the chord.

âˆ AFO and âˆ DEO are right angles Definition of perpendicular lines

â–³AFO and â–³DEO are right triangles Definition of right triangle

AO = DO All radii of a circle are equal in measure.

â–³AFO â‰… â–³DEO HL

AF = DE Corresponding sides of congruent triangles are equal in length.

F is the midpoint of \(\overline{A B}\), and E is the

midpoint of \(\overline{C D}\). Given

AB = CD AF = DE and F and E are midpoints of \(\overline{A B}\) and \(\overline{C D}\).

Exercise 5.

A central angle defined by a chord is an angle whose vertex is the center of the circle and whose rays intersect the circle. The points at which the angleâ€™s rays intersect the circle form the endpoints of the chord defined by the central angle.

Prove the theorem: In a circle, congruent chords define central angles equal in measure.

Use the diagram below.

Answer:

We are given that the two chords (\(\overline{A B}\) and \(\overline{C D}\)) are congruent. Since all radii of a circle are congruent, \(\overline{A O}\) â‰… \(\overline{B O}\) â‰… \(\overline{C O}\) â‰… \(\overline{D O}\). Therefore, â–³ABO â‰… â–³DCO by SSS. âˆ AOB â‰… âˆ DOC since corresponding angles of congruent triangles are equal in measure.

Exercise 6.

Prove the theorem: In a circle, if two chords define central angles equal in measure, then they are congruent.

Answer:

Using the diagram from Exercise 5, we now are given that mâˆ AOB = mâˆ COD. Since all radii of a circle are congruent, \(\overline{A O}\) â‰… \(\overline{B O}\) â‰… \(\overline{C O}\) â‰… \(\overline{D O}\). Therefore, â–³ABOâ‰… â–³DCO by SAS. \(\overline{A B}\) â‰… \(\overline{D C}\) because corresponding sides of congruent triangles are congruent.

### Eureka Math Geometry Module 5 Lesson 2 Problem Set Answer Key

Question 1.

In this drawing, AB = 30, OM = 20, and ON = 18. What is CN?

Answer:

\(\sqrt{301}\) â‰ˆ 17.35

Question 2.

In the figure to the right, \(\overline{A C}\)âŠ¥\(\overline{B G}\), \(\overline{D F}\)âŠ¥\(\overline{E G}\), and EF = 12. Find AC.

Answer:

24

Question 3.

In the figure, AC = 24, and DG = 13. Find EG. Explain your work.

Answer:

5, â–³ABG is a right triangle with hypotenuse = radius = 13 and AB = 12, so BG = 5 by Pythagorean theorem. BG = GE = 5.

Question 4.

In the figure, AB = 10, and AC = 16. Find DE.

Answer:

4

Question 5.

In the figure, CF = 8, and the two concentric circles have radii of 10 and 17. Find DE.

Answer:

9

Question 6.

In the figure, the two circles have equal radii and intersect at points B and D. A and C are centers of the circles. AC = 8, and the radius of each circle is 5. \(\overline{B D}\) âŠ¥ \(\overline{A C}\). Find BD. Explain your work.

Answer:

6

BA = BC = 5 (radii)

AG = GC = 4

BG = 3 (Pythagorean theorem)

BD = 6

Question 7.

In the figure, the two concentric circles have radii of 6 and 14. Chord (BF) Ì… of the larger circle intersects the smaller circle at C and E. CE = 8. \(\overline{A D}\)âŠ¥\(\overline{B F}\).

a. Find AD.

2\(\sqrt{5}\)

b. Find BF.

Answer:

8\(\sqrt{11}\)

Question 8.

In the figure, A is the center of the circle, and CB = CD. Prove that \(\overline{A C}\) bisects âˆ BCD.

Answer:

Let \(\overline{A E}\) and \(\overline{A F}\) be perpendiculars from A to \(\overline{C B}\) and \(\overline{C D}\), respectively.

CB = CD Given

AE = AF If two chords are congruent, then the center is equidistant from the two chords.

AC = AC Reflexive property

mâˆ CEA = mâˆ CFA = 90Â° Definition of perpendicular

â–³CEA â‰… â–³CFA HL

mâˆ ECA = mâˆ FCA Corresponding angles of congruent triangles are equal in measure.

\(\overline{A C}\) bisects âˆ BCD Definition of angle bisector

Question 9.

In class, we proved: Congruent chords define central angles equal in measure.

a. Give another proof of this theorem based on the properties of rotations. Use the figure from Exercise 5.

Answer:

We are given that the two chords (\(\overline{A B}\) and \(\overline{C D}\)) are congruent. Therefore, a rigid motion exists that carries \(\overline{A B}\) to \(\overline{C D}\). The same rotation that carries \(\overline{A B}\) to \(\overline{C D}\) also carries \(\overline{A O}\) to \(\overline{C O}\) and \(\overline{B O}\) to \(\overline{D O}\). The angle of rotation is the measure of âˆ AOC, and the rotation is clockwise.

b. Give a rotation proof of the converse: If two chords define central angles of the same measure, then they must be congruent.

Answer:

Using the same diagram, we are given that âˆ AOB â‰…âˆ COD. Therefore, a rigid motion (a rotation) carries âˆ AOB to âˆ COD. This same rotation carries \(\overline{A O}\) to \(\overline{C O}\) and \(\overline{B O}\) to \(\overline{D O}\). The angle of rotation is the measure of âˆ AOC, and the rotation is clockwise.

### Eureka Math Geometry Module 5 Lesson 2 Exit Ticket Answer Key

Question 1.

Given circle A shown, AF = AG and BC = 22. Find DE.

Answer:

22

Question 2.

In the figure, circle P has a radius of 10. \(\overline{A B}\)âŠ¥\(\overline{D E}\).

a. If AB = 8, what is the length of \(\overline{A C\)?

Answer:

4

b. If DC = 2, what is the length of \(\overline{A B}\)?

Answer:

12