## Engage NY Eureka Math Geometry Module 5 Lesson 16 Answer Key

### Eureka Math Geometry Module 5 Lesson 16 Exploratory Challenge Answer Key

Exploratory Challenge 1

Measure the lengths of the chords in centimeters, and record them in the table.

Answer:

Exploratory Challenge 2

Measure the lengths of the chords in centimeters, and record them in the table.

Answer:

### Eureka Math Geometry Module 5 Lesson 16 Exercise Answer Key

Opening Exercise

Identify the type of angle and the angle/arc relationship, and then find the measure of x.

a.

Answer:

x = 58; the inscribed angle is equal to half intercepted arc.

b.

Answer:

x = 86; angle formed by secants intersecting inside the circle is half the sum of arcs intercepted by angle and its vertical angle.

c.

Answer:

x = 18.5; angle formed by secants intersecting outside of the circle has a measure of half the difference of larger and smaller intercepted arcs.

d.

Answer:

x = 42.5; central angle has measure of intercepted arc.

Closing

The Inscribed Angle Theorem and Its Family

Answer:

### Eureka Math Geometry Module 5 Lesson 16 Problem Set Answer Key

Question 1.

Answer:

x = 8

Question 2.

Answer:

x(x + 1) = 2(2 + 4)

x^{2} + x – 12 = 0

(x + 4)(x – 3) = 0

x = 3

Question 3.

DF < FB, DF â‰ 1, DF < FE, and all values are integers; prove DF = 3.

Answer:

7 â‹… 6 = 42, so DF â‹… FE must equal 42. If DF < FE, DF could equal 1, 3, or 6. DF â‰ 1 and DF < FB, so DF must equal 3.

Question 4.

CE = 6, CB = 9, and CD = 18. Show CF = 3.

Answer:

6 â‹… 9 = 54 and 18 â‹… CF = 54. This means CF = 3.

Question 5.

Find x.

Answer:

x = 2\(\sqrt{13}\)

Question 6.

Find x.

Answer:

x = 11.25

Question 7.

Find x.

Answer:

(x – 27)x = 8(20)

x = 32

Question 8.

Find x.

Answer:

x(x + 7) = (x + 3)^{2}

x = 9

Question 9.

In the circle shown, DE = 11, BC = 10, and DF = 8. Find FE, BF, FC.

Answer:

x(10 – x) = (3)(8)

FE = 3, BF = 4, FC = 6

Question 10.

In the circle shown, \(m\widehat{D B G}\) = 150Â°, \(m\widehat{D B}\) = 30Â°, mâˆ CEF = 60Â°, DF = 8, DB = 4, and GF = 12.

a. Find mâˆ GDB.

Answer:

60Â°

b. Prove â–³DBF ~ â–³ECF

Answer:

mâˆ DBF = mâˆ CEF Inscribed angle âˆ DBF is half the measure of intercepted arc, \(\widehat{D C}\) and âˆ CEF formed by a tangent line and a secant line is also half the measure of the same intercepted arc \(\widehat{D C}\).

mâˆ DFB = mâˆ EFC Vertical angles are equal in measure.

â–³DBF ~ â–³ECF AA

c. Set up a proportion using \(\overline{C E}\) and \(\overline{G E}\).

Answer:

\(\frac{8}{GE + 12}\) = \(\frac{4}{CE}\) , or 2CE = GE + 12

d. Set up an equation with CE and GE using a theorem for segment lengths from this section.

Answer:

CE^{2} = GE(GE + 20)

### Eureka Math Geometry Module 5 Lesson 16 Exit Ticket Answer Key

Question 1.

In the circle below, \(m \widehat{G F}\) = 30Â°, \(m \widehat{D E}\) = 120Â°, CG = 6, GH = 2, FH = 3, CF = 4, HE = 9, and FE = 12.

a. Find a (mâˆ DHE).

Answer:

a = 75Â°

b. Find b (mâˆ DCE), and explain your answer.

Answer:

b = 45Â°; b is an angle with its vertex outside of the circle, so it has a measure half the difference between its larger and smaller intercepted arcs.

c. Find x (HD), and explain your answer.

Answer:

x = 6; x is part of a secant line intersecting another secant line inside the circle, so 2âˆ™9 = 3âˆ™x.

d. Find y (\(\overline{D G}\)).

Answer:

y = \(\frac{14}{3}\) = 4 \(\frac{2}{3}\)