# Eureka Math Geometry Module 5 Lesson 16 Answer Key

## 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.

Exploratory Challenge 2
Measure the lengths of the chords in centimeters, and record them in the table.

### 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.

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

b.

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

c.

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.

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

Closing
The Inscribed Angle Theorem and Its Family

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

Question 1.

x = 8

Question 2.

x(x + 1) = 2(2 + 4)
x2 + 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.

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.

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

Question 5.
Find x.

x = 2$$\sqrt{13}$$

Question 6.
Find x.

x = 11.25

Question 7.
Find x.

(x – 27)x = 8(20)
x = 32

Question 8.
Find x.

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.

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.
60Â°

b. Prove â–³DBF ~ â–³ECF
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 $$\overline{G E}$$.
$$\frac{8}{GE + 12}$$ = $$\frac{4}{CE}$$ , or 2CE = GE + 12

d. Set up an equation with CE and GE using a theorem for segment lengths from this section.
CE2 = 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).
a = 75Â°

d. Find y ($$\overline{D G}$$).
y = $$\frac{14}{3}$$ = 4 $$\frac{2}{3}$$