# Eureka Math Geometry Module 5 Lesson 11 Answer Key

## Engage NY Eureka Math Geometry Module 5 Lesson 11 Answer Key

### Eureka Math Geometry Module 5 Lesson 11 Exercise Answer Key

Exercises
Exercise 1.
$$\overline{C D}$$ and $$\overline{C E}$$ are tangent to circle A at points D and E, respectively. Use a two-column proof to prove a = b. Draw radii $$\overline{A D}$$ and $$\overline{A E}$$and segment $$\overline{A C}$$.
CD = a, CE = b Given
∠ADC and ∠AEC are right angles. Tangent lines are perpendicular to the radius at the point of tangency.
△ADC and △AEC are right triangles. Definition of a right triangle
AD = AE Radii of the same circle are equal in measure.
AC = AC Reflexive property
△ADC ≅ △AEC HL
CD = CE Corresponding sides of congruent triangles are equal in length.
a = b Substitution

Exercise 2.
In circle A, the radius is 9 mm and BC = 12 mm. a. Find AC.
AC = 15 mm

b. Find the area of △ACD.
A = 54 mm2

c. Find the perimeter of quadrilateral ABCD.
P = 42 mm

Exercise 3.
In circle A, EF = 12 and AE = 13. AE : AC = 1 : 3. a. Find the radius of the circle.
5

b. Find BC (round to the nearest whole number).
39

c. Find EC.
51

### Eureka Math Geometry Module 5 Lesson 11 Problem Set Answer Key

Question 1.
If AB = 5, BC = 12, and AC = 13, is $$\overleftrightarrow{B C}$$ tangent to circle A at point B? Explain. Yes. △ABC is a right triangle because the Pythagorean theorem holds: 52 + 122 = 132. Angle B is right, so $$\overleftrightarrow{B C}$$ is tangent to circle A at point B.

Question 2.
$$\overleftrightarrow{B C}$$ is tangent to circle A at point B. DC = 9 and BC = 15. a. Find the radius of the circle.
r = 8

b. Find AC.
AC = 17

Question 3.
A circular pond is fenced on two opposite sides ($$\overline{C D}$$, $$\overline{F E}$$) with wood and the other two sides with metal fencing.
If all four sides of fencing are tangent to the pond, is there more wood or metal fencing used? There is an equal amount of wood and metal fencing because the distance from each corner to the point of tangency is the same.

Question 4.
Find x if the line shown is tangent to the circle at point B. 67°

Question 5.
$$\overleftrightarrow{P C}$$ is tangent to the circle at point C, and CD = DE. a. Find x ($$m \widehat{C D}$$).
∠CDE is an inscribed angle, so $$m \widehat{C F E}$$ is two times the measure of the intercepted arc; $$m \widehat{C F E}$$ = 152° and $$m \widehat{C D E}$$ = 208°. Since CD = DE, then $$m \widehat{C D}$$ = m $$\widehat{D E}$$. Therefore, 2x = 208° and x = 104°.

b. Find y (m∠CFE).
$$m \widehat{C D E}$$ = 208°, so m∠CFE must be one half this value since it is an inscribed angle that intercepts the arc. Therefore, y = 104°.

Question 6.
Construct two lines tangent to circle A through point B.  Question 7.
Find x, the length of the common tangent line between the two circles (round to the nearest hundredth).  x = 12.17

Question 8.
$$\overline{E F}$$ is tangent to both circles A and C. The radius of circle A is 9, and the radius of circle C is 5. The circles are 2 units apart. Find the length of $$\overline{E F}$$, or x (round to the nearest hundredth). Draw radius $$\overline{A E}$$ and radius $$\overline{C F}$$. Label the intersection of $$\overline{E F}$$ and $$\overline{A C}$$ as Z. Triangles AEZ and CFZ are similar since both have right angles (∠E and ∠F) and a pair of vertical angles equal in measure (∠AZE and ∠CZF).
$$\frac{AZ}{AE}$$ = $$\frac{CZ}{CF}$$
$$\frac{9}{5}$$ = $$\frac{9 + a}{7-a}$$
a = $$\frac{9}{7}$$
EZ = $$\sqrt{\left(9 + \frac{9}{7}\right)^{2}-9^{2}}$$
ZF = $$\sqrt{\left(5 + \frac{5}{7}\right)^{2}-5^{2}}$$
EF = EZ + ZF
EF = $$\sqrt{\left(9 + \frac{9}{7}\right)^{2}-9^{2}} + \sqrt{\left(5 + \frac{5}{7}\right)^{2}-5^{2}}$$
EF ≈ 7.75
The length of $$\overline{E F}$$, or x, is approximately 7.75 units.

### Eureka Math Geometry Module 5 Lesson 11 Exit Ticket Answer Key

Question 1.
If BC = 9, AB = 6, and AC = 15, is $$\overleftrightarrow{B C}$$ tangent to circle A? Explain. No. △ABC is not a right triangle because 92 + 62 ≠ 152. This means $$\overline{A B}$$ is not perpendicular to $$\overline{B C}$$. 