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.
Engage NY Math Geometry Module 5 Lesson 11 Exercise Answer Key 1
Answer:
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.
Engage NY Math Geometry Module 5 Lesson 11 Exercise Answer Key 2
a. Find AC.
Answer:
AC = 15 mm

b. Find the area of △ACD.
Answer:
A = 54 mm2

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

Exercise 3.
In circle A, EF = 12 and AE = 13. AE : AC = 1 : 3.
Engage NY Math Geometry Module 5 Lesson 11 Exercise Answer Key 3
a. Find the radius of the circle.
Answer:
5

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

c. Find EC.
Answer:
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.
Eureka Math Geometry Module 5 Lesson 11 Problem Set Answer Key 1
Answer:
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.
Eureka Math Geometry Module 5 Lesson 11 Problem Set Answer Key 2
a. Find the radius of the circle.
Answer:
r = 8

b. Find AC.
Answer:
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?
Eureka Math Geometry Module 5 Lesson 11 Problem Set Answer Key 3
Answer:
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.
Eureka Math Geometry Module 5 Lesson 11 Problem Set Answer Key 4
Answer:
67°

Question 5.
\(\overleftrightarrow{P C}\) is tangent to the circle at point C, and CD = DE.
Eureka Math Geometry Module 5 Lesson 11 Problem Set Answer Key 5
a. Find x (\(m \widehat{C D}\)).
Answer:
∠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).
Answer:
\(m \widehat{C D E}\) = 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.
Eureka Math Geometry Module 5 Lesson 11 Problem Set Answer Key 6
Answer:
Eureka Math Geometry Module 5 Lesson 11 Problem Set Answer Key 7

Question 7.
Find x, the length of the common tangent line between the two circles (round to the nearest hundredth).
Eureka Math Geometry Module 5 Lesson 11 Problem Set Answer Key 8
Answer:
Eureka Math Geometry Module 5 Lesson 11 Problem Set Answer Key 9
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).
Eureka Math Geometry Module 5 Lesson 11 Problem Set Answer Key 10
Answer:
Draw radius \(\overline{A E}\) and radius \(\overline{C F}\). Label the intersection of \(\overline{E F}\) and \(\overline{A C}\) as Z.
Eureka Math Geometry Module 5 Lesson 11 Problem Set Answer Key 11
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.
Eureka Math Geometry Module 5 Lesson 11 Exit Ticket Answer Key 1
Answer:
No. △ABC is not a right triangle because 92 + 62 ≠ 152. This means \(\overline{A B}\) is not perpendicular to \(\overline{B C}\).

Question 2.
Construct a line tangent to circle A through point B.
Eureka Math Geometry Module 5 Lesson 11 Exit Ticket Answer Key 2
Answer:
Answers will vary.

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