## Engage NY Eureka Math Geometry Module 5 Lesson 6 Answer Key

### Eureka Math Geometry Module 5 Lesson 6 Exercise Answer Key

Opening Exercise

In a circle, a chord \(\overline{D E}\) and a diameter \(\overline{A B}\) are extended outside of the circle to meet at point C. If mâˆ DAE = 46Â°, and mâˆ DCA = 32Â°, find mâˆ DEA.

Let mâˆ DEA = yËš, mâˆ EAB = xËš

In â–³ABD, mâˆ DBA =

Reason:

Answer:

In â–³ABD, mâˆ DBA = yËš

Reason: angles inscribed in same arc are congruent

mâˆ ADB =

Reason:

Answer:

mâˆ ADB = 90Â° Reason: angle inscribed in semicircle

âˆ´46 + x + y + 90 =

Reason:

Answer:

âˆ´46 + x + y + 90 = 180 Reason: sum of angles of triangle is 180Â°

x + y =

Answer:

x + y = 44

In â–³ACE, y = x + 32

Reason:

Answer:

In â–³ACE, y = x + 32

Reason: Exterior angle of a triangle is equal to the sum of the remote interior angles

x + x + 32 =

Reason:

Answer:

x + x + 32 = 44

Reason: substitution

x =

Answer:

x = 6

y =

Answer:

y = 38

mâˆ DEA =

Answer:

mâˆ DEA = 38Â°

Exercises

Find the value of x in each figure below, and describe how you arrived at the answer.

Exercise 1.

Hint: Thalesâ€™ theorem

Answer:

mâˆ BEC = 90Â° inscribed in a semicircle

mâˆ EBC = mâˆ ECB = 45Â° base angles of an isosceles triangle are congruent and sum of angles of a triangle = 180Â°

mâˆ EBC = mâˆ EDC = 45Â° angles inscribed in the same arc are congruent

x = 45

Exercise 2.

Answer:

mâˆ BAD = 146Ëš, if parallel lines cut by a transversal, then interior angles on the same side are supplementary. Then the \(m \widehat{B D}\) = 146Ëš, because âˆ BAD is a central angle intercepting \(\widehat{B D}\). Then remaining arc of the circle, \(\widehat{B C D}\), has a measure of 214Â°. Then mâˆ BED = 107Ëš since it is an inscribed angle intercepting \(\widehat{B C D}\). The angle sum of a quadrilateral is 360Â°, which means x = 73.

Exercise 3.

Answer:

mâˆ BEC = mâˆ CFB = \(\frac{1}{2}\) mâˆ BAC = 52Â°

Inscribed angles are half the measure of the central angle intercepting the same arc.

mâˆ DEG = 128Â° linear pair with âˆ BEC

mâˆ GFD = 128Â° linear pair with âˆ CFB

mâˆ EGF = 74Â° sum of angles of a quadrilateral

x = 74 vertical angles

Exercise 4.

Answer:

The measures of arcs \(\widehat{D E}\), \(\widehat{E F}\), and \(\widehat{F C}\) are each 60Ëš, since the intercepted arc of an inscribed angle is double the measure of the angle. This means \(\widehat{m D E} C\) = 180Ëš, or \(\widehat{D E C}\) is a semicircle. This means x is 90, since âˆ DBC is inscribed in a semicircle.

### Eureka Math Geometry Module 5 Lesson 6 Problem Set Answer Key

In Problems 1â€“5, find the value x.

Question 1.

Answer:

x = 40.5

Question 2.

Answer:

x = 57

Question 3.

Answer:

x = 15

Question 4.

Answer:

x = 34

Question 5.

Answer:

x = 90

Question 6.

If BF = FC, express y in terms of x.

Answer:

y = 90 – \(\frac{x}{2}\)

Question 7.

a. Find the value of x.

Answer:

x = 90

b. Suppose the mâˆ C = aÂ°. Prove that mâˆ DEB = 3aÂ°.

Answer:

mâˆ D = aÂ° (alternate angles are equal in measure), mâˆ A = 2aÂ° (inscribed angles half the central angle), aÂ° + 2aÂ° + mâˆ AED = 180Â° (the sum of the angles of triangle is 180Â°), mâˆ AED = (180 – 3a)Â°, mâˆ AED + mâˆ DEB = 180Â° (angles form line), (180 – 3a)Â° + mâˆ DEB = 180Â° (substitution), mâˆ DEB = 3aÂ°

Question 8.

In the figure below, three identical circles meet at B, F, C, and E, respectively. BF = CE. A, B, C and F, E, D lie on straight lines.

Prove ACDF is a parallelogram.

PROOF:

Join BE and CF.

BF = CE Reason: ______________________________

a = __________ = __________ = __________ = d Reason: ______________________________

__________ = __________ Alternate interior angles are equal in measure.

\(\overline{A C}\) âˆ¥ \(\overline{F D}\)

__________ = __________ Corresponding angles are equal in measure.

\(\overline{A F}\) âˆ¥ \(\overline{B E}\)

__________ = __________ Corresponding angles are equal in measure.

\(\overline{B E}\) âˆ¥ \(\overline{C D}\)

\(\overline{A F}\) âˆ¥ \(\overline{B E}\)âˆ¥\(\overline{C D}\)

ACDF is a parallelogram.

Answer:

Join BE and CF.

BF = CEÂ Â Â Â Reason: Given

a = b = f = e = dÂ Â Â Â Reason: Angles inscribed in congruent arcs are equal in

mâˆ CBE = mâˆ FEBÂ Â Â Â Alternate interior angles are equal in measure.

\(\overline{A C}\) âˆ¥ \(\overline{F D}\)

mâˆ A = mâˆ CBEÂ Â Â Â Â Corresponding angles are equal in measure.

\(\overline{A F}\) âˆ¥ \(\overline{B E}\)

mâˆ D = mâˆ BEFÂ Â Â Â Â Â Corresponding angles are equal in measure.

\(\overline{B E}\) âˆ¥ \(\overline{C D}\)

\(\overline{A F}\) âˆ¥ \(\overline{B E}\)âˆ¥\(\overline{C D}\)

ACDF is a parallelogram.

### Eureka Math Geometry Module 5 Lesson 6 Exit Ticket Answer Key

Question 1.

Find the measure of angles x and y. Explain the relationships and theorems used.

Answer:

mâˆ EAC = 42Â° (linear pair with âˆ BAE). mâˆ EFC = \(\frac{1}{2}\) mâˆ EAC = 21Â° (inscribed angle is half measure of central angle with same intercepted arc). x = 21.

mâˆ ABD = mâˆ EAC = 42Â° (corresponding angles are equal in measure). y = 42.