# Eureka Math Geometry Module 5 Lesson 1 Answer Key

## Engage NY Eureka Math Geometry Module 5 Lesson 1 Answer Key

### Eureka Math Geometry Module 5 Lesson 1 Exploratory Challenge Answer Key

Exploratory Challenge
Choose one of the colored points (C,D, …) that you marked. Draw the right triangle formed by the line segment connecting the original two points A and B and that colored point. Take a copy of the triangle, and rotate it 180˚ about the midpoint of $$\overline{A B}$$.
Label the acute angles in the original triangle as x and y, and label the corresponding angles in the rotated triangle the same.
Todd says ACBC’ is a rectangle. Maryam says ACBC’ is a quadrilateral, but she is not sure it is a rectangle. Todd is right but does not know how to explain himself to Maryam. Can you help him out?
a. What composite figure is formed by the two triangles? How would you prove it?
A rectangle is formed. We need to show that all four angles measure 90°.

i. What is the sum of the measures of x and y? Why?
90°; the sum of the measures of the acute angles in any right triangle is 90°.

ii. How do we know that the figure whose vertices are the colored points (C,D, …) and points A and B is a rectangle?
All four angles measure 90°. The colored points (C,D, …) are constructed as right angles, and the angles at points A and B measure x + y, which is 90°.

b. Draw the two diagonals of the rectangle. Where is the midpoint of the segment connecting the two original points A and B? Why?
The midpoint of the segment connecting points A and B is the intersection of the diagonals of the rectangle because the diagonals of a rectangle are congruent and bisect each other.

c. Label the intersection of the diagonals as point P. How does the distance from point P to a colored point (C,D, …) compare to the distance from P to points A and B?
The distances from P to each of the points are equal.

d. Choose another colored point, and construct a rectangle using the same process you followed before. Draw the two diagonals of the new rectangle. How do the diagonals of the new and old rectangle compare? How do you know?
One diagonal is the same (the one between points A and B), but the other is different since it is between the new colored point and its image under a rotation. The new diagonals intersect at the same point P because diagonals of a rectangle intersect at their midpoints, and the midpoint of the segment connecting points A and B has not changed. The distance from P to each colored point equals the distance from P to each original point A and B. By transitivity, the distance from P to the first colored point, C, equals the distance from P to the second colored point, D.

e. How does your drawing demonstrate that all the colored points you marked do indeed lie on a circle?
For any colored point, we can construct a rectangle with that colored point and the two original points, A and B, as vertices. The diagonals of this rectangle intersect at the same point P because diagonals intersect at their midpoints, and the midpoint of the diagonal between points A and B is P. The distance from P to that colored point equals the distance from P to points A and B. By transitivity, the distance from P to the first colored point, C, equals the distance from P to any other colored point.

By definition, a circle is the set of all points in the plane that are the same distance from a given center point. Therefore, each colored point on the drawing lies on the circle with center P and a radius equal to half the length of the original line segment joining points A and B

### Eureka Math Geometry Module 5 Lesson 1 Example Answer Key

Example
In the Exploratory Challenge, you proved the converse of a famous theorem in geometry. Thales’ theorem states the following: If A, B, and C are three distinct points on a circle, and $$\overline{A B}$$ is a diameter of the circle, then ∠ACB is right.
Notice that, in the proof in the Exploratory Challenge, you started with a right angle (the corner of the colored paper) and created a circle. With Thales’ theorem, you must start with the circle and then create a right angle.
Prove Thales’ theorem.
a. Draw circle P with distinct points A, B, and C on the circle and diameter (AB) ̅. Prove that ∠ACB is a right angle.
Sample image shown to the right.

b. Draw a third radius ($$\overline{P C}$$). What types of triangles are △APC and △BPC? How do you know?
They are isosceles triangles. Both sides of each triangle are radii of circle P and are, therefore, of equal length.

c. Using the diagram that you just created, develop a strategy to prove Thales’ theorem.
Look at each of the angle measures of the triangles, and see if we can prove m∠ACB is 90°.

d. Label the base angles of △APC as b° and the base angles of △BPC as a°. Express the measure of ∠ACB in terms of a° and b°.
The measure of ∠ACB is a° + b°.

e. How can the previous conclusion be used to prove that ∠ACB is a right angle?
2a + 2b = 180 because the sum of the angle measures in a triangle is 180°. Then, a + b = 90, so ∠ACB is a right angle.

### Eureka Math Geometry Module 5 Lesson 1 Exercise Answer Key

Exercises

Exercise 1.
$$\overline{A B}$$ is a diameter of the circle shown. The radius is 12.5 cm, and AC = 7 cm.

a. Find m∠C.
90°

b. Find AB.
25 cm

c. Find BC.
24 cm

Exercise 2.
In the circle shown, $$\overline{B C}$$ is a diameter with center A.

a. Find m∠DAB.
144°

b. Find m∠BAE.
128°

c. Find m∠DAE.
88°

### Eureka Math Geometry Module 5 Lesson 1 Problem Set Answer Key

Question 1.
A, B, and C are three points on a circle, and angle ABC is a right angle. What is wrong with the picture below? Explain your reasoning.

Draw in three radii (from O to each of the three triangle vertices), and label congruent base angles of each of the three resulting isosceles triangles. See diagram to see angle measures. In the big triangle (△ABC), we get 2a + 2b + 2c = 180. Using the distributive property and division, we obtain 2(a + b + c) = 180, and a + b + c = 90. But we also have 90 = m∠B = b + c. Substitution results in a + b + c = b + c, giving a a value of 0-a contradiction.

Question 2.
Show that there is something mathematically wrong with the picture below.

Draw three radii ($$\overline{O A}$$, $$\overline{O B}$$, and $$\overline{O C}$$). Label m∠BAC as a° and m∠BCA as c°. Also label m∠OAC as x˚ and m∠OCA as x° since △AOC is isosceles (both sides are radii). If m∠ABC is a right angle (as indicated on the drawing), then a° + c° = 90°. Since △AOB is isosceles, m∠ABO = a° + x°. Similarly, m∠CBO = c° + x°. Now adding the measures of the angles of △ABC results in a° + a° + x° + c° + x° + c° = 180°. Using the distributive property and division, we obtain a° + c° + x° = 90°. Substitution takes us to a°s + c° = a° + c° + x°, which is a contradiction. Therefore, the figure above is mathematically impossible.

Question 3.
In the figure below, $$\overline{A B}$$ is the diameter of a circle of radius 17 miles. If BC = 30 miles, what is AC?

16 miles

Question 4.
In the figure below, O is the center of the circle, and $$\overline{A D}$$ is a diameter.

a. Find m∠AOB.
48°

b. If m∠AOB∶m∠COD = 3∶4, what is m∠BOC?
68°

Question 5.
$$\overline{P Q}$$ is a diameter of a circle, and M is another point on the circle. The point R lies on $$\overleftrightarrow{\boldsymbol{M Q}}$$ such that RM = MQ. Show that m∠PRM = m∠PQM. (Hint: Draw a picture to help you explain your thinking.)
Since RM = MQ (given), m∠RMP = m∠QMP (both are right angles, ∠QMP by Thales’ theorem and ∠RMP by the angle addition postulate), and MP = MP (reflexive property), then △PRM ≅ △PQM by SAS. It follows that ∠PRM ≅ ∠PQM (corresponding sides of congruent triangles) and that m∠PRM = m∠PQM (by definition of congruent angles).

Question 6.
Inscribe △ABC in a circle of diameter 1 such that $$\overline{A C}$$ is a diameter. Explain why:
a. sin(∠A) = BC.
$$\overline{A C}$$ is the hypotenuse, and AC = 1. Since sine is the ratio of the opposite side to the hypotenuse, sin(∠A) necessarily equals the length of the opposite side, that is, the length of $$\overline{B C}$$.

b. cos(∠A) = AB.
$$\overline{A C}$$ is the hypotenuse, and AC = 1. Since cosine is the ratio of the adjacent side to the hypotenuse, cos(∠A) necessarily equals the length of the adjacent side, that is, the length of $$\overline{A B}$$.
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