## Engage NY Eureka Math 7th Grade Module 6 Lesson 15 Answer Key

### Eureka Math Grade 7 Module 6 Lesson 15 Example Answer Key

Example 1.
A triangular fence with two equal angles, ∠S = ∠T, is used to enclose some sheep. A fence is constructed inside the triangle that exactly cuts the other angle into two equal angles: ∠SRW = ∠TRW. Show that the gates, represented by $$\overline{S W}$$ and $$\overline{W T}$$, are the same width. There is a correspondence △SRW ↔ △TRW that matches two pairs of angles of equal measurement, ∠S = ∠T and ∠SRW = ∠TRW, and one pair of sides of equal length shared, side $$\overline{R W}$$. The triangles satisfy the two angles and side opposite a given angle condition. From the correspondence, we can conclude that SW = WT, or that the gates are of equal width.

Example 2.
In △ABC, AC = BC, and △ABC ↔ △B’ A’ C’. John says that the triangle correspondence matches two sides and the included angle and shows that ∠A = ∠B’. Is John correct? We are told that AC = BC. The correspondence △ABC ↔ △B’A’C’ tells us that BC ↔ A’C’, CA ↔ C’B’, and ∠C↔∠C’, which means △ABC is identical to △B’A’C’ by the two sides and included angle condition. From the correspondence, we can conclude that ∠A = ∠B’; therefore, John is correct.

### Eureka Math Grade 7 Module 6 Lesson 15 Exercise Answer Key

Exercise 1.
Mary puts the center of her compass at the vertex O of the angle and locates points A and B on the sides of the angle. Next, she centers her compass at each of A and B to locate point C. Finally, she constructs the ray $$\overrightarrow{O C}$$. Explain why ∠BOC = ∠AOC. Since Mary uses one compass adjustment to determine points A and B, OA = OB. Mary also uses the same compass adjustment from B and A to find point C; this means BC = AC. Side $$\overrightarrow{O C}$$ is common to both the triangles,
△OBC and △OAC. Therefore, there is a correspondence △OBC ↔ △OAC that matches three pairs of equal sides, and the triangles are identical by the three sides condition. From the correspondence, we conclude that
∠BOC = ∠AOC. Exercise 2.
Quadrilateral ACBD is a model of a kite. The diagonals $$\overline{A B}$$ and $$\overline{C D}$$ represent the sticks that help keep the kite rigid.
a. John says that ∠ACD = ∠BCD. Can you use identical triangles to show that John is correct? b. Jill says that the two sticks are perpendicular to each other. Use the fact that ∠ACD = ∠BCD and what you know about identical triangles to show ∠AEC = 90°.
c. John says that Jill’s triangle correspondence that shows the sticks are perpendicular to each other also shows that the sticks cross at the midpoint of the horizontal stick. Is John correct? Explain.
a. From the diagram, we see that AC = BC, and AD = BD. $$\overline{C D}$$ is a common side to both triangles, △ACD and △BCD. There is a correspondence △ACD ↔ △BCD that matches three pairs of equal sides; the two triangles are identical by the three sides condition. From the correspondence, we conclude that ∠ACD = ∠BCD. John is correct.

b. Since we know that AC = BC and ∠ACD = ∠BCD, and that △ACE and △BCE share a common side, $$\overline{C E}$$, we can find a correspondence that matches two pairs of equal sides and a pair of equal, included angles. The triangles are identical by the two sides and included angle condition. We can then conclude that ∠AEC = ∠BEC. Since both angles are adjacent to each other on a straight line, we also know their measures must sum to 180°. We can then conclude that each angle measures 90°.

c. Since we have established that △ACE and △BCE are adjacent to each other, we know that AE = BE. This means that E is the midpoint of $$\overline{A B}$$, by definition.

Exercise 3.
In △ABC, ∠A = ∠B, and △ABC ↔ △B’A’C’. Jill says that the triangle correspondence matches two angles and the included side and shows that AC = B’C’. Is Jill correct? We are told that ∠A = ∠B. The correspondence △ABC ↔ △B’A’C’ tells us that ∠A = ∠B’, ∠B = ∠A’, and AB = B’A’, which means △ABC is identical to △B’A’C’ by the two angles and included side condition. From the correspondence, we can conclude that AC = B’C’; therefore, Jill is correct.

Exercise 4.
Right triangular corner flags are used to mark a soccer field. The vinyl flags have a base of 40 cm and a height of 14 cm.
a. Mary says that the two flags can be obtained by cutting a rectangle that is 40 cm×14 cm on the diagonal. Will that create two identical flags? Explain. b. Will measures the two non-right angles on a flag and adds the measurements together. Can you explain, without measuring the angles, why his answer is 90°?
a. If the flag is to be cut from a rectangle, both triangles will have a side of length 40 cm, a length of 14 cm, and a right angle. There is a correspondence that matches two pairs of equal sides and an included pair of equal angles to the corner flag; the two triangles are identical to the corner flag as well as to each other.

b. The two non-right angles of the flags are adjacent angles that together form one angle of the four angles of the rectangle. We know that a rectangle has four right angles, so it must be that the two non-right angles of the flag together sum to 90°.

### Eureka Math Grade 7 Module 6 Lesson 15 Problem Set Answer Key

Question 1.
Jack is asked to cut a cake into 8 equal pieces. He first cuts it into equal fourths in the shape of rectangles, and then he cuts each rectangle along a diagonal.
Did he cut the cake into 8 equal pieces? Explain.  Yes, Jack cut the cake into 8 equal pieces. Since the first series of cuts divided the cake into equal fourths in the shape of rectangles, we know that the opposite sides of the rectangles are equal in length; that means all 8 triangles have two sides that are equal in length to each other. Each of the triangular pieces also has one right angle because we know that rectangles have four right angles. Therefore, there is a correspondence between all 8 triangles that matches two pairs of equal sides and an equal, 90° non-included angle, determining 8 identical pieces of cake.

Question 2.
The bridge below, which crosses a river, is built out of two triangular supports. The point M lies on $$\overline{B C}$$ . The beams represented by $$\overline{A M}$$ and $$\overline{D M}$$ are equal in length, and the beams represented by $$\overline{A B}$$ and $$\overline{D C}$$ are equal in length. If the supports were constructed so that ∠A and ∠D are equal in measurement, is point M the midpoint of $$\overline{B C}$$? Explain. Yes, M is the midpoint of $$\overline{B C}$$. The triangles are identical by the two sides and included angle condition. The correspondence △ABM ↔ △DCM matches two pairs of equal sides and one pair of included equal angles. Since the triangles are identical, we can use the correspondence to conclude that BM = CM, which makes M the midpoint, by definition. 