Answer Key

Lesson 5.1

Study Guide

1. MN 5 53 cm, ZY 5 14 cm

2. DF 5 7 in., BC 5 12 in.

3. }

ST i } BA , ST 5 1 }

2 BA

4. (0, k)

Answer Key

Lesson 5.2

Study Guide

1. 23

2. 10

3. ###$ PQ bisects }

RS , so PR 5 PS. Because Q is on the perpendicular bisector of }

RS , QR 5 QS by Theorem 5.2.

4. No; If T were on @##$ PQ , then T would be equidistant from R and S. T is 14 units from R and 15 units from S.

5. 9

Answer Key

Lesson 5.3

Study Guide

1. 7

2. 5

3. 9

4. 7

5. 3

Answer Key

Lesson 5.4

Study Guide

1. RU 5 6, RS 5 4

2. 26

3. (5, 2)

4. (25, 6)

5. (2, 6)

6. (3, 0)

Answer Key

Lesson 5.5

Study Guide

1. m∠A < m∠C < m∠B; BC < AB < AC

2. m∠E < m∠F < m∠D; DF < DE < EF

3. m∠H < m∠ I < m∠G; GI < GH < HI

4. m∠J < m∠L < m∠K; KL < JK < JL

5. 6.

7. greater than 3 cm and less than 7 cm

8. greater than 5 in. and less than 19 in.

9. greater than 6 ft and less than 14 ft

10. greater than 1 m and less than 21 m

11. greater than 16 in. and less than 34 in.

12. greater than 7 mi and less than 9 mi

Answer Key

Lesson 5.6

Study Guide

1. 5; Two sides of the first triangle are congruent to two sides of the second triangle, and the third side of the first triangle is neither longer nor shorter than thethird side of the second triangle. So, by the Converse of the Hinge Theorem, the included angle of the firsttriangle is neither longer nor shorter than the included angle of the second triangle.

2. <; Because }

AD > }

BC , } BD > } BD , and m∠ADB < m∠CBD, then by the Hinge Theorem, AB < CD.

3. You and your friend are the same distance from the tree.

4. y ≥ 0

5. m∠B Þ m∠C