Practice WKBK Answers

Answer Key
Lesson 5.1
Practice Level B
}
}
}
1. 14 2. 8 3. 17 4. JL 5. JK 6. RT
} }
} }
} }
7. KS, RT 8. KR, ST 9. LT, RS
10. Sample answer: (0, 0), (5, 0), (0, 3)
11. Sample answer: (0, 0), (7, 0), (7, 4), (0, 4)
12. (0, 0), (6, 0), (6, 6), (0, 6)
13. (0, 0), (12, 0), (0, 12)
14. 17 15. 37 16. 46
1
2 1
2
3h k
h k
17. B(2h, k); Proof 18. S }, } , T }, } ; Proof
2 2
2 2
19. AB 5 17, BC 5 10, AC 5 21; 48; 24
20. 72 in.; The crossbar is the midsegment of
the legs. 21. no; no; yes; yes; no; 14 < LM < 28
Answer Key
Lesson 5.2
Practice Level B
1. 4 2. 26 3. 35 4. yes 5. no 6. yes 7. 44
8. 44 9. 36 10. 36 11. 31 12. 62 13. 25
14. 20 15. 24 16. 25 17. 15 18. 48
19. Check student’s drawing; AC 5 BC 5 1 in.
20. Check student’s drawing; AC 5 BC 5 1.125 in.
21. Check student’s drawing; AC 5 BC 5 1.4375 in.
22. Because a point on the ⊥ bisector is
} }
} }
equidistant to the endpoints, AC > BC. By the Reﬂexive Property of >, CD > CD. By the
} }
deﬁnition of bisector, AD > BD. By the SSS
Congruence Postulate, n ACD > n BCD.
23. Because corresponding parts of > n’s
} }
are >, GJ > FJ and ∠ GJH > ∠ FJH. By the Vertical ∠’s Theorem, ∠ GJH > ∠ EJF and
∠ FJH > ∠ EJG. By the Transitive Property,
∠ EJF > ∠ EJG. By the Reﬂexive Property,
} }
EJ > EJ. By the SAS > Postulate,
n EJG > n EJF. Because corresponding parts
} }
of > n’s are >, EF > EG.
24. The post is the ⊥ bisector of the segment
between the ends of the wires.
Answer Key
Lesson 5.3
Practice Level B
1. 19
2. 288
3. 868
}
}
4. No; you don’t know that DA and DC are ⊥ to the rays.
5. No; you don’t know that D is ⊥ to rays.
6. Yes; Converse of Angle Bisector Theorem
7. 7
8. 3
9. 8
10. Yes; x 5 9 by Angle Bisector Theorem.
11. No; you need to know that the congruent segments are ⊥ to the rays.
12. No; you need to know that the two segments are congruent.
13. 16
14. 7
15. 5
16. 8
}
17. Directly between points L and R so that SG bisects ∠ LSR; the distance between you and each goalpost is
equal which minimizes the amount you have to move in either direction.
18. 35 ft
Answer Key
Lesson 5.4
Practice Level B
1. 8
2. 16
3. 5
4. 15
5. 12
6. 6
2
7. a. M(2, 4); P(2, 1) b. N(0, 1); KP 5 4 and KN 5 6 therefore KP 5 } KN. 8. (23, 21)
3
9. (5, 22)
10. yes; yes; yes
11. no; no; no
12. no; yes; no
13. 12; 788
14. 6.5; 15
1
15. }
3
1
16. }
2
2
17. }
3
18. 7
19. 5
20. 3
21. sometimes
22. sometimes
23. always
24. sometimes
2
25. 36 in.; By Theorem 5.8, the distance from the vertex to the centroid is } times the length of the median
3
}
( AB ).
26. 8 mm; yes
Answer Key
Lesson 5.5
Practice Level B
1–3. Check student’s drawings. Longest side and largest angle are opposite each other, shortest side and
smallest angle are opposite each other.
} } }
} } }
4. DF, FE, DE; ∠ E, ∠ D, ∠ F 5. ST, RT, RS;
} } }
∠ R, ∠ S, ∠ T 6. XY, YZ, XZ; ∠ Z, ∠ X, ∠ Y
} } }
} } }
7. JK, JL, KL; ∠ L, ∠ K, ∠ J 8. AC, AB, BC;
} } }
∠ B, ∠ C, ∠ A 9. QR, PR, PQ; ∠ P, ∠ Q, ∠ R
10.
12.
11.
13. yes 14. yes
15. No; 1 1 4 < 6. 16. No; 22 1 26 < 65.
17. yes 18. No; 7 1 45 < 54.
19. 3 in. < x < 15 in. 20. 8 ft < x < 16 ft
21. 9 m < x < 27 m 22. 5 yd < x < 37 yd
23. 2 in. < x < 46 in. 24. 12 in. < x < 60 in.
25. yes; ∠ S, ∠ R, ∠ T 26. no 27. 2 < x < 7
28. 2 < x < 6 29. The building is taller than
200 ft. 30. m∠ ABC < m∠ BAC and
m∠ BAD < m∠ ABD 31. 70 mi < d < 1350 mi
32. Think of the 60- and 24-ft distances as two sides of a triangle. Then the unknown distance d is
36 ft < d < 84 ft. This doesn’t account for the cases when the ball lands straight forward (d 5 36 ft) or
straight backward (d 5 84 ft).
Answer Key
Lesson 5.6
Practice Level B
1. >; Hinge Thm. with m∠ R > m∠ U
2. <; Hinge Thm. with m∠ DGE < m∠ EGF
3. <; Hinge Thm. with m∠ JMK < m∠ LKM
4. >; Converse of Hinge Thm. with the side
opposite ∠ 1 longer than the side opposite ∠ 2.
5. >; Converse of Hinge Thm. with the side
opposite ∠ 1 longer than the side opposite ∠ 2.
6. <; Converse of Hinge Thm. with the side
opposite ∠ 1 shorter than the side opposite ∠ 2.
7. >; Converse of Hinge Thm. with the side
opposite ∠ 1 longer than the side opposite ∠ 2.
8. 5; The triangles are > by SAS. 9. x < 34
10. x > 4 11. Assume temporarily that the two parallel lines contain two sides of a triangle.
12. Assume temporarily that the transversal is not perpendicular to the parallel lines.
13. a. Because m∠ 3 < m∠ 1, by the Hinge Thm, the far side of the table is lower than the near side. b. By
the Converse of the Hinge Thm., ∠ 4 will
be larger than ∠ 2 until the distance between the tops of each pair of legs is the same.
14. the second angler; The included ∠ for the
second angler is 968 and for the ﬁrst angler is 908.
15. F, E, B, A, D, C 16. Temporarily assume that AB > AC. The steps of the proof correspond to the steps
of the proof in Ex. 15.

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