Solution Key Geometry [New ed.] 0395323231, 9780395323236
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Solution Key
GEOMETRY /
new edition
Solution Key
GEOMETRY new edition
Ray C. Jurgensen
Richard G. Brown Alice M. King Editorial Adviser
Albert E. Meder, Jr. Teacher Consultant
Anita M. Cooper
HOUGHTON MIFFLIN COMPANY ∙ BOSTON Atlanta Dallas Geneva, III. Hopewell, N.J. Palo Alto
Toronto
1983 Impression Copyright © 1980 by Houghton Mifflin Company ;hts reserved. No part of this work may be reproduced or trans in any form or by any means, electronic or mechanical, includ)tocopying and recording, or by any information storage or ral system, without permission in writing from (he publisher. Printed in (he United Slates of Λn∣eriru
CONTENTS Page
'hapter
.
1
1.
Points, Lines, Planes, and Angles .
2.
Parallel Lines and Planes........................
3.
Congruent Triangles....................................
4.
Using Congruent Triangles........................
'.>.
Similar Polygons...........................................
6.
Right Triangles...........................................
7.
Circles .............................................................
H.
Constructions and Loci..............................
∣ι.
Areas of Plane Figures..............................
.
.
12
26 46 63
136
.
Areas and Volumes of Solids
11
Coordinate Geometry....................................
Γ.,.
Transformations ...........................................
.
98
113
10.
.
77
End-of-Book Examinations........................ Appendix: Logic...........................................
154
165
184 201 203
CHAPTER 1 , Points, Lines, Planes, and Angles Pages 3-4 ■ WRITTEN EXERCISES
A
L τrue
2. True
3. False
4. False
5. True
8. True —
—9. True 15. False
10. True —
11. True —
12. False —
14. True
B 17-29. Sketches may vary.
16. True
6. False 13. True —
7. True
2
Key :: Chapter 1, pages 7-8
Pages 7-8 ∙ WRITTEN EXERCISES
A
1. Yes 9. True
15. False
2. No 10. True
21. 9
C 30. 2
31. 1.75
4. Yes
11. True
16. True
B 20. 4
c. True
3. No
5. Yes
12. True
17. True
22. 4
32. 0.9
d. May be false
23. 15
33. 112
35. a. 3
18. True 24. MY
6. No 13. False
14. False
19. False 25-29. Sketches may vary.
34. a. True
b. 6
7. Yes
c. 10
b. May be false d. 15
e. 21
8. No
Key to Chapter 1, pages 11-16
3
Pages 11-12 • WRITTEN EXERCISES
A
1. E
—>> _ >. £. AT and AS
2-8. Various answers are possible.
4. ALAT
5. A LEA
11. Right
12. Acute
c. 180°
16. a. AEB
6. A2
7. A 7
13. Straight
b. BEC
8. A 6
£. Acute
14. Obtuse c. ACE
3. A ELS
15. a. LPN
Aθ∙ Acute
b. OPG
17. 180°; yes
B 22. A BEA, ABEC, ADEA, A DEC 23-25. Various answers are possible. 23. ABAC, ABAD, ACAD, AADB, ABDC, ADCA, AACB, ACBD, ADBA 24. A ADC = 100o, ADCB = 120°, ACBA = 100°
25. A ADC and ACBA
26. AABE = 60°, A ABF = 30°, A FBC = 60° 27. A ABE = 57°, A ABF = 33°, A FBC = 57°; AEBD = A ABF
28. 2x + 3x = 4x + 7; 5x - 4x = 7; x = 7
29. 30 + 4x = 6x; 30 = 2x; x = 15
30. (60 - x) + x = 3x; 60 = 3x; x = 20 C 31. a. 6; 10
b. 15
c. n^2^
Page 12 ∙ CHALLENGE
Pages 15-16 ∙ WRITTEN EXERCISES
A
1. Addition Postulate
2 Division Postulate
3. a. Subtraction Postulate
b. Division Postulate
4. a. Multiplication Postulate
b. Division Postulate
Key to Chapter 1, pages 19-21
4
5 a. Addition Postulate
b. Subtraction Postulate
c. Division Postulate
6. a. Multiplication Postulate
b. Distributive Property
c. Subtraction Postulate
7. Addition Postulate
8. Subtraction Postulate
9. Addition Postulate
12. Subtraction Postulate
13. Addition Postulate
10. Subtraction Postulate B 11. Addition Postulate
14. Subtraction Postulate
15. Transitive Property or Substitution Property 17. Subtraction Postulate
16. Transitive Property or Substitution Property C 18. a. Not an equivalence relation
b. Equivalence relation
c. Not an equivalence relation
d. Not an equivalence relation
Pages 19-21 ∙ WRITTEN EXERCISES
A
ι. δpba, δpbc
2. ΔQBC
3. ΔABC
6. ΔCAE, ΔACB
7. ΔADB
8. BD
4. ΔQBA, ΔPBQ 9. ΔCDB, ΔBDA∙, or
5. ΔQBA, ΔQBC δeac, δacb
a. 2. AM = x - a; MB = b - x
2. Ruler Post.
3. M is the midpoint of AB.
3. Given
4. x-a=b-x
4. Def. of midpoint
5. 2x = a + b
5. Properties of algebra
+ cq
x>
I I
X
C D
14.
Coordinate of A
0
6
-12
-3
Coordinate of B
17
24
-28
Coordinate of M 15.
6. Div. Post.
8.5
15
-20
X
2
-16
13
y
J38
4
5
χ + y 2
20
-6
(1) A line contains at least two points; (2) Given and def. of collinear; (3) Through
any three noncollinear points there is exactly one plane; (4) If two points are in a plane, then the line joining the points is in that plane.
K
Key to Chapter 1, pages 37-38
C 16.
9
Reasons
Statements
1. Lines j and k intersect in point T.
1. Given
2. Line j contains at least two points,
2. A line contains at least two points.
A and T; line k contains at least
two points, B and T. 3. A, B, and T are not collinear.
3. If two lines intersect, then they intersect in exactly one point.
4. There is exactly one plane, M, which contains A, B, and T.
4. Through any three noncollinear points
5. M contains j and k.
5. If two points are in a plane, then the
there is exactly one plane.
line joining the points is in that plane. 17.
Since OX bisects L POQ, ΔPOX = ΔXOQ! Δ POX = p - x; ΔXOQ = x - q;
q
p - x = x - q; 2x = p + q; x = p
Pages 37-38 ∙ CHAPTER REVIEW
5.
Two
6. A, B, C, or E
11. ΔABC
7. 14
10.
∆ABD
13.
Transitive Property or Substitution Property
15. Substitution Property
18.
∆l, ∆3 (or ∆2, ∆4)
21.
postulates
26.
12. 65°
Definition of midpoint
28. Subtraction Postulate
14. Addition Postulate
16. Subtraction Postulate 19. Δ2 or Δ4
17. EB, DG
20. ΔFAB or ∆CAE
23. Supplements of equal angles are equal.
22. theorems
24. Vertical angles are equal.
9. ΔADC, ∆ADB, ΔBDC
8. 6
25. Definition of angle bisector
27. Complements of equal angles are equal.
29. False
30. False
31. True
32. False
10
Key to Chapter 1, pages 38-39
Pages 38-39 ∙ CHAPTER TEST
1.
AD (or DA)
5.
ΔCAD
7. B, A, E
6. ΔBAD
10.
ΔCAD, ∆l
15.
∆2, ∆4, ∆6, ∆8
17.
9
19.
3. A is the midpoint of BE.
2. A
11. ΔDAB
4. AD and AE
8. CA, BA
12. 45°
9. ΔBAC, Δ CAE
14. 56°
13. 56°
16. ∆l, Δ 3; ∆2, Δ 4; ∆5, ∆7j ∆6, ∆8
18. -2.5 Statements
Reasons
1. BD bisects AC.
1. Given
2. AD = DC
2. Def. of bisector
3. AD = BD
3. Given
4. BD = DC
4. Sub st. Prop.
20. False
21. False
Page 39 ∙ ALGEBRA REVIEW 1. a + 6 = 23; a = 17
2. 7 + b = 15; b = 8
3. c - 12 = 8; c = 20
4. 19 = d - 2; d = 21
5. 3e = 12; e = 4
6. 8f = 4; f = ∣
7. 2g + 5 = 15; 2g = 10;
g = 5
8. 3h - 2 = 10; 3h = 12; h = 4
9. 6i = 5i + 11; i = 11
11. 20 + k = 7k; 6k = 20;
10. 12 - j = 2j; 3j = 12; j = 4 k = 3∣
13. 2 - 3n = 4n - 12; 7n = 14;
n = 2
12. 6m + 12 = 4m + 18; 2m = 6; m - 3
14. 30 + 2p = 180 - 3p; 5p = 150; p = 30
15. q + 1 > 5; q>4
16. 4r > 12; r > 3
17. 2s - 3 > 7; 2s > 10; s > 5
18. 5t + 9 < 19; 5t < 10; t < 2
19. 1 + u < 2u; u > 1
20.
21.
7v - 5 > 4v + 16; 3v > 21; v > 7
3(w - 5) = 12; 3w - 15 = 12; 3w = 27; w = 9
22. 4x = 2(x + 11); 4x = 2x + 22; 2x = 22; x = 11
23. 3(3y + -∣) = 5y + 19; 9y + 1 = 5y + 19; 4y = 18; y = 4∣ 24. ∣(8z - 4) = 3(8 - 3z); 4z - 2 = 24 - 9z; 13z = 26; z = 2
25.
3(90 - x) = 180 - x; 270 - 3x = 180 - x; -2x = -90; x = 45
26.
180 - (x + 10) = 70; 180 - x - 10 = 70; -x = -100; x = 100
27.
2(90 - x) = 3(90 - 2x); 180 - 2x = 270 - 6x; 4x = 90; x = 22∣
Key to Chapter 1, page 39
28. 180 - 2(x + 20) = x - 10; 180 - 2x - 40 = x - 10; -3x = -150; x = 50 29. (n - 2)180 = 165n; 180n - 360 = 165n; 15n = 360; n = 24 30. n(n + 5) = n2 + 10; n2 + 5n = n2 + 10; 5n = 10; n - 2
11
CHAPTER 2 ∙ Parallel Lines and Planes Pages 44-45 ∙ WRITTEN EXERCISES
A
1. AD, BC; AC
2. AB, DC; AC
3. PT, QV; TQ
4. PT, QU; PR
5. Corr. zL
6. Alt. int. Λ
7. S-s. int. zL
8. S-s. int. zi.
9. Corr. zL
10. Alt. int. zL
11. Alt. int. zL
13. Corr. zL
14. S-s. int.zL
12. Corr. ∕s
16. Corr. zL are =.
15. Answers may vary. Example:
17. Alt. int. zL are =.
are supp.
18. S-s. int.
20. always
B 19. never
22. sometimes
21. never
23. never
24. sometimes
25. sometimes
26. always
27. sometimes
28. sometimes
29. sometimes
30. always
31. always
2. Vert. zL are =.
32. 1. Given
3. Subst. Prop. or Trans. Prop.
33. 1. Δ6 is supp. to ∆5. (Def. of supp. zL)
2. ∆3 is supp. to ∆5. (Given)
3. ∆3 = ∆6 (Supps. of the same angle are =.)
c 34. 1. ∆l = Δ 3 (Given) or Trans. Prop.) of supp. ∕s. )
2. ∆3 = Δ10 (Vert, ∕s are =.) 4. Δ 10 is supp. to ∆ll. (Given)
6. ∆l + ∆ll = 180° (Subst. Prop.)
8. ∆l + Δ6 = 180° (Subst. Prop.)
3. ∆l = L10 (Subst. Prop. 5. Δ10 + ∆ll = 180o (Def. 7. ∆ll = Δ6 (Vert, ∕s are =.)
9. Δ 1 is supp. to ∆6. (Def. of supp. zL )
Pages 48-49 ∙ WRITTEN EXERCISES A
2. ∆2, ∆5, ∆4, ∆7, Δ10, ∆13, ∆12, ∆15
1. ∆3, ∆9, ∆ll, ∆6, Δ8, ∆14, Δ16
3. 130°
4. 48o
5. 125°
6. xo
12
7. (180 - y)o
8. x = 60; y = 90
Key to Chapter 2, pages 53-54 9.
13
x = ∣(180 - 100) = 40; y = 180 - 40 = 140
10. x = 60; y = 60
11. y = 180 - (110 + 30) = 40; x = y = 40
12. x = 90; 5y + 4y = 180, 9y = 180, y = 20 14.
2. Corr, ∕s are =.
1. Given 5.
13. x = 61; y = 60 4. Subst. Prop.
3. Def. of straight L
Def. of supp. zk
B 15. 1. k ∣∣ m (Given)
2. L 1 = L 2 (If 2 ∣∣ lines are cut by a trans., then alt. int. zk
are =.)
3. ∆2 + ∆4 = 180o (Def. of straight Δ)
Prop.)
5. ∆l is supp. to ∆4. (Def. of supp. zk )
16γ 1. k ∣∣ m (Given)
3. ∆5 = ∆7 (If 2 ∣∣ lines are
2. ∆2 = ∆5 (Vert, zk are =.)
cut by a trans., then corr. zk are =.)
17. 1. k ∣∣ m (Given)
4. ∆l + ∆4 = 180o (Subst.
4. Δ2 = ∆7 (Trans. Prop, or Subst. Prop.)
2. Δ1 = Δ 3 (If 2 ∣∣ lines are cut by a trans., then corr. zk are =.) 4. Δl + ∆7 = 180o (Subst. Prop.)
3. ∆3 + ∆7 = 180o (Def. of straight Δ) 5. ∆l is supp. to ∆7. (Def. of supp. zk )
18. 1. PQ ∣∣ SR; PS ∣∣ QR (Given) int. zk are supp.)
2. ΔP
is supp. to ΔS∙, ΔR is supp. to ∆S. (S-s.
3. ΔP = ΔR (Supp. of the same Δ are =.)
19. 1. BA ∣∣ ED; BC ∣∣ EF (Given)
2. ΔABC = ΔDGC∙, ΔDGC = ΔDEF (If 2 ∣∣ lines
are cut by a trans., then corr. zk are =.)
3. ∆ABC = ΔDEF (Trans. Prop, or
Subst. Prop.) C 20. 1. AE ∣∣ BD (Given)
are =.)
2. Δ1 = Δ 4 (If 2 ∣∣ lines are cut by a trans., then alt. int. zk
4. ∆l = ∆3 (Subst. Prop.)
3. Δ 3 = Δ 4 (Given)
lines are cut by a trans., then corr. zk are =.)
7.
5. Δ 2 = Δ 3 (If 2 ∣∣
6. ∆l = ∆2 (Subst. Prop.)
BD bisects ΔEBC. (Def. of Δ bisector)
21. 1. AE ∣∣ BD (Given)
2. ∆5 = ΔABD (If 2 ∣∣ lines are cut by a trans., then corr.
zk are =.) 3. ∆5 = ∆6 + ∆l (Subst. Prop.)
4. BD bisects ΔEBC. (Given)
6. ∆5=∆6 + ∆2 (Subst. Prop.)
5. ∆l = ∆2 (Def. of Δ bisector)
Pages 53-54 ∙ WRITTEN EXERCISES 1. DA, CB
2. DA, CB
6. DC, AB
7. None
11. BY, ZC
12. SR, PQ
3. None 8. None
4, None 9 None
5. DC, AB
10. DA, CB
13. 4x - 20 = 3x; x = 20
14. 80 + 4x = 180; 4x = 100; x = 25
15. (2x + 10) + 90 = 180; 2x = 80; x = 40
Key to Chapter 2, pages 53-54
14
16.
2. Vert, ∕s are =.
1. Given
3. Subst. Prop.
4. If 2 lines are cut by a trans,
and corr. ∕⅞ are -, then the lines are ∣∣. 2. Δ6 is supp. to Δ 3; ∆2 is supp.
B 17. 1. ∆3 + ∆6 = 180°; ∆2 + ∆3 = 180o (Given)
4. BC ∣∣
3. Δ6 = ∆2 (Supps. of the same Δ are =.)
to ∆3. (Def. of supp. zk)
EF (If 2 lines are cut by a trans, and corr. zk are =, then the lines are ∣∣.)
18. 1. BC ∣∣ EF (Given) are =.)
2. Δ 2 = Δ 6 (If 2 ∣∣ lines are cut by a trans., then corr. zk
3. ∆6 = ∆7 (Given)
4. ∆2 = ∆7 (Trans. Prop, or Subst. Prop.)
5. AE ∣∣ DC (If 2 lines are cut by a trans, and corr. zk are =, then the lines are ∣∣.)
19. Prove: PQ ∣∣ RS. 3.
Proof: 1. ∆l = ∆2 (Given)
∆4 = ∆5 (Given)
2. ∆2 = ∆4 (Vert, zk are =.) 5. PQ ∣∣ RS
4. ∆l = ∆5 (Trans. Prop, or Subst. Prop.)
(If 2 lines are cut by a trans, and alt. int. zk are =, then the lines are ∣∣.)
Proof: 1. ∆2 = ∆4 (Vert. zk are =.)
20. Prove: ∆2 = ∆4∙, ∆l = ∆5. (Given)
2. ∆3 = ∆6
3. PQ ∣∣ RS (If 2 lines are cut by a trans, and alt. int. zk are =, then the
lines are ∣∣.)
4. Δ 1 = Δ 5 (If 2 ∣∣ lines are cut by a trans., then alt. int. zk are =.)
21. 1. Trans, t cuts k and m. (Given) — 2. ∆2 + ∆3 = 180° (Def. of straight Δ)
Y -------- γr------------
m
3. ∆3 is supp. to ∆2. (Def. of supp. zk )
∖3 ∖
4. ∆l is supp. to ∆2. (Given) 5. ∆l = Δ 3 (Supps. of the same ∆ are =.)
6. k ∣∣ m (If 2 lines are cut by a trans,
and alt. int. zk are =, then the lines are ∣∣.) 2. ∆3 = 90°; Δ 4 = 90o (Def. of J_ lines)
22. 1. Trans, t J_ /; tin (Given)
3.
4. / ∣∣ n (If 2 lines are cut by a trans, and corr. zk
∆3 = ∆4 (Subst. Prop.)
are =, then the lines are ∣∣.) 23. 1. BC ∣∣ ED (Given)
are =.)
2. Δ 2 = Δ 3 (If 2 ∣∣ lines are cut by a trans., then alt. int. zk 4. Δl + ∆2=∆3 + ∆4 (Add. Post.)
3. ∆l = Δ4 (Given)
ΔBDF (Subst. Prop.)
5. ∆ABD =
6. AB ∣∣ DF (If 2 lines are cut by a trans, and alt. int. zk
are =, then the lines are ∣∣.) 24. 1. ΔABD = ΔFDB (Given)
(Given)
2. Δl + ∆2 = ∆3 + ∆4 (Subst. Prop.)
4. ∆2 = ∆3 (Subtr. Post.)
3. ∆l = ∆4
5. BC ∣∣ ED (If 2 lines are cut by a trans,
and alt. int. zk are =, then the lines are ∣∣.) C 25. 1. ∆3 is supp. to ∆ABR. (Given) 3.
∆ABR = ΔDBR (Given)
2. ∆3 + ∆ABR = 180o (Def. of supp. zk)
4. Δ 3 + ΔDBR = 180o (Subst. Prop.)
Key to Chapter 2, pages 54-60
6. BC ∣∣ ED (If 2 lines are cut by a
5. ∆3 is supp. to ΔDBR. (Def. of supp. ∕⅞ )
trans, and s-s. int. ∕s are supp., then the lines are ∣∣.) 26.
Given: AB ∣∣ CD; GE bisects ΔIGB,, HF bisects ΔGHD. Prove: GE ∣∣ HF
Proof: 1. AB ∣∣ CD (Given)
2. ΔIGB = ΔGHD
(If 2 ∣∣ lines are cut by a trans., then corr. zL are =.) 3. ∆l + ∆2 = ∆3 + ∆4 (Subst. Prop.)
4. GE
bisects ΔIGB∙, HF bisects ΔGHD. (Given)
8.
/
υ
*J
5. ∆l=∆2∙,∆3=∆4 (Def. of Δ 7. ∆l = ∆3 (Div. Post.)
6. 2(∆l) = 2(∆3) (Subst. Prop.)
bisector)
c
GE ∣∣ HF (If 2 lines are cut by a trans, and corr. zL are -, then the lines are ∣∣.) Given: GE bisects ΔIGBj HF bisects ΔGHDj GE ∣∣ HF.
27. Refer to the art for Ex. 26.
Prove: AB ∣∣ CD.
Proof: 1. GE ∣∣ HF (Given)
by a trans., then corr. ∕s are =.)
3. GE bisects ΔIGB', HF bisects ΔGHD. (Given)
4. ∆2 = ∆lj ∆3=∆4 (Def. of Δ bisector)
Prop.)
2. ∆l = Δ 3 (If 2 ∣∣ lines are cut
6.Δ1 + ∆2=∆3 + ∆4 (Add. Post.)
5. ∆2 = ∆4 (Subst. Prop, or Trans. 7. ΔIGB = ΔGHD (Subst. Prop.)
8. AB ∣∣ CD (If 2 lines are cut by a trans, and corr. A. are =, then the lines are ∣∣.)
28. x2 + 3x = 180; x2 + 3x - 180 = 0; (x + 15)(x - 12) = 0; x = -15 (reject) or
x = 12; x = 12
Page 54 ∙ CHALLENGE
The result is one band that is twice as long as the original band.
Pages 59-60 ∙ WRITTEN EXERCISES
A
1-3, 6, 7. Sketches may vary. 1.
1∙
4. Not possible
5. Not possible
Λ
1∙
8. Not possible
Key to Chapter 2, pages 59-60
16
9.
x = 70 + 60 = 130; y = 180 - 130 = 50
10. x = 40 + 30 = 70; y = 180 - 70 = 110 11. x = 35 + 40 = 75; y = 180 - 75 = 105
12. x = 90 - 46 = 44; y = 90 - 44 = 46
13. x = 40; y = 48 + 40 = 88
14. x = 132; y = 180 - [(180 - 132) + 68] = 180 - 116 = 64
B 15. The sum of the three angles is 180o.
Since the three angles are equal, each has
measure 180o ÷ 3, or 60°.
16.
If there were 2 right angles or 2 obtuse angles in a triangle, then the sum of the measures of the three angles would be greater than 180o.
17.
Given: ∆l = ∆4∙, ∆2 = ∆5 Prove: ∆3 = ∆6
Proof: 1. Δl+∆2 + ∆3 = 180°; ∆4 + ∆5 + ∆6 = 180o (The sum of the ∕s of a Δ is 180o.)
2. Δl + ∆2 + ∆3=∆4 + ∆5 + ∆6
(Subst. Prop.) 4.
18.
3. ∆l = ∆4∙, ∆2 = ∆5 (Given)
Δl + ∆2 + ∆3=∆! + ∆2 + ∆6 (Subst. Prop.)
5. ∆3 = ∆6 (Subtr. Post.)
Given: ∆3 is a right angle. Prove: ∆l and ∆2 are comp. zL.
Proof: 1. Δl + ∆2 + ∆3 = 180o (The sum of the zL of a Δ is 180o.)
2. Δ3 is a right Δ. (Given) 4. ∆l + ∆2 = 90o (Subtr. Post.)
3. ∆3 = 90o (Def. of right Δ)
5. ∆l and ∆2
are comp, A.. (Def. of comp. zL)
19.
Let x represent the measure of the smaller acute angle.
measure of the larger acute angle,
Then 4x represents the
x + 4x = 90; 5x = 90; x = 18; 4x = 72;
180, 72°
20. Let x represent the measure of the first angle.
measures of the other angles,
Then x - 30 and x + 15 represent the
x + (x - 30) + (x + 15) = 180; 3x - 15 = 180;
3x = 195; x = 65; x - 30 = 35; x + 15 = 80; 650 , 35o, 80o
21. a. If ΔC = 220, then ∆A = 680 and ∆ABD = 22o. b.
If ∆C = 23o, then ∆A = 670 and ∆ABD = 23o.
c.
ΔABD and ∆C are complements of the same angle, ΔA.
17
Key to Chapter 2, pages 59-60 22.
a. If ΔEFG = 40o, then L EGF = 180o - (80o + 40o) = 60°.
Then ΔIFG = 20o,
ΔIGF = 30°, and L FIG = 130o.
b. If ΔEFG = 50o, then ΔEGF = 180o - (80o + 50o) = 50o.
Then ΔIFG = 250,
ΔIGF = 25o, and ΔFIG = 130o.
c. ΔFIG + ΔIFG + ΔIGF = 180°; Δ FIG + ∣ΔEFG + ∣ΔEGF = 180°; ΔFIG + ⅛(ΔEFG + ΔEGF) = 180°; Δ FIG + ⅜(180o - 80°) = 180°; ΔFIG = 180o - 50° = 130o Δ £ c. AC A DC
d. Given: AB ∣∣ DE; AC bisects ΔBAD∙, Prove: AC A DC.
DC bisects ΔADE.
Proof: 1. AB ∣∣ DE (Given)
2. ΔBAD
is supp. to ΔADE. (If 2 ∣∣ lines are cut
by a trans., then s-s. int. ∕s are supp.)
supp. A)
4. (Δ1 + ∆2) + (∆3 + ∆4) = 180° (Subst. Prop.)
ΔBAD∙, DC bisects ΔADE. (Given)
5. AC bisects
6. ∆l=∆2j∆3 = ∆4 (Def. of Δ bisector)
7. 2(∆2) + 2(∆3) = 180o (Subst. Prop.) 9.
3. ΔBAD + ∆ADE = 180o (Def. of
8. ∆2 + ∆3 = 90o (Div. Post.)
10. Δ5 = 90o (Subtr.
∆2 + Δ3 + ∆5 = 180o (The sum of the 1 of a Δ is 180°.)
Post.)
11. AC 1 DC (Def. of 1 lines)
24. 1. ΔA =∆A (Reflexive Prop.)
of one Δ
2. ΔABD = ΔAED (Given)
3.∆C=ΔF(If2A-
are = to 2 A of another Δ, then the third A are =.)
25. 1. Δ JGI = ΔH + ΔI (If one side of a Δ is extended, then the ext. Δ formed equals the sum of the two remote int. A.)
bisects ΔJGI. (Given)
6.
2. Δ1 + ∆2=ΔH + ΔI (Subst. Prop.)
4. Δ 1 = ∆2 (Def. of Δ bisector)
2(∆2) = 2(∆I) (Subst. Prop.)
3. GK
5. ΔH = ΔI (Given)
7. ∆2 = ΔI (Div. Post.)
8. GK ∣∣ HI (If 2 lines
are cut by a trans, and alt. int. ∕s are =, then the lines are ∣∣.) C 26. 1. AB 1 BD; ED 1 AD (Given)
3.
ΔB = ΔADE (Subst. Prop.)
(Def. of Δ bisector)
then the third ∕s are =.)
2. ΔB = 90°; ∆ADE = 90° (Def. of J_ lines)
4. AE bisects ΔBAD. (Given)
5. ΔBAC = ΔEAD
6. Δ BCA = ΔE (If 2 A of one Δ are = to 2 A of another Δ, 7. ΔBCA = ΔECD (Vert. A are =.)
8. ΔE = ΔECD
(Subst. Prop.)
27.
Let A and B be the vertices of the s.-s. int. A. (3x + 3y)o = 180o (S-s. int. A are supp.); (x + y)o = 60°;
ΔSAB + ΔSBA + ΔASB = 180°; 60° + ΔASB = 180°; ∆ASB = ΔRSP = 120°
Key to Chapter 2, pages 64-65
18
(x + y)0 = 60°; (2x + 2y)o = 120°; ΔQAB + ∆QBA + ΔAQB = 180°; 120o + ΔAQB = 180°; ∆AQB = ΔRQP = 60o
Since ΔQRS is an exterior L of ΔARB, ∆QRS = ARAB + ∆ABR = (2x + y)o. Since ∆QPS is an exterior L of ΔAPB, ΔQPS = ΔPAB + ΔABP = (x + 2y)o. Since 120o + 60o = 180o, ΔRSP and ΔRQP are supp. zL∙ Since (2x + y)0 + (x + 2y)0 = (3x + 3y)o = 180o, ∆QRS and ΔQPS are supp. zL .
Hence, in quad. PQRS, opposite angles are supp.
28.
29. 3(180o) = 540o
2(180°) = 360o
30. 4(180o) = 720°
Pages 64-65 ∙ WRITTEN EXERCISES A
1. (3 - 2)180° = 180°
4.
(5 - 2)180° = 540o
7.
360o
2. (8 - 2)180o = 1080°
3. (10 - 2)180° = 1440°
5. (6 - 2)180o = 720o
6. (7 - 2)180° = 900°
c. No
9
15
30
18
36
45
24
180
Each ext. L
40°
24°
12°
20°
10°
8o
15°
2°
Each int. L
140°
156°
16 8o
160°
170°
172°
165°
178°
Number of sides
10.
Sum of int. A - (10 - 2)180o = 1440°; each int. L has measure 1440o ÷ 10 - 144°.
B H. 18°
12. Sum of int. ∕s = (6 - 2)180o = 720°; each int. L has measure 720o ÷ 6 = 120o. (Another solution: 360o ÷ 3 = 120o) 13. a. Sum of int. ∕s
b.
= (8 - 2)180° = 1080°; each int. L has measure 1080° ÷ 8 = 135o.
No
14. a. Sum of int. A = (5 - 2)180° = 540°; each int. L has measure 540° ÷ 5 = 108°.
b.
No
∙.÷y to Chapter 2, pages 68-69 17.
Extend one side of a hexagon as shown. By the definition of a straight angle, ∆2 + L 3 = 180°
if the pentagons and hexagons interlock.
180° - 120° = 60o, so L2 = 60o.
∖___ /
/
But L1 =
'---- '
Also, ∆3 = 108o.
Then ∆2 + ∆3 = 168o and the figures could not interlock to tile a floor. 2500 < (n - 2)180 < 2600; 2500 < 180n - 360 < 2600;
18. Let n be the number of sides.
8 4 2860 < 180n < 2960; 15∣ < n < 16^; n = 16 19. 1. Extend ED and BC to meet at G.
are not parallel intersect.)
(Two lines which
£
2. ΔGDC and ΔGCD are
ext. zk of ABCDEF. (Def. of ext. zk )
is a regular hexagon. (Given)
3. ABCDEF
p
∖ /
/
f∖∖
/7c
4. ΔGDC
360o ÷ 6 = 60°; ΔGCD = 60o (The sum of the ext. zL of a convex polygon, one Δ at
each vertex, is 360o.)
5. ΔG + ∆GDC + ΔGCD = 180o (The sum of the zk of a Δ
6. ΔG = 60o (Subtr. Post.)
is 180o.)
7. ΔB = [(6 - 2)180o] ÷ 6 = 120o (The
sum of the ∕s of a convex polygon with n sides is (n - 2)180°.)
(Add. Post.)
9. ΔB is supp. to ΔG. (Def. of supp. zk )
8. ΔB + ∆G = 180° 10. AB ∣∣ DE (If 2 lines
are cut by a trans, and s-s. int. zk are supp., then the lines are ∣∣.)
20.
1. Extend AB and DC to meet at J. (Two lines which
are not parallel intersect.)
F
E
2. ΔJCB and ΔJBC are
ext. zk of ABCDEFGH. (Def. of ext. zk )
3. ABCDEFGH
'∖p
nk
4. ΔJCB = 360° ÷ 8 = 45°;
is a regular octagon. (Given)
√C
j
Δ JBC = 45° (The sum of the ext. zk of a convex polygon, one ∆ at each vertex, is 360o.)
6.
21. —
5. Δ JCB + Δ JBC + ΔJ = 180° (The sum of the zk of a Δ is 180°.)
7. AB 1. DC (Def. of J_ lines)
ΔJ = 90° (Subtr. Post.)
nx = 360; my = 360; nx = my; — = y; — = ’ j ’ j’ m j’ m x
Pages 68-69 ∙ WRITTEN EXERCISES
A
b. 5x - 9 = 6
£. If 5x - 9 = 6, then x = 3.
2. a. You won't
b. I will
£. If I will, then you won't.
3. a. AB J_ BC
b. ΔABC
1.. a. x = 3
4. a. Connie doesn't go
doesn't go.
= 90°
£. If ΔABC = 90°, then AB J_ BC.
b.Ann shouldn't go
£. If Ann shouldn't go,
then Connie
Key to Chapter 2, pages 68-69
20 5. a. a line contains two points of a plane
b. the line is in the plane
c. If a line is
in a plane, then the line contains two points of the plane.
6. If a figure is a rectangle, then it is a quadrilateral. 7. If a car has poor brakes, then it is a menace on the highway. 8. If a number is positive, then it has two square roots. 9. If x2 - 9, then x = 3 or x = -3.
10. If two numbers are odd, then their product is odd. 11. If an integer is divisible by 2, then it is even.
12. If two angles are complementary, then their sum is 90o.
If the sum of two angles is
90o, then the angles are complementary.
13. If ab > 0, then a and b are both positive or both negative.
If a and b are both positive
or both negative, then ab > 0. 14. If (x - 5)(x + 7) = 0, then x - 5 or x = -7.
If x = 5 or x = -7, then
(x - 5)(x + 7) = 0.
15. If Z-1 and ∆2 are called equal angles, then their measures are equal.
If the
measures of Δ 1 and ∆2 are equal, then ∆l and ∆2 are called equal angles. 16. If the corresponding angles formed by two lines and a transversal are equal, then the
lines are parallel; yes.
17. If two lines are parallel, then alternate interior angles formed by the two lines and a transversal are equal; yes. B 18. a. If a - c = b - d, then a = b and c = d.
b. Answers may vary.
Example: a = 5, b=7, c = l, d = 3
19. a. If two angles are equal, then they are vertical angles.
b. Sketches may vary.
Example: If two lines are perpendicular, then two adjacent angles formed are equal,
but not vertical.
Prove: ΔB = LC
Prove: AB = AC
∙.≡y to Chapter 2, pages 72-74
21
True
Given: Quad. ABCD;
Given: Reg. pent. ABCDE
AB ∣∣ DC; AD ∣∣ BC
Prove: AC = AD - BD
Prove: AB = DC; AD = BC
= BE = CE D l
27.
C
True
Given: Quad. ABCD; AB = BC = CD = DA Prove: AC _L BD Z 28. False
29. True
30. False
31. False
.-^ges 72-74 ■ WRITTEN EXERCISES
2
L If 5a - 17 = 43, then a = 12.
If 5a - 17 / 43, then a / 12.
If a / 12, then
5a - 17 / 43. 2. If AB = CD, then AB ∣∣ CD.
If AB / CD, then AB is not parallel to CD.
If AB is not parallel to CD, then AB / CD. 3. If a quad, is a parallelogram, then its diagonals bisect each other. parallelogram, then its diagonals do not bisect each other.
If a quad, is not a
If the diagonals of a quad,
do not bisect each other, then the quad, is not a parallelogram. 4. If Juanita is not well, then she is not here.
If Juanita is well, then she is here.
Juanita is here, then she is well. 5.
True; true; true; true
7.
False; false; false; false
6. True; true; true; true
If
Key to Chapter 2, pages 72-74
22
B 12. If a person is a senator, then he or she is at least 30 years old.
14. a. No conclusion possible b. Joan Baker is at least 30 years old,
c. No conclusion possible
d. Hank Foster is not a senator.
15. a. No conclusion possible
b. I am not guilty.
c. I was at the scene of the crime
d. No conclusion possible 16. a. Juan loves geometry.
b. No conclusion possible
c. No conclusion possible
d. Walter is not one of my students. C 17. a. Ed is at least 16 years old.
b. No conclusion possible
c. Jose isn't eligible.
d. No conclusion possible 18. contrapositive
19. inverse
20. converse
b. Yes; a point that is not inside p cannot be inside q. c. Yes
22. Given: Δl + ∆2 + ∆3- 180 .
(Given)
Prove: / ∣∣ m.
Proof: 1. Δl + ∆2 + ∆3- 180
2. Δl + ∆2 + ∆4 = 180o (The sum of the 1 of a Δ is 180o.)
3. Δ1 + ∆2 + ∆3=Δ1 + ∆2 + ∆4 (Subst. Prop.)
4. ∆3 - ∆4 (Subtr. Post.)
5. / ∣∣ m (If 2 lines are cut by a trans, and corr. A. are -, then the lines are ∣∣.) 23. Given: a is even. (Def. of even number) (Def. of even number)
Prove: a8 is even.
Proof: 1. For some integer n, a = 2n.
2. a8 = 4n8 = 2(2n8) (Def. of square)
3. a8 is even.
Key to Chapter 2, pages 75-76
23
Pages 75-76 ■ EXTRA 1. Many wordings are possible for the following postulates and theorems.
P : A committee has at least two persons; a club has at least three persons not all in
one committee; the set of all persons has at least four persons not all in one club. Pp: There is exactly one committee containing any two persons.
P3: There is at least one club containing any three persons, and there is exactly one club containing any three persons not all in one committee.
P4: If two persons are in a club, then the committee that contains the persons is
contained in the club. P5: If two clubs have at least one member in common, then the member(s) in common form a committee.
T. 1-4: If two committees have at least one person in common, then they have exactly one person in common.
T. 1-5: If there is a committee and a person not in the committee, then exactly one club contains the committee and the person.
T. 1-6: If two committees have at least one person in common, then exactly one club contains the two committees.
2. a. Postulates 1-5 are all satisfied.
b. A three-person committee: Bea, Xavera, Carlos.
Four-member clubs: Ann,
Bea, Xavera, Carlos; Bea, Xavera, Carlos, Dick 3. Given: Five points not all in one plane.
five planes.
Prove: There are at least eight lines and
Key steps of proof: Let A, B, C, and D be four of the five points not
all in one plane.
Let X be the fifth point.
Case 1: X lies on one of the lines determined by A, B, C, and D, suppose line (BC).
Then there are eight lines: (AB), (AC), (AD), (AX), (BC), (BD), (CD), and (DX).
There are five planes: (ABC), (ABD), (ACD), (ADX), and (BCD).
Case 2: X lies on one of the planes, but not on any line, determined by A, B, C, and D, suppose plane (ABC).
lines (BX) and (CX).
Then there are the eight lines listed for Case 1, and also
There are the five planes listed in Case 1, and also planes
(BDX) and (CDX).
Case 3: X does not lie on any plane determined by A, B, C, and D. the ten lines referred to in Case 2.
Then there are
There are the seven planes referred to in Case 2,
and also planes (ABX), (ACX), and (BCX).
24
Key to Chapter 2, pages 77-78
4. If there are five persons not all in one club, then there are at least eight committees
and five clubs. 5. The geometry pictured at the left, in the text, satisfies the statement.
For example,
if you consider line (AB) and point D outside (AB), then (CD) is the one line parallel to (AB).
The geometry pictured at the right, in the text, does not satisfy the For example, if you consider line (AB) and point D outside (AB), then
statement.
there are two lines, (DX) and (DC), parallel to (AB).
Pages 77-78 ∙ CHAPTER REVIEW
1.
∆7, Δ 3; ∆2, ∆6
2. ∆l, ∆3∙, ∆2, ∆4j ∆5, ∆7∙, ∆6, ∆8
3.
∆2, ∆3j ∆6, ∆7
4. Yes
6.
∆3 = 180o - 1150 = 65o
9.
Since ΔE and ΔEAB are supplementary, ED ∣∣ AB.
5. ∆5 = ∆6 = 78°
7. ∆3 = 180o - ∆2 = 180o - Δ 1 = 70o
8. 90°
10. ΔADE = 180o - (105° + 38o) = 180o - 143o = 37° 11. ΔABC = 180° - (380 + 25° + 12o) = 180o - 75o = 105o 12. ∆ADB = 90o - (25o + 12°) = 90o - 370 = 53° 13. ∆l = 180° - (690 + 40o) = 180o - 109o = 710 14. ∆3 = ∆l + ΔB = 690 + 69o =
1380
15. Δ1=∆3-ΔB = 141o - 690 = 720
16. Δl + ∆2 + ∆3 = 180°; (2x +7) + (x + 23) + 69 = 180;
3x + 99 = 180;
3x = 81; x = 27
17. Sum of int. zL = (10 - 2)180o = 1440°; each int. Δ has measure 1440° ÷ 10 = 144o. 18. Sum of ext. zL, one Δ at each vertex, is 360°; each ext. Δ has measure
360o ÷ 10 = 36°. 19. Sketches may vary.
∕↑ ∖
/
20. Since Δ FAE and ΔAEF are exterior angles of ABCDE, ΔFAE = ∆AEF = 360o ÷ 5 = 72°.
ΔF + ΔFAE +
∆AEF = 180°; ΔF = 180o - (720 + 720) = 36°
21. If a polygon is a rectangle, then it is a square.
22. False.
23. a number is greater than 20
24. If a whole number is divisible by 5, then its last digit is 0 or 5.
If the last digit of a
whole number is 0 or 5, then the number is divisible by 5. 25. No conclusion possible 28.
No conclusion possible
26. I am not hungry.
27. I am grumpy.
:: Chapter 2, pages 78-79
25
: 7o-79 ∙ CHAPTER TEST
!_ always
3. sometimes
2. never
:. sometimes
7. sometimes
4. sometimes
5. never
8. never
9. 4x + llx = 180; 15x = 180; x = 12
10. _1 = 180o - 1250 = 55°; Δ 2 = 180o - (410 + 55°) = 84°; ∆3 = 180o - L2 = 96°;
∆4 = ∆l = 55°; ∆5 = ∆2 = 840
n. δi
= 180o - 40o = 140°; ∆2 = 40°; ∆3 = ∆l = 140°; ∆4 = ∆2 = 40°; Δ 5 = 40o
12. ∆1 = 90o -• 51o = 39°; Δ2 = 51°; ∆3 = 90o - ∆2 = 39°; ∆4 = 90o - ∆3 = 51°; ∆5 = 90°; ∆6 = ∆l = 390 13. ∆l = 120°; ∆5 = 90°; ∆3 = 120° - ∆5 = 30°; ∆2 = 180o - (∆l + ∆3) = 30°; ∆4 = 120o - L2 = 90o
14. a. x = 4
b. x2 =16
1_5. a. Jill's car is shiny.
d.
c. If x8 = 16, then x = 4.
b. No conclusion possible
d. If x2
16, then x /
c. No conclusion possible
Marcia's car isn't new.
16. 1. BF bisects ∆ABCj CE bisects ∆BCD. (Given) ∣∆BCD (Def. of L bisector)
3. AB ∣∣ CD (Given)
2. ΔFBC = ∣ΔABCj ΔBCE
4. ΔABC = ΔBCD (If 2 ∣∣
lines are cut by a trans., then alt. int. zL are =.) 5. ^ΔABC = ∙∣∆BCD (Mult.
Post.)
6. L FBC = ΔBCE (Subst. Prop.)
7. BF ∣f CE(If2 lines are cut by a
trans, and alt. int. zL are =, then the lines are ∣∣.)
CHAPTER 3 ∙ Congruent Triangles
Pages 83-84 ■ WRITTEN EXERCISES A
1. AB
2. AT
4. AT
3. BU
7. Corr, parts of ≡= A are =.
5. FA
6. AG J). ΔFTA
8γ ΔUBG
10-15. ΔS = AD; ΔL = AW; AO = AN; SL = DW; LO = WN; SO = DN 16. Yes; A**Y; B 8x = 7(15 - x),
o
ID “ X
8x = 105 - 7x, 15x = 105, x = 7; CE = x = 7; EB = 15 - x = 8
23. Let CD = x; then DA = 18 - x.
x
yθ-— =
R
9x3
'fg ~∑ x = 4’ 4x = 3(18 - x),
R
9
4x = 54 - 3x, 7x = 54, x = 7|; CD = x = 7|; DA = 18 - x = 10∣∙
24. EB = 28 - 12 = 16.
Let CD = x; then DA = 21 - x.
--
4x = 3(21 - x), 4x = 63 - 3x, 7x = 63, x = 9. 25. e = c + d; a
9
∙ --o or -r Δ — = ⅛ 3z = 36; z = 12. z 2’ ’
21
16 or -z∙ 4 16. The scale factor is γx, " 26 . 4. 4z . ,8. z lβl.3
20 = ⅜ 4x = 60; x = 15. X
= ⅛ 3y = 72; y = 24.
12 = ⅛ 8x = 60; x = 7∙∣∙ X 0
40 = |; 8y = 200; y = 25. y
z
0
X
3x = 42; x = 14.
=
0
2
48 8 17. The scale factor is -z77, or -z∙
⅛ = |; 5z = 192; z = 38∣∙ 35 7 18. The scale factor is 2θ> or -ξ∙ y = 16y∙
19. Since
0
28 X
=
7x = 112 ; x = 16.
⅛≈⅞ 4z ≡ 182; z = 45∣. , = -∣> the scale factor is
7 29 = ⅛ 7y = 116; y
C'(15, 11); D,(ll, 17); E,(5, 11) or
C,(15, -1); D,(ll, -7); E'(5, -1) 20. Since
= -∣, the scale factor is -∣∙
C,(-ll, 6); D,(-7, 12); E'(-l, 6) or
C'(-ll, -6); D'(-7, -12); E,(-l, -6)
21. Trap. QRON ~ trap. TSOP.
2. ΔRON = 70o (If 2 ∣∣ lines are cut by a trans., then s-s. int. A.
110o (Given)
are supp.)
Key steps of proof: 1. ΔQ = ΔTj ΔN = Z.P; ΔR =
3. ΔSOP = 180° - 110o = 70o (Def. of straight Δ)
(If 2 ∣∣ lines are cut by a trans., then s-s. int. A. are supp.) / p∩M - /
ςp
9
9
g⅛ = 7; SP = 2SO - 2SP; 03SP = 2SO; SP = ⅛SO. 1
- or
Similarly, RP = 7RM and QP = -7QN. O o
28. 1. AB, A'B,, and CO are J_ to BB'; AC ∣∣ BO (Given)
2. AB ∣∣ CO (If a trans,
is .L to 2 lines, then the 2 lines are ∣∣.) 3. Quad. ACOB is a parallelogram. (Def. of O)
of I. lines)
6. ΔA'B'0 ~ Δ ABO (AA Sim. Post.)
ΔCA'O = ΔCA'0 (Refl. Prop.)
trans., then corr. A are =.)
∩R'
= -θg- (Corr, sides
11. Δ FOA' ~ Δ CAA' (AA Sim. Post.)
■^7 (Corr, sides of ~ A are in proportion.)
15.
ΩA'
10. ΔA'FO = ∆A'CA (If 2 ∣∣ lines are cut by a
OA'
property of proportions)
7.
8. OA' ∙ OB = OA ∙ OB' (A property of proportions)
of ~ A are in proportion.)
9.
5. ΔB = ΔB' = 90° (Def.
4. ΔAOB = ΔA'OB, (Vert. A. are =.)
12. ~ς =
13. AA' * OF = OA’ ∙ AC (A
14. AC = OB (Opp. sides of a t∑j are =.)
AA’ ∙ OF = OA' ∙ OB (Subst. Prop.)
(Trans. Prop, or Subst. Prop.)
16. OA, ∙ OB = OA , OB' = AA' ∙ OF
17. OA' + OA = AA' (Given)
10 OA' OA_________ AA' ,tλ. ., 18∙ OA’ ∙ OB + OA ∙ OB' AA’ ∙ OF ^D1V‘ P°St^
19∙ ⅛+ ⅛ = ⅛ (By a⅛ebra>
70
Key to Chapter 5, pages 185-189
Page 185 ∙ CALCULATOR KEY-IN
To the nearest millimeter, 4⅞ = v? ≈ 1.5714286, ⅛f = ⅛Γ = 1∙75, and TF = H AB 14 BC 8 ’ AC 22 1.6363636 2. Approximately 1.618034
3. a. 1 1, 2, 3, S,8, 13, 21,34, 55, 1
b. 7
6765 34 21
—
1
3
R
1
Z
o
=1; 7 = 2; -5= 1.5;
144, 233, 377, 610, 987,1597, 2584,
89,
?
—O
4181,
91
19
⅛ = 1.6; ⅞ = 1.6; ⅛ = 1.625; 4⅛ =1.6153846; □
o
lo
15Q7
9R7
9Sft4
1.6190476; ... ; ~ = 1.6180328; -⅛gy = 1.6180344; γ~ = 1.6180338;
⅛∣i = 1.6180341; ⅛∣∣ = 1.618034
c. The values of the ratios approach the
golden ratio
4. Approximately 0.61803398
Pages 188-189 ∙ WRITTEN EXERCISES δ
a∙
⅞ = I= i; 3x = 12; x = 4
r
9 5 _ 10 - x 18
θ5 5x = 45; x = 9
o x 10 2 o n. o 3∙ 12 = 15 = 3’ 3x = 24; x = 8
R 4 2L _ 15 _ -; 7x = 70; x = 10 - 14 21 7’
*∙ ⅛ = It = 2’ 2x = 20; x = 10
θ∙ 1 = ⅛ =
3x = 42; x = 14
7. ΔCAB ~ ΔGTEj If an L of one Δ is = to an Δ of another Δ and the sides incl. those
∕s are in proportion, then the A are 8. ΔCAB ~ ΔKDRj If the sides of 2 A are in proportion, then the A are 9. Δ CTM ~ Δ CAB; If an Δ of one Δ is = to an L of another Δ and the sides incl. those
∕⅛ are in proportion, then the A are 10. ΔOYZ ~ AOXW; If an Δ of one Δ is = to an L of another Δ and the sides incl. those ∕s are in proportion, then the A are
11. No; if the A are ~, the sides must be in proportion. ., 6 3 7 . 8 sides are -z = -τ ’ „ > and 777 - i 8 4 9 10 5 6 _ 2 9 12. v Yes’ 9 3’ 13.5
B 13. a. ΔABC ~ ΔAED
The ratios of the corresponding
2 12 _ 2 2. 3’ 18 = 3’ ^ιe sides are in proportion.
b. SAS Sim. Thm.
_ BC, 8 _ 5 e q r¼τ7, — iλa∙ c. AB AE ^ DE’ 32 ^ DE’ 8DE ^ 16°,
DE = 20
14. Answers may vary. EF = 6
Examples: ∆EFD ~ ABCD; ΔACB ~ ABCD; ΔACB ~ ΔEFD∙,
71
to Chapter 5, pages 188-189
3.4h - 10.71; h = 10.71 ÷ 3.4 =
-py =
15. Let h be the height in meters of the pole,
3.15; the pole is 3.15 m tall.
16. Let h be the height in centimeters of the image on the film.
500h = 300;
=
h = 0.6; the image on the film is 0.6 cm tall.
17.
--
∩∆, LzA
∩R'
8 1
= ⅛ = 7» so
∆A'OB, = ∆AOB and
VzJd
ΔA'OB, ~ ΔAOB by the SAS Sim. Thm.,
A,B, = -θp OA' = y∙ 3 and, -yy
o∙ ∙1 1 Similarly,
ΔB'OC' ~ ΔBOC, ΔC,OD' ~ ΔCOD, and
ΔA'OD, ~ ΔAOD, so
and DA - -3. DA 1
= p pp = p
The sides of A'B'C'D' and Also, ΔOAB =
ABCD are in proportion.
Z-OA'B, and ΔOAD = ΔOA'D, (Def. of ~A).
By the Addition Postulate, ΔA = ∆A'.
ΔD = ∆D'. 18.
Similarly, ΔB = ΔB,, ΔC = ∆C', and
Then, by definition, A’B’C'D’ ~ ABCD.
1. Choose X on DE so that DX = AB. (The Ruler Post, guarantees such a point.) 2.
Draw XY ∣∣ EF. (Through a point outside a line, exactly one ∣∣ can be drawn to the
3. ∆l = ∆E (If 2 ∣∣ lines are cut by a trans., then corr. A. are =.)
line.) 4.
Uh∣
(Subst. Prop.)
= uh
uh
Dr
uh
AC ∙ DF = DY ∙ DF (A prop, of proportions)
12.
ΔA = ΔD (Given)
16.
19.
Uh∣
9 ∙ ⅛S = S⅞ (Trans. Prop, or Subst. Prop.)
10.
parts of ≡ A are =.)
DV
6.Dll κ⅜ =Uh ⅛≡
AB AC 7. •=-=; = —= (Given)
(Corr, sides of ~ A are in proportion.)
8.
DX
5. Δ DXY ~ ΔDEF (AA Sim. Post.)
ΔD = ΔD (Refl. Prop.)
11. AC = DY (Div. Post.)
13. ΔABC ≡= ΔDXY (SAS Post.)
14. ΔB = ∆l (Corr.
15. ΔB = ΔE (Subst. Prop, or Trans. Prop.: Steps 3 and 14)
ΔABC ~ ΔDEF (AA Sim. Post.)
1. Choose X on DE so that DX = AB. (The Ruler Post, guarantees such a point.) 2.
Draw XY ∣∣ EF. (Through a point outside a line, exactly one ∣∣ can be drawn to
the line.)
3. ΔDXY = Z.E; ΔDYX = ΔF (If 2 ∣∣ lines are cut by a trans., then
corr. zL are =.)
4. ΔDXY ~ ΔDEF (AA Sim. Post.)
sides of ~ A are in proportion.) 7* DE = EF = DF (Given)
5.
DX uh∣
=
XV h∣h
DV
= uh ALL (Corr.
AB XY DY 6. 7= = -== = κ≡ (Subst. Prop.) uh∣
h∣h
uh7
8∙ EF = EF5 DF = DF (TrajlS- PrOp* °r ≡ubst∙ Pr°P>)
9. XY ∙ EF = BC ∙ EF; DY ∙ DF = AC ∙ DF (A prop, of proportions)
72
Key to Chapter 5, pages 192-195
10.
XY = BC; DY = AC (Div. Post.)
12.
ΔB = ADXY; ΔC = ΔDYX (Corr, parts of ⅛ A are =.)
11. ΔABC ≡ ΔDXY (SSS Post.)
14. ΔABC ~ ΔDEF (AA Sim. Post.)
ΔC = ΔF (Trans. Prop, or Subst. Prop.) 1. AB ∣∣ DC; AD ∣∣ BC (Def. ofO)
20.
2. Δ PAX = AXCQ; ∆APX = ΔXQC (If 2 ∣∣
lines are cut by a trans., then alt. int. ∕s are =.)
Post.)
4.
DY
13. ΔB = Z.E;
YΔ
= zτ77 (Corr, sides of ~ XL
3. ΔAPX ~ ΔCQX (AA Sim.
are in prop.)
5. ΔASX = ΔXRCβ,
ΔSAX = ∆RCX (If 2 ∣∣ lines are cut by a trans., then alt. int. Λ are =.)
6. ΔASX ~ ΔCRX (AA Sim. Post.) 8.
PX
SX 7. nA =ττ =XA τ777 (Corr, sides of ~ A are in prop.) XL
ςJX
9. ΔPXS = ΔQXR (Vert. A are =.)
7w =rtΛ ≡S7 (Trans. Prop, or Subst. Prop.)
10.
ΔPXS ~ ΔQXR (SAS Sim. Thm.)
Pages 192-195 ∙ WRITTEN EXERCISES
A
I* _∙
Incorrect
£. Incorrect
b. Incorrect
d . Correct
e. Correct
f. Correct 2. a. Correct — —
c. Incorrect —
b. Incorrect —
= τ⅞ = 4; 3x = 24; X = 8 3 -∙ 12 10 0
d. Correct —
4.
X = 20
= |; 7x = 60; x = 8γ
5. 24,"^ x = 12 = 3’ 3x = 120 " 5x; 8x = 120; x = 15
1 = ∣ = 2χ =i2;χ =6
6.
8. ∣ =
7x = 5x + 10; 2x
2∙ ⅞ = ⅛ = ⅛ 2x =12; x =6
= 10; x = 5
12∙ ∣ -1∏p- ∣ - τ ’ r *= 18
12. ∣ = ⅞ =
9. ∣ = ~ = |; 2x = 30;x = 15 n∙ ⅛ = l>8x = 30;x ≈10
x = lβ
13. 2l~~ = 2⅛A 19x + 19 = 1≡x + 52; 6x = 33; x = 5∣ — a5 χ x “ T = l; 70 - 2x = 5x'> 70 =7x'> χ = ιθ B 15. 1. j ∣∣ k ∣∣ λf (Given)
2. Draw the auxiliary line shown in the diagram on page 193.
(Through any two points there is exactly one line.) Proportionality Thm.)
0 a xx c ,rτ, . , 3∙ b = ? y = d (Tna∏glθ
b = d (Trans∙ proP∙ or Sub st. Prop)
AX AY AX AY 16. 777 = 777 is equivalent to — ---- — - — ---- — ’ which is equivalent to AX (AY + YC) 24.XJ AU AX + 2xxj AY ÷ YU
----
= AY(AX + XB); AX ∙ AY + AX ∙ YC = AX ∙ AY + AY ∙ XB; AX ∙ YC = AY ∙ XB. AX
AY
77 = X777 is equivalent to AX ∙ YC = AY ∙ XB. ∙ΛX3 AX _ AY XB YC '
AX
AV
Then zAχ5 7-7 = AU 777 is equivalent to
Key to Chapter 5, pages 192-195
73
17-22. Lengths which are not given cannot be found based on given information.
17. — —
OQ = 8 + 5 = 13; ∣ = ⅛ 8JK = 50, JK = 6⅛ OK = 10 + 6⅛ = 16⅜ 5 JK 4 4 4 OP
14
12 = 21 =
9
30p = 24> op = 8; PQ = 12 - 8 = 4
— ⅛ = I= 1, PQ "6; °Q =6 +6 = 12; oκ = 8 + 8 = 16; ⅛⅛ = ⅞ = I’ qk = 2o
20. OK = 36 + 24 = 60; ξ½ = ⅛ = ∣, 5 PJ = 150, PJ = 30 --
DU
21. OQ = 10 + 8 = 18;
OU
D
= ∣∣ = |» 9OJ = 135, OJ = 15; JK = 27 - 15 = 12;
1∣ = If = ∣, 5QK ≈ 135, QK = 27 1R
9∩
R
P¼)
1O
4
22∙ ⅛K = Tc = 7’ 5 PQ = 60, PQ = 12; OQ = 15 + 12 = 27; OK = 20 + 16 = 36;
--
¾hH =1’ 5PJ = 200, PJ = 22∣
23∙
—
λ∩ jκ∏Λ = ⅛ = ⅜ 3JK = 80 - 2 JK; 5JK = 80; JK = 16 ztU - JK
Jo
0
ai∙ io ≈lδ-⅛ 1SPQ≈1OO1 PQ = 10∣ 25∙ ⅛B = ⅛ = r, .^abλd = ⅛5 5AB = 468 - 4AB; 9AB = 468; AB = 52 m; BD = BD oU D 111 - AB D
—
117 - 52 = 65.
⅛ ≈ H = ∣! 65⅛ =
2BC = 195 - SBC; 5BC = 195;
BC = 39 m; CD = 65 - 39 = 26 m
26. Given: Three parallel planes as shown on page 194.
—
Prove: =r≈ = -== ∙ dL
h∣i,
Proof: 1. Draw DC intersecting the "middle" plane atX; draw AD, BX, XE, and CF. (Through any two points there is exactly one line.)
2. There is exactly one plane
which contains AB and DX; there is exactly one plane which contains CX and FE. (If two lines intersect, then exactly one plane contains them.)
3. AD ∣∣ BX;
XE ∣∣ CF (If 2 ∣∣ planes are cut by a third plane, then the lines of intersection are ∣∣.) 4, BC = XC; XC = EF (Triangle Proportionality Thm.)
Prop, or Subst. Prop.)
Given: ΔABC', ⅛∣ = fg. Prove: DE ∣∣ BC.
= ^g (Given)
Proof: 1. 3.
2. ΔA = ΔA (Refl. Prop.)
ΔADE ~ ΔABC (SAS Sim. Thm.)
(Def. of ~
)
4. ΔADE = ΔB
5. DE ∣∣ BC (If 2 lines are cut by a
trans, and corr. ∕s are =, then the lines are ∣∣.)
5* BC = EF (TranS-
74
Key to Chapter 5, page 197
= γ∣ ∙
28.
=
Key steps of proof: 1.
(If 3 lines intersect 2 transversals,
2. ~p =
then they divide the transversals proportionally.) n
JM _ 7 (Subst. Prop.) ' MQ 12
JQ = 12 19 ∕(A λ prop, off MQ
r MQ _ 12 °∙ JQ 19 (A prop, of proportions)
proportions) Sim. Post.)
4 JM + MQ = 7 + 12 MQ 12
(Given)
6. Δ QMN ~ Δ QJL (AA
7. MN = MQ = 12 19 (Corr, sides oft JL JQ
are in proportion.)
and BM ∣∣ CX. (Through a point outside a line exactly one ∣∣ can be
29. 1. Draw AN
9 AX = AP pM (Triangle Proportionality Thm.) • XB
drawn to the line.)
ΔMPB (Vert. A. are =.)
line are ∣∣ to each other.)
alt. int. ∕s are =.)
4. AN ∣∣ BM (Page 148: In a plane, 2 lines ∣∣ to a third 5. ΔANP = Z-MBP (If 2 ∣∣ lines are cut by a trans., then
6. ΔAPN ~ ΔMPB (AA Sim. Post.)
7. ΔBYM = ΔCYP
8. Z.MBY = L PCY (If 2 ∣∣ lines are cut by a trans., then alt.
(Vert. ∕s are =.)
9. ΔBYM ~ ΔCYP (AA Sim. Post.)
int. A. are =.)
10. ΔCZP = ΔAZN
11. ΔCPZ = ΔANZ (If 2 ∣∣ lines are cut by a trans., then alt.
(Vert. A. are =.) int. A. are =.)
13. ⅛∣ = ⅛⅛
12. ΔCZP ~ ΔAZN (AA Sim. Post.)
1V1XD
1V1-tr
CZ PC AZ = AN (C0rr* sides of ~
are in proportion.) 15 AX XB
Subst. Prop.: Steps 2 and 13)
BY YC
Post, and Algebra)
Page 197 ∙ EXTRA 1.- c
2. d
3. b
4. a
6. {0, 4, 6, 9}; [1, 2, 3, 5, 7}; {8}
7. Drawings may vary.
b.
3. ΔAPN =
5. d
14.
CZ _ AN ZA MB
τ∖
= MB PC
,A,TQ
= v√
X
Jrv
(Trans. Prop, or
PC _ 1 1, AN 1 (Mult‘
75
Key to Chapter 5, pages 198-201 Pages 198-199 ∙ CHAPTER REVIEW 3
—3. 5x = 28; x - 5τ5
2. 3:2:4
1. 7x:4y
x + 4x = 180; 5x = 180; x = 36;
4. Let x and 4x represent the measures.
4x = 144; 36°, 1440 5. 9a
7. -∣
6. §
— ⅛4 =
8. γ
10. ΔMJK
9. GA
12∙ ⅞ = ⅞ = I 5x = 72; x = 14t
3AC = 48; AC = 16
13. Yes, by the AA Sim. Post.
14. ED; EC; DC
15. ∣∣ =
16. Yes, by the SAS Sim. Post.
AB ∙ EC = AC ∙ ED; AC
17. No, corr. sides are not in proportion.
18. No, corr. sides are not in proportion.
19. Yes, by the SSS Sim. Thm.
20. AB
21. W = ⅞ = ⅜ 9CD = 120'> CD = 1⅛ Δλ± 1O «7 o
—∙ XU = t = ⅜ 2AC = 30; AC = 15
------
Pages 200-201 ∙ CHAPTER TEST 11. 32
22. _A , .,., ALCL a,b, Λ
4.
5x = 84; x = 16∣
9.
1. Given
33. 6y = 2θ, y 5.
1, z6 = 42 _ 2
9.
6. AA Sim. Post.
12. 7. CB
9Q
r7
40 = 10
2. If 2 ∣∣ lines are cut by a trans., then alt. int. zL are =. 5. Corr, sides of ~ A are in proportion.
4. AA Sim. Post.
3.
Vert, ∕s are =.
6.
A property of proportions
7. A property of proportions
10. = ?? = ⅛ 90 = 72 + 4x; 4x = 18; x = 4∣ — 18 + x 25 5’ 2 12∙ τ = B = ⅜ 3x = 72; x = 24
∏. f = ⅛ 2∙8x = ?; χ = 2.5
13.
a. ΔGDC, ΔDFA
b. Δ CED
14. 1. WX ∣∣ PQ; TS ∣∣ PR (Given)
c. Δ CEG 2.
ΔXSY = AQRP; ΔSXY = ΔRQP (If 2 ∣∣ lines
are cut by a trans., then corr. zL are =.)
3. ΔSYX ~ ΔRPQ (AA Sim. Post.)
Page 201 ∙ ALGEBRA REVIEW L
(x - 3)(x - 1) = 0; x
= 3 or x = 1
2.
(x - 6)(x + 1) = 0; x = 6 or x = -1
3.
(x - 8)(x - 1) = 0; x
= 8 or x - 1
4.
(x - 5)(x - 3) = 0; x = 5 or x = 3
5.
t≡ + 3t - 54 = 0; (t -
6)(t + 9) = 0; t6
6 or t = -9
6. r≡ - 4r - 32 = 0; (r - 8)(r + 4) = 0; r - 8 or r = -4
Key to Chapter 5, page 201
76
7.
z2 - 6z + 9 = 0; (z - 3)(z - 3) = 0; z = 3
8. v2 - 2v - 24 = 0; (v - 6)(v + 4) = 0; v = 6 or v - -4 9. (2x - l)(x - 2) = 0; x = ∣ or x = 2 1
10. (3y - l)(y - 3) = 0; y =
or y = 3
3
11.. (3p + l)(2p - 3) = 0; p = - 2 or p = 2 12.
5
3
20n2 - 13n - 15 = 0; (4n - 5)(5n + 3) = 0; n = -ξ or n = - g
13. x
-2 ± √4 + 12 = -2 ± /16 = -2 ± 4. x=-3orx=l 2 2 2 ’
14. x
-4 ± √16 - 12 _ -4 ± √^4 = -4 ± 2. x = -3 or x = -1 2 2 2
15.
-7 ± √ 49 - 12 -7 ± /37 x ≈ ------- 2--------- = ----- 2
7 ± √ 49 - 12 _ 7 ± /37 2 2
,r7 17.
-5 ± √25 - 8 -5 ± /17 x =------- 3-------- =----- 4-----
3 ± √9 + 120 = 3 ± /129 6 ' 6
19.
-9 ± √ 81 - 24 -9 ÷ √57 x =-------- θ--------- =----- θ-----
3 ± √9 + 32 = 3 ⅛ ∕4T 16 16
CHAPTER 6 ∙ Right Triangles Page 206 ∙ WRITTEN EXERCISES A
1. √6~÷^6 =6
2. 5∕3^÷^3 = 5(3) = 15
3. j⅜∙ ∣ = ∣
4. ∣√5 ∙ 5 = ∣(5) = 3 — 3 3
5. √5 ∙ 2 ∙ 2 = 2√5^
6. √3 ∙ 2 ∙ 2 = 2√^3
5√3 ∙ 2 ∙ 2 = 10√3^
7.
8. ∣√ 3 ∙ 10 ∙ 10 = 5√3^
m /2 —∙ √ 5
. 5 _ ∕Tθ = √10 ' 5 √ 25 5
3 √7 _ 3√7 —- √7 ' √7 ^ ?
19 5 — √8
√2^ _ 5√^2 _ 5√^2 √2^√16 4
13. - = ⅛ x2 = 36; x = 6 --- X 1Z
14. = ⅜ x2 = 100; x = 10 --- X
α ∕T I _ /1 _ √1 — √3*3 v9~3
15. - = ⅜ x2 = 7 ■ 36; x = √7 ∙ 36 = 6√7^ 16. - = #; x2 = 35; x = √35 ι-■ X ∂O X *
18. - = ⅛ x2 = 75; x = √ 3 ∙ 25 = 5√3^ — x 15’
17. θ = ⅛ x≡ = 84; x = √21 ∙ 4 = 2√21 — x 14’ ’ B 19.-= x2 = 3 ∙ 9; x = √3"÷^9 = 3√3. — — x 9’
— y
⅜ y2 = 36; y = 6.
12 z
z. 9’
z2 = 108; z = √3 ∙ 36 = 6√^3 20. - = k; x2 = 7 ∙ 9; x = √T~r^9 = 3√7^. — x 9’ ’
— = £; y2 = 7 ∙ 16; y = √7 ∙ 16 = 4√T. y 7’ j ’ j
— = ⅜5 z2 = 16 ∙ 9; z = 4 ∙ 3 = 12 z 9’ ’ 21. — = ⅛ — x 0 ~x y = √100
x≡ = 36 ∙ 64; x = √ 36 ∙ 64 = 6 ∙ 8 = 48. ∙ 36 = 10 ∙ 6 = 60.
j
22. ∣ =
3x = 36; x = 12.
23. ⅞ = ¾ 4x = 64; x = 16. — 8 4
~ = ⅛ y2 = 100 ∙ 36; y uu
— = ⅛ z2 = 100 ∙ 64; z = √ 100 ∙ 64 = 10 ∙ 8 = 80 z 64’
y = 12 - 3 = 9. y = 16 - 4 = 12.
|; z2 = 108; z = √ 3 ∙ 36 = 6√^3
— = ⅛ zs = 192; z lz
z = √3 ∙ 64 = 8∕3^ 24. = -; 9x = 36; x = 4. — 6 x’ ’
y = 9 - 4 = 5. j
C 25. (Use the figure for Thm. 6-1.) Prove: AB ∙ DC = AC ∙ CB.
- = |; z2 = 5 ∙ 4; z = √5 ∙ 4 = 2√^5 z 5’
Given: Rt. ΔABC with rt. Z.ACB; CD J_ AB.
Proof: 1. ΔACB is a rt. L∖ CD J_ AB. (Given)
ΔACB ~ Δ ADC (If the altitude is drawn to the hypotenuse of a rt. Δ , then the AB CB 2 A formed are ~ to the original Δ and to each other.) 3. (Def. of 2.
4. AB ∙ DC = AC ∙ CB (A prop, of proportions)
77
78
Key to Chapter 6, pages 208-210
26. a. CM = AM; CM - MB (The midpt. of the hypotenuse of a rt. Δ is equidistant from the 3 vertices.); 2(CM) = AM + MB = AB = AH + BH; CM = -- + —; CM is the
arithmetic mean between AH and BH.
CH is the geometric mean between AH and BH.
(When the altitude is drawn to the hypotenuse of a rt. Δ, the altitude is the geometric CH J_ AB and the perpendicular
mean between the segments of the hypotenuse.)
segment from a point to a line is the shortest segment from the point to the line.
Therefore, CM > CH. b. (r - s)*2 >0; r8 - 2rs + s8 > 0; r8 + 2rs + s8 > 4rs; (r + s)s > 4rs;
(r + s)2 . (r + s)s r + s /---—4-^ > rs> 2s > rs"> ~2~ > √ r≡
Pages 208-210 ♦ WRITTEN EXERCISES
A
1. 25 = 9 + b2; b2 = 16; b = √^16^ = 4 2. 132 = as + 123; 169 = a≡ + 144; a≡ = 25; a = √^25^ = 5
3. cs = 62 + 8s; c8 = 36 + 64 = 100; c = √lθθ^ = 10
5. c8 = 8s + 152 = 64 + 225 = 289; c = √289 = 17
6. 618 = 11s + b8; 3721 = 121 + b2; b2 = 3600; b = √3600 = 60 7. c2 = 4s + (4√3)8 = 16 + 48 = 64; c = √64 = 8
a2 = 25; a = √^25
8. 102 = a≡ + (5√^3)35 100 = a2 + 75;
= 5
9. 138 = x2 + 53; 169 = x8 + 25; x2 = 144; x = √144^ = 12
10. 108 = x3 + 68; 100 = x3 + 36; x2 = 64; x = √64 = 8 11. 172 = x3 + 158; 289 = x2 + 225; x2 = 64; x
= √64 = 8
12. χ2 = 72 + 78; x2 = 49 + 49 = 98; x = √98 = 7√T 13. (5√2)8 = x3 + x8; 2xs = 50; x2 = 25; x = √25^ = 5
14. 112 = x3 + x2; 121 = 2x3; x3 = “ x = Λp = A∣l ∙ -∣ = /^.1 B 15-16. Let x be the length of the altitude.
The altitude to the base of an isos. Δ forms
2 - rt. A .
15. 108 = x8 + 88; 100 = x8 + 64; x2 = 36; x = √^36 = 6
16. 7s = x8 + 32; 49 = x3 + 9; x3 = 40; x = √40 = 2√Iθ 17. 2x + 10 = 28; 2x = 18; x = 9. h = /144 = 12
= ⅛√2
158 = h8 + 92; 225 = h2 + 81; h2 = 144;
79
Key to Chapter 6, pages 208-210
18.
78 = hs + 52; 49 = h2 + 25; h8 = 24; h = Vr24 = 2√6.
58 = x3 + h2; 25 = x8 + 24;
x2 = 1; x = √T = 1
19.
AF ∣∣ SB; ΔXBS ≡= AXAF; AX = XB = |; (SX)2 = 28 + (∣)≡ = 4 + ∣ = / OR
SX = y∣
20.
R
~ 2> SF = 2(SX) = 5; Pierre is 5 km from his starting point.
Quad. ABCF is a O; BA = CF = 2, FA = CB = 1; AS = 6 - BA = 4;
(FS)2 = 18 + 42 = 1 + 16 = 17; FS = /17; Sofia is vrΓ7 km from her starting point.
21-24. Note: The diagonals of a rhombus are perpendicular and bisect each other. 21. 108 = (ψ^)2 + 62; 100 =
+ 36; (BD)s = 4(64) = 256; BD = √256 = 16
22. (AB)8 = 5s + 12a = 25 + 144 = 169; AB = √169 = 13
23. (AB)8 = 28 + (2√3)a = 4 + 12 = 16; AB = √T6 = 4 24. (7√^2)8 = (^)S + 72; 98 =
+ 49;
= 49; (AC)8 = 196; AC = √196^ = 14
25. Let y be the length of the side common to the 2 Δ. y8 = 25.
138 = 12δ + y8;
169 = 144 + y2;
x8 = 128 + y8 = 144 + 25 = 169; x = √169 = 13
26. Let ybe the base of the Δ with hyp. 25 and z be the base of the Δ with hyp. 17.
25s = ys + 15s; 625 = y2 + 225; y8 = 400.
z2 = 64.
178 = za + 15a;
289 = z8 + 225;
x8 = y8 + z8 = 400 + 64 = 464; x = √^464^ = √ 16 ∙ 29 = 4√29
27-30. Let j be a diagonal of the base and d be a diagonal of the box.
27. j2 = 38 + 48 = 9 + 16 = 25; d2 = 128 + j8 = 144 + 25 = 169; d = √169 = 13 28. j8 = 58 + 53 = 25 + 25 = 50; d8 = 38 + j8 = 9
+50 = 59; d = √59^
29. j3 = 7a + 62 = 49 + 36 = 85; d8 = 5a + j8 = 25 + 85 = 110; d = √T10^ ∏Λ ∙≡ Ds 8 .2 .2 .2 12 /)2 3 J ∕T^2 7≡ 30 . j = / + w;d = h + j = h + / + w ; d = √h + / + w
C 31. Let RS = x; ST = 21 - x.
108 = h8 + x8; h8 = 100 - x8.
(100 - x8) + (21 - x)8; 289 = 100 -x8 + 441
h8 = 100 - x8 = 64; h = √64 = 8
3^
178 = h2 + (21
- 42x + x8; 42x = 252; x
- x)8 = = 6.
80
Key to Chapter 6, pages 208-210
= x + 9. 102 =h2 + x2; h2 = 100 - x2.
32. Let DE = x; DF (x + 9)2 + (100 -
172 = (x + 9)2 + h2 =
x2); 289 = x2 + 18x + 81 + 100 - x2; 108= 18x; x
= 6.
h2 = 100 - x2 = 64; h = 8
33. (r + s)2 = 202 +
152 = 400 + 225 = 625; r + s = √^625^ = 25.
15 r’
r+s = 15 25 15 r’ 15
25r = 225r = 9 ’
225 = 81 + t2; t2 = 144; t = √ 144 - 12.
By Cor. 2 to Thm. 6-1,
152 = r2 + t2;
s = 25 - 9 = 16.
’
t, u; 20, v; and s, r are pairs of corr.
. . oft two . λ . 't-=-s and, — 20 = — s 12 = -yr; 16 1t, ino „3 20 16 sides ~⅛ — ur vr;,— u9 16u = 108;’ u = 6-τ∙ 4 v = -yr; 9’ 1
16v = 180;’ v = II7∙ 4
1
r = 9, s = 16, t = 12, u = 6-y, v = ll-y ’ ’ ’ 4 4
34. ΔDBC and ΔADC are isos. ∕⅞∖ and BQ and AQ are medians, so ΔDBQ and ΔADQ are
rt. As .
(BD)2 = (BQ)2 + (DQ)2; 102 = (BQ)2 + 52; (BQ)2 = 75; BQ = √75 = 5 + a
b3 - a 3 ,b2 - a3 λ≡ s ~~2~ -—> -a
c2
- T
c3 b3 - a3 4 + 2
4a3 - c3 + 2b3 - 2a2 2a2 + 2b2 - c3. .. I 2a≡ + 2b2 - cs √ 2a3 + 2bs - c3 -------------4-------------- =--------- 4--------- , m = y -------- 4- ----------- =--------- 2-----------
Page 211 ∙ CHALLENGE
Picture the room opened flat, as shown at the right. (You may want to copy the model, draw AB, and fold the model up to convince yourself that the
model is correct and to see in three dimensions the path taken by the spider.)
and the spider is at A. in 10 ÷ 21 + 13 — ιn 10g2 ∙
The fly is at B
AC = 8 and BC =
∕(AB) λt5∖≡ _— o≡ 8 + ∕32 ( 3 x) ≡ , so
AB = 1⅛ m. 0
Pages 213- 214 ∙ WRITTEN EXERCISES A
1. 132 = 52 + 122; right
2. 182 > 82 + 152; obtuse
3. 14s < 92 + 122; acute
4. 23 > 9 + 13; no triangle
5. 12 :> (0.7)2 + (0.7)2; obtuse
6. (6.1)s = (1.1)2 + (6.0)2; right
7. 12 =: (∣)3 + (|)3; right
8. (2√15)2 < 52 + 62; acute
9. (√15)2 = (√7j2 + ( (YE)2 + (EU)2, so L E is obtuse.
trans., then s-s.int. A are supp.)
Then ΔU is acute. (If 2 ∣∣ lines are cut by a Suppose ME ≥ YU.
(YE)2 + (EU)3, and, since MU = YE, AU is obtuse.
that AU is acute.
Then (ME)2 ≥ (YU)2 >
This contradicts the fact
What we supposed, that ME ≥ YU, must be false.
It follows
that ME < YU.
17. Let X be the point in which the diagonals intersect. each other, QX = XI = 4 and ZX = XU = 3.
Since the diagonals of a O bisect
(QU)2 = (QX)s + (XU)2 and ΔQXU is a
82
Key to Chapter 6, pages 213-214
rt. Δ.
Then QI _L ZU.
IZ = ZQ.
ΔQXU ≡ ΔIXU ⅛ ΔIXZ ≡ ΔQXZ (SAS Post.) and QU = UI =
By def., QUIZ is a rhombus. 2. t2 > a2 + b2 (Given)
18. 1. c3 = a3 + b2 (Pyth. Thm.)
(Subst. Prop.)
4. Δ J is an obtuse L . (SSS Inequality)
19. Draw two triangles as in Ex. 18.
Given: t2 < a2 + b3. Prove: ΔJ is an acute Δ.
2. a3 + b2 > t2 (Given)
Key steps of proof: 1. c3 = a3 + b3 (Pyth. Thm.)
3.
3. t3 > c3 and t > c
c3 > t2 and c > t (Subst. Prop.)
a = 2x
4. ΔJ is an acute Δ. (SSS Inequality)
b = x2 - 1
c = x3 + 1
x = 2
4
3
5
x = 3
6
8
10
x = 4
8
15
17
x = 5
10
24
26
b. Since c3 = a3 + b3, a, b, and c are the lengths of the sides of a right triangle.
c. c2 = (x2 + 1)3 = x4 + 2x3 + 1 = x4 + 4x3 - 2x2 + 1 = 4x3 + x4 - 2x3 + 1 =
(2x)2 + (x3 - 1)2 = a2 + b2.
Therefore a triangle with sides 2x, x2 - 1, and x3 + 1
is a right triangle when x > 1.
(When x s 1, b ≤ 0 and cannot be the length of a side.)
C 21. (AP)s = 92 + 132 = 81 + 169 = 250; (PB)2 =
92 + 12s = 81 + 144 = 225; (AB)2 = 252 = 625;
(AP)2 + (PB)2 = 250 + 225 = 475; since (AB)2 > (AP)2 + (PB)2, ΔAPB is obtuse. (PC)2 = 122 + 162 = 144 + 256 = 400;
(BC)2 = 625; (PB)2 + (PC)2 = 225 + 400 = 625 = (BC)2; ΔBPC is a right Δ.
(PD)2 = 162 + 132 = 256 + 169 = 425; (DC)2 = 625; (PC)2 + (PD)2 = 400 + 425 = 825;
since (DC)2 < (PC)2 + (PD)2, Δ CPD is acute.
(AD)2 = 625; (AP)≡ + (PD)2 =
250 + 425 = 675; since (AD)2 < (AP)s + (PD)2, ΔDPA is acute.
22.
(a + 1)3 + (b + 1)2 = a2 + 2a + 1 + b2 + 2b + 1 = a2 + b2 + 2a + 2b + 2; but a2 + b2 = c2, 2a + 2b > 2c (The sum of the lengths of any two sides of a Δ is greater than the length of the third side, so a + b > c and 2a + 2b > 2c), and 2 > 1.
Therefore a3 + b2 + 2a + 2b + 2 > c2 + 2c + 1; Then (a + 1)2 + (b + 1)2 > (c + 1)2,
so the triangle is acute.
83
Key to Chapter 6, pages 215-219 ΔEFD ~ ΔCBD by the AA Sim. Post.
23.
(DF)23*7+ x89*14= 25; (DF)2 = 25 - x8; DF =
√oκ
„2 ,
EF _ DF. x _ √25 - x8. BC BD’ 4 8
8x = 4√25 - x8; 64x≡ = 16(25 - x8) = 400 - 16x8; 80x≡ = 400; x8 = 5; x = √^5.
DC = CB. DC ED EF’ 5
4 . √5", /5 (DC) = 20;
DC = ⅛ = ⅛ ∙ 2S = = 4/5; ΔAEB ~ ΔDEC by the SAS Sim. Thm.; √5 √5 √5 t> AB = AE. y = 3 ; 5y = 12/5; y = 12/5 5 DC DE’ 4√^5 5
Page 215 ∙ COMPUTER KEY-IN 2. c = √1 ∙ 1 + 1 ∙ 1 = ∕Σ
1.1
3. It goes to line 60 and increases the value of b to 2.
4. No
Ei. Change line 20 as follows: 20 FOR B = A TO 21
6.
3, 4, 5; 5, 12, 13; 8, 15, 17
Pages 218-219 ■ WRITTEN EXERCISES A
1. ST = 5; RS = 5√2^
2. RT = γj RS = -2τp
3. RT = /13; RS = ∕13^ ∙ √^2 = /26 ς rt - 15 _ 15 . √2 _ 15/2. ςτ _ 15/2 ∕Σ ^ /2 ∕Σ ^ 2 ’ bτ 2
4. RT = √2
= 13; ST = 13
6 rt _ ∩0 _ √M . √1 _ /20 _ 2√J _ rτ. ςτ _ √5 - RT ’ √2 ’ √"2 √2 ^ 2 ’ 2-V5,ST-√5
r— 7. ME = 5/3; GM = 2 ∙ 5 = 10
7 ,— 7 14 8. ME = 4/3; GM = 2 ∙ 7O = γ u v
-
—
10. GE = ∣(19) =
9. GE = ∣(20) = 10; ME = lθ∕3
11. GE =
ME = ~~~
= 8; GM = 2(8) = 16
12. GE = ~= = -⅛ ∙
= ⅛A GM = 2
5/3 _ 10∕^3 3 3
13. a. The altitude is the longer leg of a 30-60-90 Δ with shorter leg 4 and hyp. 8.
Therefore the altitude is 4/3 units long.
b. Let x be the length of the altitude.
x2 + 42 = 8s; x2 = 64 - 16 = 48; x = /48 = 4/3; the altitude is 4/3 units long. 14. a. The diagonal is the hyp. of a 45-45-90 Δ with legs 10 units long. diagonal is 10/2 units long.
Therefore the
b. Let y be the length of the diagonal,
= 100 + 100 = 200; y = √ 200 = lθ∕∑j the diagonal is 10/2 units long.
y2 = 102 + 102
Key to Chapter 6, pages 218-219
84
B 15. RS = 20 - 5 = 15, RT = 2(10) = 20; RU = 10/3; TU = 2(5) = 10; SU = 5/3 16. RS = 5/3; ST =-⅛ =-⅛. S = ψ. RT = 5β t ψ = ⅛ — √ 3 /3 √3 3 3 3’
τu = 2 . ψ «5 = 1^3. O su
wZ = 5
17. sr=X≈X.≤^RT=2∙i⅛3=^RU=2( 8) = lβi
!
τu = 2 ∙ ψ = ιφ, su = φ . 8
33 /3 18. RS = 2/3(/3) = 6; ST = ∣ = 2; RT = 6 + 2 = 8; RU = 2(2/3) = 4√-3j SU = 2/3 19. XZ = ∣(10) = 5; ZY = 5/3; WZ = 5/3; WY = 5∕3(√2) = 5/6; (WX)2 = 58 + (5∕3)2 = 25 + 75 = 100; WX = /100 = 10
20. DC = ∣(6) = 3; AD = 3/3; AB = 6/1; (BD)s = 3s+6s = 9+ 36 = 45;
BD = /45 = 3/5
21. AC J_ BD; AC and BD bisect each other; AC bisects ΔA and LC; BD bisects Z.B and ΔD.
BX = ∙∣(12) = 6 and AX = 6√^3, so BD = 12 and AC = 12/3; the diagonals
are 12 cm and 12/3 cm long.
22.
Extend AX to Y and BR to Q as shown. Therefore ΔY = ΔQ = 90o.
Each ext. L has measure 360° ÷ 8 - 450.
AY = BQ so AYQB is a O and AB = YQ.
XZ = 4
and PR = 4, so YZ = PQ = -~= = -j= ∙ ⅛∣ = ½⅛½ = 2/2; AB = YZ + ZP + PQ = 2/2 + 4 + 2/2 = 4 + 4/2 (cm)
C 23. (See diagram top of next page.)
(6 - 2)180
Extend AB and ED so they intersect at C.
= 12θo and δbde iθ isos
0
so ΔBED = 30°; L CBD and L CDB are both
ext. ∕s and have measure 360° ÷ 6 = 60°, so L C = 60o. Δ and BE = BC(√^3) = √^3.
ΔBDE =
AB ∣∣ FE so x = BE = /3.
Then Δ EBC is a 30-60-90
Key to Chapter 6, pages 222-223
85
24. Each side of the octagon is x√^2 cm.
Then 2x + x/2 = 6; x(2 + V^2) = 6;
_ 6________ 6 2 - √2 _ 12 - 6√2 _ 12 - 6√^2 _ fi X ^ 2 + √2 ^ 2 + √2 ' 2 - √7 ^ 4-2 2
∏r
z5*
In 30-60-90 ΔAEF, AE = ∣(10) = 5 and FE = 5√^3.
25. Draw AE L FC and BD 1FC.
In 30-60-90 ∆BDC, BD = AE = 5.
Perimeter = 10 + 5V^T + 6 +
„
o
DC = -⅜∑ = -⅛ ∙ y3 = √3 ΔTXS (If a = b + c and c > 0, then a > b.)
ΔTXS = ∙∣RS (An inscribed Δ is = to half its intercepted arc.) z (Subst. Prop.)
4.
17. ΔABD ~ ΔDEB.
z
Proof: 1. AB is tangent to the circle at B. (Given)
5. ΔRTS > ⅛RS
2. ΔABD =
^BCD (An Δ formed by a chord and a tangent is = to half the intercepted arc.)
106
Key to Chapter 7, pages 270-272
3. ΔE = ^BCD (An inscribed L is = to half its intercepted arc.) (Trans. Prop, or Subst. Prop.)
5. BE ∣∣ CD (Given)
lines are cut by a trans., then alt. int. ∕⅞ are =.)
4. ΔABD = ΔE
6. ΔADB = ∆DBE (If 2 ∣
7. Δ ABD ~ ΔDEB (AA Sim.
Post.) 18.
ZB = ZP = ZC.
---- ~∕k∖
Proof: 1. AC, AB, and BC are
Let R, S, and T be the points of __ __ tangency. (Def. of Θ inscribed in a Δ) 2. PR J_ AC; _ __ __ __ PS -L AB; PT _L BC (If a line is tangent to a ©, then
tangent to ΘP.
∖ ∕∖Q Kj∖∖ / ∖∙ / √sΛ∕ C÷------- ^∕B ∖∖ / Sy
/
the line is L to the radius drawn to the point of tangency.)
4. P is equidistant from AC, AB, and BC.
3. PR = PS = PT (Radii of a Θ are =.) (Def. of distance from a point to a line)
2
5. AP and BP are the bisectors of ∆CAB
and ∆ ABC. (If a point is equidistant from the sides of an Δ, then the point lies on the bisector of the Δ.)
6. L CAZ = ΔZAB-, ΔABP = ΔCBP (Def. of L bisector)
7. ΔCAZ = ^-ZC; ΔZAB = 2ZB
inscribec^ z- is = to half its intercepted arc.)
8. ZC = ZB (Subst. Prop., Mult. Prop.)
have equal chords.) then the zL are =.)
and 10)
9. ZC = ZB (In the same Θ, = arcs
10. ΔCAZ = L CBZ (If 2 inscribed Λ intercept the same arc,
11. ΔCBZ = ΔZAB (Trans. Prop, or Subst. Prop.: Steps 6
12. ΔCBZ + ΔCBP = Δ ZAB + ΔABP (Add. Post.: Step 6)
13. Δ ZAB -
ΔABP = ΔZPB (An ext. Δ of a Δ is = to the sum of the 2 remote int. zL.)
14.
Δ CBZ + ΔCBP = ΔZBP (Subst. Prop.)
15. ΔZBP = ΔZPB (Subst. Prop.)
16. ZP = ZB (If 2 zL of a Δ are =, then the sides opp. those zL are =.) 17. ZB = ZP = ZC (Trans. Prop.: Steps 9 and 16)
Pages 270-272 ∙ WRITTEN EXERCISES A
1. Δ 1 = 40°
2. ∆2 = ∣(30o) = 150
3. Δ 3 = ∣(50o) = 25o
4. Δ 4 = ∣(30o + 50o) = 40o
5.
∆5 = j(130o) = 650
7. ∆7 = ∣(50o + 40o) = 45o 9.
11.
∆9 = ∣(50o) = 250
∆l = ∣(60° + 80o) = 70o
6. ∆6 = ∣(200o) = 100° 8. Δ 8 = ∣(90o - 30o) = 30o
10. Δ10 = ∣(130o) = 65o
Key to Chapter 7, pages 270-272
107
12. ∆l = ∣[360o - (130o + 100o)] = ∣(130o) = 650 13. 60o = -∣(50o + RT); 50o + RT = 120°; 'RT = 70o
14. 63o = ∣(3 ∙ US); 126° = 3 ∙ US; US' = 42°; RUT = 360o - 84o = 276o 16. ΔN= ∣(260o -
15. ΔN
= ∣(210o - 150o) = 30o
17. 90o
= ∣((360o - AB) - AB); 360o - 2 ∙ AB = 180°; -2 ∙ AT =-180°; AB = 90o
18. ΔS
= ∣(110o - 30o) = 40o
19. 40o
= -∣(UT - 40°); UT - 40o = 80°; “UT = 120o
20. 380
= ∣(120o - VT); 120o - VT = 76°; VT = 44°
B 21. AG
100o) = 80°
= ∣(360°) = 90°; 'co' = ∣(360o) = 45°; -∣(90o + 45o) = 67.5o
22. ∣(180o - 90o) = 450
23. 90° = ∣[(360o - AB) -
AB]; 180o = 360o - 2 ∙ AB;
AB = 90o.
94o = ∣[(360o - ^BC) -
BC]; 188o = 360o - 2 ∙ BC; 'BC' = 86°.
96o = ∣[(360o - CD) -
CD]; 192o = 360o - 2 ∙ CD;
80o
= ∣[(360o - SA) -DA];
160o
= 360o - 2 ∙ DA;
RP = 160o.
DA = 100°. 80° = ∣(SP + SR - RP);
24. 70° = ∣(SR + RT - SP); SR? + RP -'sF = 140°. SP + SR -
CD = 84o.
Adding the two equations:
(SR + RT - SP) + (SP + SR - RP)
= 300°; 2 ∙ SR = 300°; SR = 150°; ΔSPR = ∣(150o) = 750 . SP + RT -
SR = 60°.
Adding, (SP + SR - RP) + (SP + RP - SR) = 220°;
2 ∙ sF = 220°;sF = 110°; ΔSRP = ∣(110o) = 55o. 25.
30° = ∣(SP + RP - SR);
ΔPSR = 180o - (7 50 + 55o) = 50°
1. Draw TV. (Through any two points there is exactly one line.) are tangent to the circle at T and V. (Given)
2. Lines t and v
3. ∆2 - -∣TXV∙, Δ 3 = ^τF (An angle
formed by a chord and a tangent is = to half the intercepted arc.)
4. ∆2 = Δ 3 + ∆l
108
Key to Chapter 7, pages 270-272
(An exterior L of a Δ is = to the sum of the 2 remote interior zL∙) (Subtr. Post.)
6. ∆l = ∣TXV - ∣TV (Subst. Prop.)
5. ∆1=∆2-∆3
7. ∆l = -∣(TXV - 'TV)
(Distr. Prop.)
26. 1. Draw TD. (Through any two points there is exactly one line.)
tangent to the circle at T. (Given)
3. ∆2 = ^∣TD (An angle formed by a chord and
a tangent is - to half the intercepted arc.) half its intercepted arc.)
4. Δ 3 = ^TC (An inscribed Δ is = to
5. ∆3=∆2 + ∆1 (An exterior Δ of a Δ is = to the sum
of the 2 remote interior A_.)
6. Δl=∆3-∆2 (Subtr. Post.)
∣TC - ∣TD (Subst. Prop.)
8. Δ 1 = ∣(TC - TD) (Distr. Prop.)
27. x = ∣(155 - 2y); x = 77.5 - y; y = 77.5 - x.
y = -77.5 + 4x.
2. Line t is
7. ∆l =
4x = ∣(155 + 2y) = 77.5 + y;
77.5 - x = -77.5 + 4x; 5x - 155; x = 31; y = 46.5
28. n = ∣(x - y); 2n = x - y.
2n = ∣(x + y).
x - y = ∣(x + y); ∣x = |y;
x = 3y; ½ - 3; x : y = 3:1 C 29. Call one of the acute angles formed by the two chords of the outer circle ΔA. ∆A = ⅜(j + n). Since ΔA is also an inscribed angle of the inner circle, ∆A = ⅜k. Z δ Then -∣(j + n) = -∣k and j + n = k.
30. Δ A = ∣SNT -∣VT < -40°.
∣VT.
VT > 80° so SNT < 190o and ∣SNT < 950.
ΔA< 95° - 40o = 550.
SNT > 160o so VT < 110o and -∣VT > -550.
∆A = ∣SNT - -∣VT∙, ∆A > 80° - 550 = 250.
31. ∆A = ∣SNT -∣VT > -55°;
∣VT.
∣VT > 40° so
25o < ΔA < 550
VS < VT < 110o so VT + VS < 220o and 'SNT > 140°; also
∆A >70° - 55o = 15o.
VT > VS > 100o so VT + VS > 200o and
SNT < 160°; also -⅜VT < -50°; ΔA = ⅜SNT - ⅜VTj ∆A < 80° - 50o = 30o. z z z
150 < ΔA < 30o
32. Let X be the intersection of OA and CP, and Y be the intersection of OB and DP.
Key to Chapter 7, pages 274-276 1.
2. ΔOAP ⅛ AOBP; Δ PCO ≡ ΔPDO (SSS
Draw OP, XY, OC, OD, PA, and PB.
3. ΔAOP = ΔBOP∙, Z-CPO = ΔDPO
Post.)
5.
109
XO = YO; XP = YP
6. ΔXOY ~ Δ JOK; ΔXPY ~ ΔGPH (SAS Sim. Thm.)
7∙≡=≡'≡=≡
PA
OC
8. Δ OCX ~ Δ PAX (AA Sim. Post.)
OJ
9.g=⅛
11. y~ = p¾ (Substituting from Step 7 in Step 10)
10. ≤y =
≡ = -gτ
4. ΔXOP ⅛ Δ YOP (ASA Post.)
JΛ
F 750 (if a = b + c and c > 0, then a > b.); 10 is the
first positive integer whose cube is greater than 750, so try 10; 1000 = 250 + 750; x = 10
Page 277 ∙ CALCULATOR KEY-IN
Answers may vary. 1
Those given use the terms or factors actually shown in the exercises. 10
1. 3γ≈ 3.1428571; 3∣γ ≈ 3.1408451; yes
3. τι ≈ 2.972154
2. π ≈ 3.1413088
4. π ≈ 3.1214452
5. τr ≈ 2.9760 462
Pages 279-280 ∙ CHAPTER REVIEW
2. RS
1. PR, PS 7. 90o
8. 8 cm
3. RT, TS, RS
9. 'χγ = WY - WX = 91° - 48o = 43°
10. WZY = 360° - WY = 360o - 910 = 269° 12.xWZ = 180° - WX = 132o
17. -∣(380) = 19°
6. 4cm
5. 8 cm
4. ΔRTS
11. ΔZPY = ZY = 180° - 43° = 1370
13. 120°
18. -∣(1220) = 61°
14. 3
15. 60°
16. 3√^3
20. ΔJKL
19. ΔJLK
21. ∆AEC = ∣ (66° + 52o) = 59° 22. 62o = ∣(AC + 47°); 1240 = AC + 47°; AC = 77o
23. Z-K = ⅜(86o - 290) = ⅛(57o) = 28.5o 24. 25o = ⅜(FJ^ - 25°); 50o = τ7- 25°; ^FJ = 75o
—
z
25. 6x = 32; x = 5⅜
—
o
Key to Chapter 7, pages 280-281
111
26. 10(x + 10) = 12 ∙ 20; x + 10 = 24; x = 14
27. x(x + 7) = 144; x123*7+ 7x = 144; x89*1+1 7x - 144 = 0; (x + 16)(x - 9) = 0; x = -16 (reject) or x = 9
Pages 280-281 ∙ CHAPTER TEST
6.
supplementary
12. ⅜(80o) = 40° — z
Z
10. 90o
9. 12 cm
14. ∣(260o - 100°) = 80o
13. 100o —
—
61° = ∣(73° +xKL)∙, 122° = 73° + KL; KL = 49°
15.
J), its central angle
4. points
8. diameter
7. right
⅜(100°) = 50o
11. —
3. concentric
2. chord
1. inscribed
Δ
16. 12 ∙ NL = 90; NL = 7.5
17. 8 ∙ AE = 320; AE = 40
18. 3 ∙ AC = 36; AC = 12
19. —
20. L CAE = ⅜(75o - 150) = 30° — z
ΔFAE = ⅜(90o - 50°) = 20° z
2. ΔC = ∣AB∙, ΔG = ∣EFj ∆A = |bC; ΔE = ∣FG
21. 1. AB = EF; BC = FG (Given)
3. ΔC = AG; Δ A = AE (Trans.
(An inscribed L is - to half its intercepted arc.)
4. AB = EF (In = circles, = arcs have =
Prop, or Subst. Prop.; Mult. Post.)
chords.)
5. ΔABC ≡= ΔEFG (AAS Thm.)
22. 1. OE bisects AD. (Given)
Prop.)
6.
2. EA = ED (Def. of bisect)
4. OA = OD (All radii of a © are =.)
5. ΔAEO ≡= ΔDEO (SSS Post.)
AAEO = ΔDEO = 90° (Corr, parts of ≥ ⅛ are
tangent to 0 O. (Given)
8. BC
3. OE = OE (Refl.
Def. of J_ lines)
7. BC is
OB (If a line is tangent to a © , then the line is
J_ to the radius drawn to the point of tangency.)
9. ΔABC = 90° (Def. of _1_ lines)
10.
ΔAEO = ΔABC (Trans. Prop, or Subst. Prop.)
12.
ΔAEO~ ΔABC (AA Sim. Post.)
11. ΔA = ΔA (Refl. Prop.)
Page 281 ∙ ALGEBRA REVIEW 1. dx - e = f; dx = f + e; x = * ⅞..θ
—
2.
k = bx; x =b r
o o, λ , b b 3. 2(x + a) = b; x + a = -%, x = τj - a
4.
-∣(x - j) = k; x - j = 2k; x = 2k + j
5. ⅞ = t; r = 5t
6. — = k; n = gk - g
d
—
—
5
t-
,
ι
cl
7. — -c;a -be; b=— b ’ ’ c
8.
⅛ = c; a = 2bc; b = ⅛
—
9. A = bh; h = ⅜ b
10. —
V = ⅜Bh∙, 3V = Bh; B = ⅛ 3 n
11.
J =πj C= πd
12.
C = 2πr∙, r = ■£-
— d
112
Key to Chapter 7, page 281
13. S= (n -
2)180;
γ∣θ = n - 2; n = yfθ + 2
14. A
= 2^(b + c);= b + c; c = ~ - b
15. a2
+ b2 = (2a)2; b2 - 4as - a2 = 3a8; b = ±avr3
16. a2
+ b2 = c2; b2 = c8 - a8; b = ±-∕c2 - a2
CHAPTER 8 ∙ Constructions and Loci
113
114
24. a.
Key to Chapter 8, pages 286-287
ι
b. ΔG + ΔH + ΔI = 180o and ∣ΔG + ∣ΔH + 2∆I = 180°;
∣ΔG + ∣ΔH - ΔI = 0; ΔG + ΔH = 2£I;
ΔG + ΔH + ΔI = 3∆Ij 180o = 3Z.I; LI = 60o
Key to Chapter 8, pages 290-292
115
C 25. Draw a circle with radius n and draw any chord AB. At B, construct an angle equal to L 3, as shown. Draw AC and construct an angle, at A, equal to L 1,
as shown.
Z.ADC = Δ 3 since the two angles
intercept the same arc.
Ll + Z_ 3 + ∆ACD = 180o,
so Z.ACD = L2. 26. Construct LA = Z.4 and AB = s.
Then construct
a corresponding angle equal to L 4 at B as shown.
With A as center and radius d, draw an arc to determine point C.
With A as center and radius
BC, draw an arc to determine point D.
Since AD = BC and AD ∣∣ BC, ABCD is a O.
Pages 290-292 ■ WRITTEN EXERCISES A
1. Construction 4
2. Construction 5
3. Construction 6
of ΔKMN.
4. Construction 7
116
Key to Chapter 8, pages 290-292
19-20. Method I: Draw a line and construct a segment AB equal to a. angle at A.
Construct AD equal to b.
Construct BC equal to b. equal to a.
Draw DC.
Construct an no
Construct the parallel to AD through B.
Method II: Draw a line and construct AB
Construct an no angle at A.
Construct AD equal to b.
Then draw
an arc with center D and radius a, and an arc with center B and radius b. arcs intersect at C.
Draw DC and BC.
(Other methods are possible.)
The
117
Key to Chapter 8, pages 290-292
21. Draw a line and construct AB equal to a.
Construct AD and BC equal to b.
Draw DC.
22. Draw a line and construct AB equal to a. AB at A.
Construct perpendiculars to AB at A and B. (Other methods are possible.)
Construct the± bisector of AB and a_L to
Construct AE and CD equal to AC.
(Other methods are
Draw ED.
possible.) 23. Draw lines and construct RS equal to b and TW equal to a. of RS and TW.
Then construct XY and XZ equal to ∙∣b.
Construct the J_ bisectors Draw rhombus TZWY.
24. Draw two lines which intersect to form two acute angles and two obtuse angles.
center J and radius -∣b (from Ex. 23), construct GY. XA
= TT ' GA
Suppose Δ 1 = ∆2.
Since YX = XA, GY = GA.
Then GX bisects ΔAGY and
But this contradicts the fact that GA > GY.
What we supposed, that L 1 = L 2, must be false. angle is not trisected.
Then GX = GY.
It follows that L 1 ≠ L 2 and the
Key to Chapter 8, pages 305-306
124 Pages 305-306 ∙ WRITTEN EXERCISES A
The perpendicular bisector of AB
2. A line parallel to j and k, midway between j and k
3. A circle with center O and radius 2 cm 4. A pair of lines parallel to, and 2 cm from, h
J>. A segment parallel to AB and CD, midway between AB and CD
6. AC
7. BD
8. The intersection of AC and BD
9. A plane parallel to the two planes, midway between them
10. The plane which bisects CD and is perpendicular to CD 11. A sphere with center E and radius 3 cm
12. A cylindrical surface with axis m and radius 3 cm B 13. a. The locus is the bisector of ΔHEX.
b. The locus is the bisectors of the vertical angles formed by j and k.
14. The locus is lines s and t, parallel to n, each a distance of DE from n.
15. The locus is a circle whose center is the midpoint of AB and whose radius is ^AB. A and B are not included.
16. The locus is the perpendicular bisector of CD.
The midpoint of CD is not included.
17. The locus is a circle concentric with the given circle and with radius equal to one-half the radius of the given circle.
Key to Chapter 8, pages 305-306
125
18. The locus is a line which is perpendicular to the plane of the square, and which passes through the intersection of the diagonals. 19. The locus is a line which is perpendicular to the plane of the triangle, and which
passes through the intersection of the perpendicular bisectors of the sides of the triangle. 20. The locus is a plane which is perpendicular to the plane of the two lines and which is
midway between the two lines.
21. The locus is a sphere concentric with the given sphere and with radius equal to onehalf the radius of the given sphere.
22. The locus is a region bounded by adjoining portions of five circles: half of ©A with radius 5 m, quarters of ΘB and ©E with radii 4 m, and quarters of 0C and ©D with radii 2 m.
23. The locus is a 90o arc of a circle with center at the point where the wall meets the
ground, and with radius equal to ∙∣AB. 24. The locus is a circle with center O and radius 10.
25. The locus is four 90o arcs of circles with centers at the vertices of the square and radii of 1.5 cm, together with the segments joining the arcs.
26. The locus is a sphere whose center is the midpoint of CD, and whose radius is 77 CD. C and D are not included.
126
Key to Chapter 8, pages 308-31C 27. The locus is a circle whose diameter is AO.
Justification: Let Q be the midpoint of some chord passing through A.
Δ OQA is a rt. Δ.
Then ΔOQA is a rt. L, and
Let M be the midpoint of OA.
Since the midpt. of the hypotenuse of a rt. Δ is
equidistant from the three vertices, MQ = MO, and Q is a point on 0M with radius MO (and
diameter AO).
Pages 308-310 ∙ WRITTEN EXERCISES
A
1. a. ΘD with radius 1 cm
b. ΘE with radius 2 cm
d. The locus is no points, one point, or two points, depending on the intersection of (1) a circle with center D and radius 1 cm and (2) a circle with center E and radius 2 cm. 2. a. 0A with radius 3 cm
b. A pair of lines parallel to, and 1 cm from, k.
d. The locus is no points, one point, two points, three points, or four points, depending on the intersection of (1) a circle with center A and radius 3 cm and (2) two lines parallel to, and 1 cm from, line k. 3. The locus is the two-point intersection of ΘA and OB, each with radius 2 cm.
Ex. 3
Ex. 4
Ex. 5
Ex. 6
Key to Chapter 8, pages 308-310
127
4. The locus is the four-point intersection of ©P with radius 2 cm and the angle bisectors of the vertical angles formed by j and k.
5. The locus is the four-point intersection of ©Q with radius 5 cm and two lines parallel
to, and 3 cm from, ./. 6. The locus is the one-point intersection of ©A with radius 2 cm and the bisector of ∆A.
8. Let d be the distance from E to line k.
11. The locus is the two circles in which two planes parallel to, and 3 cm from, Z intersect the sphere with center A and radius 5 cm.
12. The locus is no points, two points, or two circles, depending on the intersection of (1) two planes parallel to, and d cm from, Z, and (2) a sphere with center A and
radius 5 cm. 13. The locus is the two circles in which two planes parallel to, and 2 cm from, Q intersect
the cylindrical surface with axis AB and radius 2 cm.
128
Key to Chapter 8, pages 308-31Σ
14. The locus is no points, one point, two points, or a circle, depending on the intersection of (1) two planes parallel to, and 2 cm from, Y, and (2) a sphere with center B and
radius 2 cm.
The locus is no points, one point, two points, three points, or four points, depending on the intersection of (1) the bisectors of the vertical angles formed by j and k, and
(2) a circle with center P and radius the given distance. 16. The locus is the intersection of the three circles
(and their interiors) whose centers are the given points and whose radii are 2 cm. 17. Consider the four planes determined by ΔVAB, ΔVBC, ΔVCA, and ΔABC.
In each
triangle, the locus of points equidistant from the endpoints is a line perpendicular to the plane of the triangle and passing through the point of intersection of the perpendicu
bisectors of the sides.
The point of intersection of these four lines is equidistant from
V, A, B, and C.
18. a. In the shaded region of the diagram below b. On the heavily outlined arc of OF in the middle diagram
129
'z'.~ to Chapter 8, pages 312-313
-.ges 312-313 ∙ WRITTEN EXERCISES 1.
2.
Construct j and k, which are parallel to, and h units from, AB.
a. Construct t, the perpendicular at E.
E is not included in the locus.
b. Construct a circle whose center is the midpoint of DE and whose radius is ∙^DE. Points D and E are not included in the locus.
3-4. Note: Some students may draw rectangles for which the locus is one point or no points.
3.
Construct the bisector of LB and QD with radius v.
4. Construct the perpendicular bisector of AD and BC, and OD with radius v.
Points
Y and Z satisfy the requirements. 5. Construct two lines that are parallel to, and v units from, k.
radius v.
OR and OS pass through N and are tangent to k.
Construct ON with
130
Key to Chapter 8, pages 312-31c
6. Construct the bisector of ΔXYZ.
Construct a line parallel to, and v units from, YZ.
OW is the required circle.
7-16. Constructions may vary. B
7. Construct ΔA equal to no.
Construct a line parallel to, and s units from, AX.
center C and radius CA, locate point B.
Since CB - CA, ΔB = ∆A = no.
8. Proceed as in Ex. 7 to construct ∆A and AC.
9. Construct AB equal to s. Construct AC equal to t.
Ex. 9
With
Construct BC J_ AC.
Construct line j parallel to, and r units from, AX.
(Note: It is possible to construct ΔABC so that LA is obtuse.
Ex. 10
131
•ley to Chapter 8, pages 312-313
10.
Construct AB equal to t.
Construct the J_ bisector of AB, and OM with radius s.
Construct line k parallel to, and r units from, AB.
ΔABC and Δ ABD satisfy the
requirements.
11.
Construct AB equal to r and the perpendicular at A. CD equal to s.
Construct BC equal to s and
Δ BCD is the required triangle.
12. Construct a ©R with radius s, intersecting line a in S and T. parallel to, and r units from, line a. semicircle, each is a right angle.
Construct line b
Since ΔSYT and ΔSZT are inscribed in a
In right triangles SYT and SZT, R is the midpoint
of hypotenuse RS, so RS = RT = RY = RZ.
Then ΔSYT and ΔSZT satisfy the
requirements.
C 13. Choose a point P and construct chord PQ, with PQ = r.
With O as center and OX as
With X as center and YP as radius, -- >∙ Extend ZX to intersect ©O at W. ZW = r.
radius, mark off an arc intersecting PQ at Y.
mark off an arc intersecting ©O at Z.
(ΔOYP - ΔOXZ by the SSS Post. from O to XZ.
Then the altitude from O to YP equals the altitude
Thus PQ and ZW are equally distant from the center, O.
Ex. 13
Ex. 14
Then
132
Key to Chapter 8, pages 313-31: 14. Construct LC, a right angle,and j, the bisector of ∆C.
Construct the perpendicular to CA at A. tangent to ΘB at D.
Construct CA = r as shown.
Draw ΘB with radius r.
Δ CEF is the required triangle.
Construct a
(ΔCDE - ΔCDF by the ASA
Post., so CE = CF and rt. Δ CEF is isos.)
15. Construct AB equal to r and BC equal to s.
©M with radius MA.
so AP - ^CR,
Draw PB.
Construct the L bisector of AC.
ΔADC is the required triangle.
Draw
(∆AMP - Δ CMP.
Then ∆ADP - ∆CDP and DP bisects rt. ∆ADC.)
16. Various solutions are possible.
Example: Construct XY _L AB and XZ J_ CD.
Construct the J_ bisectors of XY and XZ.
XT contains
Label their intersection T.
the intersection of AB and CD.
Page 313 ∙ CHALLENGE The locus of the cat is point C.
The cat does not move at all as the mouse moves
from A to J.
Page 315 ∙ EXTRA
3. The center of the circle is H, the intersection of the altitudes.
The midpoints of the
sides, L, M, and N, are also the points where the altitudes meet the sides, R, S,
and T.
4. Three.
If ΔC is the right angle, then S, T, and Z coincide with C.
H is also
coincident with C.
5. 1. NM = ∣AB = XY; NM ∣∣ AB; XY ∣∣ AB; NX = ∣CH = MY; NX ∣∣ CH; MY ∣∣ CH
(The segment that joins the midpoints of two sides of a triangle (1) is parallel to the third side, and (2) has a length equal to half the length of the third side.)
0; (x +
3)(x - 2) > 0; x + 3 > 0 and x - 2 > 0,
or x + 3 < 0 and x - 2 < 0; x>2orx DB + BC;
AE + EC > AB + BC.
Key to Chapter 12, pages 434-436
189
Pages 434-436 ∙ WRITTEN EXERCISES
A
1-4. Answers may vary.
1. R
-
3. R
2. R 0,-240
~
0,355
4. R 0,210
~
0,80
5. A → D; B - A; C → B; D - C
6. A → B; B → C; C → D; D - A
7. A - C; B - D; C → A; D → B
8. A
21.
10. rotation
9. B
c. AB = ■/ (2 - 5)2 +(4- 0)2 = ∙J(-3)2 + 4s = √ 25 = 5;
a, b. See art below.
A'B, = 7(-2 - (-5))≡ + (-4 - 0)2 = √3≡ + (-4)2 = √25 = 5; AB = A’B’
d.
Midpoint of AA' = ∖~2”’ —=
= (0, 0)
bb' = (—2—’ —2—
midPoint
e. Parallelogram; the diagonals bisect each other. ■y 1 P
/
.1J ∖
/
!
I
■y
∖
∖
/
∖
7
/
0l /
X
A
/
1
^^7
4'∙
∖
(
/
lB
0
/ _J L
Ex. 22a, b
Ex. 21a, b
→ →1-3 -2 1 22. a, b. See art above. c. Slope of AB = 4 _ 'θ = “4" = “ 2 ’ s*oPe 4-0 4 -*-* ^*—*—j----(—2j = 2 = 2; AB _L A'B' d. A rotation is an isometry. 23. a. P
24. a. B; E
b. (6, 4)
X
c. (3, 4)
d. (5, 0)
e. (8, 3)
-*—*" a∙'b' =
L (6 - x, 4 - y)
b. isometry
C 25. With centers S and P and radius SP, mark off arcs intersecting at a point G on
ΔSGP is the desired triangle.
190
Key to Chapter 12, pages 438-440
26. Let x, y, and z be the three parallel lines.
Choose a point P on x.
1/
∣
Construct /1 x at P.
∖∕,
∖
Let F be the point in which / intersects z.
With centers P and F and radius PF, draw arcs intersecting at F'; ΔFPF' = 60o.
Draw PF'.
At F' construct ∕' J_ PF'.
Let S be the point in which y and
intersect.
∖1
x∖i''',
y
∖U0°zz^ ∖' ∣’ ,''t>∣ ∖P ' _____ ∖χ ''' 1
Just as Rp _60 maps F into F', it maps every other point of z into a point of ∕,. particular, there is some point, T, of z that is mapped into point S.
draw arcs with centers P and S and radius PS.
Draw Δ PST.
Δ PST is the desired triangle.
Pages 438-440 ∙ WRITTEN EXERCISES A —
—1. —a. One
b. No —
—c. None
2. a. None
b. No
3. a. Five
b. None
c. 72°; 144°; 216°; 2880
4. a. Six
b. Yes
c. 60°; 120°; 180°; 240°; 300°
c. 720, 144o, 216°, 288o
5-7. Answers may vary, depending on how the letters are formed.
-ABCDEMTUVWY ⅛∙ H I 0× -HINOSXZ
To locate T,
T lies on line z;
In
Key to Chapter 12, pages 443-444
191
21. Not possible
22. Not possible 23. A regular pentagon
25. a. 90o, 180o, 270o
b. 60°: 6; 45°: 8; 30°: 12
26. a. At the midpoints of the sides C 27. The quadrilateral is a rectangle.
28. The quadrilateral is a rectangle.
b. Translational
a. x - 0; y = 0
b. (0, 0)
b. (-2, -1)
a. x = -2; y = -1
29. The quadrilateral is a rhombus; its diagonals are the lines of symmetry.
a. One line passes through (2, 5) and (-2, -3): y - 5 = 2(x - 2) or y + 3 = 2(x + 2); the other line passes through (-2, 2) and (2, 0): y = - -∣(x - 2) or y - 2 = -∙∣(x + 2).
Δ
b. The midpoint of the diagonals: 30.
The quadrilateral is a parallelogram.
- , —
Δ
θ) = (0, 1)
a. None
b. The midpoint of the diagonals:
(2÷s∙ 2→Λ = (3, 2)
Pages 443-444 ∙ WRITTEN EXERCISES A
1. C
2. A
3. B
4. C
5. A
6. C
2∙ Since the diagonals of quad. ABCD bisect each other, quad. ABCD is a parallelogram.
192
Key to Chapter 12, pages 443-444 8. a.
A
B
b. A
B
9. a.
b. 0 j
0
B 11. (-3, -1)
12. (-4, 2)
16. (6, -1)
17. (5, 1)
21. a, b.
13. (1, -2) 18. A(3, 1)
14. (-3, -2) 19. (3, 0)
15. (9, 2) 20. (-5, 2)
22. a, b. The images are A"(10, 8); B"(7, 12); and Cn(6, 8).
F'
F
Key to Chapter 12, pages 443-444 23.
193
25.
(x + 6, y + 7)
a, b. The images are A"(-4, -1);
24.
a, b. The images are A"(-4, 1);
B"(-l, -5); and C,'(0, -1).
B"(-l, 5); and C"(0, 1).
27.
26. (-x, y)
(-x, -y)
28. Hq or My o Mχ
29. A product of reflections in perpendicular lines is a half-turn. C 30. Theorems 12-4 and 12-5 imply that M. ° M. is
either a translation or a rotation.
In this case,
M. o M, appears to be a rotation. The center J £ of the rotation is the intersection of the perpendicular bisectors of any two of AA", BB",
and CC" (point O in the diagram).
The amount
of rotation is the number of degrees in ∆AOA",
∆BOB", or ∆COC,'. rotation x0.
Call this amount of
According to Theorem 12-5,
lines j and k can be any lines through O such that the measure of the angle between j
and k is ∙∣x°.
For example, let j be OB, and let k be the perpendicular bisector of
BB”, as shown in the diagram. 31. Methods of solution may vary. is given.
One example
Let j be AC, and let ΔA,B,C,
be the image of ΔABC in line j. that A = A' and C = C'.
Note
Now we can
rotate ΔA,B,C, to ΔA,,'B,"C," by a rotation Mk ° M^, using the method
described in the solution to Ex. 30. The center of the rotation (point O in the diagram) is the intersection of the
Key to Chapter 12, pages 446-451
194
perpendicular bisectors of any two of A'A"', B'B",, and C'C"; and the amount of rotation is xo, where xo = ∆A'OA"' = ΔB,OB," = ΔC'OC"'.
Lines k and / can be
any lines through O such that the measure of the angle between k and / is ∣x0.
For
example, let k be OB', and let / be the perpendicular bisector of B'B"', as shown
in the diagram.
Pages 446-448 ■ WRITTEN EXERCISES
Δ 1∙⅛
3.∣ — z
2 — -∙ 9
10. B
9. C
15. I
12. B
11. A
8. D
13. Rεo = R48 + ≡∙, C
18.
17.
lβ∙ ⅝
7. A
6. D
5. C
4. 5
14. P
19. (x + 6, y + 8)
(x + 4, y)
21.
S'1: (x, y) → (x + 3, y + 1)
22. s^1∙. (X, y) - (y ⅞)
23.
S'1: (x, y) - (4x, 4y)
24. S'1: (x, y) → (x - 4, j∣)
25.
S'1: (x, y) - (y, x)
B 20. S-1: (x, y) - (x - 5, y - 2)
C ----26. —a. M J. o M,κ ∙ two units to the right; Mk o My. two units to the left
---
K-
b. (M o Mk) o (Mk o Mj) = Mj o (Mfc O M.) o M. = M. o I o M. = M. o M. = I J
J
J
J
J
Mj o Mfc and Mk ° M . are inverses. J 27. a. The length of the segment that joins the midpoints of two sides of a triangle is equal
to half the length of the third side. 28. a. S and T are inverses.
-----
—
Haa
o
d. Translation
b. T o S = (H ABBA ° H ) 0 (H o H ) =
—
(H d °Hπ)oH. = H°IoH = H o H. =1; T and S are inverses. ' B B A A A A A
29. a. Yes
b. A, B, C and D are the vertices of a parallelogram.
30-32. Mk o χ = m√, Mk o Mk o χ = Mk o Mjj I ° X = Mk ° Mj; X = Mk o Mj.
30. When kJ. j, M, o M. is a half-turn. (See Ex. 28, Section 12-5)
—
*c
J
31. When k ∣∣ j, M, ° M . is a translation.
—
κ
J
(See Theorem 12-4, p. 441.)
32. When k is the x-axis and j is the line y - x, Mk ° M . is the rotation Rθ gθ . (See Theorem 12-5, p. 441.)
Pages 450-451 ∙ WRITTEN EXERCISES A
1. A(12, 0); B(8, 4); C(4, -4) 3.
A(3, 0); B(2, 1); C(l, -1)
2. A(18, 0); B(12, 6); C(6, -6)
4. A(-3, 0); B(-2, -1); C(-l, 1)
5. A(-12, 0); B(-8, -4); C(-4, 4)
6. A(6, 0); B(4, 2); C(2, -2)
7. A(6, 0); B(7, -1); C(8, 1)
8. A(6, 0); B(2, 4); C(-2, -4)
Key to Chapter 12, pages 450-451 9.
4; expansion
12. -
contraction
195 10. 2; expansion
11. 4; contraction --- u
13. 4; expansion
14. -3; expansion
B 15. P,(-3, 3); Q'(0, -3); R,(12, 0); S'(6, 6) ■y ∖
■y —
∖
s∕
Λ VP
or, ∖
R
R'
X
> L_
Each division = 1 unit
17.
18. P,(l, 1); Q'(0, 0); R,(-2, 0);
P'(-6, 0); Q'(-6, -8); R'(-12, -12); S,(-10, 2)
Each division = 2 units
19.
21. a. ∙≤9
20.
b. 1:9
a. 1:3
1 b. |
a. 1:4
b. 1: 8
S∙dg,.1
22. a. Slope of PQ =
slope ol W = ⅛H~j⅛ = ⅞ - aj = J÷S
⅛ paraUel
C 23. P'Q' = 7(kc - ka)2 + (kd - kb)2 = √k2(c - a)2 + k2(d - b)2 =
√k2[(c - a)2 + (d - b)2] = ∣k1√(c - a)2 + (d - b)2 = ∣k∣ PQ
24.
Da 2 0 Dβ i_ is a translation; da 2 0 db X: p "* p" anc* da 2 ° Db 2
____
,
’2
’
’2
Q" > .
by Thm. 12-6, PQ ∣∣ P'Q’ and P’Q’ ∣∣ P,,Q", so PQ ∣∣ P"Qni also PQ = ∣P,Q, =
-∣(2P"Q") = P"Q"; then PPnQ"Q is a O and PP” = QQ”; Da 2 o Dβ i is a translation.
Key to Chapter 12, pages 453-457
196
Page 453 ∙ COMPUTER KEY-IN
2. Answers may vary,
n = 50: 0.3434; n = 100: 0.33835; n = 300: 0.335001;
n = 500: 0.334329; estimates should approach ∙∣∙
3. Answers may vary.
a. Examples: (1) Change lines 50 and 80 to
”50 LET A=2*I*2*I" and ”80 LET S=2*S∕Nf3".
(2) Change line 80 to
”80 LET S=8*S∕Nf3" or ”80 LET S=2 f 3*S∕N f 3".
(3) Add a line 15 and
change lines 40 and 90 as follows:
15 PRINT ” IN EACH UNIT INTERVAL”; 40 FOR 1=1 TO 2*N 90 PRINT ’’AREA OF";2*N;" RECTANGLES IS";S;”." g b. Estimates should approach ∙ 4. a. Change lines 50 and 80 to ”50 LET A=lf3" and ”80 LET S=S∕Nf4".
b. Answers may vary.
Estimates should approach ∙∣∙
Page 455 ∙ EXTRA
1-4. Tessellations may vary.
Examples: See art below.
Pages 456-457 ♦ CHAPTER REVIEW
1.
(5, 9)
2. (0, 0)
3. Yes; yes
Key to Chapter 12, pages 458-461
10.
Example: Rθ 45
Answers may vary.
15. G
14. Yes
23. (j. ∣)
18. a. (8, 3)
21. (4, 0)
20. (4, 2)
13. No
12. Yes
11. No
17. ΔABC
16. AB
b. T^1
19. a. (x, y)
197
b. (5, 3)
22. (2, -1)
24. (6, 3)
Page 458 ∙ CHAPTER TEST
1. T,
7. True —
—8. False
13. False
14. True
18. (1, 2)
19. (-1, -2)
Ji. Reflection
4. Rotation
3. S'T'
2. RS
11. False ---
10. True —
9. False —
12. True ---
17. (10, 20)
16.. True
15. True
6. Translation
20. (1, 2)
Pages 459-463 ∙ CUMULATIVE REVIEW Pages 459-460 ∙ TRUE-FALSE EXERCISES
1. F
2. T
3. T
4. F
5. F
6. T
7. F
8. F
9. F
10. T
11. T
12. T
13. T
14. T
15. T
16. T
B 17. T
18. T
19. T
20. F
21. T
22. F
23. F
24. T
25. T
26. T
27. T
28. F
29. T
A
Pages 460-■461 ∙ MULTIPLE CHOICE EXERCISES
3. a
1. b
2. d
9. b
10. b
11. d
B 13. d
14. c
15. b
A
4. b
12. c
5∣. a
6. c
7. d
8∣. c
Key to Chapter 12, pages 461-463
198 C 16. Plot A(0, 0), B(10, 0), and C(5, 12).
The perpendicular bisector of AB has equation
---5 5 The perpendicular bisector of AC has equation y - 6 = "f2^x ~
x = 5.
119 perpendicular bisectors intersect at P(5, -^-).
ΛZ m≡ ∕Hθ√5 - 169∙ √(5 - 0) + (-δr)
The
The radius = AP =
¼ b
17. The locus of points 4 cm from Q is a sphere with center Q and radius 4 cm;
the locus
of points equidistant from P and Q is a plane that is perpendicular to PQ, and 3 cm
from P and Q; the loci intersect in a circle; d
Pages 461-463 ∙ COMPLETION EXERCISES
A
1.. perpendicular
7.
—
2. rectangle 9. 27:64
8. B
—
—
0
18.
64 :169 13. √^13^ 1 -| 19. (-5, -1)
24.
266
12.
5. (-2, ⅛y)
4. 0
10. diameter
6. 164°
11. (-4, -3)
14. point
15. 7
16. 9
20. 2520
21. 8 ∙ 17 = 136
17. 16π 22. 4
4 23. ∣
25. ⅛Bh = ⅛ ∙ 100 ∙ 12 = 400
-----
------
0
0
B 26. ∣p∕ = ∣ ∙ 40 ∙ 13 = 260
28.
3. -∣
π(3r⅛πrsh = 9:1
27. 2π ∙ 6 ∙ h + 2π ∙ 62 = 228τr, 12πh = 156tt; h = 13
29. (2x, y + 3)
30. a line, perpendicular to the plane of the points, that passes through the intersection of the perpendicular bisectors of the segments joining the points
31. ∣d = ∕103 - 82 = 6; d = 12 32. 2 ∙ 23 + 2h = 64; h = 9; A = 23 ∙ 9 = 207
33. Since the distance between (5, 4) and (-7, 4) is 12, the center of the circle is (5 2-7 > 4 2 4)> 0r (-1> 4)-> r =6; (x
34. ∣τrr3
= 36π∙, r3 = 27; r = 3
+ l)s + (y - 4)2 = 36.
35. r = √3≡ + 4s = 5; d = 10
36. -∣h(12 + 16) = 98; h = 7 37. ph +
2B = 12√2^ ∙ 12 + 2 ∙ ∣ ∙ 3√2^ ∙4√Σ = 144√2^ +
38. 4πr2
= ⅛7rr3-, r2 = ⅞r3∙, 3 = r O
—
24
0
Page 463 ■ CONSTRUCTION EXERCISES A
L Draw AB. C.
With centers A and B and radius AB mark off arcs intersecting at a point
ΔABC is equilateral.
Key to Chapter 12, page 463
199 Draw AC and bisect ΔCAB (Construction 3).
2. Proceed as in Ex. 1 to locate point C.
3. Draw a scalene triangle.
Use Construction 11 to inscribe a circle in the triangle.
Construct segments of length 2(AB), 3(AB), and 4(AB).
4. Draw AB.
With X as center and radius 3(AB), draw an arc.
With Y as center and radius 4(AB),
draw an arc that intersects the other arc at a point Z. j>. Draw ΔRST.
ΔXYZ is the required triangle.
Construct ΔABC - ΔRST (Construction 2).
bisector of ∆ABC (Construction 3).
Let XY = 2(AB).
Construct BD, the
Construct L EBA = ∆ABD (Construction 2).
2
ΔEBC = ^ΔABC. B
6. Draw OO and point P outside the circle.
Construct PX and PY tangent to 0O
(Construction 9); extend OP to intersect ©O at M. Construct a tangent to ΘOatM -- >-- >(Construction 8), intersecting PX at Q and PY at R. Δ PQR is the required triangle.
7. Divide AB into 3 equal segments: AX, XY, and YB (Construction 12). BZ - AX on AB and ZC = AX on AB (Construction 1).
8γ Draw AB = j and CD = k.
AB:AC =
Construct
3:5.
Construct AX = 3(AB) (Construction 1).Construct the
segment whose length, m, is the geometric mean of AX and CD (Construction 14).
= 9.
m2 = AX ∙ CD = 3jk; m = √ 3jk.
Construct AB = t.
midpoint, M.
Construct the _L bisector of AB (Construction 6) to locate
the —*^ 3 Construct XY = t, and YZ = AM on XY; XZ = -^t; with centers A
and B and radius XZ, mark off two pairs of arcs intersecting at points E and F. 3 Quad. AFBE is a rhombus with perimeter = 4(-^t) = 6t and diagonal AB = t. 10. Draw scalene ΔXYZ and construct ZW, the altitude to XY (Construction 5); area of ΔXYZ =
∙ XY ∙ ZW.
the midpoint, M.
Construct the _L bisector of XY (Construction 6) to locate
Construct a segment whose length t is the geometric mean of
XM and ZW (Construction 14).
t
=
z∣ w
; t8 = XM ∙ ZW.
Construct perpendiculars to AB at A and B (Construction 4). BC = t with AD and BC on the same side of AB.
Draw CD.
Construct AB = t. Construct AD = t and Quad. ABCD is a
square with area t8 = XM ∙ ZW = ∣ ∙ XY ∙ ZW = area of ΔXYZ. 11. Construct an equilateral ΔABC as in Ex. 1.
Construct CX, BZ, and AY, the _L
bisectors of AB, AC, and BC (Construction 6).
The center of each circle is
equidistant from the sides of an angle and from the J_ bisector of two sides.
Bisect
Key to Chapter 12, page 463
200
∆CZB, intersecting CX at R; bisect ∆AZB, intersecting AY at S; bisect ΔBXC,
intersecting BZ at T (Construction 3).
Construct SV J_ AB (Construction 5).
With
R, S, and T as centers and SV as radius, draw ΘR, ΘS, and ΘT, the required circles.
End-of-Book Examinations Page 464 ∙ CHAPTER 1
—1. c
—2. d
—3. b
9. a
10. b
11. a
—4. d
—5. a
6. c —
—7. d
—8. b
—5. d
6. c —
7. b —
—8. b
Page 465 ∙ CHAPTER 2 —1. d 9. c
—2. a
—3. b
—4. c
10. d
11. a
12. c
Page 466 ∙ CHAPTER 3 1. ASA or AAS
2. AAS or ASA
6. AAS or ASA
7 . SAS
12. c
3. HL
8. HL
4. SSS
5. ASA or AAS
10. a
9. d
11. d
13. a
Page 467 ∙ CHAPTER 4
ι,. a
2. a
3. d
9. c
10. b
11. c
4. d
5. b
6. c
7. b
8. a
4. d
5. b
6. a
7. c
8. b
5. c
6. d
7. c
8. a
6. c
7. b
8. d
6. c
7. a
8. d
Page 468 ∙ CHAPTER 5 1. d
2. b
3. b
9. a
10. c
11. d
Page 469 ∙ CHAPTER 6
L a
2. b
3. c
4. c
9. b
10. d
11. d
12. a
Page 470 ∙ CHAPTER 7
1. b
2. c
3. c
4. c
5. a
9. a
10. b
11. d
12. a
13. d
5. d
Page 471 ∙ CHAPTER 8 1. b
2. c
3. b
4. d
9. d
10. d
11. c
12. a
201
202
End-of-Book Examinations, pages 472-475
Page 472 ∙ CHAPTER 9
1. b
2. a
3. d
4. b
5. c
9. c
10. d
11. d
12. b
13. b
(5. a
7. d
8. c
6. b
7. a
8. b
Page 47 3 ■■ CHAPTER 10 1.. a
2. c
3. b
4. b
5. d
9. a
10. a
11. d
12. d
13. c
5. d
6. b
7. c
8. c
6. c
7. b
8. b
Page 47 4 ∙ CHAPTER 11 J,. a
2. c
3. d
4. d
9. c
10. a
11. b
12. d
Page 475 • CHAPTER 12 1. c
2. b
3. c
4. c
5. a
9. d
10. d
11. b
12. c
13. c
Appendix: Logic
Page 478 ∙ EXERCISES 1.
I like the city and you like, the country.
2. I do not like the city.
3. You do not like the country.
4. I like the city or you like the country.
5. I like the city or you do not like the country.
6. "I like the city and you like the country" is not true. 7. I do not like the city and you like the country. ∣8. "I like the city or you like the country" is not true.
14. ~ (p a q) 17.
19.
21.
11. ~ p ^ ~ q
10. ~ q
9. p v q
15. Yes
12. ~ (p v q)
13. ~ p v ~ q
16. Yes
P
q
~P
~p v q
τ
T
τ
F
T
T F
τ F
τ
F
F
T
F T
T
T
τ
F
F
T
T
P
q
P v q
~(p v q)
τ
T
τ
F
T
F
T T
F
F F
τ
F
T F
T
P
~p
p λ ~p
P
q
~q
p v ~q
T
τ
F
T
F
F
τ
F
F
P
~p
~(~ P)
τ
F
F
T
18.
20.
22.
F
F F
~p
P v ~P
T
F
T
T
F
F
F
T
T
F
T
F
P
P
q
r
q v r
p A (q v r)
T
τ
T
T
T
T
τ
F
T
T
τ
T
F
T F
τ
τ
F F
F
T
T
T
F
F F
T
F
τ
F
F
T
τ
F
F
F
F
F
F
F
203
204
Appendix: Logic, pages 479-480
24.
P
q
r
p λ q
p ^ r
(p λ q) v (p λ r)
τ
T
τ
T F
T
T
T F
τ
τ τ τ
F
T
F
T
F
F
F
F
T F
F
T
F
F
F
F
T
T F
F
F
F
F
T
F
F
F F
F
F
F
F
F
F
T
Pages 479-480 ∙ EXERCISES
L If you love bananas, then you are a monkey's uncle. 2. If you are a monkey's uncle, then your nephew is a chimp.
3. If you are not a monkey's uncle, then your nephew is not a chimp. 4. It is not true that if you love bananas, then you are a monkey's uncle. J). If you love bananas and you are a monkey's uncle, then your nephew is a chimp.
6. You love bananas, and if you are a monkey’s uncle, then your nephew is a chimp. 7. If your nephew is a chimp or you are a monkey's uncle, then you love bananas. 8γ Either your nephew is a chimp, or if you are a monkey's uncle, then you love bananas.
9. b - w
10. w - ~ s
13. ~ (b → s) 15.
11. (~ b v ~ w) - s
14. ~ b → (w a s) ~p → ~q
q
~P
T
τ
F
F
T
T
F
F
τ
T
F
τ
T
F
F
F
F
T
τ
T
P
12. s ^ (b - w)
~q
b. Yes; yes
Yes; no
P
q
~q
p → ~q
P
q
p → q
τ
τ
F
F
T
τ
T
F
τ
F
T
T
F
F
T
F
τ
T F
T
F
τ
T
F
F
F
T
T
F
F
T
F
~(p → q)
205
Appendix: Logic, pages 481-482 18.
~q
19. ~ (p -, q) and p ^ ~ q are
p a~q
P
q
τ τ
τ
F
F
F
T
T
F
τ
F
F T
F
F
P
q
p - q
q → p
(p → q) ^ (q → p)
T
τ
T
T
T
logically equivalent.
F
T
F
F
T
F
F
τ
T
F
F
F
F
T
T
T
Pages 481-482 ∙ EXERCISES
1. (1) Given; (2) Step 1 and Simplification; (3) Given; (4) Steps 2 and 3 and Modus Ponens
2. (1) Given; (2) Given; (3) Steps 1 and 2 and Modus Ponens; (4) Given; (5) Steps 3 and 4 and Modus Ponens 3.
4.
5>.
Reasons
Statements
1. Given
ι. p v q 2. ~ P 3. q
2. Given 3. Steps 1 and 2 and Disjunctive Syllogism
4. q - s
4. Given
5. s
5. Steps 3 and 4 and Modus Ponens
Reasons
Statements
1. a → b
1. Given
2. ~b
2. Given
3. ~ a
3. Steps 1 and 2 and Modus Tollens
4. a v c
4. Given
5. c
5. Steps 3 and 4 and Disjunctive Syllogism
Reasons
Statements
1. a a b
1. Given
2. a
2. Step 1 and Simplification
3. a -, ~ c
3. Given
4. ~ c
4. Steps 2 and 3 and Modus Ponens
5. c v d
5. Given 6. Steps 4 and 5 and Disjunctive Syllogism
6. d
Appendix: Logic, pages 483-484
206
6.
Statements
Reasons
1. p λ q
1. Given
2. p
2. Step 1 and Simplification 3. Given
3. p - ~ s 5. r - s
4. Steps 2 and 3 and Modus Ponens 5. Given
6. ~ r
6. Steps 4 and 5 and Modus Tollens
4. ~ s
7. Given: j - d; d - c; j ^ w.
Prove: c.
Proof:
Reasons
Statements
1. j λ w
1. Given
2. j 3. j - d 4. d
2. Step 1 and Simplification 3. Given 4. Steps 2 and 3 and Modus Ponens
5. d -, c
5. Given
6. c
6. Steps 4 and 5 and Modus Ponens
8. Given: a ^ r; b v c; a - ~ b.
Prove: c.
Proof:
Reasons
Statements
1. a ^ r
1. Given
2. a
2. Step 1 and Simplification
3. a - ~ b
3. Given
4. ~ b 5. b v c 6. c
4. Steps 2 and 3 and Modus Ponens 5. Given 6. Steps 4 and 5 and Disjunctive Syllogism
9. Given: t v r; t - p; r - s; ~ p.
Statements
Prove: s.
Proof:
Reasons
1. t → p
1. Given
2. ~p
2. Given
3. ~ t
3. Steps 1 and 2 and Modus Tollens
4. t v r
4. Given
5. r
6. r - s
5. Steps 3 and 4 and Disjunctive Syllogism 6. Given
7. s
7. Steps 5 and 6 and Modus Ponens
Pages 483-484 ∙ EXERCISES 1.
(1) Given; (2) Step 1 and Double Negation; (3) Given; (4) Steps 2 and 3 and Modus Tollens
207
Appendix: Logic, pages 483-484 2.
(1) Given; (2) Step 1 and Distributive Rule; (3) Step 2 and Simplification; (4) Step 3 and Commutative Rule; (5) Given; (6) Steps 4 and 5 and Disjunctive Syllogism
3.
4.
1. a ^ (b ^ c)
1. Given
2. (a ^ b) ^ c
2. Step 1 and Associative Rule
3. c λ (a ^ b)
3. Step 2 and Commutative Rule
4. c
4. Step 3 and Simplification
1. (p λ q) - s 3. ~ (p ^ q) 4. ~ p v ~ q
6.
1. Given 2. Given 3. Steps 1 and 2 and Modus Tollens 4. Step 3 and DeMorgan's Rule
Reasons
Statements 1. Given
1. P v (~ q) 2. (~ q) v p
2. Step 1 and Commutative Rule
3. q
3. Given
4. ~ (~ q)
4. Step 3 and Double Negation
5. p
5. Steps 2 and 4 and Disjunctive Syllogism
Reasons
Statements
1. ~ q → ~ p
1. Given
2. p - q
2. Step 1 and Contrapositive Rule
3. p 4. q
3. Given
5. q - r
6. r 7.
Reasons
Statements
2. ~ s
5.
Reasons
Statements
4. Steps 2 and 3 and Modus Ponens 5. Given 6. Steps 4 and 5 and Modus Ponens
Reasons
Statements 1. Given
1. p v (q λ s) 2. (p v q) a (p v s) 3. (p v s) a (p V q)
2. Step 1 and Distributive Rule
4. p v s
4. Step 3 and Simplification
3. Step 2 and Commutative Rule
Appendix: Logic, page 486
208
8.
Reasons
Statements 1. t v (r v s) 2. (r v s) v t
1. Given
3. ~ r ^ ~ s 4. ~(r v s)
2. Step 1 and Commutative Rule 3. Given 4. Step 3 and DeMorgan’s Rule
5. t
5. Steps 2 and 4 and Disjunctive Syllogism
9. Given: s - w; ~ s - p; ^ P∙
Prove: w.
Reasons
Statements 1. ~ s - p 2. ~ P 3. ~ (~ s) 4. s 5. s - w
Proof:
1. Given 2. Given
3. Steps 1 and 2 and Modus Tollens 4, Step 3 and Double Negation 5. Given 6. Steps 4 and 5 and Modus Ponens
6. w
10. Given: e v c; b v m; b - s; m - ~ e; ~ s.
Prove: c.
Proof:
Reasons
Statements
1. b - s
1. Given
2. ~ s
2. Given
3. ~ b 4. b v m
4. Given
5. m 6. m - ~ e 7. ~ e 8. e v c
6. Given 7. Steps 5 and 6 and Modus Ponens 8. Given
9. c
9. Steps 7 and 8 and Disjunctive Syllogism
3. Steps 1 and 2 and Modus Tollens 5. Steps 3 and 4 and Disjunctive Syllogism
Page 486 ∙ EXERCISES
1. p ^ r
2. r v s
5. (t v s) ^ (~t v s)
4. (r a p) v (~r λ q)
3. s ^ (t v p)
6. (r v s) v ~ r Electricity can always pass through the circuit p ⅛' ~p and
7. .-------- p----------- p.
can never pass through the
p∕∖~p
circuit p ^ ~ p.
.p.
8. ∙------- p---------- q--------- ∙
P^7
. Two planes, X and Y, are L if a plane Δ of the dihedral ∆ formed by X and Y
is a rt. L .
c. Y 1 X
d. that contains the line is J_ to the given plane.
7. 1. If 2 ∣∣ planes are cut by a third plane, alt. int. dihedral z⅛ are =. planes are cut by a third plane, same-side dihedral ∕⅜ are supp.
J_ to one of two ∣∣ planes, it is _L to the other also.
210
2. If 2 ∣∣
3. If a plane is
211
Appendix: Projections; Dihedral Angles, pages 489-490
C
b. Draw the altitude. XY, and draw YZ.ΔXYZ
8. a. Answerswill vary.
rt. Δ with YZ = 3√^Σ and XZ = 6, so ∆XYZ must be a45r-45^-90° lateral edge makes an
is a
Δ. Thus
a
t_of 450 with the base.
b. Draw XA L ZD and YA.
9. a. Answers will vary.
∆XZD is an equilateral Δ so XA = 3√^3.
YA = 3
2
and cos ∆XAY = —⅛z ≈ 0.5774; ∆A Z.A ≈ 55o. The The 33√^3 √3 measure of the dihedral L formed by a lateral face
c
and the base is ≈ 55°.
z Exs. 8-10
10. a. Answers will vary.
b. Draw CB and DB, both J_ to XZ. Δ CBY and ΔDBY o λ∕^o^ are ≡ rt. Δ. CY = 3√T, CB = 3√"3, so sin L CBY = -2⅜⊃ ≈ 0.8165 and ∕B≈ 3√3 54.70. Then L CBD ≈ 109o . ∆CBD is a plane L of dihedral ∆C-XZ-D, so
dihedral ∆C-XZ-D ≈ 109o. 11. a. Answers will vary.
y
b. Draw XY, the altitude of
The base is an equilateral Δ, so YZ = 2√3^ Consider AXYZ: cos z,XZY = ≈ 0.5774
the pyramid.
2√-3.
0
and L Z ≈ 55°.
Exs. 11, 12 ____
12.
a. Answers will vary. 0.3333 and ∆A ≈ 710.
Γq
____
b. Draw XA L ZB.
Consider ΔXYZ.
cos A =---- ≈
∆XAY is a plane L of dihedral ∆X-BZ-C, so dihedral
∆X-BZ-C ≈ 71°. 13.
Suppose the A shown are both plane A of the given
dihedral ∆.
All four rays are J_ to ST.
the pair of rays in each plane are ∣∣.
AJ ∣∣ CK.
Then
Choose A,
C, J and K so that AJ and CK are ∣∣ to ST.
λ
&
Since AJEB and CKEB are ti√, AJ = BE = CK, so AJKC is a ZZ7.
Then AB = JE and BC = EK.
Thus ∆ABC ≥ ΔJEK, so ∆ABC = Δ JEK.
212
Appendix: Projections; Dihedral Angles, pages 489-490
a. Consider AXAQ: cos 70° = AR ≈ 4.23; sin 65° =
AQ ≈ 3.42.
Consider AXAR: cos 65° = jγy>
XR ≈ 9.06.
b. Consider ∆AQS, a rt. Δ with ∆QAS = 50°: cos 50° ≈ Then RS = AS - AR ≈ 1.09.
tan 40° = —
Draw AP.
PR ≈ 0.92.
sin ∆PAX =
PY
AA
Ao
AS ≈ 5.32.
Now consider ΔPRS, a rt. Δ with ∆PSR = 40°: Consider rt. APRX: PX = √(XR)8 - (PR)8 ≈ 9.01. Q ∩1 —l ≈ 0.90, ∆PAX ≈ 640. Since AP is the
projection of AX into plane ABC, L PAX is the L of intersection of AX and
plane ABC.
CDEFGHIJ-A-8987654
14.