1 2

Chapter 1,
continued
2. H; Use the Midpoint Formula.
x1 1 x2 y1 1 y2
M }
,}
2
2
1
1
5. D;
2
24 1 6 5 1 (25)
2
2
AB 5 18.7 km
BC 5 2AB 5 2(18.7) 5 37.4 km
2
AC 5 AB 1 BC 5 18.7 1 37.4 5 56.1 km
M }, } 5 M (1, 0)
AB 1 BC 1 CA 5 18.7 1 37.4 1 56.1 5 112.2 km
The coordinates of point E are (1, 0).
To find the coordinates of point D, substitute the
coordinates of point C into the midpoint formula, and
set each coordinate in the midpoint formula equal to the
corresponding coordinate from the midpoint E.
21x
}51
2
81y
}50
2
21x52
81y50
x50
Jill travels 112.2 kilometers.
6. G;
}
BC represents the intersection of planes ABC and BFE.
7. 23x 1 5 5 25x 2 4
5 5 2x 2 4
9 5 2x
4.5 5 x
y 5 28
PQ 5 23(4.5) 1 5 1 25(4.5) 2 4
The coordinates of point D are (0, 28).
5 103.5 1 5 1 112.5 2 4
3. A;
5 217
}
The length of PQ is 217 inches.
B(3, 1), C(3, 2), D(3, 5), E(7, 5), F(10, 5)
BD 5 {5 2 1{ 5 {4{ 5 4
8. d 5 rt
DF 5 {10 2 3{ 5 {7{ 5 7
r 5 2.4 km/h
Using the existing roads, a trip from B to F is
4 1 7 5 11 miles.
t 5 45 min 5 0.75 h
d 5 2.4(0.75) 5 1.8
BC 5 {2 2 1{ 5 {1{ 5 1
}}
}
5.4 2 1.8 5 3.6
}
CE 5 Ï(7 2 3)2 1 (5 2 2)2 5 Ï16 1 9 5 Ï25 5 5
The length to the end of the trail is 3.6 kilometers.
EF 5 {10 2 7{ 5 {3{ 5 3
Using the new road where possible, a trip from B to F is
1 1 5 1 3 5 9 miles, which is 11 2 9 5 2 miles shorter.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
4. H;
Lesson 1.4
1.4 Guided Practice (pp. 24–28)
1. ŽPQR
T (21, 3) y
ŽRQS
S
ŽPQS
(24, 1) 1
1
ŽPQS is a right angle.
x
2.
(3, 23)
Perimeter 5 QR 1 RS 1 ST 1 TQ
}}}
QR 5 Ï(0 2 3)2 1 [25 2 (23)]2
}
5 Ï9 1 4 5 Ï13 ø 3.6
}}}
RS 5 Ï(24 2 0)2 1 [1 2 (25)]2
}
}
5 Ï16 1 36 5 Ï52 ø 7.2
}}}
ST 5 Ï[21 2 (24)]2 1 (3 2 1)2
}
}
5 Ï9 1 4 5 Ï13 ø 3.6
}}}
TQ 5 Ï[3 2 (21)]2 1 (23 2 3)2
}
P
The rays form a straight angle.
R(0, 25)
}
R
}
5 Ï16 1 36 5 Ï52 ø 7.2
Perimeter ø 3.6 1 7.2 1 3.6 1 7.2 5 21.6
The perimeter of rectangle QRST is about 21.6 units.
3. mŽKLN 1 mŽNLM 5 1808
10x 2 5 1 4x 1 3 5 180
14x 2 2 5 180
14x 5 182
x 5 13
10(13) 2 5 5 125
mŽKLN 5 1258
14(13) 1 3 5 55
mŽNLM 5 558
4. mŽEFM 1 mŽHFG 5 908
2x 1 2 1 x 1 1 5 90
3x 1 3 5 90
3x 5 87
x 5 29
mŽEFM 5 2(29) 1 2 5 608
mŽMFG 5 29 1 1 5 308
Geometry
Worked-Out Solution Key
9
Chapter 1,
continued
5. ŽQPT > ŽQRS
16. Another name for ŽABC is ŽCBA. The angle is an acute
angle because its measure is less than 908.
ŽPTS > ŽRST
17. Another name for ŽBFD is ŽDFB. The angle is a
6. If mŽQRS 5 848 and ŽQRS > ŽQPT, then
mŽQPT 5 848. If mŽTSR 5 1218 and ŽTSR > ŽSTP,
then mŽSTP 5 1218.
straight angle because its measure is 1808.
18. Another name for ŽAEC is ŽCEA. The angle is an
obtuse angle because its measure is between 908 and
1808.
7.
19. Another name for ŽBDC is ŽCDB. The angle is an
acute angle because its measure is less than 908.
M
N
20. Another name for ŽBEC is ŽCEB. The angle is an acute
P
angle because its measure is less than 908.
mŽMNQ 5 908
21. B; mŽACD is between 908 and 1808.
mŽPNQ 5 908
22. mŽQST 5 mŽQSR 1 mŽRST 5 528 1 478 5 998
23. mŽADC 5 mŽADB 1 mŽCDB 5 218 1 448 5 658
1.4 Exercises (pp. 28–32)
24. mŽNPM 5 mŽLPM 2 mŽLPN 5 1808 2 798 5 1018
Skill Practice
25. x 1 5 1 3x 2 5 5 80
1. Answers may vary. Sample answers:
4x 5 80
A
x 5 20
D
obtuse
B
mŽYXZ 5 3(20) 2 5 5 558
26. 6x 2 15 1 x 1 8 5 168
C
acute
E
L
H
7x 2 7 5 168
F
M
straight
7x 5 175
x 5 25
N
mŽFJG 5 6(25) 2 15 5 1358
right
27. A; 2x 1 6 1 80 5 140
2x 5 54
x 5 27
K
2. The measure of ŽPQR is equal to the absolute value of
###$ and
the difference between the degree measures of QP
###$
QR.
28. Ž AED > Ž ADE > Ž BDC > Ž BCD > Ž DAB >
Ž ABD; Ž EAD > Ž DBC > Ž ADB
mŽ BDC 5 348
mŽ ADB 5 1128
3. ŽABC, ŽCBA, ŽB
BA and ###$
BC are the sides.
B is the vertex, and ###$
29. mŽ ZWY 5 mŽ XWY 5 528
mŽ XWY 5 2(52) 5 1048
4. ŽNQT, ŽTQN, ŽQ
###$ and ###$
QT are the sides.
Q is the vertex, and QN
5. ŽMTP, ŽPTM, ŽT
1
30. mŽ ZWX 5 }mŽ XWY 5 688
2
mŽ XWY 5 2(68) 5 1368
###$ and ###$
T is the vertex, and TM
TP are the sides.
1
31. mŽ ZWY 5 } Ž XWY 5 35.58
2
6. ŽQRS, ŽTRS, ŽQRT
7. Straight
8. Acute
9. Right
10. Obtuse
11–14.
H
J
G
32.
J
F
K
L
308
M
11. mŽJFL 5 908, Right
K
12. mŽGFM 5 608, Acute
mŽ JKL is twice the measure of Ž JKM, not half of it.
mŽ JKL 5 608
13. mŽGFK 5 1358, Obtuse
14. mŽGFL 5 1808, Straight
15. Another name for ŽACB is ŽBCA. The angle is a right
angle because it is labeled with a red square.
10
mŽ XWZ 5 718 2 35.58 5 35.58
Geometry
Worked-Out Solution Key
L
33. a8 5 180 2 142 5 388
34. b8 > a8 5 388
35. c8 5 1428
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
J
Chapter 1,
continued
36. d 8 5 180 2 53 2 90 5 378
ŽABC is obtuse.
y
47.
37. e8 5 538
B
1
38. f 8 > d > 378
1
A
x
39. For a ray to bisect Ž AGC, the endpoint of the ray must
C
be at point G.
Sample answer: Point (2, 21) lies in the interior of
the angle.
A
0 < (2x 2 12)8 < 908
48.
0 1 12 < 2x < 90 1 12
G
40.
C
12 < 2x < 112
4x 2 2 5 3x 1 18
6 < x < 51
x 5 20
mŽABC 5 4(20) 2 2 1 3(20) 1 18 5 1568
49. 688
Sample answer: Since mŽVSP 5 178, mŽRSP 5 348.
41. 2x 1 20 5 4x
Since mŽRSP 5 348, mŽRSQ 5 688, which is equal to
mŽTSQ.
20 5 2x
10 5 x
mŽABC 5 2(10) 1 20 1 4(10) 5 808
42.
1
mŽAEB 5 }2 + mŽCED
50.
1
2
} + mŽCED 1 90 1 mŽCED 5 180
x
2
} 1 17 5 x 2 33
1
x 1 34 5 2x 2 66
1}2 + mŽCED 1 90 5 180
100 5 x
1
1}2 + mŽCED 5 90
100
mŽABC 5 }
1 17 1 100 2 33 5 1348
2
mŽCED 5 608
43. ###$
QP lines up with the 758 mark. The new mark for ###$
QR is
The angle is acute.
y
44.
1
mŽAEB 5 }2 (60) 5 308
Problem Solving
51. mŽLMP 5 mŽLMN 2 mŽPMN
mŽLMP 5 79 2 47
A
mŽLMP 5 328
1
22
B
C
52.
x
Malcolm Way
et
Stre
ney
Syd
87°
Sample answer: point (2, 1) lies on the interior of
the angle.
y
A
d
Roa
45.
162°
k
Par
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
58 less than before. The difference between the marks
###$ and ###$
that QR
QP line up with on the protractor must
remain the same.
ŽABC is acute.
B
3
1628 2 878 5 758
Malcom Way intersects Park Road at an angle of 758.
22
x
53. a. mŽDEF 5 mŽABC 5 1128
C
Sample answer: point (22, 2) lies in the interior of
the angle.
46.
ŽABC is obtuse.
y
A
1
21
x
B
C
Sample answer: point (21, 21) lies in the interior of the
angle.
1
b. Because ###$
BG bisects ŽABC, mŽABG 5 }2 + mŽABC
1
5 }2 (1128) 5 568.
1
c. Because ###$
BG bisects ŽABC, mŽCBG 5 }2 + mŽABC
1
5 }2 (1128) 5 568.
1
d. Because ###$
BG bisects ŽDEF, mŽDEG 5 }2 + mŽDEF
1
5 }2 (1128) 5 568.
Geometry
Worked-Out Solution Key
11
Chapter 1,
continued
54. Because ŽDGF is a straight angle, mŽDGF 5 1808.
Investigating Geometry Activity 1.4 (p. 34)
mŽDGE 5 908
1. Sample answer: Draw a segment more than twice as long
mŽFGE 5 908
as the given segment. Set your compass to the length of
the given segment. Using your compass, mark off two
adjacent line segments on the line segment you drew.
55. Sample answer: Acute: ŽABG, Obtuse: ŽABC,
Right: ŽFGE, Straight: ŽDGF
56. about 1588
57. about 1408
58. about 1678
59. about 628
60. about 398
61. about 1078
2. Sample answer:
Step 1
Step 1
62. a. ŽAFB, ŽBFC, ŽCFD, ŽDFE are acute.
F
ŽAFD, ŽBFD, ŽBFE are obtuse.
ŽAFC and ŽCFE are right.
A
D
b. ŽAFB > DFE, ŽAFC > ŽCFE, ŽBFC > ŽDFC
c. mŽAFB > mŽDFE 5 268
Step 2
mŽBFC 5 908 2 mŽAFD 5 908 2 268 5 648
mŽBFC > mŽCFD 5 648
E
D
F
Step 2
mŽAFC 5 908
mŽAFD 5 908 1 mŽCFD 5 908 1 648 5 1548
C
D
E
mŽBFD 5 mŽBFC 1 mŽCFO 5 648 1 648 5 1288
ŽDFE > ŽAFB, so they both have an angle
measure of 268.
A
ŽBFC and ŽAFB form a 908 angle so their
measurements add up to 908, making mŽBFC 5 648.
ŽCFD > ŽBFC, so it also has an angle measurement
of 648. ŽAFC is a right angle, so mŽAFC 5 908.
mŽAFD 5 mŽAFC 1 mŽCFD 5 908 1 648 5 1548,
mŽBFD 5 mŽBFC 1 mŽCFD 5 648 1 648 5 1288.
B
Step 3
F
D
E
Step 3
D
C
E
Step 4
63. Sample answer: In your drawer you have 4 pairs of
A
brown socks, 4 pairs of black socks, 4 pairs of gray
socks, 6 pairs of white socks, and 6 pairs of blue socks.
B
F
1
The brown, black and gray socks each represent }6 , and
D
D
1
the white and blue socks each represent }4.
E
E
Step 4
Mixed Review for TAKS
64. B; y 5 2.6x 2 2 3.4x 1 1.2
When x 5 8:
y 5 2.6(8)2 2 3.4(8) 1 12
5 2.6(64) 2 3.4(8) 1 1.2
5 166.4 2 27.2 1 1.2
5 140.4
After they are in business for 8 years, the company’s
profit is $140.4 million.
65. F;
d
4
r 5 }2 5 }2 5 2
h 5 7 2 0.5
F
D
E
Lesson 1.5
1.5 Guided Practice (pp. 35–37)
1. Because 418 1 498 5 908, ŽFGK and ŽGKL are
complementary.
Because 1318 1 498 5 1808, ŽHGK and ŽGKL are
supplementary.
Because ŽFGK and ŽHGK share a common vertex and
side, they are adjacent.
2. No, they do not share a common vertex.
V 5 Bh
5 :r 2h
5 :(2 )(7 2 0.5)
2
No, they do have common interior points.
3. mŽ1 1 mŽ2 5 908
mŽ1 1 88 5 908
mŽ1 5 828
12
Geometry
Worked-Out Solution Key
4. mŽ3 1 mŽ4 5 808
1178 1 mŽ4 5 1808
mŽ4 5 638
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
P