Midterm Review Packet Answers

Transcription

Extra Practice
Chapter 1 (pp. 896–897)
15.
AB 1 BC 5 AC
(4x 2 5) 1 (2x 2 7) 5 54
1. Sample answer: Points A, F, and B are collinear. A name
for the line is @##$
AB.
6x 2 12 5 54
@##$ is F.
2. The intersection of plane ABC and EG
6x 5 66
###$ are opposite rays, ###$
3. Sample answer: ###$
FE and FG
FA and ###$
FB
x 5 11
are opposite rays.
AB 5 4x 2 5 5 4(11) 2 5 5 39
4. Yes; because A, C, and G are noncollinear, there is a
BC 5 2x 2 7 5 2(11) 2 7 5 15
}
}
Because AB 5 39 and BC 5 15, AB and BC
are not congruent.
plane that contains them.
5. Sample answer: @##$
AB intersects plane AFD at more than
one point.
6. PQ 1 QT 5 PT
7. PS 5 PQ 1 QS
PQ 1 42 5 54
16.
PS 5 12 1 31
PQ 5 54 2 42
AB 1 BC 5 AC
(14x 1 5) 1 (10x 1 15) 5 80
24x 1 20 5 80
PS 5 43
24x 5 60
PQ 5 12
8. QR 1 RS 5 QS
x 5 2.5
9. PR 5 PQ 1 QR
QR 1 17 5 31
AB 5 14x 1 5 5 14(2.5) 1 5 5 40
PR 5 12 1 14
QR 5 31 2 17
BC 5 10x 1 15 5 10(2.5) 1 15 5 40
}
}
are congruent.
PR 5 26
QR 5 14
10. QS 1 ST 5 QT
11. RT 5 RS 1 ST
31 1 ST 5 42
RT 5 17 1 11
ST 5 42 2 31
RT 5 28
17.
AB 1 BC 5 AC
(3x 2 7) 1 (2x 1 5) 5 108
5x 2 2 5 108
ST 5 11
12.
5x 5 110
AB 1 BC 5 AC
x 5 22
(x 1 3) 1 (2x 1 1) 5 10
AB 5 3x 2 7 5 3(22) 2 7 5 59
3x 1 4 5 10
BC 5 2x 1 5 5 2(22) 1 5 5 49
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
3x 5 6
}
}
are not congruent.
x52
AB 5 x 1 3 5 2 1 3 5 5
1
}
}
Because AB 5 5 and BC 5 5, AB and BC are congruent.
13.
AB 1 BC 5 AC
2
1
2
11
23 2 8 22 1 4
19. M }, } 5 M 2}, 1
2
2
2
1
2
1
2
2
22.3 1 3.1 21.9 2 9.7
20. M }, } 5 M(0.4, 25.8)
2
2
(3x 2 7) 1 (3x 2 1) 5 16
6x 2 8 5 16
1
2
3 2 1 27 1 9
21. M }, } 5 M(1, 1)
2
2
6x 5 24
x54
1
2
5
412 312
22. M }, } 5 M 3, }
2
2
2
AB 5 3x 2 7 5 3(4) 2 7 5 5
BC 5 3x 2 1 5 3(4) 2 1 5 11
}
}
Because AB 5 5 and BC 5 11, AB and BC are
not congruent.
14.
1
3
2 1 7 24 1 1
9
18. M }, } 5 M }, 2}
2
2
2
2
BC 5 2x 1 1 5 2(2) 1 1 5 5
AB 1 BC 5 AC
(11x 2 16) 1 (8x 2 1) 5 78
19x 2 17 5 78
19x 5 95
x55
AB 5 11x 2 16 5 11(5) 2 16 5 39
BC 5 8x 2 1 5 8(5) 2 1 5 39
1
1
2
2
1.7 1 8.5 27.9 2 8.2
23. M }, } 5 M(5.1, 28.05)
2
2
}}
}
24. ZM 5 Ï (7 2 0) 1 (1 2 1)2 5 Ï 49 5 7
2
So, the length of the segment with endpoint Z and
midpoint M is 2 p ZM 5 2 p 7 5 14.
}}
}
25. YM 5 Ï (1 2 4)2 1 (7 2 3)2 5 Ï 25 5 5
So, the length of the segment with endpoint Z and
midpoint M is 2 p YM 5 2 p 5 5 10.
}}}
}
26. XM 5 Ï (12 2 0)2 1 (4 2 [21])2 5 Ï 169 5 13
So, the length of the segment with endpoint X and
midpoint M is 2 p XM 5 2 p 13 5 26.
}
}
are congruent.
Geometry
Worked-Out Solution Key
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7/11/06 11:47:38 AM
continued
}}}
}
27. WM 5 Ï (210 2 5)2 1 (25 2 3)2 5 Ï 289 5 17
So, the length of the segment with endpoint W and
midpoint M is 2 p WM 5 2 p 17 5 34.
}}}
}
28. VM 5 Ï [9 2 (23)] 1 [5 2 (24)] 5 Ï 225 5 15
2
2
So, the length of the segment with endpoint V and
midpoint M is 2 p VM 5 2 p 15 5 30.
}}
}
29. UM 5 Ï (11 2 3)2 1 (24 2 2)2 5 Ï 100 5 10
41. The figure is a concave polygon.
42. The figure is not a polygon because at least one side
intersects more than two other sides.
43. The figure is not a polygon because part of the figure is
not a line segment.
44. The figure is a convex polygon.
45. DFHKB and ABCDEFGHJK are equilateral polygons.
DFHKB is a pentagon because it has five sides.
ABCDEFGHJK is a decagon because it has ten sides.
So, the length of the segment with endpoint U and
midpoint M is 2 p UM 5 2 p 10 5 20.
30. m∠QPS 5 m∠QPR 1 m∠RPS 5 578 1 648 5 1218
46. Sample answer:
31. m∠LMN 5 m∠LMJ 1 m∠JMN 5 368 1 688 5 1048
ABK is a triangle.
32. ∠WXY > ∠ZWY, so m∠ZWY 5 438.
AEJK is a quadrilateral.
BDFHK is a pentagon.
m∠XWZ 5 m∠XWY 1 m∠ZWY 5 438 1 438 5 868
ADGHJK is a hexagon.
33. m∠ABC 5 m∠ABD 1 m∠CBD
JCDEFGH is a heptagon.
1338 5 (7x 1 4)8 1 (3x 1 9)8
133 5 10x 1 13
47.
91 5 l(7)
120 5 10x
13 5 l
12 5 x
The length l is 13 centimeters.
So, m∠ABD 5 (7x 1 4)8 5 (7[12] 1 4)8 5 888.
34. m∠GHK 5 178
4x 2 3 5 17
A 5 lw
1
1
48. A 5 } bh 5 }(6)(8) 5 24
2
2
The area of the triangle is 24 square feet.
4x 5 20
x55
So, m∠KHJ 5 (3x 1 2)8 5 (3[5] 1 2)8 5 178.
35. Because they share a common vertex and side, but
have no common interior points, ∠1 and ∠2 are
adjacent angles.
49.
36. 418 1 m∠1 1 498 1 m∠2 5 1808
50.
1
A 5 }2 bh
1
66 5 }2(12)h
11 5 h
The height h of the triangle is 11 meters.
y
908 1 m∠1 1 m∠2 5 1808
B(3, 6)
m∠1 1 m∠2 5 908
So, ∠1 and ∠2 are complementary.
37. The sides of ∠1 and ∠2 form two pairs of opposite
rays. So ∠1 and ∠2 are vertical angles. Also, ∠1
forms a linear pair with a right angle. The angles in a
linear pair are supplementary, so m∠1 1 908 5 1808,
or m∠1 5 908. Similarly, m∠2 5 908. So ∠1 and ∠2
are supplementary because m∠1 1 m∠2 5
908 1 908 5 1808.
38. Sample answer: ∠ACE and ∠ACB form a linear pair,
so they are supplementary. ∠ACB > ∠BCD, so
m∠ACB 5 m∠BCD. Using substitution,
m∠ACE 1 m∠BCD 5 1808. So, ∠ACE and ∠BCD
are supplementary.
39. Sample answer:∠ACE and ∠BCF are vertical angles that
cannot be complementary, because each angle is obtuse.
40. Because ∠ACF is a straight angle, m∠DCF 5 908, and
∠ACB > ∠BCD, ∠ACB and ∠BCD are complementary
angles and adjacent angles. Because ∠BCE is a straight
angle, m∠DCF 5 908, and ∠ECF > ∠BCD, ∠ECF and
∠BCD are complementary angles, and neither vertical
nor adjacent angles. Because ∠ECF > ∠BCD, ∠ECF
and ∠ACB are complementary angles and vertical
angles.
462
1
A(2, 1)
C(6, 1)
x
1
}}
}
}}
}
AB 5 Ï(3 2 2)2 1 (6 2 1)2 5 Ï26 ø 5.1
BC 5 Ï(1 2 6)2 1 (6 2 3)2 5 Ï34 ø 5.8
AC 5 6 2 2 5 4
P 5 AB 1 BC 1 AC ø 5.1 1 5.8 1 4 5 14.9
The height h is the vertical distance from B to the
}
horizontal base AC, so h 5 6 2 1 5 5.
1
1
A 5 }2 bh 5 }2 (4)(5) 5 10
So, the perimeter of nABC is about 14.9 units and the
area is 10 square units.
Extra Practice,
Geometry
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Extra Practice,
51.
continued
6. Starting with the third number, each number is the sum of
y
the previous two numbers.
F(6, 5)
The next number in the pattern is 16 1 26 5 42.
214
1
D (1, 1) E(3, 1)
2,
x
1
2,
4,
212
DE 5 3 2 1 5 2
}}
}
}}
}
EF 5 Ï(6 2 3)2 1 (5 2 1)2 5 Ï25 5 5
P 5 DE 1 DF 1 EF ø 2 1 6.7 1 5 5 13.7
The height h is the vertical distance from F to the
}
horizontal line containing base DE, so h 5 5 2 1 5 4.
1
1
A 5 }2 bh 5 }2(2)(4) 5 4
So, the perimeter of nDEF is about 13.7 units and the
area is 4 square units.
Chapter 2 (pp. 898–899)
15, 21, 13, 19, . . .
The pattern alternates between adding 6 to the previous
number and subtracting 8 from the previous number.
The next number in the pattern is 19 2 8 5 11.
0.5,
1
0.25, 0.125, 0.0625, . . .
1
3}2
1
3}2
3}2
Each number is }12 the previous number.
The next number in the pattern is }12 (0.625) 5 0.03125.
The numbers are consecutive prime numbers.
The next prime number in the pattern is 17.
7.5,
8.0,
8.5, . . .
Each number is 0.5 more than the previous number.
The next number in the pattern is 8.5 1 0.5 5 9.0
1
3
1
9
},
1
},
1
3}3 3}3
If x 5 0, then 2x 5 0, so 2x ò x. Because a
counterexample exists, the conjecture is false.
9. Sample answer:
AB can be drawn so that it divides
Suppose m∠A 5 408. ###$
∠A into two 208 angles. So ∠A can be bisected, but
∠A is not obtuse. Because a counterexample exists, the
conjecture is false.
20�
20�
B
10. If-then form: If two lines intersect, then they form two
pairs of vertical angles.
Inverse: If two lines do not intersect, then they do not
form two pairs of congruent angles.
Contrapositive: If two lines do not form two pairs of
vertical angles, then the two lines do not intersect.
regular polygon.
Converse: If a figure is a four-sided regular polygon, then
it is a square.
Inverse: If a figure is not a square, then it is not a
four-sided regular polygon.
10.5 10.5 10.5
5. 1,
8. Sample answer:
11. If-then form: If a figure is a square, then it is a four-sided
3. 2, 3, 5, 7, 11, 13, . . .
4. 7.0,
4 1 6 10 1 16
Converse: If two lines form two pairs of vertical angles,
then the two lines intersect.
1
3}2
10, 16, 26, . . .
The difference of 10 and 12 is 10 2 12 5 22 and 22 is
not between 10 and 12. Because a counterexample exists,
the conjecture is false.
A
16 28 16 28 16
2. 1,
6,
7. Sample answer:
DF 5 Ï(6 2 1)2 1 (5 2 1)2 5 Ï45 ø 6.7
1. 17, 23,
6 1 10
1
27
}, . . .
Contrapositive: If a figure is not a four-sided regular
polygon, then it is not a square.
12. False; Not all hexagons are regular.
Counterexample:
1
3}3
Each number is }13 the previous number.
1
}
The next number in the pattern is }13 1 }
27 2 5 81 .
1
13. True; By definition, two angles are complementary angles
if the sum of their measures is 908.
14. By the Law of Syllogism, the given conditional
statements lead to the following conditional statement:
“If a triangle is equilateral, then it is regular.”
Geometry
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Extra Practice,
continued
15. By the Law of Syllogism, the given conditional
statements lead to the following conditional statement:
“If two coplanar lines are not parallel, then they form
congruent vertical angles.”
16. John only does his math homework when he is in study
hall. John is doing his math homework. So, John is in
study hall.
17. May sometimes buys pretzels when she goes to the
supermarket. May is at the supermarket. So, she might
buy pretzels.
@##$, but
18. Cannot be determined; The marking shows @##$
SV ⊥ TW
that does not guarantee that @##$
SV is perpendicular to every
line in plane Z that it intersects.
31. If m∠JKL 5 m∠GHI and m∠GHI 5 m∠ABC, then
m∠JKL 5 m∠ABC.
Transitive Property of Equality
32. If m∠MNO 5 m∠PQR, then m∠PQR 5 m∠MNO.
Symmetric Property of Equality
33. m∠XYZ 5 m∠XYZ
Reflexive Property of Equality
34. Given: Point C is in the interior of ∠ABD.
∠ABD is a right angle.
Prove: ∠ABC and ∠CBD are complementary.
A
C
19. True; The diagram shows that point Y lies in plane Z and
20.
B
@##$ lies in plane Z.
True; The diagram shows that TW
21. False; ∠SYT and ∠WYS are adjacent because they share
side ###$
YS .
22. False; ∠SYT and ∠TYV are both right angles, so they are
Statements
Reasons
1. ∠ABD is a right angle. 1. Given
2. m∠ABD 5 908
2. Definition of right angle
###$ of ∠TYU and
23. True; The noncommon sides ###$
YT and YW
3. Point C is in the
interior of ∠ABD.
3. Given
24. Cannot be determined; The diagram does not give
4. m∠ABD 5 m∠ABC 1 4. Angle Addition Postulate
m∠CBD
supplementary.
∠UYW are opposite rays.
enough information.
25. 4x 1 15 5 39
Write original equation.
4x 5 24
Subtraction Property of Equality
x56
Division Property of Equality
26. 6x 1 47 5 10x 2 9
47 5 4x 2 9
56 5 4x
Addition Property of Equality
14 5 x
Divison Property of Equality
27. 2(27x 1 3) 5 250
214x 1 6 5 250
214x 5 256
x54
54 1 9x 5 21x 1 18
54 5 12x 1 18
36 5 12x
35x
5. 908 5 m∠ABC 1
m∠CBD
5. Substitution Property of
Equality
6. ∠ABC and m∠CBD
are complementary.
6. Definition of
complementary angles
}
}
}
35. Given: XY > YZ > ZX
Prove: The perimeter of nXYZ is 3 p XY.
X
Distributive Property
28. 54 1 9x 5 3(7x 1 6)
Z
Y
Statements
} }
1. XY > YZ > ZX
2. XY 5 YZ 5 ZX
2. Definition of congruent
segments
3. Perimeter of nXYZ 5
XY 1 YZ 1 ZX
3. Perimeter formula
4. Perimeter of nXYZ 5
XY 1 XY 1 XY
4. Substitution
5. Perimeter of
nXYZ 5 3 p XY
5. Simplify.
}
29. 13(2x 2 3) 2 20x 5 3
26x 2 39 2 20x 5 3
6x 2 39 5 3
Simplify.
6x 5 42
Addition Property of Equality
x57
30. 31 1 25x 5 7x 2 14 1 3x
31 1 25x 5 10x 2 14
Simplify.
31 1 15x 5 214
15x 5 245
x 5 23
464
D
Reasons
1. Given
@##$ passes through point Z.
XU
36. If m∠CGF 5 1588, the m∠EGD 5 1588. ∠CGF and
∠EGD are vertical angles, so they are congruent by the
Vertical Angles Congruence Theorem.
Geometry
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Extra Practice,
continued
37. If m∠EGA 5 678, the m∠FGD 5 238.
Because m∠AGD 5 908, m∠AGC 5 1808 2 908 5 908
by the Linear Pair Postulate. Then, m∠EGC 5
m∠AGC 2 m∠EGA 5 908 2 678 5 238.
Because ∠EGC and ∠FGD are vertical angles, they are
congruent. So, m∠EGC 5 m∠FGD 5 238.
38. If m∠FGC 5 1498, then m∠EGA 5 598.
By the Vertical Angles Congruence Theorem,
m∠BGC 5 908. So, m∠FGB 5 m∠FGC 2 m∠BGC 5
1498 2 908 5 598. Because ∠FGB and ∠EGA are vertical
angles, they are congruent. So, m∠FGB 5 m∠EGA 5 598.
39. m∠DGB 5 908
Because m∠AGD 5 908, m∠DGB 5 1808 2 908 5 908
by the Linear Pair Postulate.
40. m∠FGH 5 908
∠AGD and ∠BGD form a linear pair, and
m∠AGD 5 908. So, m∠BGD 5 908. m∠BGD 5
m∠FGB 1 m∠FGD. It is given that ∠FGD > ∠BGH,
so m∠FGD 5 m∠BGH. Using substitution, m∠FGB 1
m∠BGH 5 908. So, m∠FGH 5 908.
41. Given: ∠UKV and ∠VKW are complements.
Prove: ∠YKZ and ∠XKY are complements.
U
Z
V
9. ∠CJD and ∠FKL are alternate exterior angles.
∠CJD and ∠AML are alternate exterior angles.
10. ∠LMJ and ∠MLK are consecutive interior angles.
∠LMJ and ∠MJK are consecutive interior angles.
@##$ is a transversal of @##$
@##$.
11. BG
AD and HE
@##$ is a transversal of @##$
@##$.
CF
AD and HE
12. m∠1 5 448, m∠2 5 1368; m∠2 5 1368 because when
two parallel lines are cut by a transversal, the alternate
exterior angles are congruent; m∠1 5 448 because it is a
linear pair with ∠2 (1808 2 1368 5 448).
13. m∠1 5 688, m∠2 5 1128; m∠1 5 688 because when
two parallel lines are cut by a transversal, the alternate
interior angles are congruent; m∠2 5 1128 because it is a
linear pair with ∠1 (1808 2 688 5 1128).
14. m∠1 5 1068, m∠2 5 1068; m∠1 5 1068 because
when two parallel lines are cut by a transversal, the
corresponding angles are congruent; m∠2 5 1068
because vertical angles are congruent.
15. Because alternate interior angles are congruent,
9x8 5 818
x59
Because two angles that form a linear pair are
supplementary,
K
(100 2 y)8 1 818 5 1808
W
Y
8. ∠AML and ∠MLK are alternate interior angles.
181 2 y 5 180
X
181 5 y 1 180
Statements
1. ∠UKV and ∠VKW
are complements.
Reasons
1. Given
3x8 1 908 5 1808
2. m∠UKV 1 m∠VKW 2. Definition of
5 908
3. ∠UKV > ∠XKY,
∠VKW > ∠YKZ
15y
16. Because consecutive interior angles are supplementary,
3. Vertical angles are
congruent.
4. m∠UKV 5 m∠XKY, 4. Definition of congruent
m∠VKW 5 m∠YKZ
angles
(13y 1 5)8 1 (5y 2 5)8 5 1808
3x 5 90
18y 5 180
x 5 30
y 5 10
17. Because alternate exterior angles are congruent,
(7y 2 18)8 5 (6y 1 1)8
y 2 18 5 1
y 5 19
5. m∠YKZ 1 m∠XKY
5 908
5. Substitution
Because two angles that form a linear pair are
supplementary,
6. ∠YKZ and ∠XKY
are complements.
6. Definition of
(6y 1 1)8 1 (3x 2 10)8 5 1808
6(19) 1 1 1 3x 2 10 5 180
114 1 1 1 3x 2 10 5 180
Chapter 3 (pp. 900–901)
1. ∠6 and ∠2 are corresponding angles.
2. ∠7 and ∠2 are alternate exterior angles.
3x 1 105 5 180
3x 5 75
x 5 25
3. ∠5 and ∠3 are consecutive interior angles.
18. No, there is not enough information to prove that min.
4. ∠4 and ∠5 are alternate interior angles.
19. Yes; You would use the Consecutive Interior Angles
5. ∠1 and ∠5 are corresponding angles.
6. ∠3 and ∠6 are alternate interior angles.
7. ∠AMB and ∠HLM are corresponding angles.
Converse Theorem. If two lines are cut by a transversal
so that a pair of consecutive interior angles are
supplementary, then the lines are parallel.
∠AMB and ∠MJC are corresponding angles.
Geometry
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7/11/06 11:47:56 AM
continued
20. Yes; You would use the Consecutive Interior Angles
Converse Theorem. If two lines are cut by a transversal
so that a pair of consecutive interior angles are
supplementary, then the lines are parallel.
21. Yes; If two lines are cut by a transversal so that alternate
interior angles are congruent, then the lines are parallel.
22. Yes; Sample answer: If two lines are cut by a transversal
so the alternate interior angles are congruent, then the
lines are parallel.
23. Yes; If two lines are cut by a transversal so the
consecutive interior angles are supplementary, then the
lines are parallel.
524
1
24. Slope of Line 1: m 5 } 5 }
10 2 7
3
31. y 5 mx 1 b
2
When m 5 }3 , x 5 23, and y 5 0:
2
0 5 }3(23) 1 b
0 5 22 1 b
25b
2
So, y 5 }3x 1 2.
32. y 5 mx 1 b
1
When m 5 2}3, x 5 9, and y 5 4:
1
4 5 2}3 (9) 1 b
523
2
1
5 }6 5 }3
Slope of Line 2: m 5 }
822
4 5 23 1 b
Parallel; The lines are parallel because the slopes
are equal.
521
4
25. Slope of Line 1: m 5 } 5 } 5 4
1
22 2 (23)
22 2 (23)
1
5 }6
5 2 (21)
75b
1
So, y 5 2}3 x 1 7.
33. y 5 mx 1 b
When m 5 22, x 5 1, and y 5 22:
Neither; The lines are not parallel because the slopes are
not equal and the lines are not perpendicular because the
product of the slopes is not 21.
720
7
1
26. Slope of Line 1: m 5 } 5 } 5 }
14
2
8 2 (26)
22 5 22(1) 1 b
22 5 22 1 b
05b
So, y 5 22x.
34. y 5 mx 1 b
224
22
5}
5 22
221
1
When m 5 2}3 , x 5 6, and y 5 3:
Perpendicular; The lines are perpendicular because the
product of their slopes is 21.
3 5 2}3 (6) 1 b
29 2 (26)
23
3
27. Slope of Line 1: m 5 } 5 } 5 }
24 2 0
24
4
925
4
3 5 22 1 b
55b
So, y 5 2}3 x 1 5.
3
Line 2 is steeper because }3  >  }4.
3 2 (25)
21 2 (21)
1
1
5 }3
1 2 (22)
4
1
8
0
28. Slope of Line 1: m 5 } 5 } undefined
0
424
35. y 5 mx 1 b
When m 5 1, x 5 27, and y 5 3:
3 5 1(27) 1 b
3 5 27 1 b
5}
50
22
25 2 (23)
10 5 b
Line 1 is steeper because Line 1 is a vertical line and
Line 2 is a horizontal line.
So, y 5 x 1 10.
621
5
29. Slope of Line 1: m 5 } 5 } 5 5
221
1
10 2 1
9
When m 5 4, x 5 0, and y 5 3:
1
5 }2 5 4}2
321
1
Line 1 is steeper because 5 > 4}2 .
30. y 5 mx 1 b
When m 5 2, x 5 4, and y 5 7:
36. y 5 mx 1 b
3 5 4(0) 1 b
3501b
35b
So, y 5 4x 1 3.
Extra Practice,
7 5 2(4) 1 b
7581b
21 5 b
So, y 5 2x 2 1.
466
Geometry
ngws-EP.indd 466
7/11/06 11:47:59 AM
Extra Practice,
continued
45. Given: ###$
BA ⊥ ###$
BC,
37. y 5 mx 1 b
###$ bisects ∠ABC.
BD
2
When m 5 }5, x 5 29, and y 5 4:
2
B
18
4 5 2}
1b
5
Reasons
###$ ⊥ BC
###$
1. BA
38
5
1. Given
2. ∠ABC is a right angle. 2. Definition of
perpendicular lines
38
So, y 5 }5 x 1 }
.
5
38. y 5 mx 1 b
3. m∠ABC 5 908
3. Definition of right angle
23 5 1(8) 1 b
###$ bisects ∠ABC.
4. BD
4. Given
23 5 8 1 b
5. m∠ABD 5 m∠DBC
5. Definition of angle
bisector
6. m∠ABC 5 m∠ABD
1 m∠DBC
6. Angle Addition Postulate
7. m∠ABD 1 m∠DBC
5 908
7. Transitive Property of
Equality
8. m∠ABD 1 m∠ABD
5 908
8. Substitution Property of
Equality
9. 2(m∠ABD) 5 908
9. Simplify.
10. m∠ABD 5 458
10. Division Property of
Equality
When m 5 1, x 5 8, and y 5 23:
211 5 b
So, y 5 x 2 11.
39. m∠ADB 1 m∠BDC 5 m∠ADC
m∠ADB 1 218 5 908
m∠ADB 5 698
40. m∠ADB 5 908 because vertical angles are congruent.
41. m∠ADB 1 178 5 908
m∠ADB 5 738
2x8 1 (x 1 12)8 5 908
2x 1 x 1 12 5 90
3x 1 12 5 90
3x 5 78
C
Statements
}5b
Chapter 4 (pp. 902–903)
1.
y
B
x 5 26
m∠ADB 5 2x8 5 (2 p 26)8 5 528
(3x 1 32)8 1 26x8 5 908
3x 1 32 1 26x 5 90
29x 1 32 5 90
29x 5 58
x52
m∠ADB 5 (3x 1 32)8 5 (3(2) 1 32)8
5 (6 1 32)8 5 388
44.
D
Prove: m∠ABD 5 458
4 5 }5 (29) 1 b
2
A
m∠ADB 1 m∠BDC 5 m∠ADC
(4x 2 1)8 1 (2x 1 1)8 5 908
4x 2 1 1 2x 1 1 5 90
6x 5 90
x 5 15
C
1
x
22
A
}}}
AB 5 Ï(21 2 (21))2 1 (2 2 (22))2
}
}
5 Ï0 1 16 5 Ï16 5 4
}}}
}
}
AC 5 Ï(4 2 (21))2 1 (2 2 (22))2 5 Ï25 1 16 5 Ï41
}}
}
}
BC 5 Ï(4 2 (21))2 1 (2 2 2)2 5 Ï25 1 0 5 Ï25 5 5
Because no sides are congruent, the triangle is scalene.
2 2 (22)
4
}
Slope of AB: m 5 }
5 }0 undefined
21 2 (21)
0
222
}
Slope of BC: m 5 }
5 }5 5 0
4 2 (21)
}
}
} }
Because AB is vertical and BC is horizontal, AB ⊥ BC.
So, the triangle is a right triangle.
m∠ADB 5 (4x 2 1)8 5 (4(15) 2 1)8 5 (60 2 1)8 5 598
Geometry
ngws-EP.indd 467
467
7/11/06 11:48:03 AM
Extra Practice,
2.
continued
7. nDFG > nFDE; you can prove this by either using
y
the SAS Congruence Postulate or the ASA Congruence
Postulate.
B
1
x
22
8. nJNM > nKML; using the information given in the
A
C
}}}
}
}
AB 5 Ï(3 2 (21))2 1 (1 2 (21))2 5 Ï16 1 4 5 Ï20
}}}
}
}
AC 5 Ï(2 2 (21))2 1 (22 2 (21))2 5 Ï9 1 1 5 Ï10
}}
}
}
BC 5 Ï(2 2 3)2 1 (22 2 1)2 5 Ï1 1 9 5 Ï10
Because two sides are congruent, the triangle is isosceles.
22 2 (21)
21
1
5}
5 2}3
Slope of AC: m 5 }
3
2 2 (21)
sides are congruent.
11. (7x 2 5)8 5 448
5x 1 85 5 180
23
22 2 1
}
Slope of BC: m 5 }
5}
53
223
21
7x 5 49
5x 5 95
1
} }
Because 2}3 p 3 5 21, AC ⊥ BC. So, the triangle is a
right triangle.
y
9. STWX > UTWV; all pairs of corresponding angles and
10. 5x8 1 368 1 498 5 1808
}
3. A
diagram, you can use either the SAS Congruence
Postulate or the SSS Congruence Postulate. Because
} }
JN i KM you know that ∠N > ∠KML by the
Corresponding Angles Postulate. So you can now
also use the ASA Congruence Postulate.
x57
x 5 19
12. No; the labels are not in the appropriate order to
match the sides that are congruent. A true congruence
statement would be nPQR > nTVU.
B
13. No; the labels are not in the appropriate order to match
1
x
C
}}
}
}
AB 5 Ï(2 2 (23))2 1 (4 2 4)2 5 Ï25 1 0 5 Ï25 5 5
}}}
AC 5 Ï(5 2 (23)) 1 (22 2 4)
2
}
2
2
}
Because no sides are congruent, the triangle is scalene.
0
424
5 }5 5 0
Slope of AB: m 5 }
2 2 (23)
}
26
3
22 2 4
}
Slope of AC: m 5 }
5}
5 2}4
8
5 2 (23)
}
22 2 4
522
26
3
Slope of BC: m 5 } 5 } 5 22
Because there are not any negative reciprocals, there are
no perpendicular lines. So, the triangle is not a
right triangle.
4. x8 1 3x8 1 568 5 1808
}
}
}
}
BC 5 Ï(6 2 2)2 1 (2 2 6)2 5 Ï16 1 16 5 Ï 32
}}}
}
}
PQ 5 Ï(3 2 (21))2 1 (3 2 (22))2 5 Ï16 1 25 5 Ï41
}}}
PR 5 Ï(7 2 (21))2 1 (21 2 (22))2
}
}
5 Ï64 1 1 5 Ï65
}}
}
}
QR 5 Ï(7 2 3)2 1 (21 2 3)2 5 Ï16 1 16 5 Ï32
Because all 3 pairs of corresponding sides are congruent,
nABC > nPQR.
}}
}
}
16. AB 5 Ï (2 2 (24))2 1 (6 2 5)2 5 Ï 36 1 1 5 Ï 37
}}}
}
}
AC 5 Ï(22 2 (24))2 1 (3 2 5)2 5 Ï4 1 4 5 Ï8
}}
}
}
BC 5 Ï(22 2 2)2 1 (3 2 6)2 5 Ï16 1 9 5 Ï25 5 5
}}
}
}
PQ 5 Ï(8 2 2)2 1 (2 2 1)2 5 Ï36 1 1 5 Ï37
}}
}
}
}}
}
}
PR 5 Ï(5 2 2)2 1 (21 2 1)2 5 Ï9 1 4 5 Ï13
4x 1 56 5 180
4x 5 124
x 5 31
The angles of the triangle are 318, (3 3 31)8 5 938,
and 568. So, the triangle is an obtuse triangle.
5. x8 1 (x 1 1)8 1 (x 1 5)8 5 1808
3x 1 6 5 180
3x 5 174
x 5 58
The angles of the triangle are 588, (58 1 1)8 5 598, and
(58 1 5)8 5 638. So, the triangle is an acute triangle.
6. Because the angles form a linear pair, 608 1 x8 5 1808.
So, x 5 120. The angles of the triangle are 908, 608,
and (180 2 90 2 60)8 5 308. So, the triangle is a right
triangle.
468
}
}}
}
BC 5 Ï(5 2 2) 1 (22 2 4) 5 Ï9 1 36 5 Ï45
2
}
}}
AC 5 Ï(6 2 (22))2 1 (2 2 1)2 5 Ï64 1 1 5 Ï65
}
5 Ï64 1 36 5 Ï100 5 10
}}
}}
15. AB 5 Ï (2 2 (22))2 1 (6 2 1)2 5 Ï 16 1 25 5 Ï 41
QR 5 Ï(5 2 8)2 1 (21 2 2)2 5 Ï9 1 9 5 Ï18
} }
} }
Because AC À PR and BC À QR, nABC À nPQR.
} }
17. nXUV > nVWX; because XV > XV, the triangles are
congruent by the HL Congruence Theorem.
18. nNRM > nPRQ; because ∠NRM > ∠PRQ by the
Vertical Angles Congruence Theorem, nNRM > nPRQ
by the SAS Congruence Postulate.
19. nHJL > nKLJ; ∠HJL > ∠JLK by the Alternate
} }
Interior Angles Theorem. Because JL > LJ, nHJL >
nKLJ by the SAS Congruence Postulate.
21
the sides that are congruent. A true congruence statement
would be nJKM > nLKM.
} }
14. Yes; by the Segment Addition Postulate, AC > BD. Also,
} }
CD > DC. So nACD > nBDC by the SSS Congruence
Postulate.
20. Yes; because ∠HLG > ∠KLJ by the Vertical Angles
Congruence Theorem, nGHL > nJKL by the ASA
Congruence Postulate.
Geometry
ngws-EP.indd 468
7/11/06 11:48:08 AM
Extra Practice,
}
continued
}
21. Yes; because QN > QN, nMNQ > nPNQ by the AAS
31. 9x 1 12 5 12x 2 6
Congruence Theorem.
23x 5 218
22. No; you can only show that all 3 angles are congruent.
x56
23. Yes; nABC > nDEF by the ASA Congruence Postulate.
(9x 2 12)8 1 (12x 2 6)8 1 y8 5 1808
24. No; there is no SSA Congruence Postulate.
9x 1 12 1 12x 2 6 1 y 5 180
25. State the given information from the diagram, and state
} }
that AC > AC by the Reflexive Property of Congruence.
Then use the SAS Congruence Postulate to prove
nABC > nCDA, and state ∠1 > ∠2 because
corresponding parts of congruent triangles are congruent.
21x 1 6 1 y 5 180
y 5 174 2 21x
y 5 174 2 21(6) 5 48
32. 2x 2 3 5 11
26. State the given information from the diagram. Prove
nDEF > nGHJ by the HL Congruence Theorem, and
state ∠1 > ∠2 because corresponding parts of congruent
triangles are congruent.
x57
5x 5 10
}
that SR > SR by the Reflexive Property of Congruence.
Then use the Segment Addition Postulate to show
} }
that PR > US. Use the SAS Congruence Postulate to
prove nQPR > nTUS, and state ∠1 > ∠2 because
corresponding parts of congruent triangles are congruent.
}}
}
}}
}
}
}}
}
}
x52
34. 2(x 1 1)8 5 628
2
2(x 1 1)8 1 628 1 y8 5 1808
2x 1 2 5 62
2x 1 2 1 62 1 y 5 180
2x 5 60
2x 1 64 1 y 5 180
}
x 5 30
28. AB 5 Ï (6 2 0) 1 (0 2 8) 5 Ï 36 1 64 5 Ï 100 5 10
2
y57
33. 6x 2 5 5 x 1 5
27. State the given information from the diagram, and state
}
y 1 4 5 11
2x 5 14
y 5 116 2 2x
y 5 116 2 2(30)
AC 5 Ï(0 2 0)2 1 (0 2 8)2 5 Ï0 1 64 5 Ï64 5 8
BC 5 Ï(0 2 6) 1 (0 2 0) 5 Ï36 1 0 5 Ï36 5 6
2
2
}}
DE 5 Ï(9 2 3)2 1 (2 2 10)2
}
5 56
35. Because the triangle is a right triangle with two
congruent sides, the 2 remaining angles must measure
}
5 Ï36 1 64 5 Ï100 5 10
}}
}
} 5 } 5 458.
DF 5 Ï(3 2 3)2 1 (2 2 10)2 5 Ï0 1 64 5 Ï64 5 8
}}
}
}
EF 5 Ï(3 2 9)2 1 (2 2 2)2 5 Ï36 1 0 5 Ï36 5 6
nABC > nDEF by the SSS Congruence Postulate.
Then, ∠A > ∠D because corresponding parts of
congruent triangles are congruent.
908
2
1808 2 908
2
}
(2x 2 11)8 5 458
1
29. AB 5 Ï (22 2 (23)) 1 (3 2 (22))
}
2
P9
}
P
S9 S
}}}
AC 5 Ï(2 2 (23))2 1 (2 2 (22))2
37.
y
21
5 Ï1 1 25 5 Ï26
}
y 5 29
x 5 28
36.
}}}
2
(y 1 16)8 5 458
2x 5 56
Q9
S9
}
}}
P9
P
}
S
}
DE 5 Ï(6 2 5) 1 (6 2 1) 5 Ï1 1 25 5 Ï26
2
2
}}
}
}}
}
}
DF 5 Ï(10 2 5)2 1 (5 2 1)2 5 Ï25 1 16 5 Ï41
}
EF 5 Ï(10 2 6) 1 (5 2 6) 5 Ï16 1 1 5 Ï17
2
2
nABC > nDEF by the SSS Congruence Postulate.
Then, ∠A > ∠D because corresponding parts of
congruent triangles are congruent.
30. x8 1 y8 1 1328 5 1808
x8 1 x8 1 1328 5 1808
2x8 5 488
x 5 24
y 5 x 5 24
x
R9 R
}
BC 5 Ï(2 2 (22)) 1 (2 2 3) 5 Ï16 1 1 5 Ï17
2
1
Q
21
5 Ï25 1 16 5 Ï41
2
Q9
x
}
}}}
y R9
Q
R
38.
y P9
S9
Q9
R9
1
P
21
S
x
Q
R
Geometry
ngws-EP.indd 469
469
7/11/06 11:48:13 AM
Extra Practice,
39.
continued
12.
y
A
1
AB 5 BC
8x 1 7 5 11x 2 5
B
12 5 3x
1
x
45x
D
So, AB 5 8(4) 1 7 5 39.
C
}
}
Yes, CD is a rotation of AB; the rotation is 1808.
40.
y
13. KN 5 JN 5 12
14.
LJ 5 LK
12x 2 4 5 6x 1 8
A
6x 5 12
C B
x52
1
So, LJ 5 12(2) 2 4 5 20.
x
21
15. Because x 5 2, KP 5 7(2) 1 10 5 24.
}
}
No, CD is not a rotation of AB.
16. JP 5 KP 5 24
}
}
17. Yes. Because ###$
LN is the perpendicular bisector of JK and
Chapter 5 (pp. 904–905)
} } }
}
}
1. LN i AB; LN connects the midpoints of AC and CB, so it
2.
3.
4.
5.
}
is parallel to the third side, AB.
} } }
}
}
CB i LM; LM connects the midpoints of AC and AB, so it
}
is parallel to the third side, CB.
}
}
} } }
MN i AC; MN connects the midpoints of AB and CB, so
}
it is parallel to the third side, AC.
}
}
AM 5 LN 5 MB; Midsegment LN is half as long as AB
which is divided into congruent segments by
midpoint M.
}
}
MN 5 AL 5 LC; Midsegment MN is half as long as AC
which is divided into congruent segments by midpoint L.
6. Sample answer:
7. Sample answer:
y
y
R(a, b)
C(0, 4)
P is equidistant from the endpoints J and K of JK, by the
Converse of the Perpendicular Bisector Theorem, P is
on ###$
LN.
}
} ###$
###$ bisects
18. Because AC ⊥ ###$
BA, CD ⊥ BD
, and AC 5 CD, BC
∠ABD by the Converse of the Angle Bisector Theorem.
So, m∠ABC 5 m∠CBD 5 328.
19. Because ∠EFH > ∠GFH, ###$
FH bisects ∠EFG. Then
} ###$
}
because HE ⊥ ###$
FH and HG ⊥ FH
, EH 5 GH 5 15 by the
Angle Bisector Theorem.
20. m∠JKM 1 m∠LKM 5 m∠JKL
258 1 m∠LKM 5 508
m∠LKM 5 258
####$ bisects ∠JKL. Then because }
####$ and
JM ⊥ KM
So, KM
} ####$
LM ⊥ KM, LM 5 JM 5 13 by the Angle Bisector
Theorem.
21. No; Not enough information is given to apply the
Converse of the Perpendicular Bisector Theorem or to
use congruent triangles.
D(4, 0) x
E(0, 0)
8. Sample answer:
T(0, 0)
S(6, 0) x
9. Sample answer:
L(0, 5)
M(5, 5)
23. Yes, x 5 17; By the Angle Bisector Theorem.
24. By the Concurrency of Medians of a Triangle Theorem,
2
FP 5 }3 TF.
y
y
22. Yes, x 5 4; The triangles are congruent by ASA.
F(0, s)
2
FP 5 }3 TF
K(0, 0)
10.
N(5, 0) x
AB 5 BC
2x 1 3 5 x 1 7
x54
So, AB 5 2(4) 1 3 5 11.
11.
AB 5 BC
5x 2 1 5 3x 1 5
2x 5 6
x53
So, AB 5 5(3) 2 1 5 14.
470
H(0, 0)
G(t, 0) x
2
14 5 }3 TF
21 5 TF
25. By the Concurrency of Medians of a Triangle Theorem,
2
1
1
DP 5 }3 DS, so PS 5 }3 DS 5 }2 DP.
1
2
} DP 5 PS
1
2
}DP 5 8.5
D
DP 5 17
26. DS 5 DP 1 PS 5 17 1 8.5 5 25.5
1
1
27. PR 5 } RE 5 } (24) 5 8
3
3
Geometry
ngws-EP.indd 470
7/11/06 11:48:20 AM
Extra Practice,
continued
}
28. Because BD is the segment from vertex B that is
29.
30.
31.
32.
33.
34.
}
}
perpendicular to side AC of nABC, BD is an altitude of
nABC.
}
Because BD divides ∠ABC into the two congruent
}
angles, ∠ABD and ∠CBD, BD is an angle bisector
of nABC.
} }
}
Because AD > CD, D is the midpoint of AC.
}
Then because BD is the segment from vertex B of nABC
} }
to the midpoint D of the opposite side AC, BD is a
median of nABC.
} } }
Because BD ⊥ AC, BD is an altitude of nABC.
} } }
Because AD > CD, BD is a median of nABC.
} } } }
} }
Because BD ⊥ AC, AD > CD, and BD > BD,
}
nABD > nCBD by SAS. So, ∠ABD > ∠CBD and BD
is an angle bisector of nABC.
} } }
} }
Because BD ⊥ AC and AD > CD, BD is a perpendicular
bisector of nABC.
} } } }
Because nABD > nCBD, BD ⊥ AC, AD > CD, and
∠ABD > ∠CBD. So, by the same reasons used in
}
Exercise 31, BD is a perpendicular bisector, an angle
bisector, a median, and an altitude of nABC.
} } } }
Sample answer: Because AB > CB, BD ⊥ AC, and
} }
BD > BD, nABD > nCBD. Then ∠ABD > ∠CBD and
} }
}
AD > CD. By the same reasons used in Example 31, BD
is a perpendicular bisector, an angle bisector, a median,
and an altitude of nABC.
} } }
Sides from smallest to largest: PQ, PR, QR
By Theorem 5.10, angles from smallest to largest:
∠R, ∠Q, ∠P
35. m∠L 1 m∠J 1 m∠K 5 1808
m∠L 1 608 1 618 5 1808
m∠L 5 598
Angles from smallest to largest: ∠L, ∠J, ∠K
By Theorem 5.11, sides from smallest to largest:
} } }
JK, LK, LJ
36. m∠G 5 908 2 728 5 188
Angles from smallest to largest: ∠G, ∠F, ∠E
By Theorem 5.11, sides from smallest to largest:
}} }
EF, EG, FG
37. Let x be the length of the third side.
918>x
x18>9
17 > x
x>1
The length of the third side must be greater than 1 inch
and less than 17 inches.
24 1 13 > x
x 1 13 > 24
37 > x
x > 11
The length of the third side must be greater than 11 feet
and less than 37 feet.
913>x
x13>9
12 > x
x>6
The length of the third side must be greater than 6 inches
40. 1 ft 5 12 in.
12 1 17 < x
x 1 12 > 17
x>5
29 < x
The length of the third side must be greater than 5 inches
41. 2 yd 5 6 ft
416>x
x14>6
10 > x
x>2
The length of the third side must be greater than 2 feet
and less than 10 feet.
42. 2 yd 5 6 ft
616>x
x16>6
12 > x
x>0
The length of the third side must be less than 12 feet.
} }
} }
MN > QR, but included angle ∠M is greater than the
included angle ∠Q. So, by the Hinge Theorem, LN > PR.
} }
} }
44. VU < ST; In nVSU and nTUS, VS > TU and SU > US,
but m∠VSU < m∠TUS. So, by the Hinge Theorem,
VU < ST.
} }
45. m∠WYX > m∠WYZ; In nWXY and nWZY, XY > ZY
} }
and WY > WY. But WX > WZ. So, by the Converse of the
Hinge Theorem, m∠WYX > m∠WYZ.
43. LN > PR; In nLMN and nPQR, LM > PQ and
46. m∠1 5 m∠2; The triangles are conguent by SSS.
47. JK 5 MN; nJKL > nMNP by SAS.
}
}
48. BC < DE; In nABC and nCDE, AB > CD and
}
}
AC > CE, but the included ∠A is smaller than the
included ∠DCE, so by the Hinge Theorem, BC < DE.
} }
49. GH > QR; In nFGH and nPQR, FG > PQ and
} }
FH > PR, but the included ∠F is larger than the
included ∠P. So, by the Hinge Theorem, GH > QR.
50. m∠3 > m∠4; There are two triangles with two pairs of
congruent sides. But the side opposite included ∠3 is
longer than the side opposite included ∠4. So, by the
Converse of the Hinge Theorem, m∠3 > m∠4.
51. m∠5 < m∠6; There are two triangles with two pairs of
congruent sides. But the side opposite included ∠5 is
shorter than the side opposite included ∠6. So, by the
Converse of the Hinge Theorem, m∠5 < m∠6.
Chapter 6 (pp. 906–907)
1. x 1 3x 1 5x 5 1808
9x 5 1808
x 5 208
The angle measures are 208, 3(208) 5 608, and
5(208) 5 1008.
Geometry
ngws-EP.indd 471
471
7/11/06 11:48:23 AM
Extra Practice,
continued
x11
1
812
2
18. If } 5 }, then } 5 }, because you can apply the
x
1
8
2
2. x 1 5x 1 6x 5 1808
12x 5 1808
Reciprocal Property of Proportions and then add the
value of each ratio’s denominator to its numerator.
x 5 158
The angle measures are 158, 5(158) 5 758, and
6(158) 5 908.
19.
3. 2x 1 3x 1 5x 5 1808
NJ
NK
NL
NM
6
NK
6 1 15
14
}5}
}5}
10x 5 1808
6 p 14 5 NK(6 1 15)
x 5 188
84 5 21 p NK
The angle measures are 2(188) 5 368, 3(188) 5 548, and
5(188) 5 908.
4. 5x 1 6x 1 9x 5 1808
4 5 NK
20.
20x 5 1808
The angle measures are 5(98) 5 458, 6(98) 5 548, and
9(98) 5 818.
15 p 4 5 y p 20
21x 5 84
60 5 20y
x54
35y
8.
6a 2 18 5 4a 2 2
20 5 2z
2a 5 16
10 5 z
a58
6
3
x18
21
10.
x25
2
x16
3
}5}
6(21) 5 3(x 1 8)
(x 1 6)2 5 3(x 2 5)
26 5 3x 1 24
2x 1 12 5 3x 2 15
230 5 3x
27 5 x
x22
4
x 1 10
10
}5}
12
8
12.
51t
t23
}5}
(x 2 2)10 5 4(x 1 10)
12(t 2 3) 5 8(5 1 t)
10x 2 20 5 4x 1 40
12t 2 36 5 40 1 8t
6x 5 60
4t 5 76
x 5 10
}
t 5 19
}
13. x 5 Ï 4 p 9 5 Ï 36 5 6
}
14. x 5 Ï 3 p 48 5 Ï 144 5 12
The geometric mean of 3 and 48 is 12.
}
}
15. x 5 Ï 9 p 16 5 Ï 144 5 12
}
BA 1 CB
CB
EF 1 DE
DE
CA
10
12 1 8
8
}5}
CA(8) 5 10(12 1 8)
CA 5 25
21. The diagram shows ∠R > ∠S, ∠Q > ∠T, ∠P > ∠U,
and ∠N > ∠V.
RQ
ST
11 QP
20 TU
RN
SV
8.8
16
8.8
16
88
160
11 PN
20 UV
11
20
} 5 }, } 5 } 5 } 5 }, } 5 } and
88
160
11
20
}5}5}5}
Because corresponding angles are congruent and
corresponding side lengths are proportional,
RQPN , STUV. The scale factor of RQPN to STUV
is equal to the ratio of any two corresponding lengths,
or 11 : 20.
}
16. x 5 Ï 7 p 11 5 Ï 77
}
The geometric mean of 7 and 11 is Ï77 ø 8.8.
y
9
7
x
17. If } 5 }, then } 5 } by the Reciprocal Property of
y
9
7
x
Proportions.
∠F > ∠K.
DE
JL
6
3
DF
JK
8
4
EF
LK
3
1.5
} 5 } 5 2, } 5 } 5 2, } 5 } 5 2
Because corresponding angles are congruent and
corresponding side lengths are proportional,
nDEF , n JLK. The scale factor of nDEF to nJLK
is equal to the ratio of any two corresponding lengths,
or 2:1.
QR
36
3
23. The scale factor of nPQR to nLMN: } 5 } 5 }
12
1
MN
}
EF
DE
22. The diagram shows ∠D > ∠J, ∠E > ∠L, and
210 5 x
11.
BA
CB
}5}
8CA 5 200
2a 2 1
a23
}5}
6
2
(a 2 3)6 5 2(2a 2 1)
21 5 2z 1 1
}5}
DE
EF
}5}
}5}
x p 21 5 14 p 6
3 p 7 5 (2z 1 1) p 1
9.
20
4
15
y
6.
1
3
7. } 5 }
7
2z 1 1
CB
BA
24. m∠P 1 m∠Q 1 m∠R 5 1808
x8 1 908 1 22.68 5 1808
x 5 67.4
y
3
}5}
1
13
}5}
y 5 13 p 3 5 39
15 5 z p 3
15
z
3
1
6
21
x
14
}5}
BA
EF
}5}
x 5 98
5.
CB
DE
}5}
55z
25. Perimeter of nPQR: 15 1 36 1 39 5 90
Perimeter of nLMN: 5 1 12 1 13 5 30
472
Geometry
ngws-EP.indd 472
7/11/06 11:48:26 AM
Extra Practice,
continued
26. The blue special segments are altitudes of triangles.
y18
27
}5}
18
y
35. Apply the Triangle Proportionality Thoerem.
( y 1 8)18 5 y(27)
5x 5 15
18y 1 144 5 27y
x53
144 5 9y
7.5
7.5
27. The blue special segments are angle bisectors at
6
5y 5 42
42
(4y 1 2)30 5 (3y 1 4)36
y5}
5 8.4
5
120y 1 60 5 108y 1 144
36. Because three parallel lines intersect two transversals,
12y 5 84
apply Theorem 6.6.
y57
24
6
x
5
}5}
28. In nPQR, 638 1 788 1 m∠R 5 1808
m∠R 5 398
6x 5 120
So, ∠R > ∠V and ∠P > ∠W.
x 5 20
Therefore, nPQR , nWUV by AA Similarity Postulate.
29. In nBFG, 338 1 1108 1 m∠G 5 1808
37. (x, y) → (3x, 3y)
A(1, 1) → D(3, 3)
m∠G 5 378
B(4, 1) → E(12, 3)
So, nABC is not similar to nFBG, because ∠C À ∠G.
} }
} } } }
30. Because VW ⊥ WX and XY ⊥ WX, VW i XY by the Lines
Perpendicular to a Transversal Theorem. So, ∠1 > ∠3 by
the Corresponding Angles Postulate. Also ∠W > ∠Z by
the Right Angles Congruence Theorem. So,
nVWX ,nXZY by the AA Similarity Postulate.
} }
} }
31. Because JK i NP and KL i PM, ∠J > ∠PNM and
∠L > ∠PMN by the Corresponding Angles Postulate.
Therefore, nJKL , nNPM by the AA Similarity
Postulate.
32. In nVXW and nZXY, ∠VXW and ∠ZXY are vertical
C(1, 2) → F(3, 6)
y
F
D
C
1
E
A
B
x
1
38. (x, y) → (5x, 5y)
angles, so ∠VXW > ∠ZXY.
A(2, 2) → E(10, 10)
3
1
4
1
VX
WX
} 5 } 5 } and } 5 } 5 }
6
2
8
2
ZX
YX
B(22, 2) → F(210, 10)
Because an angle of nVXW is congruent to an angle of
nZXY, and the lengths of the sides including these angles
are proportional, nVXW , nZXY by the SAS Similarity
Theorem.
D(2, 21) → H(10, 25)
C(21, 21) → G(25, 25)
y
F
33. In nHJK and nSRT, by comparing the corresponding
B
sides in order from smallest to largest you have
3 JK
5 RT
5
So, }7 5 }y .
4y 1 2
36
}5}
30
3y 1 4
18
30
5
triangles are similar by the SAS Similarity Theorem.
5
corresponding vertices.
HJ
SR
5
Because }
5}
5 }7 and }
5 }7, the two
10.5
3 1 7.5
512
16 5 y
7.5
x
5
2
}5}
24
40
3
5
HK
ST
27
45
3
5
C
} 5 } 5 }, } 5 } 5 }, and } 5 } 5 }.
Because the corresponding side lengths are proportional,
nHJK , nSRT by the SSS Similarity Theorem.
G
4
E
A
D
6
x
H
34. An angle of the triangle is bisected, so Theorem 6.7
applies.
a
17
21
34
}5}
34a 5 357
a 5 10.5
Geometry
ngws-EP.indd 473
473
7/11/06 11:48:30 AM
Extra Practice,
1
1 1
39. (x, y) → } x, } y
2 2
continued
2
4.
25 m
A(2, 2) → D(1, 1)
24 m
B(8, 2) → E(4, 1)
1
h2 5 49
24 m
h57
1
The area is 168 m2.
y
C
5.
A
E
1
1 1
40. (x, y) → } x, } y
3 3
26 ft
6. 17 cm
2
h
15 cm
A(3, 26) → E(1, 22)
17 cm
152 1 h 2 5 172
15 cm
B(6, 26) → F(2, 22)
1
1
A 5 }2 bh 5 }2 (30)(8) 5 120
C(6, 9) → G(2, 3)
The area is 120 cm2.
D(23, 9) → H(21, 3)
7.
C
h58
402 0 242 1 322
1600 5 1600 ✓
3
The triangle is a right triangle.
G
8.
x
2
F
A
752 0 212 1 722
5625 0 441 1 5184
5625 5 5625 ✓
B
length of B
6
41. Enlargement; Scale factor: } 5 } 5 3 or 3 : 1
2
length of A
length of B
5
1
42. Reduction; Scale factor: } 5 } 5 } or 1 : 2
10
2
length of A
Chapter 7 (pp. 908–909)
1. A Pythagorean triple is 7, 24, 25. Notice that if you
multiply the integers of the Pythagorean triple by 2, you
get the lengths of the legs of the triangle: 7 p 2 5 14 and
24 p 2 5 48. So, the length of the hypotenuse is 25 p 2 5
50.
2. (hypotenuse)2 5 (leg)2 1 (leg)2
512 5 x 2 1 242
2601 5 x 2 1 576
2025 5 x 2
45 5 x
The length of the leg is 45.
3. A Pythagorean triple is 5, 12, 13. Notice that if you
multiply 12 and 13 by 12, you get the lengths of the
longer leg and the hypotenuse of the triangle: 12 p 12 5
144 and 13 p 12 5 156. So, the length of the shorter leg
is 5 p 12 5 60.
9. 272 0 112 1 252
729 0 121 1 625
729 Þ 746
The triangle is not a right triangle.
10. 132 0 72 1 112
169 0 49 1 121
169 Þ 170
The triangle is not a right triangle.
}
11. (5Ï 26 )2 0 172 1 192
25 p 26 0 289 1 361
650 5 650 ✓
}
12. (Ï 181 )2 0 92 1 102
181 0 81 1 100
181 5 181 ✓
E
h2 5 64
1600 0 576 1 1024
y
D
H
h 5 24
The area is 240 ft2.
x
1
h2 5 576
1
1
A 5 }2 bh 5 }2 (20)(24) 5 240
B
D
102 1 h 2 5 262
26 ft
h
10 ft
10 ft
F
474
242 1 h 2 5 252
25 m
A 5 }2 bh 5 }2 (48)(7) 5 168
C(2, 6) → F(1, 3)
1
h
Geometry
ngws-EP.indd 474
7/11/06 11:48:35 AM
Extra Practice,
continued
13. 14 1 21 5 35
14 1 25 5 39
35 > 25
39 > 21
}
18. 4Ï 21 ø 18.33
}
4Ï21 1 31 ø 49.33
43.33 > 31
21 1 25 5 46
49.33 > 25
25 1 31 5 56
46 > 14
}
56 > 4Ï21
The segment lengths form a triangle.
252 ? 142 1 212
625 ? 196 1 441
312 ?
961 ? 336 1 625
The segment lengths 14, 21, and 25 form an
acute triangle.
14. 32 1 60 5 92
961 5 961
}
32 1 68 5 100
92 > 68
The segment lengths 4Ï21 , 25, and 31 form a right
triangle.
100 > 60
19. nABD , nBCD , nACB
60 1 68 5 128
128 > 32
682 ? 322 1 602
20. nGHK , nGJH , nHJK
HJ
JG
KJ
HJ
}5}
4624 5 4624
The segment lengths 32, 60, and 68 form a right triangle.
21. nPQR , nPSQ , nQSR
RQ
PR
SR
RQ
}5}
15. 11 1 19 5 30
30 < 32
The segment lengths do not form a triangle.
5
5
22. } 5 }
x
5
x55
}
16. 3Ï 11 ø 9.95
}
3 1 3Ï11 ø 12.95
3 1 9 5 12
}
12 > 3Ï11
12.95 > 9
y
1
23. } 5 }
4
y
4 5 y2
}
9 1 3Ï11 5 18.95
25y
18.95 > 3
}
BC
BD
AB
AD
}5}
4624 ? 1024 1 3600
(3Ï11 )2
}
(4Ï21 )2 1 252
961 ? 16 p 21 1 625
625 < 637
}
4Ï21 1 25 ø 43.33
24.
5
3
x13
5
}5}
? 32 1 92
3x 1 9 5 25
9 p 11 ? 9 1 81
3x 5 16
99 > 90
}
The segment lengths 3, 9, and 3Ï11 form an obtuse
triangle.
}
17. 3Ï 40 ø 18.97
}
}
27 > 3Ï40
30.97 > 15
}
15 1 3Ï40 ø 33.97
6
8
y
2
h 56 18
h2 5 36 1 64
10
h2 5 100
10
6
8
y
}5}
}
2
h 5 10
33.97 > 12
(3Ï40 )2
the large triangle.
2
12 1 3Ï40 ø 30.97
12 1 15 5 27
x ø 5.3
25. Find the hypotenuse of
? 122 1 152
10y 5 48
y 5 4.8
9 p 40 ? 144 1 225
360 < 369
}
The segment lengths 12, 15, and 3Ï40 form an acute
triangle.
9
x
26. } 5 }
7
9
7x 5 81
x ø 11.6
Geometry
ngws-EP.indd 475
475
7/11/06 11:48:38 AM
Extra Practice,
continued
27. Find the hypotenuse of
3
the smallest triangle.
34
h 2 5 32 1 52
y
h 2 5 9 1 25
h 2 5 34
}
h 5 Ï34
}
Ï34
3
}5}
y ø 9.7
28. leg 5 leg
37.
x57
}
hypotenuse 5 leg p Ï2
12 p tan 278 5 x
}
y 5 7Ï2
6.1 ø x
}
So, x 5 7 and y 5 7Ï2 .
12
Check: tan 638 ? }
6.1
29. hypotenuse 5 2 p shorter leg
18 5 2 p g
1.9626 ø 1.97 ✓
95g
}
longer leg 5 shorter leg p Ï3
38.
}
h 5 9Ï3
25
tan 698
x5}
}
So, g 5 9 and h 5 9Ï3 .
x ø 9.6
}
30. hypotenuse 5 leg p Ï 2
9.6
}
9Ï2 5 aÏ2
Check: tan 218 ? }
25
95a
0.3839 ø 0.384
So, a 5 9 and b 5 9.
}
31. longer leg 5 shorter leg p Ï 3
39.
}
x
tan 418 5 }
19
19 p tan 418 5 x
m 5 5Ï3
hypotenuse 5 2 p shorter leg
16.5 ø x
n 5 10
19
Check: tan 498 ? }
16.5
}
So, m 5 5Ï3 and n 5 10.
1.1504 ø 1.1515
}
32. hypotenuse 5 leg p Ï 2
}
15 5 sÏ2
40.
15
x
sin 448 5 }
14
14 p sin 448 5 x
}
} 5 s
Ï2
9.7 ø x
}
15Ï2
}5s
y
2
}
cos 448 5 }
14
}
15Ï2
15Ï2
and t 5 }
.
So, s 5 }
2
2
}
33. longer leg 5 shorter leg p Ï 3
}
25
tan 698 5 }
x
x p tan 698 5 25
}
h 5 gÏ3
}
x
tan 278 5 }
12
}
10Ï3 5 wÏ3
10 5 w
hypotenuse 5 2 p shorter leg
v 5 2w
v 5 2(10)
14 p cos 448 5 y
10.1 ø y
So, x ø 9.7 and y ø 10.1.
y
5
5
opp. ∠A
18
2
34. tan A 5 } 5 } 5 } ø 0.6667
27
3
adj. to ∠A
opp. ∠B
27
3
tan B 5 }
5}
5 }2 5 1.5000
18
adj. to ∠B
opp. ∠A
60
3
35. tan A 5 } 5 } 5 } 5 0.6000
5
100
adj. to ∠A
opp. ∠B
100
5
tan B 5 }
5}
5 }3 ø 1.6667
60
adj. to ∠B
opp. ∠A
3
24
36. tan A 5 } 5 } 5 } ø 0.4286
7
56
adj. to ∠A
opp. ∠B
56
7
tan B 5 }
5}
5 }3 ø 2.3333
24
adj. to ∠B
v 5 20
So, v 5 20 and w 5 10.
476
Geometry
ngws-EP.indd 476
7/11/06 11:48:42 AM
Extra Practice,
41.
continued
8
47. (GH)2 5 (GJ)2 1 (JH)2
sin 328 5 }y
(GH)2 5 62 1 72
y p sin 328 5 8
(GH)2 5 36 1 49
8
sin 328
y5}
(GH)2 5 85
}
GH 5 Ï85
y ø 15.1
GH ø 9.2
x
cos 328 ø }
15.1
7
tan G 5 }6
15.1 p cos 328 5 x
7
m∠G 5 tan211 }6 2 ø 49.48
12.8 ø x
So, x ø 12.8 and y ø 15.1.
42.
43.
40.68 ø m∠H
The side lengths are 6 units, 7 units, and about 9.2 units.
The angle measures are 908, about 49.48, and about 40.68.
y p cos 778 5 Ï3
}
Ï3
cos 778
17
x5}
cos 268
y5}
x ø 18.9
y ø 7.7
y
x
sin 268 ø }
18.9
sin 778 ø }
7.7
48.
1808 5 908 1 258 1 m∠B
658 5 m∠B
AB
sin 258 5 }
25
18.9 p sin 268 5 y
7.7 p sin 778 5 x
25 p sin 258 5 AB
8.3 ø y
7.5 ø x
10.6 ø AB
So, x ø 18.9 and y ø 8.3.
5.7
So, x ø 7.5 and y ø 7.7.
45.
cos 548 5 }
y
y p cos 548 5 5.7
5.7
cos 548
Ï3
cos 778 5 }
y
}
x p cos 268 5 17
44.
1808 5 908 1 49.48 1 m∠H
}
17
cos 268 5 }
x
4
sin 148 5 }y
25 p cos 258 5 AC
y p sin 148 5 4
22.7 ø AC
4
sin 148
y5}
y5}
y ø 9.7
y ø 16.5
x
sin 548 ø }
9.7
x
cos 148 ø }
16.5
9.7 p sin 548 5 x
16.5 p cos 148 5 x
7.8 ø x
16.0 ø x
So, x ø 7.8 and y ø 9.7.
2
2
AC
cos 258 5 }
25
So, x ø 16.0 and y ø 16.5.
2
46. (DE) 5 (DF) 1 (EF)
(DE)2 5 122 1 52
2
(DE) 5 144 1 25
The side lengths are about 10.6 units, about 22.7 units,
and 25 units. The angle measures are 908, 658, and 258.
Chapter 8 (pp. 910–911)
1. Quadrilateral; (4 2 2) p 1808 5 3608
x8 1 598 1 1288 1 618 5 3608
x 5 112
2. Pentagon; (5 2 2) p 1808 5 5408
x8 1 1378 1 828 1 1408 1 918 5 5408
x 5 90
3. Heptagon; (7 2 2) p 1808 5 9008
x8 1 1548 1 1158 1 1228 1 1498 1 1538 1 908 5 9008
(DE)2 5 169
x 5 117
DE 5 13
4. x8 1 1468 1 1368 5 3608
5
tan D 5 }
12
m∠D 5
x 5 78
5
tan21 }
1 12 2 ø 22.68
1808 5 908 1 22.68 1 m∠E
67.48 ø m∠E
The side lengths are 5 units, 12 units, and 13 units. The
angle measures are 908, about 22.68, and about 67.48.
5. x8 1 468 1 948 1 358 1 (1808 2 1488) 1 858 5 3608
x 5 68
6. Pentagon; (5 2 2) p 1808 5 5408
x8 1 1018 1 1078 1 x8 1 1008 5 5408
2x 5 232
x 5 116
Geometry
ngws-EP.indd 477
477
7/11/06 11:48:44 AM
Extra Practice,
continued
7. (6 2 2) p 1808 5 7208
21. 2 p ZV 5 ZX
20. WZ 5 YX
22.
7208
Interior angle: }
5 1208
6
y
A
3608
Exterior angle: }
5 608
6
B
The measure of an interior angle of a regular hexagon
is 1208. The measure of an exterior angle of a regular
hexagon is 608.
1
D
x
21
8. (9 2 2) p 1808 5 12608
C
12608
}}
}
AB 5 Ï(7 2 5)2 1 (3 2 6)2 5 Ï13
Interior angle: }
5 1408
9
}}
}
CD 5 Ï(3 2 5)2 1 (1 2 (22))2 5 Ï13
} }
So, AB > CD.
3608
Exterior angle: }
5 408
9
The measure of an interior angle of a regular 9-gon is
1408. The measure of an exterior angle of a regular 9-gon
is 408.
3
} 326
Slope of AB 5 }
5 2}2
725
3
} 1 2 (22)
Slope of CD 5 }
5 2}2
325
9. (17 2 2) p 1808 5 27008
} } } }
} }
Slopes are equal, so AB i CD. AB > CD and AB i CD,
so ABCD is a parallelogram.
27008
Interior angle: }
ø 158.88
17
3608
23.
Exterior angle: }
ø 21.28
17
y
B
C
A
The measure of an interior angle of a regular 17-gon
is about 158.88. The measure of an exterior angle of a
regular 17-gon is about 21.28.
1
D
21
x
10. a 5 7, b 5 12
}
322
1
}
Slope of AB 5 }
5 }2
26 2 (28)
b55
122
1
}
Slope of CD 5 }
5 }2
23 2 (21)
So, a 5 5 and b 5 5.
a 5 18
} }
Slopes are equal, so AB i CD.
} }
} }
AB > CD and AB i CD, so ABCD is a parallelogram.
2
3
}a 5 b
24.
2
3
}(18) 5 b
y
B
A
12 5 b
C
D
So, a 5 18 and b 5 12.
13. b 5 63
2
a 5 180 2 63
21
a 5 117
}}
15. a 5 7
4a 5 180
2b 1 4 5 b 1 7
a 5 45
b53
b 5 3a
}
}
AB 5 Ï(2 2 (21))2 1 (14 2 11)2 5 Ï18 5 3Ï2
So, a 5 117 and b 5 63.
14. a8 1 3a8 5 1808
x
}}}
So, a 5 7 and b 5 3.
b 5 3(45)
}
}
CD 5 Ï(3 2 6)2 1 (8 2 11)2 5 Ï18 5 3Ï2
} }
So, AB > CD.
14 2 11
}
Slope of AB 5 }
51
2 2 (21)
} 8 2 11
Slope of CD 5 }
51
326
} } } }
} }
b 5 135
So, a 5 45 and b 5 135.
16. ∠WXV > ∠YZV
17. ∠ZWV > ∠ XYV
18. ∠WVX > ∠YVZ
19. WV 5 YV
a55
b1156
478
}
}}}
CD 5 Ï(23 2 (21))2 1 (1 2 2)2 5 Ï5
} }
So, AB > CD.
2a 5 10
12.
}}}
AB 5 Ï(26 2 (28))2 1 (3 2 2)2 5 Ï5
11. 2a 1 4 5 14
Geometry
ngws-EP.indd 478
7/11/06 11:48:49 AM
Extra Practice,
25.
2
continued
33. The diagonals of a rhombus are perpendicular, so
y
m∠LQM 5 908.
x
21
34. The four sides of a rhombus are congruent, so MN 5 5.
B
A
1
35. x 5 }(19 1 31)
2
C
x 5 25
D
}}}
}
AB 5 Ï(4 2 (21)) 1 (24 2 (25)) 5 Ï26
2
2
}}}
}
CD 5 Ï(1 2 6)2 1 (210 2 (29))2 5 Ï26
} }
So, AB > CD.
} 24 2 (25) 1
Slope of AB 5 }
5 }5
4 2 (21)
} 210 2 (29) 1
Slope of CD 5 }
5 }5
126
} } } }
} }
}
26. Draw PR to form nPQR and nRSP. Show that
nPQR > nRSP. Then show that ∠QPR > ∠SRP and
∠QRP > ∠SPR. Use the Alternate Interior Angles
} }
} }
Converse to show that PS i RQ and PQ i RS. Then by
definition, PQRS is a parallelogram.
27. Show that nPQR > nRSP by the AAS Congruence
} }
Theorem. Show that ∠QPR > ∠SRP, PQ > RS and
}
QR > SP. Use the Alternate Interior Angles Converse
} }
} }
to show that PS i RQ and PQ i RS. Then by definition,
PQRS is a parallelogram.
}
28. ∠PTQ > ∠RTS because they are vertical angles. Show
that nPTQ > nRTS by the AAS Congruence Theorem.
} }
Show that PT > RT and ∠PTS > ∠RTQ. Use the SAS
Congruence Postulate to show that nPTS > nRTQ.
Show that ∠TRQ > ∠TPS. Finally show that
} }
} }
PS i RQ and PQ i RS using the Alternate Interior Angles
Converse. Then by definition, PQRS is a parallelogram.
29. By the AAS Congruence Theorem, each of the four
triangles formed by the diagonals are congruent to each
} } } }
other. So, AB > BC > CD > DA. Because two side
lengths are equal, the angles opposite them are equal
in measure. So, the 8 angles formed at the vertices are
all congruent to each other. The interior angles of a
quadrilateral measure 3608. So, each of these 8 angles
measures 458. So,
m∠A 5 m∠B 5 m∠C 5 m∠D 5 908. Because the four
sides are congruent and there are four right angles, the
quadrilateral is a square.
30. m∠PTQ 5 908 by the Triangle Sum Theorem.
So, the diagonals are perpendicular. PQRS is a rhombus
by definition.
31. Since the quadrilateral is a parallelogram with congruent
diagonals, it is a rectangle.
32. The diagonals of a rhombus are perpendicular, so
m∠LQM 5 908.
m∠LMQ 1 m∠QLM 1 m∠LQM 5 1808
1
36. 34 5 } (x 1 43)
2
1
37. 0.5 5 } (0.6 1 x)
2
68 5 x 1 43
1 5 0.6 1 x
0.4 5 x
25 5 x
38. m∠V 5 m∠S 5 758
39. m∠S 5 m∠V
m∠S 1 m∠T 1 m∠V 1 m∠R 5 3608
m∠V 1 1048 1 m∠V 1 608 5 3608
2(m∠V) 1 1648 5 3608
2(m∠V) 5 1968
m∠V 5 988
40. m∠T 5 m∠R 5 908
m∠V 1 m∠R 1 m∠S 1 m∠T 5 3608
m∠V 1 908 1 808 1 908 5 3608
m∠V 5 1008
41. The diagonals bisect each other. By Theorem 8.10,
ABCD is a parallelogram.
42. m∠A 1 m∠B 1 m∠C 1 m∠D 5 3608
1198 1 m∠B 1 518 1 618 5 3608
m∠B 5 1298
Because m∠A 1 m∠D 5 1808 and m∠B 1 m∠C 5
1808, by the Consecutive Interior Angles Converse,
} }
AB i CD. So, ABCD is a trapezoid.
}
}
43. Because one pair of opposite sides, AD and BC, are
congruent and parallel, ABCD is a parallelogram by
Theorem 8.9. Because the diagonals of the parallelogram
are perpendicular, by Theorem 8.11 ABCD is a rhombus.
}
}
44. Because the diagonals AC and BD bisect each other,
ABCD is a parallelogram. ABCD is also a rectangle
because its diagonals are congruent and a rhombus
because its diagonals are perpendicular. Because ABCD
is a rectangle and a rhombus, ABCD is a square.
} }
45. Because AB i CD, m∠D 1 m∠A 5 1808 and
m∠B 1 m∠C 5 1808. So, m∠D 5 658 and
m∠B 5 1158. Because the base angles are congruent and
ABCD has one pair of parallel opposite sides, ABCD is
an isosceles trapezoid.
46. By the ASA Congruence Postulate nAGD > nBGC,
} }
} }
so AG > BG and GD > GC. Using the Converse of the
} }
} }
Base Angles Theorem, AG > GD and BG > GC. By
} } } }
substitution, AG > GC > BG > GD. The diagonals of
ABCD are congruent and bisect each other but are not
perpendicular. By Theorem 8.13, ABCD is a rectangle.
m∠LMQ 1 308 1 908 5 1808
m∠LMQ 5 608
Geometry
ngws-EP.indd 479
479
7/11/06 11:48:54 AM
Extra Practice,
47.
continued
49.
y
y
E(9, 12)
E(14, 4)
D(6, 8)
F(12, 8)
D(10, 3)
2
G(9, 6)
G(12, 0)
2
2
F(20, 2) x
x
2
}}
423
1
}
Slope of DE 5 }
5 }4
14 2 10
DE 5 Ï(9 2 6)2 1 (12 2 8)2 5 5
}}
EF 5 Ï(12 2 9) 1 (8 2 12) 5 5
2
2
}}
224
1
}
Slope of EF 5 }
5 2}3
20 2 14
}
FG 5 Ï(9 2 12)2 1 (6 2 8)2 5 Ï13
}}
}
GD 5 Ï(6 2 9)2 1 (8 2 6)2 5 Ï13
022
1
}
Slope of FG 5 }
5 }4
12 2 20
DEFG is a kite because two pairs of consecutive sides are
congruent and the opposite sides are not congruent.
48.
320
3
}
Slope of GD 5 }
5 2}2
10 2 12
} }
So, DE i FG. DEFG is a trapezoid because it has one pair
of parallel sides.
y
D(1, 2)
E(4, 1)
1
50.
x
1
y
E(1, 13)
F(5, 13)
D(�2, 10)
G(0, �1)
F(3, �2)
G(�2, 6)
2
}
DE 5 Ï(4 2 1)2 1 (1 2 2)2 5 Ï10
}}
2
}
EF 5 Ï(3 2 4)2 1 (22 2 1)2 5 Ï10
}}}
}
FG 5 Ï(0 2 3)2 1 (21 2 (22))2 5 Ï10
}}
}
GD 5 Ï(1 2 0) 1 (2 2 (21)) 5 Ï10
2
2
1
} 122
Slope of DE 5 }
5 2}3
421
} 22 2 1
Slope of EF 5 }
53
324
1
} 21 2 (22)
Slope of FG 5 }
5 2}3
023
} 2 2 (21)
Slope of GD 5 }
53
120
So, ∠D, ∠E, ∠F, and ∠G are right angles because the
segments that form them are perpendicular. DEFG is a
square because the four sides are congruent and the four
angles are right angles.
x
13 2 10
}
Slope of DE 5 }
51
1 2 (22)
} 13 2 13
Slope of EF 5 }
50
521
13 2 6
}
Slope of FG 5 }
51
5 2 (22)
10 2 6
4
}
Slope of GD 5 }
5 }0 5 undefined
22 2 (22)
} }
So, DE i FG.
}}
EF 5 Ï(5 2 1)2 1 (13 2 13)2 5 4
}}}
GD 5 Ï(22 2 (22))2 1 (10 2 6)2 5 4
} }
So, EF > GD. DEFG is an isosceles trapezoid because it
has one pair of parallel sides and one pair of non-parallel
congruent sides.
Chapter 9 (pp. 912–913)
1. To go from A to A9, move 4 units right and 2 units down.
So, a rule for the translation is (x, y) → (x 1 4, y 2 2). Use
the SAS Congruence Postulate. Notice that
}
AC 5 A9C9 5 3 and BC 5 B9C9 5 2. The slopes of AC and
}
}
}
A9C9 are 0, and the slopes of BC and B9C9 are undefined,
so the sides are perpendicular. Therefore,
∠C and ∠C9 are congruent right angles. So,
nABC > nA9B9C9. The translation is an isometry.
480
}}
Geometry
ngws-EP.indd 480
7/11/06 11:49:02 AM
Extra Practice,
continued
2. To go from A to A9, move 3 units left and 1 unit down. So,
a rule for the translation is (x, y) → (x 2 3, y 2 1).
}}
}
11.
}
AB 5 Ï(3 2 1)2 1 (0 2 4)2 5 Ï20 5 2Ï5
}}
}
}}
}
BC 5 Ï(21 2 3)2 1 (1 2 0)2 5 Ï17
2
1
}}}
}
}}}
2
}
}}}
}
1
F
2
}
By the SSS Congruence Postulate, nABC > nA9B9C9.
The translation is an isometry.
6.
7.
8.
FG FG F G F G
F GF GF
F G
F GF G
F
G
F G
F
GF
G
F
G
2
3
1
7
5 23
714
0
2
4
29
213
5
4
1
5
4 21
5
9.
7 23
2 21
5
6
9
5
12.
23 2 1
4 2 (21)
5(21) 1 9(8)
64
0
1
0
23 211
1
22
1
22
3 25
7
22 22
1
5
2
G
D
x
A9
B9
C9
3 3
3
4 4
4
GF
1
7 23
6
0
8 24
GF
5
10
0
3
10
12
0
G
A9
4
G
C9
22
x
C
13.
F
24 24 24 24 24
1
5
67
0
6
B
5
5(2) 1 9(6)
5
y
5 24
7(21) 1 (23)(8)
3
4
B9
8
26 26 26
5
F
24 231
5
10.
520
29 2 4
7(2) 1 (23)(6)
4
A
11
213
9
D9
5
5
1
B
21
####Y 5 〈1 2 7, 21 2 (23)〉 5 〈26, 2〉
PP9
####Y 5 〈3 2 7, 2 2 (23)〉 5 〈24, 5〉
PP9
####Y 5 〈28 2 7, 211 2 (23)〉 5 〈215, 28〉
PP9
1
C
2
So, AB 5 A9B9, BC 5 B9C9, and AC 5 A9C9.
5.
5
A
A9C9 5 Ï(24 2 (22))2 1 (0 2 3)2 5 Ï13
4.
10
GF
G
y
B9C9 5 Ï(24 2 0) 1 (0 2 (21)) 5 Ï17
####Y 5 〈23 2 7, 4 2 (23)〉 5 〈210, 7〉
3. PP9
1
22 21 22 25
A9B9 5 Ï(0 2 (22))2 1 (21 2 3)2 5 Ï20 5 2Ï5
2
1
27 27 27 27
5
AC 5 Ï(21 2 1) 1 (1 2 4) 5 Ï13
2
F
5
F
F
5
5
9
5
6
4
5
G
2
3
21 24 24 24
2
5
2
0 22 21
4
1
1
1
7
G
G
y
E9
A9
E
2
C 9 B9
D9
x
A
21
y
2
C9
C
x
22
B9
B A9
C
D
14. B�
B
y
A
B
A� A
1
C�
C
x
1
15.
B� y
C�
B
A�
A
�1
3
C
x
D� D
Geometry
ngws-EP.indd 481
481
7/11/06 11:49:09 AM
Extra Practice,
16.
y
2
A
C�
20.
B
B�
D
D�
F
0
21
x
3
E
continued
A�
0 21
GF
1
3
1
C(1, 0) → C9(0, 21)
y A�
GF
5
24 21 23
G
Q
x
R9
P9
B
21.
x
2
P9 Q9 R9
21 22 24
R
Q9 21
B(4, 2) → B9(2, 24)
C
4
P
A(22, 1) → A9(1, 2)
2
Q R
2 4
y
C
E�
17. (a, b) → (b, 2a)
A
P
1
C�
0 21
S
4
1
2 23 0
F GF
0
S9
B�
4
T V
2 1
S9 T9 V9
22 3 0
GF
5
4
2 1
G
y
18. (a, b) → (2a, 2b)
T9 S
A(23, 3) → A9(3, 23)
V9
B(1, 2) → B9(21, 22)
21
V
x
C(1, 21) → C9(21, 1)
T
D(25, 0) → D9(5, 0)
B
2
C�
D�
D
0
1
A B C
4 21 22
21
0
0 21 22 23
F GF
x
2
C9
C
B�
2
GF
5
A
21 B
x
D
23. (x, y) → (x 2 2, y 1 3)
C(6, 22) → C9(2, 6)
D(3, 23) → D9(3, 3)
A(1, 1) → A9(21, 4)
E(21, 21) → E9(1, 21)
B(4, 1) → B9(2, 4)
C�
(a, b) → (b, 2a)
B�
x
C
B
A�
C�
x
1
B�
24. Point A is one unit left of x 5 2, so its reflection A9 is one
unit right of x 5 2 at (3, 1). Also, B9 is 2 units left of
x 5 2 at (0, 1). Because point C is on x 5 2, you know that
y
C 5 C9.
(x, y) → (x 1 3, y)
C�
C� C
A9(3, 1) → A0(6, 1)
B9(0, 1) → B 0(3, 1)
2
C9(2, 4) → C 0(5, 4)
B�
A
1
482
A
C9(0, 7) → C 0(7, 0)
E�
D
C
B9(2, 4) → B 0(4, 22)
B
2
E
B�
1
A9(21, 4) → A0(4, 1)
D�
y
A�
C(2, 4) → C9(0, 7)
C�
A
G
A9
B(5, 1) → B9(21, 5)
A�
2 21
C
A(2, 1) → A9(21, 2)
2
1
24
D9
19. (a, b) → (2b, a)
y
A9 B9 C9 D9
0 21 22 23
y
B9
A�
D
1
B
A� B�
A�
22.
y
A
x
Geometry
ngws-EP.indd 482
7/11/06 11:49:19 AM
Extra Practice,
continued
25. (a, b) → (2a, 2b)
33. A9
B9
A(1, 1) → A9(21, 21)
A
B
B(4, 1) → B 9(24, 21)
E
C(2, 4) → C 9(22, 24)
y
C
D
C
C9
1
C�
B�
A
A�
B�
34.
B
x
1
A�
A
B
E
A9
C�
B9
Point A9 is one unit above y 5 22, so its reflection A0 is
one unit below y 5 22 at (21, 23). Also, B0 is one unit
below y 5 22 at (24, 23), and C 0 is two units above
y 5 22 at (22, 0).
26. (x, y) → (x 2 4, y 2 4)
y
D
C9
C
35. A9 A
B9
B
C
E
A(1, 1) → A9(23, 23)
B(4, 1) → B9(0, 23)
C(2, 4) → C9(22, 0)
(a, b) → (b, a)
C�
B�
1
C9
A
B
1
D
C�
A9(23, 23) → A0(23, 23)
C
x
36.
B9
A9
A�
A�
B�
B9(0, 23) → B 0(23, 0)
B
A
C9(22, 0) → C 0(0, 22)
E
27. 2(448) 5 888
A rotation of 888 about the intersection of lines k and m
maps A to A0.
37. A
A9
28. 2(738) 5 1468
flag has two lines of symmetry, one line passing vertically
through the center of the circle and the other passing
horizontally through the center of the circle. The flag has
rotational symmetry of 1808.
C
38.
G H J
1 3 4
F
3
4 2
G
31. The flag has line symmetry and no rotational symmetry.
2
The flag has one line of symmetry passing vertically
through the center of the rectangle.
B
E
D
B9
C
4
GF
5
y G9
The flag has one line of symmetry passing horizontally
through the center of the rectangle.
A
C9
D
30. The flag has line symmetry and no rotational symmetry.
32. A9
B
B9
E
A rotation of 1468 about the intersection of lines k and m
maps A to A0.
29. The flag has line symmetry and rotational symmetry. The
C9 C
D
G9 H9 J9
3 9 12
12 6 12
J9
J
H9
H
x
22
39.
1
}
2
G
K L M N
2 4 5 6
F
22 22 4 0
y
C9
GF
5
K9 L9 M9 N9
1 2 2.5 3
21 21
2 0
G
M
M9
1
N9
N
x
21
K9 L9
K
L
Geometry
ngws-EP.indd 483
483
7/11/06 11:49:28 AM
Extra Practice,
P Q R
23 23 21
F
4
P9 Q9
212 212
GF
5
21 23 23
2
Q9
24 212 212
G
y
P
2 x
Q
P9
R9
24
R
R9
Chapter 10 (pp. 914–915)
}
1. Sample answer: KF
2. @##$
AF is a common tangent.
@##$
3. Sample answer: CD
@##$ is a secant.
4. EH
5. Sample answer: K
6. Sample answer: A
}
}
7. GH is a chord.
8. FJ is a diameter.
SC 2 5 QC 2 1 SQ 2
9.
(r 1 3)2 5 r 2 1 52
2
r 1 6r 1 9 5 r 2 1 25
6r 5 16
16
r 5 }3
SC 2 5 QC 2 1 SQ 2
(r 1 2)2 5 r 2 1 42
r 2 1 4r 1 4 5 r 2 1 16
4r 5 12
}
r53
11. 3x 2 5 5 2x 1 7
}
x 5 12
}
12. 9x 2 1 x 1 1 5 x 1 5
9x 2 5 4
4
x 2 5 }9
}}
2
x 5 6}3
(x 2 1)2 5 x 2 2 7
}
14. 6x 1 9 5 4x 1 7
x 2 2 2x 1 1 5 x 2 2 7
22x 5 28
x54
2x 5 22
x 5 21
C
C
}
C
16. EB is a minor arc. Since BD is a diameter,
mC
DEB 5 1808. So,
C
mC
EB 5 mC
DEB 2 mDE
mC
EB 5 1808 2 308
mC
EB 5 1508
15. ED is a minor arc; mED 5 308.
484
}
}
8
13.
angles, they are congruent. So, mCD 5 758.
}
r5}
6
10.
C
C
C
C
C
mEC 5 mED 1 mDC
C 5 308 1 758
mEC
C 5 1058
mEC
C is a major arc. Since }
18. BEC
AC is a diameter,
mC
AB 1 mC
BC 5 1808. So, mC
BC 5 1058.
C 5 3608 2 mBC
C
mBEC
C 5 3608 2 1058
mBEC
C 5 2558
mBEC
C
19. BC is a minor arc. As shown in Exercise 18,
C 5 1058.
mBC
C is a semicircle. Since BCD
C is a semicircle,
20. BCD
3608
C 5 2 5 1808.
mBCD
21. mC
AED 5 3608 2 mC
AD
C 5 3608 2 508
mAED
mC
AED 5 3108
AD
C 5 mC
22. mBD
2
C 5 5082
mBD
mC
BD 5 258
C 2 mAD
C 24. mBAE
C5 mBA
C 1 mC
C 5 mAE
23. mDE
AE
C
C
mDE 5 1808 2 508
mBAE 5 258 1 1808
C 5 2058
C 5 1308
mDE
mBAE
C
3608 2 mAC
25. mC
AB 5
2
C 5 3608 22 1308
mAB
C 5 23082
mAB
mC
AB 5 1158
C 5 3608 2 12582 2 1258
26. mAB
C 5 11082
mAB
mC
AB 5 558
C 5 3608 2 13582 2 1358
27. mAB
C 5 9082
mAB
mC
AB 5 458
17. EC is a minor arc. Since ∠AGB and ∠CGD are vertical
}}
}
}
28. AB is a diameter; Theorem 10.4 states that if one chord
is a perpendicular bisector of another chord, then the first
chord is a diameter.
C C
29. AB > DE ; Theorem 10.3 states that in the same circle, the
40.
continued
minor arcs are congruent if and only if their corresonding
chords are congruent.
} }
30. Sample answer: AB > DE; Theorem 10.6 states that in the
same circle, two chords are congruent if and only if they
are equidistant from the center.
Geometry
ngws-EP.indd 484
7/11/06 11:49:32 AM
Extra Practice,
continued
31. By Theorem 10.7, the measure of the intercepted arc of the
408 inscribed angle is 808. So the measure of the arc with
the inscribed angle of y8 is 1808 2 808 5 1008. So
1
y 5 }2 (100) 5 50. Since the endpoints of the inscribed
angle of x8 form a diameter of the circle, the intercepted
1
38. By the Linear Pair Postulate, the angles adjacent to
the 1388 angle are 1808 2 1388 5 428. Then, by
Theorem 10.12,
1
428 5 }2 (x8 1 50)8
84 5 x 1 50
arc is 1808. So, x 5 }2(180) 5 90.
32. The inscribed angle of x8 has an intercepted arc of 1408.
1
So, x 5 }2(140) 5 70. The inscribed angles of 208 and
y8 intercept the same arc. By Theorem 10.8, the angles are
congruent. So, y 5 20.
33. By Theorem 10.8,
34 5 x
1
x8 5 }2 (1108)
x 5 55
(2x 2 5)8 5 (x 1 20)8
1
758 5 }2 ((12x 1 3)8 2 5x8)
x 5 25
1
By Theorem 10.7,
75 5 }2 (7x 1 3)
1
(y 1 27)8 5 }2 (4y 1 10)8
150 5 7x 1 3
147 5 7x
y 1 27 5 2y 1 5
21 5 x
22 5 y
41. By Theorem 10.13, measure of the intercepted arc is
3608 2 1108 2 928 5 1588
x8 1 758 5 1808
1
(10x 1 3)8 5 }2(1588 2 928)
x 5 105
By Theorem 10.10,
1
10x 1 3 5 }2(66)
y8 1 958 5 1808
10x 1 3 5 33
y 5 85
10x 5 30
x53
9x8 1 1178 5 1808
42. Using the tangent ratio, you find that the angle of the
9x 5 63
triangle formed by the radius and diameter shown is about
36.98. So,
x57
By Theorem 10.10,
x8 ø 3608 2 1808 2 36.98
(7y 2 1)8 1 838 5 1808
x ø 143.1
7y 1 82 5 180
7y 5 98
15x 5 6 p 5
y 5 14
36. The measure of the missing arc is
3608 2 1688 2 888 2 568 5 488.
1
13x8 5 }2 (488 1 568)
1
8y8 5 }2 (888 1 488)
1
13x 5 }2 (104)
1
8y 5 }2 (136)
13x 5 52
8y 5 68
x54
1
x8 5 }2 (558 1 358)
1
x 5 }2 (90)
x 5 45
y 5 8.5
15x 5 30
x52
10(x 1 1) 5 12 p 5
10x 1 10 5 60
10x 5 50
x55
4 p (4 1 x) 5 5(8)
3x p x 5 4 p 3
3x 2 5 12
x2 5 4
x52
122 5 8(8 1 3x 1 1)
16 1 4x 5 40
144 5 8(3x 1 9)
4x 5 24
144 5 24x 1 72
x56
72 5 24x
35x
Geometry
ngws-EP.indd 485
485
7/11/06 11:49:34 AM
Extra Practice,
continued
54.
y
2
(2x) 5 (x 1 2)(x 1 2 1 2x 1 5)
4x 2 5 (x 1 2)(3x 1 7)
2
4x 2 5 3x 2 1 13x 1 14
22
x
0 5 2x 2 1 13x 1 14
0 5 (2x 1 14)(x 1 1)
2x 1 14 5 0
x1150
14 5 x
Chapter 11 (pp. 916–917)
x 5 21
x cannot be a negative number since 2x must be positive so
x 5 14.
49. h 5 0, k 5 22, r 5 4
1. Using b 5 13 and h 5 11,
Area 5 bh 5 13(11) 5 143 square units.
2. Using b 5 16 and h 5 10,
Area 5 bh 5 16(10) 5 160 square units.
(x 2 h)2 1 ( y 2 k)2 5 r 2
3. Using b 5 15 and h 5 7.5,
(x 2 0)2 1 ( y 2 (22))2 5 42
1
1
Area 5 }2 bh 5 }2(15)(7.5) 5 56.25 square units.
x 2 1 ( y 1 2)2 5 16
50. h 5 2, k 5 23,
4. Using b 5 12 and h 5 7,
}}}
r 5 Ï(7 2 2)2 1 (28 2 (23))2
1
1
Area 5 }2 bh 5 }2 (12)(7) 5 42 square units.
}
5 Ï52 1 (25)2
}
5. Find x, the length of the other leg.
5 Ï25 1 25
252 5 202 1 x 2
}
5 Ï50
625 5 400 1 x 2
(x 2 h)2 1 ( y 2 k)2 5 r 2
(x 2 2)2 1 (y 2 (23))2 5 (Ï50 )2
225 5 x 2
(x 2 2)2 1 (y 1 3)2 5 50
15 5 x
}
Perimeter 5 15 1 20 1 25 5 60 cm
51. h 5 m, k 5 n
}}}
r 5 Ï(m 1 h 2 m) 1 (n 1 k 2 n)
2
1
2
Area 5 }2 (20)(15) 5 150 cm2
}
r 5 Ïh 2 1 k 2
512 5 242 1 x 2
}
(x 2 m)2 1 ( y 2 n)2 5 (Ïh 2 1 k 2 )2
2601 5 576 1 x 2
(x 2 m)2 1 ( y 2 n)2 5 h 2 1 k 2
52.
2025 5 x 2
y
45 5 x
Perimeter 5 45 1 24 1 51 5 120 ft
1
Area 5 }2 (24)(45) 5 540 ft2
1
x
21
1
7. A 5 } bh
2
1
53.
y
22 5 4x 1 2
14.3 5 1.1x
A 5 bh
7.2 5 3(3x)
3
7.2 5 9x
x
0.8 5 x
10.
1
14.3 5 }2 x(2.2)
55x
23
A 5 bh
22 5 }2 (2x 1 1)(4)
20 5 4x
9.
8.
1
A 5 }2 bh
13 5 x
6. Find x, the length of the other leg.
(x 2 h)2 1 ( y 2 k)2 5 r 2
276 5 1 }2 2(23)(6x)
1
276 5 69x
45x
486
Geometry
ngws-EP.indd 486
7/11/06 11:49:39 AM
Extra Practice,
1
11. A 5 } h(b1 1 b2)
2
continued
1
12. A 5 } h(b1 1 b2)
2
1
1
5 }2 (10)(12 1 21)
C 5 2πr 5 2π 1 }2 2 5 5π
5 22 square units
5 165 square units
The circumference of the red circle is 5π, or about
15.71 units.
1
14. A 5 } h(b1 1 b2)
2
1
1
5 }2 (7)(11 1 9)
5 }2 (9)(18 1 6)
5 70 square units
5 108 square units
1
15. A 5 } d1d2
2
1
16. A 5 } d1d2
2
1
1
5 }2 (9)(16)
5 }2 (11)(11)
5 72 square units
5 60.5 square units
1
17. A 5 } d1d2
2
1
18. A 5 } d1d2
2
1
1
5 }2 (3)(9)
5 }2 (4)(6)
5 13.5 square units
5 12 square units
100
a2
19. }2 5 }
81
b
10
9
} 5 }.
The ratio of the lengths of the corresponding sides is
1
2
} 5 } 5 }.
8
a2
21. }2 5 }
1
b
}
}
Ï8
2Ï 2
a
} 5 } 5 }.
1
1
b
Area of small triangle
7.5
1
22. }} 5 } 5 }
15
2
Area of large triangle
Length in small triangle
1
}} 5 }
}
Length in large triangle
Ï2
1
Ï2
}
5Ï2
1
Ï2
ST 5 }
} (CB) 5 }
} (5) 5 }
2
}
5Ï 2
ø 3.5 inches.
The length ST is }
2
Area of large rectangle
98
49
23. }} 5 } 5 }
50
25
Area of small rectangle
Length in large rectangle
7
}} 5 }
5
Length in small rectangle
7
C 5 2πr 5 2π (2) 5 4π
12.57 units.
27
27. Diameter of red circle: d 5 } 5 9
3
9
Radius: r 5 }2
C 5 2πr 5 2π 1 }2 2 5 9π
9
28.27 units.
d
8
28. Radius: r 5 } 5 } 5 4
2
2
25.13 units.
C
1208
30. Arc length of C
AB 5
p 2π(10) ø 20.94 feet
3608
308
8
31. Arc length of C
AB 5
p 2π 1 2 2 ø 2.09 inches
3608
1508
20
32. Arc length of C
AB 5
p 2π 1 2 2 ø 26.18 centimeters
3608
908
29. Arc length of AB 5 } p 2π(3) ø 4.71 meters
3608
25
a2
20. }2 5 }
100
b
5
10
d
4
26. Radius of red circle: r 5 } 5 } 5 2
2
2
C 5 2πr 5 2π (4) 5 8π
a
b
5
5 }2 (4)(4 1 7)
1
13. A 5 } h(b1 1 b2)
2
a
b
5
d
25. Radius of red circle: r 5 } 5 }
2
2
7
ST 5 }5(EF) 5 }5(10) 5 14 m
}
}
}
}
}
33. A 5 πr 2 5 π (32) 5 9π in.2
The area of a circle with a 3 inch radius is 9π square
inches. So, the area is about 28.27 square inches.
34. A 5 πr 2 5 π(2.5)2 5 6.25π
The area of a circle with a 2.5 centimeter radius is
6.25π square centimeters. So, the area is about
19.63 square centimeters.
20 2
35. A 5 πr 2 5 π } 5 100π
2
1 2
The area of a circle with a 20 foot diameter is 100π square
feet. So, the area is about 314.16 square feet.
1 2
13 2
36. A 5 πr 2 5 π }
2 5 42.25π
The area of a circle with a 13 meter diameter is
42.25π square meters. So, the area is about 132.73 square
meters.
Area of large figure
150
25
24. }} 5 } 5 }
54
9
Area of small figure
Length in large figure
Length in small figure
5
3
}} 5 }
5
5
ST 5 }3 (JK) 5 }3 (9) 5 15 in.
Geometry
ngws-EP.indd 487
487
7/11/06 11:49:41 AM
Extra Practice,
continued
C
mC
DGE 5 3608 2 458 5 3158
1
37. mDE 5 458
A 5 }2 bh
1
C
mDE
Area of small sector 5 } p πr 2
3608
458
5 } p π(52) ø 9.82 in.2
3608
mDGE
Area of large sector 5 } p πr 2
3608
3158
5 } p π(52) ø 68.72 in.2
3608
C
38. Because ∠DFE is a straight angle, both sectors have the
C
mDHE
3608
1808
22 2
5}pπ }
ø 190.07 cm2
2
3608
Area of each sector 5 } p πr 2
C
C 5 3608 2 1008 5 2608
mDGE
39. mDE 5 1008
1 2
}
5 81Ï3
ø 140.30 square units
3608
46. Measure of a central angle: } 5 728
5
The apothem bisects a 728
central angle to form a
right triangle with a 368
angle and a longer leg
that is 2 units long.
2
36�
x
x
2
tan 368 5 }
2 tan 368 5 x
The length of a side is 2x.
C
mDE
3608
1008
5 } p π (72) ø 42.76 ft2
3608
mDGE
3608
2608
5 } p π(72) ø 111.18 ft2
3608
C
C
mC
DE 5 3608 2 2408 5 1208
40. mDGE 5 2408
Perimeter: P 5 5 p 2x 5 10(2 tan 368) ø 14.53 units
1
1
Area: A 5 }2 aP ø }2 (2)(14.53) 5 14.53 square units
3608
6
right triangle with a 308
angle and a shorter leg
a 30�
4.5
C
mDE
3608
1208
5 } p π (22) ø 4.19 yd2
3608
mDGE
3608
2408
5 } p π(22) ø 8.38 yd2
3608
C
41. The measure of a central angle of a regular octagon is
3608
8
} 5 458.
42. The measure of a central angle of a regular dodecagon is
3608
12
} 5 308.
43. The measure of a central angle of a regular 20-gon is
3608
20
that is }
5 2.25 units long.
2
Perimeter: P 5 6(4.5) 5 27 units
1
Area: A 5 }2 aP 5 }2 1 } 2(27) ø 52.61 square units
1
2.25
tan 308
3608
8
right triangle with a 22.58
angle and a hypotenuse
that is 12 units long.
44. The measure of a central angle of a regular 25-gon is
3608
25
} 5 14.48.
22.5�
a
x
sin 22.58 5 }
12
12 sin 22.58 5 x
Side length: 2x 5 24 sin 22.58
Perimeter: P 5 8(24 sin 22.58) ø 73.48 units
1
45. P 5 3(18) 5 54 units
9
h 2 1 92 5 182
h
h 2 1 81 5 324
Area: A 5 }2 aP
1
ø }2(12 cos 22.58)(73.48)
18
12
x
a
cos 22.58 5 }
12
12 cos 22.58 5 a
} 5 188.
2.25
2.25
tan 308 5 }
a
2.25
a5}
tan 308
ø 407.29 square units
same area.
}
5 }2(18)9Ï 3
h 2 5 243
}
}
h 5 Ï243 5 9Ï 3
488
Geometry
ngws-EP.indd 488
7/11/06 11:49:48 AM
Extra Practice,
continued
49. The center of the circle is the point of concurrency of the
medians of the regular triangle, so the radius of the circle
2
is }3 the height of the triangle. Then the height of the
triangle is 6. Using a side length of s:
s 2
1 }2 2
s2
} 1 36 5 s 2
4
s
6
3
s
2
36 5 }4 s 2
48 5 s 2
1
}
π (42) 2 }2 (4Ï3 )6
5 }}
ø 0.587
2
)
There is about a 58.7% probability that a randomly chosen
point in the circle lies in the shaded region.
Area of sector 2 Area of triangle
50. P 5 }}}
Area of circle
908
1
2
3608
}}
} p π(152) 2 }(15)(15)
π(152)
ø 0.091
point in the circle lies in the shaded region.
51. In the large triangle, the base 5 10 5 the height.
The small triangle at the top is similar to the large triangle
and its height is 3. So its base must be equal to 3.
by polygons.
3. Yes, the solid is a polyhedron; the solid is a triangular
pyramid; the solid is bounded by polygons and has a
triangle for a base.
4. No, the solid is not a polyhedron; it is not bounded by
5. Using Euler’s Theorem,
Area of circle 2 Area of triangle
P 5 }}}
Area of circle
Sum of areas of shaded triangles
P 5 }}}
Area of large triangle
1
2
prism; the solid is bounded by polygons and has pentagons
for bases.
polygons and the bases are not polygons.
}
4Ï 3 5 s
5
1. Yes, the solid is a polyhedron; the solid is a pentagonal
2. No, the solid is not a polyhedron; it is not bounded
1 62 5 s 2
π(4
Chapter 12 (pp. 918–919)
1
2
}(3)(3) 1 }(3)(7)
5 }}
1
}(10)(10)
2
5 0.3
There is a 30% probability that a randomly chosen point in
the large triangle lies in the shaded region.
52. Base of rectangle: 4(6) 5 24
Height of rectangle: 2(6) 5 12
Area of rectangle 2 Sum of areas of circles
P 5 }}}}
Area of rectangle
24(12) 2 [π(62) 1 π(62)]
5 }}
24(12)
ø 0.215
point in the rectangle lies in the shaded region.
4.5
53. P 5 } 5 0.0375
120
6. B 5 6 p 5 5 30
F1V5E12
P 5 2(6 1 5) 5 2(11) 5 22
F 1 6 5 10 1 2
S 5 2B 1 Ph
F 1 6 5 12
S 5 2(30) 1 (22)(4)
F56
S 5 60 1 88
S 5 148 ft2
There are 6 faces.
7. Find the height of the triangle:
}
}
}
h 5 Ï52 2 (2.5)2 5 Ï25 2 6.25 5 Ï18.75 ø 4.33
1
1
B 5 }2 bh ø }2 (5)(4.33) ø 10.825
P 5 3(5) 5 15
S 5 2B 1 Ph
ø 2(10.825) 1 (15)(9)
ø 21.65 1 135
ø 156.65 cm2
8. Find the apothem of the hexagon:
a
tan 608 5 }3
3 p tan 608 5 a
P 5 6 p 6 5 36
S 5 aP 1 Ph
S 5 (3 p tan 608)(36) 1 (36)(8)
S ø 187.06 1 288
S ø 475.06 m2
9. S 5 2πr 2 1 2πrh
10. S 5 2πr 2 1 2πrh
S 5 2π(2)2 1 2π(2)(11)
S 5 2π(1)2 1 2π(1)(1)
S 5 8π 1 44π
S 5 2π 1 2π
S 5 4π ø 12.57 m2
S 5 52π
2
S ø 163.36 cm
11. S 5 2πr 2 1 2πrh
2
S 5 2π(22) 1 2π(22)(9)
12. S 5 2πr 2 1 2πrh
S 5 2π(17)2 1 2π(17)(5)
S 5 968π 1 396π
S 5 578π 1 170π
S 5 1364π
S 5 748π
S ø 4285.13 in.2
S ø 2349.91 mm2
There is a 3.75% chance that your favorite song will be
playing when you randomly turn on the radio.
Geometry
ngws-EP.indd 489
489
7/11/06 11:49:51 AM
Extra Practice,
S 5 2B 1 Ph
Find the area of the triangular base:
192 5 2(4x) 1 (2(x 1 4))(4)
1
1
}
192 5 8x 1 (2x 1 8)(4)
192 5 8x 1 8x 1 32
P 5 3(2) 5 6
160 5 16x
S 5 B 1 }2Pl
1
10 in. 5 x
1
}
S 5 Ï3 1 }2 6(5)
14. Find the height of the triangular base:
}
12 1 h2 5 3.52
S 5 Ï3 1 15
1 1 h2 5 12.25
S ø 16.73 cm2
h2 5 11.25
18. P 5 5(7) 5 35
}
h 5 Ï11.25
1
1
B 5 }2 aP 5 }2 (4.8)(35) 5 84
Find the area of the triangular base:
1
S 5 B 1 }2Pl
1
B 5 }2 bh
1
1
S 5 84 1 }2 (35)(15)
}
B 5 }2 (2)(Ï11.25 )
S 5 84 1 262.5
}
B 5 Ï11.25
P 5 2 1 3.5 1 3.5 5 9
S 5 2B 1 Ph
S 5 346.5 m2
19. S 5 πr 2 1 πrl
}
33.7 5 2(Ï11.25 ) 1 (9)(x)
}
33.7 2 2Ï11.25 5 9x
3.00 m ø x
S 5 2πr 2 1 2πrh
15.
}
B 5 }2 bh 5 }2 (2)(Ï3 ) 5 Ï3
754 5 2π(6)2 1 2π(6)(x)
21.
754 5 72π 1 12πx
754 2 72π 5 12πx
14.00 ft ø x
16. B 5 5(5) 5 25
S 5 2.7π
S ø 8.48 m2
a 2 1 b2 5 l2
S 5 π(9)2 1 π(9)(15)
81 1 144 5 l
2
S 5 81π 1 135π
2
S 5 216π
V 5 1960 cm3
25. B 5 x p x 5 x 2
V 5 π(1.15) (7.2)
V 5 Bh
V 5 π(1.3225)(7.2)
8 5 (x 2)(x)
V 5 9.522π
8 5 x3
2
V ø 29.91 mm3
12 1 h 2 5 22
1
26. B 5 }(3)(6) 5 9
2
V 5 Bh
}
h 5 Ï3
72 5 9x
8 ft 5 x
27.
1
23. B 5 }(14)(14) 5 98
2
V 5 (98)(20)
24. V 5 πr2h
17. Find the height of the triangular base:
S ø 678.58 yd2
V 5 Bh
V 5 28 ft
S 5 75 in.
S 5 πr 2 1 πrl
92 1 122 5 l2
2
2
h 53
S 5 33π
S ø 103.67 in.2
V 5 (14)(2)
S 5 25 1 50
2
S 5 π 1 1.7π
V 5 Bh
1
1 1 h2 5 4
S 5 9π 1 24π
22. B 5 (4)(3.5) 5 14
S 5 B 1 }2Pl
1
S 5 π(1)2 1 π(1)(1.7)
15 5 l
P 5 4(5) 5 20
S 5 25 1 }2(20)(5)
S 5 π(3)2 1 π(3)(8)
225 5 l
754 2 72 π
12π
}5x
20. S 5 πr 2 1 πrl
V 5 πr 2h
628 5 πx 2(8)
2 cm 5 x
13.
continued
628
8π
}�
5 x2
24.99 ø x 2
5.00 in. ø x
490
Geometry
ngws-EP.indd 490
7/11/06 11:49:56 AM
Extra Practice,
continued
28. B 5 (12)(12) 5 144
33. Find the radius of the base:
4.2
1
V 5 }3 Bh
tan 688 5 }
r
1
V 5 }3 (144)(15)
(r)(tan 688) 5 4.2
4.2
tan 688
r5}
V 5 720 in.3
29. Find the apothem and perimeter of the hexagonal base:
a 5 (2.5)(tan 608) ø 4.33013
P 5 (5)(6) 5 30
1
1
B 5 }2 aP ø }2 (4.33013)(30) ø 64.95195
1
V 5 }3 Bh
1
V ø }3 π(1.696910)2(4.2)
V ø 4.031305π
V ø 12.66 ft3
1
V ø }3 (64.95195)(8)
V ø 173.21 ft3
1
30. V 5 }πr 2h
3
1
V 5 }3 π(7.3)2(11.4)
1
3
V 5 }π(53.29)(11.4)
V 5 202.502π
V ø 636.18 m3
31. Find the radius of the base:
18
tan 458 5 }
r
(r)(tan 458) 5 18
r ø 1.696910
1
V 5 }3 πr 2h
18
r5}
tan 458
r 5 18
1
V 5 }3 πr 2h
1
V 5 }3 π(18)2(18)
1
34. S 5 4πr 2
35. S 5 4πr 2
2
S 5 4π(13)
S 5 4π(18)2
S 5 4π(169)
S 5 4π(3.24)
S 5 676π
S 5 12.96π
2
S ø 2123.72 m
S ø 40.72 in.2
4
V 5 }3 πr 3
4
4
V 5 }3 π(1.8)3
V 5 }3 π(2197)
4
V 5 }3 π(5.832)
V ø 2929.333π
V 5 7.776π
V ø 9202.77 m3
V ø 24.43 in.3
V 5 }3 πr 3
4
V 5 }3 π(13)3
4
36. S 5 4πr 2
37. S 5 4πr 2
2
S 5 4π(14)
S 5 4π(6.85)2
S 5 4π(196)
S 5 4π(46.9225)
S 5 784π
S 5 187.69π
2
S ø 2463.01 yd
S ø 589.65 cm2
4
V 5 }3 πr 3
4
4
V 5 }3 π(6.85)3
4
V 5 }3 πr 3
4
V 5 }3 π(14)3
V 5 }3 π(324)(18)
V 5 }3 π(2744)
V 5 }3 π(321.419125)
V 5 1944π
V ø 3658.667π
V ø 428.558833π
V ø 11,494.04 yd3
V ø 1346.36 cm3
V ø 6107.26 in.3
32. Find the height:
h
tan 308 5 }5
5 p tan 308 5 h
2.886751 ø h
1
3
V 5 }πr 2h
1
V ø }3 π(5)2(2.886751)
1
4
38. S 5 4πr 2
39. S 5 4πr 2
2
S 5 4π(20)
S 5 4π(17.5)2
S 5 4π(400)
S 5 4π(306.25)
S 5 1600π
S 5 1225π
2
S ø 5026.55 in.
S ø 3848.45 mm2
4
V 5 }3 πr 3
4
V 5 }3 π(17.5)3
4
V 5 }3 πr 3
V 5 }3 π(20)3
4
4
4
V ø }3 π(25)(2.886751)
V 5 }3 π(8000)
V 5 }3 π(5359.375)
V ø 24.056258π
V ø 10666.667π
V ø 7145.833π
V ø 75.57 m3
V ø 33,510.32 in.3
V ø 22,449.30 mm3
Geometry
ngws-EP.indd 491
491
7/11/06 11:49:58 AM
Extra Practice,
40. S 5 4πr 2
continued
41. S 5 4πr 2
S 5 4π(7.6)2
S 5 4π(11.5)2
S 5 4π(57.76)
S 5 4π(132.25)
S 5 231.04π
S 5 529π
S ø 725.83 m2
S ø 1661.90 ft2
4
V 5 }3 πr 3
4
4
V 5 }3 π(11.5)3
V 5 }3 π(438.976)
4
V 5 }3 π(1520.875)
V ø 585.301333π
V ø 2027.833π
V 5 }3 πr 3
4
V 5 }3 π(7.6)3
3
V ø 1838.78 m
4
V ø 6370.63 ft3
a2
Surface area of A
42. }} 5 }2
Surface area of B
b
32
324π
}} 5 }2
Surface area of B
2
9
324π
}
}} 5
4
Surface area of B
(9)(Surface area of B) 5 4(324π)
(9)(Surface area of B) 5 1296π
Surface area of B 5 144π in.2
a3
b
33
972π
}
}5 3
Volume of B
2
27
972π
}5}
8
Volume of B
Volume of A
Volume of B
} 5 }3
a2
Surface area of A
44. }} 5 }2
Surface area of B
b
42
7
16
64π
}} 5 }
49
Surface area of B
64π
Surface area of B
}} 5 }2
(16)(Surface area of B) 5 49(64π)
(16)(Surface area of B) 5 3136π
Surface area of B 5 196π cm2
a3
b
43
64π
} 5 }3
Volume of B
7
64
64π
}5}
343
Volume of B
Volume of A
Volume of B
} 5 }3
(64)(Volume of B) 5 343(64π)
Volume of B 5 343π cm3
12π
a3
45. }3 5 }�
324π
b
1
a3
}3 5 }
27
b
1
a
}5}
3
b
The scale factor of the smaller cylinder to the larger
cylinder is 1 : 3.
(27)(Volume of B) 5 7776π
Volume of B 5 288π in.3
a2
Surface area of A
43. }} 5 }2
Surface area of B
b
22
1
4
864
}
}} 5
1
Surface area of B
864
Surface area of B
}} 5 }2
(4)(Surface area of B) 5 864
Surface area of B 5 216 ft 2
a3
b
23
1728
} 5 }3
Volume of B
1
8
1728
}
}5
1
Volume of B
Volume of A
Volume of B
} 5 }3
(8)(Volume of B) 5 1728
Volume of B 5 216 ft3
492
(27)(Volume of B) 5 8(972π)
Geometry
ngws-EP.indd 492
7/11/06 11:50:00 AM

Midterm Review Packet Answers

Transcription

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