Document 1807037

Chapter 8,
continued
6. The length of the midsegment is one-half the sum of the
}
} }
}
16. EF > GH, FG > EH
lengths of the bases.
Perimeter 5 EF 1 GH 1 FG 1 EH
1
1
Midsegment 5 }2 (48 1 24) 5 }2 (72) 5 36
5 2(EF) 1 2(FG)
16 5 2(5) 1 2(FG)
The midsegment of trapezoid ABCD is 36 inches.
6 5 2(FG)
3 5 FG
}
}
}
The length of GH is 5 inches. The length of FG and EH
is 3 inches.
Chapter 8 Review (pp. 560–564)
1. The midsegment of a trapezoid is parallel to the bases.
2. A diagonal of a polygon is a segment whose endpoints
are nonconsecutive vertices.
17. Consecutive angles of a parallelogram are supplementary.
mŽ J 1 mŽ M 5 1808
3. Show the trapezoid has a pair of congruent base angles.
Show the diagonals of the trapezoid are congruent.
5x 1 4x 5 180
4. C. Rhombus; because both pairs of opposite sides are
9x 5 180
parallel and all four sides are congruent.
x 5 20
5. A. Square; there are four right angles and four congruent
mŽ J 5 5x 5 5(20) 5 1008
sides.
6. B. Parallelogram; both pairs of opposite sides
are congruent.
mŽ M 5 4x 5 4(20) 5 808
18. 2x 1 4 5 x 1 9
7. (n 2 2) + 1808 5 39608
19. 5x 2 4 5 3x 1 2
x1459
2x 2 4 5 2
x55
2x 5 6
n 2 2 5 22
x53
n 5 24
20. Both pairs of opposite sides are parallel and the diagonals
are perpendicular. So the quadrilateral is a rhombus.
y 5 21 because diagonals of a rhombus bisect opposite
angles. x8 1 y8 5 908. So x 5 90 2 21 5 69.
39608
5 1658.
each interior angle is }
24
8. x8 1 1208 1 978 1 1308 1 1508 1 908 5 (n 2 2) + 1808
x 1 120 1 97 1 130 1 150 1 90 5 (6 2 2) + 180
21. All four angles are right angles, so the quadrilateral is
a rectangle.
x 1 587 5 720
4x 2 5 5 3x 1 4
x 5 133
9.
x8 1 1608 1 2x8 1 1258 1 1108 1 1128 1 1478 5 (n 2 2) + 1808
x 1 160 1 2x 1 125 1 110 1 112 1 147 5 (7 2 2) + 180
x59
3608
11. The measure of one exterior angle is } 5 408.
9
and
}
(b 1 16)8 5 1038
a 5 28
b 5 87
15. mŽ QRS 5 1808 2 mŽ PQR 5 1808 2 1368 5 1448
Opposite sides and opposite angles of a parallelogram
are congruent.
P
1448
5 cm
368
Q
10 cm
258
10 cm
S
368
1448
R
Geometry
Worked-Out Solution Key
5 Ï169
5 13
The length of one side is 13 centimeters.
mŽ H 5 mŽ J 5 1018
24. F
19 in.
16.5 in.
M
d 5 10
and
12
5 cm
J
1
G
MN 5 }2 (FG 1 JH)
N
16.5 5 }2 (19 1 JH)
d 1 4 5 14
c56
14. a 2 10 5 18
5 Ï52 1 122
5
n2358
n 5 11
13. c 1 5 5 11
}
5
mŽ J 5 1808 2 mŽ F 5 1808 2 798 5 1018
x 5 20
and
* 5 length of one side
23. mŽ G 5 mŽ F 5 798
18x 5 360
12. m 5 10
y55
12
3x 5 246
10. 8x8 1 5x8 1 5x8 5 3608
2y 2 10 5 0
22.
3x 1 654 5 900
x 5 82
6y 2 10 5 4y
x2554
H
}
The length of JH is 14 inches.
1
33 5 19 1 JH
14 5 JH
25. All four sides of the quadrilateral are congruent, so it is
a rhombus. You do not know the angle measures, so it
cannot be determined if it is a square.
26. Because consecutive interior angles are supplementary,
} }
EF i HG, and the quadrilateral is a trapezoid. You do not
}
}
know that EH is parallel to FG.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
The polygon has 24 sides. It is a 24-gon. The measure of
Chapter 8,
continued
27. Because both pairs of opposite sides are congruent,
the quadrilateral is a parallelogram. You do not know if
the angles are right angles. So it cannot be classified as
a rectangle.
28. The quadrilateral has three right angles. Because the sum
of the measures of the interior angles is 3608, the fourth
angle is a right angle. So the quadrilateral is a rectangle.
Because consecutive sides of the rectangle are congruent,
the rectangle is a square.
1. x8 1 1038 1 1228 1 988 1 998 5 (n 2 2) + 1808
x 1 103 1 122 1 98 1 99 5 (5 2 2) + 180
x 1 422 5 540
x 5 118
2. 5x8 1 1708 1 908 1 1668 1
1508 1 1438 1 1128 1 948 5 (n 2 2) + 1808
5x 1 170 1 90 1 166 1 150 1 143 1 112 1 94
5 (8 2 2) + 180
5x 1 925 5 1080
5x 5 155
x 5 31
3. x8 1 598 1 478 1 368 1 658 1 828 5 3608
x 1 59 1 47 1 36 1 65 1 82 5 360
x 1 289 5 360
x 5 71
H
1108
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
708
708
E
1108
G
y
Q
1
P
22
x
R
S
}
320
3
}
21 2 3
24
}
22 2 (21)
21
}
0 2 (22)
2
Slope of PQ 5 }
5 }2
0 2 (22)
2
Slope of QR 5 }
5}
5 2}3
620
6
Chapter 8 Test (p. 564)
4.
12.
F
mŽ F 5 mŽ G 1 408
mŽ F 5 mŽ H
mŽ E 5 mŽ G
1808 5 mŽ F 1 mŽ G
1808 5 mŽ G 1 408 1 mŽ G
1408 5 2mŽ G
708 5 mŽ G
So, mŽ G 5 708 5 mŽ E.
mŽ F 5 mŽ G 1 408 5 708 1 408 5 1108
So, mŽ F 5 1108 5 mŽ H.
5. No; you need to know that consecutive angles are
supplementary or that both pairs of opposite angles
are congruent.
6. Because the diagonals bisect each other, the quadrilateral
is a parallelogram.
7. You need to know that one pair of sides is congruent and
parallel. The figure could be an isosceles trapezoid.
8. Only rhombuses and squares are equilateral
quadrilaterals.
9. Only rectangles and squares have four interior
right angles.
10. Only rectangles and squares have congruent diagonals.
11. Parallelograms, rectangles, rhombuses, and squares have
opposite sides that are parallel.
1
Slope of RS 5 }
5}
5 }5
25
126
Slope of PS 5 }
5 2}3
22 2 1
}
}
Sides QR and PS have the same slope so they are parallel.
Because there is exactly one pair of parallel sides, PQRS
is a trapezoid.
13. a. Sample answer:
y
K(1, 2)
L(3, 2)
1
M(4, 0) x
1
J(0, 0)
}}
}
}
JK 5 Ï(1 2 0)2 1 (2 2 0)2 5 Ï12 1 22 5 Ï 5
}}
}}
}
LM 5 Ï(4 2 3)2 1 (0 2 2)2 5 Ï12 1 ( 2 2)2 5 Ï 5
0
} 222 }
Slope of KL 5 }
5250
321
0
} 020 }
Slope of JM 5 }
5450
420
}
}
Opposite sides KJ and LM are congruent and opposite
}
}
sides KL and JM are parallel. So JKLM is an isosceles
trapezoid.
b. Sample answer:
y K(2, 5)
J(0, 3)
L(4, 3)
1
x
1 M(2, 0)
}}
}
}
}
JK 5 Ï (2 2 0) 1 (5 2 3) 5 Ï2 1 2 5 Ï8 5 2Ï2
2
2
2
2
}}
KL 5 Ï(4 2 2)2 1 (3 2 5)2
}
}
}
5 Ï22 1 (22)2 5 Ï 8 5 2Ï 2
}}
LM 5 Ï(2 2 4)2 1 (0 2 3)2
}}
}
5 Ï(22)2 1 (23)2 5 Ï13
}}
}
}
JM 5 Ï(2 2 0)2 1 (0 2 3)2 5 Ï22 1 (23)2 5 Ï13
}
}
}
}
Consecutive sides JK and KL and LM and JM are
congruent. So JKLM is a kite.
c. JKLM could be a parallelogram, trapezoid, or rectangle.
14. Trapezoid; exactly one pair of parallel sides are parallel.
15. Rhombus; mŽFJG 5 1808 2 338 2 578 5 908, so
}
}
FH > EG. The diagonals of a rhombus are perpendiuclar
but you don’t know that the angle measure of the vertices
is 908.
Geometry
Worked-Out Solution Key
259
Chapter 8,
continued
5. y 5 22(x 2 4)2 2 1
16. Kite; The diagonal forms similar triangles by the
SAS Postulate, so there are two pairs of consecutive
congruent sides.
2
x
1
17. midsegment 5 } (WX 1 YZ)
2
y
3
4
x54
y
5
6
(4, 21)
1
21
x
29 23 21 23 29
1
2.75 5 }2 (WX 1 4.25)
5.5 5 WX 1 4.25
1.25 5 WX
}
The length of WX is 1.25 centimeters.
ST 5 RS 1 3
5913
5 12
RS 1 RS 1 ST 1 ST 5 42
2RS 1 2ST 5 42
2RS 1 2(RS 1 3) 5 42
4RS 1 6 5 42
4RS 5 36
RS 5 9
}
}
The length of RS is 9 centimeters and the length of ST is
12 centimeters.
Chapter 8 Algebra Review (p. 565)
1. y 5 3x 1 5
x
y
8
3
4
5
7
7.5 3.5 3 3.5 7.5
(4, 3)
1
1
x
7. y 5 3
x
y
(2, 9)
x
y
22 21 0 1
17
y
y
2
1
x
x54
y
22 21 0 1 2
1
9
}
1
3
}
1 3 9
(21, )
(22, )
1
3
2
5 8 17
(1, 3)
2
1
9
(0, 1)
1
8. y 5 8x
x
y
(0, 5)
2
(1, 8)
x
x50
21
y
2. y 5 22x 2 1 4
x
y
x50 y
0.35
1
1 2.83 8
(20.5, 0.35)
(21, )
1
8
2
1
4 2 24
2
1
8
}
0.5
(0, 4)
22 21 0 1
24
x
21 20.5 0
x
9. y 5 2.2x
x
y
24 22
5
y
x
22
2
y
0.21 0.45 1 2.2 4.84
21 23 21 5
(2, 4.84)
x
(21, 0.45)
4. y 5 3(x 1 3) 2 3
2
(22, 0.21)
0
23
0
9
2
1
(23, 23)
x 5 23
260
Geometry
Worked-Out Solution Key
(1, 2.2)
2
(0, 1)
1
y
25 24 23 22 21
9
1
1
(0, 23)
y
0
y
4
1
x
21
x50
0
(0.5, 2.83)
(0, 1)
1
1
3. y 5 0.5x 2 2 3
2
x
x
2
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
18. RS 1 TU 1 ST 1 RU 5 42
1
6. y 5 }(x 2 4)2 1 3
2
Chapter 8,
continued
1x
10. y 5 } 2
3
1
x
y
15. y 5 2{x { 2 4
y
(22, 9)
22 21 0 1 2
9
3
1
}
3
1
x
1
}
9
y
(21, 3)
y
23.38 21 0 1 3.38
1
2
1
22 24 22 0
(22, 0)
(1, ) (2, )
1
3
x
1.5
1
(2, 0) x
(1, 22)
(0, 24)
1
9
11. y 5 x 3
21 0 1
0
0
(21, 22)
1
21.5
22 21
3
(0, 1)
x
y
16. y 5 2{x { 2 1
x
24 22
y
25 23 21 23 25
0
2
4
y
y
1
(1.5, 3.38)
2
(0, 0)
(21, 1)
(0, 21)
(1, 1)
(22, 23)
x
1
(21.5, 23.38)
(24, 25)
12. y 5 x 3 2 2
(2, 23)
(4, 25)
TAKS Practice (pp. 568–569)
x
22
21
y
210 23 22 21 6
0
1
1. B;
2
The bar graph shows that Texas has about 230,000 farms
and Missouri has about 110,000 farms. Because
230,000
110,000
y
} ø 2, Texas has about twice as many farms as
(2, 6)
Missouri.
2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x
1
x
2. H;
(1, 21)
(0, 22)
21
(21, 23)
96
5 0.24 5 24% have no
In a survey of 400 students, }
400
120
108
pets, }
5 0.3 5 30% have 1 pet, }
5 0.27 5 27%
400
400
(22, 210)
400 2 (96 1 120 1 108)
have 2 pets, and the rest, }}
5
400
13. y 5 3x 2 1
3
x
21
20.5
y
24 21.38 21 20.63 2
0
0.5
400 2 324
400
So, the circle graph in choice H shows this information.
3. B;
The histogram shows that about 28 people visited their
physician 023 times and about 16 people visited their
physician 4–7 times last year. So, 28 1 16 5 44 people,
or most of the people surveyed, visited their physician
less than 8 times last year.
y
(1, 2)
1
21
x
(20.5, 21.38)
76
400
} 5 } 5 0.19 5 19%, have 3 or more pets.
1
(0.5, 20.63)
(0, 21)
(21, 24)
14. y 5 2{x {
x
y
y
22 21 0 1 2
4
2
0 2 4
(22, 4)
(2, 4)
(21, 2)
2
(0, 0)
(1, 2)
1
x
Geometry
Worked-Out Solution Key
261
Chapter 8,
continued
Length of photo
7 in.
7
8. G; }} 5 } 5 }
5
5 in.
Width of photo
4. H;
Each time the value of x increases by 1, the value of f (x)
5.25
3.5
1
change is }2 . The function can be written in the form
f (x) 5 mx 1 b, where m is the rate of change.
5.25(5) 0 3.5(7)
26.25 Þ 24.5
The dimensions 5.25 inches by 3.5 inches are not
proportional to the dimensions of the photo, so they do
not represent an enlargement or reduction of the photo.
f (x) 5 mx 1 b
1
4 5 }2 (22) 1 b
9. C;
55b
1
So, you can use the expression }2 x 1 5 to find the values
of f (x) in the table.
In a right triangle, a2 1 b2 5 c2.
62 1 9.12 0 10.92
36 1 82.81 0 118.81
5. C;
Using the formula for the volume V of a prism with base
area A and width W,
V 5 AW
The side lengths 6, 9.1, and 10.9 are the side lengths of a
right triangle.
2x 1 y 5 25 ly 5 22x 2 5
x 2 3y 5 26
211.25
}5A
6.5
x 2 3(22x 2 5) 5 26
32.5 5 A
The base area is 32.5 square centimeters. The base is
a triangle.
1
A 5 }2 bh
x 1 6x 1 15 5 26
7x 5 221
x 5 23
y 5 22(23) 2 5
5625
2A 5 bh
51
2A
b
}5h
2(32.5)
6.5
11. Of the 640 girls in your school, 5% or 0.05(640) 5 32
are on the softball team. Of those 32 on the softball
team, 12.5% or 0.125(32) 5 4 are also on the soccer
team. So, 4 girls in your school are on both the softball
and soccer teams.
10 5 h
The height h of the prism is 10 centimeters.
6. G;
12
30
2a2c9
5b
2
5
}a(3 2 1)b21c[2 2 (27)] 5 }a2b21c9 5 }
}
27 5
7. C;
The boundary line passes through the points (3, 0) and
(0, 21). Using these points, the slope of the boundary
y2 2 y1
21 2 0
21
1
5}
5 }3
line is: m 5 }
x 2x 5}
023
23
2
1
The y-intercept, b, is 21, so the equation of the boundary
1
line is y 5 }3 x 2 1. Because the boundary line is solid,
the inequality is eitheraorq. The half-plane that
includes the point (0, 0) is shaded. Because
0q}1 (0) 2 1, the inequality isq. So, the inequality
3
1
yq}x 2 1 best describes the shaded region of the graph.
3
Geometry
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
The solution is (23, 1)
}5h
12a 3c2
30abc
118.81 5 118.81 10. G;
V
W
}5A
262
7
5
}0}
1
increases by }2 . So, f (x) is a linear function whose rate of