Chapter 11,

Chapter 11,
continued
Greatest possible side lengths: 1.4 1 0.05 5 1.45 cm
12.
17 cm
12 cm
1 cm
1 cm
Greatest
possible error
1
}(1 cm) 5 0.5 cm
2
1
} (1 cm) 5 0.5 cm
2
Relative error
} ø 0.029
Unit of measure
0.5 cm
17 cm
ø 2.9%
0.5 cm
12 cm
} ø 0.0417
ø 4.2%
5.1 1 0.05 5 5.15 cm
Greatest possible perimeter:
P 5 2(1.45) 1 2(5.15) 5 13.2 cm
Least possible side lengths: 1.4 2 0.05 5 1.35 cm
5.1 2 0.05 5 5.05 cm
Least possible perimeter:
P 5 2(1.35) 1 2(5.05) 5 12.8 cm
The precision is the same. 17 centimeters is more accurate.
Lesson 11.2
13.
18.65 ft
25.6 ft
0.01 ft
0.1 ft
Greatest
possible error
1
}(0.01 ft) 5 0.005 ft
2
1
} (0.1 ft) 5 0.05 ft
2
Relative error
0.005 ft
} ø 0.0003
18.05 ft
} ø 0.002
Unit of
measure
Investigating Geometry Activity 11.2 (p. 729)
1
1. The area of one trapezoid is } the area of the
2
parallelogram formed from two trapezoids.
0.05 ft
25.6 ft
ø 0.03%
ø 0.2%
b 5 b1 1 b2
1
1
A 5 }2 bh 5 }2 (b1 1 b2)h
2. The base of the rectangle is d1 and the height of the
rectangle is }2 d2. A 5 (d1)1 }2 d2 2 5 }2 d1d2
1
18.65 feet is more precise and more accurate.
1
1
14.
6.8 in.
13.4 ft
11.2 Guided Practice (pp. 731–732)
1
1
1. A 5 } h(b1 1 b2) 5 }(4)(6 1 8) 5 28 ft2
2
2
0.1 in.
0.1 ft
Greatest
possible error
}(0.1 in.) 5 0.05 in.
} (0.1 ft) 5 0.05 ft
Relative error
} ø 0.007
1
2
0.05 in.
6.8 in.
1
2
0.05 ft
13.4 ft
} ø 0.004
ø 0.7%
ø 0.4%
6.8 inches is more precise. 13.4 feet is more accurate.
15.
1
1
2. A 5 } d1d2 5 } (6)(14) 5 42 in.2
2
2
3. d1 5 30 m 1 30 m 5 60 m
d2 5 40 m 1 40 m 5 80 m
1
1
4. A 5 } d1d2; when A 5 80 ft2, d1 5 x, and d2 5 4x.
2
1
3.5 ft
35 in.
80 5 }2 (x)(4x)
0.1 ft
1 in.
80 5 }2 4x2
Greatest
possible error
}(0.1 ft) 5 0.05 ft
} (0.1 in.) 5 0.5 in.
Relative error
} ø 0.014
Unit of measure
1
2
0.05 ft
3.5 ft
1
2
ø 1.4%
0.5 in.
35 in.
16. Greatest possible error for 5.1 cm side:
1
40 5 x 2
}
2Ï10 5 x
} ø 0.014
ø 1.4%
35 inches is more precise. The accuracy is about the same.
}
1
1
5. A 5 } d1d2 5 }(4)(8) 5 16 units2
2
2
The area of the rhombus is 16 square units.
y
Greatest possible error for 1.4 cm side:
P 5 2(5.1 cm) 1 2(1.4 cm) 5 13 cm
Greatest possible error for perimeter:
1
2
} (1 cm) 5 0.5 cm
Geometry
Worked-Out Solution Key
}
4(2Ï10 ) 5 8Ï 10 ft.
1
2
1
} (0.1 cm) 5 0.05 cm
2
}
One diagonal is 2Ï10 ft and the other diagonal is
} (0.1 cm) 5 0.05 cm
338
1
A 5 }2 d1d2 5 }2 (60)(80) 5 2400 m2
N
M
P
1
Q
1
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Unit of
measure
Chapter 11,
continued
11.2 Exercises (pp. 733–736)
Skill Practice
1
15. B; A 5 } d1d2 when d1 5 x, d2 5 3x, and A 5 24 ft2
2
1
1. The perpendicular distance between the bases of a
trapezoid is called the height of the trapezoid.
2. The vertical diagonal is bisected by the horizontal
diagonal. You also know the angles formed by the
intersecting diagonals are right angles.
24 5 }2(x)(3x)
1
24 5 }2 (3x2)
16 5 x2
45x
d1 5 4
d2 5 3(4) 5 12
The diagonals are 4 ft and 12 ft.
1
1
3. A 5 } h(b1 1 b2) 5 } (10)(8 1 11) 5 95 units2
2
2
1
1
4. A 5 } h(b1 1 b2) 5 }(6)(6 1 10) 5 48 units2
2
2
1
1
5. A 5 } h(b1 1 b2) 5 }(5)(4.8 1 7.6) 5 31 units2
2
2
1
108 5 }2(x)(14 1 22)
108 5 18x
6 ft 5 x
1
17. A 5 } h(b1 1 b2)
2
1
300 5 }2(20)(10 1 x)
5.4 cm
6.
1
16. A 5 } h(b1 1 b2)
2
8 cm
300 5 10(10 1 x)
30 5 10 1 x
10.2 cm
1
1
A 5 }2 h(b1 1 b2) 5 }2 (8)(5.4 1 10.2) 5 62.4 cm2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1
1
7. A 5 } d1d2 5 }(50)(60) 5 1500 units2
2
2
1
8. A 5 } d1d2 5
2
1
}(16)(48) 5 384 units2
2
1
9. A 5 } d1d2 5
2
1
}(21)(18) 5 189 units2
2
1
10. A 5 } d1d2 5
2
}(10)(19) 5 95 units2
1
2
11. d1 5 12 1 12 5 24
d2 5 15 1 15 5 30
1
A 5 }2 d1d2 5
1
}(24)(30) 5 360 units2
2
12. d1 5 2 1 2 5 4
d2 5 4 1 5 5 9
1
2
1
2
A 5 } d1d2 5 }(4)(9) 5 18 units2
13. The height is 12 cm, not 13 cm.
1
A 5 }2 (12)(14 1 19) 5 198 cm2
14. The length of the one diagonal is 12 1 12 5 24, not just
12 itself.
1
20 m 5 x
1
18. A 5 } d1d2
2
1
100 5 }2(10)(x)
100 5 5x
20 yd 5 x
19. The figure is a trapezoid.
b1 5 {4 2 2{ 5 2, b2 5 {5 2 0{ 5 5
h 5 {4 2 1{ 5 3
1
1
A 5 }2h(b1 1 b2) 5 }2 (3)(2 1 5) 5 10.5 units2
20. d1 5 {3 2 (21){ 5 4
d2 5 {23 2 1{ 5 4
The figure is a rhombus.
1
1
A 5 }2 d1d2 5 }2(4)(4) 5 8 units2
21. The figure is a kite.
d1 5 {4 2 0{ 5 4
d2 5 {2 2 (23){ 5 5
1
1
A 5 }2 d1d2 5 }2(4)(5) 5 10 cm2
A 5 }2 (24)(21) 5 252 cm2
Geometry
Worked-Out Solution Key
339
Chapter 11,
continued
1
22. A 5 } h(b1 1 b2); b1 5 x, b2 5 2x
2
1
13.5 5 }2 (3)(x 1 2x)
26. Use the Pythagorean Theorem to find part of the
second base.
a2 1 202 5 292
a2 5 441
3
13.5 5 }2 (3x)
9
2
20
a 5 21
b2 5 21 1 21 5 42
9
13.5 5 }3 x
35x
1
2
a
1
2
A 5 } h (b1 1 b2) 5 }(20)(21 1 42) 5 630 units2
One base is 3 feet and the other base is 2(3) 5 6 feet.
1
23. A 5 } h(b1 1 b2); b1 5 x, b2 5 x 1 8
2
1
54 5 }2 (6)(x 1 x 1 8)
27. Height of trapezoid 5 7 2 5 5 2
Area 5 Area of trapezoid 1 Area of rectangle
1
hT 5 height of trapezoid,
bR 5 base of rectangle, and
1
h
}
5 2(2)(7 1 10) 1 (10)(5) R 5 height of rectangle
A 5 }2hT (b1 1 b2) 1 bRhR
54 5 3(2x 1 8)
18 5 2x 1 8
5 67 units2
10 5 2x
28. Use the Pythagorean Theorem to find the height of
55x
One base is 5 cm and the other base is 5 1 8 5 13 cm.
the trapezoid and half of the shorter diagonal of the
rhombus:
24. Use the Pythagorean Theorem to find the length of half
of the shorter diagonal.
20
5
h
a
4
4 1 h 5 52
2
h2 5 9
a 1 16 5 20
2
2
a2 5 144
a 5 12
d1 5 12 1 12 5 24
d2 5 16 1 30 5 46
1
1
A 5 }2 d1 1 d2 5 }2 (24)(46) 5 552 units2
25. Use the Pythagorean Theorem to find the height of
the trapezoid.
h53
b1 5 4, b2 5 4 1 4 1 4 5 12, d1 5 3 1 3 5 6,
and d2 5 4 1 4 5 8
1
Area 5 Area of trapezoid 1 }2 Area of rhombus
A 5 }2h(b1 1 b2) 1 }2 1 }2d1d1 2
1
1 1
F
G
1
1 1
5 }2(3)(4 1 12) 1 }2 1 }2 2(6)(8) 5 36 units2
29. Area 5 Area of parallelogram 2 Area of kite
9
1
1
A 5 bh 2 }2 d1d2 5 (12)(7) 2 }2(12)(7) 5 42 units2
h
15
30. Sample answer:
4
3
92 1 h2 5 152
h 5 144
h 5 12
b2 5 9 1 11 5 20
1
1
a 5 }2h(b1 1 b2) 5 }2(12)(8 1 20) 5 168 units2
3
8
2
2
1
A 5 8(3) 5 24 units2
1
A 5 }2 (3)(4 1 12) 5 24 units2
5
2
3
3
1
1
A 5 }2(3)(5 1 11) 5 24 units2
4
1
1
A 5 }2(3)(2 1 14)
5 24 units2
340
Geometry
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
16
2
2
Chapter 11,
continued
31. Use the Pythagorean Theorem to find the length of the
B
B
side of the trapezoid.
7
7
A
C
D
15 2 7
15
E
62 1 82 5 s2
F
H
170 5 BC2
10 5 s
}
Ï170 5 BC
P 5 6 1 7 1 15 1 10 5 38 units
1
To find mŽABC, use trigonometric ratios.
1
A 5 }2 h(b1 1 b2) 5 }2(6)(7 1 15) 5 66 units2
B
u1 u2
6 7
10
32. Use the Pythagorean Theorem to find the length of part
of the other diagonal.
A
13
13
a
12
170
8
C
11
8
11
tan u1 5 }6
12
13
tan u2 5 }
7
u 5 tan21 1 }3 2
u2 5 tan211 }
72
u1 ø 53.18
u2 ø 57.58
4
24
a2 1 122 5 132
a2 5 25
11
}
So, mŽABC ø 53.18 1 57.78 ø 110.68. Extend BC
and drop an altitude down from A to find the height
of the parallelogram. By the Linear Pairs Postulate,
} }
the angle made by joining extended BC to AB is
1808 2 110.68 5 69.48. Use a trigonometric function
to find the height.
a55
d1 5 5 1 5 5 10
P 5 4(13) 5 52 units
1
A 5 }2 d1d2 5 }2(10)(24) 5 120 units2
}
33. First, make a horizontal line from A to the segment BF.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
G
112 1 72 5 BC2
100 5 s2
1
C
11
s
6
6
The length of the new segment is 8 because it is parallel
}
to EF, and EF 5 8. Because BF 5 16 and AE 5 10, the
part of the segment from B to this new segment is
16 2 10 5 6. Using the Pythagorean Theorem, you can
}
find out the length of AB.
82 1 62 5 AB2
h
sin 69.48 5 }
10
10 + sin 69.48 5 h
9.36 ø h
B
69.48
h
10
A
So, the area }
of the parallelogram is
A 5 bh ø Ï 170 + 9.36 ø 122 square units.
B
100 5 AB
2
6
10 5 AB
A
8
} }
Because ABCD is a parallelogram, AB > CD and
} }
BC > DA. So, CD 5 10. Next extend a vertical line
}
}
up from GD to meet BC. Because ŽBAD > ŽBCD,
} }
AB > CD, and two more corresponding angles of
the triangles are congruent (by Alternate Interior and
Suppementary Angles Theorems), you know the triangle
}
}
}
made by AB and parts of AD and BF is congruent to
}
}
}
the triangle made by CD and parts of BC and AD by
AAS. You can now say GH 5 8 because you have
corresponding parts of congruent triangles. Now draw a
}
horizontal line connecting C to BF. You know the base is
the length of FG 1 GH 5 8 1 2 5 11 and the height is
16 2 9 5 7. Use the Pythagorean Theorem to find BC.
Problem Solving
1
1
34. A 5 } h(b1 1 b2) 5 } (35)(70 1 79) 5 2607.5 in.2
2
2
The area of the glass in the wind shield is
2607.5 square inches.
1
1
35. A 5 } d1d2 5 }(8)(5) 5 20 mm2
2
2
The area of the logo is 20 square millimeters.
Sample answer:
Geometry
Worked-Out Solution Key
341
Chapter 11,
continued
1
A 5 }2 d1d2
39. The kite was cut along the diagonal and then the other
1
432 5 }2 (36)(d2)
24 5 d2
The length of the other diagonal is 24 inches.
37. a. The two polygons are a trapezoid and a right triangle.
b. Area of field 5 Area of triangle 1 Area of trapezoid
1
1
Area of field 5 }2 bh 1 }2 h(b1 1 b2)
1
5 103,967.5
Theorem 11.5.
1
1
40. The area of nPRS is }b1h and the area of nPQR is }b2h.
2
2
1
2
1
2
The area of the playing field is about 103,968 square
feet or about 11,552 square yards.
1
2
41. By the SSS Congruence Postulate, nPQR > nPSR.
1
1
1
The area of nPQR is }2d2 }2d1 5 }4d1d2. So, the area of
1
2
kite PQRS is twice the area of nPQR, or
21 }4 2d1d2 5 }2d1d2.
1
2
1
42. a. Sample answer:
Diagram
Area, A
2
4
3
6
Rhombus
number, n
original kite
rhombus
isosceles
triangle
right triangle
many different triangles
many different
kites
many different
quadrilaterals
4
Diagram
Area, A
you can go through the same process with a rhombus,
}b1h 1 }b2h 5 }h(b1 1 b2).
103,968 ft2 1 yd2
}+}
5 11,552 yd2
1
9 ft2
Rhombus
number, n
1
The area of trapezoid PQRS is equal to
ø 103,968 ft2
1
1
1
1
Rhombus
number, n
have A 5 (d1)1 }2 d2 2, which simplifies to }2d1d 2. Because
the formula will also simplify to }2 d1d2, which is
Area of field 5 }2 (315)(322) 1 }2(179)(145 1 450)
38. a.
diagonal to make 1 big triangle and 2 smaller triangles.
The resulting figure, once the smaller triangle was
moved, was a rectangle. The formula for the area of a
rectangle is b + h. In this case, the base was the first
diagonal, and the height was half of the other diagonal.
Substituting those values into the area formula, you
8
5
b. Yes, you can make isosceles and right triangles by
moving the vertical diagonal down all the way and to
the middle or down all the way and to the left or right
all the way.
Diagram
1
1
c. Area of original kite: }d1d2 5 }(4)(6) 5 12 units2
2
2
1
1
Area, A
10
Area of rhombus: }2d1d2 5 }2(4)(6) 5 12 units2
1
1
b. The area is twice the rhombus number n. The area of
the nth rhombus is A n 5 2n.
c. The length of the other diangonal is 2n for the nth
1
1
rhombus. A 5 }2 d1d2 5 }2(2)(2n) 5 2n. The rule for
area in this part is the same as the rule for area
in part (b).
342
Geometry
Worked-Out Solution Key
Area of isosceles triangle: }2 bh 5 }2(6)(4) 5 12 units2
1
1
Area of right triangle: }2 bh 5 }2(6)(4) 5 12 units2
1
1
Area of different kiles: }2d1d2 5 }2(4)(6) 5 12 units2
All of the areas are equal, 12 square units. The lengths
of the diagonals are not being changed, just moved
around. Although the shapes made by connecting the
diagonals will be different, the area will remain the
same.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
36.
Chapter 11,
continued
Lesson 11.3
43. Looking at the trapezoid:
b1 5 a, b2 5 b1 and h 5 a 1 b
11.3 Guided Practice (pp. 738–739)
1
1
A 5 }2 h(b1 1 b2) 5 }2(a 1 b)(a 1 b)
1
1
a
4
16
1. Ratio of perimeters: } 5 } 5 }
3
b
12
1
5 }2 (a2 1 2ab 1 b2) 5 }2a2 1 ab 1 }2b2
Looking at the triangles that make up the trapezoid:
1
2
1
2
1
2
1
2
A 5 }(a)(b) 1 }(a)(b) 1 }(c)(c) 5 ab 1 }c2
1
2
1
2
1
2
64(9) 5 16(Area of nDEF)
36 5 Area of nDEF
16
1
2
20
2. Ratio of area: }
36
21 }2a2 1 }2 b2 2 5 21 }2 c2 2
1
576 5 16(Area of nDEF)
. The area of nDEF is 36 square feet.
is }
9
}a2 1 }b2 5 }c2
1
1
}
}
2Ï5
Ï20
Ï36
}
Ï5
Ratio of sides: }
} 5 } 5 }
6
3
a2 1 b2 5 c2
}
Ï5
.
The ratio of their corresponding side lengths is }
3
Mixed Review for TAKS
3. Step 1 Find the ratio of the perimeters. Use the same
44. C;
units for both lengths in the ratio.
0.25 in. j dimension in blueprint
Scale 5 }
1 ft j actual dimension
Perimeter of Rectangle I
Perimeter of Rectangle II
0.25 in. 2
0.0625 in.2
Let x 5 the area of the actual kitchen
5.5 in.2
x
66 in.
(Periemter of Rectangle I)2
(Perimeter of Rectangle II)
12
20
88 5 x
1
400
}2 5 }
The area of the actual kitchen is 88 square feet.
Step 3 Find the area of Rectangle I.
45. G;
Area of Rectangle I
Area of Rectangle II
1
400
Area of Rectangle I
(35)(20)
1
400
0
1
}} 5 }
(23)2 2 1
02 2 1
12 2 1
}} 5 }
5921
5021
5121
58
5 21
50
2(23)2 2 1
2(0)2 2 1
2(1)2 2 1
5 2(9) 2 1
5 2(0) 2 1
5 2(1) 2 1
5 18 2 1
5021
5221
23
x
Area of Rectangle I + 400 5 700
Area of Rectangle I 5 1.75 ft2
2
144 in.
Step 4 1.75 ft2 + }
5 252 in.2
2
1 ft
1
The ratio of the area is }
and the area of
400
5 17
5 21
51
2(23)2 1 1
2(0)2 1 1
2(1)2 1 1
5 29 1 1
5 20 1 1
5 21 1 1
11.3 Exercises (pp. 740–743)
5 28
5011
50
Skill Practice
51
y 5 2x 2 1 1
Area of Rectangle I
Area of Rectangle II
}}}2 5 }}
5.5 5 0.0625x
y 5 2x 2 1 1
1
Step 2 Find the ratio of the areas of the two rectangles.
0.0625 in.2
1 ft
y 5 2x 2 2 1
5.5 ft
5}
5}
5}
20
110 ft
110 ft
}5}
2
y 5 x2 2 1
66 in.
2(35 ft) 1 2(20 ft)
}} 5 }}
j area in blueprint
5}
(Scale)2 5 1 }
1 ft 2
j actual area
1 ft2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
16
9
64
Area of nDEF
The ratio of the area of nABC to the area of nDEF
}a2 1 ab 1 }b2 5 ab 1 }c2
1
2
16
}} 5 }
Because the areas are equal, you can set the area of the
trapezoid and the combined area of the triangles equal to
each other.
1
2
42
3
a2
b
Ratio of areas: }2 5 }2 5 }
9
Rectangle I is 252 square inches.
1. Sample answer:
2(23)2 1 1
2(0)2 1 1
2(1)2 1 1
5 2(9) 1 1
5 2(0) 1 1
5 2(1) 1 1
5 18 1 1
5011
5211
5 19
51
53
E
B
5
A
4
10
3
C
D
8
6
F
The function y 5 2x 2 1 best represents the mapping
shown.
2
Geometry
Worked-Out Solution Key
343