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367Geometry
Solutions Key
Chapter 11, continued
5.Scale
FactorBase Height Area
Ratio of
Area
1 3 4 12 12
} 12
= 1
3 9 12 108 108
} 12
= 9 }
1
The area of the original parallelogram is 12 square units, and the area of the similar parallelogram is 108 square units.
6. Conjecture: If the side lengths of a figure are multiplied by a factor of k, the area is multiplied by a factor of k 2 .
7. Doubling just one dimension of a rectangle doubles its area. Doubling both dimensions quadruples its area.
Lesson 11.3
11.3 Guided Practice (pp. 766–767)
1. Ratio of perimeters: 16
} 12
5 4 } 3 5
a }
b
Ratio of areas: a2
} b2 5
42
} 32 5
16 } 9
64 }}
Area of nDEF 5
16 } 9
64(9) 5 16(Area of nDEF)
576 5 16(Area of nDEF)
36 5 Area of nDEF
The ratio of the area of nABC to the area of nDEF
is 16
} 9 . The area of nDEF is 36 square feet.
2. Ratio of area: 20
} 36
Ratio of sides: Ï
}
20 }
Ï}
36 5
2 Ï}
5 } 6 5
Ï}
5 } 3
The ratio of their corresponding side lengths is Ï
}
5 }
3 .
3. Step 1 Find the ratio of the perimeters. Use the same units for both lengths in the ratio.
Perimeter of Rectangle I
}} Perimeter of Rectangle II
5 66 in. }}
2(35 ft) 1 2(20 ft)
5 66 in.
} 110 ft
5 5.5 ft
} 110 ft
5 1 } 20
Step 2 Find the ratio of the areas of the two rectangles.
(Perimeter of Rectangle I)2
}}} (Perimeter of Rectangle II)2 5
Area of Rectangle I }}
Area of Rectangle II
12
} 202 5
1 } 400
Step 3 Find the area of Rectangle I.
Area of Rectangle I
}} Area of Rectangle II
5 1 } 400
Area of Rectangle I
}} (35)(20)
5 1 } 400
Area of Rectangle I p 400 5 700
Area of Rectangle I 5 1.75 ft2
Step 4 1.75 ft2 p 144 in.2 }
1 ft2 5 252 in.2
The ratio of the area is 1 }
400 and the area of
Rectangle I is 252 square inches.
11.3 Exercises (pp. 768–771)
Skill Practice
1. Sample answer:
4
5 3
A C
B
8
610
D F
E
The side lengths of nABC are each multiplied by the same scale factor to get the lengths of each side of nDEF. AB has corresponding side length DE, BC has corresponding side length EF, and CA has corresponding side length FD. Each side in nABC is multiplied by 2 to get the length of its corresponding side in nDEF.
2. You don’t need to know the value of n to fi nd of the ratio of the perimeters or the ratio of the areas of the polygons, because you know the ratio of the side lengths, 3 : 4. This ratio is also the ratio of the perimeters. According to Theorem 11.7, the ratio of the areas is the square of the ratio of the perimeters 32 : 42, or 9 : 16.
Ratio of
corresponding
side lengths
Ratio of
perimetersRatio of areas
6 : 11 6 :11 36 : 121
5 : 9 20 : 36 5 5 : 9 25 : 81
3.
4.
5. The ratio of the perimeters is 1 : 3 and the ratio of the areas is 12 : 32, or 1 : 9.
2 }
x 5
1 } 9
18 5 x
The area of the blue triangle is 18 square feet.
6. The ratio of the perimeters is 15 : 20, or 3 : 4, and the ratio of the areas is 32 : 42, or 9 : 16.
x }
240 5
9 } 16
x 5 135
The area of the red quadrilateral is 135 square centimeters.
7. The ratio of the perimeters is 7 : 9 and the ratio of the areas is 72 : 92, or 49 : 81.
x }
210 5
49 } 81
x ø 127
The area of the red hexagon is about 127 square inches.
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368GeometrySolutions Key
Chapter 11, continued
8. The ratio of the perimeters is 5 : 3 and the ratio of the areas is 52 : 32, or 25 : 9.
40
} x 5
25 } 9
14.4 5 x
The area of the blue parallelogram is 14.4 square yards.
9. Ratio of areas 5 49 : 16
Ratio of lengths 5 Ï}
49 : Ï}
16 5 7 : 4
The ratio of the lengths of corresponding sides is 7 : 4.
10. Ratio of areas 5 16 : 21
Ratio of lengths 5 Ï}
16 : Ï}
121 5 4 : 11
The ratio of the lengths of corresponding sides is 4 : 11.
11. Ratio of areas 5 121 : 144
Ratio of lengths 5 Ï}
121 : Ï}
144 5 11 : 22
The ratio of the lengths of corresponding sides is 11 : 12.
12. C; Ratio of areas 5 18 : 24
Ratio of lengths 5 Ï}
18 : Ï}
24 5 3 Ï}
2 : 2 Ï}
6 5
3 Ï
}
2 }
2 Ï}
6 p Ï
}
6 }
Ï}
6 5
3 Ï}
12 } 12 5
6
1 Ï
}
3 } 12
2 5
Ï}
3 } 2 or Ï
}
3 : 2
13. Ratio of areas 5 7 : 281 : 4
Ratio of lengths 5 Ï}
1 : Ï}
4 5 1 : 2
4 } XY 5
1 } 2
8 5 XY
The length of XY is 8 centimeters.
14. Ratio of areas 5 198 : 88 5 9 : 4
Ratio of lengths 5 Ï}
9 : Ï}
4 5 3 : 2
XY
} 10
5 3 } 2
XY 5 15
The length of XY is 15 inches.
15. The ratio of areas is 1 : 4, so the ratio of lengths is Ï
}
1 : Ï}
4 or 1 : 2. You need to use the 1 : 2 ratio.
12
} ZY 5 1 } 2
ZY 5 24
16. Perimeter of QRSTU
}} Perimeter of VWXYZ
5 5(12 cm)
} 140 cm 5 3 } 7
So, the ratio of corresponding lengths is 3 : 7 and the ratio of areas is 32 : 72, or 9 : 49.
Area of QRSTU
}} Area of VWXYZ
5 9 } 49
248
} x 5
9 } 49
12,152 5 9x
1350 ø x
The area of regular pentagon VWXYZ is about 1350 square centimeters.
17. Area of MNPQ 5 1 } 2 d1d2 5
1 } 2 (14)(25) 5 175 ft2
Ratio of areas 5 28 : 175 5 4 : 25
Ratio of lengths 5 Ï}
4 : Ï}
25 5 2 : 5
Shorter diagonal of RSTU
}} 14
5 2 } 5
Shorter diagonal of RSTU 5 5.6 ft
Longer diagonal of MNPQ
}} 25
5 2 } 5
Longer diagonal of MNPQ 5 10 ft
The area of MNPQ is 175 square feet. The lengths of the diagonals are 10 feet and 5.6 feet.
18. Case 3, Case 1, Case 2
In Case 3, the enlargement is Ï}
5 , or approximately 2.24. Because 2.24 is less than the enlargement of 3 (Case 1) and 2.24 and 3 are both less than an enlargement of 4 (Case 2), the order of enlargement from smallest to largest is Case 3, Case 1, Case 2.
19. Doubling the side length of a square never doubles the area. Doubling the side length of a square always quadruples the area.
20. Two similar octagons sometimes have the same perimeter. The perimeters will be the same only when the octagons are congruent.
21. ratio of lengths: 4.5
} 6 5
3 } 4 ,
7.5 }
10 5
3 } 4 ,
a }
12 5
3 } 4
The ratio of lengths is 3 : 4, so the ratio of areas is 32 : 42, or 9 : 16.
You can use the proportion Area of nABC
}} Area of nDEF
5 9 } 16 to solve
for the area of nDEF.
22. First, you have to fi nd the width of ABCD.
P 5 2l 1 2w
84 5 2(24) 1 2w
36 5 2w
18 5 w
The width of ABCD is 18 feet, or
18 feet p 1 yard
} 3 feet
5 6 yards
Ratio of width 5 6 : 3 5 2 : 1
Ratio of areas 5 22 : 12 5 4 : 1
The ratio of the area ABCD to the area of DEFG is 4 : 1.
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369Geometry
Solutions Key
Chapter 11, continued
23. ∠D and ∠N are right angles, so ∠D > ∠N by the Right Angles Congruence Theorem. ∠F > ∠M is given. So, nDEF , nNLM by the AA Similarity Postulate.
Use the area formula to fi nd NL.
A 5 1 } 2 bh
294 5 1 } 2 (NL)(21)
28 5 NL
Find ML so that you can fi nd the ratio of corresponding lengths.
(MN)2 1 (NL)2 5 (ML)2
212 1 282 5 (ML)2
1225 5 (ML)2
35 5 ML
ratio of lengths: 10 : 35 5 2 : 7
ratio of areas: 22 : 72 5 4 : 49
24. ∠TUY > ∠UVW because they are corresponding angles. ∠UTY > ∠VUW because they are corresponding angles. So, nTUY , nUVW by the AA Similarity Postulate.
You have to fi nd the length of } VW to fi nd the ratio of lengths.
tan 308 5 Ï
}
3 } VW
VW 5 Ï
}
3 }
tan 308
VW 5 3
The corresponding lengths are UV and VW.
Ratio of lengths: Ï}
3 : 3
Ratio of areas: ( Ï}
3 )2 : 32 5 3 : 9 5 1 : 3
25. a. Ratio of area nAGB to nCGE 5 9 : 25
Ratio of lengths of nAGB to nCGE 5 Ï}
9 : Ï}
25
5 3 : 5
Set up proportions of corresponding sides to the ratio of lengths.
AG
} CG
5 3 } 5
GB }
GE 5
3 } 5
AG
} 10
5 3 } 5
GB }
15 5
3 } 5
AG 5 6 GB 5 9
Look at nAGF and nBGC. ∠AGF and ∠BGC are right angles, so ∠AGF > ∠BGC. ∠GFA > ∠GBC because they are alternate interior angles. So, nAGF , nBGC by the AA Similarity Postulate. Find the length of
} GF using a ratio.
GF
} GC
5 AG
} BG
GF
} 10
5 6 } 9
GF 5 20
} 3
GE 5 GF 1 FE
15 5 20
} 3 1 FE
25
} 3 5 FE
So, AG 5 6, GB 5 9, GF 5 20
} 3 , and FE 5 25
} 3 .
b. Look at n ADC and nCBA. ∠ ADC > ∠ CBA because ABCD is a parallelogram. ∠ DAC > ∠ BCA because they are alternate interior angles. So, n ADC , nCBA by the AA Similarity Postulate. Find the area of nCBA, whose height is BG and base length is AC 5 AG 1 GC.
A 5 1 } 2 bh 5
1 } 2 (6 1 10)(9) 5 72
The ratio of the area of nADC to nCBA is 72 : 72, or 1 : 1.
Problem Solving
26. ratio of longest sides 5 3 : 5
ratio of areas p 32 : 52 5 9 : 25
Area of old banner
}} Area of new banner
5 9 } 25
3(1) }}
Area of new banner 5
9 } 25
8 1 }
3 5 Area of new banner
The area of the new banner will be 8 1 }
3 square feet.
27. ratio of areas 5 360 : 250 5 36 : 25
ratio of lengths 5 Ï}
36 : Ï}
25 5 6 : 5
distance on new patio
}} distance on similar patio
5 6 } 5
distance on new patio
}} 12.5
5 6 } 5
distance on new patio 5 15
The distance on the new patio is 15 feet.
28. B;
ratio of lengths 5 90 : 60 5 3 : 2
ratio of areas 5 32 : 22 5 9 : 4
pounds for baseball diamond
}}} pounds for softball diamond
5 9 } 4
20 }}}
pounds for softball diamond 5
9 } 4
9 ø pounds for softball diamond
You need about 9 pounds for the softball diamond.
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370GeometrySolutions Key
Chapter 11, continued
29. a. Sample answer:
x
y
1
1A C
B
5
9
2 13
A 5 1 } 2 bh 5
1 } 2 (9)(4) 5 18 square units
b. A(0, 0), B(3, 4), C(9, 0)
midpoint of }
AB : S 1 0 1 3 }
2 ,
0 1 4 }
2 2 5 S 1 3 }
2 , 2 2
midpoint of }
BC : Q 1 3 1 9 }
2 ,
4 1 0 }
2 2 5 Q(6, 2)
midpoint of }
AC : R 1 0 1 9 }
2 ,
0 1 0 }
2 2 5 S 1 9 }
2 , 2 2
x
y
1
1A C
B
R
S
QR 5 Î}}
1 6 2 9 }
2 2 2 1 (2 2 0)2 5 Ï
}
2.25 1 4
5 Ï}
6.25
5 2.5
RS 5 Î}}
1 9 } 2 2
3 }
2 2 2 1 (0 2 2)2 5 Ï
}
9 1 4 5 Ï}
13
QS 5 Î}}
1 6 2 3 }
2 2 2 1 (2 2 2)2 5 Ï
}
20.25 1 0 5 4.5
AB
} QR
5 5 }
2.5 5 2,
BC }
RS 5
2 Ï}
13 }
Ï}
13 5 2,
AC }
QS 5
9 }
4.5 5 2
Because all of the ratios of corresponding lengths are equal, the triangles are similar.
Ratio of lengths of nQRS to nABC: 1 : 2
Ratio of areas of nQRS to nABC: 1 : 4
The area of the smaller triangle is 1 }
4 the area of the
larger triangle. The smaller triangle has an area of
1 }
4 (18) 5 4.5 square units.
30. Sample answer:
I IIx y
ab
Rectangle I , Rectangle II
AI 5 ax, AII 5 by
Because the rectangles are similar, the ratios of
corresponding sides are proportional. So, a }
b 5
x } y .
Ratio of areas 5 ax
} by
5 a }
b p x }
y . Because
a }
b 5
x } y , substitute
that into the previous equation.
Ratio of areas 5 a }
b p a }
b 5
a2
} b2 .
So, the ratio of corresponding sides is a : b, and the ratio of their areas is a2 : b2.
31. The graph is visually midleading because although the amount of science fi ction books read is only half the amount of mysteries read, the graph makes it seem like the amount of science fi ction books read is a quarter of the mysteries read. The student could redraw the graph by using bars of the same width for the same categories.
32. Sample answer:
I II
3
3
6
6
AI
} AII
5 9 }
6
PI
} PII 5
Ï}
9 }
Ï}
6 5
3 }
Ï}
6 5
3 Ï}
6 }
6 5
Ï}
6 }
2
33. a. ∠BFC > ∠DFE because they are vertical angles. ∠BCD > ∠DEB because they intercept C BD . So, nBFC , nDFE by the AA Similarity Postulate. Also, ∠CAD > ∠EAB because they are the same angle. So, nCAD , nEAB by the AA Similarity Postulate.
b. Sample answer:
Ratio of lengths 5 10 : 9
Ratio of areas 5 102 : 92 5 100 : 81
c. One way: Use AB p AC 5 AD p AE to solve for the length of } DE .
9 p (9 1 11) 5 10 p (10 1 DE)
180 5 100 1 10 DE
80 5 10 DE
8 5 DE
Another way: Use the ratios of corresponding sides of the similar triangles.
AD
} AC
5 AB
} AE
10
} 20
5 9 } 10 1 DE
10(10 1 DE) 5 180
10 1 DE 5 18
DE 5 8
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Chapter 11, continued
34. a. You are given JL 5 KL, so } JK > } KL . Because nJKL is in the corner of a cube, ∠L is a right angle. So, nJKL is an isosceles right triangle. ∠D is a right angle because it is the corner of a cube. A cube has equal sides, which means } MP > } NP and nMNP is an isosceles right triangle. Because both nJKL and nMNP are isosceles right triangles, their side lengths are proportional. So, nJKL , nMNP by the SAS Similarity Theorem.
b. Ratio of lengths of triangles: 1 : 3
Ratio of areas of triangles: 12 : 32 5 1 : 9
Because the area of the face PMTN is twice the area nMNP, the scale factor of nJKL to the area of one face of the cube is 1 : 9(2) 5 1 : 18.
c. The ratio of the area of nJKL to a face of the cube is 1 : 18. The area of the pentagon JKQRS is the area of a face of the cube with the triangle nJKL subtracted from it. So, the ratio of the area of nJKL to the area of pentagon JKQRS is 1 : (18 2 1), or 1 : 17.
Florida Spiral Review
35. A; A triangle is said to be obtuse when the sum of the squares of its two sides is smaller than the square of the remaining side.
Choice A: 142 + 202 � 262
596 < 676
36. H
Ready To Go On? Quiz for 11.1–11.3 (p. 771)
1. A 5 bh; h 5 3b, A 5 108 ft2
108 5 b(3b)
108 5 3b2
36 5 b2
6 5 b
The base is 6 feet and the height is 3(6) 5 18 feet.
2. Use the Pythagorean Theorem to fi nd the height of the triangle.
a2 1 122 5 132
12
13a a2 5 25
a 5 5
Area of fi gure 5 Area of rectangle 1 Area of triangle
5 bh 1 1 } 2 ba 5 (12)(6) 1
1 } 2 (12)(5)
5 102 square units
3. A 5 1 } 2 h(b1 1 b2) 5
1 } 2 (4)(3 1 6.5) 5 19 square units
4. d1 5 4 1 4 5 8 units, d2 5 5 1 8 5 13 units
A 5 1 } 2 d1d2 5
1 } 2 (8)(13) 5 52 square units
5. The ratio of the lengths of two similar heptagons is 7 : 20, so the ratio of their perimeters is also 7 : 20. The ratio of their areas is 72 : 202, or 49 : 400.
6. The ratio of the areas is 1200 : 48, or 25 : 1. The ratio of the lengths is Ï
}
25 : Ï}
1 , or 5 : 1.
50
} XY
5 5 } 1
10 5 XY
XY is 10 feet.
Problem Solving Workshop 11.3 (p. 772)
1. A 5 lw, where x 5 width and 1.5x 5 length
1.44(15 p 10) 5 1.5x p x
144 5 x2
12 5 x
The length of the third pan is 1.5(12) 5 18 inches.
2. Area of PQRS 5 1 } 2 h(b1 1 b2) 5
1 } 2 (6)(9 1 12) 5 63
The area of WXYZ is 28
} 63
5 4 } 9 of the area of PQRS. So,
each side in WXYZ is Ï
}
4 }
Ï}
9 , or
2 }
3 the size of each
corresponding side in PQRS. So,
WZ 5 2 } 3 PS 5
2 } 3 (12) 5 8 units.
3. fi rst square: A 5 s2
second square: A 5 2s2
Each side of a square is the square root of the area,
so the second length is Ï}
2s2 5 s Ï}
2 .
4. Area of nABC 5 1 } 2 bh 5
1 } 2 (8)(5) 5 20 cm2
The area of nDEF is 11.25
} 20
, or 0.5625 of the area of
nABC. So, each side in nDEF is Ï}
0.5625 , or 0.75 the size of each corresponding side in nABC.
DF 5 0.75(5) 5 3.75 cm
EF 5 0.75(8) 5 6 cm
DE2 1 EF2 5 DF2
(3.75)2 1 62 5 DF2
50.0625 5 DF2
7.08 ø DF
The length of DE is 3.75 cm and the length of DF is about 7.08 cm.
Problem Solving Connections (p. 773)
1. a. Because ∠A, ∠C, ∠E, and ∠G are right angles, they are all congruent.
} AH >
} CD > } DE >
} GH and
}
AB > }
BC > } EF > }
FG . So, nABH > nCBD > nEFD > nGEH by the SAS Congruence Postulate. Because } HB > } BD > } DF >
} FA , BDFH is a rhombus.
b. Area of rectangle 5 bh 5 (15)(9.5) 5 142.5 ft2
Area of rhombus 5 1 }
2 d1d2 5
1 }
2 (9.5)(15) 5 71.25 ft2
c. Area for marigolds 5 Area of rectangle 2 Area of rhombus
Area for marigolds 5 142.5 2 71.255 71.25 ft2
Total cost 5 cost for marigolds 1 cost for asters
5 $.30(71.25) 1 $.40(71.25) ø $49.88
The total cost is about $49.88.
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Chapter 11, continued
2. Sample answer:
12 m
6 m 6 m 6 m
6 m
12 m
12 m
8 m
8 m8 m 8 m
8 m
10 m
8 m208 ø 14.4 m
All triangles must satisfy A 5 1 }
2 bh and all parallelograms
must satisfy A 5 bh.
3. Area of the room: (12 foot)(21 foot) 5 252 ft2 p 144 in.2 }
1 ft2
5 36,288 in.2
a. Small tiles: 12 in. 3 12 in. 5 144 in.2, so you would need 36,288 4 144 5 252 tiles
Large tiles: 18 in. 3 18 in. 5 324 in.2, so you need 36,288 4 324 5 112 tiles
You would need 252 small tiles and 112 large tiles.
b. Cost for large tiles: $2.25(112) 5 $252
Cost for small tiles: $1.50(252) 5 $378
The cost of the fl oor using large tiles is $252 and the cost of the fl oor using small tiles is $378. If you want to spend as little as possible, use the large tiles.
c. Ratio of side lengths 5 12 : 18 5 2 : 3
Ratio of areas 5 22 : 32 5 4 : 9
Ratio of cost 5 1.50 : 2.25 5 2 : 3
The cost per tile is based on side length because the ratios of cost and side lengths are the same, 2 : 3.
4. Area of the dive pool at the Fort Lauderdale Aquatic Complex is
25 3 20 = 500.
Scale factor of the similar pool is 30
} 20
5 3 }
2 .
Let x = the area of the similar pool.
x }
500 5 1 3 }
2 2 2
x 5 500 1 9 } 4 2
5 1125
The area of the similar pool is 1125 square feet.
5. If you double the length of each diagonal of a rhombus, you quadruple the area. If you triple the length of each diagonal of a rhombus, you have an area 9 times greater than the original area. If each diagonal is multiplied by the same number n, the area of a rhombus will be n2 times greater than the original area.
6. 375; A 5 1 } 2 h(b1 1 b2)
C D3x
5 in.
AB x
1250 5 1 } 2 (5)(x 1 3x)
500 5 4x
125 5 x
The length of }
CD is 3(125) 5 375 inches.
7. a. Because nEFH is an isosceles right triangle, } FE > } HE .
Use the Pythagorean Theorem to fi nd FH.
FE 2 1 EH 2 5 FH 2
E
F
H
5 2
5 2
(5 Ï}
2 )2 1 (5 Ï}
2 )2 5 FH 2
100 5 FH 2
10 5 FH
b. FJ is half of FH, so FJ 5 5.
Use the Pythagorean Theorem to fi nd the length of } EJ .
FJ 2 1 EJ 2 5 EF 2
E J
F
5 2 5 52 1 EJ 2 5 (5 Ï
}
2 )2
EJ 2 5 25
EJ 5 5
Use tan 608 5 JG
} FJ to fi nd the length of }
JG .
tan 608 5 JG
} 5
5 tan 608 5 JG
5 Ï}
3 5 JG
EG 5 EJ 1 JG 5 5 1 5 Ï}
3 ø 13.7 units
c. d1 5 FH 5 10 and d2 5 EG ø 13.7
A 5 1 } 2 d1d2 ø
1 }
2 (10)(13.7) ø 68.5
The area of EFGH is about 68.5 square units.
Lesson 11.4
11.4 Guided Practice (pp. 775–777)
1. Circle with diameter of 5 inches:
C 5 πd 5 π(5) ø 15.71
The circumference is about 15.71 inches.
Circle with circumference of 17 feet: C 5 πd
17 5 πd
5.41 ø d
The diameter is about 5.41 feet.
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