Chapter 11
Chapter 11
Chapter 11
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<strong>Chapter</strong> <strong>11</strong>Copyright © by McDougal Littell, a division of Houghton Mifflin Company.Prerequisite Skills (p. 718)4. A 5 bhA 5 }1 2 bh 5 }1 2 (12)(5) 5 30 square units1. radius 5 3 2. diameter 5 6153 5 b(17)3. m C ADB 5 3608 2 708 5 29089 5 b4. P 5 2l 1 2w24 5 2(9) 1 2w24 5 18 1 2wThe base is 9 inches.5. The area of the window is 5 p 10 5 50 square feet.Because you don’t paint over the window, you need topaint about 637 2 50 5 587 square feet.6 5 2w3 5 wThe width is 3 centimeters.5. AC 2 1 CB 2 5 AB 2<strong>11</strong>.1 Exercises (pp. 723–726)Skill Practice1. Either pair of parallel sides of a parallelogram can beAC 2 1 6 2 5 14 2called its bases, and the perpendicular distance betweenthese sides is called the height.AC 2 1 36 5 196AC 2 5 160AC 5 4 Ï } 102. Two formulas for the area of rectangle are base p heightor length p width. Because the base is the length andthe height is the width, the two formulas give the same6. cos A 5 }AC results.AB3. A 5 bh 5 7(4) 5 28 square unitscos 358 5 }AC 4. A 5 bh 5 14(2) 5 168 square units255. A 5 s 2 5 15 2 5 225 square inches25(cos 358) 5 AC20.5 ø AC7. BC 5 x and AC 5 x Ï } 3 in a 308-608-908 triangle.Because BC 5 5, AC 5 5 Ï } 3 .8. Parallelograms, rhombuses, rectangles, and squares have6. A 5 }1 2 bh 5 }1 (10)(13) 5 65 square units27. Pythagorean Theorem: 18 2 1 b 2 5 30 2b 2 5 576b 5 24diagrams that bisect each other.9. }UVXY 5 }VW A 5 }1 2 bh 5 }1 (24)(18) 5 216 square units2YZ5}XY 5 }8 8. A 5 }1 2 bh 5 }1 (15)(9) 5 67.5 square units2129. Method 1 Use } DC as the base. The base is extended to60 5 8XYmeasure the height } AE . So, b 5 10 and15} 5 XY or 7.5 5 XY2h 5 16. A 5 bh 5 10(16) 5 160 square unitsMethod 2 Use } AD as the base. Then the height is DF } .So, b 5 20 and h 5 8.Lesson <strong>11</strong>.<strong>11</strong>1.1 Guided Practice (pp. 721–722)1. P 5 17 1 10 1 21 5 48 unitsA 5 }1 2 bh 5 }1 p 21 p 8 5 84 square units22. p 5 2l 1 2w 5 2(30) 1 2(20) 5 100 unitsA 5 bh 5 30(17) 5 510 square units3. Pythagorean Theorem: 5 2 1 b 2 5 13 2A 5 bh 5 20(8) 5 160 square units.10. The height of the parallelogram is 4 units, not 5.A 5 bh 5 6(4) 5 24 square units<strong>11</strong>. The base of the parallelogram is 3 units, not 7.A 5 bh 5 3(4) 5 12 square units12. a 2 1 b 2 5 c 212 2 1 b 2 5 15 2b 2 5 81b 2 5 144b 5 9135b 5 12P 5 15 1 12 1 9 5 36 in.12P 5 5 1 13 1 12 5 30 unitsA 5 }1 2 bh 5 }1 (9)(12) 5 54 in.22GeometryWorked-Out Solution Key351
<strong>Chapter</strong> <strong>11</strong>, continued13. a 2 1 b 2 5 c 216 2 1 b 2 5 34 2b 2 5 900b 5 30P 5 34 1 16 1 30 5 80 ft21. Sample answer:10 m10 m 16 m10 mA 5 }1 2 bh 5 }1 (30)(16) 5 240 ft2214. a 2 1 b 2 5 c 284 2 1 b 2 5 85 2b 2 5 169b 5 1385 1 84 1 13 5 182 m1}2 bh 5 }1 (13)(84) 5 546 m2215. a 2 1 b 2 5 c 220 2 1 b 2 5 29 2b 2 5 441b 5 21P 5 29 1 20 1 21 5 70 cmA 5 }1 2 bh 5 }1 (21)(20) 5 210 cm2216. A 5 1 }2 bh36 5 }1 2 (12)(x)6 in. 5 x17. A 5 bh 18. A 5 bh276 5 x(12) 476 5 x(17)23 ft 5 x 28 cm 5 x19. A 5 4 ft 2 h 5 1 }2 bA 5 1 }2 bh4 5 1 }2 b 1 1 }2 b 24 5 }1 4 b216 5 b 24 5 bThe base of the triangles is 4 inches, and the height is1}(4) 5 2 inches.220. A 5 507 cm 2 h 5 3bA 5 bh507 5 b(3b)507 5 3b 2169 5 b 213 5 bThe base of the parallelogram is 13 centimeters, and theheight is 3(13) 5 39 centimeters.8 m16 m 16 m10 m10 m 10 m8 m 8 m22. Area 5 Area of rectangle 1 Area of triangle5 17(8) 1 }1 (17)(5) 5 178.5 ft2223. Area 5 Area of first triangle 1 Area of parallelogram1 Area of second triangle5 }1 2 (9)(13) 1 18(13) 1 }1 2 (<strong>11</strong>)(13)5 364 cm 224. Base of triangle 5 16 2 base of parallelogram5 16 2 <strong>11</strong> 5 5 mArea 5 Area of triangle 1 Area of parallelogram5 }1 (5)(10) 1 (<strong>11</strong>)(10)25 135 m 225. Height of triangle: 15 2 1 h 2 5 25 2h 2 5 400h 5 20 in.Area 5 Area of triangle 1 Area of rectangle5 }1 (15)(20) 1 (25)(19)25 625 in. 226. Height of triangles and parallelogram: 10 2 1 h 2 5 26 2h 2 5 576h 5 24 mArea 5 Area of first triangle 1 Area of parallelogram1 Area of second triangle5 }1 2 (10)(24) 1 (40)(24) 1 }1 (20)(24) 5 1320 m2227. Height of triangle 5 8 2 5 5 3 in.Area 5 Area of square 2 Area of triangle5 8 2 2 }1 (8)(3) 5 52 in.22Copyright © by McDougal Littell, a division of Houghton Mifflin Company.352GeometryWorked-Out Solution Key
<strong>Chapter</strong> <strong>11</strong>, continued28.1yAB1 x34. h 2 1 4 2 5 5 2 5h 2 1 16 5 25h 2 5 94h 5 38h45uDCA 5 }1 2 bh 5 }1 (8)(3) 5 12 ft22The area of the triangle is 12 square feet.Copyright © by McDougal Littell, a division of Houghton Mifflin Company.29.A 5 bh 5 7(6) 5 42 square unitsE2<strong>11</strong>yGA 5 }1 2 bh 5 }1 (5)(3) 5 7.5 square units230. B; 2 ft 3 in. 5 24 in. 1 3 in. 5 27 in.4 ft 2 in. 5 48 in. 1 2 in. 5 50 in.A 5 bh 5 50(27) 5 1350 in. 2F31. Check students’ work; The base, height, and area ofnABC remains the same no matter how you movepoint C.32. sin DAB 5 h }ABxWhen m∠ DAB 5 208: When m∠ DAB 5 508:sin 208 5 h }8sin 508 5 h }88 sin 208 5 h 8 sin 508 5 h2.74 ø h 6.13 ø hA 5 bh ø 15(2.74) ø 41 A 5 bh ø 15(6.13) ø 92The height is about 2.74 units and the area is about 41square units when m∠DAB 5 208. The height is about6.13 units and the area is about 92 square units whenm∠DAB 5 508.33. A 5 }1 2 bh 5 }1 (12)(35) 5 210 cm2212 cmua37 cm35 cmtan u 5 opp }adjtan u 5 35 }12u ø 718sin 718 5 a }1212 sin 718 5 a<strong>11</strong>.3 ø aThe area of the right triangle is 210 cm 2 . The length ofthe altitude drawn to the hypotenuse is about <strong>11</strong>.3 cm.sin u 5 }3 5a5u ø 36.985usin 36.98 5 }a 988 sin 36.98 5 a4.8 ft ø aBecause the original triangle is isosceles, the altitudefrom the vertex where u is at is also 4.8 ft. The height ofthe original triangle is an altitude. So, the lengths of thealtitudes are 3 ft, 4.8 ft, and 4.8 ft.35. Construct a rectangle so that A, B, C, and D are all on thesides of the rectangle. The area of ABCD is equal to thearea of the retangle minus the areas of the right trianglesformed at each corner of the rectangle that are not partof ABCD.D34C1y1A 5 A rect.2 A n12 A n22 A n32 A n4A15 (<strong>11</strong>)(7) 2 }1 2 (4)(6) 2 }1 2 (1)(7) 2 }1 2 (4)(3) 2 }1 2 (7)(4)5 41.5The area of ABCD is 41.5 square units.Problem Solving36. Sail A: A 5 }1 2 bh 5 }1 (65)(35) 5 <strong>11</strong>37.5 ft22Sail B: A 5 }1 2 bh 5 }1 (29.5)(10.5) 5 154.875 ft222Sail A}Sail B 5 }<strong>11</strong>38155 ø 7.3The area of Sail A is about <strong>11</strong>38 ft 2 and the area of Sail Bis about 155 ft 2 . The area of Sail A is about 7.3 times asgreat as the area of Sail B.BxGeometryWorked-Out Solution Key353
<strong>Chapter</strong> <strong>11</strong>, continued37. Area of triangle plot: A 5 }1 2 bh 5 }1 (24)(25) 5 300 yd2The dimensions of rectangle you need to start with are2If you mow 10 yd 2 300 yd2/min:}p }1 min 5 30 min1210 ydArea of rectangular plot: A 5 bh 5 (24)(36) 5 864 yd 2 42. a. By the way the segments were drawn, } RV > } QTand ∠ RVS is a right angle. Because PQRS is aIf you mow 10 yd 2 864 yd2/min:}p }1 min 5 86.4 minparallelogram, } PQ > } SR . By the Hypotenuse-Leg1210 yd Congruence Theorem, n PQT > n SRV.It takes you 30 minutes to mow the triangular plot and86.4 minutes to mow the rectangular plot.38. Area of old table: A 5 bh 5 (24)(15) 5 360 in. 2Area of new table: A 5 bh360 5 (20)h18 5 hThe height of the new table should be 18 inches.39. A 4 inch square does not have an area of 4 square inches. the area of one triangle, nXYW, is }1 2A 2 inch square has an area of 4 square inches. Thebh.formula for the area of a square is s 2 , and 2 2 5 4, so the 44. a. h 2 1 1 }1 2 s 2 2 5 s 2side length should be 2 inches to produce an area of 4square inches.h 2 5 s 2 2 }s2440. The dimensions of a 308-608-908 triangle are:h 2 5 }3526082x 6084xx308308h 5 s Ï} 3}26 ftx 312 ftA 56 5 x Ï } }1 23bh 5 }1 2 (s) 1 s Ï} 3}22 5 s2 Ï } 3}46 Ï } 3}3 5 xb. s2 Ï } 3} 5 442 Ï } s 2 Ï } 3 5 163 5 xArea 5 Area of triangle 1 Area of rectangles 2 5 16 Ï} 3}35 }1 2 (12)(2 Ï} 3 ) 1 12(6.5) ø 98.8 ft 2s ø 3.04Two sides 5 2(98.8) 5 197.6 ft 2The side lengths of the triangle is about 3 cm.If you paint 200 ft 2 197.6 ft2per hour: } p }1 hour45. The perimeter cannot be determined because the base1 200 ft 5 0.988and height are not necessarily side lengths. The least2possible perimeter for the parallelogram isIf you charge $20 per hour: $20(0.988) 5 $19.767 1 3 1 7 1 3 5 20 feet. The greatest possibleYou will get paid $19.76 for painting those two sides ofperimeter cannot be determined.the shed.41. height 5 6 1 14 1 14 5 34 cmbase 5 3 1 17 1 3 5 23 cmArea of paper used 5 Area of small rectangle1 Area of large rectangleb. Area of MNPQ 5 Area of MRVU 1 Area of UVTQ1 Area of triangle1 Area of RNSV 1 Area of VSPT5 17(14) 1 23(14) 1 }1 2 (17)(6)b 2 1 2bh 1 h 2 5 A 1 b 2 1 h 2 1 A5 6<strong>11</strong> cm 22bh 5 2Abh 5 AArea of paper that is cut off 5 Area of original rectangle2 Area of paper used5 34(23) 2 6<strong>11</strong> 5 171 cm 234 cm by 23 cm, the area of the paper that is actuallyused in the envelope is 6<strong>11</strong> cm 2 , and the area of the paperthat is cut off is 171 cm 2 .b. Because QRVT is a rectangle, its area is bh. RectangleQRVT is composed of nSRV and trapezoid QRST.Because nPQT > nSRV, their areas are equal. So, thearea of nPQT plus the area of trapezoid QRST is equalto the area of QRVT, which is bh.43. Opposite sides are congruent, so XYZW is aparallelogram. The area of a parallelogram is bh. Becausethe parallelogram is made up of two congruent triangles,46. a. RNSV is a square because its base and height are bothh. UVTQ is a square because its base and height areboth b. MNPQ is a square because its base and heightare both b 1 h. Area of RNSV 5 h 2 , Area of UVTQ 5b 2 , and Area of MNPQ 5 (b 1 h) 2 5 b 2 1 2bh 1 h 2 .Copyright © by McDougal Littell, a division of Houghton Mifflin Company.354GeometryWorked-Out Solution Key
<strong>Chapter</strong> <strong>11</strong>, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.47. Let } AB > } BC . To get from A to B move 2 units up and2 units to the right, which is (4, 4). To get from B to C,move 2 units down and 2 units to the right, which is(6, 2). The height of nABC is 2 and the base is 4, so thearea of nABC is 4 square units. Find the equation of theline through B(4, 4) and C(6, 2).m 5 }4 2 24 2 6 5 }2 22 5 21y 5 2x 1 b4 5 24x 1 b8 5 by 5 2x 1 8Now, let } BC > } AC and choose } BC to be a vertical line.So, the base and height of nABC is the same. Since thearea has to be 4 square units, find the measure of thebase.A 5 1 }2 bhA 5 1 }2 b(b)4 5 }1 2 b28 5 b 22 Ï } 2 5 bSo, the base and height of nABC are 2 Ï } 2 units. So,point C is located at (2 1 2 Ï } 2 , 2) and the equation ofthe line x 5 2 1 2 Ï } 2 . The other line that could fit thisrequirement is when } AB > } AC .Mixed Review48. MN 5 }1 (8 1 18) 5 132The length of the midsegment is 13 units.49. MN 5 }1 (13 1 27) 5 202The length of the midsegment is 20 units.50. MN 5 }1 (29 1 46) 5 37.52The length of the midsegment is 37.5 units.51. (x, y) → (x 1 1, y 1 4)yP(24, 1) → P9(23, 5)Q(2, 5) → Q9(3, 9)P9QR(1, 24) → R9(2, 0)52. (x, y) → (x 1 3, y 2 5)P(24, 1) → P9(21, 24)Q(2, 5) → Q9(5, 0)R(1, 24) → R9(4, 29)P24P2424RP926Q9R9yQRxQ9xR953. (x, y) → (x 2 3, y 2 2)P(24, 1) → P9(27, 21)Q(2, 5) → Q9(21, 3)R(1, 24) → R9(22, 26)54. (x, y) → (x 2 2, y 1 3)P(24, 1) → P9(26, 4)Q(2, 5) → Q9(0, 8)R(1, 24) → R9(21, 21)<strong>11</strong>.1 Extension (p. 728)P9P9P24P24R9y6 QQ96RyQ9QR9R1. The precision of a measurement depends only on theunit of measure. The accuracy of measurement dependson both the unit of measure and on the size of the objectbeing measured.Sample answer:Pencil (14.8 cm)Textbook (21 cm)Unit of measure 0.1 cm 1 cmGreatestpossible errorRelative error1}2 (0.1) 5 0.05 cm }1 20.05 cm ø 0.0037}14.8 cm ø 0.37%(1 cm) 5 0.5 cm0.5 cm}21 cmxxø 0.0240ø 2.4%The measurement of the pencil is more precise becausethe unit of measure is smaller and the greatest possibleerror is smaller. The pencil also has the smaller relativeerror, so it is more accurate.2. Unit of measure: }1 inch; greatest possible error:101}2 1 }102 1 5 }1 5 0.05 inch203. Unit of measure: 1 meter; greatest possible error:1}(1) 5 0.5 meter214. Unit of measure: } kilometer; greatest possible error:10001 1 1}2 1 }10002 5 } 5 0.0005 kilometer20005. Unit of measure: }1 yard; greatest possible error:161}2 1 }162 1 5 }1 5 0.03125 yard326. Greatest possible error: 1 }2 1 1 }10 cm 2 5 0.05 cmgreatest possible error 0.05 cmRelative error: }} 5 }measured length 4 cm 5 0.01255 1.25%7. Greatest possible error: }1 (1 in.) 5 0.5 in.2greatest possible error 0.5 in.Relative error: }} 5 }measured length 28 in. ø 0.0179ø 1.8%GeometryWorked-Out Solution Key355
<strong>Chapter</strong> <strong>11</strong>, continued8. Greatest possible error: }1 (0.1 m) 5 0.05 m2greatest possible errorRelative error: }} 5 }0.05 mmeasured length 4.6 m ø 0.0109ø 1.1%9. Greatest possible error: }1 (0.01 mm) 5 0.005 mm2greatest possible error 0.005 mmRelative error: }} 5 }measured length 12.16 mmø 0.0004 ø 0.04%10. 1; You are estimating, so the greatest possible errorof }1 inch is precise enough.2<strong>11</strong>. Such a measurement could be more precise whenmeasuring larger objects (like a running track) thansmaller objects (like a pencil). Because the actualmeasured length goes in the denominator whendetermining the relative error, a bigger measured lengthwhen the greater possible error is constant yields asmaller relative error. Because the larger length has thesmaller relative error, it is more accurate.12.17 cm 12 cmUnit of measure 1 cm 1 cmGreatestpossible errorRelative error1}2 (1 cm) 5 0.5 cm }1 20.5 cm}17 cm ø 0.029ø 2.9%(1 cm) 5 0.5 cm0.5 cm}12 cm ø 0.0417ø 4.2%The precision is the same. 17 centimeters is more accurate.13.Unit ofmeasureGreatestpossible errorRelative error18.65 ft 25.6 ft0.01 ft 0.1 ft1}2 (0.01 ft) 5 0.005 ft }1 20.005 ft}18.05 ft ø 0.0003ø 0.03%18.65 feet is more precise and more accurate.14.Unit ofmeasureGreatestpossible errorRelative error(0.1 ft) 5 0.05 ft0.05 ft}25.6 ft ø 0.002ø 0.2%6.8 in. 13.4 ft0.1 in. 0.1 ft1}2 (0.1 in.) 5 0.05 in. }1 (0.1 ft) 5 0.05 ft20.05 in.0.05 ft} ø 0.007 }6.8 in.13.4 ft ø 0.004ø 0.7%ø 0.4%6.8 inches is more precise. 13.4 feet is more accurate.15.3.5 ft 35 in.Unit of measure 0.1 ft 1 in.Greatestpossible errorRelative error1}2 (0.1 ft) 5 0.05 ft }1 20.05 ft} ø 0.0143.5 ftø 1.4%(0.1 in.) 5 0.5 in.0.5 in.}35 in. ø 0.014ø 1.4%35 inches is more precise. The accuracy is the same.16. Greatest possible error for 5.1 cm side:1} (0.1 cm) 5 0.05 cm2Greatest possible error for 1.4 cm side:1} (0.1 cm) 5 0.05 cm2P 5 2(5.1 cm) 1 2(1.4 cm) 5 13 cmGreastest possible error for perimeter:1} (1 cm) 5 0.5 cm2Greatest possible side lengths: 1.4 1 0.05 5 1.45 cm5.1 1 0.05 5 5.15 cmGreatest possible perimeter:P 5 2(1.45) 1 2(5.15) 5 13.2 cmLeast possible side lengths: 1.4 2 0.05 5 1.35 cm5.1 2 0.05 5 5.05 cmLeast possible perimeter:P 5 2(1.35) 1 2(5.05) 5 12.8 cmLesson <strong>11</strong>.2Investigating Geometry Activity <strong>11</strong>.2 (p. 729)1. The area of one trapezoid is }1 the area of the2parallelogram formed from two trapezoids.b 5 b <strong>11</strong> b 2A 5 }1 2 bh 5 }1 2 (b 1 1 b 2 )h2. The base of the rectangle is d 1and the height of therectangle is 1 }2 d 2 . A 5 (d 1 ) 1 1 }2 d 2 2 5 1 }2 d 1 d 2<strong>11</strong>.2 Guided Practice (pp. 731–732)1. A 5 }1 2 h(b 1 1 b 2 ) 5 }1 (4)(6 1 8) 5 28 ft222. A 5 }1 2 d 1 d 2 5 }1 (6)(14) 5 42 in.223. d 15 30 m 1 30 m 5 60 md 25 40 m 1 40 m 5 80 mA 5 }1 2 d 1 d 2 5 }1 (60)(80) 5 2400 m22Copyright © by McDougal Littell, a division of Houghton Mifflin Company.356GeometryWorked-Out Solution Key
<strong>Chapter</strong> <strong>11</strong>, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.<strong>11</strong>. d 15 12 1 12 5 244. A 5 }1 2 d 1 d 2 ; when A 5 80 ft2 , d 15 x, and d 25 4x.2 (10)(19) 5 95 units2 A 5 }180 5 }1 d2 (x)(4x)25 15 1 15 5 30A 5 }180 5 }1 2 d 1 d 2 5 }1 22 4x212. d 15 2 1 2 5 440 5 x 2d 25 4 1 5 5 92 Ï } 10 5 xA 5 }1 2 dOne diagonal is 2 Ï } 1 d 2 5 }1 210 ft and the other diagonal is4(2 Ï } 10 ) 5 8 Ï } 10 ft.5. A 5 }1 2 d 1 d 2 5 }1 (4)(8) 5 16 units22A 5 }1 2The area of the rhombus is 16 square units.yNA 5 }1 2MP1Q1x24 5 }1 2 (x)(3x)24 5 }1 2 (3x2 )16 5 x 21. The perpendicular distance between the bases of a4 5 xtrapezoid is called the height of the trapezoid.d2. The vertical diagonal is bisected by the horizontal15 4diagonal. You also know the angles formed by thed 25 3(4) 5 12intersecting diagonals are right angles.16. A 5 }1 2 h(b 1 1 b 2 )108 5 }1 23. A 5 }1 2 h(b 1 1 b 2 ) 5 }1 (10)(8 1 <strong>11</strong>) 5 95 units22108 5 18x6 ft 5 x4. A 5 }1 2 h(b 1 1 b 2 ) 5 }1 (6)(6 1 10) 5 48 units2217. A 5 }1 2 h(b 1 1 b 2 )5. A 5 }1 2 h(b 1 1 b 2 ) 5 }1 (5)(4.8 1 7.6) 5 31 units22300 5 }1 2300 5 10(10 1 x)6. 5.4 cm30 5 10 1 x20 m 5 x8 cm18. A 5 }1 2 d 1 d 210.2 cm100 5 }1 2 (10)(x)A 5 }1 2 h(b 1 1 b 2 ) 5 }1 (8)(5.4 1 10.2) 5 62.4 cm22 100 5 5x7. A 5 }1 2 d 1 d 2 5 }1 (50)(60) 5 1500 units2220 yd 5 x8. A 5 }1 2 d 1 d 2 5 }1 (16)(48) 5 384 units229. A 5 }1 2 d 1 d 2 5 }1 (21)(18) 5 189 units22h 5 ⏐4 2 1⏐ 5 3<strong>11</strong>.2 Exercises (pp. 733–736)Skill Practice10. A 5 1 }2 d 1 d 2 5 1 }(24)(30) 5 360 units2(4)(9) 5 18 units213. The height is 12 cm, not 13 cm.(12)(14 1 19) 5 198 cm214. The length of the one diagonal is 12 1 12 5 24, not just12 itself.(24)(21) 5 252 cm215. B; A 5 }1 2 d 1 d 2 when d 1 5 x, d 5 3x, and A 5 24 ft22The diagonals are 4 ft and 12 ft.(x)(14 1 22)(20)(10 1 x)19. The figure is a trapezoid.b 15 ⏐4 2 2⏐ 5 2, b 2 5 ⏐5 2 0⏐ 5 52 h(b 1 1 b 2 ) 5 }1 (3)(2 1 5) 5 10.5 units22GeometryWorked-Out Solution Key357
<strong>Chapter</strong> <strong>11</strong>, continued20. d 15 ⏐3 2 (21)⏐ 5 425. Use the Pythagorean Theorem to find the height ofa 5 12h 5 3bd 15 12 1 12 5 2415 4, b 25 4 1 4 1 4 5 12, d 15 3 1 3 5 6,and d 25 4 1 4 5 8d 25 16 1 30 5 46A 5 }1 2 d 1 1 d 2 5 }1 2 (24)(46) 5 552 Area 5 Area of trapezoid 1 }1 Area of rhombusunits2 2A 5 }1 2 h(b 1 1 b 2 ) 1 }1 2 1 }1 2 d 1 d 12d 25 ⏐23 2 1⏐ 5 4the trapezoid.9The figure is a rhombus.A 5 }1 2 d 1 d 2 5 }1 (4)(4) 5 8 units22 15h21. The figure is a kite.d 15 ⏐4 2 0⏐ 5 4d 25 ⏐2 2 (23)⏐ 5 59 2 1 h 2 5 15 2h 2 5 144A 5 }1 2 d 1 d 2 5 }1 (4)(5) 5 10h 5 12cm22b 25 9 1 <strong>11</strong> 5 2022. A 5 }1 2 h(b 1 1 b 2 ); b 1 5 x, b 2 5 2xa 5 }1 2 h(b 1 1 b 2 ) 5 }1 (12)(8 1 20) 5 168 units2213.5 5 }1 (3)(x 1 2x)26. Use the Pythagorean Theorem to find part of the2second base.13.5 5 }3 2 (3x)a 2 1 20 2 5 29 213.5 5 }9 3 xa 2 5 441a 5 2120293 5 xb 25 21 1 21 5 42aOne base is 3 feet and the other base is 2(3) 5 6 feet.23. A 5 }1 A 5 }12 h(b 1 1 b 2 ); b 1 5 x, b 2 5 x 1 82 h (b 1 1 b 2 ) 5 }1 (20)(21 1 42) 5 630 units2227. Height of trapezoid 5 7 2 5 5 254 5 }1 (6)(x 1 x 1 8)2Area 5 Area of trapezoid 1 Area of rectangle54 5 3(2x 1 8)A 5 }1 2 h T (b 1 1 b 2 ) 1 b R h hR T5 height of trapezoid,18 5 2x 1 8b R5 base of rectangle, andh10 5 2x5 }1 (2)(7 1 10) 1 (10)(5) R5 height of rectangle25 5 xOne base is 5 cm and the other base is 5 1 8 5 13 cm.24. Use the Pythagorean Theorem to find the length of halfof the shorter diagonal.5 67 units 228. Use the Pythagorean Theorem to find the height ofthe trapezoid and half of the shorter diagonal of therhombus:20ah5164a 2 1 16 2 5 20 2a 2 5 1444 2 1 h 2 5 5 2h 2 5 95 }1 2 (3)(4 1 12) 1 }1 2F 1 }1 22 (6)(8) G 5 36 units 229. Area 5 Area of parallelogram 2 Area of kiteCopyright © by McDougal Littell, a division of Houghton Mifflin Company.A 5 bh 2 }1 2 d 1 d 2 5 (12)(7) 2 }1 (12)(7) 5 42 units22358GeometryWorked-Out Solution Key
<strong>Chapter</strong> <strong>11</strong>, continued30. Sample answer:83A 5 8(3) 5 24 units 2 A 5 }1 (3)(4 1 12) 5 24 units225233<strong>11</strong>A 5 }1 2 (3)(5 1 <strong>11</strong>) 5 24 units2 A 5 }1 (3)(2 1 14)23412145 24 units 231. Use the Pythagorean Theorem to find the length of theside of the trapezoid.76 66 2 1 8 2 5 s 2100 5 s 210 5 s15 15 2 7P 5 6 1 7 1 15 1 10 5 38 unitssBecause ABCD is a parallelogram, } AB > } CD and}BC > } DA . So, CD 5 10. Next extend a vertical lineup from } GD to meet } BC . Because ∠BAD > ∠BCD,}AB > } CD , and two more corresponding angles ofthe triangles are congruent (by Alternate Interior andSuppementary Angles Theorems), you know the trianglemade by } AB and parts of } AD and } BF is congruent tothe triangle made by } CD and parts of } BC and } AD byAAS. You can now say GH 5 8 because you havecorresponding parts of congruent triangles. Now draw ahorizontal line connecting C to } BF . You know the base isthe length of FG 1 GH 5 8 1 2 5 <strong>11</strong> and the height is16 2 9 5 7. Use the Pythagorean Theorem to find BC.ABDE F G H<strong>11</strong> 2 1 7 2 5 BC 2170 5 BC 2Ï } 170 5 BCTo find m∠ABC, use trigonometric ratios.BC7B<strong>11</strong> CCopyright © by McDougal Littell, a division of Houghton Mifflin Company.A 5 }1 2 h(b 1 1 b 2 ) 5 }1 (6)(7 1 15) 5 66 units2232. Use the Pythagorean Theorem to find the length of partof the other diagonal.24a 2 1 12 2 5 13 2a 2 5 25a 5 5d 15 5 1 5 5 1012P 5 4(13) 5 52 units1313A 5 }1 2 d 1 d 2 5 }1 (10)(24) 5 120 units2233. First, make a horizontal line from A to the segment } BF .The length of the new segment is 8 because it is parallelto } EF , and EF 5 8. Because BF 5 16 and AE 5 10, thepart of the segment from B to this new segment is16 2 10 5 6. Using the Pythagorean Theorem, you canfind out the length of } AB .8 2 1 6 2 5 AB 2100 5 AB 210 5 ABA 8a1312B6Au10 1 u 2617078tan u 15 8 }6<strong>11</strong>Ctan u 25 <strong>11</strong> }7u 5 tan 21 1 }4 32 u 25 tan 21 1 }<strong>11</strong> 7 2u 1ø 53.18 u 2ø 57.58So, m∠ABC ø 53.18 1 57.78 ø <strong>11</strong>0.68. Extend } BCand drop an altitude down from A to find the heightof the parallelogram. By the Linear Pairs Postulate,the angle made by joining extended } BC to } AB is1808 2 <strong>11</strong>0.68 5 69.48. Use a trigonometric functionto find the height.sin 69.48 5 }h 10Bh 69.4810 p sin 69.48 5 h109.36 ø hASo, the area of the parallelogram isA 5 bh ø Ï } 170 p 9.36 ø 122 square units.Problem Solving34. A 5 }1 2 h(b 1 1 b 2 ) 5 }1 (35)(70 1 79) 5 2607.5 in.22The area of the glass in the wind shield is2607.5 square inches.GeometryWorked-Out Solution Key359
<strong>Chapter</strong> <strong>11</strong>, continued35. A 5 }1 2 d 1 d 2 5 }1 (8)(5) 5 20 mm22The area of the logo is 20 square millimeters.Sample answer:36. A 5 1 }2 d 1 d 2432 5 }1 2 (36)(d 2 )24 5 d 2The 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 trapezoidArea of field 5 1 }2 bh 1 1 }2 h(b 1 1 b 2 )Area of field 5 }1 2 (315)(322) 1 }1 (179)(145 1 450)2Area of field 5 103,967.5 ft 2103,967.5 ft 2} p }1 yd2 ø <strong>11</strong>,552 yd2129 ftThe area of the playing field is 103,967.5 square feet orabout <strong>11</strong>,552 square yards.38. a. Rhombusnumber, nRhombusnumber, nDiagramDiagram1 2Area, A 2 43 4Area, A 6 8Rhombusnumber, nDiagram5b. The area is twice the rhombus number n. The area ofthe nth rhombus is An 5 2n.c. The length of the other diangonal is 2n for the nthrhombus. A 5 }1 2 d 1 d 2 5 }1 (2)(2n) 5 2n. The rule for2area in this part is the same as the rule for areain part (b).39. The kite was cut along the diagonal and then the otherdiagonal to make 1 big triangle and 2 smaller triangles.The resulting figure, once the smaller triangle wasmoved, was a rectangle. The formula for the area of arectangle is b p h. In this case, the base was the firstdiagonal, and the height was half of the other diagonal.Substituting those values into the area formula, youhave A 5 (d 1) 1 }1 2 d 22 , which simplifies to }1 2 d 1 d 2 . Becauseyou can go through the same process with a rhombus,the formula will also simplify to }1 2 d 1 d , which is2Theorem <strong>11</strong>.5.40. The area of nPRS is }1 2 b 1 h and the area of nPQR is }1 2 b 2 h.The area of trapezoid PQRS is equal to1}2 b 1 h 1 }1 2 b 2 h 5 }1 2 h(b 1 1 b 2 ).41. By the SSS Congruence Postulate, nPQR > nPSR.The area of nPQR is }1 2 d 21 }1 2 d 12 5 }1 4 d 1 d . So, the area of2kite PQRS is twice the area of nPQR, or21 }1 42 d 1d 25 }1 2 d 1 d 2 .42. a. Sample answer:original kite rhombus isoscelestriangleright trianglemany different trianglesmany different many differentkitesquadrilateralsb. Yes, you can make isosceles and right triangles bymoving the vertical diagonal down all the way and tothe middle or down all the way and to the left or rightall the way.Copyright © by McDougal Littell, a division of Houghton Mifflin Company.Area, A 10360GeometryWorked-Out Solution Key
<strong>Chapter</strong> <strong>11</strong>, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.c. Area of original kite: }1 2 d 1 d 2 5 }1 (4)(6) 5 12 units22Area of rhombus: }1 2 d 1 d 2 5 }1 (4)(6) 5 12 units22Area of isosceles triangle: }1 2 bh 5 }1 (6)(4) 5 12 units22Area of right triangle: }1 2 bh 5 }1 (6)(4) 5 12 units22Area of different kiles: }1 2 d 1 d 2 5 }1 (4)(6) 5 12 units22All of the areas are equal, 12 square units. The lengthsof the diagonals are not being changed, just movedaround. Although the shapes made by connecting thediagonals will be different, the area will remain thesame.43. Looking at the trapezoid:b 15 a, b 25 b 1and h 5 a 1 bA 5 }1 2 h(b 1 1 b 2 ) 5 }1 (a 1 b)(a 1 b)25 }1 2 (a2 1 2ab 1 b 2 ) 5 }1 2 a2 1 ab 1 }1 2 b2Looking at the triangles that make up the trapezoid:A 5 }1 2 (a)(b) 1 }1 2 (a)(b) 1 }1 2 (c)(c) 5 ab 1 }1 2 c2Because the areas are equal, you can set the area of thetrapezoid and the combined area of the triangles equal toeach other.1}2 a2 1 ab 1 }1 2 b2 5 ab 1 }1 2 c21}2 a2 1 1 }2 b2 5 1 }2 c22 1 1 }2 a2 1 1 }2 b2 2 5 2 1 1 }2 c2 2Mixed Reviewa 2 1 b 2 5 c 244. d 5 rt Original equationd}r 5 tDivision Property of Equality45. A 5 }1 bh2Original equation2A 5 bh Multiplication Property of Equality2A}b 5 hDivision Property of Equality46. P 5 2l 1 2w Original equationP 2 2l 5 2wSubtraction Property ofEqualityP 2 2 l}25 w Division Propertyof Equality47. 4x8 1 4x8 1 x8 5 18089x 5 180x 5 20The angle measures are 208, 808, and 808.48. }h 18 5 }20 3030h 5 360h 5 12Perimeter of nPQR}}Perimeter of nSTU 5 }PQ STPerimeter of nPQR}}815 20 }3030 p Perimeter of nPQR 5 1620Perimeter of nPQR 5 54The height of nPQR is 12 inches and the perimeter is54 inches.Lesson <strong>11</strong>.3<strong>11</strong>.3 Guided Practice (pp. 738–739)1. Ratio of perimeters: }16 12 5 }4 3 5 }a bRatio of areas: }a2b 5 422 }3 5 16 2 }964}}Area of nDEF 5 }16 964(9) 5 16(Area of nDEF)576 5 16(Area of nDEF)36 5 Area of nDEFThe ratio of the area of nABC to the area of nDEFis }16 . The area of nDEF is 36 square feet.92. Ratio of area: 20 }36Ratio of sides: Ï} 20Ï } 36 5 2 Ï} 5}6 5 Ï} 5}3The ratio of their corresponding side lengths is Ï} 5}3 .3. Step 1 Find the ratio of the perimeters. Use the sameunits for both lengths in the ratio.Perimeter of Rectangle I}}Perimeter of Rectangle II 566 in.}}2(35 ft) 1 2(20 ft)66 in.5 }<strong>11</strong>0 ft5.5 ft5 }<strong>11</strong>0 ft 5 }1 20Step 2 Find the ratio of the areas of the two rectangles.(Periemter of Rectangle I) 2}}}(Perimeter of Rectangle II) 5 Area of Rectangle I2 }}Area of Rectangle II1}220 5 1 2 }400Step 3 Find the area of Rectangle I.Area of Rectangle I}}Area of Rectangle II 5 }1 400Area of Rectangle I}} 5 }1 (35)(20) 400Area of Rectangle I p 400 5 700Area of Rectangle I 5 1.75 ft 2GeometryWorked-Out Solution Key361
<strong>Chapter</strong> <strong>11</strong>, continuedStep 4 1.75 ft 2 144 in.2p } 5 252 in.221 ftThe ratio of the area is }1 and the area of400Rectangle I is 252 square inches.<strong>11</strong>.3 Exercises (pp. 740–743)Skill Practice1. Sample answer:A54BC3D108The side lengths of nABC are each multiplied by thesame scale factor to get the lengths of each side ofnDEF. AB has corresponding side length DE, BC hascorresponding side length EF, and CA has correspondingside length FD. Each side in nABC is multiplied by 2 toget the length of its corresponding side in nDEF.2. You don’t need to know the value of n to find of the ratioof the perimeters or the ratio of the areas of the polygons,because you know the ratio of the side lengths, 3 : 4. Thisratio is also the ratio of the perimeters. According toTheorem <strong>11</strong>.7, the ratio of the areas is the square of theratio of the perimeters 3 2 : 4 2 , or 9 : 16.3.4.Ratio ofcorrespondingside lengthsEF6Ratio ofperimetersRatio of areas6 : <strong>11</strong> 6 :<strong>11</strong> 36 : 1215 : 9 20 : 36 5 5 : 9 25 : 815. The ratio of the perimeters is 1 : 3 and the ratio of theareas is 1 2 : 3 2 , or 1 : 9.2}x 5 }1 918 5 xThe area of the blue triangle is 18 square feet.6. The ratio of the perimeters is 15 : 20, or 3 : 4, and the ratioof the areas is 3 2 : 4 2 , or 9 : 16.x}240 5 }9 16x 5 135The area of the red quadrilateral is 135 square centimeters.7. The ratio of the perimeters is 7 : 9 and the ratio of theareas is 7 2 : 9 2 , or 49 : 81.x}210 5 }49 81x ø 127The area of the red hexagon is about 127 square inches.8. The ratio of the perimeters is 5 : 3 and the ratio of theareas is 5 2 : 3 2 , or 25 : 9.40}x 5 }25 914.4 5 xThe area of the blue parallelogram is 14.4 square yards.9. Ratio of areas 5 49 : 16Ratio of lengths 5 Ï } 49 : Ï } 16 5 7 : 4The ratio of the lengths of corresponding sides is 7 : 4.10. Ratio of areas 5 16 : 21Ratio of lengths 5 Ï } 16 : Ï } 121 5 4 : <strong>11</strong>The ratio of the lengths of corresponding sides is 4 : <strong>11</strong>.<strong>11</strong>. Ratio of areas 5 121 : 144Ratio of lengths 5 Ï } 121 : Ï } 144 5 <strong>11</strong> : 22The ratio of the lengths of corresponding sides is <strong>11</strong> : 12.12. C; Ratio of areas 5 18 : 24Ratio of lengths 5 Ï } 18 : Ï } 24 5 3 Ï } 2 : 2 Ï } 6 53 Ï } 22 Ï } 6 p Ï} 6Ï } 6 5 3 1 } Ï} 12}12 5 6 Ï 3}12 5 Ï} 3}2 or Ï} 3 : 2213. Ratio of areas 5 7 : 281 : 4Ratio of lengths 5 Ï } 1 : Ï } 4 5 1 : 24}XY 5 }1 28 5 XYThe length of XY is 8 centimeters.14. Ratio of areas 5 198 : 88 5 9 : 4Ratio of lengths 5 Ï } 9 : Ï } 4 5 3 : 2XY}10 5 }3 2XY 5 15The length of XY is 15 inches.15. The ratio of areas is 1 : 4, so the ratio of lengthsis Ï } 1 : Ï } 4 or 1 : 2. You need to use the 1 : 2 ratio.12}ZY 5 }1 2ZY 5 24Perimeter of QRSTU 5(12 cm)16. }} 5 }Perimeter of VWXYZ 140 cm 5 }3 7So, the ratio of corresponding lengths is 3 : 7 and theratio of areas is 3 2 : 7 2 , or 9 : 49.Area of QRSTU}}Area of VWXYZ 5 }9 49248}x 5 }9 4912,152 5 9x1350 ø xThe area of regular pentagon VWXYZ is about1350 square centimeters.Copyright © by McDougal Littell, a division of Houghton Mifflin Company.362GeometryWorked-Out Solution Key
<strong>Chapter</strong> <strong>11</strong>, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.17. Area of MNPQ 5 }1 2 d 1 d 2 5 }1 (14)(25) 5 175 ft22Ratio of areas 5 28 : 175 5 4 : 25Ratio of lengths 5 Ï } 4 : Ï } 25 5 2 : 5Shorter diagonal of RSTU}}145 2 }5Shorter diagonal of RSTU 5 5.6 ftLonger diagonal of MNPQ}}255 2 }5Longer diagonal of MNPQ 5 10 ftThe area of MNPQ is 175 square feet. The lengths of thediagonals are 10 feet and 5.6 feet.18. Case 3, Case 1, Case 2In Case 3, the enlargement is Ï } 5 , or approximately2.24. Because 2.24 is less than the enlargement of 3(Case 1) and 2.24 and 3 are both less than an enlargementof 4 (Case 2), the order of enlargement from smallest tolargest is Case 3, Case 1, Case 2.19. Doubling the side length of a square never doublesthe area. Doubling the side length of a square alwaysquadruples the area.20. Two similar octagons sometimes have the same perimeter.The perimeters will be the same only when the octagonsare congruent.21. ratio of lengths: }4.56 5 }3 4 , }7.510 5 }3 4 , }a 12 5 }3 4The ratio of lengths is 3 : 4, so the ratio of areas is 3 2 : 4 2 ,or 9 : 16.Area of nABCYou can use the proportion }}Area of nDEF 5 }9 to solve16for the area of nDEF.22. First, you have to find the width of ABCD.P 5 2l 1 2w84 5 2(24) 1 2w36 5 2w18 5 wThe width of ABCD is 18 feet, or18 feet p }1 yard3 feet 5 6 yardsRatio of width 5 6 : 3 5 2 : 1Ratio of areas 5 2 2 : 1 2 5 4 : 1The ratio of the area ABCD to the area of DEFG is 4 : 1.23. ∠D and ∠N are right angles, so ∠D > ∠N by the RightAngles Congruence Theorem. ∠F > ∠M is given. So,nDEF , nNLM by the AA Similarity Postulate.Use the area formula to find NL.A 5 1 }2 bh294 5 }1 2 (NL)(21)28 5 NLFind ML so that you can find the ratio of correspondinglengths.(MN) 2 1 (NL) 2 5 (ML) 221 2 1 28 2 5 (ML) 21225 5 (ML) 235 5 MLratio of lengths: 10 : 35 5 2 : 7ratio of areas: 2 2 : 7 2 5 4 : 4924. ∠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 find the length of } VW to find the ratio oflengths.tan 308 5 Ï} 3}VWÏ } 3VW 5 }tan 308VW 5 3The corresponding lengths are UV and VW.Ratio of lengths: Ï } 3 : 3Ratio of areas: ( Ï } 3 ) 2 : 3 2 5 3 : 9 5 1 : 325. a. Ratio of area nAGB to nCGE 5 9 : 25Ratio of lengths of nAGB to nCGE 5 Ï } 9 : Ï } 255 3 : 5Set up proportions of corresponding sides to the ratioof lengths.AG}CG 5 }3 5AG}10 5 }3 5GB}GE 5 }3 5GB}15 5 }3 5AG 5 6 GB 5 9GeometryWorked-Out Solution Key363
<strong>Chapter</strong> <strong>11</strong>, continuedLook at nAGF and nBGC. ∠AGF and ∠BGC areright angles, so ∠AGF > ∠BGC. ∠GFA > ∠GBCbecause they are alternate interior angles. So, nAGF, nBGC by the AA Similarity Postulate. Find thelength of } GF using a ratio.GF}GC 5 }AG BGGF}10 5 }6 9GF 5 20 }3GE 5 GF 1 FE15 5 20 }3 1 FE25}3 5 FESo, AG 5 6, GB 5 9, GF 5 }20 3 , and FE 5 }25 3 .b. Look at n ADC and nCBA. ∠ ADC > ∠ CBA becauseABCD is a parallelogram. ∠ DAC > ∠ BCA becausethey are alternate interior angles. So, n ADC , nCBAby the AA Similarity Postulate. Find the area ofnCBA, whose height is BG and base length isAC 5 AG 1 GC.A 5 }1 2 bh 5 }1 (6 1 10)(9) 5 722The ratio of the area of nADC to nCBA is 72 : 72,or 1 : 1.Problem Solving26. ratio of longest sides 5 3 : 5ratio of areas p 3 2 : 5 2 5 9 : 25Area of old banner}}Area of new banner 5 }9 253(1)}}Area of new banner 5 }9 258 }1 5 Area of new banner3The area of the new banner will be 8 }1 square feet.327. ratio of areas 5 360 : 250 5 36 : 25ratio of lengths 5 Ï } 36 : Ï } 25 5 6 : 5distance on new patio}}distance on similar patio 5 }6 5distance on new patio}} 5 }6 12.55distance on new patio 5 15The distance on the new patio is 15 feet.28. B;ratio of lengths 5 90 : 60 5 3 : 2ratio of areas 5 3 2 : 2 2 5 9 : 4pounds for baseball diamond}}}pounds for softball diamond 5 }9 420}}}pounds for softball diamond 5 }9 49 ø pounds for softball diamondYou need about 9 pounds for the softball diamond.29. a. Sample answer:1Ay15B92 13A 5 }1 2 bh 5 }1 (9)(4) 5 18 square units2b. A(0, 0), B(3, 4), C(9, 0)midpoint of } AB : S 1 }0 1 32 , }0 1 42 2 5 S 1 }3 2 , 2 2midpoint of } BC : Q 1 }3 1 92 , }4 1 02 2 5 Q(6, 2)midpoint of } AC : R 1 }0 1 92 , }0 1 01AyS1BRCCx2 2 5 S 1 9 }x2 , 2 2QR 5 Î }}16 2 }9 22 2 1 (2 2 0) 2 5 Ï } 2.25 1 45 Ï } 6.255 2.5RS 5 Î }}1 }9 2 2 }3 22 2 1 (0 2 2) 2 5 Ï } 9 1 4 5 Ï } 13QS 5 Î }}16 2 }3 22 2 1 (2 2 2) 2 5 Ï } 20.25 1 0 5 4.5AB}QR 5 }5 2.5 5 2, }BC RS 5 2 Ï} 13Ï } 13 5 2, }AC QS 5 }9 4.5 5 2Because all of the ratios of corresponding lengths areequal, the triangles are similar.Ratio of lengths of nQRS to nABC: 1 : 2Ratio of areas of nQRS to nABC: 1 : 4The area of the smaller triangle is }1 the area of the4larger triangle. The smaller triangle has an area of1}(18) 5 4.5 square units.4Copyright © by McDougal Littell, a division of Houghton Mifflin Company.364GeometryWorked-Out Solution Key
<strong>Chapter</strong> <strong>11</strong>, continued30. Sample answer:Rectangle I , Rectangle IIA I5 ax, A II5 byxIaBecause the rectangles are similar, the ratios ofcorresponding sides are proportional. So, }a b 5 }x y .Ratio of areas 5 }axby 5 }a b p }x y . Because }a b 5 }x y , substitutethat into the previous equation.Ratio of areas 5 }a b p }a b 5 }a2b . 2So, the ratio of corresponding sides is a : b, and the ratioof their areas is a 2 : b 2 .31. The graph is visually midleading because although theamount of science fiction books read is only half theamount of mysteries read, the graph makes it seem likethe amount of science fiction books read is a quarter ofthe mysteries read. The student could redraw the graphby using bars of the same width for the same categories.32. Sample answer:3IIIy6IIb34. a. You are given JL 5 KL, so } JK > } KL . Because nJKLis in the corner of a cube, ∠L is a right angle. So,nJKL is an isosceles right triangle. ∠D is a rightangle because it is the corner of a cube. A cube hasequal sides, which means } MP > } NP and nMNP isan isosceles right triangle. Because both nJKL andnMNP are isosceles right triangles, their side lengthsare proportional. So, nJKL , nMNP by the SASSimilarity Theorem.b. Ratio of lengths of triangles: 1 : 3Ratio of areas of triangles: 1 2 : 3 2 5 1 : 9Because the area of the face PMTN is twice thearea nMNP, the scale factor of nJKL to the areaof 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 is1 : 18. The area of the pentagon JKQRS is the area ofa face of the cube with the triangle nJKL subtractedfrom it. So, the ratio of the area of nJKL to the area ofpentagon JKQRS is 1 : (18 2 1), or 1 : 17.Mixed Review35. C 5 πd 5 (3.14)(4 cm) 5 12.56 cm36. C 5 πd 5 (3.14)(10 ft) 5 31.40 ft37. C 5 2πd 5 2(3.14)(2.5 yd) 5 15.70 yd38. C 5 2πr 5 2(3.14)(3.1 m) 5 19.47 mCopyright © by McDougal Littell, a division of Houghton Mifflin Company.3A I} 5 }9 A II6P I} 5 Ï} 9P II Ï } 6 5 3Ï } 6 5 3 Ï} 6}6 5 Ï} 6}233. 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 : 9Ratio of areas 5 10 2 : 9 2 5 100 : 81c. One way: Use AB p AC 5 AD p AE to solve for thelength of } DE .9 p (9 1 <strong>11</strong>) 5 10 p (10 1 DE)180 5 100 1 10 DE80 5 10 DE8 5 DEAnother way: Use the ratios of corresponding sides ofthe similar triangles.AD}AC 5 }ABAE10}20 5 9}10 1 DE10(10 1 DE) 5 18010 1 DE 5 18DE 5 8639. }1 2 x 5 10 40. 85 5 }1 2 xx 5 20170 5 x41. x 5 }1 (180 2 88)2x 5 }1 2 (92)x 5 46Quiz <strong>11</strong>.1–<strong>11</strong>.3 (p. 743)1. A 5 bh; h 5 3b, A 5 108 ft 2108 5 b(3b)108 5 3b 236 5 b 26 5 bThe base is 6 feet and the height is 3(6) 5 18 feet.2. Use the Pythagorean Theorem to find the height of thetriangle.a 2 1 12 2 5 13 213aa 2 5 25a 5 512Area of figure 5 Area of rectangle 1 Area of triangle5 bh 1 }1 2 ba 5 (12)(6) 1 }1 2 (12)(5)5 102 square units3. A 5 }1 2 h(b 1 1 b 2 ) 5 }1 (4)(3 1 6.5) 5 19 square units2GeometryWorked-Out Solution Key365
<strong>Chapter</strong> <strong>11</strong>, continued4. d 15 4 1 4 5 8 units, d 25 5 1 8 5 13 unitsb. Area of rectangle 5 bh 5 (15)(9.5) 5 142.5 ft 2A 5 }1 2 d 1 d 2 5 }1 (8)(13) 5 52 square units25. The ratio of the lengths of two similar heptagons is 7 : 20,so the ratio of their perimeters is also 7 : 20. The ratio oftheir areas is 7 2 : 20 2 , or 49 : 400.6. The ratio of the areas is 1200 : 48, or 25 : 1. The ratio ofthe lengths is Ï } 25 : Ï } 1 , or 5 : 1.50}XY 5 }5 <strong>11</strong>0 5 XYXY is 10 feet.Problem Solving Workshop <strong>11</strong>.3 (p. 744)1. A 5 lw, where x 5 width and 1.5x 5 length1.44(15 p 10) 5 1.5x p x144 5 x 212 5 xThe length of the third pan is 1.5(12) 5 18 inches.2. Area of PQRS 5 }1 2 h(b 1 1 b 2 ) 5 }1 (6)(9 1 12) 5 632The area of WXYZ is }28 63 5 }4 of the area of PQRS. So,9each side in WXYZ is Ï} 4Ï } 9 , or }2 the size of each3corresponding side in PQRS. So,WZ 5 }2 3 PS 5 }2 (12) 5 8 units.33. first square: A 5 s 2second square: A 5 2s 2Each side of a square is the square root of the area,so the second length is Ï } 2s 2 5 s Ï } 2 .4. Area of nABC 5 }1 2 bh 5 }1 (8)(5) 5 20 cm22The area of nDEF is }<strong>11</strong>.25 , or 0.5625 of the area of20nABC. So, each side in nDEF is Ï } 0.5625 , or 0.75 thesize of each corresponding side in nABC.DF 5 0.75(5) 5 3.75 cmEF 5 0.75(8) 5 6 cmDE 2 1 EF 2 5 DF 2(3.75) 2 1 6 2 5 DF 250.0625 5 DF 27.08 ø DFThe length of DE is 3.75 cm and the length of DF isabout 7.08 cm.Mixed Review of Problem Solving (p. 745)1. a. Because ∠A, ∠C, ∠E, and ∠G are right angles, theyare 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.Area of rhombus 5 }1 2 d 1 d 2 5 }1 (9.5)(15) 5 71.25 ft22c. Area for marigolds 5 Area of rectangle 2 Area ofrhombusArea for marigolds 5 142.5 2 71.255 71.25 ft 2Total cost 5 cost for marigolds 1 cost for asters5 $.30(71.25) 1 $.40(71.25) ø $49.88The total cost is about $49.88.2. Sample answer:8 m6 m12 m8 m12 m10 m8 m 8 m208 ø 14.4 m12 m6 m 6 m 6 m8 m8 mAll triangles must satisfy A 5 }1 bh and all parallelograms2must satisfy A 5 bh.3. Area of the room: (12 foot)(21 foot) 5 252 ft 2 144 in.2p }1 ft 25 36,288 in. 2a. Small tiles: 12 in. 3 12 in. 5 144 in. 2 , so you wouldneed 36,288 4 144 5 252 tilesLarge tiles: 18 in. 3 18 in. 5 324 in. 2 , so you need36,288 4 324 5 <strong>11</strong>2 tilesYou would need 252 small tiles and <strong>11</strong>2 large tiles.b. Cost for large tiles: $2.25(<strong>11</strong>2) 5 $252Cost for small tiles: $1.50(252) 5 $378The cost of the floor using large tiles is $252 and thecost of the floor using small tiles is $378. If you wantto spend as little as possible, use the large tiles.c. Ratio of side lengths 5 12 : 18 5 2 : 3Ratio of areas 5 2 2 : 3 2 5 4 : 9Ratio of cost 5 1.50 : 2.25 5 2 : 3The cost per tile is based on side length because theratios of cost and side lengths are the same, 2 : 3.4. If you double the length of each diagonal of a rhombus,you quadruple the area. If you triple the length of eachdiagonal of a rhombus, you have an area 9 times greaterthan the original area. If each diagonal is multiplied bythe same number n, the area of a rhombus will be n 2times greater than the original area.5. a. A 5 }1 2 bh 5 }1 (40)(41) 5 820 square feet2Copyright © by McDougal Littell, a division of Houghton Mifflin Company.366GeometryWorked-Out Solution Key
<strong>Chapter</strong> <strong>11</strong>, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.b. Use the Pythagorean Theorem to find the length of UT.US 2 1 ST 2 5 UT 2 U 40 ft S40 2 1 41 2 5 UT 23281 5 UT 257.3 ø UT41 ftPerimeter of nSTU ø 40 1 41 1 57.3 ø 138.3 ftPerimeter of nPQR}}Perimeter of nSTU ø }1.31Perimeter of nPQR}}138.3ø 1.3 }1Perimeter of nPQR ø 180 feetc. Ratio of perimeters ø 1.3 : 1Ratio of areas ø 1.3 2 : 1 2 ø 1.69 : 1Area of nPQR}}Area of nSTU ø }1.691Area of nPQR}}820ø 1.69 }1Area of nPQR ø 1386 square feetArea of path around pool 5 Area of nPQRT2 Area of nSTUø 1386 2 820ø 566 square feet.The area of nPQR is about 1386 square feet. The areaof the path around the pool is about 566 square feet.6. 375; A 5 }1 2 h(b 1 1 b 2 ) B x A5 in.1250 5 }1 (5)(x 1 3x)2C 3x D500 5 4x125 5 xThe 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 find FH.FE 2 1 EH 2 5 FH 2(5 Ï } 2 ) 2 1 (5 Ï } 2 ) 2 5 FH 2100 5 FH 210 5 FHb. FJ is half of FH, so FJ 5 5.E5 25 2Use the Pythagorean Theorem to find the length of } EJ .FJ 2 1 EJ 2 5 EF 2 E JF5 2 1 EJ 2 5 (5 Ï } 2 ) 2EJ 2 5 25EJ 5 55 2FH5Use tan 608 5 JG }FJ to find the length of } JG .tan 608 5 JG }55 tan 608 5 JG5 Ï } 3 5 JGEG 5 EJ 1 JG 5 5 1 5 Ï } 3 ø 13.7 unitsc. d 15 FH 5 10 and d 25 EG ø 13.7A 5 }1 2 d 1 d 2 ø }1 (10)(13.7) ø 68.52The area of EFGH is about 68.5 square units.Lesson <strong>11</strong>.4<strong>11</strong>.4 Guided Practice (pp. 747–749)1. Circle with diameter of 5 inches:C 5 πd 5 π(5) ø 15.71The circumference is about 15.71 inches.Circle with circumference of 17 feet: C 5 πd17 5 πd5.41 ø dThe diameter is about 5.41 feet.2. Circumference: C 5 πd 5 π(28) ø 87.96 in.Distance traveled 5 Number of revolutions pcircumference500 ft 5 Number of revolutions p 87.96 in. p }1 ft12 in.500 ft 5 Number of revolutions p 7.33 ft68 ø Number of revolutionsThe tire makes about 68 revolutions while traveling500 feet.3. Arc length of C DQ 5 }758 p π(9) ø 5.89 yards3608Arc length of C C4. }}LM 5 }m LMc360861.26m} 5 }2708c 360861.26m} 5 }3 c 481.68 Cø cC5. Arc length of }EF2πr 5 }m EF360810.5 ft} 5 }15082πr 36083780 5 300πr4.01 ft ø r6. Distance 5 2(84.39) 1 2 p 1 1 }2 p 2π p 44.02 2Distance ø 445.4 metersThe runner on the blue path travels about 445.4 meters.GeometryWorked-Out Solution Key367
<strong>Chapter</strong> <strong>11</strong>, continued<strong>11</strong>.4 Exercises (pp. 749–752)Skill PracticeArc length of C C1. }}AB 5 }m AB2πr 36082. The arc measure is the number degrees of a circle thearc is bounded by. The arc length is the part of thecircumference of the circle that the arc occupies.3. C 5 2πr 5 2π(6) ø 37.70 in.4. C 5 πd 5 π(17) ø 53.41 cm5. C 5 2πr 6. C 5 πd 5 π(5) 5 5π in.63 5 2πr10.03 ft ø r7. C 5 2πr28π 5 2πr14 m 5 r8. C 5 πd 5 π(14) ø 43.98 units9. C 5 2πr 5 2π(3 1 2) 5 2π(5) ø 31.42 units10. C 5 πd 5 π(10 4 2) 5 π(5) ø 15.71 units<strong>11</strong>. Arc length of C C AB 5 }m AB 408p 2πr 5 }3608 3608 p 2π(6)ø 4.19 m12. Arc length of C C AB 5 }m AB 1208p 2πr 5 }3608 3608 p 2π(14)ø 29.32 cm13. Arc length of C C AB 5 }m AB 458p πd 5 } p π(8) ø 3.14 ft3608 360814. Two arcs from different circles have the same length onlyif the circles have the same circumference.15. m QRS 5 3608 2 m C QSm C QRS 5 3608 2 608m C QRS 5 300816. Arc length of CCQRS 5 }m QRS3608ø 41.89 ft17. ∠QPR > ∠RPS3608 2 608m∠QPR 5 }2m C QR 5 1508p 2πr 5 3008 }3608 p 2π(8)5 3008 }2 5 150818. m RS 5 1508 (from Exercise 17)m C RSQ 5 m C RS 1 m CSQ 5 1508 1 608 5 210819. Arc length of C CQR 5 }m QR 1508p 2πr 5 }3608 3608 p 2π(8)ø 20.94 ft20. Arc length of C CRSQ 5 }m RSQ p 2πr 5 }21083608 3608 p 2π(8)ø 29.32 ftArc length of C C21. }}AB 5 }m AB2πr 3608C8.73}2π(10) 5 }m AB36083608 p }8.73 C20π 5 m AB508 ø m C ABArc length of CD22. }}C23.5 m CD }36087.5}C 5 }76836087.5}C 5 }19 9035.53 units ø CArc length C C }}LM 5 }m LM2πr 360838.95}2πr 5 }2608360838.95}2πr 5 }13 18701.1 5 26πr8.58 units ø r24. P 5 2 lengths 1 circumference of the circle5 2(13) 1 π(6) ø 44.85 units25. P 5 2 lengths 1 }1 (circumference of the circle)25 2(6) 1 }1 (2π(3)) ø 21.42 units226. x 2 1 y 2 5 16 has a radius of Ï } 16 5 4.C 5 2πr 5 2π(4) 5 8π27. (x 1 2) 2 1 ( y 2 3) 2 5 9 has a radius of Ï } 9 5 3.C 5 2πr 5 2π(3) 5 6π28. x 2 1 y 2 5 18 has a radius of Ï } 18 5 3 Ï } 2 .C 5 2πr 5 2π(3 Ï } 2 ) 5 6 Ï } 2 π29. C 5 2πr C 5 πdC}2π 5 r C }π 5 dWhen C 5 26π: When C 5 26π:26 π}2 π 5 r 26 π} 5 dπ13 5 r 26 5 d30. When m C AB 5 458 and length of C AB 5 4:4 5 }4583608 p 2πr5.09 ø rWhen r 5 2 and m C AB 5 608:Arc length of C AB 5 }608 p 2π(2) ø 2.093608Copyright © by McDougal Littell, a division of Houghton Mifflin Company.368GeometryWorked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.<strong>Chapter</strong> <strong>11</strong>, continuedWhen r 5 0.8 and length Cof C AB 5 0.3:0.3}2π(0.8) 5 }m AB36080.3C3608 p }2π(0.8) 5 m AB21.498 ø m C ABWhen r 5 4.2 and m C AB 5 1838:Arc length of C AB 5 }1838 p 2π(4.2) ø 13.413608When m C AB 5 908 and length of C AB 5 3.22:3.22 5 908 }3608 p 2πr2.05 ø rWhen r 5 4 Ï } 2 and length of CCAB 5 2.86:2.862π(4 Ï } 2 ) 5 }m AB36082.86C3608 p2π(4 Ï } 2 ) 5 m AB28.978 ø m C ABRadius 5.09 2 0.8 4.2 2.05 4 Ï } 2m C AB 458 608 21.498 1838 908 28.988Length ofC AB4 2.09 0.3 13.41 3.22 2.8631. When m C EF 5 x8 and r 5 r: Arc length of C EF 5 }x8πr1808x8a. double the radius: Arc length of C EF 5 }3608 p 2π(2r)Arc length of EF 5 }x8πr908Because }x8πrx8πris twice as large as }, the length of C908 1808 EFis twice as large when the radius is doubled.b. If you double the measure of C EF :Arc length of C EF 5 }2x8360 p 2πrArc length of C EF 5 }x8πr908Because }x8πrx8πris twice as large as }, the length of C908 1808 EFis twice as large when the measure of C EF is doubled.32. A;m C XY 1 m C YZ 5 1808 because } XZ is the diameter of thecircle.1408 1 m C YZ 5 1808m C TZ 5 408The diameter is 6, so the radius is 6 4 2 5 3.Arc length of C C YZ 5 }m YZ3608p 2πr 5408}3608 p 2π(3) 5 2 }3 π 33.12 cm6 cmr8 cm16 cmA rhombus has diagonals that bisect each other. Therhombus can also be split into 4 congruent triangles.Look at one such triangle.1 with legs }16 2 5 8 and }12 2 5 6 2 . Use thePythagorean Theorem to find the side length.6 2 1 8 2 5 s 2 s6 cm100 5 s 2r10 5 s8 cmNow look at the radius of the circle and break theoriginal right triangle into two smaller right triangles.Use the Pythagorean Theorem and a system of equationsto find the length of the radius.r 2 1 (10 2 x) 2 5 64x6 cm r 10 2 x2 r 2 1 x 2 5 36r(10 2 x) 2 2 x 2 5 288 cm100 2 20x 1 x 2 2 x 2 5 2872 5 20x3.6 5 xNow substitute 3.6 for x to find r.r 2 1 (3.6) 2 5 36r 2 1 12.96 5 36r 2 5 23.04r 5 4.8You have the radius, so you can now find thecircumference.C 5 2πr 5 2π(4.8) ø 30.16The circumference of a circle inscribed in a rhombuswith diagonals that are 12 centimeters and 16 cenitmeterslong is about 30.16 centimeters.34. Because you know the measure of the shaded red angleis 308 and the arc length is 2, you can set up a proportionfor the circumference of the blue circle.Arc length C Cblue}} 5 }m blueC blue 3608In Exercise 31, you found that when you double theradius you double the arc length. Because the radius ofthe blue circle is double that of the red circle, the arclength is twice as big, or 2(2) 5 4. Because the anglemeasure of the blue angle and the shaded red angle arethe same, m C blue 5 308. So, substituting those values inthe proportion, you get:4}C blue 5 }308360848 5 C blueSo, the circumference of the blue circle is 48 units.GeometryWorked-Out Solution Key369
<strong>Chapter</strong> <strong>11</strong>, continuedProblem Solving35. The rope length, 21 feet 8 inches, representsthe circumferencec of the tree. If you divide thecircumference by π, you will get the diameter ofthe tree.21 feet 8 inches 5 21 }2 3 feet21 }2 feet 4 π ø 6.93The diameter of the tree is about 7 feet.36.6dTo find the diameter, use the Pythagorean Theorem.6 2 1 6 2 5 d 272 5 d 2d66 Ï } 2 5 dC 5 πd 5 π(6 Ï } 2 ) ø 26.666The circumference of the circle is about 26.66 units.37. Find the circumference of the tire:C 5 πd 5 π(8 in.) ø 25.13 in.length of path 5 number of rotations p circumference5 87 p 25.<strong>11</strong>3 ø 2186.31The length of the path is about 2186 inches long.38. a. The chain touches about half of each sprocket and hasan additional two lengths of 6 }9 inches. Find half of16each sprocket’s circumference, add them together andthen add 216 }162 9 to that.larger sprocket: C 5 πd 5 π 16 }1 82 ø 19.24So, half of the circumference of the larger sprocket isabout }1 (19.24) 5 9.62 in.2smaller sprocket: C 5 πd 5 π <strong>11</strong> }162 7 ø 4.52So, half of the circumference of the smaller sprocket isabout }1 (4.52) 5 2.26 in.2length of chain 5 9.62 1 2.26 1 216 }162 9 ø 25The length of the chain is about 25 inches.b. The chain touches about half of each sprocket, so onlyhalf of the teeth on each sprocket are gripping thechain at any given time.Half of the teeth on the larger sprocket: }1 (76) 5 382Half of the teeth on the smaller sprocket: }1 (15) 5 7.52Total teeth 5 38 1 7.5 5 45.5There are about 46 teeth gripping the chain at anygiven time.39. Because l 1⏐⏐ l 2, ∠1 > ∠2 by the Alternate InteriorAngles Theorem. So, m∠1 5 7.28.distance from Alexandra to Syene}}} 5 }m∠1C3608575}C 5 }7.28360828,750 ø CThe Earth’s circumference is about 28,750 miles.40. Because each arc is a half-circle, the measure of each arcis 1808.m each arclength of each arc 5 } p 2πr3608length of each arc 5 }18083608 p 2πrlength of each arc 5 πrBecause there are 4 arcs, the sum of the arc lengthsis 4πr.m each arc41. 8 segments: length of each arc 5 } p πd where3608d 5 rlength of each arc 5 }18083608 p πrlength of each arc 5 }1 2 πrBecause there are 8 arcs, the sum of all of their arclengths is 8 1 1 }2 πr 2 or 4πr.m each arc16 segments: lengths of each arc 5 } p πd where3608d 5 1 }2 rlength of each arc 5 1808 }3608 p π 1 1 }2 r 2length of each arc 5 }1 4 πrBecause there are 16 arcs, the sum of all their arc lengthsis 1 16 1 }4 π 2 r, or 4πr.The sum of the arc lengths for 8 segments is 4πr, thesum for the arc lengths for 16 segments is 4πr, and thesum of the arc legnths for n segments is 4πr. The lengthwill be the same, you just have to allocate the radiusdifferently according to how many segments you have.Mixed Review42. A 5 πr 2 5 π(6) 2 ø <strong>11</strong>3.10 cm 243. A 5 πr 2 5 π(4.2) 2 ø 55.42 in. 244. A 5 πr 2 5 π 18 }3 42 2 ø 240.53 mi 245. A 5 πr 2 5 π <strong>11</strong> }3 82 2 ø 5.94 in. 246. 4 p x 5 5 p 84x 5 40x 5 10Copyright © by McDougal Littell, a division of Houghton Mifflin Company.370GeometryWorked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.<strong>Chapter</strong> <strong>11</strong>, continued47. 5 p (5 1 <strong>11</strong>) 5 4 p (4 1 x)25 1 55 5 16 1 4x64 5 4x16 5 x48. x 2 5 8 p (8 1 24)x 2 5 8(32)x 2 5 256x 5 16<strong>11</strong>.4 Extension (p. 754)1. The equator and the longitude lines are great circles.The latitude lines are not great circles. Earth’s center isthe center for the equator and lines of longitude. Earth’scenter is not the center for lines of latitude.2. Sample answer:More than 1 linethrough endpoints ofdiameter on a sphereOnly 1 line through 2points not endpointsof diameter3. If two lines intersect, then their intersection is exactly 2points. In Spherical Geometry, lines are great circles. Allgreat circles intersect in two points, one point being 1808from the other.Arc length of C C4. }}AB 5 }m AB2πr 3608x 1208} 5 }20π 3608 5 }1 35.x 5 6 }2 3 πThe distance are 6 }2 3 π and 13 }1 3 π.Arc length of C CACB}} 5 }m ACB2πr3608x 3608 2 1208} 5 }20π 3608x 5 13 }1 3 πArc length of C C }}AB 5 }m AB2πr 3608x 908} 5 }16π 3608 5 }1 4x 5 4πArc length of C CACB}} 5 }m ACB2πr 3608x 3608 2 908} 5 }16π 3608x 5 12πThe distances are 4π and 12π.5 }24083608 5 }2 35 }27083608 5 }3 4Arc length of C C6. }}AB 5 }m AB2πr 3608x 1408} 5 }30π 3608 5 }7 18x 5 <strong>11</strong> }2 3 πArc length of C CACB}} 5 }m ACB2πr 3608x 3608 2 1408} 5 }30π 3608x 5 18 1 }3 πThe distances are <strong>11</strong> 2 }3 π and 18 1 }3 π.Lesson <strong>11</strong>.55 2208 }3608 5 }<strong>11</strong>18<strong>11</strong>.5 Guided Practice (pp. 756-757)1. A 5 πr 2 5 π(14) 2 ø 615.75The area of (D is about C615.75 square feet.2. Area of red sector 5 }m FE3608 p πr25 }12083608 p π p 142 ø 205.25The area of the red sector is Cabout 205.25 square feet.3. Area of the blue sector 5 }m FGE p πr360823608 2 12085 } p π p 1436082 ø 410.50The area of the blue sector Cis about 410.50 square feet.4. Area of sector FJG 5 }m FG p Area of (H3608214.37 5 }858 p Area of (H3608907.92 5 Area of (HThe area of (H is 907.92 square centimeters.5. Area of figure 5 Area of semicircle 1 Area of triangle5 1808 }3608 p 1 π p 1 7 }22 2 2 1 1 }2 (7)(7)5 6.125π 1 24.5 ø 43.74The area of the figure is about 43.74 square meters.6. Yes, you can find the measure of the intercepted arc ifyou know the area and radius of a sector of the circle.The formula for the area of a sector ismeasure of arcArea of sector 5 }} p πr36082 . If you solve forthe measure of the arc, you get:Area of circle measure of arc}} 5πr 2 }}36083608 p Area of circle}} 5 measure of arcπr 2GeometryWorked-Out Solution Key371
<strong>Chapter</strong> <strong>11</strong>, continued<strong>11</strong>.5 Exercises (pp. 758–761)Skill Practice1. A sector of a circle is the region bounded by two radii ofthe circle and their intercepted arc.2. Doubling the arc measure of a sector in a given circlewill double its area. When you double the arc measure,you make the sector twice as big, which means that thearea also doubles.3. A 5 πr 2 5 π p 5 2 5 25π ø 78.54The area of the circle is exactly 25π square inches, whichis about 78.54 square inches.4. The radius of a circle is half of the diameter, so the radiusis }1 (16) 5 8 ft.2A 5 πr 2 5 π p 8 2 5 64π ø 201.06The area of the circle is exactly 64π square feet, which isabout 201.06 square feet.5. The radius of a circle is half of the diameter, so the radiusis }1 (23) 5 <strong>11</strong>.5 cm.2A 5 πr 2 5 π(<strong>11</strong>.5) 2 5 132.25π ø 415.48The area of the circle is exactly 132.25π squarecentimeters, which is about 415.48 square centimeters.6. A 5 πr 2 5 π p (1.5) 2 5 2.25π ø 7.07The area of the circle is exactly 2.25π square kilometers,which is about 7.07 square kilometers.7. A 5 πr 2 8. A 5 πr 2154 5 πr 2 380 5 πr 2154}π 5 380r2}π 5 r27.00 ø r <strong>11</strong>.00 ø rThe radius is aboutThe radius is about7 meters. <strong>11</strong> meters.9. A 5 πr 2676π 5 πr 2676 π} 5 rπ226 5 rThe radius is 26 centimeters, so the diameter is 52centimeters.10. The area of sector XZY should be divided by the area ofthe circle, not 3608. Also, the right side should be themeasure of the sector divided by 3608, or }7583608 .n}48 5 }7583608n 5 10The area of sector XZY is 10 square feet.C<strong>11</strong>. Area of small sector 5 }m ED3608 p πr25 }6083608 p π p 102 ø 52.36CArea of large sector 5 }m EGD p πr360823608 2 6085 } p π p 1036082 ø 261.80The areas of the small and large sectors are about 52.36square inches and 261.80 Csquare inches, respectively.12. Area of small sector 5 }m ED3608 p πr23608 2 25685 } p π p 1436082ø 177.88CArea of large sector 5 }m EGD p πr360825 }25683608 p π p 142 ø 437.87The areas of the small and large sectors are about 177.88square centimeters and 437.87 square centimeters,respectively.C13. Area of small sector 5 }m DE3608 p πr25 }13783608 p π p 282ø 937.31CArea of large sector 5 }m EGD p πr360823608 2 13785 } p π p 2836082 ø 1525.70The areas of the small and large sectors are about 937.31square meters and 1525.70 Csquare meters, respectively.14. Area of sector JK 5 }m JK p Area of (M360838.51 5 }1658 p Area of (M360884.02 ø Area of (MThe area of (M is about C84.02 square meters.15. Area of sector KLJ 5 }m JLK p Area of (M36083608 2 50856.87 5 } p Area of (M360866.04 ø Area of (MThe area of (M is about C66.04 square centimeters.16. Area of sector JK 5 }m JK3608 p πr212.36 5 }8983608 p πr250}π 5 r23.99 ø rThe radius of (M is about 3.99 meters.Copyright © by McDougal Littell, a division of Houghton Mifflin Company.372GeometryWorked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.<strong>Chapter</strong> <strong>11</strong>, continued17. Area of shaded region 5 Area of square2 2(Area of semicircle)5 6(6) 2 2 1 1808 }3608 (π p 32 ) 25 36 2 9π ø 7.73The area of the shaded region is about 7.73 squaremeters.18. Area of shaded region 5 Area of trapezoid19. A;2 Area of semicircle5 }1 (8)(16 1 20)22 1 1808 }3608 p (π p 42 ) 25 144 2 8π ø <strong>11</strong>8.87The area of the shaded region is about <strong>11</strong>8.87 squareinches.Area of the putting green 5 Area of square1 Area of sector of circle1 Area of rectangle5 3.5 2 1 1 908 }3608 p (π(3.5)2 ) 21 7(3.5)ø 12.25 1 9.62 1 24.5 ø 46.37The area of the putting green is about 46 square feet.20. A 5 πr 2260.67 5 πr 2260.67} 5 rπ29.<strong>11</strong> ø rThe radius of (M is about 9.<strong>11</strong> inches.21. C 5 2πr ø 2π(9.<strong>11</strong>) ø 57.24The circumference of (M Cis about 57.24 inches.22. Area of sector KML 5 }m KL p Area of (M3608 C 42 5 }m KL3608 p 260.67588 ø m C KLm C KL is about 588.23. Perimeter of blue region 5 arc length of C KNL1 C2(radius)5 }m KNL p 2πr 1 2r36083608 2 588ø } p 2π(9.<strong>11</strong>) 1 2(9.<strong>11</strong>)3608ø 48.02π 1 18.22 ø 66.24The perimeter of the blue region is about 66.24 inches.C24. Arc length of KL 5 }m KL3608 p 2πrø }588 p 2π(9.<strong>11</strong>) ø 9.223608The length of C KL is about 9.22 inches.25. Perimeter of red region 5 length of C KL 1 2(radius)ø 9.22 1 2(9.<strong>11</strong>) ø 27.44The perimeter of the red region is about 27.44 inches.26. Use the Pythagorean Theorem to find the radius of thecircle and base and height of the triangle.r 2 1 r 2 5 5 22r 5 25r 5 5 Ï} 7}2rr5 in.Area of shaded region 5 Area of circle 2 Area of triangle5 π p 1 5 Ï} 2}22 2 2 }1 }2221 5 Ï} 2ø 12.5π 2 12.5 ø 26.77The area of the shaded region is about 26.77 squareinches.27. Area of shaded region 5 2(Area of 1 shaded sector)5 2 1 C }m shaded arc p πr36082 21808 2 10985 21 } p π p (5.2) 2 36082ø 2(16.75) ø 33.51The area of the shaded region is about 33.51 square feet.28. Area of shaded region 5 Area of square2 4(Area of 1 circle)5 20 2 2 4(π p (5) 2 )5 400 2 100π ø 85.84The area of the shaded region is about 85.84 square inches.29. Area of shaded region 5 Area of bigger semicircle2 Area of smaller semicircle5 F }1808G3608 p (π p (17 1 17)22 F }18083608 p (π p 172 ) G5 578π 2 144.5π ø 1361.88The area of the shaded region is about 1361.88 squarecentimeters.30. Area of shaded region 5 Area of largest circle2 Area of second largest circle1 Area of third largest circle2 Area of smallest circle5 π p (2 1 2 1 2 1 2) 22 π p (2 1 2 1 2) 21 π p (2 1 2) 2 2 π(2) 25 64π 2 36π 1 16π2 4π ø 125.66The area of the shaded region is about 125.66 square feet.GeometryWorked-Out Solution Key373
<strong>Chapter</strong> <strong>11</strong>, continued31. Both triangles are right triangles by Theorem 10.9.Use the Pythagorean Theorem to find the length of thediameter of the circle3 2 1 4 2 5 d 225 5 d 25 5 dArea of shaded region 5 Area of circle p 2(Area of triangles)5 π 1 }22 5 2 2 21 }1 2 (3)(4) 25 6.25π 2 12 ø 7.63The area of the shaded region is about 7.63 squaremeters.C32. Area of the red region 5 }m RS3608 p πr25 }10883608 p π p (4 1 4)2 5 19.2πArea of the blue region 5 πr 2 5 π p 4 2 5 16πArea of the yellow region 5 Area of (P2 Area of red region2 Area of blue region5 π p (4 1 4) 2 2 19.2π2 16π 5 28.8πThe area of the red region is 19.2π square units, the areaof the blue region is 16π square units, and the area of theyellow region is 28.8π square units.33. Rewrite of the Perimeters of Similar Polygons Theorem:For any two circles, the ratio of their circumferences isequal to the ratio of their corresponding radii. Rewriteof the Area of Similar Polygons Theorem: For any twocircles, if the length of their radii is in the ratio ofa : b, then the ratio of their areas is a 2 : b 2 . All circles aresimilar, so you do not need to add that the circles mustbe similar.34. These sectors are not similar, therefore the studentshouldn’t have used Theorem <strong>11</strong>.7 (Areas of SimilarPolygons). The correct ratio of the arcs of the sectorsbounded by these areas is 2 : 1.35.rxsradius of large circle 5 r, radius of small circle 5 x, sideof square 5 sUse the Pythagorean Theorem to find the radius of thesmall circle in terms of r.r 2 1 r 2 5 s 22r 2 5 s 2r Ï } 2 5 s1 r Ï} 2}22 2 1 x 2 5 r 2x 2 5 r 2 2 r2 }2x 2 5 1 }2 r2x 5 Î } 1 }2 r2rrrx 5 r Ï} 2}2Area of large circle}}Area of small circle 5 }πr2πx 5 πr 22 }π 1 r Ï} 2}22 5 πr22 }15 }1 5 }2}2 πr2 1 1}2The ratio of the area of the large circle to the area of thesmall circle is 2 : 1.36. Let r be the radius of the smaller circle.Length of C C FG 5 }m FG3608 Cp 2πr10 5 }m FG3608 Cp 2πr5}πr 5 }m FG3608Length of C C EH 5 }m EH3608 Cp 2πr30 5 }m EH p 2π(8 1 r)3608 C15}π(8 1 r) 5 }m EH3608Because m C FG 5 m C EH , set their equations equal to eachother to find the length of the radius.5}πr 5 15}π(8 1 r)5π(8 1 r) 5 15πr8 1 r 5 3r8 5 2r4 5 rNow substitute C4 5 r to find m C FG .10 5 }m FG3608 p 2π(4)C10}2(π)4 5 }m FG3608143.28 ø m C FGSo, m C FD 5 m C EH ø 143.28.Area of shaded region 5 Area of larger sector2 Area of smaller sector5 }143.28 p π p (8 1 4)36082 2 }143.28 p π p 436082ø 160So, the area of the shaded region is about 160 squaremeters.xss2Copyright © by McDougal Littell, a division of Houghton Mifflin Company.374GeometryWorked-Out Solution Key
<strong>Chapter</strong> <strong>11</strong>, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.Problem Solving37. The radius for the eye of the hurricane is }1 (20) 5 102miles.A 5 πr 2 5 π p 10 2 ø 314.16The area of land underneath the eye is about 314.16square miles.38. A 5 πr 213,656 5 πr 213,656}π5 r2210.08 ø rThe distance you walk around the edge is thecircumference of the pond.C 5 2πr ø 2π(210.08) ø 1319.97You walk about 1320 feet.39. a. A circle graph is appropriate because the data isalready in percentages.b. percentage p (3608) 5 central angleBus: 0.65 p (3608) 5 2348Walk: 0.25 p (3608) 5 908Other: 0.10 p (3608) 5 368The central angle for the bus is 2348, the central anglefor walking is 908, and the central angle for othermodes of transportation is 368.Bus65%Walk25%Other10%Measure of arcc. Area of sector 5 }} p πr36082Bus: 2348 }3608 p π(2)2 ø 8.17Walk: 908 }3608 p π(2)2 5 πOther: 368 }3608 p π(2)2 ø 1.26The area of the bus sector is about 8.17 square inches,the area of the walking sector is π square inches, andthe area of the sector for other modes of transportationis about 1.26 square inches.40. The area of a tortilla with a 6 inch diameter isπ 1 }6 22 2 5 9π. The area of a tortilla with a 12 inch diameteris π 1 }12 2 2 2 5 36π. Because the 12 inch tortilla is 4 timeslarger, you need 4 times as much dough. So, you need41 }1 42 5 1 cup of dough to make a tortilla with a 12 inchdiameter.41. a. Sample answer:old “a” 5 }1 2 (π p 82 ) 1 }1 (12 1 19)(10 1 16) ø 3742new “a” 5 π(14) 2 1 3(22) ø 682The area of the old “a” is about 374 square units, thearea of the new “a” is about 682 square units, and thepercentage increase in interior area is about 82%.b. area of old “a”: 75.5(66) 5 4983area of new “a”: 85(76) 5 64606460 2 4983percent increase 5 } ø 0.296 ø 30%4983Sample answer: No; the percent increase of the area of“a” is about 30%, which is much less than the percentincrease of the interior area.42. Area of 12-inch diameter pizza: π 1 12 }2 2 2 5 36π ø <strong>11</strong>3.10Because a 12-inch diameter pizza has an area of about<strong>11</strong>3 square inches and feeds you and two friends, eachperson eats about <strong>11</strong>3 4 3 5 38 square inches of pizza.Area of 10-inch diameter pizza: π 1 10 }2 2 2 5 25π ø 78.54Area of 14-inch diameter pizza: π 1 14 }2 2 2 5 49π ø 153.94If you want to feed yourself and 7 friends, you needabout 8(38) 5 304 square inches of pizza.a. In order to have 304 square inches of pizza you couldbuy four 10-inch pizzas, two 14-inch pizzas, or two10-inch pizzas and one 14-inch pizza.Price of four 10-inch pizzas 5 4($6.99) 5 $27.96Price of two 14-inch pizzas 5 2($12.99) 5 $25.98Price of two 10-inch pizzas and one 14-inch pizza5 2($6.99) 1 $12.99 5 $26.97To spend as little money as possible, buy two 14-inchpizzas.b. You should buy two 10-inch pizzas and one 14-inchpizza. Two 10-inch pizzas and one 14-inch pizza isthree pizzas total, the amount of pizza you need, and ischeaper than than three 14-inch pizzas.Circumference ofc.four 10-inch pizzas 5 4 12 p π p }10 2 2 5 40π ø 125.66Circumference oftwo 14-inch pizzas 5 2 1 2 p π p 14 }2 2 5 28π ø 87.96Circumference oftwo 10-inch pizzasand one 14-inchpizza5 2 1 2 p π p 10 }2 2 1 1 2π p 14 }2 25 34π ø 106.81You should buy four 10-inch pizzas because thecircumference is greatest, which means more crust.43. a. The height is equal to the radius, the base is equal tohalf of the circumference and the area is the base timesthe height.h 5 r, b 5 }1 (2πr) 5 πr, r p πr 5 πr22GeometryWorked-Out Solution Key375
<strong>Chapter</strong> <strong>11</strong>, continuedb. Sample answer: When you cut the circle into 16congruent sectors, you are not losing area. Whenyou rearrange the 16 pieces of the circle to make aparallelogram, you can determine that the area is πr 2 .Because no area is lost when you make the circle intoa parallelogram, the area of the circle will be the sameas the area of the parallelogram, or πr 2 .44. Let black diameter 5 x, blue diameter 5 y, and reddiameter 5 z. Using Pythagorean Theorem, you can sayx 2 1 y 2 5 z 2 .base 5 ⏐9 2 2⏐ 5 7 unitsheight 5 ⏐16 2 2⏐ 5 14 unitsA 5 }1 2 bh 5 }1 (7)(14) 5 49 square units251.ABy221 xArea of triangle 5 1 }2 xyArea of semicircle z 5 }18083608 p 1 π p 1 }22 z 2 2 5 }πz28Area of semicircle y 5 }18083608 p 1 π p 1 }22 y 2 25 }πy28Area of semicircle x 5 }18083608 p 1 π p 1 }22 x 2 2 5 }πx28Area of non-shaded areas 5 Area of semicircle z2 Area of triangle5 }πz28 2 }1 2 xyArea of shaded crescent 5 Area of semicircle y1 Area of semicircle x2 Area of non-shaded areas5 }πy28 1 }πx28 2 1 }πz28 2 }1 2 xy 25 π(y2 1 x 2 )}82 1 πz2 }8 1 1 }2 xy 25 }πz28 2 }πz28 1 }1 2 xy 5 }1 2 xySo, because }1 2 xy 5 }1 xy, the sum of the area of the two2shaded crescents equals the area of the triangle.Mixed Review45. m∠DFG 5 908 by the definition of an altitude.46. m∠FDG 5 198 because the altitude bisects ∠EDG,which means m∠FDE 5 m∠FDG.47. FG 5 3 cm because the altitude bisects } EG ,so EF 5 FG.48. 49.50.yCbase 5 ⏐28 2 (23)⏐ 5 5 unitsheight 5 ⏐210 2 3⏐ 5 13 unitsCA 5 }1 2 bh 5 }1 (5)(13) 5 32.5 square units2Quiz <strong>11</strong>.4–<strong>11</strong>.5 (p. 761)1. Arc length of C C AB 5 }m AB 788p 2πr 5 }3608 3608 p 2π 1 }142 2 ø 9.53The length of C AB is about 9.53 meters.Arc length of C C2. }}GE 5 }m GEC 360836}C 5 }10283608127.06 ø CThe circumference of (F is about 127.06 inches.Arc length of C C3. }}JK 5 }m JK2πr 360829}2πr 5 }658360810,440 5 130πr25.56 ø rThe radius of (L is about 25.56 feet.4. Area of shaded region 5 Area of larger circle2 Area of smaller circle5 π p 33 2 2 π p <strong>11</strong> 25 968π ø 3041.06The area of the shaded region is about 3041.06 squaremeters.5. Area of shaded region 5 2(Area of each shaded sector)5 2 1 638 }3608 p π p (8.7)2 2 ø 83.23The area of the shaded region is about 83.23 squareinches.6. Area of shaded region 5 Area of circle2 Area of rhombus5 π p 1 }6 22 2 2 }1 2 (3)(6)5 9π 2 9 ø 19.27Copyright © by McDougal Littell, a division of Houghton Mifflin Company.2A1BxThe area of the shaded region is about 19.27 squarecentimeters.376GeometryWorked-Out Solution Key
<strong>Chapter</strong> <strong>11</strong>, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.Lesson <strong>11</strong>.6<strong>11</strong>.6 Guided Practice (pp. 762–764)1. The center of the polygon is P, a radius is } PY or } PX ,the apothem is } PQ , and the central angle is ∠XPY.2. m∠XPY 5 }36084 5 908, m∠XPQ 5 }1 m∠XPY 5 458,2and m∠PXQ 5 1808 2 908 2 458 5 4583. Use the Pythagorean Theorem to find the base of thetriangle.6.58ss 2 1 6.5 2 5 8 2s 2 5 21.75s ø 4.66The base of the triangle is about 4.66, so the length ofone side of the pentagon is 2(4.66) ø 9.32. A pentagonhas 5 sides, so P ø 5(9.32) ø 46.6 units.A 5 }1 2 aP 5 }1 (6.5)(46.6) ø 151.452The perimeter of the regular pentagon is about 46.6 unitsand the area is about 151.5 square units.4. A decagon has 10 sides, so P 5 10(7) 5 70 units.Because each of the 10 triangles that make up the polygonis isosceles, each apothem bisects the side length andcentral angle. Each side length is 7 units, so the bisectedside length is }1 (7) 5 3.5 units. Each central angle is2is }360810 5 368, so the bisected angle is }1 (368) 5 188.2To find the length of the apothem a use the sine functionwith one of the bisected isosceles triangles.188atan 188 5 }3.5aa ø 10.773.5A 5 }1 2 aP ø }1 (10.77)(70) ø 376.952The perimeter for the regular decagon is 70 units and thearea is about 377.0 square units.5. The measure of the central angle is }3608 5 1208. The3bisected angle is 608. To find the side length of thetriangle, use trigonometric ratios.6085stan 608 5 }s 5s p tan 608 5 s5 Ï } 3 5 5The regular triangle has side length 2(5 Ï } 3 ) 5 10 Ï } 3 .So, the perimeter is P 5 3s 5 3(10 Ï } 3 ) 5 30 Ï } 3 .A 5 }1 2 aP ø }1 2 (5)(30 Ï} 3 ) ø 129.9 square units.The perimeter for the regular triangle is 30 Ï } 3 units andthe area is about 129.9 square units.6. You can use a special right triangle to solve Exercise 5.<strong>11</strong>.6 Exercises (p. 765–768)Skill Practice1. F 2. ∠AFE3. 6.8 units 4. 5.5 units5. To find the measure of a central angle of a regularpolygon with n sides, divide 3608 by the number ofsides n of the polygon.6. }360836085 368 7. }10 18 5 2088. }3608245 158 9.3608}7 ø 51.4810. ∠GJH is a central angle. So m∠GJH 5 3608 }8 5 458.<strong>11</strong>. } JK is an apothem, which makes it an altitude of isoscelesnGJH. So, } JK bisects ∠GJH andm∠GJK 5 }1 m∠GJH 5 22.58.212. m∠KGJ 5 1808 2 908 2 22.58 5 67.5813. ∠EJH is 3 central angles combined.So m∠EJH 5 3(458) 5 1358.14. P 5 3(12) 5 36 unitsA 5 }1 2 aP 5 }1 2 (2 Ï} 3 )(36) ø 62.4The area of the regular triangle is about 62.4 squareunits.15. P 5 9(6.84) 5 61.56 units. Use the PythagoreanTheorem to find the apothem a.10 a 103.426.84a 2 1 3.42 2 5 10 2a 2 5 88.3036a ø 9.4Area 5 }1 2 aP ø }1 (9.4)(61.56) ø 289.32The area of the regular nonagon is about 289.3 squareunits.GeometryWorked-Out Solution Key377
<strong>Chapter</strong> <strong>11</strong>, continued16. Use the Pythagoream Theorem to find the length of aside of the regular heptagon.2.52.7720. The measure of the central angle is }3608 5 728. The5bisected angle is }1 (728) 5 368. To find the side length2of the pentagon, use a trigonometric ratio.4.1 368 sss 2 1 2.5 2 5 2.77 2s 2 5 1.4229s ø 1.19The length of the base of the triangle is about 1.19,so the length of one side of the polygon is about2(1.19) 5 2.38.A 5 }1 2 a p ns ø }1 (2.5)(7)(2.38) ø 20.82The area of the regular polygon is about 20.8 squareunits.17. 7.5 is not the length of one side of the hexagon, it is thelength of half of one side (or the base of the triangle). 7.5should be doubled to get the actual side of the hexagon.A 5 }1 2 a p ns 5 }1 (13)(6)(15) 5 585 units218. B; The measure of the central angle of a dodecagonis }360812 5 308. The bisected angle is }1 (308) 5 158. You2know the side length is 8, so the base of the triangle is1}(8) 5 4. Use a trigonomic ratio to find the apothem.2tan 158 5 }4 a4a 5 }tan 15819. The measure of the central angle is }3608 5 458. The8bisected angle is }1 (458) 5 22.58. Use trigonometric2ratios to find the apothom a and the side length s ofthe triangle.sin 22.58 5 }s 2020 p sin 22.58 5 scos 22.58 5 }a 2020 p cos 22.58 5 a22.5820a 20sThe regular octagon has side length2(20 p sin 22.58) 5 40 sin 22.58.P 5 8(40 p sin 22.58) ø 122.5 unitstan 368 5 }s 4.14.1 p tan 368 5 sThe regular pentagon has side length 5 2(4.1 p tan 368)5 8.2 p tan 368. So, the perimeter is 5(8.2 p tan 368)ø 29.8 units, and the area is A 5 }1 2 aP 5 }1 2 (4.1)(29.8)ø 61.1 square units.21. The measure of the central angle is }3608 ø 51.48. The7bisected angle is }1 (51.4) ø 25.78. The side of the2heptagon is 9, so the base of the triangle is }1 (9) 5 4.5.2To find the apothem use a trigonometric ratio.a25.784.5tan 25.78 5 }4.5aa ø 9.35So, the perimeter is 7(a) 5 63 units, and the area isA 5 }1 2 aP 5 }1 (9.35)(63) ø 294.5 square units.222. There is enough information to find the area. Themeasure of the central angle is }3608 5 408. The bisected9angle is }1 (408) 5 208. Because the perimeter is 18 inches2and a nonagon has 9 sides, each side length is18 4 9 5 2 inches. The side length of the nonagon is2 inches, so the base of the triangle is }1 (2) 5 1 inch.2To find the apothem, use a trigonometric ratio.a2081tan 208 5 }1 a ø 2.8The area is A 5 }1 2 aP 5 }1 (2.8)(18) ø 24.8 square inches.2Copyright © by McDougal Littell, a division of Houghton Mifflin Company.A 5 }1 2 aP 5 }1 (20 p cos 22.58)(122.5) ø <strong>11</strong>31.8 units22The perimeter of the regular octagon is about 122.5 unitsand the area is about <strong>11</strong>31.8 square units.378GeometryWorked-Out Solution Key
<strong>Chapter</strong> <strong>11</strong>, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.23. You need to know the apothem and side length to findthe area of the regular square. You can use the methodsof special right triangles or trigonometry to find theapothem length. The measure of the central angle is3608}4 5 908 and the bisected angle is }1 (908) 5 458.2To find the length of the apothem and side length, use aspecial triangle14x Ï } 2 5 14458 45814xxx 2x 5 14Ï } 2 p Ï} 2Ï } 2 5 14 Ï} 2} 5 7 Ï } 22The regular square has side length s 5 2(7 Ï } 2 ) 5 14 Ï } 2 .So, the area is A 5 }1 2 a p ns 5 }1 2 (7 Ï} 2 )(4)(14 Ï } 2 )5 392 square units.24. You need to know the apothem to find the area ofthe regular hexagon. You can use the methods of thePythagorean Theorem, special triangles, or trigonometry.The measure of the central angle is }3608 5 608 and the6bisected angle is }1 (608) 5 308. The side of the hexagon2is 10, so the base of the triangle is }1 (10) 5 5. To find the2apothem, use a special triangle.106085308a2x608xSo, x 5 5. The apothem is 5 Ï } 3 .308x 3The area is A 5 }1 2 a p ns 5 }1 2 (5 Ï} 3 )(6)(10) ø 259.8 squareunits.25. You need to know the side length to find the area of thedecagon. You can use the methods of the PythagoreanTheorem or trigonometry. To find the side length, use thePythagorean Theorem.8.4x8x 2 1 8 2 5 8.4 2x 2 5 6.56x 5 Ï } 6.56The regular decagon has side length s 5 2 Ï } 656 .So, the area isA 5 }1 2 a p ns 5 }1 2 (8)(10)(2 Ï} 6.56 ) ø 204.9 square units.45826. Area of unshaded region 5 Area of circle2 Area of square5 π p 14 2 2 3925 196π 2 392 ø 223.8The area of the unshaded region in Exercise 23 is about223.8 square units.27. The measure of the central angle is }3608 5 728 and the5measure of the bisected angle is }1 (728) 5 368. The side2length of the pentagon is 12, so the base of the triangleis }1 (12) 5 6. Use trigonometric ratios to find the length2of the apothem and the radius of the circle.r6368 asin 368 5 6 }r6r 5 }sin 368tan 368 5 6 }a6a 5 }tan 368Area of shaded region 5 Area of circle2 Area of pentagon5 π p r 2 2 1 1 }2 a p ns 265 π p 1 }sin 3682 22 F }1 62 1 } (5)(12)tan 3682 Gø 79.6 square unitsThe area of the shaded region is about 79.6 square units.28. The measure of the central angle is }3608 5 1208 and the3measure of the bisected angle is }1 (1208) 5 608. Use a2special triangle to find the apothem and size length ofthe triangle.a6088308b608x2x308x 3So, x 5 4. The apothem a is 4 and the side length b ofthe triangle is 4 Ï } 3 .The length of a side of the equilateral triangle is2(4 Ï } 3 ) 5 8 Ï } 3 .Area of shaded region 5 Area of circle 2 Area of triangle5 π p r 2 2 1 }2 p a p ns5 π p 8 2 2 1 }2 (4)(3)(8 Ï} 3 )5 64π 2 48 Ï } 3ø <strong>11</strong>7.9 square unitsThe area of the shaded region is about <strong>11</strong>7.9 square units.GeometryWorked-Out Solution Key379
<strong>Chapter</strong> <strong>11</strong>, continued29. The measure of the central angle is 608, so the bisectedangle is }1 (608) 5 308. Use a special triangle to find the2radius of the circle and base of the smaller triangle.34.hs308x 32xr3082 3608 b2x308x 3608 xSo, the base b of the triangle is 2 and the radius r is2(2) 5 4.The base of the larger triangle is 2(2) 5 4.Area of shaded region 5 Area of sector2 CArea of triangle5 }m arc3608 p πr22 F }1 (base of larger n)(a) G25 608 }3608 p π p 42 2 F 1 }2 (4)(2 Ï} 3 ) G5 }8 3 π 2 4 Ï} 3ø 1.4 square unitsThe area of the shaded region is about 1.4 square units.30. The circle has a radius of Ï } 25 5 5. The central angle ofthe pentagon is }3608 5 728 and the bisected angle is51}(728) 5 368. Use trigonometric ratios to find the side2length and apothem of the pentagon.2y22 (4, 22)5x3685acos 368 5 }a sin 368 5 }b 555 cos 368 5 a 5 sin 368 5 bThe regular pentagon has side length2b 5 2(5 p sin 368) 5 10 p sin 368. So, the area isA 5 }1 2 a p ns 5 }1 (5 p cos 368)(5)(10 p sin 368)2ø 59.4 square units.31. True. Because the radius is fixed and the circle aroundthe n-gons also stays the same, more and more of thecircle gets covered up as n gets larger.32. True. Because the hypotenuse of the right trianglerepresents the radius and the leg represents the apothem,the apothem must always be less than the radius.33. False. The radius can be equal to the side length, like in aregular hexagon.bb or ss2608xUse a special triangle to find the height of the triangle.So, x 5 }b 2 , s 5 b, and h 5 b Ï} 3}2 .Because b 5 s, substitute that into the height you justfound: h 5 b Ï} 3}2 5 s Ï} 3}2 . Use the formula A 5 }1 bh to find2the area in terms of s. A 5 }1 2 (s) 1 s Ï} 3}22 5 Ï} 3 s}24 .aaaIn an equilateral triangle, each altitude creates twocongruent triangles. So, each altitude is also a median.Looking at the figure the apothem is }1 the altitude of3the equilateral triangle. You found the height of thetriangle above, s Ï} 3}2 , so the apothem is }1 31 s Ï} 3}22 .Substituting into the formula A 5 }1 a p ns, you get2A 5 1 }2 1 1 }31 s Ï} 3}222 p 3 p s 5 Ï} 3 s}24square units.35. Shaded area 5 Area of hexagon 2 Area of pentagon1 Area of square2 Area of triangleHexagon: The central angle measures }3608 5 608 and the6bisected angle measures }1 (608) 5 308. The side2length of the hexagon is 8, so the base of thetriangle is }1 (8) 5 4. Use a special triangle2to find the apothem.6084308a2x608xSo, x 5 4 and a 5 4 Ï } 3 .So, the area of the hexagon is308x 3A 5 1 }2 a p ns 5 1 }2 (4 Ï} 3 )(6)(8) 5 96 Ï } 3 square units.Copyright © by McDougal Littell, a division of Houghton Mifflin Company.380GeometryWorked-Out Solution Key
<strong>Chapter</strong> <strong>11</strong>, continuedPentagon: The central angle measures }3608 5 728 and thesbisected angle measures }1 (728) 5 368. Use a2trigonometric ratio to find the apothem.368 a37. The apothem of the octagon is 1 1 0.2 5 1.2 centimeters.The measure of the central angle is }36088 5 458and the bisected angle measures }1 (458) 5 22.58.2Use a trigonometric function to find the side length ofthe octagon.b41.2 cmCopyright © by McDougal Littell, a division of Houghton Mifflin Company.tan 368 5 }4 a4a 5 }tan 368So, the area of the pentagon isA 5 }1 2 a p ns 5 }1 42 1 } (5)(8) ø <strong>11</strong>0 squaretan 3682units.Square: The area of the square is A 5 s 2 5 8 2 5 64square units.Triangle: Look at the smaller triangle used for thehexagon. The height of the triangle is theapothem of the smaller triangle used for thehexagon. So, thearea 5 A 5 }1 2 bh 5 }1 2 (8)(4 Ï} 3 )5 16 Ï } 3 square units.Problem Solving36.8 in.Shaded area 5 96 Ï } 3 2 <strong>11</strong>0 1 64 2 16 Ï } 3ø 92.6The area of the shaded region is about 93square units.a. The measure of the central angle is }3608 5 608 and the6bisected angle measures }1 (608) 5 308. Use a special2triangle to find the apothem.8 in.608308 a 2x608x308 x 3So, x 5 4. The apothem is 4 Ï } 3 .b. A regular hexagon is made up of 6 equilateraltriangles, so the side length of the hexagon is the sameas the radius of the hexagon, or 8. So, the perimeter isP 5 6(8) 5 48 inches and the area isA 5 1 }2 aP 5 1 }2 (4 Ï} 3 )(48) ø 166 square inches.22.58tan 22.58 5 }b 1.21.2 p tan 22.58 5 bThe regular octagon has side lengths 5 2b 5 2(1.2 p tan 22.58) 5 2.4 p tan 22.58.Area of silver border 5 Area of octagon 2 Area of circle5 }1 a p ns 2 πr225 }1 (1.2)(8)(2.4 p tan 22.58) 2 π(1)22ø 1.6 square centimetersThe apothem is 1.2 centimeters, the area of the octagonis about 4.8 square centimeters, and the area of the silverborder is about 1.6 square centimeters.38. Sample prediction: The pentagon will have the greatestarea and the circle will have the smallest area.a. Area of circle 5 πr 2 5 π 1 13 }2 2 2 ø 132.7 in. 2b. Area of triangle 5 }1 2 bh 5 }1 (18)(15) 5 135 in.22c. The measure of the central angle is }3608 5 728 and the5bisected angle measures }1 (728) 5 368. The side length2of the pentagon is 9 inches, so the base of the triangleis }1 (9) 5 4.5 inches. Use a trigonomic ratio to find2the apothem.4.5 in.368 atan 368 5 }4.5a4.5a 5 }tan 368So, the area is A 5 }1 2 a p ns 5 }1 4.52 1 } (5)(9)tan 3682ø 139.4 in. 2The area of the circle is about 132.7 square inches, thearea of the triangle is 135 square inches, and the areaof the pentagon is about 139.4 square inches.GeometryWorked-Out Solution Key381
<strong>Chapter</strong> <strong>11</strong>, continued39. A regular pentagon has 5 sides of equal length. So, theside length of the smaller pentagon is 15 4 5 5 3 inches.The measure of the central angle is }3608 5 728 and the5bisected angle measures }1 (728) 5 368. Because the side2length of the smaller pentagon is 3 inches, the base of thetriangle is }1 (3) 5 1.5 inches. Use a trigonometric ratio2to find the apothem.1.5 in.368 atan 368 5 }1.5a1.5a 5 }tan 368So, the area of the small trivet isA 5 }1 2 aP 5 }1 1.52 1 } (15) ø 15.5 square inches.tan 3682area of small trivet (perimeter of small trivet)2}} 5 }}area of large trivet (perimeter of large trivet) 215.5}}area of large trivet 5 }15225 243.0 ø area of large trivetThe area of the small trivet is about 15.5 square inchesand the area of the large trivet is about 43.0 square inches.40. a. A B1 in.b. The measure of the central angle is }3608 5 608 and the6bisected angle measures }1 (608) 5 308. Because the2side length is 1 inch, the base of the triangle is1}(1) 5 0.5 inch. Use a trigonometric ratio to find2the apothem.308 a0.5 in.tan 308 5 }0.5a0.5a 5 }tan 308So, the area of the hexagon isA 5 }1 2 a p ns 5 }1 0.52 1 } (6)(1) ø 2.6 square inches.tan 3082Area of shaded region 5 Area of circle2 Area of hexagonø π p 12 2 2.6ø 0.54 square inchThe area of the hexagon is about 2.6 square inchesand the area of the shaded region is about 0.54 squareinches.c. Draw } AB with length 1 inch. Open compass to 1 inchand draw a circle with that radius. Using the samecompass setting, mark off equal parts along the circle.Connect every other intersection or connect 2 adjacentpoints with the center of the circle.41. Let x be the side length of a regular hexagon. A hexagonhas 6 central angles, so 6 traingles are formed. Themeasure of a central angle is }3608 5 608. The bisected6angle has a measure of }1 (608) 5 308. Draw one of the26 triangles formed by the central angles and find the sidelength s and an expression in terms of a for x.s1x2308axcos 308 5 a }sas 5 }cos 308a5 }1 Ï} 3}1}2tan 308 5x}a2a p tan 308 5 x2a 1 Ï} 3}2 25 a p 2 }Ï } 35 2a Ï} 3}332 5 x2a Ï } 3} 5 x3The side length x of the hexagon is equal to the distances from the center of the hexagon to a vertex. So, eachof the 6 triangles is an equilateral triangle.42. Sample answer: a regular hexagon can be broken into6 equilateral triangles, 2 isosceles trapezoids, and3 parallelograms.Area of hexagon 5 }1 382 a p ns 5 }1 ( 2 2 Ï} 3 )(6)(2) 5 6 Ï } 3Area of 6 equilateral triangles 5 6 F }1 G 2 bh5 6 1 1 }1 22 (2)( Ï } 3 ) 25 6[ Ï } 3 ]5 6 Ï } 3Copyright © by McDougal Littell, a division of Houghton Mifflin Company.382GeometryWorked-Out Solution Key
<strong>Chapter</strong> <strong>11</strong>, continued32Area of 2 isosceles trapezoids 5 2 F }1 G2 h(b 1 1 b 2 )5 2 F 1 }2 ( Ï} 3 )(2 1 (2 1 2)) G5 2 F 3 Ï } 3 G5 6 Ï } 3The length of a side of the hexagon is2b 5 2(0.26 p tan 308) 5 0.52 p tan 308 ø0.3 centimeter. So, the area of the cell isA 5 }1 2 a p ns 5 }1 2 (0.26)(6)(0.3)ø 0.234 square centimeter.c. 100 cells p 0.234 square centimeter/cell 523.4 square centimeters(1 decimeter)223.4 square centimeters p }}(10 centimeter) 25 0.234 square decimeterd. From part (c), 100 cells have an area of about0.234 square decimeter. To find how many cells are in1 square decimeter, divide 100 by 0.234. So, there areapproximately 427 cells per square decimeter.45. a. triangle:332Area of 3 parallelograms 5 3(bh) 5 3(2)( Ï } 3 ) 5 6 Ï } 33 3320 cmh20 cm60810 cm20 cmtan 608 5 }h 1010 p tan 608 5 hA 5 }1 2 bh ø }1 (20)(10 p tan 608) ø 173.2 cm22square:Copyright © by McDougal Littell, a division of Houghton Mifflin Company.243. Because PC is a radius, then P is the circumcenter ofn ABC. Let E be the midpoint of } AB . Because P is thecircumcenter, } CE and } BD are perpendicular bisectors andmedians of n ABC. By the Concurrency of Medians of aTriangle Theorem, CP 5 }2 3 CE and BP 5 }2 3 BD.So, DP 5 }1 3 BD. Because } CE and BD } are perpendicularbisectors, BP 5 CP. So, CP 5 }2 3 BP and 2DP 5 }2 3 BD.So, radius CP is twice the apothem DP.44. a. The average distance across a cell is }2.655 0.52 centimeter.b. The apothem is }1 2 diameter 5 }1 2 (0.52)5 0.26 centimeter. The measure of the central angleis }3600 5 608 and the bisected angle measures61} (608) 5 308. Use a trigonometric ratio to find the2length of the base of the triangle.btan 308 5 }0.260.26 p tan 308 5 b0.26 cm308b15 cm15 cmA 5 s 2 5 15 2 5 225 cm 2regular pentagon:tan 368 5 }6 a6a 5 }tan 368a3686 cm12 cm12 cmA 5 }1 2 aP ø }1 62 1 } (60) ø 247.7 cmtan 3682 2GeometryWorked-Out Solution Key383
<strong>Chapter</strong> <strong>11</strong>, continuedregular hexagon:3810 cma1 cm0.5 cmtan 308 5 5 }aa10 cm5a 5 }tan 3083085 cm215A 5 }1 2 aP ø }1 } (60) ø 259.8 cmtan 3082 2regular decagon:Not drawn to scaletan 38 5 }0.5a0.5a 5 }tan 38The area of a 60-gon is A 5 }1 2 aP 5 }1 2 1 }0.5 (60)tan 382ø 286.2 square centimeters.120-gon: The central angle for 120-gon measures3608}128 5 38, the bisected angle measures }1 (38) 5 1.58,2and the side length is 60 4 120 5 0.5 centimeter.Because the side length is 0.5 centimeter, the baseof the triangle is }1 (0.5) 5 0.25 centimeter. Use a2trigonometric ratio to find the apothem.6 cm1.58tan 188 5 }3 a3a 5 }tan 188188a6 cm 3 cmA 5 }1 2 aP ø }1 32 1 } (60) ø 277tan 1882The area of the equilateral triangle is about 173.2 squarecentimeters. The area of the square is 225 squarecentimeters. The area of the regular pentagon is about247.7 square centimeters. The area of the regularhexagon is about 259.8 square centimeters. The area ofthe regular decagon is about 277 square centimeters.The area increases as the number of sides increase.b. 60-gon: The central angle for a 60-gon measures3608}60 5 68, the bisected angle measures }1 (68) 5 38,2and the side length is 60 4 60 5 1 centimeter.Because the side length is 1 centimeter, the base ofthe triangle is }1 (1) 5 0.5 centimeter. Use a2trigonometric ratio to find the apothem.a0.25 cm0.5 cmNot drawn to scaletan 1.58 5 }0.25a0.25a 5 }tan 1.58The area of a 120-gon is A 5 }1 2 aP 5 }1 }21 0.25 (60)tan 1.582ø 286.4 square centimeters.c. AArea3002401801206000 2 4 6 8 10Number of sides of polygonA circle will have the greatest area.C 5 2πr60 5 2πr30}π 5 rA 5 πr 2 5 π 1 }30 π 2 2 5 }900 ø 286.5 cmπ2The area of the circle will be about 286.5 squarecentimeters.nCopyright © by McDougal Littell, a division of Houghton Mifflin Company.384GeometryWorked-Out Solution Key
<strong>Chapter</strong> <strong>11</strong>, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.46. The inscribed n-gon would have a central angle of3608}n and a bisected angle of }1 2 1 }3608n 2 , or }1808n . Whenbroken down into reference triangles, the radius of thepolygon would equal the radius of the circle.rba1808nsin 1 }1808n 2 5 }b rr p sin 1 1808 }cos 1 }1808n 2 5 }a rn 2 5 b r p cos 1 }180n 2 5 aThe side length of the inscribed n-gon iss 5 2b 5 2r p 1 sin 1808 }n 2 .So, the area of the inscribed n-gon isA 5 1 }2 a p ns 5 1 }5 r 2 n cos 1 1808 }2 1 r p cos 1 }1808n 22 (n) 12r p sin 1 }1808n 22n 2 sin 1 }1808n 2 .The circumscribed n-gon would have a central angle of3608}n and a bisected angle of }1 2 1 }3608n 2 , or }1808n . Whenbroken down into reference triangles the apothem wouldequal the radius of the circle.br1808ntan 1 }1808n 2 5 }b rr p tan 1 }1808n 2 5 bThe side length of the circumscribed n-gon iss 5 2b 5 2r p tan 1 }1808n 2 . So, the area of thecircumscribed n-gon isA 5 1 }2 a p ns 5 1 }2 (r)(n) 12r p tan 1 }1808n 22 5 r 2 n tan 1 }1808n 2 .The area between the two polygons is equal to the areaof the inscribed polygon subtracted from the area ofthe circumscribed polygon. So, the area between thepolygons isr 2 n tan 1 }1808n 2 2 r 2 n cos 1 }1808n 2 sin 1 }1808n 2 , which factorsto r 2 n F tan 1 }1808n 2 2 cos 1 }1808n 2 sin 1 }1808n 2 G .Mixed Review47. They are independent events, so P(A and B)5 P(A) p P(B).10}18 p }10 18 5 }100324 5 }25 8148. They are dependent events, so P(A and B)5 P(A) p P(B⏐A).6}18 p }2 17 5 }12 306 5 }2 5149. }widthlength 5 }9 ft18 ft 5 }1 250. }widthlength51. }widthlength52. AD12 cm5 }42 cm 5 }2 736 in.5 }45 in. 5 }4 52<strong>11</strong>yBCxThe slopes of } AD and } BC are undefined, so } AD i } BC .Slope of } 1 2 3AB :}1 2 (23) 5 }224 5 }212 5 2 }1 2Slope of } 23 2 (21)DC :}1 2 (3)5 22 }4 5 }212 5 2 }1 2The slopes of } AB and } DC both equal 2 1 }2 , so}AB i } DC . Because } AD i } BC and } AB i } DC , ABCD is aparallelogram.Spreadsheet Activity <strong>11</strong>.6 (p. 769)1. The regular n-gons approach the shape of a circle as thevalue of n gets very large. The more sides there are, thesmaller the central angle is, which makes the shapemore circular.2. As n gets very large, the perimeter approaches 2π. Theradius is 1 and as n increases, the value of n p sin 1 1808 }n 2approaches π, which suggests that if the number of sideswere infinitely large, the circumference would be 2πr.3. As n gets very large, the areas appraoch π. The radiusis 1, so r 2 5 1. As n increases, the value ofn p sin 1 }1808n 2 p cos 1 }1808n 2 approaches π, which suggeststhat if the number of sides were infinitely large, the areawould be πr 2 .Lesson <strong>11</strong>.7Investigating Geometry Activity <strong>11</strong>.7 (p. 770)1. Answers will vary.2. Answers will vary.3. As the number of tosses increases, the experimentalprobability gets closer to the theoretical probability.GeometryWorked-Out Solution Key385
<strong>Chapter</strong> <strong>11</strong>, continued<strong>11</strong>.7 Guided Practice (pp. 772–773)1. P(point is on } RT ) 5 Length of } RT}}Length of } PQ5 }1 ; 0.1, 10%102. P(Point is on } TS ) 5 Length of } TS}}Length of } PQ5 }1 ; 0.5; 50%23. P(Point is on } PT ) 5 Length of } PJ}}Length of } PQ5 }2 ; 0.4, 40%54. P(Point is on } RQ ) 5 Length of } RQ}}Length of } PQ5 }7 ; 0.7, 70%105⏐21 2 (22)⏐}}⏐5 2 (25)⏐⏐4 2 (1)⏐5 }⏐5 2 (25)⏐ 5 }5 10⏐21 2 (25)⏐5 }}⏐5 2 (25)⏐ 5 }4 105⏐5 2 (22)⏐}⏐5 2 (25)⏐5. The longest you can wait is 8:49 2 8:43 5 6 minutes.P(You get to the station by 8:58)Favorable waiting time5 }}maximum waiting time 5 }6 12 5 }1 2The probability that you will get to the station by8:58 is }1 , or 50%.26. Area of black region 5 Area of larger black circle 2Area of large white circle 1 Area of smaller black circle2 Area of smaller white circle5 π(8 1 8 1 8 1 8 1 8) 2 2 π(8 1 8 1 8 1 8) 21 π(8 1 8 1 8) 2 2 π(8 1 8) 25 1600π 2 1024π 1 576π 2 256π 5 896πArea of black regionP(arrow lands in black region) 5 }}Area of target5 }896π1600π 5 }56 100 5 }14 25The probability that the arrow lands in the black regionis }14 or 56%.257. Area of woods 5 Total area of park 2 Area of field5 58.5 2 30 5 28.5Area of woodsP(ball in woods) 5 }}Total area of park ø }28.558.5 5 }285585 5 }19 39The probability that the ball is in the field is about 19 }39 , or48.7%.<strong>11</strong>.7 Exercises (pp. 774–777)Skill Practice1. If an event cannot occur, its probability is 0. If an event iscertain to occur, its probability is 1.2. Geometric probabilities are determined by comparinggeometric measures. A probability, found by dividingthe number of favorable outcomes by the total numberof possible outcomes, deals with events, not geometricmeasures. In a geometric probablity, you divide the“favorable measure” by the total measure.3. P(Point k is on } AD ) 5 Length of } AD}}Length of } AE⏐3 2 (212)⏐5 }}⏐12 2 (212)⏐ 5 }15 245 }5 ; 0.625; 62.5%84. P(Point k is on } BC ) 5 Length of } BCLength of AE }5⏐23 2 (26)⏐}}⏐12 2 (212)⏐5 }3 24 5 }1 ; 0.125; 12.5%85. P(Point k is on } DE ) 5 Length of } DE}}Length of AE } 5 ⏐12 2 3⏐}}⏐12 2 (212)⏐5 }9 24 5 }3 ; 0.375; 37.5%86. P(Point k is on } AE ) 5 Length of } AE ⏐12 2 (212)⏐Length of AE } 5 }}⏐12 2 (212)⏐5 }24 5 1; 1.0; 100%247. AD 1 DE 5 AE, so }5 8 1 }3 5 1. The sum of their8probabilities will add up to 1, or 100%.8. Area of shaded region 5 Area of circle 2 Area of square5 πr 2 2 s 2 5 π p 2 2 2 (2 Ï } 2 ) 25 4π 2 8Area of shaded regionP(Point is in shaded region) 5 }}Area of entire figure5 }4π 2 8 ø 0.36 or 36%.4πThe probability that a randomly chosen point lies in theshaded region is about 36%.Area of smaller triangle9. P(Point is in shaded region) 5 }}Area of larger triangle1}25(6)(7)} 5 }21 1}2 (12)(14) 84 5 }1 4The probability that a randomly chosen point lies in theshaded region is }1 , or 25%.410. Area of shaded region 5 Area of trapezoid2 Area of rectangle5 1 }2 h(b 1 1 b 2 ) 2 b 1 h5 }1 (5)(8 1 20) 2 (8)(5)25 70 2 40 5 30Area of shaded regionP(Point is in shaded region) 5 }}Area of trapezoid5 }30 70 5 }3 7The probability that a randomly chosen point lies in theshaded region is }3 , or about 43%.7Copyright © by McDougal Littell, a division of Houghton Mifflin Company.386GeometryWorked-Out Solution Key
<strong>Chapter</strong> <strong>11</strong>, continued<strong>11</strong>. In the numerator, the area of the unshaded rectangle atthe bottom should be subtracted. This rectangle has aheight of 7 2 5 5 2.20. Use The Pythagorean Theorem to find the height of thetriangle.Copyright © by McDougal Littell, a division of Houghton Mifflin Company.10(7) 2 }1 2 π(5)2 2 2(10)50 2 12.5π}} 5 } ø 0.15310(7)70The probability that a randomly chosen point in thefigure lies in the shaded region is about 15.3%.12. Sample answer: The area of the north side of the island isabout 31.5 square units. The area of the south side of theisland is about 32.5 square units. The area of the wholeisland is about 64 square units.Area of north side of island13. P(location is on north side) 5 }}Total area of island5 }31.564 5 }315640 5 }63 128The probability that a randomly chosen location on theisland lies on the north side is }63 , or about 49.2%.128Area of south side of island14. P(location is on south side) 5 }}Total area of island5 }32.564 5 }325640 5 }65 128The probability that a randomly chosen location on theisland lies on the south side is }65 , or about 50.8%.12815. The shaded triangle is similar to the whole triangle bythe AA Similarity Postulate. The ratio of sides is }6 12 5 }1 2and }7 14 5 }1 . Because you know the ratio of sides, you2can use the Area of Similar Polygons Theorem to find thedesired probability. The ratio of areas is 1 2 : 2 2 5 1 : 4,which is the desired probability.16. x 2 6 ≤ 1x ≤ 7P(x ≤ 7) 5 }5 717. 1 ≤ 2x 2 3 ≤ 54 ≤ 2x ≤ 82 ≤ x ≤ 4P(2 ≤ x ≤ 4) 5 }2 7x18. }2 ≥ 7x ≥ 14P(x ≥ 14) 5 019. 3x ≤ 27x ≤ 9P(x ≤ 9) 5 1h453 2 1 h 2 5 5 2h 2 5 16h 5 4Find the area of the entire figure.Area of entire figure 5 Area of shaded region1 Area of trapezoid5 1 }2 b π h t 1 1 }2 h 2 (b 1 1 b 2 )5 }1 2 (3)(4) 1 }1 (3)(5 1 7)25 6 1 18 5 24Write a ratio of the areas to find the probability.Area of shaded regionP(Point is in shaded region) 5 }}Area of entire figure5 }6 24 5 }1 4The probability that a randomly chosen point in thefigure lies in the shaded region is }1 , or 25%.421. Set up a ratio of side lengths to find the base length ofthe smaller triangle.base of smaller triangle height of smaller triangle}} 5 }}base of larger triangle height of larger trianglebase of smaller triangle (12 2 8)}} 5 }1412base of smaller triangle 5 }14 3Find the area of the shaded region.Area of shaded region 5 Area of larger triangle2 Area of smaller triangle5 1 }2 b i h i 2 1 }2 b s h s5 }1 2 (14)(12) 2 }1 2 14 }2 32 (4)5 84 2 }28 3 5 }2243Write a ratio of the areas to find the probability.Area of shaded regionP(Point lies in shaded region) 5 }}Area of larger triangle224}35 }84 5 }8 9The probability that a randomly chosen point in thefigure lies in the shaded region is }8 , or about 88.9%.9GeometryWorked-Out Solution Key387
<strong>Chapter</strong> <strong>11</strong>, continued22. Use a special triangle to find the radius of the circle.8458r458x458x 2x458So, r 5 8 Ï } 2 . The base of the triangle formed by the radiiis 8 1 8 5 16.Find the area of the shaded region.Area of shaded region 5 Area of sector2 CArea of triangle5 }m arc3608 p πr2 2 }1 2 bh5 }9083608 p π p (8 Ï} 2 ) 2 2 }1 2 (16)(8)5 32π 2 64Write a ratio of the areas to find the probability.Area of shaded regionP(Point lies in shaded region) 5 }}Area of circle32π 2 64 32π 2 645 π (8 Ï } 5 }2 )2 128πø 0.091The probability that a randomly chosen point in thefigure lies in the shaded region is about 0.091, or 9.1%23. D;Area not in A Area of U 2 Area of AP(x not in A) 5 } 5 }}Total Area Area of UCircumference of24. P(Point is an arc) 5 }}C arccircumference of cirlcem C arc}3608 p 2πr5 } 5 }8082πr 3608 5 }2 9m C arc}Area of sectorP(Point is in sector) 5 }}Area of circle 5 3608 p πr2}πr 25 }8083608 5 }2 9The probability that a randomly chosen point on thecircle lies on the arc is }2 , or about 22.2%. The probability9that a randomly chosen point in the circle lies in thesector is }2 , or about 22.2%. The probabilities do not9depend on the radius because the circumference and thearea of a circle end up canceling with the values in thedenominator.25. C 5 2πr188.5 5 2πr30 ø rArea of circle: A 5 πr 2 5 π(30) 2 5 900πThe measure of the central angle of a regular hexagonis }36086 5 608. The bisected angle measures }1 (608) 5 308.2Use a 308-608-908 triangle to find the apothem a of thehexagon and base b of the triangle.308ab30308x 32x608xSo, a 5 15 Ï } 3 and b 5 15. The side length of thehexagon is s 5 2b 5 2(15) 5 30. The area of thehexagon isA 5 }1 2 a p ns 5 }1 2 (15 Ï} 3 )(6)(30) 5 1350 Ï } 3 .Area of hexagonP(Point is in the hexagon) 5 }}Area of circle5 1350 Ï} 3} ø 0.827900πThe probability that a randomly chosen point in the circlelies in the inscribed regular hexagon is about 0.827,or 82.7%.26. Area of circle 5 πr 2The measure of the central angle for a regular octagonis }3608 5 458 and the bisected angle measures81} (458) 5 22.58. Use trigonometric ratios to find the2apothem and side length.a22.58brsin 22.58 5 b }rr p sin 22.58 5 bcos 22.58 5 a }rr p cos 22.58 5 aThe side length of the regular octagon iss 5 2b 5 2r p sin 22.58. So, the area of the octagon isA 5 }1 2 a p ns 5 }1 (r p cos 22.5)(8)(2r p sin 22.58)25 8r 2 p cos 22.58 p sin 22.58Area of octagonP(Point is in polygon) 5 }}Area of circle5 8r2 cos 22.58 p sin 22.58}}πr 2 ø 0.90The probability that a randomly chosen point in the circlelies in the inscribed polygon is about 0.90, or 90%.Copyright © by McDougal Littell, a division of Houghton Mifflin Company.388GeometryWorked-Out Solution Key
<strong>Chapter</strong> <strong>11</strong>, continued27. Because points A and B are end points of a diameterof (D, the angle they form when connected to pointC will always be a right angle, by Theorem 10.9. So,the probability that nABC is a right triangle is 100%.m∠CAB will be less than or equal to 458 half the time,so the probability that m∠CAB ≤ 458 is 50%.28.yProblem SolvingArea of inner square30. P(Dart hits inner square) 5 }}Total area of board 5 }6218 25 }36 324 5 }1 9Area outside inner square but inside circle5 Area of circle 2 Area of inner squareCopyright © by McDougal Littell, a division of Houghton Mifflin Company.2<strong>11</strong>b 15 ⏐3 2 2⏐ 5 1, b 25 ⏐3 2 0⏐ 5 3,and h 5 ⏐2 2 0⏐ 5 2xArea of trapezoid 5 }1 2 h(b 1 1 b 2 ) 5 }1 (2)(1 1 3) 5 42x 2 1 y 2 ≥ 4 is the equation of a circle with r 5 Ï } 4 5 2.1} of the circle is in the shaded region, so the area of the8shaded circle is 1 }8 (π(2)2 ) 5 1 }2 π.P(the point (x, y) in the solution region is in x 2 1 y 2 ≥ 4)Area of trapezoid 2 Area of shaded circle5 }}}}Area of trapezoid5 4 2 }1 2 π} ø 0.607 or 60.7%4The probability that a point (x, y) in the solution region isin x 2 1 y 2 ≥ 4 is about 60.7%.29. An expression for each step isstep n 2 <strong>11</strong> 0.5(1 2 step ), which simplifies ton 2 10.5 step n 2 <strong>11</strong> 0.5, where n is the current step number.Step 1: 0.5 paintedStep 2: 0.5(0.5) 1 0.5 5 0.75 5 75%Step 3: 0.5(0.75) 1 0.5 5 0.875 5 87.5%Step 4: 0.5(0.875) 1 0.5 5 0.9375 5 93.75%Step 5: 0.5(0.9375) 1 0.5 5 0.96875 5 96.875%Step 6: 0.5(0.96875) 1 0.5 5 0.984375 5 98.4375%Step 7: 0.5(0.984375 1 0.5 5 0.9921875 5 99.21875%Step 8: 0.5(0.9921875) 1 0.5 5 0.996093755 99.609375%Step 9: 0.5(0.99609375) 1 0.5 5 0.9980468755 99.8046875%Step 10: 0.5(0.998046875) 1 0.5 5 0.99902343755 99.90234375%Step <strong>11</strong>: 0.5(0.9990234375) 1 0.5 5 0.9995<strong>11</strong>718755 99.95<strong>11</strong>71875%After <strong>11</strong> steps, the painted portion of the stick is greaterthan 99.95%. So, n ≥ <strong>11</strong> steps.5 π p 1 18 }2 2 2 2 6 2 5 81π 2 36P(Dart hits outside inner square but inside circle)Area outside inner square but inside circle5 }}}}Total area of board81π 2 365 } ø 0.674324The probability that it hits inside the inner square is }1 9 , orabout <strong>11</strong>.1%. The probability that it hits outside the innersquare but inside the circle is about 0.674 or 67.4%.wait time31. a. P(bus waiting) 5 }}time between arrivals 5 }4 10 5 }2 5The probability that there is a bus waiting when apassenger arrives at a random time is }2 , or 40%.5time between arrivals 2 waiting timeb. P(bus not waiting) 5 }}}time between arrivals5 }10 2 4 5 }6 10 10 5 }3 5The probability that there is not a bus waiting when apassenger arrives at a random time is }3 , or 60%.5amount of time before lunch32. P(fire drill for lunch) 5 }}}total amount of time at school 5 }4 7The probability that the fire drill begins before lunch is 4 }7 ,or about 57.1%.33. P(You miss the call)amount of overlap while practicing drums5 }}}}Interval of time for phone call⏐7:10 2 7:00⏐5 }}⏐8:00 2 7:00⏐10 minutes5 }60 minutes 5 }1 6The probability that you missed your friend’s call is }1 6 ,or about 16.7%.34. a. First count all the whole squares. There are about 64whole squares. Next make groups of partially coveredsquares, so the combined area of each group is about1 square unit. There are about 14 more squares, whichmakes a total of 64 1 14 5 78 squares. Each side ofeach square is 2 kilometers, so the area of each squareis 2 2 5 4 square kilometers. 78 squares p 4 km 2 /square5 312 square kilometers. The area of the plannedlanding region was about 312 square kilometers.GeometryWorked-Out Solution Key389
<strong>Chapter</strong> <strong>11</strong>, continued35.Area of craterb. P(Beagle 2 landing in crater) 5 }}Area of landing regionπ p 1 1 }22 25 } ø 0.0025312The probability that Beagle 2 landed in the crater isabout 0.0005 or 0.25%.AC 608 F 6081.5 cmB3 cmArea of (F 5 πr 2 5 π p 3 2C5 9πArea of sector DEF 5 }m DE3608 p πr2 5 }6083608 p π p 325 1.5πArea of (C 5 πr 2 5 π p (1.5) 2C5 2.25πArea of sector ABC 5 }m AB3608 p πr25 }6083608 p π(1.5)2 5 0.375πArea of sectorSmaller circle: P(Point in sector) 5 }}Area of circle5 0.375π }2.25π5 }1 or about 16.7%6Area of sectorLarger circle: P(Point in sector) 5 }}Area of circle5 }1.5π9π5 }1 or about 16.7%6The probability of a randomly selected point being inthe sector stays the same when the central angle staysthe same and the radius of the circle doubles. As theradius doubles, the area of the sector and entire circlequadruples, but they are still praportional by the Areasof Similar Polygons Theorem. Because they are stillpraportional, the probabilities remain equal.36. P(both pieces are at least 1 in.) is the same as1 2 P(one piece is less than 1 inch) because both piecesmay not be less than 1 inch. The probability of producinga piece less than 1 inch is }1 6 . So, 1 2 }1 6 5 }5 ø 83.3% is6the probability we want to find.DE37. a. From Exercise 30, the probability that one dart hits theyellow square is }1 . Because the throws are9independent, you multiply the probabilities.[P(Dart hits yellow square)] 2 5 1 }1 92 2 5 }1 81The probability that both darts hit the yellow squareis }1 , or about 1.2%.81b. P(Dart hits outside the circle)Area of outer square 2 Area of circle5 }}}Area of outer square5 182 2 π p 9}218 5 324 2 81π2 }324Because the throws are independent, you multiply theprobabilities.P(Dart hits yellow square)p P(Dart hits outside the circle) 5 }1 324 2 81πp }9 324ø 0.024The probability that the first dart hits the yellow squareand the second hits outside the circle is about 2.4%.c. From Exercise 30. The probability that one dart hitsinside the circle but outside the yellow square isabout 0.674. Because the throws are independent, youmultiply the probabilities.[P( Dart hits inside the circle but outside the yellowsquare)] 2 ø (0.674)(0.674) ø 0.454The probability that both darts hit inside the circle butoutside the yellow square is about 45.4%.38. P(Part of bird call erased) 5 Time of silence1 Time of bird callTime of erased data1 }}Total time of tape8 min 1 5 min 1 10 min5 }}60 min5 }23 or about 38.3%60P(All of the bird call erased)Time of bird call 1 Time of erased data5 }}}Total time of tape5 }5 1 10 5 }15 60 60 5 }1 or 25%4The probability that part of the bird call was erasedis }23 , or about 38.3%. The probability that all of the bird60call was erased was }1 , or 25%.4Mixed Review39. Sample answer:Copyright © by McDougal Littell, a division of Houghton Mifflin Company.concavehexagonconcavepentagon390GeometryWorked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.<strong>Chapter</strong> <strong>11</strong>, continued40. The intersection of plane DCH and Plane ADE is @##$ DH .41. Sample answer: A plane that appears to be parallel toplane ADH is plane BCF.42. Use the Pythagorean Theorem to find the height of theparallelogram.0.3 m0.3 2 1 h 2 5 1.5 2h 2 5 2.16h ø 1.471.5 mA 5 bh ø (0.6)(1.47) ø 0.88hThe area of the parallelogram is about 0.88 square meter.43. A 5 }1 2 h(b 1 1 b 2 ) 5 }1 (7)(9 1 12) 5 73.52The area of the trapezoid is 73.5 square feet.44. The central angle of the pentagon measures 3608 }5 5 728and the bisected angle measures }1 (728) 5 368. Use a2trigonometric ratio to find the apothem a.4 in.368acos 36 5 a }44 p cos 368 5 aA 5 }1 2 a p ns 5 }1 (4 p cos 368)(5)(4.6) ø 37.22The area of the pentagon is about 37.2 square inches.Quiz <strong>11</strong>.6–<strong>11</strong>.7 (p. 777)1. The central angle of the regular pentagon measures3608}5 5 728 and the bisected angle measures }1 (728) 5 368.2Use a trigonometric ratio to find the apothem a.a368 17 cmcos 368 5 a }1717 p cos 368 5 aA 5 }1 2 a p ns 5 }1 (17 p cos 368)(5)(20) ø 687.662The area of the regular pentagon is about 687.7 squarecentimeters.2. The central angle of the regular octagon measures3608} 5 458 and the bisected angle measures81} (458) 5 22.58. Use a trigonometric ratio to find the2apothem a and base b.a22.58b25 msin 22.58 5 b }25cos 22.58 5 a }2525 p sin 22.58 5 b 25 p cos 22.58 5 aThe side length of the octagon is s 5 2b 5 50 p sin 22.58.A 5 }1 2 a p ns 5 }1 (25 p cos 22.5)(8)(50 p sin 22.5)2ø 1767.8The area of the regular octagon is about 1767.8 squaremeters.Area of shaded circle3. P(point is in shaded region) 5 }}Area of large circle22 25 π π 1 3 9}}}rs2π rl 5 2 }π 1 }102 2 5 4 π2 }100}4 π5 9 }100The probability that a randomly chosen point in thefigure lies in the shaded region is }9 , or 9%.100Area of shaded region4. P(Point is in shaded region) 5 }}Area of rectangleArea of rectangle 2 Area of trapezoid5 }}}Area of rectangle5 bh 2 }1 2 h t (b 1 1 b 2 )}} 5 8(5) 2 }1 (2)(3 1 5)2}} 5 }32bh8(5) 40 5 }4 5The probability that a randomly chosen point in thefigure lies in the shaded region is }4 , or 80%.5Mixed Review of Problem Solving (p. 778)1. a. The central angle measures }3608 5 608 and the6bisected angle measures }1 (608) 5 308. Use a special2triangle to find the length of the apothem.a308 308x 30.25 cm0.5 cm2x608xThe apothem of a small mirror is Ï} 3}4 meters.GeometryWorked-Out Solution Key391
<strong>Chapter</strong> <strong>11</strong>, continuedb. A 5 }1 2 a p ns 5 }1 21 Ï} 3}42 (6) 1 }1 22 ø 0.65The area of one of the small mirrors is about0.65 square meter.c. 91(0.65) ø 59.15The area of the primary mirror is about 59.2 squaremeters.2. 24.1; The central angle measures }3608 5 728 and the5bisected angle measures }1 (728) 5 368. Use a2trigonometric ratio to find the side length of the pentagon.Area of red dish4. a. P(It will land it in the dish) 5 }}Area of jar22 25 π π 1 }5}rd2π rj 5 2 }π 1 }202 2 225}45π} 5 }1 400}4 π 16The probability that it will land in the dish is }1 16 ,or 6.25%.b. Because the probability is }1 16 , multiply 400 coins by }1 16to find out how man prizes you would expect people towin.7 368 btan 368 5 }b 77 p tan 368 5 bThe side length of the pentagon is s 5 2b 5 14 p tan 368Area of shaded region 5 Area of pentagon 2 Area of circle5 }1 2 a p ns 2 πr2 5 }1 (7)(5)(14 p tan 368) 2 π(7)225 245 p tan 368 2 49π ø 24.1The area of the shaded region is about 24.1 Csquare units.3. a. Area of illuminated water surface 5 }m arc3608 p πr25 }21683608 p π p 202ø 754The area of the water’s surface that is illuminated bythe lighthouse is about 754 square miles.b. The shortest distance between the lighthouse andthe boat is the height of the triangle formed whenthe endpoints of the boat and illuminated water areconnected to the lighthouse (which are the radii of thecircle). The base of the original triangle is 31, so thebase of the reference triangle is }1 (31) 5 15.5. Use the2Pythagorean Theorem to find the height of the triangle.15.5 mi20h20 2 5 h 2 1 15.5 2159.75 5 h 212.64 ø hThe closest distance between the lighthouse and theboat is about 12.6 miles.4001 }162 1 5 25You would expect people to win 25 prizes if 400 coinsare dropped.5. Because the triangle is a right isosceles triangle, you canuse a special triangle to find out the base lengths of thetriangle, which are the diameters of the two smallersemi circles. From the diagram, x 5 4. So, the diameterof the small semicircles is 4.458x 2xx458P 5 }1 (perimeter of large semicircle)21 2(perimeter of small semicircle)5 1 }2 (πd) 1 2 1 1 }2 πd s 2 5 1 }2 π(4 Ï} 2 ) 1 π(4)5 2 Ï } 2 π 1 4πA 5 }1 (Area of large semicircle)21 2(Area of small semicircle) 1 Area of triangle2225 }1 2 π 1 d 2L}22 1 21 }1 2 π 1 d c}2 1 1 }2 x25 }1 2 π 1 4 Ï} 2}22 2 1 π 1 }4 22 2 1 }1 2 (4)2 5 8π 1 8The perimeter of the figure is 2 Ï } 2 π 1 4π units and thearea of the figure is 8π 1 8 square units.6. Sample answer: The area of the given fan isA 5 }12083608 p π p 92 ø 84.8 cm 2 .A fan with a radius of 8 cm and an arc of 1608 wouldhave an area of A 5 }16083608 p π p 82 ø 89.4 cm 2 . Becausethe area of the new fan is greater than the area of thegiven fan, it does a better job of cooling you.Copyright © by McDougal Littell, a division of Houghton Mifflin Company.392GeometryWorked-Out Solution Key
<strong>Chapter</strong> <strong>11</strong>, continued<strong>Chapter</strong> <strong>11</strong> Review (pp. 780–783)1. A sector of a circle is the region bounded by two radii ofa circle and their intercept arc.2. Either pair of parallel sides can be used as the basesof a parallelogram and the height is the perpendiculardistance between them.3. An apothem of the square is } XZ .4. A radius of the square is } XY .5. A 5 bh 5 10(6) 5 60 square units6. Use the Pythagorean Theorem to find the base of thetriangleb10. R25QTS1yd 15 QS 5 ⏐21 2 (23)⏐ 5 2 andd 25 RT 5 ⏐3 2 (22)⏐ 5 5A 5 }1 2 d 1 d 2 5 }1 2 (2)(5) 5 5The area of the kite is 5 square units.<strong>11</strong>.yDEx32681Copyright © by McDougal Littell, a division of Houghton Mifflin Company.32 2 1 b 2 5 68 2b 2 5 3600b 5 60A 5 }1 2 bh 5 }1 (32)(60) 5 960 square units27. Height of triangle: 40 2 16 5 24A 5 Area of square 1 Area of triangle5 s 2 1 1 }2 bh5 16 2 1 }1 (16)(24) 5 448 square units28. A 5 147 in. 2 and h 5 1.5bA 5 1 }2 bh147 5 }1 2 b(1.5b)147 5 0.75b 2196 5 b 214 5 bThe base of the triangle is 14 inches and the height is1.5(14) 5 21 inches.9. yPN<strong>11</strong>LMxb 5 LM 5 ⏐6 2 2⏐ 5 4 and h 5 ⏐4 2 2⏐ 5 2A 5 bh 5 (4)(2) 5 8The area of the parallelogram is 8 square units.21GFxb 15 GF 5 ⏐3 2 1⏐ 5 2, b 25 DE 5 ⏐5 2 (21)⏐ 5 6,and h 5 ⏐4 2 (22)⏐ 5 6A 5 }1 2 h(b 1 1 b 2 ) 5 }1 (6)(2 1 6) 5 242The area of the trapezoid is 24 square units.12. Ratio of sides 5 3 : 4Ratio of perimeters 5 3 : 4 and ratio of areas 5 3 2 : 4 2Area of red}Area of blue 5 }9 164.5}Area of blue 5 }9 165 9 : 168 5 Area of blueThe ratio of the perimeters is 3 : 4, the ratio of the areas is9 : 16, and the area of the blue triangle is 8 square feet.13. Ratio of sides 5 10 : 13Ratio of perimeters 5 10 : 13 and ratio of areas5 10 2 : 13 2 5 100 : 169Area of red}Area of blue 5 }10016990}Area of blue 5 }100169152.1 5 Area of blueThe ratio of the perimeters is 10 : 13, the ratio of the areasis 100 : 169, and the area of the blue quadrilateral is 152.1square centimeters.14. The ratio of the lengths of corresponding sides isÏ } 144 : Ï } 49 , or 12 : 7.15. C 5 πd94.24 5 πd94.24} 5 dπ30 ø dThe diameter of (F is about 30 feet.GeometryWorked-Out Solution Key393
<strong>Chapter</strong> <strong>11</strong>, continuedArc length of C C16. }}GH 5 }m GHC36085.5}C 5 }35836085.5(3608) 5 358(C)56.57 5 CThe circumference of (F is about 56.57 centimeters.Arc length of C CGH17. }} 5 }m GH2πr 3608Arc length of C GH}} 5 }<strong>11</strong>582π(13) 3608Arc length of C GH 5 }<strong>11</strong>58(26π)3608Arc length of C GH ø 26.09The length of C GH is about 26.09 inches.C18. Area of sector TWU 5 }m TWU p πr360825 2408 }368 p π p 92 ø 169.65The area of the blue shaded region is about 169.65 squareinches.19. Area of blue shaded region 5 Area of rectangle2 Area of semicircle5 bh 2 1 }2 πr25 6(4) 2 }1 2 π 1 }4 22 25 24 2 2π ø 17.72The area of the blue shaded region is about 17.72 squareinches.C20. Area of red sector 5 }m RQ3608 p πr227.93 5 }5083608 p π p r28 ø rFind the measure of the major arc.m C RQ 5 50, so m C RTQ 5 C3608 2 508 5 3108.Area of blue sector 5 }m RTQ p πr36082ø }31083608 p π(8)2 ø 173.14The area of the blue shaded region is about 173.14 squarefeet.21. The central angle is }3608 5 458 and the bisected angle8is }1 (458) 5 22.58. Use a trigonometric ratio to find the2side length.22.586 in.btan 22.58 5 }b 66 p tan 22.58 5 bThe side length s 5 2b 5 12 p tan 22.58An octagon has 8 sides, so the perimeter isP 5 8(12 p tan 22.58) 5 96 p tan 22.58 ø 39.8.A 5 }1 2 a p ns 5 }1 (6) p 8(12 tan 22.58) ø<strong>11</strong>9.32The perimeter of the platter is about 39.8 inches andthe area is about <strong>11</strong>9.3 square inches.22. The pentagon’s side length is 20 centimeters, so thelength of the base of the triangle is }1 (20) 5 102centimeters. Use the Pythagorean Theorem to findthe apothem.17 cm10 cma10 2 1 a 2 5 17 2a 2 5 189a 5 Ï } 189A 5 }1 2 a p ns 5 }1 ( 2 Ï} 189 )(5)(20) ø 687.39The area of the jigsaw puzzle is about 687.4 squarecentimeters.23. P(k is on } AB ) 5 length of } AB ⏐2 2 (22)⏐length of } 5 }AC ⏐5 2 (22)⏐ 5 }4 7The probability that k is on } AB is }4 7 .Area of shaded region24. P(Point is in shaded region) 5 }}Area of semicircle258}3608 p π p 1 }12 2 2 2 258}5 }}1}2 p π p 1 }122 2 5 36082 } ø 0.1391}2The probability that a randomly chosen point in thefigure lies in the shaded region is about 13.9%.Copyright © by McDougal Littell, a division of Houghton Mifflin Company.394GeometryWorked-Out Solution Key
<strong>Chapter</strong> <strong>11</strong>, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.25. P(Point is in shaded area)Area of triangle5 }}}Area of semicircle 1 Area of triangle1}25(6)(15)45}} 5 }1}2 (6)(15) 4.5π 1 45 ø 0.7612 π p 1 6 }22 2 1 1 }The probability that a randomly chosen point in thefigure lies in the shaded region is about 76.1%.26. P(Point is in shaded area)Area of rectangle 1 Area of semicircle5 }}}Area of square4 1 4 }22 1 1 }2 p π p 1 4 }22 25 }}4 5 8 1 2π2 } ø 0.89316The probability that a randomly chosen point in thefigure lies in the shaded region is about 89.3%<strong>Chapter</strong> <strong>11</strong> Test (p. 784)1. A 5 bh 5 7(4.7) 5 32.9The area of the parallelogram is 32.9 square centimeters.2. Use the Pythagorean Theorem to find the base of thetriangle.5 ft13 ftb5 2 1 b 2 5 13 2b 2 5 144b 5 12A 5 }1 2 bh 5 }1 (12)(5) 5 302The area of the triangle is 30 square feet.3. b 25 18 cm 1 9 cm 5 27 cmA 5 }1 2 h(b 1 1 b 2 ) 5 }1 (10)(18 1 27) 5 2252The area of the trapezoid is 225 square centimeters4. A 5 }1 2 h(b 1 1 b 2 ) 5 }1 (9)(8 1 15) 5 103.52The area of the trapezoid is 103.5 square meters.5. A 5 }1 2 d 1 d 2 5 }1 (32)(40) 5 6402The area of the rhambus is 640 square inches.6. A 5 }1 2 d 1 d 2 5 }1 (41)(67) 5 1373.52The area of the kite is 1373.5 square centimeters.7. b 5 3h and A 5 108 in 2A 5 bh108 5 3(h)(h)108 5 3h 236 5 h 26 5 hThe height of the parallelogram is 6 inches and the baseis 3(6) 5 18 inches.8. Ratio of perimeters 5 40 : 16 5 5 : 29. Ratio of corresponding side lengths5 ratio of perimeters 5 5 : 210. Ratio of areas 5 5 2 : 2 2 5 25 : 4<strong>11</strong>. Arc length of C C AB 5 }m AB 1088p 2πr 5 }360 3608 p 2π(17)ø 32.04The length of C AB is about 32.04 centimeters.Arc length of C C12. }}EHD 5 }m EHDC360864}C 5 }2108360864(3608) 5 C(2108)109.71 ø CThe circumference of (F is about 109.71 inches.Arc length of C C13. }}GH 5 }m GH2πr 3608C35}2π (27) 5 }m GH360835C3608 p }2π (27) 5 m GH74.278 ø m C GHThe measure of C GH is about 74.38.14. Find the measure of the major arc.m C QTR 5 3608 2 1058 5 2558CArea of sector QTR 5 }m QTR3608p πr 25 2558 }3608 p π p 82 ø 142.42The area of the shaded sector is about 142.42 squareinches.C15. Area of shaded sector ∠NM 5 }m LM p Area of (N360849 5 }688 p Area of (N3608259.41 ø Area of (NThe area of (N is about 259.41 square meters.GeometryWorked-Out Solution Key395
<strong>Chapter</strong> <strong>11</strong>, continuedC16. Area of sector RPS 5 }m RPS p πr3608236 5 }<strong>11</strong>483608 p πr26.02 ø rThe radius of (P is about 6.02 centimeters17. A hexagon has 6 sides. The perimeter is 18 inches, so theside length is 18 4 6 5 3 inches. The central angleis }36086 5 608 and the bisected angle is }1 (608) 5 308.2Use a special triangle to find the apothem.308a6081.5 in.308x 32x608xThe apothemis about 1.5 Ï } 3 ø 2.6 inches.A 5 }1 2 a p ns ø }1 2 (1.5 Ï} 3 )(6)(3) ø 23.38The area of the tile is about 23.4 square inches.18. P(Point is in red region)2(area of red semicircle)5 }}}Area of square 1 2(Area of seimicircle)21 }1 2 p π p 1 }1025π5 }}10 2 1 21 }1 2 p π p 1 }10 5 }2 2 2 100 1 25π2ø 0.44The probability that a randomly chosen point in thefigure lies in the red region is about 44%.19. P(Point is in blue region)Area of square 2 2(Area of red semicircle)5 }}}}Area of square 1 2(Area of semicircle)2 2 2 210 2 2 21 }1 2 p π p 1 }10100 2 25π5 }}10 2 1 21 }1 2 p π p 1 }10 5 }2 2 2 100 1 25π2ø 0.12The probability that a randomly chosen point in thefigure lies in the blue region is about 12%.2 2 2 2<strong>Chapter</strong> <strong>11</strong> Algebra Review (p. 785)1 hr1. 90 min p } 5 1.5 hour60 mind 5 rt when d 5 14.25 and t 5 1.514.25 5 r(1.5)14.25}1.5 5 rd 5 rt 5 1 14.25 }1.5 2 (2) 5 19The algebraic model is d 5 1 14.25 }1.5 2 (2). You can bike19 miles in 2 hours.2. 25% off means she paid 75% of the original price forthe jacket.Let j represent the original price of the jacket.12 1 0.75j 5 390.75j 5 27j 5 36The algebraic model is 12 1 0.75j 5 39. The originalcost of the jacket is $36.3. 29.50 1 0.25m 5 32.75 where m is the number ofadditional minutes over 2000.25m 5 3.25m 5 13The algebraic model is 29.50 1 0.25m 5 32.75.You used 13 additional minutes.4. 7.6(20) 1 12.1r ≥ 400 where r is the number of minutesspent running12.1r ≥ 248r ≥ 20.5The algebraic model is 7.6(20) 1 12.1r ≥ 400. Jaimeneeds to run 20.5 minutes or more to meet his goal.5. If the value of the car decreases 10% each year, it keeps90% (or 0.9) of its value per year.18,000(0.9) 5 5 A18,000(0.59049) 5 A10,628.82 5 AThe algebraic model is 18,000(0.9) 5 5 A. The value ofthe car after 5 years is $10,628.82.6. 5s 1 8a 5 2065 and s 5 a 1 62 where s 5 number ofstudent tickets sold and a 5 number of adult tickets sold.5(a 1 62) 1 8a 5 20655a 1 310 1 8a 5 206513a 5 1755a 5 135s 5 a 1 62 5 135 1 62 5 197The algebraic models are 5s 1 8a 5 2065 ands 5 a 1 62. There were 197 student tickets soldand 135 adult tickets sold.7. 0 5 216t 2 1 47t 1 6Quadratic formula:247 6 Ï }}47 2 2 4(216 p 6)}}} 5 247 6 Ï} 2209 1 384}}2 p (216)232ø 3.06 or 20.12The algebraic model is 0 5 216t 2 1 47t 1 6. It takesabout 3.06 seconds for the tennis ball to reach theground.Standardized Test Preparation (p. 787)1. a. The student should receive full credit because thesolution is correct and all the work is shown.Copyright © by McDougal Littell, a division of Houghton Mifflin Company.396GeometryWorked-Out Solution Key
<strong>Chapter</strong> <strong>11</strong>, continuedb. The student should receive partial credit because thereasoning and steps taken are correct, but the areas ofthe squares are incorrect (they should be 20 2 5 400and 36 2 5 1296, not 20 and 36).Standardized Test Practice (pp. 788–789)1. a. m∠region 5 3608 2 C908 5 270. So, m C region 5 2708Area of sector 5 }m region p πr360825 }27083608 p π p 182 ø 763.41The area of the sector with radius 18 feet is about763.41 square feet.b. The angle of the smaller sector arc is 1808 2 9085 908. The radius of the smaller sector is 18 2 125 6 feet because the radius is the difference betweenthe radius of the large sector Cand the width of the shed.Area of smaller sector 5 }}m smaller region p πr360825 }9083608 p π p 62 ø 28.27The area of the small sector is about 28.27 square feet.c. The area over which the dog can move is the areas ofboth sectors added together.A ø 763.41 1 28.27 ø 791.68 square feet.2. a. y<strong>11</strong> Cx3. a. Set the ratios of the actual areas of the trays equal tothe corresponding ratios of the areas. Then solve forthe unknown area.Area of small tray}}Area of medium tray 5 }2 3483}}Area of medium tray 5 }2 3724.5 5 Area of medium trayArea of small tray}}Area of large tray 5 }2 4483}}Area of large tray 5 }2 4966 5 Area of large trayThe area of the medium-sized tray is 724.5 squarecentimeters and the area of the large tray is 966 squarecentimeters.b. The ratio of the areas of the trays is 2 : 3 : 4.So, the ratio of the perimeters of the trays isÏ } 2 : Ï } 3 : Ï } 4 5 Ï } 2 : Ï } 3 : 2. Set the ratios of the actualperimeters of the trays equal to the corresponding ratiosof the perimeters. Then solve for the unknown area.Perimeter of small tray}}Perimeter of medium tray 5 Ï} 2Ï } 380}}Perimeter of medium tray 5 Ï} 2Ï } 398.0 ø Perimeter of medium trayPerimeter of small tray}}Perimeter of large tray 5 Ï} 2}2Copyright © by McDougal Littell, a division of Houghton Mifflin Company.The coordinates of its center are (6, 0).b. y<strong>11</strong> CThe center of the dilated circle is (12, 0). So, thecoordinates of the dilated circle are twice thecoordinates of the center of the original circle.c. Circumference of original circle 5 2πr 5 2π(3) 5 6πCircumference of dilated circle 5 2πr 5 2π(6) 5 12πArea of original circle 5 πr 2 5 π p 3 2 5 9πArea of dilated circle 5 πr 2 5 π p 6 2 5 36πThe circumference is twice as large because theradius doubled. Using the Areas of Similar PolygonsTheorem, the area is 4 times greater because the radiusdoubled.x80}}Perimeter of large tray 5 Ï} 2}2<strong>11</strong>3.1 ø Perimeter of large trayThe perimeter of the medium-sized tray is about 98.0centimeters and the perimeter of the large tray is about<strong>11</strong>3.1 centimeters.4. a. Because EFGH is a rhombus, the diagonals bisecteach other. Because } EG is a diagonal that is 6 unitslong with J as the center, EJ 5 }6 5 3 units. Because2}FH is a diagonal that is 8 units long with J as thecenter, FJ 5 }8 5 4 units. You know ∠ EJF is a right2angle because EFGH is a rhombus, so use thePythagorean Theorem to find EF.EF3J43 2 1 4 2 5 EF 225 5 EF 25 5 EFEF is 5 units.GeometryWorked-Out Solution Key397
<strong>Chapter</strong> <strong>11</strong>, continuedb. A 5 }1 2 d 1 d 2 5 }1 (6)(8) 5 242The area of EFGH is 24 square units.c. 24 5 EF p XYd. You know that EF 5 5 units, so use the equation24 5 EF p XY to solve for XY.24 5 5 p XY4.8 5 XYXY is 4.8 units. Because XY is a diameter of the circle,the radius is }1 (4.8) 5 2.4. So, the area of the inscribed2circle isA 5 π p r 2 5 π p (2.4) 2 5 5.76π ø 18.1 square units.5. D; JL 5 JK 1 KL6 5 JK 1 24 5 JKLength of JKRatio of lengths 5 }Length of JL 5 }4 6 5 }2 3 , or 2 : 3Ratio of areas 5 2 2 : 3 2 5 4 : 96. A; P(Point lies inside nMNR)Area of nMNR5 }}}}Area of nMNR 1 Area of MRST 1 Area of NPQR1}25(s)(s)}}<strong>11</strong> }5 2 s2}}2 (s)(s) 1 s2 1 s 2 5}2s25 0.27. 126; The area of the lawn is A 5 bh 5 (300)(150)5 45,000 square feet. Because each bag covers 5000square feet, you need 45,000 4 5000 5 9 bags. Becauseeach bag costs $14, you will spend 9($14) 5 $126.8. 1.5; Because ACDE is a square, } AC > } CD > } DE> } EA , so AC, CD, DE, and EA all equal 2 units and∠A is a right angle. Because AB 5 BC, AF 5 FE,AB 1 BC 5 AC, AF 1 FE 5 AE, and both AC andAE equal 2 units, AB, BC, AF, and FE equal 1 unit.Area of shaded region 5 Area of nAEC2 Area of nABF5 }1 2 (2)(2) 2 }1 (1)(1) 5 1.52The area of the shaded region is 1.5 square units.9. 96; C 5 πd8π 5 πd8 5 dBecause the diameter of each circle is 8 units, the radiusis }1 (8) 5 4 units. The length of the rectangle is2d 1 r 5 8 1 4 5 12 and the width of the rectangle isd 5 8. So, the area of the rectangle isA 5 bh 5 (12)(8) 5 96 square units.10. a. There are two red sectors of the circle that are 458, so908 of the circle is covered by red sectors. 908 is }1 4 of acircle. So, the probability that the arrow points to a redsector is 25%.b. Red should cover }2 of the spinner and blue should3cover }1 of the spinner.3x 5 }2 3 p 3608 5 2408 y 5 }1 p 3608 5 12083So, x should be 2408 and y should be 1208.<strong>11</strong>. a. Quadrilateral JKLM is a kite.}JL and } KM are diagonals and JL 5 3 p KM.A 5 1 }2 d 1 d 254 5 1 }2 (KM)(JL)54 5 }1 2 (KM)(3KM)36 5 KM 26 5 KMThe length of } JL is 3(KM) 5 3(6) 5 18 centimetersand the length of } KM is 6 centimeters.b. The ratio of the area of NPQR to the area of JKLMis }486 or 9 : 1. Because the ratio of areas is 9 : 1, the54ratio of lengths is Ï } 9 : Ï } 1 5 3 : 1.PNRBecause } NQ is similar to } JL and the ratio of lengthsis 3 : 1, } NQ is 3 times as large as } JL . SoNQ 5 3(18)5 54 centimeters. The length of } NQ is54 centimeters.QCopyright © by McDougal Littell, a division of Houghton Mifflin Company.398GeometryWorked-Out Solution Key
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