Chpter 1 Prerequisite Skills (p. 790) 1. The re of regulr polygon is given by the formul A 5 1 p P, where is the pothem nd P is the perimeter.. Two polygons re similr if their corresponding ngles re congruent nd the lengths of corresponding sides re proportionl.. tn 58 5 x 0 x 5 0 tn 58 x ø 1.99. cos 708 5 5 x 5 x 5 cos 708 x ø 1.6 5. sin 508 5 x 0 x 5 0 sin 508 x ø.98 6. r 5 m C 5 πr A 5 πr 5 π() 5 π() 5 π 5 π ø 1.57 ø 1.57 The circumference is bout 1.57 m nd the re is bout 1.57 m. 7. d 5 in., r 5 in. C 5 πr 5 π 1 5 π ø 9. A 5 πr 5 π 1 5 9 π ø 7.07 The circumference is bout 9. in. nd the re is bout 7.07 in.. 8. r 5 Ï 5 cm C 5 πr 5 π( Ï 5 ) 5 π Ï 5 ø 8.10 A 5 πr 5 π( Ï 5 ) 5 0π ø 6.8 The circumference is bout 8.1 cm nd the re is bout 6.8 cm. Lesson 1.1 Investigting Activity 1.1 (p. 79) 1.. fces meet t ech vertex. b. 5 fces meet t ech vertex.. The ngle mesures of ll the ngles in n equilterl tringle re 608, so for 6 of these ngles to meet mens the totl ngle mesure is 6 p 608 5 608. A solid cnnot hve vertex with 6 equilterl tringles becuse they would form plne.. Three congruent regulr hexgons meeting t vertex would result in plne being formed. The sum of the ngles ( p 108) is 608 which will not form convex vertex.. F 1 V 5 E 1 1.1 Guided Prctice (pp. 795 797) 1. The solid is formed by polygons, so it is polyhedron. The bse is squre, so it is squre pyrmid. It hs 5 fces, 5 vertices, nd 8 edges.. The solid hs curved surfce, so it is not polyhedron.. The solid is formed by polygons, so it is polyhedron. The two bses re congruent tringles, so it is tringulr prism. It hs 5 fces, 6 vertices, nd 9 edges.. The dodechedron hs 1 fces, 0 vertices, nd 0 edges. F 1 V 5 E 1 1 1 0 5 0 1 5 5. The cross section is tringle. 6. The cross section is circle. 7. The cross section is hexgon. 1.1 Exercises (pp. 798 801) Skill Prctice 1. The five Pltonic solids nd their fces re: regulr tetrhedron, fces; cube, 6 fces; regulr octhedron, 8 fces; regulr dodechedron, 1 fces; regulr icoshedron, 0 fces.. Euler s Theorem sttes the sum of the number of fces nd vertices of polyhedron is equl to the number of edges of the polyhedron plus two.. The solid is formed by polygons, so it is polyhedron. The bse is pentgon, so it is pentgonl pyrmid.. The solid is formed by polygons, so it is polyhedron. The two bses re congruent hexgons, so it is hexgonl prism. 5. The solid hs curved surfce, so it is not polyhedron. 6. The bses of the prism re tringles, not rectngles. The solid is tringulr prism. 7. 8. 99
Chpter 1, continued 9. 10. 11. n 1 1 5 18 1 1. 5 1 n 5 8 1 n 1 1 5 0 5 1 n 5 10 n 5 8 n 5 5 There re 8 fces. There re 5 vertices. 1. 10 1 16 5 n 1 1. n 1 1 5 0 1 6 5 n 1 n 1 1 5 5 n n 5 0 There re edges. There re 0 fces. 15. There re fces, vertices, nd 6 edges. 1 5 6 1 8 5 8 16. There re 5 fces, 5 vertices, nd 8 edges. 5 1 5 5 8 1 10 5 10 17. There re 5 fces, 6 vertices, nd 9 edges. 5 1 6 5 9 1 11 5 11 18. There re 5 fces, 6 vertices, nd 9 edges. 5 1 6 5 9 1 11 5 11 19. There re 8 fces, 1 vertices, nd 18 edges. 8 1 1 5 18 1 0 5 0 0. There re 8 fces, 1 vertices, nd 18 edges. 8 1 1 5 18 1 0 5 0 1. A cube is solid formed by six congruent fces, so by nme it is regulr hexhedron.. concve. concve. convex 5. 6. 7. circle rectngle tringle 8. A; The cross section prllel to the squre bse is lso squre. 9. Euler s Theorem proves this is incorrect. 1 6 0 1 10 Þ 8 Insted there should be fces, vertices, nd 6 edges. 1 5 6 1 8 5 8 0. C; A tringulr prism nd squre pyrmid both hve 5 fces. 1. D; An octgonl prism hs 10 fces, 16 vertices, nd edges.. 1 v 5 90 1 1 v 5 9 v 5 60 Euler s Theorem cn be used. The solid hs 60 vertices.. The plne should intersect the cube t n ngle so tht it touches ech of the six fces. The figure will look like: Problem Solving.. There re 10 vertices. b. 7 1 10 5 E 1 17 5 E 1 15 5 E There re 15 edges. 5. There re 18 edges nd 1 vertices. 8 1 1 5 18 1 0 5 0 6. The cross section is circle. 7. The cross section is squre. 8. The cross section is rectngle. 9. A polyhedron with vertices nd 6 edges is tringulr pyrmid. No, Euler s Theorem mens tht ll polyhedrons with vertices nd 6 edges hve the sme number of fces. 0.. The cross section is rectngle. b. The length of the rectngle is the hypotenuse of the right tringle formed by two sides of the cube. The length is Ï 6 1 6 5 Ï 7 5 6 Ï. P 5 6 1 6 1 6 Ï 1 6 Ï 5 1 1 1 Ï ø 8.97 The perimeter is pproximtely 8.97 inches. c. A 5 l p w 5 6 Ï p 6 5 6 Ï The re of the cross section is 6 Ï inches or bout 50.9 inches. 00
Chpter 1, continued 1.. The cross section is trpezoid. b. Yes; Smple nswer: 7. Yes; c. The cross section is squre. d. Yes; Smple nswer: 8. Yes;. Pltonic Solid Fces Vertices Edges F 1 V 5 E 1 Tetrhedron 6 Cube 6 8 1 Octhedron 8 6 1 Dodechedron 1 0 0 Icoshedron 0 1 0. No. Yes; 5. Yes; 6. Yes; 1 5 6 1 6 1 8 5 1 1 8 1 6 5 1 1 1 1 0 5 0 1 0 1 1 5 0 1 9.. There would be 7 fces, 10 vertices, nd 15 edges. b. There would be 7 fces, 10 vertices, nd 15 edges. c. The number of fces, vertices, nd edges would remin the sme. d. There would be 9 fces, 1 vertices, nd 1 edges. 50.. There would be 5 fces, 6 vertices, nd 9 edges. b. There would be 5 fces, 6 vertices, nd 9 edges. c. The number of fces, vertices, nd edges would remin the sme. d. Cutting two edges would mke 6 fces, 7 vertices, nd 11 edges. 51. The first solid hs 8 vertices nd ech vertex hs three 908 ngles. D 5 608 (908) 5 908 DV 5 90(8) 5 708 The second solid hs 10 vertices nd ech vertex hs two 908 ngles nd one 1088 ngle. D 5 608 (908) 1088 5 78 DV 5 7(10) 5 708 The third solid hs 1 vertices nd ech vertex hs two 908 ngles nd one 108 ngle. D 5 608 (908) 108 5 608 DV 5 60(1) 5 708 Mixed Review 5. x8 5 1 (78 1 818) 5 1 (158) 5 778 The vlue of x is 77. 5. 1808 68 5 1 (1098 1 x8) 1 5 1 (109 1 x) 68 5 109 1 x 159 5 x The vlue of x is 159. 01
Chpter 1, continued 5. 1808 x8 5 1 (588 1 1978) 180 x 5 1 (55) 180 x 5 17.5 x 5 5.5 The vlue of x is 5.5. 55. r 5 11 cm C 5 πr 5 π(11) 5 π ø 69.1 A 5 πr 5 π(11) 5 11π ø 80.1 The circumference of the circle is bout 69.1 cm nd the re is bout 80.1 cm. 56. d 5 8 in., r 5 8 5 1 in. C 5 πr 5 π(1) 5 8π ø 87.96 A 5 πr 5 π(1) 5 196π ø 615.75 The circumference of the circle is bout 87.96 in. nd the re is bout 615.75 in.. 57. d 5 15 ft, r 5 15 ft 5 7.5 ft C 5 πr 5 π (7.5) 5 15π ø 7.1 A 5 πr 5 π (7.5) 5 56.5π ø 176.71 The circumference of the circle is bout 7.1 ft nd the re is bout 176.71 ft. 58. 608 6 5 608 The segment whose length is the pothem bisects the centrl ngle. So, 08-608-908 tringle is formed by the segment nd rdius of the hexgon. 17 r 5 cos 08 r 17 17 08 r 5 cos 08 r 5 Ï units x 5 tn 08 17 x 5 17 tn 08 x 5 17 Ï units s 5 x 5 1 17 Ï 5 Ï P 5 ns 5 6 1 Ï 5 68 Ï ø 117.78 units A 5 1 P 5 1 (17)(68 Ï ) 5 578 Ï ø 1001.1 sq. units 59. 608 8 5 58 A segment whose length is the pothem bisects the centrl ngle. So, x 5 sin.58 x 5 9(sin.58). One side 9 of the octgon 5 x 5 58 sin.58. 9 x.58 P 5 ns P 5 8(58 sin.58) 5 6 sin.58 ø 177.57 units 5 cos.58 9 5 9(cos.58) A 5 1 P 5 1 (9 cos.58)(6 sin.58) ø 78.71 sq. units 60. P 5 ns P 5 () 5 7 units A 5 1 Ï s A 5 1 Ï () 5 576 Ï ø 9. sq. units Lesson 1. Investigting Activity 1. (p. 80) STEP The polyhedron is rectngulr prism. The polyhedron is not regulr, but it is convex. STEP 1 sq. units 1. A 5 p 7 5 1 5 1 sq. units P 5 () 1 (7) 5 0 units h 5 5 units. A 1 Ph 5 (1) 1 0(5) 5 1 sq. units The vlues re the sme.. The surfce re of prism cn be found by summing the re of ll the fces, or by using the formul S 5 A 1 Ph.. Smple nswer: F B C D A E S 5 (5) 1 (5) 1 () 1 (5) 1 () 1 (5) 5 76 units S 5 (10) 1 1() 5 76 units 0
Chpter 1, continued 1. Guided Prctice (pp. 80 806) 1. Smple nswer: 1. Exercises (pp. 806 809) Skill Prctice 1. lterl fce lterl edge lterl edge bse.. b. B in. in. A C F in. D in. E in. Are 5 () 1 (7) 1 (7) 1 (7) 1 (7) 1 () in. 5 1 sq. inches. 7 in. in. B 5 () 5 1 P 5 () 1 () 5 1 Are 5 B 1 Ph 5 (1) 1 1(7) 5 1 in.. S 5 πr 1 πrh 5 π(10) 1 π(10)(18) 5 00π 1 60π 5 560π ø 1759.9 in.. S 5 πr 1 πrh 08π 5 πr 1 πr(5) 0 5 πr 1 10πr 08π 0 5 r 1 5r 10 0 5 (r 1 1)(r 8) r 1 1 5 0 or r 8 5 0 r 5 1 or r 5 8 A cylinder cnnot hve negtive rdius, so the rdius is 8 feet. bse lterl fce lterl edge lterl fce The bses re the two prllel tringulr fces. The lterl fces re the rectngulr fces tht connect the two bses. The lterl edges re the three edges where the lterl fces meet.. The formul sttes the surfce re is equl to two times the re of the bse plus the perimeter of the bse times the height. The formul is true for cylinders s well s right prisms becuse πr is the re of the bse nd πr is the circumference, or perimeter of the bse. So, πr 1 πrh is relly B 1 Ph.. d 5 in., r 5 5 in. S 5 πr 1 πrh 5 π() 1 π()(10) 5 8π 1 0π 5 8π ø 150.80 The surfce re is bout 150.80 in... S 5 B 1 Ph 5 (16 Ï ) 1 (8)(0) 5 Ï 1 80 ø 55. The surfce re is bout 55. cm. 5. P 5 ns 5 6(0) 5 0 ft B 5 1 P 5 1 (.6)(0) 5 156.8 ft S 5 B 1 Ph 5 (156.8) 1 0(80) 5 81.6 1 19,00 5 7,51.6 The surfce re is bout 7,51.6 ft. 6. B 5 (8) 5 P 5 (8) 1 () 5 S 5 B 1 Ph 5 () 1 () 5 9 The surfce re is 9 ft. 7. S 5 B 1 Ph 5 1 1 ( Ï 61.75 )() 1 19(9.1) 5 Ï 61.75 1 17.9 ø 196.7 The surfce re is bout 196.7 m. 0
Chpter 1, continued 8. S 5 B 1 Ph 15. S 5 B 1 Ph 5 1 1 (tn 58)()(5) 1 (5)(.5) 5 10 tn 58 1 5 ø 8.76 The surfce re is bout 8.76 in.. 9. S 5 πr 1 πrh 5 π(0.8) 1 π(0.8)() 5 1.8π 1.π 5.8π ø 1.07 The surfce re is bout 1.07 in.. 10. S 5 πr 1 πrh 5 π(1) 1 π(1)(0) 5 88π 1 960π 5 18π ø 90.71 The surfce re is bout 90.71 mm. 11. S 5 πr 1 πrh 5 π(8) 1 π(8)(8) 5 18π 1 18π 5 56π ø 80.5 The surfce re is bout 80.5 in.. 1. The dimeter of the cylinder ws used in plce of the rdius. S 5 π( ) 1 π()(8) 5 π(9) 1 π() 5 66π ø 07.5 The surfce re is bout 07.5 cm. 1. S 5 B 1 Ph 606 5 (7 p x) 1 (1 1 x)(15) 606 5 1x 1 10 1 0x 606 5 x 1 10 96 5 x 9 5 x The vlue of x is 9 yd. 1. S 5 πr 1 πrh 1097 5 π(8.) 1 π(8.)(x) 1097 5 1.8π 1 16.πx 67.5 5 16.πx 1.09 ø x The vlue of x is bout 1.09 m. 616 5 1 1 (8)(17) 1 (5 1 Ï 5 )(x) 616 5 16 1.79x 80 5.79x 10.96 ø x The vlue of x is bout 10.96 in. 16. 8 9 15 Use the Pythgoren Theorem to find the height. h 1 9 5 15 h 5 1 S 5 B 1 Ph S 5 1 1 (9)(1) 1 (9 1 15 1 1)(8) 5 108 1 88 5 96 The surfce re is 96 squre units. 17. C; S 5 B 1 Ph S 5 B 1 Ph 5 (l ) 1 (l)(l) 5 (l) 1 ((l))(l) 5 l 1 l 5 18l 1 6l 5 6l 5 5l 5 9(6l ) The surfce re is 9 times the originl surfce re. 18. S 5 πr 1 πrh S n 5 π 1 r Ï 5 1 π 1 r Ï 5 1 h Ï 5 5 1 5 (πr ) 1 1 5 (πrh) 5 1 5 (πr 1πrh) The new surfce re is 1 the originl re. 5 19. S 5 B 1 Ph 5 1 1 5 1 (6 p 10) tn 08 1 (6 p 10)(10) ø 519.6 1 600 ø 1119.6 The surfce re of the hexgonl prism is bout 1119.6 in.. 0
Chpter 1, continued 0. S 5 πr 1 πrh 108π 5 π(h) 1 π(h)(h) 108π 5 8πh 1 πh 108π 5 1πh 9 5 h 5 h The height of the cylinder is meters. 1. l 1 (l Ï ) 5 8 l 1 l 5 6 l 5 6 l 5 6 S 5 B 1 Ph 5 l 1 l 5 1 6 1 1 6 5 18 The surfce re of the cube is 18 sq. units. Problem Solving. S 5 πr 1 πrh 5 π(10) 1 π(10)(8) 5 00π 1 160π 5 60π ø 110.97 The surfce re is bout 110.97 in.... S 5 B 1 Ph 5 (1 p 6) 1 6(6) 5 1 1 16 5 60 The minimum mount of pper needed is 60 in.. b. The net of the box includes res tht re covered by the top, so there is extr mteril tht is not touched by the pper. c. The mount in prt () is the exct mount of pper. It is better to hve more so tht there is some overlp of wrpping pper... S 5 πr 1 πrh S 5 π() 1 π()(10) 5 π 1 80π 5 11π ø 51.86 The surfce re is bout 51.86 ft. b. Doubling the rdius will crete greter surfce re. 8 c. Double rdius: Double height: S 5 πr 1 πrh S 5 πr 1 πrh S 5 π(8) 1 π(8)(10) S 5 π() 1 π()(0) 5 18π 1 160π 5 π 1 160π 5 88π 5 19π ø 90.78 ø 60.19 When the rdius is doubled, the surfce re is bout 90.78 ft. When the height is doubled, the surfce re is bout 60.19 ft. Doubling the rdius cretes greter surfce re. 5. A; The white rectngle is touching the bse tht contins the rrow but not the edge where the rrow strts. 6. Surfce re of prism: S 5 B 1 Ph 5 (1)(1) 1 (5)(6) 5 168 1 187 5 00 Surfce re of cylinder: S 5 πr 1 πrh S 5 π(6) 1 π(6)(6) 5 6π 1 π 5 68π ø 170.7 The rectngulr prism hs greter surfce re, so it will tke more mteril to mke the rectngulr prism recycle bin. 7.. Surfce re 5 the number of exposed fces 5 5 sq. units b. Surfce re 5 the number of exposed fces 5 5 sq. units c. When the red cubes re removed, inner fces of the cubes remining replce the re of the red cubes tht re lost. When the blue cubes re removed, there re still fces of the blue cubes whose re is not replced by inner fces of the remining cubes. Therefore, the re of the solid fter removing blue cubes is squre units less thn the solid fter removing red cubes. 8.. S 5 π(r 1 ) 1 π(r 1 )(h) π(r ) 1 π(r )(h) 5 π(1) 1 π(1)(8) π(6) 1 π(6)(8) 5 88π 1 19π 7π 1 96π 5 50π ø 158.6 The surfce re is bout 158.6 m. b. S 5 π(r 1 ) 1 π(r 1 )(h) π(r ) 1 π(r )(h) 05
Chpter 1, continued 9. in. 1 ft 5. 608 5 908 A segment whose length is the pothem bisects the centrl ngle. So, tn 58 5 x 6. 1 ft 1 ft S 5 B 1 Ph πr 1 πrh 5 (1 p 1) 1 (1 p )(1) π() 1 π()(1) 5 88 1 576 8π 1 8π 5 86 1 0π ø 989.66 The surfce re is bout 989.66 in.. 0.. A C 5 B E 5 6 F D x 5 6 tn 58 5 6 One side of the squre 5 x 5 1. A 5 (x) 5 1 5 1 The re is 1 squre units. 6. 608 6 5 608 A segment whose length is the pothem bisects the centrl ngle. So, 08-608-908 tringle is formed. K L b. S 5 6(5 p 5) 1 8 1 1 Ï (5) 5 150 1 50 Ï ø 6.60 The surfce re is bout 6.60 mm. Mixed Review 1. 1608 5 (n ) p 1808 7 5 n 9 5 n The polygon hs 9 sides. It is nongon.. 10808 5 (n ) p 1808 6 5 n 8 5 n The polygon hs 8 sides. It is n octgon.. 708 5 (n ) p 1808 5 n 6 5 n The polygon hs 6 sides. It is hexgon.. 18008 5 (n ) p 1808 10 5 n 1 5 n The polygon hs 1 sides. It is dodecgon. J P 0 H M 1 N 6 5 tn 08 6 5 tn 08 5 6 Ï P 5 ns 5 6(1) 5 7 A 5 1 P 5 1 (6 Ï )(7) 5 16 Ï ø 7.1 The re is bout 7.1 squre units. 7. 608 5 5 78 The segment whose length is the pothem bisects the centrl ngle with the rdius drwn, so 68-58-908 tringle is formed. Use trigonometric rtios to find the length of the pothem ( 5 9 cos 68) nd side. The length of side is (9 sin 68) 5 18 sin 68. U T 6 V W 9 x S R P 5 ns 5 5(18 sin 68) 5 90 sin 68 ø 5.90 A 5 1 P 5 1 (9 cos 68)(18 sin 68)(5) 5 05(cos 68)(sin 68) ø 19.59 The re is bout 19.59 squre units. 06
Chpter 1, continued Lesson 1. 1. Guided Prctice (pp. 81 81) 1. l 5 h 1 l 5.8 1 5.5 5 5.9 l 5 7. The re of ech lterl fce is A 5 1 bl 5 1 (8)(7.) 5 9. squre meters.. S 5 B 1 1 Pl 5 1 (5.5)(8 p 5) 1 1 (8 p 5)(7.) 5 110 1 16 5 56 The surfce re is 56 m.. l 5 h 1 r Lterl re 5 πrl l 5 0 1 15 5 65 5 π(15)(5) l 5 5 5 75π ø 1178.10 The lterl re is bout 1178.10 yd.. S 5 πr 1 πrl 5 π(15) 1 π(15)(5) 5 5π 1 75π 5 600π ø 188.96 The surfce re is bout 1885 yd. 1. Exercises (pp. 81 817) Skill Prctice 1. height bse slnt height. The height of right cone is the perpendiculr distnce from the vertex of the cone to its bse. The slnt height of the cone is the height of its lterl surfce.. A 5 1 bl 5 1 (8)(10) 5 0 The re of lterl fce is 0 cm.. A 5 1 bl 5 1 (10)(15) 5 75 The re of lterl fce is 75 in.. 5. l 5 h 1 1 1 b l 5 1 1 0 5 81 l 5 9 6. S 5 B 1 1 Pl 5 ( p ) 1 1 ( p )() 5 1 1 5 16 The surfce re is 16 ft. 7. S 5 B 1 1 Pl 5 5 1 1 (6.9)(10) 1 1 (5 p 10)(0) 5 17.5 1 500 5 67.5 The surfce re is 67.5 mm. 8. S 5 B 1 1 Pl 5 1 (5) 1 5 Ï 1 1 ( p 5)(8) ø 10.8 1 60 ø 70.8 The surfce re is bout 70.8 in.. 9. The slnt height l is 5 ft, not ft. S 5 B 1 1 Pl 5 6 1 1 ()(5) 5 96 The surfce re is 96 ft. 10. l 5 r 1 h Lterl re 5 πrl l 5 7.5 1 5 ø π(7.5)(6.10) l 5 681.5 ø 195.75π l ø 6.10 ø 61.98 The lterl re is bout 61.98 cm. 11. l 5 r 1 h Lterl re 5 πrl l 5 1 1 5 π(1)( Ï 17 ) l 5 17 5 Ï 17 π l 5 Ï 17 ø 1.95 The lterl re is bout 1.95 in.. 1. l 5 r 1 h Lterl re 5 πrl l 5.5 1 1 5 π(.5)(1.5) l 5 156.5 5.75π l 5 1.5 ø 17. The lterl re is bout 17. in.. 1. S 5 πr 1 πrl 5 π() 1 π()(15) 5 16π 1 60π ø 8.76 The surfce re is bout 8.76 in.. 1. S 5 πr 1 πrl 5 π(1) 1 π(1)(0) 5 169π 1 60π ø 17.7 The surfce re is bout 17.7 cm. Are 5 1 bl 5 1 (0)(9) 5 580 The re of lterl fce is 580 ft. 07
Chpter 1, continued 15. l 5 h 1 r S 5 πr 1 πrl l 5 8 1 5 5 π(5) 1 π(5)( Ï 89 ) l 5 89 ø 5π 1 7.17π l 5 Ï 89 ft ø 6.7 The surfce re is bout 6.7 ft. 16. The formul for the surfce re of cone is wrong. The r in the second term should not be squred. S 5 πr 1 πrl 5 π(6) 1 π(6)(10) 5 96π The surfce re is 96π cm. 17. B; S 5 πr 1 πrl 00π 5 π(8) 1 π(8)(l) 00π 5 6π 1 8πl 16π 5 8πl 17 5 l The slnt height is 17 ft. 18. 15 ft 0 ft S 5 πr 1 πrl 5 π(15) 1 π(15)(0) 5 5π 1 00π 5 55π ø 169. The surfce re is bout 169. ft. 19. 0. 16 m 0 m l 5 r 1 h S 5 πr 1 πrl l 5 8 1 0 5 π(8) 1 π(8)( Ï 96 ) l 5 96 ø 6π 1 8.9π l 5 Ï 96 ø 981.9 The surfce re is bout 981.9 m. 10 in. in. S 5 B 1 1 Pl 5 1 (10)(5 Ï ) 1 1 ( p 10)() 5 5 Ï 1 60 ø 0.0 The surfce re is bout 0.0 in.. 1. 6 cm 9 cm 0 B 5 1 ()(6)(6) 5 18 5 18 1 5 5 Ï tn 08 S 5 B 1 1 Pl 5 5 Ï 1 1 (6 p 6)(9) 5 5 Ï 1 16 ø 55.5 The surfce re is bout 55.5 cm.. l 5 r 1 h l 5 5 1 l 5 1 l 5 Ï 1 5 cm S 5 πr 1 πrh 1 πrl 5 π(5) 1 π(5)(1) 1 π(5)( Ï 1 ) 5 5π 1 10π 1 (5) Ï 1 π ø 556.11 The surfce re is bout 556.11 cm.. l 5 h 1 1 1 b l 5 1 1 5 l 5 15.5 l 5 Ï 15.5 S 5 B 1 Ph 1 1 Pl in. h 5 cm h 51 cm 5 in. 5 in. 5 (5 p 5) 1 ( p 5)(5) 1 1 ( p 5)( Ï 15.5 ) 5 5 1 100 1 10 Ï 15.5 5 15 1 10 Ï 15.5 ø 16.05 The surfce re is bout 16.05 in... (l 1 ) 5 1 (l 1 ) 5 5 yd l 1 5 5 l (l ) 5 1 8 (l ) 5 7 l 5 Ï 7 S 5 πrl 1 1 πrl 5 π()(5) 1 π()( Ï 7 ) 5 15π 1 π Ï 7 ø 17.65 The surfce re is bout 17.65 yd. 5 in. yd 8 yd 08
Chpter 1, continued 5. 6. S 5 p lterl re 5 1 1 ()( Ï ) 5 ( Ï ) 5 16 Ï ø 7.71 The surfce re is bout 7.71 cm. 5 in. in. x Cone: S 5 πr 1 πrl 5 π() 1 π()(5) 5 16π 1 0π 5 6π Pyrmid: S 5 B 1 1 Pl 5 in. 5 x 1 1 (x)(5) 5 x 1 10x 6π 5 x 1 10x 0 5 x 1 10x 6π x 5 b 6 Ï b c 5 10 6 Ï 10 (1)(6π) (1) 10 6 Ï 100 1 1π 5 x ø 6.75 or x ø 16.75 The length of bse edge is bout 6.75 inches. Problem Solving 7. l 5 h 1 1 1 b l 5 1 l 5 5 l 5 5 S 5 B 1 1 Pl 5 6 1 1 ( p 6)(5) 5 6 1 60 5 96 The surfce re is 96 in.. 8.. lterl re 5 1 1 b l 5 1 1 (8)(1) 5 The lterl re is bout in.. b. The lterl fces re not tringles becuse the top of the pyrmid hs been cut off. The lmpshde is pproximtely the shpe of regulr squre pyrmid, but not exctly. 9. Regulr squre pyrmid; l 5 6 l 5 7 l 5 Ï 7 S 5 B 1 1 Pl 5 6 1 1 (6 p ) Ï 7 ø 6 1 6.5 ø 98.5 The surfce re is bout 98.5 cm. 0. Right cone; S 5 πr 1 πrl 5 π() 1 π()(9) 5 9π 1 7π 5 6π ø 11.10 The surfce re is bout 11.10 in.. 1.. A > D by the Right Angle Congruence Theorem nd C > C by the Reflexive Property of Congruence. So, by the AA Similrity Postulte, n ABC, ndec. b. AC 1 AB 5 BC 1 5 BC 5 5 BC 5 5 BC AB AC 5 DE DC 5 DE DE 5 6 DE 5 6 5 AC BC 5 DC EC 5 5 EC EC 5 10 EC 5 10 5 5 09
Chpter 1, continued c. Smller cone: S 5 πr 1 πrl 5 π 1 1 π 1 1 5 5 9 π 1 15 π 5 6π Lrger cone: S 5 πr 1 πrl 5 π() 1 π()(5) 5 9π 1 15π 5 π The surfce re of the smller cone is 1 5 5% of the surfce C re of the lrger cone... A 5 m AB 608 p πr b. A 5 0 0 5 1508 608 p πl 0 5 5 1 p πl 8 π 5 l Î 8 π 5 l.91 ø l The slnt height is pproximtely.91 m. c. Lterl re 5 πrl 0 5 πr Î 8 π 1.6 ø r h 1 r 5 l h 1 1.6 5.91 h ø 1.61 h ø.55 The rdius is bout 1.6 m nd the height is bout.55 m.. Originl rdius to height 5 1.08 1.08 5 r 1.08 0.8 r ø 0.69 Originl volcno: l 5 r 1 h lterl re 5 πrl l 5 1 1.08 5 π()(.188) l ø.188 ø 0.0507 Destroyed top: l 5 r 1 h l 5 0.69 1 0.5 l ø 0.777 Lterl re 5 πrl 5 π(0.69)(0.777) ø 1.608 0.0507 1.608 5 8. The lterl re of the volcno fter 1980 is bout 8. mi.. 5 16 in. 6 in. The height of the originl cone 5 8 tn 58 nd the slnt height of the originl cone 5 Ï 8 1 (8 tn 58) ø 1.9 in.. The originl cone s lterl re is πrl 5 π(8)(1.9) ø.01 in.. The cone tht is cut off the top is similr to the originl cone by the AA Similrity Postulte. So, the height of the cone cut off the top is tn 58 nd the slnt height is Ï 1 ( tn 58) ø.98 in.. The lterl re of the top cone is πrl ø π()(.98) ø 6.9 in.. The surfce re of the dog s cone is the surfce re of the originl cone minus the surfce re of the cone tht ws cut off the top. S ø.01 6.9 ø 87.07 The surfce re of the collr is pproximtely 87.07 in.. Mixed Review 5. 11x 5 10x 6. x8 5 (6x 5)8 5 x x 5 5 5 x x 5 15 7. A 5 s 5 7 5 9 The re is 9 mi. 8. A 5 bl 5 ( Ï )( Ï ) 5 The re is yd. 9. A 5 1 h(b 1 1 b ) 5 1 (8)(9 1 10) 5 76 The re is 76 mm. Quiz 1.1 1. (p. 817) 1. F 1 V 5 E 1 F 1 8 5 1 1 F 1 8 5 1 F 5 6 The polyhedron hs 6 fces.. S 5 B 1 Ph 66 5 (11x) 1 ( 1 x)8 66 5 x 1 176 1 16x 66 5 8x 1 176 190 5 8x 5 5 x The vlue of x is 5 ft.. S 5 πr 1 πrh 717 5 π(6.1) 1 π(6.1)x 717 5 7.π 1 1.πx 8.0 ø 1.πx 1.61 ø x The vlue of x is bout 1.61 in. 10
Chpter 1, continued. S 5 B 1 Ph 567 5 1 1 (9)(1) 1 ( 1 Ï 50 )x 567 5 117 1 ( 1 5 Ï 10 )x 50 5 ( 1 5 Ï 10 )x 11.90 ø x The vlue of x is bout 11.90 m. 5. S 5 B 1 1 Pl c. The visible portion of the pencil is the lterl re of the cone-shped end plus the lterl re of the hexgonl prism shft. Lterl re of Lterl re of hexgonl prism: cone end: L 1 5 6r(x 1) l 5 r 1 1 L 1 5 6rx 6r l 5 Ï r 1 1 L 5 πrl L 5 πr( Ï r 1 1 ) 5 10 1 1 ( p 10)(1) 5 100 1 60 5 60 The surfce re is 60 cm. 6. S 5 πr 1 πrl 5 π() 1 π()(9) 5 16π 1 6π 5 5π ø 16.6 The surfce re is bout 16.6 ft. 7. l 5 h 1 r S 5 πr 1 πrl l 5 16 1 10 5 π(10) 1 π(10)( Ï 56 ) l 5 56 5 100π 1 10π Ï 56 l 5 Ï 56 ø 906.91 The surfce re is bout 906.91 m. Mixed Review of Problem Solving (p. 818) 1. F 1 V 5 E 1 F 1 6 5 7 1 F 1 6 5 9 F 5 For polyhedron to hve 6 vertices nd 7 edges it would need to hve only fces. Becuse it is not possible for polyhedron to hve only fces, it is not possible for polyhedron to hve 6 vertices nd 7 edges.. Method 1: Add the res of ll 6 sides of the rectngulr solid. Method : Use the formul S 5 A 1 Ph where A is the res of the bse, P is the perimeter of the bse, nd h is the height of the solid... S represents the totl surfce re; r represents the rdius of the circumscribed circle for the bse nd the length of one side of the hexgonl bse; h represents the length of the pencil. b. The new length of the pencil is x nd only one end is visible, so the formul for the visible portion is S 5 Ï r 1 6rx. S 5 L 1 1 L S 5 6rx 6r 1 πr( Ï r 1 1 ) d. Let D be the difference in the two surfce res nd r be the rdius of the circumscribed circle for the bse nd the length of one side of the hexgonl bse. D 5 Ï r 1 6rx (6rx 6r 1 πr( Ï r 1 1 )) 5 Ï r 1 6r πr( Ï r 1 1 ). Lterl re 5 πrh 5 π 1.8 () 5 11.π ø 5.19 The lterl re is 5.19 in.. 5. The only vribles in the formul for the surfce re of right cone re the rdius nd the slnt height. The rdius is hlf the dimeter. If the dimeter nd slnt height re known, the surfce re cn be found. S 5 π 1 d 1 π 1 d l 6. Answers will vry. 7.. The cross section is squre. b. P 5 s 5 (10) 5 0 The perimeter is 0 feet. A 5 s 5 10 5 100 The re is 100 squre feet. c. Isosceles right tringles re formed. 8. d. 10 10 Are 5 1 bh 5 1 (10)(10) 5 50 The re is 50 squre feet. Slnt height 5 Ï (x) 1 (x) x 5 Ï 16x 1 9x 5 Ï 5x x 5 5x Surfce re 5 πr 1 πrl 19π 5 π(x) 1 π(x)(5x) 19π 5 9πx 1 15πx 19π 5 πx 81 5 x 9 5 x The vlue of x is 9 units. 11
Chpter 1, continued Lesson 1. 1. Guided Prctice (pp. 81 8) 1. V 5 (1)(1)(1) 1 ()(1)() 5 1 1 6 5 7 The volume is 7 units.. V 5 Bh 5 (5) (1) 5 00 The volume is 00 ft.. V 5 πr h 68π 5 πr (18) 8 5 r Ï 8 5 r The rdius is Ï 8 inches.. V 5 Bh 5 1 (9 p 5)(8) 5 180 The volume is 180 m. 5. B 5 Are of tringle re of squre 5 1 (10)(5 Ï ) 5 5 Ï 9 V 5 Bh 5 (5 Ï 9)(6) ø 05.81 The volume is bout 05.81 ft. 1. Exercises (pp. 8 85) Skill Prctice 1. The volume of solid is mesured in cubic units.. Not necessrily. For exmple: 10 S 5 B 1 Ph S 5 B 1 Ph 5 (0) 1 ()() 5 (16) 1 (0)() 5 11 squre units 5 11 squre units V 5 Bh V 5 Bh 5 0() 5 16() 5 60 cubic units 5 6 cubic units Two solids cn hve the sme surfce re nd not hve the sme volume.. A; Volume of box 5 Bh 5 (15 p 9)() 5 05 cubic inches Volume of cube 5 s 5 5 7 cubic inches Number of cubes 5 05 7 5 15 15 three inch cubes cn fit in the box.. V 5 ()(1) 1 7()(1) 5 6 1 1 5 0 unit cubes The volume is 0 cubic units. 5. V 5 1()() 1 5(1)() 5 8 1 10 5 18 unit cubes The volume is 18 cubic units. 6. V 5 (1)() 1 5()() 5 1 1 60 5 7 unit cubes The volume is 7 cubic units. 7. V 5 Bh 5 1 1 (7 p 10) p 5 5 175 The volume is 175 in.. 8 8. V 5 Bh 5 ( p )(1.5) 5 1 The volume is 1 m. 9. V 5 Bh 5 6 1 1 (7.5)(.75 Ï ) (18) 5 1518.75 Ï ø 60.55 The volume is bout 60.55 cm. 10. V 5 πr h 5 π(7) (1) 5 588π ø 187.6 The volume is bout 187.6 ft. 11. V 5 πr h 5 π 1 10 (16) 5 00π ø 156.6 The volume is bout 156.6 in.. 1. V 5 πr h 5 π 1 6.8 (9.8) 5 1759.688π ø 558. The volume is bout 558. cm. 1. The volume of right cylinder is πr h not πrh. V 5 πr h 5 π() () 5 8π The volume is 8π ft. 1. 11 in. 7 ft ft In cubic inches, V 5 Bh 5 (7 p 1)( p 1)(11) 5,6. The volume is,6 in. 5 19.5 ft. 15. V 5 Bh 1000 5 (x p x) p x 1000 5 x 10 5 x The vlue of x is 10 inches. 16. B 5 1 bh 5 1 (5)( Ï 9.5 ) ø 1.615 V 5 Bh 5 ø 1.615x x ø.08 The vlue of x is bout.08 cm. 17. V 5 πr h 18π 5 π 1 x (8) 18π 5 πx 6 5 x 8 5 x The vlue of x is 8 inches. 18. V 5 πr h πr h 5 π() (7) π(1)(7) 5 6π 7π 5 56π ø 175.9 The volume is bout 175.9 m. 19. V 5 Bh 5 [(7.8)(1.) (1.8)()] p 9 5 (96.7 5.) p 9 5 (91.)9 5 81.88 The volume is 81.88 ft. 1
Chpter 1, continued 0. V 5 Bh 1 1 (πr h) 7. 5 ( p )() 1 1 (π() ()) 5 6 1 8π ø 89.1 The volume is bout 89.1 in. 1. A; V 5 πr h. 6π 5 π() h 6π 5 16πh 5 h The height of the cylinder is feet. h 5 in. 6 m 8 m s 6 m 8 m 8 m s 5 Ï 6 1 8 5 Ï 6 1 6 5 Ï 100 5 10 Lterl re 5 P p h 5 (10) p 8 5 0 The lterl re is 0 m. B 5 re of rhombus 5 1 d 1 d 5 1 (1)(16) 5 96 Surfce re 5 B 1 Ph 5 (96) 1 0 5 51 The surfce re is 51 m. Volume 5 Bh 5 96 p 8 5 768 The volume is 768 m. Problem Solving in. in. V 5 Bh 96 5 1 ( p )h 8. B 5 Are of rectngulr fce Are of cylindricl end 5 9.5 π 1 5 5 9 π 96 5 6h 16 5 h The height of the prism is 16 inches.. V 5 πr h 1005.5 5 πr (8) 0 ø r 6. ø r d 5 r ø (6.) ø 1.65 The dimeter is bout 1.65 cm.. V 5 Bh 5 ( p 7)(6) 5 168 The volume is 168 in. 5. V 5 πr h 5 π(8) (1) 5 896π ø 81.87 The volume is bout 81.87 ft. 6. sin 608 5 h 18 18 sin 608 5 h 9 Ï 5 h V 5 πr h 5 π 1 1 (9 Ï ) 5 π Ï ø 176.01 The volume is bout 176.01 m. V 5 Bh 5 1 5 9 π (17) ø 6.8 The volume is bout 6.8 mm. 9.. V 5 Volume of lrge prism Volume of two smller prisms V 5 15.75(8)(8) (.5)()(8) (.5)()(8) 5 1008 1 1 5 70 The volume is 70 in.. b. B 5 Are of lrge rectngle p 5 15.75(8) (.5 p ) 5 90 V 5 Bh 5 90(8) 5 70 The volume is 70 in.. c. The nswers re the sme. 0.. V 5 πr h 5 π 1 1000 (00) Are of smll rectngle 5 100,000,000π ø 1,159,65 The volume of Blue Hole is bout 1,159,65 ft. b. Gllons of wter 5 1,159,65 ft 7.8 1 ft ø,9,911,0 There re bout,9,911,0 gllons of wter in the Blue Hole. 1
Chpter 1, continued 1. C 5 πr 10 5 πr 5 π 5 r V 5 πr h 5 π 1 5 π (0) ø 159.15 The volume of the column is bout 159.15 ft.. The first cylinder hs volume V 5 πr h 5 π() (5) 5 5π The second cylinder hs volume V 5 πr h 5 π(5) () 5 75π The second cylinder hs greter volume becuse 75π is greter thn 5π... V 5 Bh 5 (0 p 10)(0) 5 500 The volume of the wter is 500 in.. b. The volume of the rock is equl to the volume the wter rose. V 5 Bh 5 (0 p 10)(0.5) 5 75 The volume of the rock is 75 in.. c. The cpcity of the tnk is (0)(10)(0) 5 6000 in.. Let n represent the number of rocks tht cn be plced in tnk. Then, 6000 5 500 1 75n 1500 5 75n 0 5 n So, 0 rocks cn be plced in the qurium before the wter spills out. Are of Are of. V 5 Bh 5 rectngulr 1 tringulr 1 bottom top h 907 5 F (8)(18) 1 1 (18)( Ï x 9 ) G (6) 5 5 1 1 9 Ï x 81 108 5 9 Ï x 81 1 5 Ï x 81 1 5 ( Ï x 81 ) 1 5 x 81 5 5 x Ï 5 5 x 15 5 x Ech hlf of the roof is 15 ft by 6 ft. Mixed Review 5. tn.58 5 x x 5 tn.58 ø.8 The vlue of x is bout.8. 6. sin 68 5 7 x 7 x 5 sin 68 ø 11.91 The vlue of x is bout 11.91. 7. cos 758 5 x 5 x 5 5 cos 758 ø 1.9 The vlue of x is bout 1.9. 8. A 5 πr 5 π(9.5) 5 90.5π ø 8.5 The re is bout 8.5 in.. 9. A 5 1 P 5 1 (7.5)(78) 5 9.5 The re is 9.5 m. 0. 608 5 5 78 10.6 in. x 6 The pothem bisects centrl ngle, so the rdius nd pothem form 68-58-908 tringle with one hlf of side length. So, 5 10.6 cos 68 nd x 5 10.6 sin 68. The length of ech side 5 x 5 (10.6 sin 68). Are 5 1 P 5 1 (10.6 cos 68)(5)()(10.6 sin 68) ø 67.15 The re is bout 67.15 in.. Problem Solving Workshop 1. (p. 87) 1.. Volume V of pencil holder 5 Volume of lrge prism Volume of cylinders 5 7.5 p p (π(1.5) ()) 5 10 18π ø 6.5 The volume of the pencil holder is bout 6.5 in.. b. B 5 Are of rectngle Are of two circles 5 (7.5)() (π)(1.5) 5 0.5π V 5 Bh 5 (0.5π) ø 6.5 The volume of the pencil holder is bout 6.5 in... The student ws correct tht times the bse re of the cylinders needs to be subtrcted, but they forgot tht the lterl re of ech cylinder needs to be dded. S 5 (7.5 p ) 1 (7.5 p ) 1 ( p ) (π(1.5) ) 1 (π(1.5)) S 5 15 1 15π S ø 199.1 The surfce re is bout 199.1 squre inches. 1
Chpter 1, continued. Volume of cylinder 5 πr h Volume of hole 5 1 πr h 5 πr h 1.5 Guided Prctice (pp. 80 81) 1. 608 6 5 608 πr h 5 1 πr h 0. r 5 1 R r 5 Ï 1 R r 5 Ï R For the hole to hve hlf the volume of the cylinder, r should be Ï R. Volume of solid Volume of 5 lrge prism Volume of smll prism V 5 ()()(5) (1)(1)(5) 5 0 5 5 5 The volume is 5 ft. 5. V 5 Are of sector p h 5 608 608 p π() p.5 5 1 (π) p.5 6 5 7 π ø 7. The volume is bout 7. in.. 6.. The surfce re cn be found by finding the sum of the res of ech exposed fce. b. Not ll of the fces of the bems re exposed, so dding the individul surfce res will result in totl tht is greter thn the ctul surfce re. Lesson 1.5 Investigting Activity 1.5 (p. 88) Step It tkes times to fill the prism. So, the volume of the prism is times the volume of the pyrmid. 1. The res of the bses re the sme.. The heights re pproximtely the sme.. The volume of the pyrmid is one-third the volume of the prism.. V 5 1 Bh, where B is the re of the bse nd h is the height. yd The pothem bisects the centrl ngle, so 08-608-908 tringle is formed. 5 tn08 5 Ï B 5 1 P V 5 1 Bh 5 1 1 Ï (6 p ) 5 1 1 Ï (11) 5 Ï 5 88 Ï ø 15. The volume is bout 15. yd.. h 5 l r h 5 8 5 h 5 9 h 5 Ï 9 V 5 1 Bh 5 1 1 π(5) 1 Ï 9 5 5 π Ï 9 ø 16.9 The volume is bout 16.9 m.. V 5 1 Bh 150π 5 1 (π)(18) h 150π 5 108πh 1.5 5 h The height is 1.5 meters.. tn 08 5 r 5.8 r 5 5.8 tn 08 V 5 1 1 πr h 5 1 (π)(5.8 tn 08) (5.8) ø 1.86 The volume is bout 1.86 in.. 15
Chpter 1, continued 5. Volume 5 Volume of cylinder 1 Volume of cone 5 πr h 1 1 1 πr h 5 π() (10) 1 1 1 π p 5 5 90π 1 15π cm 5 105π ø 9.87 The volume of the solid is bout 9.87 cubic centimeters. 6. Flow rte 5 volume 5 101mL ø 1.56 ml/s time. s The flow rte of the snd is bout 1.56 ml/s. 1.5 Exercises (pp. 8 86) Skill Prctice 5 cm 10 cm 1. A tringulr prism hs two congruent tringulr bses tht re prllel to ech other. The lterl fces re ll prllelogrms. A tringulr pyrmid hs single tringulr bse with lterl fces re tringles with common vertex. 10. The volume of the pyrmid is 1 Bh not 1 Bh. V 5 1 17 (10) 5 90 ø 16. The volume is bout 16. ft. 11. D; V 5 1 Bh 5 5 1 B(9) 5 5 B 15 5 B The re of the bse is 15 ft. 1. V 5 1 Bh 00 5 1 110 x 00 5 100 x 6 5 x The vlue of x is 6 cm. 1. V 5 1 Bh 16π 5 1 1πr h 16π 5 1 1π p x (18) prism pyrmid Tringulr prism Tringulr pyrmid. The volume of squre pyrmid is one third the volume of squre prism with the sme bse nd height.. V 5 1 Bh 5 1 1 5 (6) 5 50 The volume is 50 cm.. V 5 1 Bh 5 1 1 πr h 5 1 1 π p 10 (1) 5 100 π ø 161.6 The volume is bout 161.6 mm. 5. V 5 1 Bh 5 1 (.5)() 5 0 ø 1. The volume is bout 1. in.. 6. V 5 1 Bh 5 1 1 πr h 5 1 1 π p 1 () 5 π ø.09 The volume is bout.09 m. 7. V 5 1 Bh 5 1 1 1 p p () 5 6 The volume is 6 in.. 8. V 5 1 Bh 5 1 1 6 p 1 p 6 Ï p 1 (17) 5 1 Ï ø 10.0 The volume is bout 10.0 ft. 9. The slnt height ws used insted of the ctul height. h 5 Ï 15 9 5 Ï 1 5 1 V 5 1 π 19 (1) 5 π ø 1018 The volume is bout 1018 ft. 16π 5 6πx 6 5 x 6 5 x The vlue of x is 6 inches. 1. V 5 1 Bh 7 Ï 5 1 1 1 1 Ï () (x) 7 Ï 5 Ï x 7 5 x The vlue of x is 7 feet. 15. tn 608 5 opp. dj. Ï 5 r r 5 Ï V 5 1 1πr h ø 1 π1 Ï () ø 716.85 The volume is bout 716.85 ft. 16. tn 8 5 opp. dj. tn 8 5 7 h 7 h 5 tn 8 V 5 1 1πr h ø 1 π(7) 7 1 tn 8 ø 57.8 The volume is bout 57.8 yd. 16
Chpter 1, continued 17. cos 58 5 dj. hyp. cos 58 5 r 15 r 5 15 cos 58 sin 58 5 opp. hyp. sin 58 5 h 15. V 5 Volume of cube Volume of cone 5 S 1 1 πr h 5 5.1 1 1 π p.55 (5.1) ø 97.9 The volume is bout 97.9 m. 5. V 5 Volume of cylinder Volume of pyrmid 5 1 πr h 1 Bh h 5 15 sin 58 V 5 1 1 πr h 5 1 π(15 cos 58) (15 sin 58) ø 987.86 The volume is bout 987.86 cm. 18. B; tn 98 5 opp. dj. tn 98 5 5 h 5 h 5 tn 98 V 5 1 1 πr h ø 1 1 π p 5 5 1 ø 6.15 tn 98 The pproximte volume of the cone is 6.15 ft. 19. V 5 1 1 πr h 1.6 5 1 1 π p h 1.6 5 1 16 π h 8.57 ø h The height of the cone is bout 8.57 cm. 0. V 5 Volume of cylinder 1 Volume of cone 5 πr h 1 1 1 πr h 5 π p p 7 1 1 1 π p () 5 6π 1 9π 5 7π ø 6.19 The volume is bout 6.19 cm. 1. V 5 Volume of cube Volume of pyrmid 5 s 1 Bh 5 10 1 1 5 Ï (10) ø 8. The volume is bout 8. in... V 5 Volume of lrge cone 1 Volume of smll cone 5 1 1 πr H 1 1 1 πr h 5 1 1 π p 1 () 1 1 1 π 1 1 (1) 5 π 1 1 1 π 5 π ø.6 The volume is bout.6 ft.. V 5 p Volume of pyrmid 5 1 1 Bh 5 1. (.) ø 16.70 The volume is bout 16.70 cm. B 5 1 P 5 cos.58 P 5 8 1 sin.58 V 5 1 π p () 1 1 1 1 cos.58 (8) 1 sin.58 ø 6.9 The volume is bout 6.9 yd. 6. V 5 1 1 πr h 5 1 1 π p () 5 π ø 1.57 The volume of the solid is bout 1.57 cm. 7. Drwing digonl in the squre cretes two 58-58-908 tringles. The digonl hs length of 5 Ï meters, which is lso the dimeter of the bse of the cone. So, the rdius is 5 Ï. 8. V 5 1 1 πr h 5 m 5 1 1 π 1 5 Ï (7) ø 91.6 The volume is bout 91.6 m. B 0 C ft sin 08 5 opp. hyp. 0.5 5 1.5 BC C A 5 B tn 58 5 opp. dj. tn 58 5 AC 7 m BC 5 AC 5 tn 58 ø.10 tn 08 5 opp. dj. Ï 5 1.5 V 5 1 Bh 5 1.5 Ï ø 1 6 p 1 1 p 1.5 Ï p (.10) ø 16.7 The volume is bout 16.7 ft. 17
Chpter 1, continued Problem Solving 9.. V 5 1 1 πr h 5 1 1 π p (1) 5 6π ø 01 There re bout 01 cubic inches of frosting in the bg. 01 in. b. 15 flowers 5 1. in. per flower About 1. cubic inches of frosting re used per flower. 0. V 5 1 1 πr h 5 1 1 π p (8) 5 π ø 75.0 The volume of the smll cup is bout 75. in.. 1. The cone nd the cylinder hve the sme bse nd height, so the volume of the cone is 1 p Volume of cylinder. This mens you hve to buy smll cups of popcorn to hve the sme mount s the lrge cup.. The cost of the lrge cup is twice the cost of the smll cup but the volume is three times s gret. So, the lrge cylindricl cup gives you more popcorn for your money.. h 5 Ï 6 5 Ï 5 Ï V 5 1 1 πr h 5 1 1 π p 1 Ï ø.70 The volume is bout.70 in... h.5 ft 5 ft.5 ft 5 ft 5 ft l 5 Ï 5.5 5.5 Ï h 5 Ï 1.5 Ï.5 5.5 Ï 5 ft V 5 1 Bh 5 1 1 5 1.5 Ï ø 9.6 The volume is bout 9.6 ft. in. 5 ft 6 in. 5.. Originl volume 5 1 Bh 5 1 1 7 (10) 5 90 Let h 5 (10): V 5 1 Bh 5 1 1 7 (0) 5 1 90 The volume is doubled. b. Let S 5 side length 5 (7) V 5 1 Bh 5 1 ((7)) (10) 5 1 90 The originl volume is multiplied by four. c. If the height of pyrmid is multiplied by x, the volume is multiplied by fctor of x. If the side length is multiplied by x, the volume is multiplied by fctor of x. 6.. Volume 5 Volume of cylinder 1 Volume of cone 5 πr h 1 1 1 πr h 5 π 1.5 (7.5) 1 1 1 π p.5 () ø 17. in. 17. in. 1. in. ø 1.0 The continer cn hold bout 1 cups. b. p 1 5 cups dy. dys 5 5 1 p 5 18 The feeder cn go 18 dys before it needs to be refilled. 7. Volume of big prism 5 6 1 Ï S h 5 6 1 Ï (.5) (1.5) ø 7.7 Volume of smll prism 5 6 1 Ï S h 5 6 1 Ï (.5) (0.5) ø 6.86 Volume of pyrmid 5 1 p 6 1 Ï S h 5 1 p 6 1 Ï 1 () ø.8 Volume of solid 5 7.7 1 6.86 1.8 5 77.98 The volume of the deck prism is bout 77.98 in.. 8.. Averge re 5 π p 1 verge vlue of R 5 π 1 R b. V cone 5 1 Bh 5 1 πr h 5 πr p h, where B is the re of the bse of the cone, πr is the verge re of circulr cross section, nd h is the height. 18
Chpter 1, continued 9.. h 5 r r 5 h b. c. V 5 1 Bh 5 1 πr h 5 1 π 1 h (h) 5 1 π 1 h (h) 5 πh 1, where B is the re of the bse of the cone, r is the rdius, nd h is the height. Height (meters) Time (minutes) h 6 Height (meters) 1 1.90.0.7.0 5.5 0 0 6 Time (minutes) t There is not liner reltionship between the height of the wter nd time. The dt points on the grph do not form stright line, so there cnnot be liner reltionship. 0. V 5 [re of bse 1 1 re of bse 1 geometric men] 1 h 5 F π() 1 π(9) 1 Ï π() p π(9) G 1 (10) ø F 9π 1 81π 1 8.8 G 10 ø [67.57] 10 ø 15. The volume of the frustum is bout 15. cm. h 1 1 h 1.. 5 h 1 r r 1 h 1 r 5 r 1 (h 1 1 h ) h 1 r 5 h 1 r 1 1 h r 1 h 1 r h 1 r 1 5 h r 1 h 1 (r r 1 ) 5 h r 1 h r 1 h 1 5 r r 1 b. V 5 πr (h 1 1 h ) πr h 1 1 1 r 1 h 1 h r r 1 5 πr πr 1 1 r 1 h r r 1 1 r 1 h 1 h r r 1 c. V 5 πr 5 1 1 πr 1 r 1 h 1 r h r 1 h 5 π 1 r h r h 1 r r 1 r r 1 5 π h 1 r r 1 r r 1 πr 1 1 r 1 h r r 1 r r 1 π 1 r h 1 r r 1 5 πh 1 (r r 1 ) 1 r 1 r 1 r 1 r 1 r r 1 5 h 1 πr 1 πr 1 r 1 πr 1 The volume is equl to one-third of the height times the re of the lower bse plus the geometric men of the two bses plus the re of the upper bse.. V 5 1 Bh 5 1 1 6 Ï (1) 5 1 (7)(1) 5 88 h r The volume of the pyrmid is 88 cubic feet. Mixed Review..5 6 5 x 1. x 15 5 8 0 6x 5 1(.5) 0x 5 8(15) 6x 5 0x 5 10 x 5 7 x 5 6 5. 6 9 5 8 6 or 9 1 x 5 9 x 6(9 1 x) 5 7 6x 5 18 9 1 x 5 1 x 5 x 5 6. A AF is rdius, BE is dimeter, nd CD is B chord. E D F C 7. C BC is minor rc. 9. A 5 π() 5 9π ø 8.7 The re is bout 8.7 m. 50. A 5 π(.5) 5 1.5π ø 8.8 The re is bout 8.8 mi. 51. A 5 π(0.) 5 0.16π ø 0.50 The re is bout 0.50 cm. 8. C ABE is mjor rc. 19
Chpter 1, continued 5. πr 5 8π r 5 A 5 π() 5 16π ø 50.7 The re is bout 50.7 in.. Quiz 1. 1.5 (p. 86) 1. V 5 Bh 5 1 (10)(15)(7) 5 55 The volume is 55 cm.. V 5 πr h 5 π(6) (10) 5 60π ø 110.97 The volume is bout 110.97 in... V 5 πr h 5 π(.5) (16) 5 π ø 1017.88 The volume is bout 1017.88 m.. V 5 1 Bh 5 1 1 () 5 6 The volume is 6 cm. 5. V 5 1 (πr )h 5 1 1 π p 0 (0) ø 7,699.11 The volume is bout 7,699.11 ft. 6. V 5 1 (πr )h 5 1 1 π p 8 (15) 5 0π ø 1005.1 The volume is bout 1005.1 yd. 7. The cups hve the sme rdius nd height, so the volume of the cone is one-third the volume of the cylinder. When one conicl cup is poured into the cylindricl cup, one third of the cylinder will be full. The full level will be t height tht is one-third the cylinder s height or 1 (6) 5 inches. Spredsheet Activity 1.5 (p. 87) 1. The rdius should be.5 cm nd the height should be pproximtely.5 cm. The vlue in the surfce re column decreses s you pproch 95.81 nd increses fter.. The rdius should be bout.6 cm nd the height should be bout 9.9 cm. Lesson 1.6 1.6 Guided Prctice (pp. 89 81) 1. S 5 πr 5 π(0) 5 1600π ø 506.55 The surfce re is bout 506.55 ft.. S 5 πr 0π 5 πr 7.5 5 r.7 ø r The rdius is bout.7 m.. C 5 πr πr 5 6π r 5 S 5 πr 5 π() 5 6π The surfce re of the inner bll is 6π, or bout 11.10 ft.. V 5 πr 5 π(5) ø 5.60 The volume is bout 5.60 yd. 5. V 5 Volume of cone 1 Volume of hemisphere 5 1 πr h 1 1 1 πr 5 1 π(1) (5) 1 π(1) 1 m 5 5 π 1 π 5 7 π ø 7. The volume is bout 7. m. 1.6 Exercises (pp. 8 85) Skill Prctice 5 m 1. The formul for the surfce re of sphere is πr nd the formul for the volume of sphere is πr, where r is the rdius of the sphere.. The center of the sphere must be contined for the intersection to be gret circle.. S 5 πr 5 π() 5 6π ø 01.06 The surfce re is bout 01.06 ft.. S 5 πr 5 π(7.5) 5 5π ø 706.86 The surfce re is bout 706.86 cm. 5. S 5 πr 5 π(9.15) 5.89π ø 105.09 The surfce re is bout 105.09 m. 6. B; S 5 πr π 5 πr 8 5 r Ï 8 5 r The pproximte rdius is.8 m. 7. C 5 πr 9.6π 5 πr.8 5 r The rdius is.8 inches. 8. d 5 r 5 (.8) 5 9.6 The dimeter is 9.6 inches. 9. S 5 1 (πr ) 5 1 (π(.8) ) 5 6.08π ø 1.76 The surfce re of the hemisphere is bout 1.76 in.. 10. The surfce re of hemisphere is hlf the surfce re of the sphere. S 5 1 (πr ) 5 1 (π(5) ) 5 50π ø 157.08 The surfce re of the hemisphere is bout 157.08 ft. 11. C 5 πr 8.π 5 πr. 5 r S 5 πr 5 π(.) ø 759.7 The surfce re is bout 759.7 cm. 0
Chpter 1, continued 1. V 5 πr 5 π(6) 5 88π ø 90.78 The volume is bout 90.78 in.. 1. V 5 πr 5 π(0) ø 68,08.57 The volume is bout 68,08.57 mm. 1. V 5 πr 5 π(.5) ø 65.5 The volume is bout 65.5 cm. 15. The formul for the volume of sphere is πr not πr. V 5 πr 5 π(8) 5 68.66π ø 1.66 The volume is bout 1.66 ft.. S 5 πr 1 πrh 1 1 1 πr 5 π(5.8) 1 π(5.8)(1) 1 1 ()(π)(5.8) ø 87. The surfce re is bout 87. ft. V 5 πr h 1 1 1 πr 5 π(5.8) (1) 1 1 1 (π)(5.8) ø 1888.1 The volume is bout 1888.1 ft.. S 5 πrl 1 1 1 πr 16. V 5 πr 17. V 5 πr 16.76 5 πr 91.95 5 πr ø r 1.95 ø r 7 ø r.80 ø r The rdius is bout The rdius is bout 7.00 m..80 cm. 18. V 5 πr 0,81.7 5 πr 969.06 ø r 17.06 ø r The rdius is bout 17.06 in. 19. V 5 πr 6π 5 πr 7 5 r 5 r The rdius is feet, so the dimeter is () 5 6 feet. 0. A; S 5 πr V 5 πr 5 r 1 πr 5 rs The reltionship is V 5 rs. 1. S 5 πr 1 πrh 1 1 1 πr 5 π(.) 1 π(.)(7) 1 1 ()(π)(.) ø 7.78 The surfce re is bout 7.78 in.. V 5 πr h 1 1 πr 5 π(.) (7) 1 1 (π)(.) ø 16. The volume is bout 16. in.. 7. 5 π(.9) (1.5) 1 1 ()(π)(.9) ø 58.97 The surfce re is bout 58.97 cm. V 5 1 πr h 1 1 1 πr 5 1 π(.9) (1.6) 1 1 1 (π)(.9) ø 56.1 The volume is bout 56.1 cm. Rdius of sphere Circumference of gret circle. 10 ft 0π ft 5. 1 in. 6π in. 6. 5 cm 50π cm 7. 1 m π m Surfce re of sphere Volume of sphere. 00π ft 1 000 π ft 5. 676π in. 1 8788 π in. 6. 500π cm 1 6,500 π cm 7. 176π m 1,8π m 8. C; Volume of cube 5 s s 5 6 s 5 r 5 s 5 Surfce re of sphere 5 πr 5 π() 5 16π The surfce re is 16π cm. 1
Chpter 1, continued 9.. V 5 πr 5 π(1) 5 π in. Smple nswers: r 5 1 in., h 5 in... C 5 πr 7π 5 πr 1.5 5 r r 5 0.5 in., h 5 16 in. r 5 in., h 5 1 in. b. The surfce re of the cylinder will be greter thn the surfce re of the sphere when πrh 1 πr > πr. Solve to get h > r. The surfce re of the cylinder will be greter thn the surfce re of the sphere when the height of the cylinder is greter thn the rdius. Problem Solving 0. V 5 πr h 1 1 1 πr 5 π(10) (60) 1 1 1 (π)(10) ø 0,9.95 The volume of the grin silo is bout 0,9 ft. 1. C 5 πr,855 5 πr 855 π 5 r S 5 1 (πr ) 5 1 855 ()(π ) ø 98,1,1. π The surfce re of the western hemisphere is pproximtely 98,1,1 squre miles... V 5 πr 17.5 5 πr 0.8 ø r 6.99 ø r The rdius is bout 6.99 cm. b. S 5 πr 5 π(6.99) ø 61.99 The surfce re is bout 61.99 cm... C 5 πr 8 5 πr π 5 r V 5 πr 5 π 1 π ø 8.65 The volume of tennis bll is bout 8.65 in.. b. V 5 Volume of cylinder p volume of one bll 5 πr h 1 πr 5 π(1.) (8.65) 1 π 1 ø 9.7 There is bout 9.7 in. of spce not tken up. π V 5 πr 5 π(1.5) ø 10,06 The volume of the blloon is bout 10,06 cm. b. The rdius is cubed, so (r) would give 8r. This mens the volume should be eight times the originl volume. c. V 5 π(r) 5 π(.7) ø 87.96 ø 8(10,06) The volume is bout 8,7.96 cm. The prediction from prt (b) ws correct. The rtio of this volume to the originl is 8 to 1 or 8 : 1. 5.. Surfce re of Torrid Zone 5 πrh 5 π(96)(50) ø 80,95,856 mi Surfce re of Erth 5 πr 5 π(96) ø 197,59,88 mi Surfce re of Torrid Zone b. Probbility of Torrid Zone 5 Surfce re of the Erth 80,95,856 mi 5 197,59,88 mi ø 0.1 The probbility tht meterorite will lnd in the Torrid Zone is bout 1%. 6.. Solid I: S 5 πr Solid II: S 5 πr 1 πr(r) 5 6πr Solid III: S 5 F πr 1 r Ï 1 πr G ø.8πr Solid I, Solid III, Solid II b. Solid I: V 5 πr Solid II: V 5 πr (r) 5 πr Solid III: V 5 F 1 1 πr (r) G 5 πr Solid III, Solid I, Solid II 7. A sphere is formed. S 5 πr 5 π(9) ø 1017.88 The surfce re is bout 1018 in.. V 5 πr 5 π(9) ø 05.6 The volume is bout 05 in.. 8.. The digonl of the cylinder s height nd dimeter is equl to the dimeter of the sphere. r Using the Pythgoren Theorem: x r 5 8 x 8 r 5 6 x 8 x Volume of the cylinder 5 πr h r 5 π 16 x (x) 5 πx 16 x -
Chpter 1, continued b. 500 50 0 The mximum vlue occurs when x is bout.6 m. c. V 5 πx 1 6 x 5 π(.6) 1 6.6 ø 18. The mximum volume is bout 18. m. 9. The lterl edge of the cone is tngent to the sphere, so rdius drwn from the sphere is perpendiculr to the tngent lterl side. By the AA Similrity Postulte, two similr right tringles re formed by this rdius. AC hs length of cm becuse EC is rdius of the sphere nd hs length cm. By the Pythgoren Theorem, AB hs length Ï 5 Ï 1 5 Ï cm. 6 cm C E A BC EA DE 5 AB B cm BC DE 5 AC AD DE 5 Ï 6 Ï 5 AD DE 1 Ï 5 1 (AD) 5 8 Ï DE 5 Ï AD 5 Ï The rdius of the cone is Ï cm nd the lterl edge is Ï cm. S 5 πrl 1 πr 5 π 1 Ï 1 Ï 1 π 1 Ï 5 6π ø 11.10 The surfce re is bout 11.10 cm. V 5 1 πr h 5 1 π 1 Ï (6) 5 π ø 75.0 The volume is bout 75.0 cm. Mixed Review 0. rtio of sides 5 8 : 5 rtio of res 5 8 : 5 or 6 : 5 Are of nabc Are of ndef 5 6 5 Are of ndef 5 6 5 6(Are of ndef) 5 5() Are of ndef ø 16.1 Are of ndef is bout 16.1 ft. 1. rtio of sides 5 15 : 1 or 5 : 7 rtio of res 5 5 : 7 or 5 : 9 Are of JKLM Are of PQRS 5 5 9 A 195 5 5 9 9A 5 5(195) A 5 99.9 The re of PQRS is bout 99.9 cm. Are of shded region 8(1) π(7). P 5 5 ø 0.1 Totl re 8(1) The probbility tht the point lies in the shded region is bout 1%. Are of shded region 51 1 (7.85)(11.0). P 5 5 Totl re π(9.7) ø 0.76 The probbility tht the point lies in the shded region is bout 76%.. Volume of cone 5 1 p Volume of cylinder 5 1 (0) 5 110 cubic units Lesson 1.7 Investigting Activity 1.7 (p. 86) Step 1 nd Step : Scle fctor of Solid A to Solid B Pir 1 1 Surfce re of Solid A, S A Surfce re of Solid B, S B 7 88 Pir 8π 6π Pir 1 Pir 1 1 675 75 S A V A V S A V B B V B 6 88 1 8 Pir 9 0π 67.5π 8 7 Pir 9 1 97.8 6.08 7 1 Step : S A is the squre of the scle fctor while V A is the cube S B V B of the scle fctor. 1. The rtio of the surfce res of similr solids is the squre of the scle fctor of the solids. The rtio of the volumes of similr solids is the cube of the scle fctor of the solids.. rtio of surfce res 5 1 k rtio of volumes 5 1 k
Chpter 1, continued 1.7 Guided Prctice (pp. 88 89) 1. Lengths 5 1 9 5 Widths 5 16 1 5 Heights 5 The prisms re similr becuse the rtios of corresponding liner mesures re equl.. Heights 5 15 10 5 Rdii 5 10 5 5 The cones re not similr becuse the rtios of corresponding liner mesures re not equl.. C D 5 5 150 C D 5 Ï 6 5 Ï 6 C D 5 5 The scle fctor of C to D is 5. The edge length of C is Ï 5 6 5 units. 5 5 C D 7 15 5 7 D D 5 15 The volume of D is 15 cubic units.. 1 5 Price $1.50 Price 5 8($1.50) 5 $1 The price of the lrge bll should be $1 for neither bll to be better buy. 1.7 Exercises (pp. 850 85) Skill Prctice 1. Two solids re similr if they re the sme type of solid with equl rtios of ll corresponding liner mesures.. If the corresponding liner mesures hve rtio :b, then the volumes re in the rtio :b.. Rdii 5 7 Heights 5 16 10 5 8 5 The solids re not similr becuse the rtios of corresponding liner mesures re not equl.. Lengths 5 5 9 7 Widths 5 1.6 5 5 9 Heights 5 11 1.8 5 55 7 The solids re not similr becuse the rtios of corresponding liner mesures re not equl. 5. Lengths 5 6 8 5 Widths 5 1.5 18 5 Heights 5.5 6 5 The solids re similr becuse the rtios of corresponding liner mesures re equl. 6. Rdii 5 8 1 5 Heights 5 18 7 5 The solids re similr becuse the rtios of corresponding liner mesures re equl. 7. D; Lengths 5 6 15 5 5 8. 9. Widths 5 10 5 5 Heights 5 10 5 5 5 15 feet by 10 feet by 5 feet Surfce re of A Surfce re of B 5 b 150π Surfce re of B 5 1 Surfce re of B 5 600π in. Volume of A Volume of B 5 b 50π Volume of B 5 1 Volume of B 5 000π in. Surfce re of A Surfce re of B 5 b 1500 Surfce re of B 5 1 Surfce re of B ø 166.67 m Volume of A Volume of B 5 b.6 Volume of B 5 1 Volume of B 5 17.1 m
Chpter 1, continued 10. Surfce re of A Surfce re of B 5 b 56. Surfce re of B 5 5 Surfce re of B ø 77 cm Volume of A Volume of B 5 b 750.9 Volume of B 5 5 Volume of B ø 76.86 cm 11. The rtio of the volumes is, not b b. 500π Volume of B 5 1 1. b 5 8π 15 π b 5 8 15 b 5 5 The scle fctor of Solid I to Solid II is : 5. 1. b 5 7 79 b 5 9 b 5 1 The scle fctor of Solid I to Solid II is 1 :. 1. b 5 88 18 b 5 1 6 b 5 1 8 b 5 The scle fctor of Solid I to Solid II is :. 15. b 5 19 108 b 5 16 9 b 5 The scle fctor of Solid I to Solid II is :. 16. C; b 5 8π 7π b 5 8 7 b 5 b 5 5 9 17. b 5 π 16π b 5 1 8 b 5 1 b 5 1 The rtio of the surfce re of the smller sphere to the surfce re of the lrger sphere is 1 :. Surfce re of A 18. Surfce re of B 5 b 78π Surfce re of B 5 The surfce re of B or the lrge cylinder is 175.5π ø 551.5 m. 19. Scle fctor 5 8 5 1 Surfce re of Solid I 5 πr 1 πrh 1 πrl 5 π() 1 π()() 1 π() 1 Ï 1 ø 85.9 ft Volume of Solid I 5 πr h 1 1 πr h Surfce re of Solid I Surfce re of Solid II 5 5 π() () 1 1 π() () ø 6.8 ft b 85.9 Surfce re of Solid II 5 1 Surfce re of Solid II ø 1.96 ft Volume of Solid I Volume of Solid II 5 b 6.8 Volume of Solid II 5 1 Volume of Solid II ø 50.6 ft 0. Scle fctor 5 8 Surfce re of Solid I 5 B 1 Pl 1 1 1 p Ï p Volume of Solid II 5 Bh 1 1 Bh Surfce re of Solid I Surfce re of Solid II 5 5 1 1(8) 1 1 1 p Ï p ø 10.59 cm 5 1 (8) 1 1 1 1 1.5 Ï ø 78.6 cm 10.59 Surfce re of Solid II 5 8 Surfce re of Solid II ø 16.96 cm Volume of Solid I Volume of Solid II 5 b 78.6 Volume of Solid II 5 8 b Volume of Solid II ø.1 cm 5
Chpter 1, continued 1. Scle fctor 5 7 Surfce re of Solid I 5 lh 1 lw πr 1 πrh 5 ()() 1 ()() π 1 1 π()() ø 89.1 in. Volume of Solid I 5 Bh 5 ( π( ))() ø 1.75 in. Surfce re of Solid I Surfce re of Solid II 5 b 89.1 Surfce re of Solid II 5 7 Surfce re of Solid II ø 7.96 in. Volume of Solid I Volume of Solid II 5 b 1.75 Volume of Solid II 5 7 Volume of Solid II ø 7.61 in.. Scle fctor 5 8 5 Surfce re of Solid I 5 5(8) 1 6(5) 1 π 1 1 (8) 1 (1 p 8) ø 19.1 m Volume of Solid I 5 15 (8) 5 00 m Surfce re of Solid I Surfce re of Solid II 5 b 19.1 Surfce re of Solid II 5 8 5 Surfce re of Solid II ø 85.60 m Volume of Solid I Volume of Solid II 5 b 00 Volume of Solid II 5 8 5 Volume of Solid II ø 8.8 m. h 5 r S 5 πr 1 πrh 5π 5 πr 1 πr(r) 5π 5 6πr 9 5 r r 5 h 5 r 5 6 (rdius of smll cylinder) (rdius of lrge cylinder) 5 5π 8 π (rdius of lrge cylinder) 5 9 6 rdius of lrge cylinder 5 8 height of lrge cylinder 5 8() 5 16 The rdius nd height of the smll cylinder re ft nd 6 ft. The rdius nd height of the lrge cylinder re 8 ft nd 16 ft... The rtio from the smll cone to big cone is 8 : 10 or : 5. The rtio of the re is then : 5 or 16 : 5. b. :5 c. : 5 or 16 : 5 d. : 5 or 6 : 15 e. The volume of the top prt of the cone is 6 15. If 1 represents the volume of the entire cone, 1 6 represents the volume of the bottom prt of 15 the cone. 1 6 15 5 61. The rtio of the pieces is 15 6 15 : 61 or 6 : 61. 15 Problem Solving Volume of smller mug 5. Volume of lrger mug 5 b Volume of smller mug 5.5 1 Volume of smller mug ø 8.0. So, the cpcity of the smller mug is bout 8.0 fluid ounces. 6. The units should be the sme for ech prt of the rtio. Volume of smll binoculrs Volume of lrge binoculrs 5 b (0.5 feet) 5 (5 feet) 5 0.15 9115 1 5 79,000 The rtio of the volumes is 1 : 79,000. 7. Volume of smll bowl Volume of lrge bowl 5 Volume of smll bowl 6 b 5 Volume of smll bowl 5 7. So, the smll bowl requires 7 fluid ounces of lemonde. 8. Smple nswer: 0 cm 5 cm cm 8 cm 9.. Volume of smller ornge 5 π(1.6) ø 17.16 in. Volume of lrger ornge 5 π() ø.51 in. 6