CHAPTER 9. Chapter Opener. congruent faces: T PQR T PQS; congruent edges: QR **** QS **& and **** PR PS **** c 2

Transcription

1 CHAPTER 9 Chapter Opener Chapter Readiness Quiz (p. 47) 1. A; 5 1 c c. G; A 1 bh 169 c c 1 c 1 (1)(5) 0 units. C; A πr π(6) Lesson in. 9.1 Checkpoint (p. 475) 1. The solid is a polyhedron with triangular bases, so it is a triangular prism.. A cone has a curved surface, so it is not a polyhedron.. The solid is a polyhedron with a rectangular base, so it is a rectangular pyramid Guided Practice (p. 476) 1. C. A. B 4. true 5. false 6. true 7. false 8. true 9. true 10. The solid is a polyhedron with rectangular bases, so it is a rectangular prism. 11. A cylinder has a curved surface, so it is not a polyhedron. 1. The solid is a polyhedron with a pentagonal base, so it is a pentagonal pyramid. 1. hexagonal prism; 8 faces and 18 edges; congruent faces: ABCDEF UVWXYZ and AFZU FEYZ EDXY DCWX CBVW BAUV; congruent edges: **** AB BC **** CD **** DE **** EF **** FA **** UV **& VW **& WX **& **& XY YZ **** ZU **** and **** AU BV **** CW **& DX **** EY **** FZ **& 14. triangular pyramid; 4 faces and 6 edges; congruent faces: T PQR T PQS; congruent edges: QR **** QS **& and **** PR PS **** 15. cube or square prism; 6 faces and 1 edges; congruent faces: JKLM TUVW JMWT MLVW LKUV KJTU; congruent edges: JK **** KL **** LM **& MJ **& TU **** UV **& VW **& WT **& **** JT KU **& LV **** MW **& &* faces and 1 edges 9.1 Practice and Applications (pp ) 17. A cone has a curved surface, so it is not a polyhedron. 18. The solid is a polyhedron with triangular bases, so it is a triangular prism. 19. The solid is a polyhedron with a rectangular base, so it is a rectangular pyramid. 0. False; a pyramid only has one base. 1. True; prisms have two bases.. True; the bases of a prism are congruent polygons.. False; a cone only has one base. 4. False; a sphere has a rounded surface, so it is not a polyhedron. 5. F 6. D 7. A 8. E 9. B 0. C 1. triangular prism. Sample answer: a pyramid with a square base has one square face and four triangular faces, while a triangular prism has two congruent triangular faces and three rectangular faces.. The solid is a polyhedron with a rectangular base, so it is a rectangular pyramid. 4. The solid is a polyhedron with pentagonal bases, so it is a pentagonal prism. 5. The solid has a curved surface, so it is not a polyhedron. 6. rectangular pyramid; 5 faces and 8 edges; congruent faces: T ABE T ACD and T ABC T AED; congruent edges: BE **** CD ****, BC **** ED ****, and **** AB **** AC AD **** AE ****. 7. triangular prism; 5 faces and 9 edges; congruent faces: T JKL T FGH; congruent edges: FH **** **** JL, FG **** JK ****, GH **** KL ****, and FJ **** GK **& HL **& Geometry, Concepts and Skills 17

2 8. pentagonal pyramid; 6 faces and 10 edges; congruent faces: T NPT T NTS T NSR T NRQ T NQP; congruent edges: NP **** NT **** NS **** NR **** NQ **& and PT **** TS **** **** SR RQ **** QP **** 9. True; solids are three-dimensional shapes. 40. False; cylinders, cones, and spheres are not polyhedra because they have curved surfaces. 41. False; if a prism is not rectangular, then the bases are the congruent faces that are not rectangular. 4. True; all plane surfaces of a prism are faces F V E 57. F V E 5 6 E F E F E F F V E 8 1 E 0 E 18 E 9.1 Standardized Test Practice (p. 480) 59. D 60. F 61. C 9.1 Mixed Review (p. 480) Sample answer: 47. Sample answer: 48. Sample answer: 49. Sample answer: 50. Sample answer: 6. A s 6. A bh 9 (7)(4) 81 cm 8 m 64. A bh 60 6b 10 in. b 65. A s 169 s s 1 ft s 66. C πd π(1) 8 ft; A πr π(6) π(6) 11 ft 67. C πr π(5) 10π 1 cm; A πr π(5) π(5) 79 cm 68. C πr π(14) 8π 88 yd; A πr π(14) π(196) 616 yd 9.1 Algebra Skills (p. 480) 51. Sample answer: 5. F V E 5. F V E 6 V 1 8 V 1 6 V 14 8 V 14 V 8 V F V E 55. F V E 1 V 0 0 V 0 1 V 0 V V 0 V (1 9) (7 5) (5 ) (10 ) (7) (49) l w h l w h () () (5) πh π(5) 10π 78. πw h π() (5) π(4)(5) 0π 18 Geometry, Concepts and Skills

3 79. πl π() 9π 80. l w () () Lesson Activity (pp ) Step 4. rectangular prism Step 5. Rectangle Length Width Area Step 6. A Area of rectangle A l w 7 1 square units; P Perimeter of rectangle A l w (7) () units; h Height of rectangles B, C, D, and E 5 units 1. A Ph (1) 0(5) They are equal.. Surface Area A Ph, where A is the area of the base, P is the perimeter of the base, and h is the height of the prism. 4. Sample answer: F B A 7 units units 1 units B 7 units 5 units 5 units C units 5 units 15 units D 7 units 5 units 5 units E units 5 units 15 units F 7 units units 1 units Surface Area A Ph (0) 18() units C Total Area 14 units D A E 9. Checkpoint (pp ) 1. B 6 1 P () (6) h S B Ph (1) 16() in.. B P (5) (8) h 6 S B Ph (40) 6(6) ft. B P h 4 S B Ph (4) 4(4) cm 4. S πr πrh 5. S πr πrh π() π()(5) π(6) π(6)(10) 18π 0π 7π 10π 48π 19π 151 in. 60 ft 6. L πrh π(1)() 4π 1 m 9. Guided Practice (p. 487) 1. true. false. true 4. true 5. B P (6) (4) h 10 S B Ph (4) 0(10) in. 6. S πr πrh π(4) π(4)(6) π 48π 80π 51 ft Geometry, Concepts and Skills 19

4 7. B P h S B Ph (6) 1() m 9. Practice and Applications (pp ) 8. 5 in. 9. A 6 in. 10. P () () in m 1. A πr π(4) 16π 50 m 1. C πr π(4) 8π 5 m ft 15. A ft 16. P ft 17. B P (10) (9) h 6 S B Ph (90) 8(6) ft 18. B P (4) (4) h 4 S B Ph (16) 16(4) in. 19. B P h 7 S B Ph (4) 4(7) m 0. No; doubling the height does not double the surface area. Prism A: B 9 P () () h S B Ph (9) 1() m Prism B: B 9 P () () h 1 S B Ph (9) 1(1) m 1. rectangular prism. cylinder. triangular prism 4. S πr πrh 5. S πr πrh π(6) π(6)(11) π() π()(1) 7π 1π 18π 78π 04π 96π 641 ft 0 m 6. S πr πrh 7. S πr πrh π(8) π(8)(8) π(5) π(5)(7) 18π 18π 50π 70π 56π 10π 804 cm 77 in. 8. B P (16) (16) 64 h 16 S B Ph (56) 64(16) mm 9. B P (.7) 8.1 h 18 S B Ph () 8.1(18) in. 140 Geometry, Concepts and Skills

5 0. S πr πrh π(1.5) π(1.5)(1) 4.5π π 7.5π 4 in. 7. m 7 m 1. B 60 P 6(10) 60 h 11 S B Ph (60) 60(11) cm 8. S πr πrh π() π()(7) 18π 4π 60π 188 m. The formula Juanita used to find the surface area is incorrect. The radius of the cylinder is 6 inches, not 1 inches. Also, 1 4. The correct solution is as follows. S πr πrh π(6) π(6)(10) 7π 10π 19π 60 in ft 10 ft ft B 6 18 P (6) () h 10 S B Ph (18) 18(10) ft 6 in. 8 in. B P h 5 S B Ph (4) 4(5) in. 9. P () (6) h 7 L Ph 16(7) 11 m 40. L πrh π()(8) 48π 151 yd 41. P h 5 L Ph 4(5) 10 ft 4. L πrh 4. L πrh π()(90) π(1)(4) 60π 8π 111 cm 5 in. 44. L πrh π(60)(1.) 144π 45 mm 5 in. 45. P 64(4) 56 h 415 L Ph 56(415) 106,40 m Geometry, Concepts and Skills 141

6 46. B S B Ph ,40 110,6 m 9. Standardized Test Practice (p. 490) 47. B; S πr πrh π(5) π(5)(15) 50π 150π 00π 68 cm 48. H; B 5 10 P (5) () 14 S B Ph 104 (10) 14x x 84 14x 6 ft x 9. Mixed Review (p. 490) 49. A s 1 bh 50. A bh 1 bh 6 1 (11)(6) 10(4) 1 (10)(5) ft 65 cm 51. base of triangle: 5. A bh a b c 70 5b 5 b 1 14 ft b 5 b 169 b 144 b 1; A bh 1 bh (1) 1 (1) in. 9. Algebra Skills (p. 490) 5. 6 x x x 6(5) x 10(4) x 0 x 40 x x x 5x (5) 4x 7(16) 5x 105 4x 11 x 1 x x 1 x 8 11x (1) 8x (5) 11x 49 8x 160 x 9 x 0 Lesson Checkpoint (pp ) 1. B P l 9 S B 1 Pl 49 1 (8)(9) 175 in.. B 5.1 P l 1 S B 1 Pl (7)(1) cm. B P l l S B 1 Pl (40)(1) 60 ft 4. S πr πrl 5. S πr πrl π(10) π(10)(1) π(4) π(4)(8) 100π 10π 16π π 0π 48π 691 in. 151 ft 14 Geometry, Concepts and Skills

7 6. l r h l 5 5; S πr πrl π() π()5 9π 15π 4π 75 cm 9. Guided Practice (p. 495) 1. The red line segment is the height of the pyramid.. The blue line segment is the slant height of the pyramid.. The height of the lateral faces of the pyramid is the slant height. 4. The height of a pyramid is the perpendicular distance between the vertex and base in. 10 in. l l The slant height is 1 in. 6. B.9 P 9 l 5 S B 1 Pl.9 1 (9)(5) 6 m 10 in. 5 in. 7. S πr πrl π() π()(5) 4π 10π 14π 44 in. 8. B P l 10 S B 1 Pl 6 1 (4)(10) 156 ft 9. Practice and Applications (pp ) 9. height 10. height 11. slant height 1. l l l l 5 10 m 15 mm 14. l l in. 15. B B 1. P P l 8 l 11 S B 1 Pl S B 1 Pl 5 1 (0)(8) 1. 1 (1)(11) 105 m 16.7 in. 17. B P l l S B 1 Pl (96)(19.1) 1498 cm Geometry, Concepts and Skills 14

8 18. B 1.7 P 6 l 4 S B 1 Pl (6)(4) 1.7 yd 19. B 104 P 18 l l S B 1 Pl (18)(0) 04 mm 0. B P l l S B 1 Pl (56)(5) 896 cm 1. The slant height of a pyramid is the hypotenuse of a right triangle formed by the height of the pyramid and half the length of the base. Since the hypotenuse of a right triangle is always the longest side of the triangle, the slant height of a pyramid is always greater than the height.. Jamie substituted the height of the pyramid into the formula where the slant height belongs; P 4(40) 160 l l S πr πrl 4. S πr πrl π(9) π(9)() π(4) π(4)(10) 81π 198π 16π 40π 79π 56π 877 m 176 ft 5. S πr πrl 6. S πr πrl π(7) π(7)(5) π(6) π(6)(10) 49π 175π 6π 60π 4π 96π 704 m 0 cm 7. l l l l S πr πrl S πr πrl π(9) π(9)(41) π(10) π(10)(6) 81π 69π 100π 60π 450π 60π 1414 yd 111 in. 9. L 1 Pl 1 (8)(14) 196 cm 0. L πrl 1. L πrl π(4.)(.) π(4)(14) 01.5 in. 176 in.. 8 m 1 m 1 m B P 4(1) 48 l l S B 1 Pl (48)(10) 84 m S 40 1 (160)(5) m 144 Geometry, Concepts and Skills

9 . 1 ft 8 ft 8 ft 8 ft B 7.7 P (8) 4 l 1 S B 1 Pl (4)(1) 184 ft 4. 7 yd 6 yd S πr πrl π() π()(7) 9π 1π 0π 94 yd in. B 7.7 ft 10 in. l l S πr πrl π(10) π(10)(17.) 100π 17π 7π 855 in. 6. square pyramid; B P 4(6) 4 l 4 S B 1 Pl 6 1 (4)(4) 84 ft 7. cone; 8. L πrl S πr πrl π(6)(10) π() π()(6) 60π 4π 1π 188 in. 16π 50 cm 9. l l 5 5 L πrl π()(5) 15π 47 in π 15π 45π Approximately 141 square inches of material is needed to make the Elizabethan collar. 41. l l 5 0 cm 4. P 4(8) 11 l L 1 Pl 1 (11)() 188 cm 4. Lateral area of one pyramid: P 4(1) 48 l l L 1 Pl 1 (48)(10.65) 55.6 square units Surface area of solid L square units 44. l l S B Ph 1 Pl ( ) (4 )(6) 1 (4 )(.5) (9) (1)(6) 1 (1)(.5) square units Geometry, Concepts and Skills 145

10 45. l l 5 5 S πr πrh πrl π() π()(10) π()(5) 9π 60π 15π 84π 64 square units 9. Standardized Test Practice (p. 498) 46. a. Cone A: Cone B: S πr πrl S πr πrl π() π()(6) π() π()(8) 9π 18π 9π 4π 7π π 85 ft ; 104 ft ; Cone C: S πr πrl π() π()(10) 9π 0π 9π 1 ft b. The radius stayed the same. c. Cone D: Cone E: S πr πrl S πr πrl π() π()(8) π(4) π(4)(8) 4π 16π 16π π 0π 48π 6 ft ; 151 ft ; Cone F: S πr πrl π(6) π(6)(8) 6π 48π 84π 64 ft d. The slant height stayed the same. e. The radius has a greater influence on surface area. The radius is used in computing both the base area and the lateral surface area, and is squared when computing the base area, while the slant height affects only the lateral surface area. 9. Mixed Review (p. 499) 47. x (6) 48. x 6 (4) 6 (6) Geometry, Concepts and Skills 49. x () 50. 4x 10 4() 10 (9) 4(4) x 4x (6) 4(6) x 5x (5) 5(5) (5) x x x x x 18,000 60x 6840 x 50 m x 19 ft x x x 0,960 x 86 cm 9. Algebra Skills (p. 499) 56. x 6x ( 6)x 4x 57. 4y 5 4y 1 (4 4)y (5 1) (x 1) x 4 x 1 x ( 1)x (4 1) x x (x ) 7x x (7 )x ( ) 4x x 9x 6x (1 9 6)x x c (5 c) 10c 5 c (10 )c 5 1c 5 Quiz 1 (p. 499) 1. Triangular base; triangular prism; yes, this is a polyhedron with 5 faces.. Circular base; cone; no, this is not a polyhedron.. Rectangular base; rectangular pyramid; yes, this is a polyhedron with 5 faces.

11 4. S πr πrh π() π()(9) 8π 6π 44π 18 ft 5. B 7 1 P () (7) 0 h 10 S B Ph (1) (0)(10) in. 6. l l S πr πrl π(6) π(6)(10) 6π 60π 96π 0 m 7. B 1 (1)(5) 0 P h 8 S B Ph (0) (0)(8) 00 in. 8. B P l 9 S B 1 Pl (40)(9) 80 m 9. S πr πrh π(5) π(5)(7) 50π 70π 10π 77 cm Lesson Checkpoint (pp ) 1. V Bh. V Bh (9 4) 6 (5 5) ft 15 cm. V Bh in. 4. V πr h 5. V πr h π() () π(1) (5) 1π 5π 8 ft 16 in. 6. V πr h π() (10) 40π 16 m 9.4 Guided Practice (p. 50) 1. volume. surface area. surface area 4. volume 5. V Bh (6.6)(1) 76. cm 6. V Bh (4) cm 7. V Bh (.4)(1) 80.8 cm 9.4 Practice and Applications (pp ) 8. 0 unit cubes; the base is 5 units by units. So, 5, or 15 unit cubes are needed to cover the bottom layer. There are layers. So, the total number of cubes is 15, or unit cubes; the base is 4 units by 4 units. So, 4 4, or 16 unit cubes are needed to cover the bottom layer. There are layers. So, the total number of cubes is 16, or unit cubes; the base is units by units. So,, or 6 unit cubes are needed to cover the bottom layer. There are 4 layers. So, the total number of cubes is 6 4, or V Bh 1. V Bh (5 5) 4 ( ) in. 6 cm Geometry, Concepts and Skills 147

12 1. V Bh m m V Bh V Bh ( ) (7 7) 7 7 m ft 10 cm V Bh (10 10) cm m m 10 cm 4 m m 10 cm 7 m V Bh (4 4) 7 V Bh 16 7 ( 6) 8 11 m ft 19. V Bh Bh ( ) 8 (5 ) ft 0. V Bh Bh ( 1) 4 (6 1) in. 1. V Bh Bh (7 10) 4 (7 ) m 148 Geometry, Concepts and Skills ft 7 ft 6 ft 8 ft 7 ft 7 ft. smaller box: V Bh (8 ) in. larger box: V Bh (10 4) in boxes 4. The larger box gives you the most cereal for your money. Four times the volume of the smaller box would cost $8, not $6. 5. V Bh (800 80) 64,000 1,408,000 ft 6. 1,408,000 ft 7.5 gal 1 ft 10,560,000 gal 7. V πr h 8. V πr h π(4) (9) π(6) () 144π 108π 45 in. 9 m 9. V πr h 0. V πr h π(1) (15) π(10) (4) 160π 400π 6786 m 157 ft 1. V πr h. V πr h π(1) (4) π(7.5) () 576π π 1810 ft 50 ft. The pool in Exercise has the smallest volume, so it would require the least amount of water to fill it. 4. horizontal line: V πr h π() (5) 45π 141 in. vertical line: V πr h π(5) () 75π 6 in. 5. The solid on the right has a greater volume. The solid with the vertical line of rotation has a volume that is almost twice the volume of the solid with the horizontal line of rotation.

13 6. V πr h π(0) () 900π 8,90 ft 7. 8,90 ft 7.5 gal 1 ft 16,77 gal 8. V Bh (0 10) in. 00 in. 1 gal 1 in. 9.5 gal 9 or 10 fish 9. small carton: V πr h π(5) (10) 50π 785 cm jumbo carton: V πr h π(10) (0) 000π 68 cm 40. No; the jumbo carton contains about 8 times as much ice cream as the regular carton ( ). 41. V Bh Bh (6 8) (1 ) in. 4. V πr h πr h π(8) (10) π() (10) 640π 90π 550π 178 ft 4. V Bh Bh (4 4) 4 (1 1) m 44. V Bh ( x) 4 1x 45. V Bh 1 x x 7 x 7 7x 46. V Bh (x 5x) x 15x x 15x 47. V πr h 48. V πr h π() h πr (10) πh r h r 8 in. h 7 ft r 49. Let x width of prism V Bh 54 (x x) x 54 x x 54 x 7 x in. x width in., length 6 in., height in. 9.4 Standardized Test Practice (p. 507) 50. C; V πr h 51. H; V Bh π(10) (5) 168 (7 ) h 500π 168 1h 1570 in. 8 ft h 9.4 Mixed Review (p. 507) 5. 7 a b 9 49 a b 81 a 95 b 17 a 9.7 b a a a 18 a S πr πrh π() π()(7) 18π 4π 60π 188 ft Geometry, Concepts and Skills 149

14 56. B 9 18 P () (9) 4 18 h 8 S B Ph (18) (8) m 57. S πr πrl π(5) π(5)(1) 5π 60π 85π 67 yd 9.4 Algebra Skills (p. 507) 58. x m 1 1 x m x 7 m c 10 c c b b 4 4 b 6. 14d 14d d n 10 6 n n 17 Lesson Activity (pp ) 1. The base areas are the same.. The heights are the same.. The rectangular prism has the greater volume; the pyramid fits inside the prism with extra room around the pyramid. 4. times; the volume of the prism is times the volume of the pyramid. 5. Volume of the prism times the volume of the pyramid; Volume of the pyramid 1 times the volume of the prism 6. Volume of a pyramid: V 1 Bh 9.5 Checkpoint (pp ) 1. V 1 Bh. V 1 Bh 1 (6 6) 7 1 (8 9) 5 84 in. 10 ft. V 1 Bh cm 4. V 1 πr h 1 π(5) (9) 75π 6 in. 5. V 1 πr h 1 π(5) (7) 17 5 π 18 ft h h 676 h 576 h 4 V 1 πr h 1 π(10) (4) 800π 51 m 7. V 1 πr h 1 π(4) (6) π 101 in h h 89 h 5 h 15 V 1 πr h 1 π(8) (15) 0π 1005 ft 150 Geometry, Concepts and Skills

15 9.5 Guided Practice (p. 51) 1. D. C. B 4. A 5. V 1 Bh 1 ( 5) 7 5 m 6. V 1 πr h 1 π(9) (10) 70π 848 in h 1 5 h 169 h 144 h 1 V 1 Bh 1 (10 10) ft 8. 9 h h 5 h 144 h 1 V 1 πr h 1 π(9) (1) 4π in. 9.5 Practice and Applications (pp ) 9. B ft 10. B in. 11. B π(8) 64π 01 cm 1. V 1 Bh 1. V 1 Bh 1 ( ) 5 1 (8 1) 7 15 ft 4 in. 14. V 1 Bh 15. V 1 Bh 1 (48)(5) 198 yd 80 ft V 1 Bh 1 (4 4) 16 in. 17. V 1 πr h 18. V 1 πr h 1 π() (10) 1 π(1) (7) 0π 6π 94 yd 1056 cm 19. V 1 πr h 0. V 1 Bh 1 π(9) (8) 1 (16)() 16π 16 in. 679 m 1. V 1 πr h. V 1 Bh 1 π(5) (8) 1 (46,55)(144) 0 0 π,,680 m 09 cm. V 1 Bh 1 (6 6) 8 96 ft V 1 Bh 1 (6 6) ft The volume is doubled. 4. V 1 Bh 1 (1 1) 8 84 ft The volume is multiplied by The student used the slant height instead of the height of the pyramid. 5 h 1 5 h 169 h 144 h 1 V 1 Bh 1 (10 10) in. Geometry, Concepts and Skills 151

16 6. The student used the diameter instead of the radius of the base. V 1 πr h 1 π(4) (6) cm 10 cm 4 m 8 m 7. V 1 Bh π ft 0 1 (15)h 0 5h 4 in. h 8. V 1 πr h 9. V 1 Bh 8π 1 πr (6) 10 1 s (10) 8π πr 6 s 4 r 6 m s ft r 0. 8 r h r h 89 r 6 h 5 r 6 h 15 V 1 πr h V 1 Bh 1 π(6) (8) 1 (16 16) 15 96π 180 m 0 ft. 7 h 5 49 h 65 h 576 h 4 V 1 πr h 1 π(7) (4) 9π 1 in h 10 4 h 8 6 h h 64 h 64 h 48 h 8 h 4 V 1 πr h 1 π(6) (8) V 1 πr h 1 π(4) (4 ) 96π 6.95π 0 cm 116 m 6 ft h 5 9 h 5 h 16 h 4 V 1 Bh 1 (6 6) 4 48 ft 6. small container: V 1 πr h 1 π() (6) 18π 57 in. large container: V πr h π() (6) 54π 170 in. 5 ft 6 ft containers 8. The large container gives you more popcorn for your money; Sample answer: You need to buy three small containers for $6 to equal the amount of popcorn in the large container which costs $4. 15 Geometry, Concepts and Skills

17 9. V πr h 1 πr h 49. B π(.5) (7.5) 1 π(.5) (4) π 5 π 17.4 in cups 14.4 in 7 in. 1 cup. 41. Yes; the feeder holds about 17.4 cubic inches and only 7 cubic inches are needed for five days. 4. V 1 πr h 1 π() (1.8) 5.49π 17.5 mi 4. V 1 πr h 1 π(0.4) (0.5) 7 15 π 0.04 mi 44. V mi P (5) (8) h 7 S B Ph (40) 6(7) in. 50. S πr πrh π(5) π(5)(9) 50π 90π 140π 440 ft l l 5 l 15 l S πr πrl π(9) π(9)(15) 81π 15π 16π 679 in. The volume of Mount St. Helens today is about 17.1 cubic miles. 9.5 Standardized Test Practice (p. 516) 45. A; V 1 Bh 96 1 B(9) 9.5 Algebra Skills (p. 516) 5. y 5. (4, ) 1 (0, 0) x y (5, 5) 4 (1, ) x 96 B B 4 8 B 9.5 Mixed Review (p. 516) 46. C πd π(6) 19 m; A πr π() 9π 8 m 47. C πr π(14) 8π 88 in.; A πr π(14) 196π 616 in. 48. C πd π(1) 8 cm; A πr π(6) 6π 11 cm The slope is positive because the line is rising. 54. y (, ) (6, ) x The slope is zero because the line is horizontal. 56. y 57. ( 4, 7) The slope is positive because the line is rising. y (0, 4) 1 1 x (, ) The slope is positive because the line is rising. y (, 5) ( 4, 1) x (4, 1) x The slope is undefined because the line is vertical. The slope is negative because the line is falling. Geometry, Concepts and Skills 15

18 Lesson Checkpoint (pp ) 1. S 4πr. S 4πr 4π(4) 4π(6) 64π 144π 01 in. 45 ft. S 4πr 4π(7) 196π 616 in. 4. V 4 πr 4 π(4) 5 6 π 68 in. 5. V 1 4 πr 6. V 4 πr 1 4 π() 4 π(9) 18π 97π 57 cm 054 ft 9.6 Guided Practice (p. 519) 1. A sphere can be divided into two hemispheres, so a hemisphere is half a sphere.. S 4πr. S 4πr 4π(1) 4π() 4π 6π 1 in. 11 ft 4. S 4πr 4π(9) 4π 1018 m 5. V 4 πr 4 π() 6π 11 ft 6. V 4 πr 7. V 1 4 πr 4 π(11) 1 4 π(10) 5 4 π 40 π cm 094 yd 154 Geometry, Concepts and Skills 9.6 Practice and Applications (pp. 50 5) 8. S 4πr 9. S 4πr 4π(7) 4π(16) 196π 104π 616 cm 17 in. 10. S 4πr 4π(15) 900π 87 m 11. Bob wrote V instead of S for surface area, used the wrong formula, used the diameter rather than the radius, and wrote the answer in cubic units rather than square units; S 4πr 4π(5) 100π 14 mm 1. S 4πr 1. S 4πr 4π(4.) 4π(.) 7.96π 4.56π in. 17 cm 14. S 4πr 15. S 4πr 4π(10.9) 4π(0.85) 475.4π.89π 149 cm 9 in. 16. S 4πr 17. S 4πr 4π(4.75) 4π(4.8) 90.5π 9.16π 84 in. 90 cm 18. No; the surface area of the sphere is 4π() 6π and if you double the radius, the surface area is 4π(6) 144π. So, the surface area will quadruple if you double the radius (144π 6π 4). 19. S 4πr 0. S 4πr 4π(960) 4π(1080) 6,76,400π 4,665,600π 197,000,000 mi 14,700,000 mi , 000, 000 mi , 700, 000 mi The surface area of Earth is approximately 1.4 times larger than the surface area of its moon.. (0.70)(197,000,000 mi ) 18,000,000 mi. V 4 πr 4. V 4 πr 4 π(8) 4 π(4) π π 145 m 68 ft

19 5. V 4 πr 6. V 4 πr 4 π(10) 4 π(11) π π 4189 cm 5575 yd 7. V 4 πr 8. V 4 πr π(7) 4 π(.5) π π 147 ft 180 in. Comparing Spheres A B C 1 Radius, Surface area, Surface area of new sphere r 4π r Surface area of original sphere When the radius is doubled, the surface area is quadrupled. When the radius is tripled, the surface area is nine times greater. When the radius is quadrupled, the surface area is 16 times greater. 0. The surface area is changed by a greater amount than the radius because the radius is squared in the formula S 4πr. 1. Sample answer: If the radius of a sphere doubled, the volume would be, or 8 times greater. If the radius of a sphere is tripled, the volume would be, or 7 times greater.. Comparing Spheres A B C 1 Radius, Volume, Volume of new sphere r 4 π r Volume of original sphere a. When the radius is doubled, the volume is, or 8 times greater. When the radius is tripled, the volume is, or 7 times greater. When the radius is quadrupled, the volume is 4, or 64 times greater. b. The volume is changed by a greater amount than the radius because the radius is cubed in the formula V 4 πr.. S 4πr 4π(4.5) 7569π,779 ft 4. V 4 πr 4 π(4.5) 109,750.5π 44,791 ft 5. V s ,75 ft 6. S B Ph, but excluding the ground and roof (the bases) leaves S Ph. P h 95 S Ph (80)(95) 6,100 ft 7. V 1 4 πr 8. V 1 4 πr 1 4 π(7) 1 4 π(15) 1 7 1, 500 π π cm 7069 m 9. V 1 4 πr 1 4 π(9) 486π 157 in. 40. V πr h 1 4 πr π(10) (18) 1 4 π(10) 1800π 0 00 π π 7749 cm 41. V 1 πr h 1 4 πr 1 π(6) (1) 1 4 π(6) 144π 144π 88π 905 ft Geometry, Concepts and Skills 155

20 4. V πr h 1 4 πr π(5 )(9) 1 4 π(5) 5π 5 0 π 4 5 π 445 in. 4. V 1 4 πr 1 4 π(5.) 64, π 6,917 ft 44. S 1 (4πr ) 45. S 1 (4πr ) 1 (4π(5.) ) 1 (4π(4) ) π 115π 40 ft 619 ft 9.6 Standardized Test Practice (p. 5) 46. A; S 4πr 4π(16) 104π 17 in. 9.6 Mixed Review (p. 5) 47. l l S πr πrl π() π()(1.4) 9π 7.π 46.π 145 m 48. B P l S B 1 Pl 16 1 (16)() 40 ft 49. S πr πrh π(9) π(9)(9) 16π 16π 4π 1018 cm Algebra Skills (p. 5) 58. P w l (6) (11) 1 4 ft 59. A s 100 s s 10 s P 4s 4(10) 40 in. 60. A lw 61. P 4s 40 8w 44 4s 5 m w 11 yd s Quiz (p. 5) 1. V Bh (9 4) 4 (6) in.. V Bh ft. V πr h π(5) (7) 175π 550 m 4. V 1 Bh 1 (10 10) ft 156 Geometry, Concepts and Skills

21 5. V 4 πr 4 π(8) 0 48 π 145 m 6. h V 1 πr h 1 π() (4) 1π 8 in in. 4 in. V πr h S 4πr 9 cm π(4) (4) 4π(9) 64π 4π 01 in cm Chapter 9 Summary and Review (pp ) 1. Polyhedra are named by the shape of their base(s).. The surface area of a polyhedron is the sum of the areas of its faces.. A prism is a polyhedron with two congruent faces that lie in parallel planes. 4. The lateral faces of a prism are the faces of the prism that are not bases. 5. The volume of a solid is the number of cubic units contained in its interior. 6. The solid is a polyhedron with pentagonal bases, so it is a pentagonal prism. 7. A sphere has a curved surface, so it is not a polyhedron. 8. The solid is a polyhedron with a hexagonal base, so it is a hexagonal pyramid. 9. S B Ph ( 5) ( 5)() (15) (16)() ft 10. S πr πrh π() π()(6) 18π 6π 54π 170 in. 11. S B Ph (6 8 10)(9) (4) (4)(9) m 1. S B 1 Pl (5 5) 1 (4 5)(7) 5 1 (0)(7) m 1. S πr πrl π(8) π(8)(11) 64π 88π 15π 478 in. 14. l 4 5 l 5 5 S B 1 Pl (6 6) 1 (4 6)(5) 6 1 (4)(5) ft 15. l l S πr πrl π(6) π(6)(10) 6π 60π 96π 0 cm 16. S B 1 Pl ( 4)(6) ft Geometry, Concepts and Skills 157

22 17. l l S πr πrl π(5) π(5)(1) 5π 65π 90π 8 ft 18. V Bh (6 8)() (48)() 144 in. 19. V Bh () (10)() 0 m 0. V πr h 1. V πr h π() (9) π(9) (8) 81π 648π 54 cm 06 ft. V Bh. V πr h (4 4)(6) π(10) (14) (16)(6) 1400π 96 yd 498 in. 4. V 1 Bh 1 ( 5)(6) 0 m 5. V 1 πr h 1 π(5) (1) 100π 14 in. 6. h 9 15 h 81 5 h 144 h 1 V 1 πr h 1 π(9) (1) 4π 1018 cm 7. S 4πr 8. S 4πr 4π(6.5) 4π(18) 169π 196π 51 m ; 407 cm ; V 4 πr 4 π(6.5) 1150 m 9. S 4πr 4π(.85) 59.9π 186 ft ; V 4 πr 4 π(.85) 9 ft Chapter 9 Chapter Test (p. 58) V 4 πr 4 π(18) 7776π 4,49 cm 1. The solid is a polyhedron with hexagonal bases, so it is a hexagonal prism.. The solid is a polyhedron with a square base, so it is a square pyramid.. A cylinder has a curved surface, so it is not a polyhedron. 4. The solid is a polyhedron with triangular bases, so it is a triangular prism. 5. S B Ph (4 9) ( 4 9)(5) (6) (6)(5) ft 6. S B Ph 7. S πr πrh ( ) (4 )() π(4) π(4)(5) (9) (1)() π 40π π 54 in. 6 cm 8. S πr πrh π(5) π(5)(7) 50π 70π 10π 77 m 158 Geometry, Concepts and Skills

23 9. l l S πr πrl π(8) π(8)(10) 64π 80π 144π 45 in. 10. S B 1 Pl (6 6) 1 (4 6)(5) yd 11. cylinder; S πr πrh π() π()(10) 18π 60π 78π 45 ft ; V πr h π() (10) 90π 8 ft 1. V Bh (7) (16)(7) 11 yd 1. V Bh (6 )() (1)() 6 in. 14. V 1 Bh 1 (6 6)(9) 1 (6)(9) 108 m 15. V πr h π() (5) 0π 6 ft 16. V 4 πr 4 π(9) 97π 054 ft 17. V 1 πr h 1 π() (5) 0 π 1 cm 18. The volume is quadrupled. V π(r) h 4πr h 19. The volume is doubled. V πr h(h) πr h 0. The volume is multiplied by 8. V 4 π(r) 8 4 πr 1. V Bh (14 6)(8) (84)(8) 67 in.. B P (14) (6) 40 h 10 There is only one base of the aquarium, so S B Ph 84 (40)(10) 484 in. Chapter 9 Standardized Test (p. 59) 1. B. G; S πr πrh. C; π() π()(7) S 4πr 8π 8π 4π(4) 6π 64π 11 in. 01 cm Geometry, Concepts and Skills 159

24 4. G; B h h V 1 Bh 1 (144)(8) 84 ft 5. C; When s : S B Ph ( ) (4 )() (9) (1)() ft When s 6: S B Ph (6 6) (4 6)(6) (6) (4)(6) ft J; V 1 πr h 48π 1 π(6) h 48π 1πh 4 8π h 1π 4 in. h 7. B; V Bh πr h (6 6)(8) π(1) (6) (6)(8) 6π 88 6π 69 ft 8. V Bh 1 Bh (6 6)(4) 1 (6 6)(4) (6)(4) 1 (6)(4) cm 9. S B Ph 1 Pl (6 6) (4 6)(4) 1 (4 6)(5) 6 (4)(4) 1 (4)(5) cm 10. The volume of the prism section is times the volume of the pyramid section. Sample answer: The prism and the pyramid have the same base and height; the formulas for the volumes of a prism and a pyramid tell us that the volume of the prism is times the volume of the pyramid. Chapter 9 Algebra Review (p. 51) Chapters 1 9 Cumulative Practice (pp. 5 5) 1. B( 1, ) 8 A(4, 6) 8 x C(5, 1) aabc is acute because its measure is less than 90.. By the Linear Pair Postulate, ma ma1 54. By the Vertical Angles Theorem, ma ma1 54 and ma 16.. Perpendicular lines form 4 right angles, so ma1 ma Geometry, Concepts and Skills

25 4. By the Vertical Angles Theorem, ma 118. By The Corresponding Angles Theorem, ma4 ma By the Linear Pair Postulate, ma By the Alternate Interior Angles Theorem, ma6 81. By the Alternate Exterior Angles Theorem, ma7 ma maa mab mac x x 180 x All three angles of T ABC are acute, so T ABC is acute. 8. maa mab mac, so BC AC AB. The sides of the triangle from longest to shortest are BC ****, **** AC, and **** AB. 9. Yes; T ABC T DCB by the SSS Congruence Postulate because **** AB DC **** (Given), **** AC DB **** (Given), and BC **** BC **** (Reflexive Property of Congruence). 10. Yes; T JLK T NLM by the AAS Congruence Theorem because ak am (Given), aklj amln (Vertical Angles Theorem), and **** JL NL **** (Given). 11. Yes; T PQR T SRT by the HL Congruence Theorem because aq ar (Corresponding Angles Theorem) and aq is a right angle, so ar is a right angle. In right triangles PQR and SRT the hypotenuses PR **** and ST **** are congruent and the legs QR **& and RT **** are congruent (Given). 1. Two acute angles are sometimes complementary. 1. The sides of a rhombus are always congruent. 14. A rectangle always has exactly two lines of symmetry. 15. Two squares are sometimes congruent. 16. UR 1 (PQ TS) x 1 (7 1) 1 (0) AD BC 18. JK ML x 1 8 x 5 4x 5 x 9 5 x 5 x, 10 x maa mac 5 x, 70 y, maj 90 maa mab 180 y z 180 z AA Similarity Postulate; aabc aade and aacb aaed by the Corresponding Angles Theorem. 0. D B EC DA E A AB A 1. C B D CE x x 4 6 4x 1 x. The polygon has 5 sides so it is a pentagon.. Convex; Sample answer: None of the extended sides pass through the interior. 4. (n ) 180 (5 ) The sum of the exterior angles is 60 by the Polygon Exterior Angles Theorem. 5. (n ) 180 (8 ) Square; Sample answer: The sum of the measures of one interior angle in each of octagons and in one of the yellow figures is x x 60 x 90 So, each interior angle of the yellow figure must measure 90. Since the sides of the yellow figure are congruent, the yellow figure is a square. 7. A s m 8. A 1 bh 1 (8)(17) 68 ft 9. A bh (14)(8) 11 cm 0. A 1 h(b 1 b ) 1 (4)(4 9) 1 (4)(1) 6 in. Geometry, Concepts and Skills 161

26 1. S B Ph (4 10) ( 4 10)(8) (40) (8)(8) m ; V Bh (4 10)(8) (40)(8) 0 m. l l S 4πr 4π(16) 104π 17 in. 6. V 4 πr h 4 π(16) 17,157 in. 1 V 1 (17,157) 8579 in. S B 1 Pl (10 10) 1 (4 10)(1) (40)(1) ft ; V 1 Bh 1 (10 10)(1) 400 ft. S πr πrh π() π()(9) 18π 54π 7π 6 cm ; V πr h π() (9) 81π 54 cm 4. S πr πrl π(1) π(1)(0) 144π 40π 84π 106 cm ; h 1 0 h h 56 h 16 V 1 πr h 1 π(1) (16) 768π 41 cm 16 Geometry, Concepts and Skills