Chapter 8 Exercise 8.1
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- Collin Hodges
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1 Chapter 8 Exercise 8.1 Q. 1. (i) 9 cm (vi) 4(7) + (8)(7) = 56 cm (ii) 45.5 cm (vii) 7.(8.7) = 6.64 cm (iii) 90 cm (viii) ( )(4) = 3,19 cm (iv) 100 cm (ix) (4)(6) = 1 cm (v) 1 ( ) (5 + 40) = 390 cm (x) 10(1) = 10 cm Q.. (i) Area = 5(30) + 15(4) = 10 cm Perimeter = = 100 cm (ii) Area = 33(35) 7(13) = 1,064 cm Perimeter = = 150 cm (iii) Area = 11(1.5) = 36.5 cm Perimeter = = 54 cm (iv) Area = 13(8) + 1(17) + (5)(1) = 338 mm. (v) Area = 1(15) + 1(15) + 6(1) = 36 cm Perimeter = 4(1) + (15) + (6) + (14) = 74 cm (vi) m n 0 1 m Area = 16(3.5) + 1() + 4(0) + (1)(16) = 56 m Perimeter = () (3.5) (4) + 0 = 67 m (vii) Area = 4(1.5) + 9(.3) + 1.( ) = cm Perimeter = = 47.6 cm 1 Q. 3. (i) (1)(10) = 60 cm (ii) = 00,000 cm 00, = 1, Answer = 1,667 tiles (iii) 100,000 cm or 10 m 1 Q. 4. (i) (0)(8) = = 40 Clear portion (0)(8) = = 16 Blue portion Total cost = 56 (ii) 4.48 m vs 80 cm 80 cm = (0)(8) = (0.)(0.08) m = m 4.48 = 560 panels Active Maths 3 Book Strands 1 5 Ch 8 Solutions 1
2 Q (3)(1) = 18 cm (1 + 4)(1) = 96 cm = 78 cm % = 81.5% = = 5 = Q. 6. (a) (i) = 81 = ( )(1)(9) = 16 m (ii) (1 + 9) 16 = 5 m (b) 441(19) = 8,379 5(10)(1.1) =, (1.5)(1.1) = (5) =,05 8,379 (, ,05) = 3, ,059 Exercise 8. Q. 1. Area Circumference (i) 3.14(10) = 314 cm (3.14)(10) = 6.8 cm (ii) 3.14(1) = 3.14 m (3.14)(1) = 6.8 m (iii) 3.14(10) = 45,16 mm (3.14)(10) = mm (iv) 3.14(160) = 80,384 cm (3.14)(160) = 1,004.8 cm Q.. Area Circumference (i) 7 (3) = 8 _ 7 cm ( 7 )(3) = 18 6_ 7 cm (ii) 7 (140) = 61,600 cm ( 7 )(140) = 880 cm (iii) 7 (89) = 4,894 4_ 7 cm ( 7 )(89) = 559 3_ 7 cm (iv) 7 (5.6) = cm ( 7 )(5.6) = 35. cm Active Maths 3 Book Strands 1 5 Ch 8 Solutions
3 Q. 3. Area Circumference (i) p(10) = 100p cm p(10) = 0p cm (ii) p(4) = 16p mm p(4) = 8p mm (iii) p(1) = 144p m p(1) = 4p m (iv) p(0.5) = 0.065p cm p(0.5) = 0.5p cm Q. 4. Area Arc Length Perimeter (i) = 31.4 cm = 6.8 cm 6.8 cm (ii) = 35.5 cm = 31.4 cm 61.4 cm (iii) = = 4. cm 37.4 cm (iv) = 905 1_ 30 cm = 100 4_ 7 cm 136 4_ 7 cm (v) = 1,039.5 cm = 99 cm 141 cm Q. 5. (i) Area = 7(15) + (3.14)(3.5) = cm Perimeter = 7 + (15) = cm (ii) Area = (33)(56) + (3.14)(3.5) =, cm (65 = 3.5) Perimeter = 3.14 ( 65 ) = cm = 45 = 65 (iii) Area = (0 + 14)(0 + 8) (3.14)(14) = 1,34.8 m Perimeter = ()(3.14)(14) = m (iv) Area = 3 4 (3.14)(15) + 15 = cm Perimeter = ()(3.14)(15) = cm (v) Area = 34(0) (3.14)(10) (0)(14) = 383 cm Perimeter = (34) + ()(3.14)(10) + ( 96 ) cm = = cm 0 cm cm Q. 6. (3.14)(3,100) = 19,468 km = 19,468,000 m Time = 19,468,000 = 3,44 _ 6,000 3 seconds = minutes (= 54 minutes 4 _ 3 seconds) 54 minutes Active Maths 3 Book Strands 1 5 Ch 8 Solutions 3
4 Q. 7. (i) 0 = x 0 00 = x 00 = x 10 = x radius = cm (ii) Circumference (large) = (3.14)(10) x x Circumference (small) = ( 3.14)(5 ) = 6.8 cm = cm = minutes = Q = 150,000 cm 7 minutes 4 seconds (3.14)(14) + (3.14)(7) (3.14)(3.5) = cm (i) 150, = parts 433 complete parts (ii) ()(3.14)(14) + ()(3.14)(7) + (3.14)(3.5) = 87.9 cm Exercise 8.3 Q. 1. (i) 30 5 = 6 (ii) = 0 0 = 10 (iii) x = 100 x = 10 Q.. p r Area Circumference p 7 49p 14p , (iv) ( )(x) = x = 99 x = 1 (v) (9)(x) = x = 180 x = 40 (vi) 133 (10)(5) = = 9 (vii) 6.5x + ( )(x) = x x = x = x = 4.3 p 35 1,5p 70p _ p 9 Q. 3. (i) 7p = 6 p r 1 81 p 9 p 1 = r Ans: r = 1 cm (ii) r = 1,413 r = 3,600 r = 60 cm 4 Active Maths 3 Book Strands 1 5 Ch 8 Solutions
5 q (iii) = 78 4_ q = 78 4_ 7 q = 90 q (iv) p 30 = 3.8p q = 3.8 q = 14.8 (v) Exercise 8.4 q (4) = q = q = Q. 1. (i) (30 + 1) = 10 cm (ii) 30(1) = 630 cm Q.. (i) (5 + 8) = 66 m (ii) 5(8) = 00 m Q. 3. (i) 1 4 = 3 m (ii) = 5 m Q. 4. (i) 1 = = = 5 m (ii) = 13 m Q. 5. (i) (x + x + 10) = 100 x + 10 = 50 x = 40 x = 0 Area = 0(30) = 600 cm Q. 6. (i) (150 6)(75 6) = 9,936 m (ii) 150(75) 9,936 = 1,314 m (iii) 1,314 = = 3,449.5 Q. 7. (i) ( 7 )(r) = 308 r = 49 cm (ii) 308(10,000) = 3,080,000 cm = 30,800 m = 30.8 km Q. 8. pr = 64p r = 64 r = 8 cm Q. 9. (i) pr = 5p r = 5 r = 15 m (ii) p(15) = 30p = 94. m Q. 10. (i) 7 (15) = 707 1_ 7 m 7 (16.5) = Ans = m 14 m (ii) = Q. 11. Yellow: 1 p(4) cm Red: 15 p(6) cm Blue: 10 ( 4 ( p(7) + p(3) + p() ) ) = cm Red has largest shaded area Pick red Q (3.14)(30) + 30x = x = x = x = 1.01 m (1.01) + (30) + 4 ()(3.14)(30) = m Q. 13. (i) (134) + ( 7 )(1) = 400 m (ii) ( 7 )(8) = = 4 4 = 11 m Active Maths 3 Book Strands 1 5 Ch 8 Solutions 5
6 Q. 14. (x + y) = 34 x + y = 17 y = 17 x xy = 30 x(17 x) = 30 17x x = 30 x 17x + 30 = 0 (x )(x 15) = 0 x = or x = 15 Width = m y = 15 or y = } Length = 15 m Q. 15. (i) (p)(4) : (p)(16) = 8 : 3 = 1 : 4 (ii) p(4) : p(16) = 16 : 56 = 1 : 16 Q. 16. (i) (10)(5) = 50 cm (ii) p(5) 50 = cm x Q. 17. Radius of inner circle = x Area inner circle = ( x ) p = p x 4 Radius of inner shaded region = x + 3x = 7 x Area of inner shaded region = ( 7x ) p px 4 = 1px Radius of outer shaded region = x + 5x = 11 x Area of outer shaded region = ( 11 x ) p ( 9 x ) p = 10px The inner shaded region has the greater area Q. 18. (i) 64p = pr r = 8 m r tarpaulin = = 850 cm = 8.5 m (ii) p(10) p(8.5) = 7.75p = 7.75(3.14) = m Exercise 8.5 Q. 1. Volume Surface Area (i) 1,680 cm cm (ii) 60 cm 3 94 cm (iii) m m (iv) 13,000 mm 3 3,80 mm (v) 4.14 m m (vi) 15 mm mm 6 Active Maths 3 Book Strands 1 5 Ch 8 Solutions
7 Q.. Volume Surface Area (i) 105 cm 3 (7)(5) + (3)(5) + (3)(7) = 14 cm (ii) 64 cm 3 6(4) = 96 cm (iii) 450 m 3 (10)(3) + (15)(3) + 10(15) = 300 m Q. 3. (i) 4 cm 4 cm 4 cm (ii) 10 cm 7 cm 4 cm Q. 4. (i) Figure 4 (ii) Q. 5. (i) (6)(10) + (6)(4) + (10)(4) = 48 cm (ii) 6(10)(4) 3 = 40 = 30 cubes 8 Q. 6. (i) (55)(36)(10) = 9,900 cm3 9,900 1,000 = 9.9 litres (ii) 55(36) + (55)(10) + (36)(10) = 3,800 cm Exercise 8.6 Q. 1. (i) Volume = [45(5) + 5(0)] 30 = 48,750 cm 3 Surface Area = [45(5) + 5(0)] + 5(30) + 5(30) + 50(30) + 0(30) + 5(30) + 45(30) = 8,950 cm (ii) Volume = [4(15) + 10(8) + 6(15)] 14 = 3,0 cm 3 Surface Area = [4(15) + 10(8) + 6(15)] + 0(14) + 15(14) + 7(14) = 1,636 cm (iii) Volume = [(4)(11) + (15)(18)] 6 = 3,04 cm 3 Surface Area = [4(11) + 15(18)] + 9(6) + 4(6) = 1,704 cm (iv) Volume = (15 + 5)(1) 30 = 3,600 m3 Surface Area = (15 + 5)(1) + 13(30) + 15(30) + 5(30) = 1,60 m Active Maths 3 Book Strands 1 5 Ch 8 Solutions 7
8 (v) Volume = [4 + (4)(.1)] 5 = 101 m3 Surface Area = [4 + (4)(.1)] + 4(5) + (4)(5) +.9(5)() = 19.4 m.1 + = =.9 (vi) Volume = [18(19) (9.5) ] 50 = 156,346.5 cm 3 Surface Area = [18(19) (9.5) ] + 18(50) + (3.14)(9.5)(50) = 5, cm Q.. (i) (iii) r pr (ii) (iv) (v) Q. 3. Volume Surface Area (i) 7 cm 3 54 cm (ii) 43,00 cm 3 0,160 cm (iii) 40 cm cm = 1681 = Q. 4. ()()(8) = 16 cm3 Q. 5. (i) ( 7 ) (7) = 77 m (ii) = 3,080 m 8 Active Maths 3 Book Strands 1 5 Ch 8 Solutions
9 Q. 6. (i) (6)() + 6() = 18 m (ii) 18 8 = 144 m 3 Q. 7. (i) (60 30 ) 00 = 540,000 cm 3 Note: 1 m 3 = (100) 3 cm 3 Answer = 0.54 m 3 = 1,000,000 cm 3 (ii) = 47.5 Exercise 8.7 Q. 1. (i) 3.14(10) (10) = cm 3 (ii) 7 (1.) (4) = (iii) p(9) (8) = 648p m cm3 (iv) 7 (5) (1) = 41,50 m 3 (v) 3.14(5) (14) = 1,099 mm 3 Q.. CSA TSA (i) (3.14)(0.5)() = 3.14 m (3.14)(0.5) = m (ii) (p)(3)(5) = 30p cm 30p + (p)(3) = 48p cm (iii) (3.14)(5)(11) = cm (3.14)(5) = 50.4 cm (iv) ( 7 )()(44) = 6,084 4_ 7 cm _ 7 + ( 7 )() = 9,16 6_ 7 cm Q. 3. Volume TSA (i) 7 (14) () = 13 mm 3 ( 7 )(14)() + ( 7 )(14) = 1,408 mm (ii) p(.5) (5) = mm 3 (p)(.5)(5) (p)(.5) = mm Q. 4. (i) r = 0.9 cm (ii) r = 1.45 cm h = 6.4 cm h = 4.1 cm Volume = 3.14 (0.9) (6.4) Volume = 3.14(1.45) (4.1) = 16.8 cm 3 = 7.07 cm 3 CSA = (3.14)(0.9)(6.4) CSA = (3.14)(1.45)(4.1) = cm = cm Q. 5. (i) r = m (ii) r = 4.5 m h = 4.6 m h = 7 m Volume =.10p m 3 Volume = p m 3 TSA = 7.1p m TSA = 83.5p m Q. 6. (i) 3.14(14) () = 13, cm 3 (ii) (3.14)(14)() + (3.14)(14) 1 litre = 1,000 cm 3 = 3,165.1 cm Ans = litres litres Active Maths 3 Book Strands 1 5 Ch 8 Solutions 9
10 Q. 7. (i) p(16) (10) p(8) (10) =,560p cm 3 = 640p cm 3 (ii),560p : 640p =,560 : 640 = 4 : 1 Q. 8. r = 37.5 cm h = 300 cm Volume = 3.14(37.5) (300) = 1,34,687.5 cm 3 1,34, = 1,371, g (1 litre = 1,000 cm 3 TSA = (3.14)(37.5)(300) + (3.14)(37.5) 1 ml = 1 cm 3 ) = 79,481.5 cm 79, = 198, g Ans = 1,569, g 1, kg Exercise 8.8 Q. 1. (i) 3 (p)() (4) = 16p (ii) 3( 3 cm3 (iv) 3 (3.14)(150) (0.5) = 5,887 1_ cm3 7 ) (40) (1) = 35,00 mm 3 (v) 3( (iii) 3 (3.14)(40) (3) = 5,04 cm 3 Q.. (i) l = = 10 m CSA = p(6)(10) = 60p m TSA = 60p + p(6) = 96p m (ii) CSA = 3.14(11)(61) =, cm 7 ) (5) (14) = 9,166 _ 3 cm3 (vi) 3 (3.14)(5) (16) = 418 _ 3 cm3 TSA =, (11) =, cm (iii) CSA = { (8)(35) = 3, = 1 + r mm 7 TSA = 3, (8) = mm r = 35 1 r = 784 r = 8 mm (iv) CSA = p(30)(50) = 1,500p cm TSA = 1,500p + p(30) =,400p cm Q. 3. (i) Vol = 3 p(0) (0) = cm 3 TSA = p(0)(0 ) + p(0) = cm l = (ii) Vol = 3 p(1.8) (.6) = 8.8 m 3 TSA = p(1.8)( 10 ) + p(1.8) = 8.06 m l = Active Maths 3 Book Strands 1 5 Ch 8 Solutions
11 Q. 4. (i) A: 3 (3.14)(16) (1) = 3,15.36 m 3 = litres B: 3 (3.14)(4) (0) = m 3 = litres (ii) A: r = 16 m h = 1 m l = = 400 = 0 m TSA = 3.14(16)(0) (16) = 1, m B: r = 4 m h = 0 m l = = 416 = 4 6 m TSA = 3.14(4)(4 6 ) (4) m One sheet = 3 m A: = sheets (603 full sheets/panels) 3 B: = sheets (103 full sheets/panels) 3 Q. 5. (i) r = 00 = 400 m h = 4 00 = 800 m l = = 800,000 Volume = 3( 7 ) (400) (800) = 134,095,38.10 m 3 TSA = 7 (400)( (ii) r = = 0.15 km 800,000 ) + h = 0.15 km l = = Volume = 3( 7 ) (0.15) ( 3 7 (400) = 1,67, m 1,67, = ) = km km 3 TSA = 7 ( (0.15) 3 0 ) + 7 (0.15) = km 0.17 km Q. 6. l = = 1,369 = 37 TSA = (3.14)(1) (1)(37) = 1,846.3 cm 1, = 9,31.60 Q. 7. Cone: h = 5 4 = 9 = 3 m Volume = p(4) (8) + 3 (p)(4) (3) = 18p + 16p = 144p m 3 [Note: 1 m 3 = (100) 3 cm 3 = 1,000,000 cm 3 = 1000 litres] Active Maths 3 Book Strands 1 5 Ch 8 Solutions 11
12 Exercise 8.9 Q. 1. (i) 3 (p)(9)3 = 97p cm 3 (ii) 3( 7 ) (1)3 = 7,41 1_ 7 cm3 (iii) 3 (3.14)(5)3 = 53 1_ 3 mm3 (iv) 3( 7 ) (10.5)3 = 4,851 cm 3 (v) 3 (p)(4)3 = 56 3 p m3 Q.. (i) 3( 7 ) (13)3 = 4, cm3 (ii) 3 p(6)3 = 144p m 3 (iii) 3 (3.14)(1.5)3 = m 3 (iv) 3 (p)()3 = 16 3 p cm3 (v) 3 (3.14)(4)3 = 155, cm 3 Q. 3. (i) 4(p)(9) = 34p cm (ii) 4 ( 7 ) (1) = 1,810 _ 7 m Q. 4. (i) 3(3.14) 5 = 35.5 mm (ii) 3 ( 7 ) (100) = 94,85 5_ 7 km Q. 5. (i) TSA = 4p() = 16 pcm (ii) CSA = (3.14)(15) = 1,413 m TSA = 3 1,413 =,119.5 m (iii) CSA = (3.14)(7) = 4,578.1 cm TSA = 3 4,578.1 = 6, cm (iv) TSA = 4 ( 7 ) ( 14 1_ 4 ) =, cm Q. 6. (i) Volume = 3 (p)(1.5)3 = p m3 TSA = 3(p)(1.5) = p m (ii) Volume = 3 (p)(0.5)3 = 6 p km3 TSA = 3(p)(0.5) = 3 4 p km Q. 7. r = 10.8 (i) 4(3.14)(10.8) = 1, ,465 cm (ii) 3 (3.14)(10.8)3 = 5, ,74 cm 3 Q. 8. (i) 3( 7 ) (7)3 = 1,437 1_ 3 cm3 (ii) 14 3 =,744 cm 3 (iii),744 1,437 1_ 3 = 1,306 _ 3 1,306 _ 3,744 = % 47.6% 1 Q. 9. (i) ( 7 ) (6) (14) = 1,584 cm 3 (ii) ( 3 )( 7 ) (4.5)3 = 381 6_ 7 cm3 (iii) Answer (i) Answer (ii) = 1,0 1_ 7 cm3 Q. 10. (i) 3 (3.4)(3.5) cm 3 (ii) 6(3.5) = 1 cm (iii) 3.5 cm (iv) (3.14)(3.5) (1) = cm 3 (v) 3 ( 53, ) = 53,851 = cm = 3 Q ( 7 ) (3.5) (6) + 3( 7 ) (3.5)3 = 166 5_ 6 cm3 Q (3.14)()3 = cm3 167 cm = (per ball) 4(3.14)() = 50.4 cm = (per ball) Cost per ball: = Total Cost = 43, , Active Maths 3 Book Strands 1 5 Ch 8 Solutions
13 Exercise 8.10 Q. 8. (i) L (cm) B (cm) H (cm) V (cm 3 ) SA (cm ) _ 3 1_ _ Workings: (a) 180 4(5) = 9 (4)(5) + (4)(9) + (5)(9) = 0 (b) 6 (3) = 1 ()(1) + ()(3) + (1)(3) = (c) 10 (7.5) = 14 (14)(7.5) + (14)() + (7.5)() = 96 (d) 5 ( 1 1_ ) ( 3 1_ 3 ) = 1 (1) ( 1 1_ ) + (1) ( 3 1_ 3 ) + ( 1 1_ ) ( 3 1_ 3 ) = 19 _ 3 (e) (0.5)(x) + (0.5)(0.5) + (x)(0.5) = 1 x x = 1 1.5x = 0.75 x = 0.5 Volume = (0.5) (0.5) = (ii) p r (cm) h (cm) V (cm 3 ) CSA (cm ) TSA (cm ) p p 40p 90p , , , , ,418 3_ _ 7 1,046 4_ 7 Workings: (a) p(5) (h) = 100 p h = 4 (p)(5)(4) = 40p 40p + (p)(5) = 90p (b) 3.14(r) (11) =, r = 81 r = 9 (3.14)(9)(11) = (3.14)(9) = 1,130.4 p (c) (3.14)(13)(h) = h = (13) (1) = (3.14)(13) = 1,14.96 (d) ( 7 )(r)(14) = 616 r = 7 7 (7) (14) =, ( 7 ) (7) = 94 (e) 1,046 4_ 7 = ( 7 ) (9)(h) + ( 7 ) (9) (iii) r (cm) 1,046 4_ 7 = 509 1_ 7 + (56 4_ 7 )(h) 537 3_ 11 7 = (56 7 )(h) 9.5 = h 7 (9) (9.5) =,418 3_ 7 ( 7 ) (9)(9.5) = 537 3_ 7 h (cm) l V (cm 3 ) (cm) CSA (cm ) TSA (cm ) p p 0p 36p , ,391. 1, , , , , ,485 5_ 7 6,348 4_ 7 7, Workings: (a) 3 ( p)(4) 4 (h) = 16 p h = = 5 = l p(4)(5) = 0p 0p + p(4) = 36p (b) 3( 7 ) (7) (h) = 1,3 ( 51 1_ 3 )(h) = 1,3 h = = 5 = l (7)(5) = (7) = 704 Active Maths 3 Book Strands 1 5 Ch 8 Solutions 13
14 (c) prl = (3.14)(r)(41) = 1, r = 1, r = 9 l = h + r 41 = h + 9 1,600 = h 40 = h Volume = 3 (3.14)(9) (40) = cm 3 TSA = 1, (3.14)(9) = 1,413 cm (d) TSA prl + pr = 4, (3.14)(r) = 4, r = r = 169 r = 13 l = h + r l = (84) + (13) l = 75 l = 85 Volume = 3 (3.14)(13) (84) = 14, cm 3 (e) 3( 7 ) (r) (99) = 41,485 5_ 7 r = 400 r = = (0)(101) = 6,348 4_ 7 6,348 4_ (0) = 7,605 5_ 7 (iv) (a) Spheres p r (cm) V (cm 3 ) TSA (cm ) p _ 3 p 100p _ , , p 13,99 1_ 3 676p Workings: (i) 3. p. r3 = 166 _ 3 p r 3 = 15 r = 5 4(p)(5) = 100p (ii) 4 ( 7 ) (r) = 50 _ 7 r = 4 r = = (iii) 3 (3.14)(r)3 = 3,05.08 r 3 = 79 r = 9 4(3.14)(9) = 1, (iv) 4( p)(r) = 676 p r = 169 r = 13 3 (p)(13)3 =,99 1_ 3 p (b) Hemisphers p r (cm) V (cm 3 ) TSA (cm ) CSA (cm ) p _ 3 p 75p 50p _ 7 5 1_ p 13 1,464 _ 3 507p 338p Workings: (i) 3 ( p)(r)3 = (83 1_ 3 ) p r 3 = 15 r = 5 3(p)(5) (p)(5) = 50p (ii) 3 ( 7 ) (r) = 37 5_ 7 r = 4 r = 3( 7 ) ()3 = ( 7 ) () = 5 1_ 7 14 Active Maths 3 Book Strands 1 5 Ch 8 Solutions
15 (iii) (3.14)(r) = r = 81 r = 9 3 (3.14)(9)3 = 1, (3.14)(9) = (iv) ( p)(r) = 338 p r = 169 r = 13 3 (p)(13)3 = (1,464 _ 3 )p 3(p)(13) = 507p Exercise 8.11 Q = 18 cm Q.. (5)(40) + (5)(w) + (40)(w) = 5,900, w + 80w = 5, w = 3,900 w = 30 cm Q. 3. (i) 3 (p)(6)3 = 88p cm 3 Q. 4. Q. 5. Q. 6. (ii) 88 p = 3 ( p)(3) (h) 96 cm = h 3 (p)(9)3 = 97 p cm 3 97p = 3 ( p)(r) (81) 97 = 7r 36 = r 6 cm = r 3 (3.14)(r)3 = 1,56.04 r 3 = 79 r = 9 cm 3 ( p)(10) (h) = p(4) (8) ( ) h = 18 h = 3.84 cm h = 38.4 mm h 38 mm Q. 7. (4)(5) = 40 cm 3 0.9(40) = 36 cm 3 18 = 3 (3.14)r = r = r r 1.63 cm Q. 8. r = 1.5 cm p(1.5) (h) + 3 ( p)(1.5)3 = 9.5p.5 h +.5 = h = 7 h = 1 cm Q. 9. r = 6 cm 3 ( p)(6) (h) + 3 ( p)(6)3 = 64 p 1h = 64 1h = 10 (i) h = 10 cm (ii) = 136 = 34 6 p(6)( 34 ) + (p)(6) = ( )p = ( ) ( 7 ) = cm Q. 10. (a) (i) Hemisphere: 3 (p)(6)3 = 144p cm 3 (ii) Cone: 48p cm 3 48 p = 3 ( p)(6) (h) 48 = 1h 4 cm = h (iii) = 10 cm (b) p(6) (10) = 360p cm 3 Toy: 144p + 48p = 19p 19p 360p = 8 15 > More than half Active Maths 3 Book Strands 1 5 Ch 8 Solutions 15
16 Q. 11. (i) h = 3 h = 3 cm 5 (ii) 3 (p)(4) (3) = 16p cm 3 (iii) 160p = p(4) (x) 160 p = 16x p 10 = x 10 cm Q. 1. h = x l = x w = x 486 = x ( x ) + (x)(x) + ( x ) (x) 486 = x + 4x + x 486 = 6x 81 = x 9 = x h = 9 cm l = 18 cm w = 4.5 cm Q. 13. Let: h = x l = x w = 3x x(x)(3x) = 10,368 6x 3 = 10,368 x 3 = 1,78 x = 1 1, 4, 36 (cm) Q. 14. Sphere: 3 p(6)3 = 88p p(1) h = 88p 144h = 88 h = cm Q. 15. Cone: 3 (p)() (3) = 4p cm 3 4 p = p(4) (h) 4 = h 0.5 cm is answer Q. 16. (i) p(3) (7) + 3 (p)(3)3 = 63p + 18p = 81p cm 3 (ii) p(3) (h) = 81p h = 9 cm Q. 17. Per second: 7 () (35) = 440 cm 3 litres =,000 cm 3,000 = 50 seconds 440 Q. 18. Cone 1 Cone r = 3x r = x h = 5y h = 1y 3 (p)(3x) (5y) 3 (p)(x) (1y) 3 (p)(9x )(5y) 3 (p)(4x )(1y) 15x yp 16x yp The second cone has greater volume Q. 19. A B C r = 5x r = 3x r = x h = 3y h = 8y h = 19y p(5x) (3y) p(3x) (8y) p(x) (19y) p(5x )(3y) p(9x )(8y) p(4x )(19y) 75x yp 7x yp 76x yp Q C has greatest volume Q. 0. Per second: p ( 7 ) (10) = 1.5p cm 3 1.5p(60) = 7,350p (in one minute) = 7,350(3.14) = 3,079 cm 3 3,079 = cm (55)(0) Q. 1. (i) 3 (p)(1.5)3 =.5p cm 3 (ii) 4(.5p) = 54p cm 3 54 p = p(6) (h) 54 = 36h 1.5 cm = h 1.5 cm drop (iii) p(6) (.5) = 90p 90 p = 40 ladlefuls.5 p 16 Active Maths 3 Book Strands 1 5 Ch 8 Solutions
17 Exercise 8.1 Q. 1. [1 + + ( )] = 73 units Q.. 5 [ ( )] = 330 units Q [ ( )] = 405 units Q. 4. [ ( )] = 308 units Q Q [ ( )] =,60 units [ ( )] = units Q [ ( )] = 70 units Q. 8. (i) 50 [ ( )] = 67,750 m = hectares. (ii) 180 [ ( )] = 117,000 m = 11.7 hectares. Q [x ( )] = 3 [x ] = 3 [x + 73] = 3x + 19 Q. 10. Q = 3x = 3x 11 3 = x Ans: x = 3 _ 3 y [ ( )] = y [160] = 80y 1,00 = 80y 15 = y 3 [ ( x + )] = 3 [ (9 + x)] = 7 + 3x 40 = 7 + 3x 13 = 3x 13 3 = x Ans: x = 4 1_ 3 Active Maths 3 Book Strands 1 5 Ch 8 Solutions 17
18 Q. 1. h [ ( )] = h [000] = 1,000h ,000 = 1,000h 15, 000 = 1,000h h = 15 Revision Exercises Q. 1. (a) (i) 9(14) + 5() = 516 cm (ii) (100)(30) + (100)(70) = 5,000 mm (iii) 6(11) = 86 m (iv) 7(55) + ( )(13) =,089.5 cm (v) Area = (0 + 14)(8) = 136 m (b) Area Perimeter (i) p(5) = 65p cm (p)(5) = 50p cm (ii) 3.14(18) = 1, cm (3.14)(18) = cm (iii) 7 (0.5) = mm ( 7 )(0.5) = 1 7 mm (iv) p(1.5) = 156.5p cm (p)(1.5) = 5p cm Q.. (a) (i) = cm (ii) = cm (iii) 87(36) (3.14)(5) (3.14) ( 39 ) = 1, mm (iv) 3 4( 7 )( 47 ) = 1, mm (b) (i) (4 + x) = 4 (ii) 1x = 64 (iii) px = 169p 4 + x = 1 x = 5 3 x = 13 x = 8 (iv) 7 x = 5,08 7 (v) x = 110 x = 1,600 ( 13 1 ) x = 110 x = 40 x = 4 18 Active Maths 3 Book Strands 1 5 Ch 8 Solutions
19 (vi) q (3) = q = q = q = 70 x = = 90 Q. 3. (a) =,700 m,700 = r r 1.5 m (b) (i) = 8 cm (ii) 3 (p)(8) (15) = 30p cm 3 Q. 4. (a) (i) 1 [ ( )] = 91 units (ii) 5 [ ( )] = 1,380 units (iii).5 [ ( )] = 30.5 units (iv) 3 [ ( )] = 31 units (b) (i) [10 + x + ( )] [x ] = 55 x + 8 = 55 x = 7 x = 13.5 (ii) 10 [ ( x + 3)] = 5[ x] =, x 10x +,945 = 3,395 10x = 450 x = 45 (iii) h [0 + + ( )] = h [46] = 13h 13h = 80 h = 6 3 Active Maths 3 Book Strands 1 5 Ch 8 Solutions 19
20 Q. 5. (a) Volume Surface Area (i) [8(8) + 3(39)] 3 (3,544) + (3 60) + (8 3) = 16,176 m = 113,408 m 3 (ii) (1)(8) 0 = 960 m3 ( ( ) (1)(8) ) + 1(0) + (10)(0) = 736 m (iii) (4 + 30)(5) 75 (900) + 75(5) + 30(75) + 4(75) = 67,500 cm 3 = 11, cm (b) (i) 14 cm 17 cm 10 cm 8 cm 10 cm 10π cm 14 cm Volume = 10(8)(17) + (p)(14) (17) 9, cm 3 Surface Area = (10)(17) + 8(17) + 14p(17) + (8)(10) + p(14), cm (ii) Volume = ( (1 + 4)(6 3 ) 35 = 13, cm 3 Surface Area = 6(35 1) + ( ) = 3,68.5 cm (c) (i) ( )(30)(0) = 30,000 cm3 = 30 litres = 1 litres 1,000 (ii) = 35 cm3 /second (iii) <Ask author>. 45 Q. 6. () (a) = cm Q. 8. () (a) p(14) (30) = 5880p cm 3 (b) 3 p 43 = ( 85 3 ) p cm3 3 p (1.5) (3.5) =.65p cm 3 ( 85 3 ) p = 5880p 3. p. =,40 cups 8. h.65p 4 cm = h (b) [4 + + ( )] Q. 7. (a) = 6 square units = 1,100 cm 3 Q. 9. (a) (i) 63(4)(1) = 55,566 cm 3 (b) = 1. m 3 (ii) r = 1 = = 4.8 m 3 3( 7 ) (10.5)3 = 4,851 cm 3 Possible: m.4 m 1 m 0 Active Maths 3 Book Strands 1 5 Ch 8 Solutions
21 (iii) 7 (1) (1) = 9, cm 3 (b) h [ ( )] = 100h = 500 h = 5 m Q. 10. (a) = cm (Area) (6) 6 = 18.8 cm (perimeter) (b) (i) 3. p. (4.5)3 = 60.75p cm 3 (ii) 60.75p 4 = 1,458p cm cm drop 6 (iii) 4.5 = 1 1_ p = p(18) (h) h = 4.5 cm 1 1_ 3 4 = 3 ladlefuls. Q. 11. (a) (i) 10 [ ( )] =,300 m (ii),300 7 = 16,100 m 3 (b) (i) 8(11)(6) = 58 cm 3 (ii) = 66 cm 3 66 = 7. r = r 1.87 = r r 1.9 cm Q. 1. (i) 6(8)(10) = 480 cm 3 (ii) r = 3 3 (p)(3)3 = 36p cm 3 (iii) p(3) (10) = 90p cm 3 OR p(4) (6) = 96p cm (b) (i) p + 3p + 5p = 10p cm (ii). p. 5. p p. = ( ) p = 10p cm Q. 13. (a) r = 3x r = x 3. p. (3x)3 3. p. 7x3 36px 3 Ratio is: 36 : 3 = 108 : 4 = 7 : 1 (b) r = x 3. p. x3 3. p. x3 = 3. p. y3 3. p. x3 r = y 3. p. y3 Active Maths 3 Book Strands 1 5 Ch 8 Solutions 8 1 x3 y 3 = 8 1 x 3 = 8y 3 x = y radii have ratio = : 1 Q. 14. (a) ( )(1.8) = 6.75 cm Length l 6.75l = volume of bar 6.75l = 54 cm l = 8 cm (b) (i) Pink: r Green: r 4r + 4r 8r 8 r r r Diameter (green) = r radius (green) = r (ii) Pink: pr Green: p ( r) = p.. r = pr ratio is pr : pr = : 1 Q. 15. (a) 3. p : 3. p = 1,00 : 150 = 8 : 1 (b) p = 135p cm 3 per second 3. p. (60) (90) = 108,000p 108,000 p 135p = 800 seconds = minutes 1
22 Q. 16. (a) (8 + 11)(5)(30) = 1,45 cm3 475 = p. r = 30(3.14)r = r r =.455 diameter = cm (b) (i) (90) + p(35) = m (ii) (90) + p(36) = m (iii) (90) + p(37) = m B : 6m C : 1 m Q. 17. (a) 1.4(1.1)(1.5) =.31 m 3 7 (0.0175) h = h m cm ( = 1.75 cm = m ) =,400 seconds h =.31.31h =.31 h = 1 (m) Rate is 100 cm/s (b) 3. p.. 7 = 4.5p cm p = p. 6. h h = 8 cm (i) cm drop 8 (ii) 4.5 p =. p. r = r3 h = 100 cm 1.5 cm = r Q. 18. (a) (i) p. (3.5) (10) = 1.5p cm 3 = 110.5p cm 3 (ii) 4.5 p = 3. p. (3.5). h 6 cm = h (iii) h in cylinder = 7 cm p. (3.5) (7) + 4.5p (iv) p = p. (3.5) h h = 9 cm Active Maths 3 Book Strands 1 5 Ch 8 Solutions (b) h [ ( )] = 1,000h 15,000 = 1,000h 15 m = h Q. 19. (i) p. (1.5) (7) + 3. p. (1.5)3 = 18p cm 3 (ii) 9p 3. p (1.5)3 = 6.75p cm p = p(1.5) (h) h = 3 Ans = = 4.5 cm. (b) A B r = 11x r = 7x h = y p. 11x. y = 4x y p h = 5y p. 49x. 5y = 45x y p Second cylinder has greater volume. Q. 0. (a) cylinder: r = r h = r CSA = prh = pr(r) = 4pr = CSA of the sphere (b) 3. p. 63 = 144p cm 3 Cone Cylinder 3. p p = 18p cm 3 = 54p cm 3 144p + 54p = 7p cm 3 144p 7p = 7p cm 3 % = 7p 100% = 50% 144p Q. 1. (a) (i) V = p r h = p(6) (0) = 70p mm 3 (ii) V = 70p mm 3 = = mm 3 (d.p) (b) Volume of bock = = 4,000 mm 3 Volume of one cylindrical hole = mm 3 Total number of holes = 4, Greater than half the block requires 10 holes to be drilled Textbook answer: 10 holes
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