08[A Math CD]

1 © Penerbitan Pelangi Sdn. Bhd.

Paper 2

1. Volume of the cone

= 1—3

× 22–––7

× 32 × 7

= 66 cm3

2. Volume of the right prism

= 1—2

× 12 × 5 × 10

= 300 cm3

3. Volume of the cylinder

= 22–––7

× 52 × 14

= 1100 cm3

4. Volume of the hemisphere

= 1—2

4—3

× 22–––7

× 7—2

3

= 89 5—6

cm3

5. Volume of the pyramid

= 1—3

× 6 × 6 × 8

= 96 cm3

Paper 2

1. Volume of the right prism= area of ∆PQU × PS

= 1—2

× 12 × 8 × 7

= 336 cm3

Volume of the half cylinder

= 1—2

22–––7

× 7—2

2 × 12

= 231 cm3

Volume of the solid= 336 + 231= 567 cm3

2. Volume of the prism

= 1—2

× (4 + 7) × 5 × 9

= 247.5 cm3

Volume of the cuboid= 4 × 10 × 10= 400 cm3

Volume of the solid= 247.5 + 400= 647.5 cm3

3. Volume of the quarter cylinder

= 1—4

× 22–––7

× 72 × 20

= 770 cm3

Volume of the cuboid= 20 × 7 × h= 140h cm3

Volume of the solid = 770 + 140h 1890 = 770 + 140h 140h = 1890 – 770

h = 1120140

= 8

CHAPTER

8 Solid GeometryCHAPTER

2

Mathematics SPM Chapter 8

© Penerbitan Pelangi Sdn. Bhd.

4. Volume of the cone

= 13

× 227

× 72

2 × 4

= 51 13

cm3

Volume of the hemisphere

= 12

× 43

× 227

× 73

= 718 23

cm3

Volume of the solid

= 51 13

+ 718 23

= 770 cm3


= 13

× 10 × 10 × 12

= 400 cm3

Volume of the cylinder

= 227

× 72

2 × 5

= 192.5 cm3

Volume of the remaining solid= 400 − 192.5= 207.5 cm3

6. Volume of the cuboid= 11 × 9 × 12= 1188 cm3

Volume of the cone

= 13

× 227

× 72

2 × 9

= 115.5 cm3

Volume of the remaining solid= 1188 − 115.5= 1072.5 cm3

7. (a) Volume of the cone

= 13

× 227

× 92 × 7

= 594 cm3

(b) Volume of the hemisphere

= 12

× 43

× 227

× 73

= 718 23

cm2

Volume of the solid = 594 + 718 2

3 = 1312 2

3 cm3

Paper 2

1. Volume of the half cylinder

= 12

227

× 32 × 14

= 198 cm3




= 13

× 8 × 8 × 9

= 192 cm3

Volume of the cube= 8 × 8 × 8= 512 cm3


3. Volume of the hemisphere

= 12

43

× 227

× 33

= 56 4—7

cm3


= 227

× 32 × 7

= 198 cm3

Volume of the solid

= 56 47

+ 198

= 254 47

cm3

3



4. Let the length of the cylinder be l, in cm.Volume of the cylinder

= 22–––7

× 42 × l

= 352––––7

l cm3

Volume of the two hemispheres

= 4—3

× 22–––7

× 43

= 5632–––––21

cm3

Volume of the solid = 352––––7

l + 5632–––––21

771 = 352––––7

l + 5632–––––21

352––––7

l = 771 − 5632–––––21

352––––7

l = 10 559––––––21

l = 10 559––––––21

× 7––––352

= 10 cm


= 22–––7

× 72 × 30

= 4620 cm3

Volume of the cone

= 1—3

× 22–––7

× 7—2

2 × 9

= 115.5 cm3

Volume of the remaining solid= 4620 − 2(115.5)= 4389 cm3

6. Let the radius of the cylinder be r, in cm.Volume of the cylinder

= 22–––7

× r2 × 14

= 44r2 cm3

Volume of the cone

= 1—3

× 22–––7

× r2 × 7

= 22–––3

r2 cm3

Volume of the remaining solid = 44r2 – 22–––3

r2

330 = 44r2 – 22–––3

r2

132––––3

r2 – 22–––3

r2 = 330

110––––3

r2 = 330

r2 = 330 × 3–––––––110

= 9 r = 3 cm


= 1—2

22–––7

× 7—2

2 × 20

= 385 cm3

Volume of the right prism= area of trapezium ABGH × BC

= 1—2

× (7 + 13) × 8 × 20

= 1600 cm3


8. Let the height of the cone be h, in cm.Volume of the cone

= 1—3

× 22–––7

× 72 × h

= 154––––3

h cm3


= 1—2

4—3

× 22–––7

× 73

= 2156–––––3

cm3

Volume of the solid = 154––––3

h + 2156–––––3

1129 1—3

= 154––––3

h + 2156–––––3

154h + 2156–––––––––––3

= 3388–––––3

154h + 2156 = 3388 154h = 3388 − 2156

h = 1232–––––154

= 8 cm

4



9. Volume of the right prism= 30 × 10= 300 cm3


= 22–––7

× 22 × 10

= 125 5—7

cm3

Volume of the remaining solid

= 300 − 125 5—7

= 174 2—7

cm3


= 22–––7

× 42 × 7

= 352 cm3


= 1—2

4—3

× 22–––7

× 43

= 134 2–––21

cm3


= 352 − 134 2–––21

= 217 19–––21

cm3

11. Let the height of the cylinder be h, in cm.Volume of the cylinder

= 22–––7

× 72 × h

= 154h cm3

Volume of the two hemispheres

= 4—3

× 22–––7

× 73

= 1437 1—3

cm3

Volume of the remaining solid = 154h − 1437 1—3

872 2—3

= 154h − 1437 1—3

154h = 872 2—3

+ 1437 1—3

= 2310

h = 2310–––––154

= 15 cm


= 1—2

22–––7

× 52 × 14

= 550 cm3

Volume of the prism= area of trapezium ABCD × BJ

= 1—2

× (6 + 10) × 5 × 14

= 560 cm3



= 22–––7

× 42 × 12

= 603 3—7

cm3

Volume of the two cones

= 2 1—3

× 22–––7

× 42 × 6

= 201 1—7

cm3


= 603 3—7

− 201 1—7

= 402 2—7

cm3


= 22–––7

× 72 × 10

= 1540 cm3

Volume of the cone

= 1—3

× 22–––7

× 72 × (22 − 10)

= 616 cm3


5



15. Let the height of the water be h, in cm.

8 cm

4 cm

E F

GH

h

h

V

Volume of the conical container

= 1—3

× 22–––7

× 82 × 2h

= 2816–––––21

h cm3

Volume of the empty space inside the container

= 1—3

× 22–––7

× 42 × h

= 352––––21

h cm3

Volume of the water = 2816–––––21

h − 352––––21

h

821 1—3

= 2816–––––21

h − 352––––21

h

2464–––––3

= 352––––3

h

352h = 2464

h = 2464–––––352

= 7 cm


= 1—2

22–––7

× 7—2

2 × 14

= 269.5 cm3

Volume of the right prism= area of the trapezium FGMN × EF

= 1—2

× (4 + 14) × 5 × 7

= 315 cm3

Volume of the solid= 269.5 + 315= 584.5 cm3

17. Let the height of water in the cuboid container be h, in cm.

0.5 m

1.3 mx

x2 = 1.32 − 0.52

x = 1.44 = 1.2 m

60 × 80 × h = 1—3

× 22–––7

× 502 × 120

h = 22 × 502 × 120–––––––––––––21 × 60 × 80

= 65 10–––21

cm


= 22–––7

× 7—4

2 × 8

= 77 cm3

Volume of the two discs

= 2 22–––7

× 7—2

2 × 2

= 154 cm3


19. EF 2 = EJ 2 + JF 2

EF = 32 + 42

= 5 cm

Volume of the prism= area of ∆EFJ × FG

= 1—2

× 3 × 4 × 8

= 48 cm3



08[A Math CD]

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