Module 2 Solid Geometry
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Module 2 Solid Geometry

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Module 2 Solid Geometry Document Transcript

  • 1. PPR Maths-nbk MODULE 2 SKIM TUISYEN FELDA (STF) MATEMATIK SPM “ENRICHMENT” TOPIC: SOLID GEOMETRY TIME: 2 HOURS 1. DIAGRAM 1 22 Diagram 1 shows a solid formed by joining a cuboid and a half-cylinder. Using π = , 7 calculate the volume, in cm3, of the solid. [4 marks] Answer: 1
  • 2. PPR Maths-nbk 2. DIAGRAM 2 Diagram 2 shows a solid pyramid. A small pyramid (shaded region) with base square 6cm and height 4 cm is taken out of the solid. Calculate the volume, in cm3, of the remaining solid. [4 marks] Answer: 2
  • 3. PPR Maths-nbk 3. Diagram 3 shows a solid formed from a cone and hemisphere. DIAGRAM 3 The diameters of the cone and the hemisphere are 21cm each. The volume of the solid 22 is 4 042.5 cm3. Using π = , calculate the height of the cone in cm. [4 marks] 7 Answer: 3
  • 4. PPR Maths-nbk 4. Diagram 4 shows a solid formed by joining a right pyramid and a cuboid. DIAGRAM 4 The volume of the solid is 1 100 cm3. Calculate the height of the pyramid. [4 marks] Answers: 4
  • 5. PPR Maths-nbk 5. Diagram 5 shows a solid formed by joining a cone and a cylinder. DIAGRAM 5 The diameter of the cylinder and the diameter of the base of the cone are both 7 cm. 22 The volume of the solid is 231 cm3 . By using π = , calculate the height, in cm, of 7 the cone. [4 marks] Answer: 5
  • 6. PPR Maths-nbk 6. Diagram 6 shows a solid cylinder of height 20cm and diameter 14 cm. A cone with radius 7 cm and height 9 cm is taken out of the solid. Calculate the volume in cm3 of 22 the remaining solid. (Use π = ). `[4 marks] 7 DIAGRAM 6 Answer: 6
  • 7. PPR Maths-nbk 7. Diagram 7 shows a solid formed by combining a right prism with a half cylinder on the rectangular plane DEFG. DIAGRAM 7 DE = 14 cm, EJ = 8 cm, ∠DEJ = 90° and the height of the prism is 6 cm. Calculate the 22 volume, in cm3, of the solid. (Use π = ) 7 [4 mark Answer: 7
  • 8. PPR Maths-nbk MODULE 2 – ANSWERS TOPIC : SOLID GEOMETRY 1. Volume of solid = Volume of cuboid + volume of half-cylinder 1 22 = 14 x 6 x 4 + x x 32 x 14 2 7 = 336 cm3 + 197.92cm3 = 533.92 cm3 ≈ 534 cm3 2. Volume of remaining solid = volume of big pyramid - volume of small pyramid 1 1 = ( x 152 x 18 ) - ( x 62 x 4 ) 3 3 1 1 = ( x 225 x 18 ) - ( x 36 x 4 ) 3 3 = (1350 - 48) = 1302 cm3 3. Volume of solid = Volume of cone + Volume of hemisphere 1 2 1 4 = πr h + × × πr 3 3 2 3 ⎡ 1 22 ⎛ 21 ⎞ 2 ⎤ ⎡ 1 4 22 ⎛ 21 ⎞3 ⎤ 4042.5 = ⎢ × × ⎜ ⎟ × h⎥ + ⎢ × × × ⎜ ⎟ ⎥ ⎢3 7 ⎝ 2 ⎠ ⎣ ⎥ ⎢2 3 7 ⎝ 2 ⎠ ⎥ ⎦ ⎣ ⎦ 4042.5 = 115.5 h + 2425.5 h = 14 cm 4. Volume of pyramid = volume of solid – volume of cuboid = 1100 cm3 − (10 × 10 × 8) = 1100 - 800 = 300 cm3 1 Volume of pyramid = x Area of base x h 3 1 = x (10 x 10) x h 3 100 = xh 3 8
  • 9. PPR Maths-nbk 100 h = 300 3 3 = 300 x 100 = 9 cm 5. Volume of cylinder = πr2h 22 = x 3.5 2 x 4 7 = 154 cm3 1 2 Volume of cone = πr h 3 1 22 = x x 3.5 2 x t 3 7 269.5 = t cm3 21 Volume of solid = 231 cm3 269.5 154 + t = 231 21 269.5 t = 1617 1617 t= 269.5 = 6 cm 9
  • 10. PPR Maths-nbk 6. Volume of remaining solid = Volume of cylinder – volume of cone 1 2 = πr2h - πr h 3 22 1 22 =[ x 7 x 7 x 20] – [ x x 7 x 7 x 9] 7 3 7 = 3080 – 462 = 2618 cm3 7. Volume of solid = Volume of half – cylinder + volume of prism 1 Volume of half – cylinder = x πr2h 2 Volume of prism = area of base x height Volume of solid = Volume of half – cylinder + volume of prism 1 22 14 2 1 =[ x x( ) x 6] + [ x 14 x 8 x 6] 2 7 2 2 = 462 + 336 = 798 cm2 10