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0580/42 Feb/Mar 2021, Question 8: Volume and Surface Area of a Cone

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 12 marks

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The question, in short

Sector problems: show a sector area with radius 9.5 cm and angle 53°, find OA from a triangle area relationship, then find the volume of a cone formed from a 24 cm, 60° sector.

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Part (ai) · 2 marks

Show the area of the sector with radius 9.5 cm and angle 53° is 41.7 cm² to 1 decimal place.

  1. 1Area = 53/360 × π × 9.5²
  2. 2= 41.74… ≈ 41.7 cm²

Answer41.74 to 41.75

Part (aii) · 4 marks

Given sector area 41.7 cm² and triangle OAB area one third of it, calculate OA.

  1. 1Sector area = 53/360 × π × 9.5² = 41.74…
  2. 2Triangle area = 1/3 × 41.74… = 13.91…
  3. 31/2 × OA × OA × sin 53 = 13.91…
  4. 4OA² = 2 × 13.91… ÷ sin 53 = 34.84…
  5. 5OA = √34.84… = 5.90

Answer5.90 or 5.899 to 5.903

Part (b) · 6 marks

Calculate the volume of the cone made by joining OP to OQ of a 24 cm, 60° sector.

  1. 1Arc length = 60/360 × 2 × π × 24 = 8π
  2. 2Base circumference 2πr = 8π so r = 4 cm
  3. 3Slant height = 24 cm
  4. 4h = √(24² − 4²) = √560 = 23.66…
  5. 5V = 1/3 × π × 4² × 23.66…
  6. 6V = 396.4… ≈ 396 cm³

Answer396 or 397 or 396.4 to 396.6

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