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4024/21 Oct/Nov 2020, Question 7: Surface Area of a Cuboid and Cylinder

Cambridge O Level Mathematics 4024 · Paper 21 (Calculator) · Calculator · 9 marks

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The question, in short

Use circle geometry with centre O and points A, B, C, D to work out angle CDO, then handle a sector-based prism and a similar solid.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (a) · 3 marks

Given angle BAD = 68° and angle CBO = 52°, determine the size of angle CDO.

  1. 1Angle BOD = 2 × 68 = 136° (angle at centre = twice angle at circumference)
  2. 2Angle BCD = 180 − 68 = 112° (opposite angles of cyclic quadrilateral ABCD)
  3. 3Angles in quadrilateral OBCD add to 360°
  4. 4Angle CDO = 360 − 136 − 112 − 52 = 60°

Answer60°

Part (bi) · 4 marks

For a prism of height 5 cm whose cross-section is a sector of radius 6 cm and angle 40°, work out its total surface area.

  1. 1Two sectors: 2 × 40/360 × π × 6² = 25.13… cm²
  2. 2Curved face: 40/360 × 2 × π × 6 × 5 = 20.94… cm²
  3. 3Two rectangles: 2 × 5 × 6 = 60 cm²
  4. 4Total = 25.13… + 20.94… + 60 = 106.07… ≈ 106.1 cm²

Answer106 to 106.1

Part (bii) · 2 marks

Given volume ratio 27 : 1 between similar solids A and B, find the surface area of solid B.

  1. 1Volume ratio 27 : 1, so length ratio = ∛27 : 1 = 3 : 1
  2. 2Area ratio = 3² : 1 = 9 : 1
  3. 3Surface area of A = 2 sectors + curved face + 2 rectangles = 106.07… cm²
  4. 4Surface area of B = 106.07… ÷ 9 = 11.78… cm², i.e. 11.7 to 11.8

Answer11.7 to 11.8

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