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4024/21 Oct/Nov 2021, Question 10: Pythagoras' Theorem

Cambridge O Level Mathematics 4024 · Paper 21 (Calculator) · Calculator · 12 marks

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The question, in short

Cuboid with measurements to nearest mm: find upper bound of surface area; square-based pyramid with EX = 17 cm and EC = 19 cm: show base length, find volume and angle CBE.

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

Part (a) · 3 marks

Use upper bounds of 6.2, 4.8 and 2.5 cm to calculate the maximum possible surface area of the cuboid.

  1. 1Upper bounds: 6.25, 4.85, 2.55
  2. 26.25 × 4.85 + 6.25 × 2.55 + 4.85 × 2.55 = 30.3125 + 15.9375 + 12.3675 = 58.6175
  3. 3Surface area = 2 × 58.6175 = 117.235 cm²

Answer117.235

Part (bi) · 4 marks

Using EX = 17 cm and EC = 19 cm, show that the square base side is 12 cm.

  1. 119² = XC² + 17² → XC² = 361 − 289 = 72
  2. 2AC = 2 × XC, so AC² = 4 × 72 = 288
  3. 3Base side s: s² + s² = AC² → 2s² = 288 → s² = 144
  4. 4s = √144 = 12 cm

Answer√144 = 12

Part (bii) · 2 marks

Calculate the volume of the pyramid using its base area and height 17 cm.

  1. 1In triangle EXC: XC² = 19² − 17² = 72
  2. 2Diagonal AC = 2 × XC, so AC² = 4 × 72 = 288
  3. 3Square base: side² = AC² ÷ 2 = 144, so side = 12 cm
  4. 4V = 1/3 × 144 × 17 = 816 cm³

Answer816

Part (biii) · 3 marks

Calculate angle CBE in the pyramid.

  1. 1Triangle CBE is isosceles with EB = EC = 19, BC = 12
  2. 2cos CBE = 6/19
  3. 3Angle CBE = cos⁻¹(6/19) = 71.59… ≈ 71.6°

Answer71.5 to 71.6

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Also searched as: 4024/21/O/N/21 · 4024_w21_qp_21 · 4024 w21 21 · question 10.
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