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4PM1/01R Jan 2023, Question 9: Surface Area of a Pyramid

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 13 marks

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The question, in short

Right pyramid OABCD has square base side 12 cm, equal edges x cm, and base-diagonal angles 30°. Find AC, show x = 4√6, then find total surface area and the obtuse angle between planes OAB and OBC.

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

Part (a) · 2 marks

Find the exact length of the square base diagonal AC.

  1. 1AC² = 12² + 12² = 288
  2. 2AC = √288 = 12√2 cm

AnswerAC = 12√2 cm

Part (b) · 2 marks

Show, using the given 30° angle and lengths, that the sloping edge x equals 4√6.

  1. 1Let X be the centre of the base: AX = 6√2
  2. 2In triangle OAX: cos 30° = AX/OA
  3. 3x = 6√2/cos 30° = 6√2 × 2/√3
  4. 4x = √96 = 4√6

Answerx = 4√6 shown

Part (c) · 5 marks

Calculate the pyramid's total surface area, giving the answer to the nearest cm².

  1. 1cos AOB = (96 + 96 − 144)/(2 × 96) → AOB = 75.522…°
  2. 2Area OAB = ½ × 96 × sin 75.522…° = 46.4758…
  3. 3(or slant height √(96 − 36) = 2√15, area = ½ × 12 × 2√15 = 12√15)
  4. 4Total = 4 × 46.4758… + 12² = 329.90…
  5. 5= 330 cm² (nearest cm²)

Answer330 cm²

Part (d) · 4 marks

Find the obtuse angle between planes OAB and OBC, to the nearest degree.

  1. 1Angle OBC = (180° − 75.522°)/2 = 52.239°
  2. 2Let Y on OB be the foot of the perpendicular from A (and from C): AY = CY = 12 sin 52.239° = 9.4868…
  3. 3cos AYC = (9.4868² + 9.4868² − 288)/(2 × 9.4868²)
  4. 4AYC = 126.87… ≈ 127°

Answer127°

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