4024/22 May/Jun 2020, Question 10: Trigonometry in 3D: Cuboids
Cambridge O Level Mathematics 4024 · Paper 22 (Calculator) · Calculator · 11 marks
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The question, in short
Rectangular-based pyramid ABCDE has base ABCD, apex E above centre F, with AD = 10, DC = 6, EC = 12; find height, volume, angle DEC and triangle DEC area.
Summarised in our own words. Use the official paper for the exact wording and the diagram.
Part (a) · 4 marks
Show that the vertical height EF is 10.5 cm to 1 decimal place.
- 1AC² = 10² + 6² = 136, so AC = √136
- 2FC = 1/2 × √136 = √34
- 3EF² + FC² = EC²: EF² = 12² − 34 = 110
- 4EF = √110 = 10.488… = 10.5 cm (1 d.p.)
Answer10.49 or 10.488…
Part (b) · 2 marks
Find the volume of the pyramid.
- 1Diagonal AC = √(10² + 6²) = √136 = 11.66… cm
- 2F is the midpoint of AC, so FC = 11.66… ÷ 2 = 5.83… cm
- 3Height EF = √(12² − 5.83…²) = √(144 − 34) = √110 = 10.49… cm
- 4Volume = 1/3 × 6 × 10 × 10.49… = 209.76… ≈ 210 cm³
Answer210 or 209.7 to 209.8
Part (c) · 3 marks
Calculate angle DEC.
- 1E is directly above F, the centre of the base, so ED = EC = 12 cm
- 2Cosine rule in triangle DEC: cos DEC = (12² + 12² − 6²)/(2 × 12 × 12)
- 3cos DEC = 252/288 = 0.875
- 4Angle DEC = 28.955… = 29.0°
Answer29.0 or 28.95 to 28.96
Part (d) · 2 marks
Calculate the area of triangle DEC.
- 1DE = EC = 12 cm, so triangle DEC is isosceles with base DC = 6 cm.
- 2The height from E meets DC at its midpoint, 3 cm from D.
- 3Height = √(12² − 3²) = √135 = 11.62… cm
- 4Area = 1/2 × 6 × 11.62… = 34.86 ≈ 34.9 cm²
Answer34.9 or 34.85 to 34.86
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