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0580/31 Oct/Nov 2023, Question 10: Pythagoras' Theorem

Cambridge IGCSE Mathematics 0580 · Paper 31 (Core) · Calculator · 9 marks

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The question, in short

Use similar triangles and right-angled triangle trigonometry to find a missing side, two lengths and an angle.

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

Part (a) · 2 marks

Triangles PQR and UVW are similar; QR = 12.8 cm, PR = 9.6 cm, UV = 21.4 cm, UW = 14.4 cm. Find VW.

  1. 1VW/14.4 = 12.8/9.6
  2. 2VW = 14.4 × 12.8 ÷ 9.6 = 19.2 cm

Answer19.2

Part (b) · 3 marks

Right-angled triangle ABC has the right angle at B, AB = 3.5 cm and AC = 9.1 cm. Calculate BC.

  1. 1BC² + 3.5² = 9.1²
  2. 2BC² = 82.81 − 12.25 = 70.56
  3. 3BC = √70.56 = 8.4 cm

Answer8.4

Part (c) · 2 marks

Right-angled triangle DEF has the right angle at E, DF = 8.4 cm and angle DFE = 35°. Calculate EF.

  1. 1cos 35 = EF/8.4
  2. 2EF = 8.4 × cos 35 = 6.880… cm

Answer6.88 to 6.881

Part (d) · 2 marks

Right-angled triangle JKL has the right angle at J, JL = 10 cm and JK = 8 cm. Calculate angle JKL.

  1. 1tan JKL = 10/8
  2. 2Angle JKL = tan⁻¹(1.25) = 51.34…°

Answer51.34…

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Also searched as: 0580/31/O/N/23 · 0580_w23_qp_31 · 0580 w23 31 · question 10.
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