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0580/42 Feb/Mar 2022, Question 6: Proving Triangles Are Similar

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 10 marks

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The question, in short

Four-part geometry question: find a regular polygon's side count from its interior angle, then use circle theorems with given angles and diagrams to calculate angles and prove two triangles similar.

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

Part (a) · 2 marks

Given an interior angle of 156° for a regular polygon, work out how many sides it has.

  1. 1Exterior angle = 180 − 156 = 24°
  2. 2Number of sides = 360 ÷ 24 = 15

Answer15

Part (b) · 2 marks

With O the centre and angle OBA = 52°, calculate angle ACB using circle angle properties.

  1. 1OA = OB, so angle OAB = 52°; angle AOB = 180 − 2 × 52 = 76°
  2. 2Angle ACB = 76 ÷ 2 = 38° (angle at the centre is twice the angle at the circumference)

Answer38

Part (c) · 2 marks

Given WSR straight and angle WSP = 112°, calculate angle PTR for points on the circle.

  1. 1Angle RSP = 180 − 112 = 68° (angles on a straight line)
  2. 2Angle PTR = angle PSR = 68° (angles in the same segment)

Answer68

Part (d) · 4 marks

Using the tangent FGH at G and MG parallel to OH, prove triangles GKM and OHG are similar, justifying each step.

  1. 1Angle KMG = 90° (angle in a semicircle, KG is a diameter)
  2. 2Angle OGH = 90° (tangent is perpendicular to the radius)
  3. 3Angle MGK = angle GOH (alternate angles, MG parallel to OH)
  4. 4Two pairs of equal angles, so triangle GKM is similar to triangle OHG

AnswerTwo pairs of equal angles identified with fully correct reasons, so the triangles are similar (e.g. angle KMG = 90° (angle in a semicircle) and angle OGH = 90° (angle between tangent and radius); angle GOH = angle MKG (alternate angles))

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Also searched as: 0580/42/F/M/22 · 0580_m22_qp_42 · 0580 m22 42 · question 6.
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