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0580/41 May/Jun 2021, Question 11: Congruent Triangles: Proof Conditions

Cambridge IGCSE Mathematics 0580 · Paper 41 (Extended) · Calculator · 11 marks

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The question, in short

Four-part geometry and similarity question: exterior angle of an 18-sided regular polygon, length BE using parallel lines in triangle ACD, height of a similar solid from volumes, and proving triangles PQN and SRN congruent.

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

Part (a) · 2 marks

Calculate the exterior angle of a regular polygon having 18 sides.

  1. 1Exterior angle = 360 ÷ 18 = 20°

Answer20

Part (b) · 2 marks

With BE parallel to CD, AE = 5.2 cm, ED = 2.6 cm and CD = 6.75 cm, work out the length BE.

  1. 1Triangles ABE and ACD are similar, AD = 5.2 + 2.6 = 7.8
  2. 2BE/6.75 = 5.2/7.8
  3. 3BE = 6.75 × 5.2 ÷ 7.8 = 4.5 cm

Answer4.5

Part (c) · 3 marks

Similar solids: smaller height 2 cm, volume 32 cm³; larger volume 780 cm³. Find the larger height.

  1. 1Volume scale factor = 780 ÷ 32 = 24.375
  2. 2Length scale factor = ∛24.375 = 2.899…
  3. 3Height = 2 × 2.899… = 5.798… ≈ 5.80 cm

Answer5.80 or 5.798 to 5.799

Part (d) · 4 marks

Given PQ parallel to RS, PNS straight and N the midpoint of RQ, explain with reasons why triangles PQN and SRN are congruent.

  1. 1QN = NR (N is the midpoint of RQ)
  2. 2Angle PQN = angle SRN (alternate angles, PQ parallel to RS)
  3. 3Angle PNQ = angle SNR (vertically opposite angles)
  4. 4So the triangles are congruent (ASA)

AnswerQN = NR (given) Two pairs of angles with reasons from: angle PQN = angle SRN (alternate); angle QPN = angle RSN (alternate); angle PNQ = angle SNR (vertically opposite) So congruent (ASA)

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Also searched as: 0580/41/M/J/21 · 0580_s21_qp_41 · 0580 s21 41 · question 11.
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