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4024/22 Oct/Nov 2020, Question 7: Similar Triangles

Cambridge O Level Mathematics 4024 · Paper 22 (Calculator) · Calculator · 10 marks

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The question, in short

Regular polygon with angles DCF and EFC both 24 degrees: find side count and angle HFG. Parallelogram with parallel line and given ratios: prove similarity, find length and area ratios.

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

Part (ai) · 2 marks

Use the 24° angles in the regular polygon to find the number of sides n.

  1. 1CF is parallel to DE, so angle CDE = 180 − 24 = 156° (co-interior angles)
  2. 2Exterior angle = 180 − 156 = 24°
  3. 3n = 360 ÷ 24 = 15

Answer15

Part (aii) · 2 marks

Find angle HFG in the regular polygon.

  1. 1Triangle FGH is isosceles (FG = GH) with angle FGH = 156°
  2. 22 × angle HFG + 156 = 180
  3. 3Angle HFG = 24 ÷ 2 = 12°

Answer12°

Part (bi) · 3 marks

Prove triangles PQS and TQX are similar, justifying each statement with a reason.

  1. 1Angle PQS = angle TQX (common angle)
  2. 2Angle PSQ = angle TXQ (corresponding angles, TU parallel to PS)
  3. 3Angle SPQ = angle XTQ (corresponding angles, TU parallel to PS)
  4. 4All three angles equal, so triangle PQS is similar to triangle TQX

Answerangle Q is common; angle PSQ = angle TXQ, corresponding angles; angle SPQ = angle XTQ, corresponding angles; hence similar

Part (bii) · 1 mark

Find the ratio SX : SQ using the parallel line and given ratios.

  1. 1TX is parallel to PS, so SX : SQ = PT : PQ
  2. 2PT : TQ = 2 : 1, so PT : PQ = 2 : 3
  3. 3SX : SQ = 2 : 3

Answer2 : 3

Part (biii) · 2 marks

Find the ratio of the area of triangle TQX to the area of parallelogram PQRS.

  1. 1TQ : PQ = 1 : 3, so area TQX : area PQS = 1² : 3² = 1 : 9
  2. 2Area PQS = 1/2 × area PQRS
  3. 3Area TQX : area PQRS = 1/9 × 1/2 : 1 = 1 : 18

Answer1 : 18

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Also searched as: 4024/22/O/N/20 · 4024_w20_qp_22 · 4024 w20 22 · question 7.
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