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0606/23 May/Jun 2026, Question 2: Radians, Arcs and Sectors

Cambridge IGCSE Additional Mathematics 0606 · Paper 23 (Calculator) · Calculator · 7 marks

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The question, in short

Shape ABCDP has square sides AB, BC, CD of 4, arc AD centred at P with MP = 5; show θ = 2.21 radians, then find the shaded area.

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

Part (a) · 2 marks

Show that angle APD, θ, equals 2.21 radians to 3 significant figures using the given lengths.

  1. 1P is 5 − 4 = 1 cm above the midpoint of AD
  2. 2Half of AD = 2
  3. 3tan(θ/2) = 2/1 = 2
  4. 4θ = 2 tan⁻¹ 2 = 2.2143 = 2.21 (3 s.f.)

Answertan(θ/2) = 2/1, so θ = 2 tan⁻¹ 2 = 2.214… = 2.21 to 3 s.f. — shown

Part (b) · 5 marks

Find the area of the shaded region between the square and the circular arc.

  1. 1Radius PA = √(2² + 1²) = √5
  2. 2Sector APD = (1/2)(√5)²θ = 2.5 × 2.2143 = 5.536
  3. 3Triangle APD = (1/2) × 4 × 1 = 2
  4. 4Segment between chord AD and arc = 5.536 − 2 = 3.536
  5. 5Shaded area = square − segment = 16 − 3.536 = 12.46 ≈ 12.5 cm²

Answer12.5 cm² (12.46)

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Also searched as: 0606/23/M/J/26 · 0606_s26_qp_23 · 0606 s26 23 · question 2.
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