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0580/41 May/Jun 2022, Question 5: Area of a Sector

Cambridge IGCSE Mathematics 0580 · Paper 41 (Extended) · Calculator · 17 marks

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The question, in short

Regular octagon ABCDEFGH has side 6 cm, centre O, M midpoint of AB. Find angle OAM, octagon area, circumscribed circle area; then a half-cylinder container radius 0.45 m, length 4 m, find volume and litres of water at depth 0.3 m.

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

Part (aia) · 2 marks

Show that angle OAM equals 67.5° using the regular octagon's geometry and triangle OAB with M the midpoint of AB.

  1. 1Interior angle of octagon = (8 − 2) × 180 ÷ 8 = 135°
  2. 2OA bisects angle HAB, so angle OAM = 135 ÷ 2 = 67.5°

Answer((8 − 2) × 180)/(8 × 2)

Part (aib) · 4 marks

Calculate the area of the regular octagon with side length 6 cm.

  1. 1AM = 6 ÷ 2 = 3 cm
  2. 2OM = 3 × tan 67.5 = 7.242… cm
  3. 3Area of triangle OAB = 1/2 × 6 × 7.242… = 21.72… cm²
  4. 4Area of octagon = 8 × 21.72… = 173.8… ≈ 174 cm²

Answer174 or 173.8....

Part (aii) · 3 marks

Find the area of the circle passing through all vertices of the octagon.

  1. 1Radius OA: cos 67.5 = 3/OA, so OA = 3 ÷ cos 67.5 = 7.839… cm
  2. 2Area = π × 7.839…² = 193.06… ≈ 193 cm²

Answer193 or 193.0 to 193.1

Part (bi) · 2 marks

Calculate the volume of the half-cylinder container with radius 0.45 m and length 4 m.

  1. 1Volume = 1/2 × π × 0.45² × 4
  2. 2= 1.272… ≈ 1.27 m³

Answer1.27 or 1.272 to 1.273

Part (bii) · 6 marks

Calculate the number of litres of water when the greatest depth is 0.3 m, to the nearest integer.

  1. 1Water surface is 0.45 − 0.3 = 0.15 m below the centre
  2. 2Angle at centre = 2 × cos⁻¹(0.15 ÷ 0.45) = 141.06°
  3. 3Segment = 141.06/360 × π × 0.45² − 1/2 × 0.45² × sin 141.06° = 0.1856 m²
  4. 4Volume = 0.1856 × 4 = 0.7426 m³
  5. 50.7426 × 1000 = 742.6, so 743 litres

Answer742 or 743

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