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0580/42 Oct/Nov 2022, Question 10: Arc Length

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 8 marks

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The question, in short

Three-part question: bound a triangle's perimeter from rounded sides, find a sector's area from radius 15.6 cm and 150°, and use a perimeter ratio to derive x = (90(π − 2))/π.

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

Part (a) · 2 marks

Sides 11.4 cm, 14.8 cm, 15.7 cm are each rounded to 1 d.p.; work out the largest possible total perimeter.

  1. 1Upper bounds: 11.45, 14.85, 15.75
  2. 2Perimeter = 11.45 + 14.85 + 15.75 = 42.05 cm

Answer42.05

Part (b) · 2 marks

For a circle of radius 15.6 cm, work out the area of the sector with central angle 150°.

  1. 1Area = 150/360 × π × 15.6²
  2. 2= 318.55… ≈ 319 cm²

Answer319 or 318.5 to 318.6

Part (c) · 4 marks

With radius r cm and sector angle x°, the major sector's perimeter is triple the minor's; prove x = (90(π − 2))/π.

  1. 1Minor perimeter = (x/360) × 2πr + 2r, major perimeter = (360 − x)/360 × 2πr + 2r
  2. 2(360 − x)/360 × 2πr + 2r = 3((x/360) × 2πr + 2r)
  3. 3Divide by r and multiply by 180: (360 − x)π + 360 = 3xπ + 1080
  4. 4360π − 720 = 4πx
  5. 5x = (360π − 720)/(4π) = (90(π − 2))/π

Answer(360 − x)/360 × 2πr + 2r = 3((x/360) × 2πr + 2r), leading to x = (90(π − 2))/π

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Also searched as: 0580/42/O/N/22 · 0580_w22_qp_42 · 0580 w22 42 · question 10.
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