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0580/43 May/Jun 2024, Question 4: Area of a Triangle Using Sine

Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 9 marks

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The question, in short

Triangle with sides r + 2 and r + 5 enclosing 30° has the same area as a square of side r + 1. Form and solve a quadratic, then find the square's perimeter.

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

Part (a) · 4 marks

Using the triangle area formula with the 30° angle and the square's area, show the resulting equation simplifies to 3r² + r − 6 = 0.

  1. 11/2(r + 5)(r + 2) × sin 30° = (r + 1)²
  2. 21/4(r² + 7r + 10) = r² + 2r + 1
  3. 3r² + 7r + 10 = 4r² + 8r + 4
  4. 40 = 3r² + r − 6

Answer1/2(r + 5)(r + 2)sin 30 = (r + 1)² leading to 3r² + r − 6 = 0

Part (b) · 3 marks

Solve 3r² + r − 6 = 0 showing full working, giving the answer to 2 decimal places.

  1. 1r = (−1 ± √(1 + 72))/6 = (−1 ± √73)/6
  2. 2r = (−1 + 8.544)/6 = 1.257… ≈ 1.26
  3. 3r = (−1 − 8.544)/6 = −1.590… ≈ −1.59

Answer−1.59 and 1.26

Part (c) · 2 marks

Using the value of r, calculate the perimeter of the square with side r + 1.

  1. 1Area equation ½(r + 2)(r + 5) sin 30° = (r + 1)² gives 3r² + r − 6 = 0.
  2. 2Positive root: r = 1.257…
  3. 3Side = r + 1 = 2.257…
  4. 4Perimeter = 4 × 2.257… = 9.03

Answer9.028 to 9.040

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Also searched as: 0580/43/M/J/24 · 0580_s24_qp_43 · 0580 s24 43 · question 4.
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