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4024/22 May/Jun 2025, Question 20: Constructing Equations

Cambridge O Level Mathematics 4024 · Paper 22 (Calculator) · Calculator · 9 marks

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The question, in short

Rectangles ABCD and EFGD with given side expressions; show a side length, derive a quadratic from an area ratio, solve it, then find the shaded area.

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

Part (ai) · 1 mark

Show that side CD equals x − 1 using the rectangle dimensions.

  1. 1DG = EF = 2x − 3 (opposite sides of rectangle EFGD)
  2. 2CD = DG − CG = 2x − 3 − (x − 2)
  3. 3= x − 1

Answer2x − 3 − (x − 2) leading to x − 1

Part (aii) · 3 marks

Given ABCD's area is one fifth of EFGD's, form an equation in x and show it reduces to x² − 6x + 3 = 0.

  1. 1AD = ED − EA = (x + 3) − x = 3, so area ABCD = 3(x − 1)
  2. 2Area EFGD = (2x − 3)(x + 3)
  3. 35 × 3(x − 1) = (2x − 3)(x + 3)
  4. 415x − 15 = 2x² − 3x + 6x − 9
  5. 52x² − 12x + 6 = 0
  6. 6x² − 6x + 3 = 0

Answer15x − 15 = 2x² − 3x + 6x − 9 leading to x² − 6x + 3 = 0

Part (bi) · 3 marks

Solve x² − 6x + 3 = 0 with the quadratic formula, showing working, giving answers to 2 decimal places.

  1. 1x = (6 ± √((−6)² − 4 × 1 × 3))/(2 × 1)
  2. 2= (6 ± √24)/2
  3. 3= (6 ± 4.898…)/2
  4. 4x = 5.45 or x = 0.55

Answer5.45 and 0.55

Part (bii) · 2 marks

Using x = 5.45 and the given side expressions, calculate the shaded area outside ABCD.

  1. 1x > 2 (since CG = x − 2 > 0), so x = 5.449…
  2. 2Shaded = (2x − 3)(x + 3) − 3(x − 1)
  3. 3= 7.898… × 8.449… − 3 × 4.449…
  4. 4= 66.74… − 13.35… = 53.39… ≈ 53.4

Answer53.4 or 53.38 to 53.41

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Also searched as: 4024/22/M/J/25 · 4024_s25_qp_22 · 4024 s25 22 · question 20.
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