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4024/21 Oct/Nov 2023, Question 6: Constructing Equations

Cambridge O Level Mathematics 4024 · Paper 21 (Calculator) · Calculator · 10 marks

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The question, in short

For rectangle ABCD of area 30 cm² with AB = x and a 2 cm by 2 cm corner rectangle DEFG removed, work with the shaded area formula y = 2x + 60/x − 4, tabulate, graph and use it to find the rectangle's dimensions when the shaded area is 24 cm² and AB < BC.

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

Part (a) · 1 mark

Give BC as an expression involving x, using the fact that the rectangle's area is 30 cm².

  1. 1Area = AB × BC = 30
  2. 2BC = 30/x

Answer30/x

Part (b) · 3 marks

Derive that the shaded region's area y equals 2x + 60/x − 4.

  1. 1EF = DG = x − 2 and DE = BC − 2 = 30/x − 2
  2. 2Removed area = (x − 2)(30/x − 2) = 30 − 2x − 60/x + 4
  3. 3y = 30 − (34 − 2x − 60/x)
  4. 4y = 2x + 60/x − 4

Answery = 2x + 60/x − 4

Part (c) · 1 mark

Work out the missing y value in the table for x = 14, giving it to 1 decimal place.

  1. 1x = 14: y = 2 × 14 + 60/14 − 4 = 28 + 4.285… − 4
  2. 2y = 28.3

Answer28.3

Part (d) · 3 marks

Plot the curve y = 2x + 60/x − 4 on the grid for x values from 2 to 14.

  1. 1Plot (2, 30), (3, 22), (4, 19), (6, 18), (8, 19.5), (10, 22), (12, 25), (14, 28.3)
  2. 2Join with a smooth curve, lowest near x = 5.5, y ≈ 17.9

AnswerCorrect smooth curve

Part (e) · 2 marks

Using your graph and a shaded area of 24 cm², with AB shorter than BC, determine the side lengths of ABCD.

  1. 1Draw y = 24 and read the two x-values where it meets the curve
  2. 2x ≈ 2.64 and x ≈ 11.36 (from x² − 14x + 30 = 0)
  3. 3AB < BC, so AB = x ≈ 2.6 cm and BC = 30 ÷ 2.64 ≈ 11.4 cm

Answer2.5 to 2.8, 11.2 to 11.5

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Also searched as: 4024/21/O/N/23 · 4024_w23_qp_21 · 4024 w23 21 · question 6.
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