The length of one side of a rectangle is 2 cm longer than its width.

If the smaller side is x cm, find the perimeter of the rectangle in terms of x.

The length of one side of a rectangle is 2 cm longer than its width.

The perimeter of a square is equal to the perimeter of the rectangle in part (a).

Determine the length of each side of the square in terms of x.

The length of one side of a rectangle is 2 cm longer than its width.

The perimeter of a square is equal to the perimeter of the rectangle in part (a).

The sum of the areas of the rectangle and the square is \(2x^2 + 4x +1\) (cm²).

(i) Given that this sum is 49 cm², find x.

(ii) Find the area of the square.

Unit penalty (UP) is applicable where indicated in the left hand column.

(UP) \({\text{P (rectangle)}} = 2x + 2(x + 2) = 4x + 4{\text{ cm}}\)     (A1)     (C1)

(UP) Simplification not required

[1 mark]

Unit penalty (UP) is applicable where indicated in the left hand column.

(UP) Side of square = (4x + 4)/4 = x + 1 cm     (A1)(ft)     (C1)

[1 mark]

(i) \(2x^2 + 4x + 1 = 49\) or equivalent     (M1)

\((x + 6)(x – 4) = 0\)

\(x = - 6\) and \(4\)     (A1)

Note: award (A1) for the values or for correct factors    

Choose \(x = 4\)     (A1)(ft)

Award (A1)(ft) for choosing positive value.     (C3)

(ii) \({\text{Area of square}} = 5 \times 5 = 25{\text{ c}}{{\text{m}}^2}\)     (A1)(ft)

Note: Follow through from both (b) and (c)(i).     (C1)

[4 marks]

a) and b) Two thirds of the candidates found the perimeter of the rectangle and the side of the square correctly, though most of them did not include units (thereby incurring a unit penalty).

a) and b) Two thirds of the candidates found the perimeter of the rectangle and the side of the square correctly, though most of them did not include units (thereby incurring a unit penalty).

c) Although a majority of candidates produced the quadratic equation many were unable to solve it correctly. This could easily be done using the GDC so it was disappointing.