Problem-solving with quadratic context (6 marks)

Question. A rectangle has perimeter 28 cm and area 45 cm². Letting the length be x cm and the width be y cm, form two equations and find the possible dimensions of the rectangle.

Mark scheme.

• B1: forming 2x + 2y = 28 (i.e. x + y = 14).
• B1: forming xy = 45.
• M1: substituting y = 14 − x into xy = 45 → x(14 − x) = 45.
• M1: rearranging to x² − 14x + 45 = 0.
• A1: factorising (x − 5)(x − 9) = 0, giving x = 5 or x = 9.
• A1: dimensions are 9 cm by 5 cm (accept either ordering as valid; recognise length ≥ width convention).

Total: 6 marks.