Higher June 2023 Paper 6 Q11

11 The diagram shows a quadrilateral, PQRS.

Not to scale

PS = 10 cm.
Angle QPS = Angle PSR = 90°.

SR is 6 cm longer than PQ.
The area of quadrilateral PQRS is \(A\) cm².

Write a simplified expression for the length PQ in terms of \(A\).
You must show your working. [5]

Mark scheme

Answer	Marks	Part marks and guidance
\(\frac{A}{10} - 3\) or \(\frac{1}{10}(A - 30)\) or \(\frac{A - 30}{10}\) with correct working or other simplified equivalents	5	B4 for \(\frac{A\text{cm}^2}{10} - 3\) etc with correct working OR The below assumes PQ = \(x\). Mark similarly use of SR = \(x\). M2 for \(10x + \frac{1}{2} \times 6 \times 10\) or \(\frac{10(x + x + 6)}{2}\) oe or for \(10x\) and 30, may be indicated on diagram A1 for [\(A\) =] \(10x + 30\) or \(10(x + 3)\) or M1 for lengths \(x\) and \(x + 6\) oe or for area \(10x\) or area 30 AND M1FT for \(10x = A - 30\) or \(x + 3 = \frac{A}{10}\) If 0 or 1 scored, instead award SC2 for \(\frac{A}{10} - 3\) or \(\frac{1}{10}(A - 30)\) or \(\frac{A - 30}{10}\) with no or insufficient working	‘Correct working’ requires evidence of at least M2 Condone use of PQ, PQ + 6 etc instead of \(x\) and \(x + 6\) Working may be on diagram For M2 accept area \(A - 30\) for area \(10x\) For A1 accept equivalents such as \(\frac{A}{5} = 2x + 6\), \(2A = 20x + 60\) For M and A marks, both lengths must be in terms of the same variable eg PQ and PQ + 6, not \(x\) and \(y\) unless \(y = x + 6\) subsequently seen FT \(ax + b = A\) or \(a(x + b) = A\) (\(a \ne \pm 1\) or 0, \(b \ne 0\))