4MA0 Higher June 2014 Paper 3R Q3

3 The diagram shows a rectangle \(PQRS\).

\(PQ = 14\) cm and \(QR = 9\) cm.
The point \(A\) lies on \(PS\) so that \(PA = 5\) cm.
The point \(B\) lies on \(SR\) so that \(BR = 8\) cm.

Diagram NOT accurately drawn

(a) Work out the area of triangle \(AQB\). (4)

(b) Work out the length of \(AQ\).
Give your answer correct to 3 significant figures. (3)

Mark scheme (a)

Scheme	Marks
\(\frac{1}{2} \times 8 \times 9\) or \(\frac{1}{2} \times 5 \times 14\) or 36 or 35	M1
\(\frac{1}{2} \times 4 \times 6\) or 12	M1
\(9 \times 14 - \frac{1}{2} \times 4 \times 6 - \frac{1}{2} \times 8 \times 9 - \frac{1}{2} \times 5 \times 14\) or \(126 - 12 - 36 - 35\)	M1
43	A1
	(4)

M1: Correct expression for area of \(RQB\) or \(PQA\).

M1: Correct expression for area of \(ABS\).

M1: Area of rectangle — their three triangles (obtained by using the correct formula for area of triangle)

Scheme	Marks
\(AB = \sqrt{52}\), (7.2…), \(BQ = \sqrt{145}\) (12.0…) \(AQ = \sqrt{221}\) (14.8… or 14.9)	M1
ABQ = 97.9434… or BQA = 28.7126… or BAQ = 53.3438…	M1
\(\frac{1}{2}(\sqrt{52})(\sqrt{145})\sin(97.9..)\) or \(\frac{1}{2}(\sqrt{145})(\sqrt{221})\sin(28.7..)\) or \(\frac{1}{2}(\sqrt{52})(\sqrt{221})\sin(53.3..)\) oe	M1
43	A1

M1: A correct method to find all 3 sides of triangle ABQ

M1: A correct method to find an angle in ABQ (cosine rule or 180 — use of trig in 2 smaller triangles)

M1: Correct use of formula \(\frac{1}{2}ab\sin C\) to find area of ABQ

A1: Must be exact answer – not from rounding.

M2: For a correct method to find 2 sides and the correct included angle (by use of trig and angles on a straight line).

M1: Correct use of formula \(\frac{1}{2}ab\sin C\) to find area of ABQ (see above)

A1: Must be exact answer – not from rounding.

Mark scheme (b)

M1: For squaring and adding

M1: (dep on M1 above) for square root

A1: For answer rounding to 14.9