1MA0 Higher November 2013 Paper 2 Q22
22 The diagram shows a trapezium.
All the measurements are in centimetres.
Diagram NOT accurately drawn
The area of the trapezium is 351 cm\(^2\).
(a) Show that \(2x^2 + x - 351 = 0\) (2)
(b) Work out the value of \(x\). (3)
| Answer | Mark | Notes |
|---|---|---|
| ‘show’ | 2 | M1 for \(\frac{1}{2} \times (x - 4 + x + 5) \times 2x\) or \(2x \times (x - 4) + \frac{1}{2} \times 2x \times 9\) A1 for completion with correct processes seen |
Notes
Corrected from the printed mark scheme, which shows \(\frac{1}{2} \times (x - 4 + x + 5) \times 2\) (the height is \(2x\)).
| Answer | Mark | Notes |
|---|---|---|
| 13 | 3 | M1 for \(\dfrac{-1 \pm \sqrt{1^2 - 4 \times 2 \times -351}}{2 \times 2}\) condone incorrect sign for 351 M1 for \(\dfrac{-1 \pm \sqrt{2809}}{4}\) A1 for 13 NB for either M mark accept + only in place of \(\pm\) OR M2 for \((2x + 27)(x - 13)\) (M1 for \((2x \pm 27)(x \pm 13)\)) A1 for 13 |