4MA0 Higher January 2012 Paper 3 Q21
21
Diagram NOT accurately drawn
The diagram shows a trapezium \(ABCD\) with \(AD\) parallel to \(BC\).
\(AB = x\) cm, \(BC = (x + 5)\) cm and \(AD = (x + 8)\) cm.
The area of the trapezium is 42 cm\(^2\).
(a) Show that \(2x^2 + 13x - 84 = 0\) (2)
(b) Calculate the perimeter of the trapezium. (5)
| Scheme | Marks |
|---|---|
| \(0.5x[(x + 5) + (x + 8)] = 42\) (trapezium formula) or \(x(x + 5) + 0.5x \times (3) = 42\) (partitioning) | M1 |
| \(x(2x + 13) = 84\) or \(x^2 + 5x + 1.5x = 42\) | M1 |
| (2) |
Notes
M1: dep on 1st M1 then needs to develop on to quadratic given.
| Scheme | Marks |
|---|---|
| \((2x + 21)(x - 4)\) (= 0) oe | B2 |
| \(x = 4\) | A1 |
| (P=) “4” +”9” +”12” + \(\sqrt{(3^2 + \text{“}4\text{”}^2)}\) | M1 |
| 30 | A1cao |
| (5) | |
| (Total 7 marks) |
Notes
B2: B1 for either factor correct or \((2x \pm 21)(x \pm 4)\)
or M1 for \(x = \dfrac{-13 \pm \sqrt{13^2 - 4 \times 2 \times -84}}{4}\) (condone 1 sign error)
then M1 for \(x = \dfrac{-13 \pm \sqrt{169 + 672}}{4}\)
A1: dep on M1 or B2
M1: i.e \(x + (x + 5) + (x + 8) + \sqrt{(3^2 + x^2)}\) in numeric form.
A1cao: (Last two marks independent)
N.B. Working for solving quadratic could be seen in (a) if not contradicted in (b).