Higher January 2019 Paper 2 Q15
15 The diagram shows a trapezium.
Diagram NOT accurately drawn
All measurements shown on the diagram are in centimetres.
The area of the trapezium is 133 cm2
(a) Show that \(8x^2 - 6x - 275 = 0\) (3)
(b) Find the value of \(x\).
Show your working clearly. (3)
Show your working clearly. (3)
| Scheme | Marks |
|---|---|
e.g. \(\dfrac{1}{2} \times (x + 5 + 3x - 2) \times (2x - 3)\) or \(0.5(4x + 3)(2x - 3)\) oe | M1 |
eg. \(\dfrac{1}{2} \times (8x^2 - 12x + 6x - 9) = 133\) or \(8x^2 - 12x + 6x - 9 = 266\) | M1 |
| shown | A1 |
| (3) |
Notes
M1: correct algebraic expression for area
M1: for correct equation with brackets expanded
A1: for completion to given equation
dep on M2
| Scheme | Marks |
|---|---|
\(\dfrac{--6 \pm \sqrt{36 - -8800}}{2 \times 8}\) or \(\dfrac{6 \pm \sqrt{36 + 8800}}{16}\) or \(\dfrac{6 \pm \sqrt{8836}}{16}\) or \((4x - 25)(2x + 11)\) (=0) | M2 |
| 6.25 oe | A1 |
| (3) | |
| (6 marks) |
Notes
M2: If not M2 then award M1 for \(\dfrac{--6 \pm \sqrt{(-6)^2 - 4 \times 8 \times -275}}{2 \times 8}\)
Condone one sign error in substitution; allow evaluation of individual terms e.g. 36 in place of \((-6)^2\) [allow \(-6^2\) or \(6^2\) in place of \((-6)^2\), throughout allow + rather than \(\pm\)]
or
\((4x \pm 25)(2x \pm 11)\) (=0)
(if student gains M1 and shows both answers the 2nd M1 can be awarded)
ft from an incorrect 3 term quadratic equation
A1: dep on M1 and 6.25 oe alone given as final answer