Higher November 2018 Paper 6 Q12
12 In the diagram, the square and the trapezium share a common side of length \(x\) cm.
Not to scale
The area of the square is equal to the area of the trapezium.
Work out the value of \(x\). [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 7.17 to 7.18 or 7.2 nfww | 6 | M3 for \(x^2 - 3x - 30 = 0\) or M2 for \(\dfrac{6}{2}(10 + x) = x^2\) oe or M1 for \(\dfrac{6}{2}(10 + x)\) oe AND M2FT for \(\dfrac{3 + \sqrt{(-3)^2 - 4 \times (-30)}}{2}\) or better or 7.17 to 7.18 and –4.18 to –4.17 or M1FT for either formula with at most two errors | Condone missing brackets for M1 FT from their 3 term quadratic Allow M2FT for \(\dfrac{3 \pm \sqrt{(-3)^2 - 4 \times (-30)}}{2}\) or better Alternative by completing the square: M2FT for \(1.5 + \sqrt{32.25}\) or \(1.5 \pm \sqrt{32.25}\) or 7.17 to 7.18 and –4.18 to –4.17 or M1FT for \((x - 1.5)^2 - 32.25\) |