Higher June 2018 Paper 1 Q19
19. The perpendicular height of a triangle is \(\sqrt{5}\) cm.
The area of this triangle is \((10 - 3\sqrt{5})\) cm2.
Find the length of the base of the triangle.
Give your answer in the form \(p\sqrt{5} + q\) where \(p\) and \(q\) are integers. [4]
| Scheme | Marks | Comments |
|---|---|---|
| \(10 - 3\sqrt5 = \dfrac12 \times base \times \sqrt5\) | B1 | Any clear notation is acceptable. |
| \((base =)\ \dfrac{10 - 3\sqrt5}{\frac12 \times \sqrt5}\) or equivalent | B1 | Allow \(\dfrac{10 - 3\sqrt5}{\sqrt5} = \dfrac12 \times base\) or \(10\sqrt5 - 3\times5 = \dfrac12 \times base \times 5\) or \(20 = \sqrt5(base + 6)\) |
| \((base =)\ \dfrac{20}{\sqrt5} - 6\) or better or \((base =)\ \dfrac{2\times10\sqrt5 - 2\sqrt5\times3\sqrt5}{\left(\sqrt5\right)^2}\) or better | M1 | Allow \(\dfrac{10}{\sqrt5} - 3 = \dfrac12 \times base\) or better or \(\dfrac{10\sqrt5 - 3\sqrt5\times\sqrt5}{\left(\sqrt5\right)^2} = \dfrac12 \times base\) or better |
| \((base =)\ 4\sqrt5 - 6\) | A1 | Allow \(-6 + 4\sqrt5\) |
| (4) |