June 2019 Paper 1 Q18
18 \(ADEF\) is a trapezium.
\(ABCD\) is a straight line.
\(BCEF\) is a square of side \(\sqrt{6}\) cm
(a) Show that \(\quad AB = \sqrt{2}\) cm [1 mark]
(b) Show that \(\quad DE = 2\sqrt{6}\) cm [1 mark]
(c) Work out the perimeter of the trapezium \(ADEF\).
Give your answer in the form \(\quad t\sqrt{2} + w\sqrt{6} \quad\) where \(t\) and \(w\) are integers.
You must show your working. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \((AB =)\ \dfrac{\sqrt{6}}{\tan 60} = \dfrac{\sqrt{6}}{\sqrt{3}} = \sqrt{2}\) | B1 | oe must see tan60 oe and some evidence of manipulation with \(\sqrt{3}\) oe as well as the final answer to award B1 |
| Alternative method 2 | ||
| Use of \(1 : 2 : \sqrt{3}\) triangle and showing that our triangle is an enlargement scale factor \(\sqrt{2}\) | B1 | oe must see the triangle drawn and labelled or the ratio clearly seen and the scale factor clearly stated |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \((DE =)\ \dfrac{\sqrt{6}}{\sin 30} = \dfrac{\sqrt{6}}{0.5} = 2\sqrt{6}\) | B1 | oe must see sin30 oe and some evidence of manipulation with 0.5 oe as well as the final answer to award B1 |
| Alternative method 2 | ||
| Use of \(1 : 2 : \sqrt{3}\) triangle and showing that our triangle is an enlargement scale factor \(\sqrt{2}\) | B1 | oe must see the triangle drawn and labelled or the ratio clearly seen and the scale factor clearly stated |
| Answer | Mark | Comments |
|---|---|---|
| \(AF = \dfrac{AB}{\cos 60} = \dfrac{\sqrt{2}}{0.5} = 2\sqrt{2}\) or \(AF = \dfrac{BF}{\sin 60} = \dfrac{\sqrt{6}}{\frac{\sqrt{3}}{2}} = 2\sqrt{2}\) or \(AF^2 = (\sqrt{2})^2 + (\sqrt{6})^2\), so \(AF = \sqrt{8}\) or \(2\sqrt{2}\) | B1 | oe allow \(2\sqrt{2}\) or \(\sqrt{8}\) for this mark seen on the diagram or clearly shown in working |
| \(CD = \sqrt{6} \times \tan 60 = \sqrt{6} \times \sqrt{3}\) \(= \sqrt{18}\) or \(3\sqrt{2}\) or \(CD = DE \cos 30^\circ\) \(= 2\sqrt{6} \times \dfrac{\sqrt{3}}{2} = \sqrt{6} \times \sqrt{3}\) \(= \sqrt{18}\) or \(3\sqrt{2}\) or \(CD^2 = (2\sqrt{6})^2 - (\sqrt{6})^2 = 18\) so \(CD = \sqrt{18}\) or \(3\sqrt{2}\) | B1 | oe allow \(\sqrt{6} \times \sqrt{3}\) or \(\sqrt{18}\) or \(3\sqrt{2}\) for this mark seen on the diagram or clearly shown in working |
| \(6\sqrt{2} + 4\sqrt{6}\) | B1dep | dependent on B1, B1 already awarded |
Additional guidance
| Condone brackets missed off if recovered | |
| \(AF\) and \(CD\) could be seen in part (a) or part (b) so could be awarded B1 in part (c) if used correctly |