1MA0 Higher June 2012 Paper 2 Q18
18 The diagram shows a quadrilateral \(ABCD\).
Diagram NOT accurately drawn
\(AB = 16\) cm.
\(AD = 12\) cm.
Angle \(BCD = 40^\circ\)
Angle \(ADB\) = angle \(CBD = 90^\circ\)
Calculate the length of \(CD\).
Give your answer correct to 3 significant figures. (5)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(BD^2 + 12^2 = 16^2\) oe \(BD = \sqrt{256 - 144}\) (=10.58...) \(\sin 40 = \dfrac{\text{‘}10.58\text{’}}{CD}\) \(CD = \dfrac{\text{‘}10.58\text{’}}{\sin 40}\) | 16.5 | 5 | M1 for \(BD^2 + 12^2 = 16^2\) oe or \(16^2 - 12^2\) or 112 seen M1 for \(\sqrt{256 - 144}\) or \(\sqrt{112}\) (=10.58...) M1 for \(\sin 40 = \dfrac{\text{‘}10.58\text{’}}{CD}\) or \(\cos 50 = \dfrac{\text{‘}10.58\text{’}}{CD}\) M1 for (\(CD =\)) \(\dfrac{\text{‘}10.58\text{’}}{\sin 40}\) or \(\dfrac{\text{‘}10.58\text{’}}{\cos 50}\) A1 for 16.4 – 16.5 OR M1 for \(BD^2 + 12^2 = 16^2\) oe or \(16^2 - 12^2\) or 112 seen M1 for \(\sqrt{256 - 144}\) or \(\sqrt{112}\) (=10.58...) M1 for (\(BC =\)) ‘10.58’ × tan 50 or \(\dfrac{\text{‘}10.58\text{’}}{\tan 40}\) (=12.6…) M1 for \(\sqrt{\text{‘}12.6\text{’}^2 + \text{‘}10.58...\text{’}^2}\) A1 for 16.4 – 16.5 |