1MA0 Higher June 2016 Paper 1 Q25
25 \(ABD\) is a right angled triangle.
Diagram NOT accurately drawn
All measurements are given in centimetres.
\(C\) is the point on \(BD\) such that \(CD = \dfrac{\sqrt{3}}{3}\)
\(AD = BD = \dfrac{\sqrt{2}}{2}\)
Work out the exact area, in cm\(^2\), of the shaded region. (3)
| Answer | Mark | Notes |
|---|---|---|
| \(\tfrac{1}{4} - \tfrac{\sqrt{6}}{12}\) | 3 | M1 for \(\tfrac{1}{2} \times \tfrac{\sqrt{2}}{2} \times \tfrac{\sqrt{2}}{2}\) or \(\tfrac{1}{2} \times \tfrac{\sqrt{2}}{2} \times \tfrac{\sqrt{3}}{3}\) M1 for \(\tfrac{1}{2} \times \tfrac{\sqrt{2}}{2} \times \tfrac{\sqrt{2}}{2} - \tfrac{1}{2} \times \tfrac{\sqrt{2}}{2} \times \tfrac{\sqrt{3}}{3}\) A1 for \(\tfrac{1}{4} - \tfrac{\sqrt{6}}{12}\) oe OR M1 for \((BC =)\ \tfrac{\sqrt{2}}{2} - \tfrac{\sqrt{3}}{3}\) M1 for \(\tfrac{1}{2} \times \left\{\tfrac{\sqrt{2}}{2} - \tfrac{\sqrt{3}}{3}\right\} \times \tfrac{\sqrt{2}}{2}\) A1 for \(\tfrac{1}{4} - \tfrac{\sqrt{6}}{12}\) oe |