June 2025 Paper 2 Q18
18 \(ABCDEF\) is a triangular prism.
\(ABCD\) is a rectangle and angle \(BCF\) is \(90^\circ\)
\(AB = 14\) cm \(\qquad BC = 8\) cm \(\qquad CF = 6\) cm
\(M\) is the midpoint of \(EF\).
Work out the size of angle \(BMD\). [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((DM^2 =)\ 6^2 + 7^2\) or 85 | M1 | oe eg \((DM =)\ \sqrt{6^2 + 7^2}\) or [9.2, 9.22] may be seen on diagram |
| \((BD^2 =)\ 14^2 + 8^2\) or 260 | M1 | oe eg \((BD =)\ \sqrt{14^2 + 8^2}\) or \(2\sqrt{65}\) or [16.1, 16.125] may be seen on diagram |
| \((BM^2 =)\ 6^2 + 8^2 + 7^2\) or \(10^2 + 7^2\) or 149 | M1 | oe eg \((BM =)\ \sqrt{6^2 + 8^2 + 7^2}\) or \(\sqrt{10^2 + 7^2}\) or [12.2, 12.21] may be seen on diagram |
| their \(260 =\) their \(85 +\) their \(149 -\) \(2 \times \sqrt{\text{their } 85} \times \sqrt{\text{their } 149} \cos BMD\) or \(\cos BMD = \dfrac{\text{their } 85 + \text{their } 149 - \text{their } 260}{2 \times \sqrt{\text{their } 85} \times \sqrt{\text{their } 149}}\) | M1dep | oe cosine rule eg \(\cos BMD = [-0.12, -0.11]\) or \(\cos^{-1}\dfrac{\text{their } 85 + \text{their } 149 - \text{their } 260}{2 \times \sqrt{\text{their } 85} \times \sqrt{\text{their } 149}}\) dep on M3 |
| [96.3, 97] | A1 |
Additional guidance
| Up to M3 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| 4th M1 Accept any indication of \(BMD\) eg \(x\) or \(M\) | |
| 4th M1 \(2\sqrt{65}^{\,2}\) must be recovered |