Higher November 2019 Paper 2 Q14
14.

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Calculate the area of \(ABCD\). [6]
| Scheme | Marks | Comments |
|---|---|---|
| \(BD^2 = 10.8^2 + 12.3^2 - 2 \times 10.8 \times 12.3 \times \cos 112^\circ\) | M1 | |
| \(BD^2 = 367.455\ldots\) | A1 | |
| \(BD = 19.169(\ldots\text{cm})\) or \(BD = 19.2\) (cm) or \(BD = 19.17\) (cm) | A1 | |
| (Area \(DAB =\)) \(\tfrac12 \times 10.8 \times 12.3 \times \sin 112^\circ\) | M1 | |
| (Area \(DBC =\)) \(\tfrac12 \times 21.5 \times 19.169\ldots \times \sin 75^\circ\) | M1 | FT ‘their derived \(BD\)’ |
| (Area \(DAB + DBC =\) \(61.58\ldots + 199.04\ldots =\)) 260.6 (cm²) to 261 (cm²) | A1 | FT provided at least one of the previous two M marks have been awarded |
| (6) |