Higher November 2017 Paper 2 Q17
17 Work out the area of the parallelogram. [3 marks]
Not drawn accurately
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\sin 72 = \dfrac{h}{12}\) or \(12 \sin 72\) or \(\cos (90 - 72) = \dfrac{h}{12}\) or \(12 \cos (90 - 72)\) or \(\dfrac{h}{\sin 72} = \dfrac{12}{\sin 90}\) or 11.4… | M1 | oe Any letter |
| \(16 \times\) their 11.4… | M1dep | |
| [182.4, 182.603] or 183 | A1 | |
| Alternative method 2 | ||
| \(h^2 + (12 \cos 72)^2 = 12^2\) or \(h^2 + (12 \sin (90 - 72))^2 = 12^2\) or \(\sqrt{12^2 - (12 \cos 72)^2}\) or \(\sqrt{12^2 - (12 \sin (90 - 72))^2}\) or 11.4… | M1 | oe Any letter |
| \(16 \times\) their 11.4… | M1dep | |
| [182.4, 182.603] or 183 | A1 | |
| Alternative method 3 | ||
| \(0.5 \times 16 \times 12 \times \sin 72\) or 91.3… | M1 | oe eg \(0.5 \times 16 \times 12 \times \sin 108\) |
| \(2 \times\) their 91.3… | M1dep | |
| [182.4, 182.603] or 183 | A1 | |
Additional guidance
| \(2 \times 16 \times 12 \times \sin 72\) | M1M0A0 |
| \(\sin = \dfrac{h}{12}\) or \(\sin \theta = \dfrac{h}{12}\) (unless recovered) | M0 |