1MA0 Higher June 2014 Paper 2 Q27
27
Diagram NOT accurately drawn
\(ABCD\) is a parallelogram.
\(AC = 9\) cm
\(DC = 11\) cm
Angle \(DAC = 100^\circ\)
Calculate the area of the parallelogram.
Give your answer correct to 3 significant figures. (5)
| Answer | Mark | Notes |
|---|---|---|
| 43.9 | 5 | M1 for \(\dfrac{11}{\sin 100} = \dfrac{9}{\sin D}\) oe M1 for \(\sin D = \dfrac{9\sin 100}{11}\) (=0.80575…) or \(D = 53.68\ldots\) M1 for angle \(DCA = 180 - 100 - \text{``}D\text{''}\) (=26.317..) M1 for area of \(ABCD = 2 \times \frac{1}{2} \times 11 \times 9 \times \sin \text{``}26.317\text{''}\) A1 for 43.8 – 43.9 OR M1 for \(\dfrac{11}{\sin 100} = \dfrac{9}{\sin D}\) oe M1 for \(\sin D = \dfrac{9\sin 100}{11}\) (=0.80575…) or \(D = 53.68\ldots\) M1 for (height=) \(9 \times \sin(180 - 100 - \text{``}D\text{''})\) or height = 3.990… M1 for area of \(ABCD = (2 \times \frac{1}{2}) \times 11 \times\) “height” A1 for 43.8 – 43.9 OR M1 for \(11^2 = AD^2 + 9^2 - 2 \times AD \times 9 \times \cos 100\) M1 for \(AD = \dfrac{18\cos 100 + \sqrt{(18\cos 100)^2 - 4(1)(-40)}}{2(1)}\) M1 for \(AD = \dfrac{18\cos 100 + \sqrt{169.7(69795\ldots)}}{2(1)}\) (= 4.95195(…)) M1 for area of \(ABCD = 2 \times \dfrac{1}{2} \times \text{``}4.95195\text{''} \times 9 \times \sin 100\) A1 for 43.8 – 43.9 |