4MA0 Higher January 2017 Paper 3R Q21
21 The shape \(OABC\) is made from a triangle and a sector of a circle.
Diagram NOT accurately drawn
\(OAB\) is a triangle.
\(OBC\) is a sector of a circle, centre \(O\).
\(OA = 12\) cm
\(AB = 16\) cm
Angle \(OAB = 60^\circ\)
Angle \(BOC = 38^\circ\)
Work out the area of \(OABC\).
Give your answer correct to 3 significant figures.
(5)
| Scheme | Marks |
|---|---|
| (\(OB^2 =\)) \(12^2 + 16^2 - 2 \times 12 \times 16 \times \cos(60^\circ)\) | M1 |
| (\(OB =\)) \(\sqrt{208}\) or \(4\sqrt{13}\) or 14.4.... or (\(OB^2\)) = 208 | M1 |
0.5 × 12 × 16 × sin(60°) (= 83.1… or \(48\sqrt{3}\)) or \(\dfrac{38}{360} \times \pi \times \text{“}14.4\text{”} \times \text{“}14.4\text{”}\) (=68.9…) or \(\dfrac{38}{360} \times \pi \times \text{“}208\text{”}\) (=68.9…) | M1 |
\(0.5 \times 12 \times 16 \times \sin(60^\circ) + \dfrac{38}{360} \times \pi \times \text{“}14.4\text{”} \times \text{“}14.4\text{”}\) (68.9…. + 83.1…) | M1 |
| 152 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: ft their 14.4 provided first M1 awarded.
M1: ft their 14.4 provided first M1 awarded.
A1: awrt 152
or M2 for \(\sqrt{12^2 + 16^2 - 2 \times 12 \times 16 \times \cos(60^\circ)}\)