Higher June 2019 Paper 1 Q9
9 The diagram shows a shape made from a right-angled triangle and a semicircle.
Diagram NOT accurately drawn
\(AC\) is the diameter of the semicircle.
\(BA = BC = 6\) cm
Angle \(ABC = 90^\circ\)
Work out the area of the shape.
Give your answer correct to 1 decimal place.
(5)
| Scheme | Marks |
|---|---|
| 0.5 × 6 × 6 (=18) | M1 |
| (\(d^2\) =) \(6^2 + 6^2\) (=72) or \(\dfrac{AC}{(\sin 90)} = \dfrac{6}{\sin 45}\) | M1 |
\(\sqrt{6^2 + 6^2}\) (= \(\sqrt{72}\) = \(6\sqrt{2}\) = 8.4(85…) or 8.5) or \(AC = \dfrac{6(\sin 90)}{\sin 45} = 6\sqrt{2}\) = 8.4(85…) or 8.5) oe | M1 |
| \(0.5 \times \pi \times \left(\dfrac{\text{``}{8.48..}\text{''}}{2}\right)^2\) (= \(9\pi\) or 28.…) | M1 |
| 46.3 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: For area of triangle, or may use \(\dfrac{1}{2} \times 6 \times 6\sqrt{2}\sin 45\) or \(\dfrac{1}{2} \times 6\sqrt{2} \times 3\sqrt{2}\) oe
A1: for 46.2 – 46.3