Higher January 2020 Paper 2 Q11
11 The diagram shows a shaded shape \(ABCD\) made from a semicircle \(ABC\) and a right-angled triangle \(ACD\).
Diagram NOT accurately drawn
\(AC\) is the diameter of the semicircle \(ABC\).
Work out the perimeter of the shaded shape.
Give your answer correct to 3 significant figures.
(5)
| Scheme | Marks |
|---|---|
| \((AC^2 =)\; 17^2 - 15^2\) | M1 |
| \((AC =)\; \sqrt{17^2 - 15^2}\;\; (= \sqrt{64} = 8)\) | M1 |
| \(\dfrac{\pi \times \text{‘}8\text{’}}{2}\) (= 4π = 12.566...) | M1 |
| ‘12.566…’ + 15 + 17 | M1 |
| 44.6 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: dep on M2 for \(\dfrac{\pi \times \text{‘}8\text{’}}{2}\) oe or 4π 12.5663…
M1: for ‘12.566’ + 15 + 17 and no additional values
A1: for awrt 44.6
Alternative mark scheme for 11
| Scheme | Marks |
|---|---|
| \(\cos^{-1}\left(\dfrac{15}{17}\right)\) (= 28.0724) or \(\sin^{-1}\left(\dfrac{15}{17}\right)\) (= 61.9275) | M1 |
| 15 × tan (‘28.0724’) (= 8) or 15 ÷ tan (‘61.9275’) (= 8) | M1 |
| \(\dfrac{\pi \times \text{‘}8\text{’}}{2}\) (= 4π = 12.566…) | M1 |
| “12.566” + 15 + 17 | M1 |
| 44.6 | A1 |
Notes
M1: for a correct method to find one of the angles
M1: dep on M2 for \(\dfrac{\pi \times \text{‘}8\text{’}}{2}\) or 12.5663… or 4π
M1: for “12.566” + 15 + 17 and no additional values
A1: for awrt 44.6