Foundation January 2020 Paper 2 Q27
27 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 |
| \(\dfrac{\pi \times \text{‘}8\text{’}}{2}\ (= 4\pi = 12.566\ldots)\) | 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
| 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 |
| Answer: 44.6 | A1 |
M1 (1st): for a correct method to find one of the angles
M1 (3rd): dep on M2 for \(\dfrac{\pi \times \text{‘}8\text{’}}{2}\) or 12.5663… or 4π
M1 (4th): for “12.566” + 15 + 17 and no additional values
A1: for awrt 44.6
(corrected from the printed mark scheme, which heads this “Alternative mark scheme for 11”)