Higher June 2025 Paper 5 Q19
19 The diagram shows a shape formed by a semicircle joined to a sector of a circle, BAC, along a common length AB.
Not to scale
The area of the sector BAC is \(16\pi\) cm\(^2\).
Work out the total area of the shape.
Give your answer in terms of \(\pi\).
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(34\pi\) final answer with correct working | 5 | M2 for \(\text{AB}^2 = \frac{16 \times 360}{40}\) or better or M1 for \(\frac{40}{360}[\pi]\text{AB}^2 [= 16[\pi]]\) oe or area of full circle = \(\frac{360}{40} \times 16\ [\pi]\) or \(144[\pi]\) | Correct working requires at least M2A1 For M2 accept AB\(^2\) = 144 Accept other variable for AB e.g. \(r\) , \(d\) |
| A1 for AB = 12 or better e.g. \(\frac{\text{AB}}{2} = 6\) M1 for \(\pi\left(\frac{\textit{their } \text{AB}}{2}\right)^2 \div 2\ [+\ 16\pi]\) oe | After M1, A1 implies M2 their AB must be clearly indicated in working or on diagram . Allow AB as the answer to their work with the sector BAC | ||
| If 0 or 1 scored instead award SC2 for answer \(34\pi\) with no or insufficient working If 0 scored, award SC1 for AB = 12 with no or insufficient working | Do not award SC2 if clearly from wrong working Do not award SC1 if clearly from wrong working | ||