Foundation June 2018 Paper 1 Q28
28. Nia and David are trying to work out the area of this sector of a circle.
They must give the answer as a multiple of \(\pi\).
Diagram not drawn to scale
Here is Nia’s answer.
| Step 1 | 360 ÷ 45 = 8 |
| Step 2 | Area of whole circle = \(\pi \times 24\) |
| Step 3 | Area of sector = \(\dfrac{1}{8}\) of \(24\pi = \dfrac{24\pi}{8}\) |
| Step 4 | Answer = \(3\pi\) cm\(^2\) |
David looks at Nia’s answer and says,
‘Your answer is wrong.’
- Explain the error that Nia has made.
- Calculate the correct answer as a multiple of \(\pi\). [3]
| Scheme | Marks | Comments |
|---|---|---|
| Any valid reason e.g. ‘Nia has found the circumference.’ or ‘Circumference circle = \(\pi \times 24\).’ or ‘She has used the diameter, not the radius squared’. or ‘Area circle = \(\pi \times 144\)’ | E1 | Allow e.g. ‘Area of circle should be \(\pi \times r^2\)’ or ‘Area circle = \(\pi \times 12^2\).’ Do not allow e.g. ‘Area circle = \(\pi \times r^2\) and Circumference = \(\pi \times d\)’ or Area = \(\pi \times r^2\) without further evidence e.g. correct method shown |
| \(\left(\dfrac{\pi \times 144}{8} = \right) 18\pi\) (cm\(^2\)) | B2 | Mark final answer B1 for (area sector =) \(\dfrac{\pi \times 12^2}{8}\) or equivalent; Allow use of \(\pi = 3.14\) for E1 and B1 |
| (3) |