Higher June 2024 Unit 1 Q10
10. The area of the shaded sector \(AOB\) shown below is \(\dfrac{132\pi}{5}\) cm2.
Diagram not drawn to scale
Find the size of angle \(x\).
In your working, you should express any areas in terms of \(\pi\). [3]
| Scheme | Marks | Comments |
|---|---|---|
| \(\dfrac{(\text{reflex})\ A\hat{O}B}{360} \times \pi \times 6^2\ \left(= \dfrac{132\pi}{5} = 26{\cdot}4\pi\right)\) or equivalent | M1 | Area of major sector. Allow \(\dfrac{x}{360} \times \pi \times 6^2\ \left(= \dfrac{132\pi}{5}\right)\). |
| (360 – \(x\) OR reflex \(A\hat{O}B\) =) \(\dfrac{132\pi \times 360}{5 \times \pi \times 6^2}\) (= 264) | A1 | |
| \(x\) (= 360 – 264) = 96(°) | A1 | From A0, FT 360 – ‘their 264’, provided M1 awarded. An unsupported answer of 96(°) is awarded M1A1A1. |
Alternative method
| Scheme | Marks | Comments |
|---|---|---|
| Area of minor sector = \(\pi \times 6^2 - \dfrac{132\pi}{5}\ \left(= \dfrac{48\pi}{5} = 9{\cdot}6\pi\right)\) | B1 | Area of minor sector. |
| \(\dfrac{x}{360} \times \pi \times 6^2 = \dfrac{48\pi}{5}\) or equivalent | M1 | FT ‘their derived \(\dfrac{48\pi}{5}\)’. |
| \(x\) = 96(°) | A1 |