Higher November 2016 Unit 2 Q14
14. Points \(E\) and \(F\) lie on a circle, centre \(O\).
The radius of the circle is 10 cm.
The area of the shaded sector is 65 cm2.
Diagram not drawn to scale
(a) Calculate the size of \(E\hat{O}F\). [3]
(b) Hence, calculate the length of the arc \(EF\). [2]
| Scheme | Marks | Comments |
|---|---|---|
| \(\theta\) = (65×360)/(102 × \(\pi\)) | M2 | Award M1 for \(\theta\)÷360 × \(\pi\) × 10² = 65 |
| \(\theta\) = 74.48(°) | A1 | Accept answers in the range 74.47 to 74.53 or 74 or 75. |
| Scheme | Marks | Comments |
|---|---|---|
| EF = (74.48/360) × 2 × \(\pi\) × 10 | M1 | FT their '74.48'. |
| = 13(cm) | A1 | Accept answers between 12.9 and 13.1, truncated or rounded. Allow in terms of \(\pi\), e.g. \(\dfrac{931}{225}\pi\) |
Alternative (the major sector arc)
| Scheme | Marks | Comments |
|---|---|---|
| Allow 285.52/360 × 2 × \(\pi\) × 10 | M1 | |
| = 49.8(3…) or 50(cm) | A1 |