Higher June 2019 Paper 2R Q19
19 The diagram shows a sector \(OAPB\) of a circle, centre \(O\).
Diagram NOT accurately drawn
\(AB\) is a chord of the circle.
Angle \(AOB = 80°\)
The area of sector \(OAPB\) is \(\dfrac{25}{2}\pi\) cm2
Work out the perimeter of the shaded segment.
Give your answer correct to 3 significant figures.
(6)
| Scheme | Marks |
|---|---|
| \(\dfrac{25}{2}\pi = \pi r^2 \times \dfrac{80}{360}\) | M1 |
| \(r = 7.5\) | A1 |
(\(APB\) =) \(2 \times \pi \times \text{``}{7.5}\text{''} \times \dfrac{80}{360}\) (= 10.471)…. (\(APB\) =) 10.471…. (=\(10\pi/3\)) | M1ft |
(\(AB^2\)) = \(\text{``}{7.5}\text{''}^2 + \text{``}{7.5}\text{''}^2 - (2 \times \text{``}{7.5}\text{''} \times \text{``}{7.5}\text{''} \times \cos 80)\) or \(\dfrac{AB}{\sin 80} = \dfrac{\text{``}{7.5}\text{''}}{\sin 50}\) or (\(AB\) =) 2 × “7.5” × sin 40 (\(AB\) =) 9.6418 | M1ft |
| “9.6418” + “10.4719” | M1ft |
| 20.1 | A1 |
| (6) | |
| (6 marks) |
Notes
M1: Equation of sector equal to \(\dfrac{25\pi}{2}\) or a calculation that leads to \(r\) or \(r^2\)
M1ft: Dep on 1st M1
Accept 10.5 or better
M1ft: Dep on 1st M1
Correct equation to find \(AB\) (= 9.6) or \(AB^2\) (= 93 or better) must use a clearly identified radius value
M1ft: Dep on 2nd and 3rd method marks
A1: awrt 20.1