Higher January 2019 Paper 2R Q22
22
Diagram NOT accurately drawn
\(OPQ\) is a sector of a circle, centre \(O\)
\(OAB\) is a sector of a circle, centre \(O\)
\(A\) is the point on \(OP\) such that \(OA : AP = 3 : 2\)
\(B\) is the point on \(OQ\) such that \(OB : BQ = 3 : 2\)
Angle \(POQ = 45^\circ\)
The area of the shaded region is \(\dfrac{81}{2}\pi\) cm²
Work out the perimeter of the shaded region.
Give your answer in terms of \(\pi\).
(6)
| Scheme | Marks |
|---|---|
\(\pi \times (5r)^2 \times \dfrac{45}{360}\) or \(\pi \times (3r)^2 \times \dfrac{45}{360}\) \(\pi \times r^2 \times \dfrac{45}{360}\) or \(\pi \times (0.6r)^2 \times \dfrac{45}{360}\) | M1 |
\(\pi \times (5r)^2 \times \dfrac{45}{360} - \pi \times (3r)^2 \times \dfrac{45}{360} = \dfrac{81}{2}\pi\) or \(\pi \times r^2 \times \dfrac{45}{360} - \pi \times (0.6r)^2 \times \dfrac{45}{360} = \dfrac{81}{2}\pi\) | M1 |
| \(r^2 = (40.5 \times 8) \div (1 - 0.36)\) or \(r^2 = 506.25\) oe (\(r = 22.5\)) \(r^2 = (40.5 \times 8) \div (25 - 9)\) or \(r^2 = 20.25\) oe (\(r = 4.5\)) | M1 |
\((AB =)\; 2 \times \pi \times \text{‘}13.5\text{’} \times \dfrac{45}{360} \left(= \dfrac{27}{8}\pi\right)\) or \((PQ =)\; 2 \times \pi \times \text{‘}22.5\text{’} \times \dfrac{45}{360} \left(= \dfrac{45}{8}\pi\right)\) oe | M1 |
| Perimeter = \(\text{‘}\dfrac{27}{8}\pi\text{’} + \text{‘}\dfrac{45}{8}\pi\text{’} + \text{‘}9\text{’} + \text{‘}9\text{’}\) | M1 |
| \(9\pi + 18\) | A1 |
| (6) | |
| (6 marks) |
Notes
M1: oe
M1: oe
M1: or 1 share = 4.5 or \(r = 22.5\) or
\(OA = 13.5\) or \(AP = 9\)
(\(r^2 = 20.25\) corrected from the printed mark scheme, which says \(r^2 = 80.25\); \(324 \div 16 = 20.25\) and \(r = 4.5\))
M1: dep on M3
or
\(2 \times \pi \times (\text{‘}13.5\text{’} + \text{‘}22.5\text{’}) \times \dfrac{45}{360} \; (= 9\pi)\)
M1: dep on M4
A1: oe
M2 for \(0.64\pi r^2 \times \dfrac{45}{360} = \dfrac{81}{2}\pi\) or \(16\pi r^2 \times \dfrac{45}{360} = \dfrac{81}{2}\pi\)