4MA0 Higher January 2016 Paper 4 Q24
24 The diagram shows sector \(OAB\) of a circle, centre \(O\).
Diagram NOT accurately drawn
Angle \(AOB\) = 50°
Sector \(OAB\) has area \(20\pi\) cm²
Calculate the perimeter of sector \(OAB\).
Give your answer correct to 3 significant figures.
(5)
| Scheme | Marks |
|---|---|
| \(\dfrac{50}{360} \times \pi \times r^2 = 20\pi\) oe | M1 |
| \(r^2 = \dfrac{360 \times 20}{50}\) or \(\dfrac{360 \times 20\pi}{50\pi}\) oe (=144) | M1 |
| \(r = 12\) | A1 |
(perimeter =) \(12 + 12 + \dfrac{50}{360} \times 2 \times 12 \times \pi\) (\(= 24 + \dfrac{10\pi}{3}\)) (=24 + 10.47...) | M1ft |
| 34.5 | A1 |
| (5) |
Notes
M1: A correct equation for area of sector
M1: Correct rearrangement in terms of \(r^2\)
A1: \(r = 12\)
M1ft: (ft dep on M2) for arc length
\(\dfrac{50}{360} \times 2 \times \text{‘12’} \times \pi\) (=10.47...)
Or \(l = \dfrac{20\pi}{12 \times 0.5}\) oe
A1: Accept 34.4 - 34.5
Misread of 20 for 20π
| Scheme | Marks |
|---|---|
| \(\dfrac{50}{360} \times \pi \times r^2 = 20\) | M1 |
| \(r^2 = \dfrac{360 \times 20}{50\pi}\) (=45.8…) | M1 |
| \(r = 6.77\) | A0 |
(perimeter =) \(6.77 + 6.77 + \dfrac{50}{360} \times 2 \times \text{‘6.77’} \times \pi\) (=19.4….) | M1ft |
| A0 |
Notes
M1: Equation for area of sector using 20
M1: Correct rearrangement in terms of \(r^2\)
M1ft: (ft dep on M2) for arc length
\(\dfrac{50}{360} \times 2 \times \text{‘6.77’} \times \pi\) (=5.91…)
(Misread maximum 3 marks)