Higher June 2024 Paper 1 Q20
20 The diagram shows a sector \(OABC\) of a circle centre \(O\)
Diagram NOT accurately drawn
Angle \(AOC = 60^\circ\)
The area of the shaded segment \(ABC\) is 38 cm\(^2\)
Work out the perimeter of the shaded segment \(ABC\)
Give your answer correct to one decimal place.
(4)
| Scheme | Marks |
|---|---|
eg \(\pi r^2 \times \dfrac{60}{360} - \dfrac{1}{2}r^2 \sin 60\) oe or \(\dfrac{\pi r^2}{6} - \dfrac{\sqrt{3}}{4}r^2\) oe | M1 |
eg \((r^2 =)\, 38 \div \left(\dfrac{\pi}{6} - \dfrac{\sqrt{3}}{4}\right)\big(= 38 \div 0.09(058)\big)\) (= 419(.490…)) oe or \((r =)\sqrt{38 \div \left(\dfrac{\pi}{6} - \dfrac{\sqrt{3}}{4}\right)}\) (= 20.4(81…)) oe | M1 |
| \(\dfrac{\pi}{6} \times \text{``}{20.4(81...)}\text{''} \times 2\) (= 21.4(48…)) oe or | M1 |
| Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: 41.9 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for a correct expression for the area of the segment
Expression may be embedded in an equation, eg
\(\pi r^2 \times \dfrac{60}{360} - \dfrac{1}{2}r^2 \sin 60 = 38\) or
\(\pi r^2 \times \dfrac{60}{360} = 38 + \dfrac{1}{2}r^2 \sin 60\) or
\(\pi r^2 \times \dfrac{60}{360} - 38 = \dfrac{1}{2}r^2 \sin 60\)
M1: dep on M1 for a correct expression for \(r^2\) or \(r\)
M1: for using the value of \(r\) to find arc length
A1: allow 41 - 42