Higher June 2024 Paper 2R Q13
13 \(OAB\) is a sector of a circle with centre \(O\) and radius \(r\) cm.
Diagram NOT accurately drawn
Angle \(AOB = 60^\circ\)
The perimeter of the sector is \(P\) cm.
Find a formula for \(P\) in terms of \(r\)
Give your answer in the form \(\;P = r(c\pi + k)\;\) where \(c\) and \(k\) are values to be found.
(3)
| Scheme | Marks |
|---|---|
\(\dfrac{60}{360} \times 2 \times \pi \times r\) oe or \(\dfrac{1}{6} \times 2 \times \pi \times r\) oe | M1 |
\(\text{``}{\dfrac{60}{360} \times 2 \times \pi \times r}\text{''} + 2r\) oe or \(\text{``}{\dfrac{1}{6} \times 2 \times \pi \times r}\text{''} + 2r\) oe | M1 |
Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: \(P = r\left(\dfrac{1}{3}\pi + 2\right)\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for finding the length of the arc
M1: dep on M1 for a complete expression from correct working for a method for the perimeter
A1: oe
eg \(P = r(0.33...\pi + 2)\) or
\(P = \left(\dfrac{1}{3}\pi + 2\right)r\) or
\(P = \left(2 + \dfrac{120}{360}\pi\right)r\) or
\(P = \left(\dfrac{120}{360}\pi + 2\right)r\)