June 2025 Paper 1 Q4
4 The diagram shows part of a circle with centre O and radius 5 cm. The circle passes through the points A and B. The length AB is 5 cm.
Calculate the area of the shaded region. [4]
| Scheme | Marks | AO |
|---|---|---|
| (Triangle OAB with OB = 5 cm and AB = 5 cm) Angle AOB \(= 60^\circ\) or \(\frac{\pi}{3}\) (equilateral triangle) | B1 | 1.1a |
| Angle in major sector is \(2\pi - \frac{\pi}{3}\ \left(=\frac{5\pi}{3}\right)\) | M1 | 3.1a |
| Area of major sector is \(\frac{1}{2}r^2\theta = \frac{1}{2}\times 5^2 \times \frac{5\pi}{3}\) | M1 | 1.1a |
| \(=\dfrac{125\pi}{6}\) | A1 | 1.1 |
| [4] |
Notes
B1: May be implied by \(\frac{1}{6}\) or \(\frac{5}{6}\) seen
M1: Uses their AOB to find the angle in the major sector soi
M1: Uses sector formula with their \(\theta\) soi
A1: Accept awrt 65.4 or 65.5
Alternative method for last 3 marks
| Scheme | Marks |
|---|---|
| Area of minor sector is \(\frac{1}{2}r^2\theta = \frac{1}{2}\times 5^2\times\frac{\pi}{3} = \frac{25\pi}{6}\) | M1 |
| Area of major sector is \(25\pi - \frac{25\pi}{6}\) | M1 |
| \(=\dfrac{125\pi}{6}\) | A1 |
M1: Uses their AOB to find the area of the minor sector soi
M1: Subtracts from \(25\pi\) soi
Do not allow for the difference between the areas of the circle and the triangle
A1: Accept awrt 65.4 or 65.5