Higher November 2020 Paper 2 Q26
26. The diagram shows a sector of a circle with radius \(r\) cm and angle \(x^\circ\).
Diagram not drawn to scale
The arc length of the sector is \(5\pi\) cm.
(a) Show that \(x=\dfrac{900}{r}\). [2]
(b) The area of the sector is \(30\pi\) cm2.
Calculate the value of \(x\). [4]
Calculate the value of \(x\). [4]
| Scheme | Marks | Comments |
|---|---|---|
| \(\dfrac{x}{360}\times2\pi r=5\pi\) | M1 | |
| \(x=\dfrac{900}{r}\) from clear correct working | A1 |
Alternative method
| Scheme | Marks | Comments |
|---|---|---|
| \(\dfrac{x}{360}=\dfrac{5}{2r}\) | M1 | |
| \(x=\dfrac{900}{r}\) from clear correct working | A1 |
| Scheme | Marks | Comments |
|---|---|---|
| \(\dfrac{x}{360}\times\pi r^2=30\pi\) | M1 | |
| \(\dfrac{\left(\frac{900}{r}\right)}{360}\times\pi\times r^2=30\pi\) | m1 | |
| (r =) 12 | A1 | |
| (x =) 75 | A1 | FT ‘their derived 12’ provided M1 previously awarded |
Alternative method
| Scheme | Marks | Comments |
|---|---|---|
| \(\dfrac{x}{360}\times\pi r^2=30\pi\) | M1 | |
| \(xr=900\) and \(xr^2=10800\) oe | m1 | |
| (r =) 12 | A1 | |
| (x =) 75 | A1 | FT ‘their derived 12’ provided M1 previously awarded |
Alternative method
| Scheme | Marks | Comments |
|---|---|---|
| \(\dfrac{x}{360}=\dfrac{30}{r^2}\) oe | M1 | |
| \(\dfrac{5}{2r}=\dfrac{30}{r^2}\) | m1 | |
| (r =) 12 | A1 | |
| (x =) 75 | A1 | FT ‘their derived 12’ provided M1 previously awarded |
Alternative method
| Scheme | Marks | Comments |
|---|---|---|
| \(\dfrac{x}{360}=\dfrac{30}{r^2}\) oe | M1 | |
| \(\dfrac{x}{360}=\dfrac{30}{\left(\frac{900}{x}\right)^2}\) | m1 | |
| (x =) 75 | A2 |
| (6) |