Foundation January 2023 Paper 1R Q25
25 \(R\) and \(T\) are points on a circle, centre \(O\)
Diagram NOT accurately drawn
\(RT = 12\) cm
\(M\) is the midpoint of \(RT\)
Angle \(ROM = 52^\circ\)
Work out the area of the circle.
Give your answer correct to 3 significant figures.
(4)
| Scheme | Marks |
|---|---|
\(\sin 52 = \dfrac{12 \div 2}{r}\) oe or \(\dfrac{r}{\sin 90} = \dfrac{6}{\sin 52}\) oe or \(\cos(90 - 52) = \dfrac{12 \div 2}{r}\) oe or \((r^2 =)(12 \div 2)^2 + \left(\dfrac{12 \div 2}{\tan 52}\right)^2\) oe \(\left[r^2 = 6^2 + 4.687...^2\right]\) or \(\dfrac{r}{\sin 38} = \dfrac{12}{\sin 104}\) oe | M1 |
\(r = \dfrac{6}{\sin 52}\;(= 7.614)\) oe or \(r = \dfrac{6}{\cos 38}\) oe or \((r =)\sqrt{(12 \div 2)^2 + \left(\dfrac{12 \div 2}{\tan 52}\right)^2}\) \(\left[r = \sqrt{6^2 + 4.687...^2}\right]\) oe or \(\dfrac{12\sin 38}{\sin 104}\) oe | M1 |
| (Area =) \(\pi \times (\text{“}7.61...\text{”})^2\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 182 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: A correct trig statement for the radius
use of tan must also include a correct Pythagoras statement.
M1: A correct method to find the radius of the circle
use of tan must also use Pythagoras to find an expression for \(r\)
M1: the radius must come from a completely correct method
A1: Accept 181 - 183