Find an expression for the shaded area in terms of $\alpha$, $\theta$ and $r$.

Show that $\alpha  = 4\arcsin \frac{1}{4}$.

Hence find the value of $r$ given that the shaded area is equal to 4.

$A = 2(\alpha  - \sin \alpha ){r^2} + \frac{1}{2}(\theta  - \sin \theta ){r^2}$    M1A1A1

Note: Award M1A1A1 for alternative correct expressions eg. $A = 4\left( {\frac{\alpha }{2} - \sin \frac{\alpha }{2}} \right){r^2} + \frac{1}{2}\theta {r^2}$.

[3 marks]

METHOD 1

consider for example triangle ADM where M is the midpoint of BD     M1

$\sin \frac{\alpha }{4} = \frac{1}{4}$    A1

$\frac{\alpha }{4} = \arcsin \frac{1}{4}$

$\alpha  = 4\arcsin \frac{1}{4}$    AG

METHOD 2

attempting to use the cosine rule (to obtain $1 - \cos \frac{\alpha }{2} = \frac{1}{8}$)     M1

$\sin \frac{\alpha }{4} = \frac{1}{4}$ (obtained from $\sin \frac{\alpha }{4} = \sqrt {\frac{{1 - \cos \frac{\alpha }{2}}}{2}}$)     A1

$\frac{\alpha }{4} = \arcsin \frac{1}{4}$

$\alpha  = 4\arcsin \frac{1}{4}$    AG

METHOD 3

$\sin \left( {\frac{\pi }{2} - \frac{\alpha }{4}} \right) = 2\sin \frac{\alpha }{2}$ where $\frac{\theta }{2} = \frac{\pi }{2} - \frac{\alpha }{4}$

$\cos \frac{\alpha }{4} = 4\sin \frac{\alpha }{4}\cos \frac{\alpha }{4}$    M1

Note: Award M1 either for use of the double angle formula or the conversion from sine to cosine.

$\frac{1}{4} = \sin \frac{\alpha }{4}$    A1

$\frac{\alpha }{4} = \arcsin \frac{1}{4}$

$\alpha  = 4\arcsin \frac{1}{4}$    AG

[2 marks]

(from triangle ADM), $\theta  = \pi  - \frac{\alpha }{2}{\text{ }}\left( { = \pi  - 2\arcsin \frac{1}{4} = 2\arcsin \frac{1}{4} = 2.6362 \ldots } \right)$     A1

attempting to solve $2(\alpha  - \sin \alpha ){r^2} + \frac{1}{2}(\theta  - \sin \theta ){r^2} = 4$

with $\alpha  = 4\arcsin \frac{1}{4}$ and $\theta  = \pi  - \frac{\alpha }{2}{\text{ }}\left( { = 2\arccos \frac{1}{4}} \right)$ for $r$     (M1)

$r = 1.69$    A1

[3 marks]