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]