${\displaystyle {\begin{array}{ll}{\text{Circle diameter}}=1&\therefore {\text{radius,}}\ r={\tfrac {1}{2}}\\\triangle {\text{OAC}}:{\tfrac {y}{r}}=\sin \theta &\therefore y={\tfrac {\sin \theta }{2}}\\\triangle {\text{OAC}}:{\tfrac {x}{r}}=\cos \theta &\therefore x={\tfrac {\cos \theta }{2}}\\{\text{Area of circle}}&=\pi r^{2}=\pi ({\tfrac {1}{2}})^{2}={\tfrac {\pi }{4}}\\{\text{Area of }}\triangle {\text{OAC}}&={\tfrac {1}{2}}xy={\tfrac {\sin \theta \cos \theta }{8}}={\tfrac {2\sin \theta \cos \theta }{16}}={\tfrac {\sin 2\theta }{16}}\\{\text{Area of sector OAD}}&={\tfrac {\theta }{2\pi }}({\text{Area of circle}})={\tfrac {\theta }{2\pi }}{\tfrac {\pi }{4}}={\tfrac {\theta }{8}}\\{\text{Area of ABED}}&=4({\text{Area of OAD}}-{\text{Area of OAC}})\\&=4({\tfrac {\theta }{8}}-{\tfrac {\sin 2\theta }{16}})={\tfrac {2\theta -\sin 2\theta }{4}}\\f={\dfrac {\text{Area of ABED}}{\text{Area of circle}}}&={\dfrac {\tfrac {2\theta -\sin 2\theta }{4}}{\tfrac {\pi }{4}}}={\dfrac {2\theta -\sin 2\theta }{\pi }}\\{\text{where }}{\tfrac {1}{2}}+x+{\tfrac {1-t}{2}}=1&\therefore {\tfrac {\cos \theta }{2}}={\tfrac {t}{2}}\therefore \theta =\arccos t\end{array}}}$

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