Area of portion of circle inside a square. Consider a square grazing field with each side of length 8 metres. There is a pillar at the
centre of the field (i.e. at the intersection of the two diagonals). A cow is tied to the pillar
using a rope of length $8\over\sqrt3$ meters. Find the area of the part of the field that the cow is allowed
to graze.


First let's calculate the central angle of each segment in the circle which is cut off by the square.

Let $\angle BAC=\theta$. It is given that $AC=4$, and $AB={8 \over \sqrt3}$.
From the figure it is easy to deduce that in $\Delta ABC$, $\cos\theta={\sqrt3 \over 2}$,
which means that $\theta=30^\circ$ and hence the central angle $\angle BAG$ is equal to $60^\circ$.
The area of a segment in a circle of radius $R$, of central angle $\alpha$ is $$area={R^2 \over 2}{\left( {\pi\alpha \over 180^\circ}-\sin\alpha \right)}$$
So now all you have to do is calculate the area cut off by the four segments together, and then subtract it from the total area of the circle.
