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.

 A: 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.
A: 
$$|OA|=\frac{8}{\sqrt{3}}\\
|OC|=4\\
|AC|=\frac{16}{\sqrt{3}}\\
\cos \angle AOC = |OC|/|OA| = \frac{\sqrt{3}}{2} = \cos 30° ⇒ \angle AOC = 30°$$
So, we arrive at the angle subtended by sector is also 30 degrees (there is another triangle like AOC (not shown in figure) with apex angle 30 degrees ⇒ angle subtended by sector $= 90° - (30°+30°) = 30°$)
$$\text{Area} = 2*\text{area of triangle $OAC$} + \text{Area of sector $OAB$}\\
\text{Area of sector $OAB$} = 30°/360° (π⋅\frac{64}{3}) = π⋅\frac{16}{9}\\
\text{Area of triangle $OAC$} = |AC|⋅|OC|/2 = \frac{32}{\sqrt{3}}$$
Final answer: $\frac{64}{\sqrt{3}} + π⋅\frac{16}{9}$
