This is similar to a number of such quesiton, but the overlapping areas are a bit tricky. Ive a attached a diagram which i think is correct (may not be) overlapping bits
The file is also available here
The answer should preferably be in a trigonometric form, and pi is to be left unsubstituted.
(edit: the rope is attached to a corner of the platform) (edit 2:I tried dividing the overlapping area into smaller triangles, but the answer is unsatisfactory [268.75pi - 14.25m^2]
Here is a graphical representation evidencing the area where the cow can graze (for some of them that can be attained in two ways)
This area is the sum of
the fraction $3/4$ of the area $\pi 15^2$ of the disk with center $D$ and radius $25$
plus twice the fraction $1/4$ of the area of the disk with center $A$ and radius $15$
minus twice the area $\mathcal{A}$ of "triangle" $CGF$ (made of two line segments and a circular arc) because of double counting.
minus (of course) the area of the square ($100$).
$\mathcal{A}$ is equal to the difference between the area of circular sector $AGF$ and the area of ordinary triangle $ACF$. For this, we need coordinates of
$$C=(0,5 \sqrt{2}), F=(0,5 \sqrt{7}), G=(-\tfrac52 \sqrt{2},-\tfrac{15}{2} \sqrt{2})$$
Up to you for the final round...