Imagine that the southwest corner of the square is $(0, 0)$, the square has side $r$, and we coordinatize in the usual way with the $x$-axis along the bottom side of the square and the $y$-axis along its left side.
The area $Area(z)$ can obviously be broken up into four congruent areas, one in each quadrant of the square. Consider the northeastern one. This is that part of the circle of radius $r$ which has $x > r/2, y > r/2$. Call its area $A$; then $Area(z) = 4A$.
This is just a sector of a circle (centered at 0), with two triangles removed. Call the area of the sector $S$ and the area of each triangle $T$; then $z = 4(S-2T) = 4S - 8T$. Now we just need to find $S$ and $T$.
$S$ is a thirty-degree sector of the circle, as you can see because the classical construction of an equilateral triangle is hidden in your figure many times over. So $S = \pi r^2/12$.
One example of $T$ is a triangle with vertices at $(0, 0), (r/2, r/2), (r \sqrt{3}/2, r/2)$; it has base $r(\sqrt{3}-1)/2$ and height $r/2$, so its area is $T = r^2 \times (\sqrt{3}-1)/8$.
Putting it all together we get
$$z = 4S - 8T = 4 \times {\pi r^2 \over 12} - 8 \times {r^2 {\sqrt{3}-1 \over 8}} = r^2 \left( {\pi \over 3} + 1 - \sqrt{3} \right)$$
Although I've expressed this in terms of coordinate geometry there's nothing about the solution that's essentially coordinate-based; I've just chosen to do it this way rather than to draw lots of pictures.
