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First, we need to see what this looks like. Below is a diagram. For this square with side length 1, the distance from center to vertex is $r=\frac{\sqrt{2}}{2}$, hence the area is composed of a semicircle of radius $r$, plus $4$ times a parallelogram (or a kite with diagonals of $(\sqrt{2}-1)$ and $r\text{ or}\frac{\sqrt{2}}{2}$) with height $\frac{1}{2}$ and base $\frac{\sqrt{2}}{2(1+\sqrt{2})}$. That is to say, the total area is $\frac{1}{2}\pi {\left(\frac{\sqrt{2}}{2}\right)}^2+4\frac{\sqrt{2}}{4(1+\sqrt{2})}=\boxed{2-\sqrt{2}+\frac{\pi }{4}}$. (To turn each dart-shaped piece into a parallelogram, cut along the dashed line and flip over one half.) Alternatively, you can move the dart-shaped piece to the other side and make a kite.