jmc

The shape created is shown below: We can decompose this area into four circular sectors, four small triangles, and four large triangles, as shown: Points $A,$ $B,$ and $O$ are marked above for convenience. Because the square was rotated $45^{\circ },$ each circular sector (shown in gray) has a central angle of $45^{\circ }$ and a radius of $AO=\frac{\sqrt{2}}{2}.$ Therefore, put together, they form a semicircle of radius $\frac{\sqrt{2}}{2},$ which has area $$\frac{1}{2}\pi {\left(\frac{\sqrt{2}}{2}\right)}^2=\frac{\pi }{4}.$$The four larger triangles (shown in blue) have area equal to half the area of the original square, so they contribute $\frac{1}{2}$ to the overall area. Finally, each of the smaller triangles (shown unshaded) has legs of length $AB=AO-BO=\frac{\sqrt{2}}{2}-\frac{1}{2},$ so their total area is $$4\cdot \frac{1}{2}{\left(\frac{\sqrt{2}}{2}-\frac{1}{2}\right)}^2=\frac{3-2\sqrt{2}}{2}.$$Thus, the area of the entire given region is $$\frac{\pi }{4}+\frac{1}{2}+\frac{3-2\sqrt{2}}{2}=\boxed{\frac{\pi }{4}+2-\sqrt{2}}.$$