8 semi-circles are drawn along the side lines inside of a unit square as shown in the following figure, with another circle centered at the center of the square and tangent to all of 8 semi-circles. What is the area of the square not covered by the circle and semi-circles (i.e. the green area)? Click here for the solution.
Lets draw line $\overline{GI}$, between the midpoint of a side of the square, and the center of the circle; $\overline{IJ}$, between the center of circle and that of a semi-circle on the same side of point $G$, thru the tangent point of the circle and the semi-circle; line $\overline{JK}$, between the center of the semi-circle and the intersect point $K$ with its adjacent semi-circle; and line $\overline{DK}$, between the corner of the square adjacent to point $G$ and $K$.
Obviously, the radius of the semi-circles is $$s=\dfrac{1}{4}$$ The area marked in blue color is $$A_{blue}=\dfrac{1}{4}\pi s^2+\dfrac{1}{2}s^2=\dfrac{2+\pi}{64}$$ Therefore, the total area of overlapping semi-circles is $$A_s=8\cdot A_{blue}=8\cdot\dfrac{2+\pi}{64}=\dfrac{2+\pi}{8}$$
Assume the radius of the circle centered at $I$ is $r$. Because $\triangle{GIJ}$ is a right triangle, we have: $$s^2+(\dfrac{1}{2})^2=(r+s)^2$$ i.e $$(\dfrac{1}{4})^2+(\dfrac{1}{2})^2=(\dfrac{1}{4}+s)^2$$
Solve the above equation and ignoring the negative $r$ value, we have: $$r=\dfrac{\sqrt{5}-1}{4}$$ Therefore, the area of the circle is $$A_{r}=\pi r^2=\pi(\dfrac{\sqrt{5}-1}{4})^2=\dfrac{3-\sqrt{5}}{8}\pi$$ And the area of not covered by the circles is $$A_{green}=1-A_{s}-A_{r}=1-\dfrac{2+\pi}{8}-\dfrac{3-\sqrt{5}}{8}\pi=\dfrac{3}{4}-\dfrac{1}{2}\pi+\dfrac{\sqrt{5}}{8}\pi \approx 0.0573$$
