I was to find the blue area in this question :
As described about how it's a square with 4 quadrants of same radius intertwined with each other, now to find the blue part area I thought about finding any possible relation in between the arcs TA, AB and BP so I proceeded further with this diagram :
It's quite clear that arcs TA and BP are equal for they subtend equal angles($45°$) at the centre point of blue part(point I), but I wasn't able to link the arc AB with former 2 arcs in any way, so I dumped that approach.
Now I thought if I could link something with arcs BC and BP, so I tried to find any relation between line segments BC and BP or more accurately between the triangles BIC and BCP, but there too I met with a dead end.
I think I am faltering in finding any links or let's say I have been devoid of any fresh approaches towards this problem, so I came up for help here.
Let the side of the square be 1. Then $AP=PS=SA=1$, so $APS$ is equilateral. Hence $\angle APS=60^o$. Similarly $\angle QPD=60^o$, so $\angle SPD=90^o-60^o=30^o$, and hence $\angle APD=60^o-30^o=30^o$. So $APD$ is an isosceles triangle with $AP=DP=1$ and $\angle APD=30^o$. Hence $\angle PAD=\angle PDA=75^o$. So $AD=2\cos75^o$.
The area of the square side $AD$ is $AD^2$. We need to add to that four times the green shaded area. That is the difference between the sector $APD$ and the triangle $APD$. The sector is one-twelfth of a circle radius 1, so has area $\frac{\pi}{12}$. The triangle has area $\frac{1}{2}AP\cdot PD\sin30^o=\frac{1}{4}$.
So the required area is $4\cos^275^o+\frac{\pi}{3}-1$. We have $-\frac{\sqrt3}{2}=\cos150^o=2\cos^275^o-1$, so $\cos^275^o=\frac{1}{2}(1-\frac{\sqrt3}{2})$ and hence the required area is $(2-\sqrt3)+\frac{\pi}{3}-1=\frac{\pi}{3}+1-\sqrt3$.
Obviously if the side of the square is $a$ rather than 1 we multiply the result by $a^2$.