# How to find the area shared by 4 quadrants inside a square?

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.

• You just need to calculate the distance AB. That gives you the area of the square ABCD. Then you have to add on the four extra bits. Each of those is easy because it is the difference of a sector (like PAD) and a triangle. – almagest Jun 14 '16 at 6:31
• you can try to use analytic geometry to solve this problem – clark Jun 14 '16 at 6:32
• Look at the triangle formed by A and the two blue points. – Empy2 Jun 14 '16 at 6:32
• @clark I did try but, the debacle there is how do I get the coordinates of A and C ? – Arnav Das Jun 14 '16 at 6:39
• @almagest for the distance of AB I need, distances of AI and IB, which I described above, how I could not find them, and also we are in dark about what could be the angles of triangle IAB, so I think we don't have any of the tools to calculate the distance of AB. – Arnav Das Jun 14 '16 at 6:45

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 $a^2$. 