Slightly inflated square

$A_{AFECDA} = \frac{\pi}{4}(4^{2}) = 4\pi$

$A_{AEGDA} = \frac{\pi}{6}(4^{2}) = \frac{8\pi}{3}$

$A_{AFEA} = \frac{\pi}{6}(4^{2}) - \frac{(4^{2})\sin{(60)}}{2} = \frac{8\pi}{3} - 4\sqrt{3}$

$A_{DGECD} = A_{AFECDA} - A_{AEGDA} - A_{\overline{AFEA}}$

$A_{DGECD} = 4\pi - \frac{8\pi}{3} - (\frac{8\pi}{3} - 4\sqrt{3}) = \frac{-4\pi}{3} + 4\sqrt{3}$

$A_{EFGHE} = A_{square} - 4A_{DGECD}$

$A_{EFGHE} = 4^{2} - 4( \frac{-4\pi}{3} + 4\sqrt{3})$

$A_{EFGHE} = \frac{16\pi}{3} + 16 - 16\sqrt{3}$

$therefore:$

$a=16 ; b=3; c=16; d=16; e=3$

$a+b+c+d+e = 16+3+16+16+3 = \boxed{54}$

We can calculate the area given by the intersection of the four quadrants as the sum of : $4Area_{DFC}+2Area_{DB}-Area_{ABCD}=Area_{EFGH}$

Consider the notation as the area enclosed by the lines (No matter if they are curved) that connect the given points corresponding the image above. We call the side of the square $r$ . Then:

$Area_{DB}=w\Rightarrow \dfrac{\pi r^2}{4}\times 2-w= r^2\Rightarrow Area_{DB}=\dfrac{\pi r^2}{2}-r^2$

In order to calculate the area of $DFC$ we note that triangle $\Delta_{ABF}$ is equilateral . So its area is :

$\dfrac{r \sqrt{3}}{2}\times\dfrac{1}{2}\times r=\dfrac{r^2 \sqrt{3}}{4}$

The sum of the area of two circular sectors of $60^{\circ}$ is: $\dfrac{\pi r^2}{3}$

The sum of the area of two quadrants is: $\dfrac{\pi r^2}{2}$

Then the area of $DFC$ is:

$r^2-\left (\dfrac{\pi r^2}{2}-\left (\dfrac{\pi r^2}{3}-\dfrac{r^2 \sqrt{3}}{4}\right )\right )=r^2-\dfrac{\pi r^2}{6}-\dfrac{r^2 \sqrt{3}}{4}$

Finally, the generalized form of the area in common of the $4$ quadrants is:

$Area_{EFGH}= x\Rightarrow x= 2\left (\dfrac{\pi r^2}{2}-r^2\right )+4\left (r^2-\dfrac{\pi r^2}{6}-\dfrac{r^2 \sqrt{3}}{4}\right)-r^2$

$=\boxed{\dfrac{\pi r^2}{3}+r^2-r^2\sqrt{3}}$

Substituting r, we get:

$\dfrac{16 \pi}{3}+16-16\sqrt{3}$

So the answer is:

$16+3+16+16+3=\boxed{\boxed{54}}$

On 6th May 2016 I had answered almost the same equation but square side of 1 instead of 4 as it is here "Blue area by Alan Enrique Ontiveros ".
The above is the copy of that solution. Since the square here has side 4, the required area is:-
$4^2*\dfrac{\pi+3-3\sqrt3} 3=\frac{16 } 3+16-16\sqrt3=\frac{a } b*\pi+c-d\sqrt3.$
So a+b+c+d+e=16+3+16+16+3=54.
This is the fifth time almost the same equation has appeared. Bob Kadylo below too has same thing to say.

At the end of January 2016, I did a problem called "Area Of Overlapping Leaves" - Geometry Level 3 - which currently awards 60 points. Lots of detailed beautiful colored solutions I learned from - that were different from the way I did it - but I love that. Anyway, here is the result: $\frac{a^{2}}{3}(\pi+3-3\sqrt{3})$

I just applied that general result to this question using "a=4" and multiplied $\frac{16}{3}$ into the brackets.

Yes 54: Using the notation in Keil Cerito's diagram: Take A as the origin, G has co-ordinates (√12,2). Then 4 x (Int - 2(√12-2)) where Int is the area under the curve x^2+y^2 = 4^2 from 2 to √12! It works! (Also interesting that the value of Int is 1/3 of the area of the quarter circle.)

Set the origin of a system of coordinates at the center of the square. Using the circumferences' equations centered at the vertices $V_1=(2, 2); V_2=(2, -2); V_3=(-2, -2) \text{ and } V_4=(-2, 2)$ one may find the intersection between the circumferences of $V_1$ and $V_2$ and integrate the function of $V_2$ over the interval $[2-\sqrt{12},0]$ (i. e. from the intersection to the origin). This is a quarter of the shaded region, thus we multiply the value achieved by four and hence the answer.