Circles are drawn with four vertices as the center and radius equal to

Circles are drawn with four vertices as the center and radius equal to the side of the square. If the square is formed by joining the mid points of another square of side \(2*\sqrt{6})\), what is the area common to all the four circles?


A. \(\frac{\sqrt{3} -1}{2}*6\pi\)

B. \(4\pi -3*\sqrt{3}\)

C. \(\frac{\pi - 3*\sqrt{3}}{2}\)

D. \(4\pi-12(\sqrt{3}-1)\)

E. \(4\pi-8(\sqrt{3}-1)\)

Please, refer to the attached drawings



From the first figure, we can deduce that the side of the smaller square is half of the diagonal of the larger square, so it is equal to \(\frac{1}{2}\cdot{2}\sqrt{6}\cdot{sqrt{2}=2\sqrt{3}}\).

We have to calculate 4 times the area of the shaded region in the first figure.
Now, please refer to the second drawing.
Since \(OA=2\sqrt{3}\) is twice \(AB=\sqrt{3}\), triangle \(OAB\) is a 30-60-90 triangle. We can see that the quarter of the circle with radius \(OA\) is divided into three equal sectors with a central angle of 30 degrees. The area of each such sector is\(\frac{\pi(2\sqrt{3})^2}{4\cdot{3}}=\pi\).

To find the area of the shaded region, we should subtract from the area of the 30 degrees sector twice the area of the triangle \(ACO\).
\(OB=\sqrt{3}AB=\sqrt{3}\sqrt{3}=3\), therefore \(x=3-\sqrt{3}\).
The area of the shaded region is \(\pi-2\cdot\frac{x\sqrt{3}}{2}=\pi-(3-\sqrt{3})\sqrt{3}=\pi-(3\sqrt{3}-3)=\pi-3(\sqrt{3}-1)\).
Thus the requested area is \(4(\pi-3(\sqrt{3}-1))=4\pi-12(\sqrt{3}-1)\).

Answer D.