Area of a quatrefoil inside a circle What is the maximum area of a quatrefoil that is inscribed in a circle of radius 6?
My first guess was to cut  the circle into small regions but that doesn't seem to work. The solution does not have to be simple.
Thank you for your time. 

 A: 
If it's a quatrefoil, then four circles have the same diameter and to be inscribed in the large circle, their diameter have to be equal to the radius of the large circle.
Here, we see the symmetrical shape of four partially overlapping circles at the point $A, B, C$ and $D$ respectively. Connect those points and you will get a square having side equal to the diameter of small circle and that is obviously equal to 6. Let denote the area of the square $ABCD = [ABCD]$.
So, the area of the quatrefoil = $[ABCD]$ + The area of 4 semi circles = $6^2 + 4*\frac{1}{2}*\pi3^2$ = $36 + 18\pi \approx 92.549$ 
And that's your answer.
A: If I understand the intended shape correctly, its outline is formed by four semicircles of radius $3$ whose centers form a square with a diagonal of $6$ units.  They intersect at their endpoints, which form a larger square with a side of $6$ units.  The total area inside the quatrefoil is then the area of the square plus the area inside the four semicircles: $6^2 + 4(\frac12\pi\cdot 3^2) =  36 + 18\pi$.
