pull down to refresh

It might help to assemble four quarter circles into a full one:
Let's say the doubly shaded right triangles have shorter sides x \gt y.
The outer square's side length is 2x+12, area 4x^2+48x+144.
We have to subtract the four white rectangles with area 1+x each.
Further add the doubly shaded triangle areas 2xy (they count twice).
Then the answer is obviously a fourth of that.
That's how far I got, but I couldn't get x and y.
I didn't proceed that way, but I imagine this may work too. I'll post the answer tomorrow.
reply