So I have a friend who is a mathematician. I thought my answer was straight forward, but too long, so I asked him how to solve this problem. He gave me the answer before I had even finished talking. If you click on the solution left you can see my long winded answer and the graphical proof he used.
My answer uses trig to calculate the area of the bounded inner circle. That is fairly obvious. The trick here is to realize what you are calculating with the trig is a hypotinuse of 8 equal sized triangles that form the outer square, four of which form the inner square. As a result the inner square is half the area of the outer.