I am trying to find the area in the overlap of two quarter-circles that are in a $10 \times 10$ square, as somewhat-crudely drawn below:
In this problem I was given a square of side length $10$, with two quarter circles going through the center of the square, meeting at the corners of the square.
So, I found the diagonal of the square, $10 \sqrt 2$, and I reasoned that this is the distance from one end of the quarter circle to the other.
I realized this formed a $45^\circ-45^\circ-90^\circ$ triangle when placed inside of a circle, and two circles' radii compose the legs of the triangle and $10 \sqrt 2$ is the length of the hypotenuse. Thus, the radius of each circle had to be $10.$
Thus, the area of each circle (as a whole) would be given by $\pi 10^2 = 100\pi$, and dividing by $4$ gave the area of each quarter circle to be $25\pi$ (totalling to $50 \pi$ for both).
This simply isn't possible, the area of two quarter circle cannot be $50 \pi$ since it wouldn't fit inside of the square, which has an area of $100$.
Therefore this problem must be impossible, but apparently it's solvable, please show me how to solve it.