Find the sum of the areas of regions $X$ and $Y$ Right triangle $ABC$ is inscribed in a circle with $AC = 6$, $BC = 8$ and $AB=10$. $AC$ and $CB$ are semi-circles. Find the sum of the areas of regions $X$ and $Y$.

This is not so obvious to me. I started off with the formula for area which is A=$\pi r^2$ and since each semi-circle, we are going to have divide A by $2$. However, how would I find $r^2$ or the radius?
Any ideas would help.

The inscribed triangle is right-angled, therefore its hypotenuse $AB$ is a diameter (by the converse to Thales' Theorem) and the large circle $C$ has radius $5$.
Let the white region between $X$ and the triangle be $X'$, and the white region between $Y$ and the triangle be $Y'$.
Then $$A(X') + A(Y') = \frac{1}{2}A(C) - A(triangle) = \frac{1}{2}(5^2\pi)-\frac{1}{2}(6\times8) = \frac{25}{2}\pi-24 $$
Finally the smaller semi-circles (that is $X\cup X'$ and $Y\cup Y'$) have area $ \frac{1}{2}(3^2\pi) $ and $ \frac{1}{2}(4^2\pi) $ respectively, so we obtain:
$$ A(X) + A(Y) = \frac{1}{2}(3^2\pi) + \frac{1}{2}(4^2\pi) - (A(X') + A(Y')) = \frac{9}{2}\pi + 8\pi - (\frac{25}{2}\pi-24) = 24 $$
