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.
 A: I do not know how to make a figure here, but I can answer your question.
Let $O$ be the centre of the bigger circle, Then $AO=BO=5$.
Join OC (thereby completing $\triangle OBC$)
In $\triangle OBC$, we must have, $$\sin \angle {\frac{COB}{2}}=\frac{4}{5}$$
$$\implies \angle \frac{COB}{2}=\sin^{-1}\frac{4}{5}$$
So you can now figure out the area of the segment $BCY$
$$=\left(\frac{\angle {COB}}{2\pi}\pi (5)^2-12\right) \space \text {sq. units}$$
And then subtract it from the semicircle having diameter as $8$, to get the area of $Y$, Do the same for $X$
A: 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:
$$ A(X) + A(Y) = \frac{1}{2}(3^2\pi) + \frac{1}{2}(4^2\pi) - (A(X') + A(Y')) = \frac{9}{2}\pi + 8\pi - (\frac{25}{2}\pi-24) = 24 $$
A: First we know by the converse of Thales' theorem that AB is the diameter of the circle triangle ABC is incribed in. Now lets call $X'$ and $Y'$ the area of the entire semi-circles $x$ and $y$ are contained in respectively. We know that
$$X'=\pi\left(\frac{AC}{2}\right)^2=9\pi$$
$$Y'=\pi\left(\frac{BC}{2}\right)^2=16\pi$$
because we are given two line segments, BC and AC, that are their diameters. Next we just need to subtract out the overlap with the large white circle. We know since AB is the diameter of that circle, that overlap can be found by taking half the area of that circle and subtracting out the area of the triangle.
$$\text {area of circle}=A_1=\pi\left(\frac{AB}{2}\right)^2=25\pi$$
$$\text {area of triangle}=A_2= \frac{AC\cdot BC}{2}=24$$
$$\therefore$$
$$\text {overlap}=O=\frac{A_1}{2} - A_2= \frac{25}{2}\pi-24$$
Then we just add up $X'$ and $Y'$ and subtract the overlap.
$$x+y = X'+Y'-O=25\pi-\frac{25}{2}\pi+24$$
$$x+y=\frac{25}{2}\pi+24$$
A: Answer is area of triangle ABC, 24 square meters.  Because total area of two smaller Half circles equals big half circle area.  So x+ y= area of ABC
