A triangle is drawn by joining a point on the semi-circle to the ends of the diameter. Two semi-circles are drawn with the other two sides as diameter.

Find the sum of the areas of shaded part.

Let AB=a, AC=b

$\therefore$ BC = $\sqrt{a^2+b^2}$ ($\Delta$ ABC is a right triangle)

$\Delta BAC={90}^{\circ }$ (angle in semi-circle)

Now, area of $\Delta$ ABC = $\frac{1}{2}ab$

Now, area of semi-circle as diameter AB = $\frac{1}{2}\pi {\left(\frac{a}{2}\right)}^2$

and , area of semi-circle as diameter AC = $\frac{1}{2}\pi {\left(\frac{b}{2}\right)}^2$

area of semi-circle as diameter BC =$\frac{1}{2}\pi {\left[\frac{\sqrt{a^2+b^2}}{2}\right]}^2$

Now, area of shaded part

= $[\frac{1}{2}\pi \frac{a^2}{4}+\frac{1}{2}\pi \frac{b^2}{4}]-[\frac{1}{2}\pi \left(\frac{b^2+a^2}{4}\right)-\frac{1}{2}ab]$

=$\frac{\pi }{8}(a^2+b^2)-\frac{\pi }{8}(a^2+b^2)+\frac{1}{2}ab$

= $\frac{1}{2}ab$ = $\frac{1}{2}\left(AB\right)\left(BC\right)$