jmc

Let $A,B$ be the areas of the semicircles on the legs of the right triangle, and let $C$ be the area of the semicircle on the hypotenuse of the right triangle. Then we see that by Pythagorean theorem $A+B=C$.

The area of the triangle plus the area of the two small semicircles is $$A+B+\frac{6\cdot 8}{2}=A+B+24.$$But this is also the area we are interested in, plus $C.$ Therefore, the answer is $A+B+24-C=\boxed{24}.$