jmc

We know that $OA$ and $OB$ are each radii of the semi-circle with center $O$. Thus, $OA=OB=OC+CB=32+36=68$. Therefore, $AC=AO+OC=68+32=100$.

The semi-circle with center $K$ has radius $AK=\frac{1}{2}(AC)=\frac{1}{2}(100)=50$. Thus, this semi-circle has an area equal to $\frac{1}{2}\pi (AK{)}^2=\frac{1}{2}\pi (50{)}^2=\boxed{1250\pi }$.