jmc

If we let $E$ be the midpoint of line segment $AB$ and $F$ be the midpoint of $CD$, then line segment $MF$ will pass through point $E$. Also, $MF$ is perpendicular to $CD$, so $△MFC$ is a right triangle. Now, if we can find the lengths of $MF$ and $FC$, we can use the Pythagorean Theorem to find the length of $MC$.



Since $F$ is the midpoint of $CD$ and $CD$ has length $6$, $FC$ has length $3$. $EF$ has length $6$, because it has the same length as the side length of the square. $ME$ is the radius of the semicircle. Since the diameter of the semicircle is $6$ (the same as the side length of the square), $ME$ has length $3$. Now, $MF=ME+EF=3+6=9$. Finally, from the Pythagorean Theorem, we have that $M{C}^2=M{F}^2+F{C}^2=9^2+3^2=90$, so $MC=\sqrt{90}=\boxed{3\sqrt{10}}$ cm.