A square whose sides measure $x$ in length will have a diagonal of length $\sqrt{2}x$, which means $\overline{AC}=20\sqrt{2}\text{ cm}$, and this agrees with what you found when you simply.
Okay, so we now know the length of $\overline{AC}$. So next, we can subtract from this the altitude of $\triangle AMN$ where $\overline{MN}$ is the base. This will be half the distance of the diagonal of a square having side lengths of $10\text{ cm}$. Or, we can simply observe this is 1/4 of the length of $\overline{AC}$. So, we are left with 3/4 of the diagonal $\overline{AC}$ which we must cut in half to get the altitude of $\triangle {CPQ}$ where $\overline{PQ}$ is the base. What do you find?