3. Given, as shown in the figure, a semicircle $O$ with a diameter of $20 \mathrm{~cm}$ has two points $P$ and $Q$, $P C \perp A B$ at $C, Q D$ $\perp A B$ at $D, Q E \perp$ $O P$ at $E, A C=4 \mathrm{~cm}$. Then $D E=$
$3.8 \mathrm{~cm}$.
Take the midpoint $M$ of $O P$ and the midpoint $N$ of $O Q$, and connect $C M, D N$, and $E N$. Then
$$
\begin{array}{c}
M C=P E=\frac{1}{2} O P \\
=\frac{1}{2} O Q=E N=D N, \\
\angle P M C=\angle M C O+ \\
\angle M O C=2 \angle M O C .
\end{array}
$$
Since $O, D, Q, E$ are concyclic, with $N$ as the center,
$$
\begin{array}{l}
\therefore \angle D N E=2 \angle D Q E=2 \angle M O C=\angle P M C . \\
\therefore \triangle P C M \cong \triangle D E N .
\end{array}
$$

Thus $D E=P C=\sqrt{O P^{2}-O C^{2}}$
$$
=\sqrt{10^{2}-(10-4)^{2}}=8 .
$$