4. As shown in Figure 2, in

Rt $\triangle A B C$,
$\angle C=90^{\circ}$, point $M$
is the intersection of the three
medians of the triangle. Perpendiculars are drawn from $M$
to $A B$, $B C$, and $A C$,
with the feet of the perpendiculars being $D$, $E$, and $F$, respectively. If $A C=3$, $B C=12$, then the area of $\triangle D E F$ is $\qquad$
4.4.
$$
\begin{array}{l}
AB = \sqrt{3^2 + 12^2} = \sqrt{153} = 3 \sqrt{17}, \\
S_{\triangle ABC} = \frac{1}{2} \times 3 \times 12 = 18.
\end{array}
$$

Draw a perpendicular from $C$ to $AB$, with the foot of the perpendicular at $H$, and let it intersect $EF$ at $G$. We have
$$
\begin{array}{l}
S_{\triangle ABC} = \frac{1}{2} \cdot AB \cdot CH = 18, \\
CH = \frac{18 \times 2}{3 \sqrt{17}} = \frac{12}{\sqrt{17}}. \\
\because ME = FC = \frac{1}{3} AC = 1, CE = \frac{1}{3} BC = 4. \\
\therefore EF = \sqrt{CF^2 + CE^2} = \sqrt{1 + 4^2} = \sqrt{17}. \\
\because EF \cdot CG = CE \cdot CF,
\end{array}
$$

i.e., $CG = \frac{CE \cdot CF}{EF} = \frac{4 \times 1}{\sqrt{17}} = \frac{4}{\sqrt{17}}$.
$$
\begin{array}{l}
\therefore GH = CH - CG = \frac{12}{\sqrt{17}} - \frac{4}{\sqrt{17}} = \frac{8}{\sqrt{17}}. \\
\text{Therefore, } S_{\triangle DEF} = \frac{1}{2} \cdot EF \cdot GH = \frac{1}{2} \times \sqrt{17} \times \frac{8}{\sqrt{17}} = 4.
\end{array}
$$