5. As shown in Figure 2, on the two legs $AC$ and $BC$ of the right triangle $\triangle ABC$, squares $ACDE$ and $CBFG$ are constructed, respectively, and line segment $DG$ is connected. The midpoints of line segments $AB$, $BF$, $FG$, $GD$, $DE$, and $EA$ are $P$, $L$, $K$, $I$, $H$, and $Q$, respectively. If $AC=14$, $BC=28$, then the area of hexagon $HIKLQP$ =.
5. 1004.5.

As shown in Figure 6, connect $D F$, $F A$, $A D$, $E G$, $G B$, and $B E$.
$$
\begin{array}{l}
\text { Then } S_{\text {hexagon } A F F G E}=1372, \\
S_{\triangle B I P}=S_{\triangle F K L}=S_{\triangle G K I}=\frac{1}{8} \times 28^{2}=98, \\
S_{\triangle D I H}=S_{\triangle E H Q}=S_{\triangle A Q P}=\frac{1}{8} \times 14^{2}=24.5 .
\end{array}
$$

Therefore, $S_{\text {hexagon HIKLPQ }}$
$$
\begin{aligned}
= & S_{\text {hexagon } A B F G D E}-S_{\triangle B L P}-S_{\triangle F K L}-S_{\triangle G K I}- \\
& S_{\triangle D I H}-S_{\triangle E H Q}-S_{\triangle A Q P} \\
= & 1372-3 \times 98-3 \times 24.5=1004.5 .
\end{aligned}
$$