Example 9 Given $a, b>0$. Find the area of the triangle with side lengths
$$
\sqrt{a^{2}+b^{2}}, \sqrt{a^{2}+4 b^{2}}, \sqrt{4 a^{2}+b^{2}}
$$
Solve as shown in Figure 9, construct rectangle $ABCD$, such that $AB$
$$
=CD=2b, AD=BC
$$
$=2a$, and $E, F$ are the midpoints of
$CD, AD$ respectively. In this case, in $\triangle DEF$,
$$
\begin{array}{l}
EF=\sqrt{a^{2}+b^{2}}, BF=\sqrt{a^{2}+4b^{2}}, \\
BE=\sqrt{4a^{2}+b^{2}} . \\
\text{Therefore, } S_{\triangle DEF}=S_{\text{rectangle } ABCD}-S_{\triangle ABF}-S_{\triangle BCE}-S_{\triangle DEF} \\
=4ab-\frac{1}{2}ab-ab-ab=\frac{3}{2}ab .
\end{array}
$$