6. Given in $\triangle A B C$, $\angle A, \angle B$ are acute angles, and $\sin A$ $=\frac{5}{13}, \tan B=2, A B=29 \mathrm{~cm}$. Then the area of $\triangle A B C$ is $\qquad$ $\mathrm{cm}^{2}$
6.145 .

Draw a perpendicular from point $C$ to $AB$. Let the foot of the perpendicular be $D$.
$\because \sin A=\frac{5}{13}=\frac{CD}{AC}$, let $m>0$,
$\therefore CD=5m, AC=13m$.
$\because \tan B=\frac{CD}{BD}=2$, we can set $n>0, CD=2n, BD=n$,
$\therefore BD=n=\frac{CD}{2}=\frac{5}{2}m$.
$\therefore AD=\sqrt{(13m)^2-(5m)^2}=12m$.
Thus, $AB=AD+BD=12m+\frac{5}{2}m=\frac{29}{2}m$.
From $29=\frac{29}{2}m$, we get $m=2$. Then $CD=5m=10$.
Therefore, $S_{\triangle ABC}=\frac{1}{2} AB \cdot CD=\frac{1}{2} \times 29 \times 10=145(\text{cm}^2)$.