C

将此长方形折叠，使点$B$与点$D$重合，$\therefore BE=ED$.
$\because AD=9cm=AE+DE=AE+BE$.
$\therefore BE=9-AE$，
根据勾股定理可知$AB^{2}+AE^{2}=BE^{2}$，
$\therefore 3^{2}+AE^{2}=\left(9-AE\right)^{2}$，
解得$AE=4$.
$\therefore \triangle ABE$的面积为$3\times 4\div 2=6(cm^{2})$.
故选：$C$.