In a triangle ABC, medians AD and BE are perpendicular to each other, and have lengths 12 cm and 9 cm, respectively. Then, the area of triangle ABC, in sq cm, is

Let the medians $AD$ and $BE$ intersect at the centroid $G$ of triangle $ABC$.

Since the centroid divides each median in the ratio 2:1, we can compute the lengths of the segments:
$AG=\frac{2}{3}\times AD=\frac{2}{3}\times 12=8\text{cm}$
$GD=\frac{1}{3}\times AD=\frac{1}{3}\times 12=4\text{cm}$

Similarly, for the median $BE$:
$BG=\frac{2}{3}\times BE=\frac{2}{3}\times 9=6\text{cm}$
$GE=\frac{1}{3}\times BE=\frac{1}{3}\times 9=3\text{cm}$

Since the medians are perpendicular, triangle $ABG$ is a right-angled triangle at $G$.
The area of right-angled triangle $ABG$ is:
$\text{Area}(\Delta ABG)=\frac{1}{2}\times AG\times BG=\frac{1}{2}\times 8\times 6=24{\text{cm}}^2$

Since $G$ is the centroid, the area of triangle $ABC$ is three times the area of triangle $ABG$:
$\text{Area}(\Delta ABC)=3\times \text{Area}(\Delta ABG)=3\times 24=72{\text{cm}}^2$