Given that, $ \triangle ABC,$ median $AD$, and $BE$ are perpendicular to each other. And, $AD = 12\;\text{cm} ,BE = 9\;\text{cm}$
We can draw the third median by $CF.$
The intersection point of median i.e., centroid $(G)$ divides each median into $2:1.$
$\therefore$ All three medians divide the triangle into $6$ equal triangles.
$AD= 12\;\text{cm}$ and $AG + GD = 12 \Rightarrow AG:GD = 2:1$
- $ AG = \frac{2}{3} \times12 =8$
- $ GD= \frac{1}{3} \times 12 = 4$
$BE = 9\;\text{cm}$ and $BG + GE=9\;\text{cm} \Rightarrow BG :GE = 2:1.$
- $BG= \frac{2}{3} \times 9 = 6$
- $GE = \frac{1}{3} \times 9 = 3$
$AD$ perpendicular to $BE. \angle AGE = 90^{\circ}$
Now, the area of $\triangle AGE = \frac{1}{2} \times \text{base} \times \text{height}$
$ \qquad \qquad \qquad \qquad =\frac{1}{2} \times 3 \times 8$
$ \qquad \qquad \qquad \qquad = 12\;\text{cm}$
So, area of $ \triangle ABC = 6 \; \ast $ area of small triangle
$ \qquad \quad = 6 \times \triangle AGE$
$ \qquad \quad = 6 \times 12\;\text{cm}$
$ \qquad \quad = 72\;\text{cm}$
Correct Answer: B
Reference: https://math.stackexchange.com/questions/2132395/prove-that-three-medians-divide-a-triangle-into-6-triangles-with-equal-surface-a