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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