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

1. $68$
2. $72$
3. $78$
4. $80$

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

$\qquad \quad = 6 \times \triangle AGE$

$\qquad \quad = 6 \times 12\;\text{cm}$

$\qquad \quad = 72\;\text{cm}$

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