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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. $80$
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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

Reference: https://math.stackexchange.com/questions/2132395/prove-that-three-medians-divide-a-triangle-into-6-triangles-with-equal-surface-a

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