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Let $\text{D}$ and $\text{E}$ be points on sides $\text{AB}$ and $\text{AC},$ respectively, of a triangle  $\text{ABC},$ such that $\text{AD}$ : $\text{BD} = 2 : 1$ and $\text{AE}$ : $\text{CE} = 2 : 3.$ If the area of the triangle $\text{ADE}$ is $8 \; \text{sq cm},$ then the area of the triangle $\text{ABC, in sq cm},$ is

Let’s first draw the diagram.

We know that, the area of $\triangle \text{ADE} = \frac{\text{AD}}{\text{AB}} \times \frac{\text{AE}}{\text{AC}} \times \text{Area of}\; \triangle \text{ABC}$

$\Rightarrow 8=\frac{2}{3} \times \frac{2}{5} \times \text{Area of}\; \triangle \text{ABC}$

$\Rightarrow \boxed{\text{Area of}\; \triangle \text{ABC} = 30\; \text{cm}^{2}.}$

Correct  Answer $:30$

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