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In the following figure, the area of the isosceles right triangle $\text{ABE}$ is $7$ sq.cm. If $\text{EC = 3BE},$ then the area of rectangle $\text{ABCD}$ id (insq.cm.)

1. $64$
2. $82$
3. $26$
4. $56$

Given: 1) $\text{Area of$\triangle$ABC =7 cm$^{2}$}$

2) $EC=3BE$

We know that area of triangle =$\frac{1}{2}*B*H$

so Area of $\triangle$ ABE $= \frac{1}{2}*B*H$

$\implies 7=\frac{1}{2}*B*H$

$\implies B*H=14$

This can be also written as  $\text{ BE*AB=14 cm$^{2}$}\cdots\cdots(i)$

Now area of rectangle $ABCD= AB*BC$

$\implies AB* (4*BE)$

$\implies 4*14=56$ sq cm

Note: $(BC=BE+EC\implies BE+3BE=4BE)$

Option $(D)$ is correct.
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