in Quantitative Aptitude edited by
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In the following figure, the area of the isosceles right triangle $\text{ABE}$ is $7$ If $\text{EC = 3BE},$ then the area of rectangle $\text{ABCD}$ id (

  1. $64$
  2. $82$
  3. $26$
  4. $56$
in Quantitative Aptitude edited by
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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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