CAT 2002 | Question: 58

Question

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.)

• A. $64$
• B. $82$
• C. $26$
• D. $56$

Answer 1

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.