0 votes

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

- 64
- 82
- 26
- 56

1 vote

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.

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.