Question

In the given figure DE $\parallel$BC,

If $DE=4cm,BC=8cm$ and area ($\Delta ADE$) = $25sq.cm$, find the area of $\Delta ABC$.

Answer 1

Given

$\Delta ABC$ in which $DE$ $\parallel BC$ and $DE=4cm$, $BC=8cm$. and ar. ($\Delta ADE$)$=25sq.cm$.

In $\Delta ADE$ and $\Delta ABC$ ,
$\angle A=\angle A$ [Common]
$\angle ADE=\angle ABC$ [Corresponding angles]
$\angle AED=\angle ACB$ [Corresponding angles]
$\therefore \Delta ADE\sim \Delta ABC$ [AAA similarity]
Since ratio of areas of two similar triangles is equal to ratio of squares on the corresponding sides,
$\therefore \frac{ar\left(\Delta ADE\right)}{ar\left(\Delta ABC\right)}=\frac{(DE{)}^2}{(BC{)}^2}$
$\Rightarrow \frac{25}{ar\left(\Delta ABC\right)}=\frac{(4{)}^2}{(8{)}^2}=\frac{16}{64}=\frac{1}{4}$
Hence, $ar\left(\Delta ABC\right)=25\times 4=100sq.cm$