Question

In figure if $DE\parallel BC$, then the find ratio of area of $\left(\Delta ADE\right)$ and area of $\left(DECB\right)$. Consider $DE=6cm$ and $BC=12cm$.

Answer 1

Given that, in $\Delta ABC$, $DE||BC$

and $DE=6cm$ anc $BC=12cm$

To find: $\frac{area\left(\Delta ADE\right)}{area\left(DECB\right)}$

Solution:

In $\Delta ABC$ and $\Delta ADE$

$DE||BC$ [GIven]

(corresponding angles)$\{\begin{array}{c}\angle 1=\angle 2\\ \angle 3=\angle 4\end{array}$

$\therefore ABC\sim \Delta ADE$ [By AA similarity criterion]

Now, $\frac{area\left(\Delta ABC\right)}{area\left(\Delta ADE\right)}={\left(\frac{BC}{DE}\right)}^2$

($\because$ Ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.)

$=\frac{area\left(DECB\right)+area\left(\Delta ADE\right)}{area\left(\Delta ADE\right)}={\left(\frac{12}{6}\right)}^2$

$\Rightarrow \frac{area\left(DECB\right)}{area\left(\Delta ADE\right)}$$+\frac{area\left(\Delta ADE\right)}{area\left(\Delta ADE\right)}=(2{)}^2$

$=\frac{area\left(DECB\right)}{area\left(\Delta ADE\right)}+1=4$

$=\frac{area\left(DECB\right)}{area\left(\Delta ADE\right)}=4-1=3$

$=\frac{area\left(\Delta ADE\right)}{area\left(DECB\right)}=\frac{1}{3}$

Hence, the required ratio is $1:3$.