Question

In the figure given below, $DE\parallel BC$ and $AD:DB=1:2$, find the ratio of the areas of $△ADE$ and trapezium $DBCE$.

Answer 1

Given : $DE\parallel BC$ and $AD:DB=1:2$

Now, $\frac{AD}{DB}=\frac{1}{2}⟶\left(1\right)$

$\Rightarrow \frac{AD}{DB}+1=\frac{1}{2}+1$

$\Rightarrow \frac{AD+DB}{DB}=\frac{3}{2}\Rightarrow \frac{AB}{DB}=\frac{3}{2}⟶\left(2\right)$

From $\left(1\right)$ & $\left(2\right)$

$\frac{AD}{AB}=\frac{1}{3}⟶\left(3\right)$

$\Delta ADE$ and $\Delta ABC$

$\angle A=\angle A$ (common)

$\angle ADE=\angle ABC$ (corresponding angles)

$\angle AED=\angle ACB$ (corresponding angles)

$\therefore$ $\Delta ADE$ is similar to $\Delta ABC$ by $AA$ criterion and if the triangles are similar then their ratios are equal to the square of the ratios of the corresponding sides.

$\therefore$ $\frac{Areaof\Delta ADE}{Areaof\Delta ABC}=\frac{{AD}^2}{{AB}^2}=\frac{1}{9}$

$\therefore$ $\Rightarrow \frac{Areaof\Delta ABC}{Areaof\Delta ADE}=\frac{9}{1}$

$\Rightarrow \frac{AreaoftrapeziumDBCE+Areaof\Delta ADE}{Areaof\Delta ADE}=\frac{9}{1}$

$\Rightarrow \frac{AreaoftrapeziumDBCE}{Areaof\Delta ADE}=9-1=8$

$\therefore$ $\frac{Areaof\Delta ADE}{AreaoftrapeziumDBCE}=\frac{1}{8}$