Question

In the figure given below, $DE\parallel BC$. If $DE=6cm,BC=9cm$ and area of $△ADE=28sq.cm$, find the area of $△ABC$.

Answer 1

Given : $DE\parallel BC$ ; Area of $\Delta ADE=28{cm}^2$ ;

Consider the $\Delta ADE$ and $\Delta ABC$

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

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

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

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

i.e, $\frac{Area\Delta ABC}{Area\Delta ADE}=\frac{{BC}^2}{{DE}^2}$

$\Rightarrow \frac{Ar.of\Delta ABC}{28}=\frac{9^2}{6^2}$

$\Rightarrow$ Area of $\Delta ABC=\frac{9^2}{6^2}\times 28=\frac{3\times 3}{2\times 2}\times 28=9\times 7=63{cm}^2$