Question

In the given figure, $DE\parallel BC$. If $DE=4cm$, $BC=6cm$ and Area $\left(△ADE\right)=20{cm}^2$, find the area of $△ABC$.

• A. $92{cm}^2$
• B. $55{cm}^2$
• C. $45{cm}^2$
• D. $30{cm}^2$

Answer 1

The correct option is C $45{cm}^2$

Given: $DE\parallel BC$
In $△ADE$ and $△ABC$
$\angle DAE=\angle BAC$ (Common)
$\angle ADE=\angle ABC$ (Corresponding angles)
$\angle AED=\angle ACB$ (Corresponding angles)
Thus, $△ABC\sim △ADE$ (AAA rule)
Hence, $\frac{A\left(△ABC\right)}{A\left(△ADE\right)}=\frac{B{C}^2}{D{E}^2}$
$\frac{A\left(△ABC\right)}{20}=\frac{6^2}{4^2}$
$A\left(△ABC\right)=9\times 5$
$A\left(△ABC\right)=45c{m}^2$