Question

In $△ABC,DF\parallel BC$ and $AD=1cm,BD=2cm$, what is the ratio of the area $△ABC$ to the area $△ADF$?

Answer 1

Given that,

$AD=1cm,BD=2cm.$

$\text{and}\text{DE}||\text{BC}$

$\text{So,}\angle ADE=\angle ABC(\text{Corrosponding}\text{Angles)}$

$\angle A=\angle A(\text{Common)}$

Therefore, by $A.A$. Similar.

$△ADE\stackrel{~}{}△ABC$

Ratio’s are.

$\therefore \frac{ar(△ABC)}{ar(△ADE)}=\frac{A{B}^2}{A{D}^2}$

$\Rightarrow \frac{ar(△ABC)}{ar(△ADE)}={\left(\frac{3}{1}\right)}^2$

$\Rightarrow \frac{ar(△ABC)}{ar(△ADE)}=\frac{9}{1}$

Hence, this is the answer.