Question

In figure, DE || BC and $\frac{AD}{DB}=\frac{5}{4}$. Find $\frac{area\left(\Delta DFE\right)}{area\left(\Delta CFB\right)}$

• A. 25/81
• B. 15/81
• C. 18/81
• D. 2/81

Answer 1

The correct option is A 25/81

In $\Delta$ ABC, DE || BC and $\frac{AD}{DB}=\frac{5}{4}$
Since DE || BC
$\therefore \angle ADE=\angle ABCand\angle AED=\angle ACB$
$\therefore \Delta ADE\sim \Delta ABC$ [AAA similarity]
$\Rightarrow \frac{AD}{AB}=\frac{DE}{BC}$
Now, $\frac{AD}{DB}=\frac{5}{4}\Rightarrow \frac{DB}{AD}=\frac{4}{5}$
$\Rightarrow \frac{DB}{AD}+1=\frac{4}{5}+1\Rightarrow \frac{DB+AD}{AD}=\frac{9}{5}$
$\Rightarrow \frac{AB}{AD}=\frac{9}{5}\Rightarrow \frac{AD}{AB}=\frac{5}{9}\therefore \frac{DE}{BC}=\frac{5}{9}$
In $\Delta$DEF and $\Delta$CFB,
we have $\angle DEF=\angle FBCand\angle FDE=\angle FCB$
$\Delta DEF\sim \Delta CFB$ (AAA criterion)
$\therefore \frac{ar\left(\Delta DEF\right)}{ar\left(\Delta CFB\right)}=\frac{D{E}^2}{B{C}^2}={\left(\frac{5}{9}\right)}^2=\frac{25}{81}$