Question

In the given figure, $DE\parallel BC$and$AD:DB=5:4$, find $\frac{area\left(\Delta DFE\right)}{area\left(\Delta CFB\right)}$.

• A. $5:9$
• B. $25:16$
• C. $25:81$
• D. $81:25$

Answer 1

Answer: (C)

$25:81$

$DE\parallel BC$ [ Given ]

Let $AD=5x$ and $DB=4x$

Then, $AB=AD+DB=5x+4x=9x$

In $△ADE$ and $△ABC$

$\Rightarrow$ $\angle ADE=\angle ABC$ [ Corresponding angles ]

$\Rightarrow$ $\angle A=\angle A$ [ Common angle ]

$\therefore$ $△ADE\sim △ABC$ [ By $AA$ similarity rule ]

$\Rightarrow$ $\frac{AD}{AB}=\frac{DE}{BC}$ [ CPCT ]

$\Rightarrow$ $\frac{5x}{9x}=\frac{DE}{BC}$

$\Rightarrow$ $\frac{DE}{BC}=\frac{5}{9}$ ----- ( 1 )

Now, In $△DEF$ and $△CFB$

$\Rightarrow$ $\angle EDF=\angle BCF$ [ Alternate angles ]

$\Rightarrow$ $\angle DFE=\angle CFB$ [ Vertically opposite angles ]

$\Rightarrow$ $△DFE\sim △CFB$ [ $AA$ similarity rule ]

$\Rightarrow$ $\frac{ar\left(△DFE\right)}{ar\left(△CFB\right)}=\frac{D{E}^2}{B{C}^2}$

$=\frac{(5{)}^2}{(9{)}^2}$

$=\frac{25}{81}$