Question

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

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

Answer 1

The correct option is C $25:81$

In $△ADE$ and $△ABC$

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

$\Rightarrow$ $\angle AED=\angle ACB$ [ Corresponding angles ]

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

$\Rightarrow$ $△ADE\sim △ABC$ [ By AAA similarity ]

$\Rightarrow$ $\frac{AD}{AB}=\frac{DE}{BC}$

$\Rightarrow$ $\frac{AD}{DB}=\frac{5}{4}$

$\Rightarrow$ $\frac{DB}{AD}=\frac{4}{5}$ [ Taking reciprocal ]

$\Rightarrow$ $\frac{DB}{AD}+1=\frac{4}{5}+1$ [ Adding $1$ on both sides ]

$\Rightarrow$ $\frac{DB+AD}{AD}=\frac{4+5}{5}$

$\Rightarrow$ $\frac{AB}{AD}=\frac{9}{5}$

$\Rightarrow$ $\frac{AD}{AB}=\frac{5}{9}$

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

In $△DEF$ and $△CFB,$ we have

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

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

$\Rightarrow$ $\angle DEF=\angle FBC$ [ Alternate angles ]

$\Rightarrow$ $△DEF\sim △CFB$ [ By AAA similarity ]

$\Rightarrow$ $\frac{ar\left(△DEF\right)}{ar\left(△CFB\right)}={\left(\frac{DE}{BC}\right)}^2$

$\Rightarrow$ $\frac{ar\left(△DEF\right)}{ar\left(△CFB\right)}={\left(\frac{5}{9}\right)}^2=\frac{25}{81}$