Question

In the figure, DE is parallel to BC and $\frac{AD}{DB}=\frac{3}{2}$. Area of $\Delta DEF$ is $81c{m}^2$. Area of $\Delta CFB$ is .

• A. $180c{m}^2$
• B. $250c{m}^2$
• C. $225c{m}^2$
• D. $195c{m}^2$

Answer 1

The correct option is C $225c{m}^2$

Here, DE || BC (given)

In $\Delta ADE$ and $\Delta ABC$,

$\angle A=\angle A$ (common)

$\angle ADE=\angle ABC$ (corresponding angles)

$△ADE\sim △ABC$ (AA similarity criterion)

$\frac{AD}{AB}=\frac{DE}{BC}$ ---- (1)

Given, $\frac{AD}{DB}=\frac{3}{2}$

$\Rightarrow \frac{DB}{AD}=\frac{2}{3}$

Adding 1 on both sides, we get

$\frac{DB}{AD}+1=\frac{2}{3}+1\frac{DB+AD}{AD}=\frac{2+3}{3}\frac{AB}{AD}=\frac{5}{3}$

$\Rightarrow \frac{AD}{AB}=\frac{3}{5}$ --- (2)

From (1) and (2)

$\frac{DE}{BC}=\frac{3}{5}$

In $\Delta DFE$ and $\Delta BFC$,

$\angle DFE+\angle BFC$ (vertically opposite angle)

$\angle FCB=\angle EDF$ (alternate angle)

$\Delta DFE\sim CFB$ (by AA similarity criterion)

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

(The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides)

$\frac{81}{ar\left(\Delta CFB\right)}=(\frac{DE}{BC}{)}^2\frac{81}{ar\left(\Delta CFB\right)}={\left(\frac{3}{5}\right)}^{2}ar(\Delta CFB)=\frac{81\times 25}{9}=225c{m}^2$