Question

$△ABC$ is an isosceles triangle in which AB = AC = 13 cm. If area of $△ADC$ is $169c{m}^2$, then area of $△EFB$ is equal to

• A. $169c{m}^2$
• B. $196c{m}^2$
• C. $396c{m}^2$
• D. $324c{m}^2$

Answer 1

The correct option is D

$324c{m}^2$

In $\Delta ADC$ and $\Delta EFB$,

$\angle ACD=\angle EBF$
(Since base angles of an isosceles triangle are equal)

$\angle ADC=\angle EFB\left(given{90}^{\circ }\right)$

$\Delta ADC\sim \Delta EFB$ (by AA similarity criterion)

So, $\frac{ar\left(\Delta ADC\right)}{ar\left(\Delta EFB\right)}=\frac{A{C}^2}{(EB{)}^2}$

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

$\Rightarrow \frac{169c{m}^2}{ar\left(\Delta EFB\right)}=\frac{{13}^2}{(EC+CB{)}^2}$

$\Rightarrow \frac{169c{m}^2}{ar\left(\Delta EFB\right)}=\frac{{13}^2}{(8+10{)}^2}$

$\Rightarrow ar\left(\Delta EFB\right)=\frac{(18\times 18)\times 169}{(13{)}^2}=324c{m}^2$