Question

In the given figure, ABC is a triangle with $\angle EDB=\angle ACB$. Prove that $\Delta ABC\sim \Delta EBD$. If BE = 6 cm, EC = 4 cm, BD = 5 cm and area of $\Delta BED=9c{m}^2$. Calculate the:

i) length of AB
ii) area of $\Delta ABC$ [3 MARKS]

Answer 1

Formula : 1 Mark
Length : 1 Mark
Area : 1 Mark

In $\Delta ABC$ and $\Delta EBD$,
$\angle ACB=\angle BDE$ [Given]
$\angle ABC=\angle DBE$ [Common]
$\therefore \Delta ABC\sim \Delta EBD$

i) Since $\Delta ABC$ and $\Delta EBD$ are similar
$\therefore \frac{AB}{BE}=\frac{BC}{BD}$
$\Rightarrow \frac{AB}{6}=\frac{6+4}{5}\Rightarrow AB=\frac{10}{5}\times 6=12cm$

ii) $\frac{Areaof\Delta ABC}{Areaof\Delta BED}={\left(\frac{AB}{BE}\right)}^2$
$\Rightarrow Areaof\Delta ABC=\frac{12\times 12}{6\times 6}\times 9=36c{m}^2$