Question

In the given figure, ABC is a triangle with $\angle EDB=\angle ACB$. Prove that $\Delta ABC\sim \Delta EBD.IfBE=6cm,EC=4CM,bd=5cmandareaof\Delta BED=9c{m}^2$, calculate: [4 MARKS]
(i) Leight of AB
(ii) Area of $\Delta ABC$.

Answer 1

Application of theorems: 2 Marks
Calculation: 2 Marks
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In $\Delta ABCand\Delta EBD$.
$\begin{array}{ccc}& \angle ACB=\angle EDB& \left[Given\right]\\ & \angle B=\angle B& \left[Common\right]\\ So,& \Delta ABC\sim \Delta EBC& \left[AAsimilarity\right]\end{array}$
Given $BE=6CM,ec=4CM,BD=5cm,andareaof\Delta BED=9c{m}^2$
$\begin{array}{cc}\left(i\right)\because \Delta ABC\sim \Delta EBD& \\ \therefore & \frac{AB}{EB}=\frac{AC}{ED}=\frac{BC}{BC}\\ \Rightarrow & \frac{AB}{6}=\frac{AC}{ED}=\frac{6+4}{5}\Rightarrow \frac{AB}{6}=\frac{10}{5}\\ \Rightarrow & AB=\frac{10\times 6}{5}=12cm\\ \left(ii\right)& \frac{Areaof\Delta ABC}{Areaof\Delta EBD}=\frac{A{B}^2}{E{B}^2}\Rightarrow \frac{Areaof\Delta ABC}{9}=\frac{(12{)}^2}{(6{)}^2}\\ \Rightarrow & \frac{Areaof\Delta ABC}{9}=\frac{144}{36}\\ \Rightarrow & Areaof\Delta ABC=\frac{9\times 144}{36}36c{m}^2\end{array}$