Question

Given $\Delta$ ABC $\sim$ $\Delta$ PQR, if $\frac{AB}{PQ}=\frac{1}{3}$, then find $\frac{ar\Delta ABC}{ar\Delta PQR}$.

Answer 1

Let $AD$ be the perpendicular dropped from $A$ to the side $BC$ and $PS$ be the perpendicular dropped from $P$ to the side $QR$ respectively.

Since, $\Delta ABC\sim \Delta PQR$, we have $\Delta ABD\sim \Delta PQS$.

Therefore, $\frac{AB}{PQ}=\frac{AD}{PS}=\frac{BC}{QR}$

$\Rightarrow \frac{ar\Delta ABC}{ar\Delta PQR}=\frac{AD\times BC}{PS\times QR}=\frac{AB}{PQ}\times \frac{AB}{PQ}=\frac{A{B}^2}{P{Q}^2}=\frac{1}{9}$