Question

$\Delta ABC$ and $\Delta PQR$ are two similar triangles shown in the figure. AM and PN are the medians on $\Delta ABC$ and $\Delta PQR$, respectively. The ratio of areas of $\Delta ABC$ and $\Delta PQR$ is 9:25. If AM = PO = 5 cm. Find the value of 3(ON).

Answer 1

We are given two triangles $\Delta ABC$ and $\Delta PQR$ such that $\Delta ABC\sim \Delta PQR$.
$\Rightarrow$ Corresponding sides will be proportional.
$\Rightarrow$ $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$
Here, $\frac{\frac{1}{2}\times BC}{\frac{1}{2}\times QR}=\frac{BM}{QN}$
$\therefore$ $\frac{AB}{PQ}=\frac{BM}{QN}$
Since both the triangles $\Delta ABC$ and $\Delta PQR$ are similar, there angles will be equal
$\Rightarrow$ $\angle$ A = $\angle$ P, $\angle$ B = $\angle$ Q and $\angle$ C = $\angle$ R
In $\Delta ABM$ and $\Delta PQN$,
$\frac{AB}{PQ}=\frac{BM}{QN}$ (Proved above)
$\angle$ B = $\angle$ Q
$\therefore$ $\Delta ABM\sim \Delta PQN$ [By SAS similarity]
Given: AM = PO = 5 cm
$\Rightarrow$ $\frac{Areaof\Delta ABC}{Areaof\Delta PQR}=\frac{A{B}^2}{P{Q}^2}=\frac{9}{25}$
$\Rightarrow \frac{AB}{PQ}=\frac{3}{5}$
$\Rightarrow$ $\frac{AB}{PQ}=\frac{AM}{PN}=\frac{3}{5}$
$\Rightarrow \frac{5}{5+ON}=\frac{3}{5}$
$\Rightarrow$ $25=15+3\left(ON\right)$
$ON=\frac{10}{3}$
$3\left(ON\right)=10cm$
Hence the length of 3(ON) is 10 cm.