Question

In $\Delta$PQR, MN is parallel to QR and $\frac{PM}{MQ}=\frac{2}{3}$. Find, Area of $\Delta$OMN:Area of $\Delta$ORQ.

Answer 1

$\frac{PM}{PQ}=\frac{PM}{PM+MQ}=\frac{2}{2+3}=\frac{2}{5}$

Since $MN||QR$, we have, $△PMN$ is similar to $△PQR$

$\Rightarrow \frac{PM}{PQ}=\frac{MN}{QR}=\frac{2}{5}$

$MN||QR$

Alternate interior angles are equal, so

$\Rightarrow \angle OMN=\angle ORQ$

$\Rightarrow \angle ONM=\angle OQR$

Vertically opposite angles are equal, so $\angle MON=\angle QOR$

Thus, we have that $△OMN\sim △ORQ$

So,

$\frac{Areaof△OMN}{Areaof△ORQ}=\frac{M{N}^2}{Q{R}^2}=\frac{4}{25}$