Question

In quadrilateral PQRS diagonal PR and QS intersects at O such that $Area\left(\Delta POS\right)$= $Area\left(\Delta QOR\right)$ .If the distance between side PQ and SR is 4 cm, PQ=3cm, SR=7 cm. Find ar (PQRS).

Answer 1

It is given that
$Area\left(\Delta POS\right)=Area\left(\Delta QOR\right)$
$Area\left(\Delta POS\right)+Area\left(\Delta SOR\right)=Area\left(\Delta QOR\right)+Area\left(\Delta SOR\right)$
$Area\left(\Delta PSR\right)=Area\left(\Delta QSR\right)$
We know that triangles on the same base having area equal to each other lie between the same parallels.
Therefore, these triangles, $\Delta PSRand\Delta QSR$, are lying between the same parallels.
i.e., $PQ||SR$
Therefore PQRS is a trapezium.
Area of $PQRS=12PQ+SR\times$ distance between PQ and $SR=123+7\times 4=20c{m}^2$