Question

Let $P(2,1), Q(4,-1)$ and $R(3,2)$ be the vertices or $\triangle PQR$. If through $P$ and $R$ lines parallel to opposite sides are drawn to intersect in $S$, then find the area of $\square PQRS$, in sq. units.

Answer 1

$| PQ |=\sqrt{(4-2)^{2}+(1+1)^{2}}=2 \sqrt{2} $
$| PR |=\sqrt{(3-2)^{2}+(2-1)^{2}}=\sqrt{2}$
$\Rightarrow$ Area of $\square PQRS =2$ area of $\triangle PQR$
$=2 \times \frac{1}{2} \times 2 \sqrt{2} \times \sqrt{2} $
$=4$