Question

Diagonals of a trapezium PQRS intersect each other at the point, O, PQIIRS and PQ = 3 RS. find the ratio of the areas of $\Delta$POQ and $\Delta$ROS.
Thinking process
Firstly, show that $\Delta$POQ, and $\Delta$ROS are similar by AAA similarity, then use property of area of similar triangle to get required ratio.

Answer 1

Given PQRS is a trapezium in which PQ II RS and PQ = 3 Rs
$\Rightarrow \frac{PQ}{RS}=\frac{3}{1}$ ....(i)



In $\Delta POQ$ and $\Delta ROS$, $\angle SOR=\angle QOP$ [vertically opposite angles]
$\angle SRP=\angle RPQ$ [ alternate angles]
$\therefore \Delta POQ\sim \Delta ROS$ [by AAA similarity criterion]
By property of area of similar triangle,
$\frac{ar\left(\Delta POQ\right)}{ar\left(\Delta SOR\right)}=\frac{(PQ{)}^2}{(RS{)}^2}={\left(\frac{PQ}{RS}\right)}^2={\left(\frac{3}{1}\right)}^2$ [from Eq. (i)]
$\Rightarrow \frac{ar\left(\Delta POQ\right)}{ar\left(\Delta SOR\right)}=\frac{9}{1}$
Hence, the required ratio is 9:1