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Diagonals of a trapezium PQRS intersect each other at the point $O, PQ \parallel RS$ and $PQ = 3RS$. Find the ratio of the areas of triangles $POQ$ and $ROS$.
Given:
Diagonals of a trapezium PQRS intersect each other at the point $O, PQ \parallel RS$ and $PQ = 3RS$.
To do:
We have to find the ratio of the areas of triangles $POQ$ and $ROS$.
Solution:
In triangles $POQ$ and $ROS$,
$\angle POQ=\angle ROS$ (Vertically opposite angles)
$\angle OPQ=\angle ORS$ (Alternate angles as $PQ \parallel RS$)
Therefore,
$\triangle POQ \sim\ \triangle ROS$ (By AA congruence)
We know that,
If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.
Therefore,
$\frac{ar(\triangle POQ)}{ar(\triangle ROS)}=\frac{PQ^2}{RS^2}$
$=\frac{(3RS)^2}{RS^2}$
$=\frac{9RS^2}{RS^2}$
$=\frac{9}{1}$
The ratio of the areas of triangles $POQ$ and $ROS$ is $9:1$.