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# Diagonals of a trapezium PQRS intersect each other at the point $ O, P Q \| R S $ and $ P Q=3 \mathrm{RS} $. 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 $9:1$.

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