Question

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 $ △POQ$ and $ △ROS$.

Answer 1

STEP 1 : Given

$PR$ and $QS$ are the diagonals of a trapezium $PQRS$ intersecting each other at the point $O$.

$PQ=3RS$

$PQ\parallel RS$

$\angle OSR=\angle OQP$ (Alternate interior angles)

$\angle ORS=\angle OPQ$ (Alternate interior angles)

STEP 2 : Proving that $△POQ$ and $△ROS$ are similar

In $△POQ$ and $△ROS$

$\angle OQP=\angle OSR$ (Alternate interior angles)

$\angle OPQ=\angle ORS$ (Alternate interior angles)

$\angle QOP\angle SOR$ (Vertically opposite angles)

By AAA similarity criterion

$△POQ~△ROS$

As we know, If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides. Therefore,

$\frac{ar(△POQ)}{ar(△ROS)}={\left(\frac{PQ}{RS}\right)}^2$

$\frac{ar(△POQ)}{ar(△ROS)}={\left(\frac{3\times RS}{RS}\right)}^2$ $[PQ=3RS]$

$\frac{ar(△POQ)}{ar(△ROS)}={\left(\frac{3}{1}\right)}^2=\frac{9}{1}$

$ar(△POQ)\mathbf{:}ar(△ROS)=9\mathbf{:}1$

Therefore, the ratio of the areas of $△POQ$ and $△ROS$ is $9\mathbf{:}1$.