Question

Diagonals of a trapezium $PQRS$ intersect each other at the point O, \[PQ||RS\] and \[PQ = 3RS\]. Find the ratio of the areas of the triangles $POQ$ and $ROS$

Answer 1

Hint: We are given a trapezium whose diagonals intersect at a point and as we know that the trapezium has two parallel sides and the measurement of one side is thrice the other. We have to find the ratio of the areas of the triangle formed by diagonals. We will use the concept of similarity of triangle first we will prove the two triangles similar in that we also use the concept of alternate angles as we are having the pair of parallel lines alternate angles are angles that are in opposite position relative to a transversal intersecting two lines after proving the similarity by AA, SAS, SSS criterion we will use the theorem of the area of triangles 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 this proves that the ratio of the area of two similar triangles is proportional to the square of the corresponding sides of both the triangles

Complete step by step answer:
Step1: We are given a trapezium $PORS$ in which $PQ$ is parallel to $RS$ and $PQ$=$3RS$.
Here we will draw the trapezium. The diagonals of trapezium intersect at 0. To find the ratio of the area of triangles $POQ$ and $ROS$, first we will prove the similarity between the triangles POQ and ROS.
Step2: In $\vartriangle POQ$ and $\vartriangle ROS$
$\angle SOR = \angle QOP$ (Vertically opposite angles)
$\angle SRP = \angle RPQ$ (Alternate angles)
$\vartriangle POQ \sim \vartriangle ROS$ (By AAA similarity criterion)
Step3: Now by using the property of the area of similar triangles we will find the ratio of the area of two similar triangles which states the ratio of the area of two similar triangles is proportional to the square of the corresponding sides of both the triangles.
$\Rightarrow \dfrac{{ar(\vartriangle POQ)}}{{ar(\vartriangle SOR)}} = \dfrac{{{{(PQ)}^2}}}{{{{(RS)}^2}}} = {\left( {\dfrac{{PQ}}{{RS}}} \right)^2}$………………….(1)
As it is given $PQ = 3RS$
$\Rightarrow \dfrac{{PQ}}{{RS}} = \dfrac{3}{1}$
Step4: Substituting the value in equation (1) we get
$\Rightarrow \dfrac{{ar(\vartriangle POQ)}}{{ar(\vartriangle SOR)}} = {\left( {\dfrac{3}{1}} \right)^2}$
$\Rightarrow \dfrac{{ar(\vartriangle POQ)}}{{ar(\vartriangle SOR)}} = \dfrac{9}{1}$

Hence the ratio of the area of $\vartriangle POQ$ and $ \vartriangle ROS$ is $9:1$

Note:
In such types of questions students mainly get confused or even don’t know which concept they have to apply. They sometimes use the concept of congruence which is wrong they should be kept in mind if they have to solve anything for the ratio of the area of the triangle then they have to simply use the property of the ratio of the area of similar triangles
Commit to memory: Theorem of the ratio of the area of similar triangles