Diagonals of a trapezium PQRS intersect each | vedclass

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(D) Given $PQRS$ is a trapezium in which $PQ \parallel RS$ and $PQ = 3 RS$.$\frac{PQ}{RS} = \frac{3}{1}$ ......$(i)$In $	riangle POQ$ and $	riangle

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$9: 1$

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(D) Given $PQRS$ is a trapezium in which $PQ \parallel RS$ and $PQ = 3 RS$.$\frac{PQ}{RS} = \frac{3}{1}$ ......$(i)$In $	riangle POQ$ and $	riangle ROS$:$\angle SOR = \angle QOP$ [Vertically opposite angles]$\angle OSR = \angle OPQ$ [Alternate interior angles,since $PQ \parallel RS$]$	herefore 	riangle POQ \sim 	riangle ROS$ [By $AA$ similarity criterion]By the property of the area of similar triangles,the ratio of their areas is equal to the square of the ratio of their corresponding sides:$\frac{	ext{ar}(	riangle POQ)}{	ext{ar}(	riangle ROS)} = \left(\frac{PQ}{RS}ight)^2 = \left(\frac{3}{1}ight)^2 = \frac{9}{1}$Hence,the required ratio is $9: 1$.

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

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