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Home » Diagonals of a trapezium PQRS intersect each other at the point 0, PQ || RS and PQ = 3 RS. Find the ratio of the areas of Δ POQ and Δ ROS.
Diagonals of a trapezium PQRS intersect each other at the point 0, PQ || RS and PQ = 3 RS. Find the ratio of the areas of Δ POQ and Δ ROS.
CBSE | Class 10 | Exercise 6.3 | Maths | NCERT Exemplar | Triangles
Diagonals of a trapezium PQRS intersect each other at the point 0, PQ || RS and PQ = 3 RS. Find the ratio of the areas of Δ POQ and Δ ROS.

Solution:

According to the given question,

The given figure, PQRS is a trapezium in which PQ || RS and PQ = 3RS

PQ/RS\text{ }=\text{ }3/1\text{ }=\text{ }3…(i)

NCERT Exemplar Solutions Class 10 Maths Chapter 6 Ex. 6.3-7

In triangles POQ and ROS,

\angle SOR\text{ }=~\angle QOP [vertically opposite angles]

\angle SRP\text{ }=~\angle RPQ[alternate angles]

∴ ∆POQ ∼ ∆ROS [using the AAA similarity criterion]

Using the property of area of similar triangle,

\frac{ar\left( \Delta POQ \right)}{ar\left( \Delta SOR \right)}=\frac{{{\left( PQ \right)}^{2}}}{{{\left( RS \right)}^{2}}}={{\left( \frac{PQ}{RS} \right)}^{2}}={{\left( \frac{3}{1} \right)}^{2}}

\Rightarrow \frac{ar\left( \Delta POQ \right)}{ar\left( \Delta SOR \right)}=\frac{9}{1}

As a result, 9:1 is the required ratio.

i

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