Maths / Areas Of Parallelograms And Triangles / Median and Altitude of Triangle

QUESTION

In the given figure, O is any point on diagonal PR of parallelogram PQRS. Prove that $\dpi{120} \fn_jvn \large ar(POS)=ar(PQO).$

EXPLANATION
Explain TypeExplanation Content
Text

Given : PQRS is a parallelogram.

Proof : SQ is a diagonal.

M is the midoint of SQ.

$\therefore$              OM is a median of $\Delta SOQ$

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$ar(SMO)=ar(QMO)$                       (Median divides the triangle in 2 equal areas)  .............. (1)

Also,

In  $\Delta SQR,\;RM$ is median.

$\therefore \: \: \: \: \: \: \: \: \: \: \: ar(SMR)=ar(QMR)\: \: \: \: \: \: \: \: \: \: ..........(2)$

Subtracting (1) and (2), we get

$\Rightarrow \: \: \: \: \: \: ar(SOR)=ar(QOR)$

Now, PR is a diagonal.

$\therefore \: \: \: \: \: \: \: \: \: \: \: ar(\Delta PSR)=ar(\Delta PQR)$             (Diagonal divides the parallelogram in 2 equal areas)

$\Rightarrow \: \: \: \: \: ar(PSO)+ar(SOR)=ar(POQ)+ar(OQR)$

$\Rightarrow \: \: \: \: \: ar(PSO)=ar(POQ)$

Hence proved

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