Maths / Areas Of Parallelograms And Triangles / Relation between Area of Parallelogram and Triangle

QUESTION

In the following figure, $\dpi{120} \fn_jvn \large AC \parallel PS\parallel QR\;and\;PQ\parallel DB \parallel SR.$ Prove that area of quadrilateral $\dpi{120} \fn_jvn \large PQRS=2 \times area$ of quadrilateral ABCD.

EXPLANATION
Explain TypeExplanation Content
Text

APSC is a parallelogram.

$\therefore \: \: \: \: \: \: \: \: \: \: \: AC \parallel PS$

Now,  $ar(\Delta ADC)=\frac {1}{2} ar(APSC) \: \: \: \ : \: \: \: \: \: \: \: \: ..........(1)$

[Parallelogram and triangle on the same base and betweem the same parallel lines]

Similarly,

$ar(\Delta ABC)=\frac {1}{2} ar(ACRQ) \: \: \: \: \: \: \: \: \: \: \: \: ...............(2)$

Adding (1) and (2), we get

$ar(\Delta ADC)+ar(\Delta ABC)=\frac {1}{2} (ar(APSC)+ar(ACRQ))$

$ar(ADCB)=\frac {1}{2} ar(PQRS)$

$\Rightarrow$             $ar(PQRS)=2ar(ABCD)$

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