Question

$\large\bold{\underline{\underline{Question:-}}}$


P and Q are any two points lying on the Sides DC and AD respectively of a parallelogram ABCD.Show that ar(APB) = ar(BQC)

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Answer 1

Hope it's helpful.............

Answer 2

$\large\bold{\underline{\underline{Given:-}}}$

In parallelogram ABCD, P & Q any two points lying on the sides DC and AD.

To show: ar (APB) = ar (BQC).

$\large\bold{\underline{\underline{Proof:-}}}$

Here, ΔAPB and ||gm ABCD stands on the same base AB and lie between same parallel AB and DC.

Therefore,

ar(ΔAPB) = 1/2 ar(||gm ABCD) — (i)

Similarly,

Parallelogram ABCD and ∆BQC stand on the same base BC and lie between the same parallel BC and AD.

ar(ΔBQC) = 1/2 ar(||gm ABCD) — (ii)

From eq (i) and (ii),we have

ar(ΔAPB) = ar(ΔBQC)

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$\large\bold{\underline{\underline{Additional \: Info:-}}}$

If a parallelogram and a triangle are on the same base and between the same parallels then area of the triangle is half the area of the parallelogram.