Question

P and Q are any two points lying on the side DC and AD respectively of a parallelograme ABCD.shown that ar (APB)=ar (BQC)​

Answer 1

${\huge{\mathfrak{Answer:-}}}$
$<u>$
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.

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$<b>$
Given: In parallelogram ABCD, P & Q any two

points lying on the sides DC and AD.

To show:

ar (APB) = ar (BQC).

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)

Answer 2

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