Question

P and Q are any two points lying on the sides DC and AD of parallelogram ABCD show that area of triangle APB =area of triangle BQC​

Answer 1

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

Step-by-step explanation:

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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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)

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