Question

"Question 3 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).

Class 9 - Math - Areas of Parallelograms and Triangles Page 159"

Answer 1

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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Hope this will help you...

 

Answer 2

It can be observed that ΔBQC and parallelogram ABCD lie on the same base BC and these are between the same parallel lines AD and BC

If a triangle and parallelogram are on the same base and have the same altitude, the area of the triangle will be half that of the parallelogram.

Area (ΔBQC) = Area (ABCD) (i)

Similarly,

If a triangle and parallelogram are on the same base and have the same altitude, the area of the triangle will be half that of the parallelogram.

ΔAPB and parallelogram ABCD lie on the same base AB and between the same parallel lines AB and DC

Area (ΔAPB) = Area (ABCD) (ii)

From equation (i) and (ii), we get

Area (ΔBQC) = Area (ΔAPB)