Question

If P and Q are any two points lying respectively on the sides DC and AD of a parallelogram ABCD then show that $ar\left(△APB\right)=ar\left(△BQC\right)$.

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.


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

To show:

$ar\left(△APB\right)=ar\left(△BQC\right)$.

Here, $△APB$ and ||gm ABCD stands on the same
base AB and lie between the same parallels AB and DC.

Therefore,

$△\left(\Delta APB\right)=\frac{1}{2}$ 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$△BQC=\frac{1}{2}ar\left(||gmABCD\right)$ — (ii)
From eq (i) and (ii),
we have
ar$△APB=ar△BQC$