Question

In the adjoining figure,MNPQ and ABPQ are parallelogram and T is any point on the side BP.Prove that

(i) ar(MNPQ)=ar(ABPQ)

(ii) $ar\left(△ATQ\right)=\frac{1}{2}ar\left(MNPQ\right)$

Answer 1

(i) We know that parallelograms on the same base and between the same parallels are equal in area.
So, ar(MNPQ) = ar(ABPQ) (Same base PQ and MB || PQ) .....(1)

(ii) If a parallelogram and a triangle are on the same base and between the same parallels then the area of the triangle is equal to half the area of the parallelogram.
So,$ar\left(△ATQ\right)=\frac{1}{2}ar\left(ABPQ\right)$ (Same base AQ and AQ || BP) .....(2)
From (1) and (2)
$ar\left(△ATQ\right)=\frac{1}{2}ar\left(MNPQ\right)$