Question

In the adjoining figure, MNPQ and ABPQ are parallelograms and T is any point on the side BP. Prove that
(i) ar(MNPQ) = are(ABPQ)
(ii) ar(∆ATQ) = $\frac{1}{2}$ar(MNPQ).

Answer 1

(i) We know that parallelograms on the same base and between the same parallels are equal in area.
So, ar(MNPQ) = are(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(∆ATQ) = $\frac{1}{2}$ar(ABPQ) (Same base AQ and AQ || BP) .....(2)
From (1) and (2)
ar(∆ATQ) = $\frac{1}{2}$ar(MNPQ)