Question

A triangle and a parallelogram on the same base and between same parallels. Prove that ar (∆) = 1/2 ar (llgm).​

Answer 1

Let ΔABP and a parallelogram ABCD be on the same base AB and between the same parallels AB and PC.

Draw BQ ||AP to obtain another parallelogram. ABQP and ABCD are on the same base AB and between the same parallels AB and PC.

There fore, ar(ABQP) =  ar(ABCD)

But ΔPAB ≅ ΔBQP( Diagonals PB divides parallelogram ABQP into two congruent triangles.

So  ar (PAB) = ar(BQP)     ------- (2)

∴ ar (PAB) = (1/2)ar(ABQP) ------- (3)

Therefore, ar (PAB) = (1/2)ar(ABCD).

#Hope it helped you.