Question

if a triangle and a parallelogram are on the same base and same parallels, then prove that the area of the triangle is equal to half the area of the parallelogram

Answer 1

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

To Prove : ar( ΔPAB ) = (1/2)ar( ABCD)

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) [ from (2)]

This gives ar (PAB) = (1/2)ar(ABCD)   [ from (1) and (3)]