Question

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

Answer 2

area of triangle ADC=1/2×DC×AD
ar(triangle ABC)=1/2×AC×BC
Ar.(ADC)=Ar.(ABC)______eq1
Because DC=AC AND AD=BC in //gram opp. are equal and //.
so, by adding two areas of triangle we get the area of //gram.
ar(triangle ADC)=ar(triangle ABC) { from eq1 }
ar(triangle ADC)+ar(triangle ABC)=ar(ABCD//gram)
ADC+ABC=ABCD
ABC+ABC=ABCD
2ABC=ABCD
ABC=1/2(ABCD)
Hence proved.