using that the triangle in a parallelogram with same base and between a same pair of parallel lines are equal and a triangle within a same base and same pair of parallel line is equal

answer:    (i)

here's your answer!

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(PLEASE REFER TO THE DIAGRAM SHOWN IN THE ATTACHMENT)

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)

(here, ar = area; written in short)

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)]

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Step-by-step explanation:

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