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Prove that area of triangle having …

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Prove that area of triangle having base with a parallelogram lie between parallels is half the area of parallelogram

Posted by Shreya Thakur 5 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Yogita Ingle 5 years, 10 months ago

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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