Question

ABCD is a parallelogram with diagonals AC and BD intersecting each other at O .Prove that ar ( AOB) +ar(DOC)= half of area ( ABCD)

Answer 1

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Your answer
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In parallelogram ABCD,

Diagonals AC and BD intersect at O.

To prove : - Ar(∆AOB) + Ar(∆COD) = 1/2 Ar(ABCD)

Now,
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We know that,
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Each diagonal divides a parallelogram into two triangles of equal area.

Then, ar(∆ABC) = ar(∆ADC) = 1/2 Ar(ABCD)


ar(∆ABC) = ar(∆ABC) + ar(∆BOC) .......(i)
Again,
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The two diagonals of a parallelogram divide it into for triangles of equal area.

Then, ar(∆AOB) = ar (∆BOC) = ar(∆COD) = ar(∆AOD)

Now,

ar(∆ABC ) = 1/2 Ar(ABCD)

=> ar(∆AOB) + ar(∆BOC) = 1/2 Ar(ABCD). 【From (i) 】

=> ar(∆AOB) + ar(∆COD) = 1/2 Ar (ABCD). [ Because ar(∆BOC) = ar (∆COD)

Hence, proved.

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Answer 2

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