Answer :

Given: ABCD is a gm in which diagonals AC and BD intersect at O and ar(gm ABCD) is 52cm2.


Here,


Ar (∆ABD) = ar(∆ABC)


( ΔABD and ΔABC on same base AB and between same parallel lines AB and CD)


Here,


ar(∆ABD) = ar(∆ABC) = 1/2 × ar(||gm ABCD)


( ΔABD and ΔABC on same base AB and between same parallel lines AB and CD are half the area of the parallelogram)


ar(∆ABD) = ar(∆ABC) = 1/2 × 52 = 26cm2


Now, consider ΔABC


Here OB is the median of AC


( diagonals bisect each other in parallelogram)


ar(∆AOB) = ar(∆BOC)


(median of a triangle divides it into two triangles of equal area)


ar(∆AOB) = 1/2 × ar(ΔABC)


ar(∆AOB) = 1/2 × 26 = 13cm2


ar(∆AOB) = 13cm2

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