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Question

In the given figure, ABCD is a trapezium, Diagonals AC and BD interset at Q Prove that the area of ΔAOD is the same as ΔBOC

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Solution

Given a trapezium ABCD where ABDC ξ diagonal AC ξ BD intersect each other at 0.
We have prove that ar(AOD)=ar(BOC)
Consider ADC ξ BDC, ;ie on the same base DC and between the same parallel AB and CD.
Area(ADC)=Area(BDC)
[ Triangle with same base and between the same parallel are equal in area .]
ar(ADC)ar(DOC)=ar(BDC)ar(DOC)
ar(AOD)=ar(BOC)
Hence, the answer is proved.

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