ABCD is a parallelogram o is any point on AC PQ parallel AB and LM parallel AD prove that area parallelogram DLOP=area parallelogram BMOQ
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FIGURE IS IN THE ATTACHMENT
Since,a diagonal of a parallelogram divides it into two Triangles of equal area.
Since, AO & OC are diagonals of a parallelogram AMOP & OQCL
ar(∆APO) = ar(∆AMO) ………..(1)
ar(∆OLC) = ar(∆OQC) ………..(2)
ar(∆ADC) = ar(∆ABC)
ar(∆APO) + ar(||gmDLOP) +ar(∆OLC) =
ar(∆AMO) + ar(||gm BMOP) +ar(∆OQC) …(3)
Subtract eq 1 & 2 from 3
ar(||gmDLOP) = ar(||gm BMOQ)
HOPE THIS WILL HELP YOU...
Since,a diagonal of a parallelogram divides it into two Triangles of equal area.
Since, AO & OC are diagonals of a parallelogram AMOP & OQCL
ar(∆APO) = ar(∆AMO) ………..(1)
ar(∆OLC) = ar(∆OQC) ………..(2)
ar(∆ADC) = ar(∆ABC)
ar(∆APO) + ar(||gmDLOP) +ar(∆OLC) =
ar(∆AMO) + ar(||gm BMOP) +ar(∆OQC) …(3)
Subtract eq 1 & 2 from 3
ar(||gmDLOP) = ar(||gm BMOQ)
HOPE THIS WILL HELP YOU...
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Since,a diagonal of a parallelogram divides it into two Triangles of equal area.
Since, AO & OC are diagonals of a parallelogram AMOP & OQCL
ar(∆APO) = ar(∆AMO) ………..(1)
ar(∆OLC) = ar(∆OQC) ………..(2)
ar(∆ADC) = ar(∆ABC)
ar(∆APO) + ar(||gmDLOP) +ar(∆OLC) =
ar(∆AMO) + ar(||gm BMOP) +ar(∆OQC) …(3)
Subtract eq 1 & 2 from 3
ar(||gmDLOP) = ar(||gm BMOQ)
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