plzz answer this ques no 4.a
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Since, triangle APD and parallelogram ABCD lie on the same base and between same parallels .
So ar(APD) = 1/2ar( ABCD)
Thus ar ABP + arDPC = 1/2 ar(ABCD)
Hence proved
So ar(APD) = 1/2ar( ABCD)
Thus ar ABP + arDPC = 1/2 ar(ABCD)
Hence proved
mahakagarwal22:
but how do we know that angleABP +angle CDP= area 1/2 of parallelogram abcd
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