p is the poiint on ab in parallelogram abcd prove that area triangle adp+area of triangle bdp =1/2 area abcd
Mathexpert:
I think, it is not triangle bdp but it should be triangle bcp
yes it is triangle bcp
ok..check the sollution.....and mark it as best
thnk u
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ΔCPD and ||gm ABCD are on the same base CD and between two parallel lines AB and CD.
Therefore, area of ΔCPD = area of ||gm ABCD
But area of ||gm ABCD = area of ΔCPD + area of ΔADP + area of ΔBCP
⇒area of ||gm ABCD = 1/2 area of ||gm ABCD + area of ΔADP + area of ΔBCP
⇒1/2 area of ||gm ABCD = area of ΔADP + area of ΔBCP
Therefore, area of ΔCPD = area of ||gm ABCD
But area of ||gm ABCD = area of ΔCPD + area of ΔADP + area of ΔBCP
⇒area of ||gm ABCD = 1/2 area of ||gm ABCD + area of ΔADP + area of ΔBCP
⇒1/2 area of ||gm ABCD = area of ΔADP + area of ΔBCP
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in llgm ABCD and triangle DPC lie on the same base DC and between same parallel's AB and DC
so area(DPC) = 1/2 area(ABCD)
ABCD - DPC = 1/2 area(ABCD)
area(ADP) + area(PCB) = 1/2 area(ABCD)
so area(DPC) = 1/2 area(ABCD)
ABCD - DPC = 1/2 area(ABCD)
area(ADP) + area(PCB) = 1/2 area(ABCD)
mrk mine as the best please
sorry bcos he gave the answer first
okay
sorry again i mreally sorry there annnd here also i couldnt mark you the best really sorry when i post another question and if you answer last also i will mark you the best
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