abcd is a trapezium such that ab is parallel to cd. ef is parallel to ab and cd such that ef divides trapezium in two equal areas. find ef
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ABCD is trapezium in which AB∥DC.
EF is parallel to side DC.
Then we have AB∥DC∥EF.
Hence we have also trapezium ABFE and trapezium EFCD.
Let AP be the perpendicular to DC and this intersects EF at Q.
AQ will be perpendicular to EF.
For △APD and △AQE we have EAAD=AQAP=2
This gives AP=2AQ
i.e, AQ=QP
Consider the area we have area ABCD= area ABFE+ area EFCD
(21)AP×(AB+DC)=(21)AQ×(AB+EF)+(21)QP×(EF+DC)
⇒AP(AB+DC)=AP×2AB+AP×2EF+AP×2EF+AP×2DC
⇒AP×2AB
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