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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​

Answer»

Answer:

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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