If E, F G and H are respectively the midpoints of the sides AB, BC, CD and AD of a parallelogram ABCD, show that ar(EFGH)=1/2ar(ABCD).
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Given that ABCD is a parallelogram .
E , F , G and H are the mid points of
the sides .
Join E , G .
Now ,
∆EFG and EBCG lie on the same base
EG and between the same parallels
EG // BC .
Therefore ,
∆EFG = ( 1/2 ) area ( EBCG ) -------( 1 )
Similarly ,
∆EHG = ( 1/2 ) Area ( EGDA ) -------( 2 )
Adding ( 1 ) and ( 2 ) , we get
∆EFG+∆EHG=(1/2)area(EBCG+(1/2)area(EGDA)
=> Area ( EFGH ) = 1/2[ area(EBCG)+(EGDA)]
Area ( EFGH ) = 1/2 Area ( ABCD )
Hence proved.
••••
E , F , G and H are the mid points of
the sides .
Join E , G .
Now ,
∆EFG and EBCG lie on the same base
EG and between the same parallels
EG // BC .
Therefore ,
∆EFG = ( 1/2 ) area ( EBCG ) -------( 1 )
Similarly ,
∆EHG = ( 1/2 ) Area ( EGDA ) -------( 2 )
Adding ( 1 ) and ( 2 ) , we get
∆EFG+∆EHG=(1/2)area(EBCG+(1/2)area(EGDA)
=> Area ( EFGH ) = 1/2[ area(EBCG)+(EGDA)]
Area ( EFGH ) = 1/2 Area ( ABCD )
Hence proved.
••••
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