1.

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).

Answer»

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.

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