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2015-05-10T11:18:27+05:30

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GIVEN :-

⇒ E,F,G and H are respectively the mid points of the sides of a parallelogram ABCD.

CONSTRUCTION ;-

⇒ Draw a line HF which should be parallel to CD and AB.

TO PROVE :-

⇒ Ar. ( EFGH ) = Half of . Ar ( ABCD )

PROOF :-

⇒ We know that  AB is parallel to HF, so we can say that ,

ABHF is a parallelogram ( opp. sides of a parallelogram is parallel and                                                     equal)

⇒ Now in , Parallelogram ABHF lies and Δ EFH lies on the same base HF and between same parallels AB and HF.


 Hence ,
 
                           
Ar.( EFH )  =  1 / 2  ar  ( ABFH )                            
 
                                                      ⇒         ( Equation - 1 )

⇒ Also, HF is parallel and equal to DC

So

,DCFH is a parallelogram ( opp.sides of a parallelogram are equal and                                                 parallel )

⇒Also ,we know that Parallelogram DCFH and Triangle GFH lies on the same base HF and between same parallels DC and HF.


Hence,
 
                      Ar.  ( GFH )  =  1 / 2 Ar. ( DCFH ) 
 
 
                                                        
⇒ { Equation - 2  }

FROM EQUATION 1 AND 2 WE GET ,


⇒Ar  .( EFH )  +   Ar.  ( GFH )  =  1 / 2  Ar. ( ABFH )  +  1 / 2  Ar .( DCFH )
               
 ⇒Ar.  ( EFGH )                       = 1 / 2 (  Ar. ( ABFH ) +  Ar. ( DCFH )
               
⇒Ar .  ( EFGH )                       = 1 / 2  ( Ar.( ABCD )
               
⇒Ar.   ( EFGH )                       = 1 / 2 Ar (ABCD)  

Therefore it is proved that ,

         
  Ar.   ( EFGH )            = 1 / 2 Ar (ABCD)   
  

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