Question

if E,F,G AND H ARE RESPECTIVELY THE MID POINT OF THE SIDES OF A PARALELLOGRAM ABCD SHOW THAT ar(EFGH) = HALF OF ar(ABCD)

Answer 1

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)   
  

Answer 2

Step-by-step explanation:

▶ Given :-

→ ABCD is a ||gm .

→ E, F G and H are respectively the midpoints of the sides AB, BC, CD and AD of a ||gm ABCD .

▶ To Prove :-

→ ar(EFGH) = ½ ar(ABCD).

▶ Construction :-

→ Join FH , such that FH || AB || CD .

Since, FH || AB and AH || BF .

→ •°• ABFH is a ||gm .

∆EFH and ||gm ABFH are on same base FH and between the same parallel lines .

•°• ar( ∆ EFH ) = ½ ar( ||gm ABFH )............. (1) .

▶ Now,

Since, FH || CD and DH || FC .

→ •°• DCFH is a ||gm .

∆FGH and ||gm DCFH are on same base FH and between the same parallel lines .

•°• ar( ∆ FGH ) = ½ ar( ||gm DCFH )............. (2) .

▶ On adding equation (1) and (2), we get

==> ar( ∆ EFH ) + ar( ∆ FGH) = ½ ar( ||gm ABFH ) + ½ ar( ||gm DCFH ) .

==> ar( ∆ EFH ) + ar( ∆ FGH) = ½ [ ar( ||gm ABFH ) + ar( ||gm DCFH ) ] .

•°• ar(EFGH) = ½ ar(ABCD).

✔✔ Hence, it is proved ✅✅.

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