Question

E,F,G and H are the mid-points of the sides of the parallelogram ABCD. Show that ar(EFGH)=1/2 ar(ABCD).​

Answer 1

Given: ABCD a parallelogram in which          E,F,G,H, are mid points of AB,BC,CD,AD. 

Then to prove: ar(EFGH) = ar(ABCD) 

Construction: Draw HF parallel to AB and CD. 

Proof:  AB is parallel and equal to HF .

Therefore, ABFH is a parallelogram Since, triangle EFH and parallelogram ABFH lies on the same base HF and between same parallels AB and HF.  

Therefore, ar(EFH) = 1/2 ar (ABFH)   -(1)                        

 Now, DC is parallel and equal to HF. Therefore,DCFH is a parallelogram Since, triangle GFH and parallelogram DCFH lies on the same base HF and between same parallels DC and HF. 

Therefore, ar(GFH) = 1/2 ar (DCFH)   -(2)                         

From 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)

at (EFGH) = 1/2 ar (ABCD)                    

  proved

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Answer 2

Answer:

Given,

ABCD is a parallelogram,

EFGH are midpoint of ABCD

construction - HF divide ABCD and EFGH,

As area of triangles on same base between same //gm

As ar(triangleHGF) =1/2 of ar(//gm HFCd) (1)

Similarly ,

ar(triangleEFH )=1/2ar(// GM ABFH) (2)

Add (1) and (2)

ar(Triangle ) HGF+FGH=1/2ar//gm ABFH+DCFH,

As, ar //gm EFGH =1/2 at //gm ABCD