Question

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

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Answer 1

Given :

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

To prove :

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

Construction :

H and F are joined.

Proof :

AD∥BC and AD=BC (Opposite sides of a parallelogram)

⇒ 1/2 AD= 1/2 BC

Also,

AH∥BF and and DH∥CF

⇒AH=BF and DH=CF ∣ H and F are mid points

Thus,

ABFH and HFCD are parallelograms.

Now,

△EFH and ||gm ABFH lie on the same base FH and between the same parallel lines AB and HF.

∴ Area of EFH= 1/2 ar(ABFH) --- (i)

Also,

Area of GHF= 1/2 ar(HFCD) --- (ii)

Adding (i) and (ii),

Area of △EFH+ area of △GHF =1/2 ar(ABFH)+ 1/2 ar(HFCD)

⇒ Area of EFGH= Area of ABFH

⇒ar(EFGH)= 1/2 ar ( ABCD )

Answer 2

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

To Prove:ar(EFGH)= 21ar(ABCD)

Construction:

H and F are joined.

Proof:

AD∥BC and AD=BC (Opposite sides of a parallelogram)⇒2

AD= 2

1

BC

Also,

AH∥BF and and DH∥CF

⇒AH=BF and DH=CF ∣ H and F are mid points

Thus,

ABFH and HFCD are parallelograms.

Now,

△EFH and ||gm ABFH lie on the same base FH and between the same parallel lines AB and HF.

∴ Area of EFH= 21 ar(ABFH) --- (i)

Also,

Area of GHF= 21ar(HFCD) --- (ii)

Adding (i) and (ii),

Area of △EFH+ area of △GHF = 2 ar(ABFH)+21ar(HFCD)

⇒ Area of EFGH= Area of ABFH

⇒ar(EFGH)=21 ar(ABCD)