Question

in E, F, G and H are respectively the midpoints of parallelogram ABCD show that ar(EFGH) =1/2ar(ABCD)​

Answer 1

Answer:

Given:

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

To Prove:

ar(EFGH)=

2

1

ar(ABCD)

Construction:

H and F are joined.

Proof:

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

⇒

2

1

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=

2

1

ar(ABFH) --- (i)

Also,

Area of GHF=

2

1

ar(HFCD) --- (ii)

Adding (i) and (ii),

Area of △EFH+ area of △GHF =

2

1

ar(ABFH)+

2

1

ar(HFCD)

⇒ Area of EFGH= Area of ABFH

⇒ar(EFGH)=

2

1

ar(ABCD)