Question

area of parallelogram abcd is x cm square .if EFG and H are mid-points of the sides,find the area of EFGH

Answer 1

Step-by-step explanation:

Given:

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

To Prove sides of a parallelogram)

$ar(efgh) = \frac{1}{2} ar = abcd$

Construction:

H and F are joined.

H and F are joined.Proof:

H and F are joined.Proof:AD∥BC and AD=BC (Opposite sides of a parallelogram)

H and F are joined.Proof:AD∥BC and AD=BC (Opposite sides of a parallelogram) →

$\frac{1}{2} ad = \frac{1}{2} bc$

Also,

Also,AH∥BF and and DH∥CF

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

Also,AH∥BF and and DH∥CF⇒AH=BF and DH=CF ∣ H and F are mid pointsThus,

Also,AH∥BF and and DH∥CF⇒AH=BF and DH=CF ∣ H and F are mid pointsThus, ABFH and HFCD are parallelograms.

Also,AH∥BF and and DH∥CF⇒AH=BF and DH=CF ∣ H and F are mid pointsThus, ABFH and HFCD are parallelograms.Now,

Also,AH∥BF and and DH∥CF⇒AH=BF and DH=CF ∣ H and F are mid pointsThus, 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=

$\frac{1}{2}$

AR(ABFH)____(1)

Also,

″→Area of GHF=

$\frac{1}{2}$

AR(HFCD)___(2)

Adding (i) and (ii),

Adding (i) and (ii),Area of △EFH+ area of △GHF =

$\frac{1}{2}$

ar(ABFH)+

$\frac{1}{2}$

ar(HFCD)

ar(HFCD)⇒ Area of EFGH= Area of ABFH

ar(HFCD)⇒ Area of EFGH= Area of ABFH⇒ar(EFGH)= 1/2

ar(ABCD)