Question

E,F,G,H are respectively the mid points of the sides of a parallelogram ABCD show that ar(EFGH) =ar(ABCD)

Answer 1

Right question : If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that

ar (EFGH) =ar (ABCD)

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Given : ABCD is a parallelogram and E, F, G and H are the mid points of llgm ABCD.

To prove : ar(EFGH) = ar(ABCD)

Construction : join H and F

Proof :

AD ll BC and AD =BC ( opposite sides are equal and parallel in llgm)

AB = CD (Opposite sides of a llgm are equal)

So we can say

1/2 AD = 1/2 BC

And AH || BF

AH = BF and AH || BF

Therefore, ABFH is a parallelogram

Now

Area of triangle HEF = 1/2 × Area (ABFH)...... (i)

Similarly

Area of triangle HGF = 1/2 × Area (HDCF)..... (ii)

Adding eq i and ii

Area of triangle HEF + Area of triangle HGF = 1/2 Area (ABFH) + 1/2 Area (HDCF)

Area (EFGH)= 1/2 [Area (ABFH) + Area (HDCF)]

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

HENCE proved