Question

If E, F G and H are respectively themidpoints of the sides AB, BC, CD and ADof a parallelogram ABCD, show thatar(EFGH) 1ar(ABCD)2=.

Answer 1

Answer:

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Answer 2

Let us join HF

In parallelogram ABCD,

AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)

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

AD = BC

And AH || BF

AH = BF and AH || BF (H and F are the mid-points of AD and BC)

Therefore, ABFH is a parallelogram

Since,

ΔHEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF

Therefore,

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

Similarly,

It can also be proved that

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

On adding (i) and (ii), we get

Area of triangle HEF + Area of triangle HGF = Area (ABFH) + Area (HDCF)

Area (EFGH)= [Area (ABFH) + Area (HDCF)]

Area (EFGH) = Area (ABCD)