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)
​

Answer 1

Step-by-step explanation:

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Answer 2

If we draw an imaginary line between point H and F, then AH = BF & AB = HF.

And AB // HF & AH // BF.

So, ABFH is a parallelogram.

In the same way, HFCD is a parallelogram.

In parallelogram ABFH,

∆HEF and //gm ABFH has same base.

If triangle and parallelogram has same base, area of triangle is half of area of parallelogram.

So, ar(∆HEF) = 1/2 × ar(ABHF) - (1)

In the same way,

ar(∆HGF) = 1/2 × ar(HFGD) - (2)

(1) + (2),

ar(∆HEF) + ar(∆HGF) = 1/2 × ar(ABHF) + 1/2 × ar(HFCD)

ar(EFGH) = 1/2 × ar(ABCD)

${\large{\bold{\red{★\:Hence\:proved}}}}$