Question -
Answer -
Given,
E, F, G and H are themid-points of the sides of a parallelogram ABCD respectively.
To Prove,
ar (EFGH) = ┬╜ ar(ABCD)
Construction,
H and F are joined.
Proof,
AD || BC and AD = BC(Opposite sides of a parallelogram)
тЗТ ┬╜ AD = ┬╜ BC
Also,
AH || BF and and DH ||CF
тЗТ AH = BF and DH = CF(H and F are mid points)
тИ┤, ABFH and HFCD areparallelograms.
Now,
We know that, ╬ФEFH andparallelogram ABFH, both lie on the same FH the common base and in-between thesame parallel lines AB and HF.
тИ┤ area of EFH = ┬╜ areaof ABFH тАФ (i)
And, area of GHF = ┬╜area of HFCD тАФ (ii)
Adding (i) and (ii),
area of ╬ФEFH┬а+area of ╬ФGHF = ┬╜ area of ABFH┬а+ ┬╜ area of HFCD
тЗТ area of EFGH = areaof ABFH
тИ┤ ar (EFGH) = ┬╜ar(ABCD)