Q. 254.0( 50 Votes )
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 :
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)
1/2 AD = 1/2 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 = 1/2 Area (ABFH) + 1/2 Area (HDCF)
Area (EFGH)= 1/2 [Area (ABFH) + Area (HDCF)]
Area (EFGH) = 1/2 Area (ABCD)
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