Step-by-step explanation:
Given
E, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.
To Prove
ar (EFGH) = ½ ar(ABCD)
Construction
Let us join HF.
Proof
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
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,
∴ Area (ΔHEF) = 1/2Area (ABFH) … (1)
Similarly, it can be proved that
On adding equations (1) and (2), we get
area of ΔEFH + area of ΔGHF = ½
⇒ area of EFGH = area of ABFH
∴ ar (EFGH) = ½ ar(ABCD)
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Verified answer
Step-by-step explanation:
Given
E, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.
To Prove
ar (EFGH) = ½ ar(ABCD)
Construction
Let us join HF.
Proof
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
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,
∴ Area (ΔHEF) = 1/2Area (ABFH) … (1)
Similarly, it can be proved that
On adding equations (1) and (2), we get
area of ΔEFH + area of ΔGHF = ½
⇒ area of EFGH = area of ABFH
∴ ar (EFGH) = ½ ar(ABCD)