Question

2. If E,F,G and Hare respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH)= 1/2 ar (ABCD).​

Answer 1

Answer:

w=12

Explanation:

Let w be the width and l the length.

Then

1) w=13l→l=3w

2) 2(w+l)=96→w+l=48

Let's substitute l=3w in w+l=48 and get:

w+3w=48→4w=48→w=12

Answer 2

Answer:

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

Area (ΔHGF) = 1/2 Area (HDCF) … (2)

On adding equations (1) and (2), we get

area of ΔEFH + area of ΔGHF = ½ area of ABFH + ½ area of HFCD

⇒ area of EFGH = area of ABFH

∴ ar (EFGH) = ½ ar(ABCD)

Hence Proved

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