Question

If E, F, G & H are respectively the midpoint of the sides of a parallelogram ABCD, show that ar(EFGH)=1/2ar(ABCD).​

Answer 1

Given :

◉ ABCD is a parallelogram.

◉ E is the mid point of AB

◉ F is the mid point of BC

◉ G is the mid point of CD

◉ H is the mid point of DA

To Prove :

▪ ar EFGH = ½ ar ABCD

Construction :

༓ HF ∥ AB ∥ CD

Proof :

⛋ As ∆HFG and parallelogram ABHF are on the same base and between same parallels.

• ar HFE = ½ ar ABHF .....[Equation (i)]

→ HF ∥ CD

• ar HFG = ½ ar HFCD (same base and between same parallels) .......[Equation (ii)]

Adding equation (i) and equation (ii) we get :

→ ar HFE + ar HFG = ½ (ABHF + HFCD)

→ ar EFGH = ½ ar ABCD

Hence, proved ! .-.

Answer 2

Given:-

• ABCD is a parallelogram

• E,F,G,H are mid points of the sides AB, BC, CD, AD respectively of parallelogram ABCD

To Find:-

• area of (EFGH) = ½ area of (ABCD)

Solution:-

→Since, FH || AB and AH || BF.

→ABFH is a ||gm.

→△EFH and ||gm ABFH are on same base FH and between the same parallel lines.

✦ar(△EFH) = ½ ar(||gm ABFH) -----》(1)✦

Now,

→Since, FH || CD and DH || FC.

→ DCFH is a ||gm.

→△FGH and ||gm DCFH are on same base FH and between the same parallel lines.

✦ar(△FGH) = ½ ar(||gm DCFH) -----》(2)✦

Adding Equation (1) and (2),

✦ar(△EFH ) + ar( △FGH) = ½ ar( ||gm ABFH) +ar(||gm DCFH)✦

✦ar(EFGH) = ½ ar(ABCD)✦

❉HENce PrOved❉