Question

If E , F , G , H are the mid points of a Parallelogram ABCD respectively , Show that the area of of EFGH = 1/2 of the area of ABCD




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Answer 1

Answer:

▶ Given :-

→ ABCD is a ||gm .

→ E, F G and H are respectively the midpoints of the sides AB, BC, CD and AD of a ||gm ABCD .

▶ To Prove :-

→ ar(EFGH) = ½ ar(ABCD).

▶ Construction :-

→ Join FH , such that FH || AB || CD .

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) .

▶ On adding equation (1) and (2), we get

==> ar( ∆ EFH ) + ar( ∆ FGH) = ½ ar( ||gm ABFH ) + ½ ar( ||gm DCFH ) .

==> ar( ∆ EFH ) + ar( ∆ FGH) = ½ [ ar( ||gm ABFH ) + ar( ||gm DCFH ) ] .

•°• ar(EFGH) = ½ ar(ABCD).

✔✔ Hence, it is proved ✅✅.

THANKS

Answer 2

Given :

• ABCD is a parallelogram .
• E , F , G and H are mid - points of AB , BC , CD and DA respectively .

To prove :

• ar( EFGH) = 1/2 ar(ABCD)

Construction:

• Joining H to F such that the line FH // AB// CD .

Theorem to use :

If a triangle and parallelogram lie on the same base and between the same parallels , the area of triangle is half the area of the parallelogram .

Proof:

In parallelogram ABFH ,

∆EFH and parallelogram ABFH lie on the same base FH and between the same parallels FH // AB .

$\therefore$ ar (∆ EFH) = 1/2 ar (ABFH) ____(1)

In parallelogram FHCD ,

∆ HFG and parallelogram FHCD are lying on the same base FH and between the same parallels FH// CD .

$\therefore$ ar (∆ HFG) = 1/2 ar (FHCD) _____(2)

Adding equation (1) and (2)

ar (∆ EFH) + ar (∆ HFG) = 1/2 ar (ABFH) + 1/2 ar (FHCD)

ar ( EFGH ) = 1/2 ar (ABCD)

Hence proved