Question

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 1

Answer:

Step-by-step explanation:

Draw the diagram as per the questions says then also join h with f now AB parallel to HF parallel to CD

Parallelogram ABFH and triangle EHF

Lie between the same parallel ab parallel to HF

Are on same base HF

Therefore ar(EHF)=1/2ar(ABFH)........(1)

Parallelogram DCFH and triangle GHF

Lie between same parallel CD parallel to HF

Are on same base HF

Therefore ar(GHF)=1/2ar(DCFH)......(2)

ADDING 1 AND 2

ar(EHF)+ar(GHF)=1/2ar(ABFH)+ar(DCFH).

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

Answer 2

Step-by-step explanation:

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 ✅✅.

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