Question

if E F G and H are respectively the midpoint of the side of a parallelogram ABCD so that
at(EFGH)=1/2ar(ABCD)/

Answer 1

given, ABCD a parallelogram in which E,F,G,H, are mid points of AB,BC,CD,AD. Then

to prove, ar(EFGH) = ar(ABCD)

Construction, Draw HF parallel to AB and CD


Proof: AB is parallel and equal to HF . Therefore, ABFH is a parallelogram
Since, triangle EFH and parallelogram ABFH lies on the same base HF and between same parallels AB

and HF
Therefore,

ar(EFH) = 1/2 ar (ABFH) ( 1)

Now, DC is parallel and equal to HF. Therefore,DCFH is a parallelogram Since, triangle GFH and parallelogram DCFH lies on the same base HF and between same parallels DC and HF.
Therefore, ar(GFH) = 1/2 ar (DCFH) (2)

From 1 and 2 we get,
ar(EFH) + ar (GFH) = 1/2 ar (ABFH) + 1/2 ar (DCFH)
ar (EFGH) = 1/2 ( ar(ABFH) + ar(DCFH))

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

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

proved that

I hope it is helped you

follow

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 ✅✅.

THANKS