Question

if E F G and H are respectively the mid points of the sides of parallelogram ABCD show that ar(EFGH)=1\2 ar(ABCD)

Answer 1

Diagram

Given : A Δ ABC, E is the mid- point of the median AD.
To prove : ar (BED) =14 ar (ABC) 

Proof :  Since AD is a median of Δ ABC and median divides a triangle into two triangles of equal area. 
Therefore ar (ABD) = ar (ADC) 
⇒ ar (ABD) = 12 ar (ABC) ... (1) 
In Δ ABD, BE is the median 
Therefore ar(BED) = ar (BAE) ... (2) 
⇒ ar(BED)=12ar(ABD)     [ar(BAE)=12ar(ABD) ]
⇒ ar (BED) = 12×12 ar (ABC)                 [Using (1)]
⇒ ar (BED) = 14 ar (ABC) .

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