Question

if E F G and H are respectively thr mid poinds of sides of parallelogram abcd show that area efgh = half of area abcd

Answer 1

E,F,G,H are the mid point of ||gmABCD
SO AH=1/2AD. AND BF=1/2BC
AD=BC ,AD||BC
SO In AHBF ||gm,AH||BF
∆EHF=1/2||GM AHBF…...1
Similarly ∆HGF=1/2||gm DHFC ......2
ADD 1 AND 2
||gm EHFG=1/2 ABCD
Hence proved

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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