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

Hey mate here is ur answer ......

GIVEN: A parallelogram ABCD , where E , F , G , H are the mid - points of AB , BC , CD , and AD respectively.

TO PROVE : Area ( EFGH) = 1/ 2 Area ( ABCD )

PROOF: join H & F

Now,

AD || BC { opposite sides of parallelogram are parallel }

DH || CF { parts of parallel lines are parallel }

Also AD = BC { Opposite sides of parallelogram are equal }

1/2 AD = 1/2 BC { H is the mid point of AD }

DH = CF { F is the mid point of BC }

In DHFC ,

DH = CF & DH || CF

since one pair of opposite sides are equal and parallel .

so ,

DHFC ia a parallelogram..

Similarly ,

HABF is a parallelogram

since DHFC is a parallelogram.

DC || HF { opposite sides of parallelogram are parallel} .

As triangle HGF and parallelogram DHFC are on same base HF & between the same parallel lines DC and HF .______________(1)

Area ( Triangle HGF ) = 1/2 ( Area DHFC )

Similarly ,

Since HABF is a parallelogram.

HF || AB

As triangle HEF and ||gm HABF are on same base HF and between the same parallel lines AB and HF ______________(2)

By adding from (1) & (2)

Area ( Triangle HGF ) + Area ( HEF ) = 1/2 Area ( DHFC ) + Area ( HFAB )

Area ( HEFG ) = 1/2 Area ( DHFC ) + Area ( HFAB )

so ,

Therefore , Area ( EFGH ) = 1/2 Area ( ABCD )

HENCE IT IS PROOVED .....

HOPE IT HELPS UUUUUUUUU....... :)

BE BRAINLY $$$$$........

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