Question

If E, F G and H are respectively themidpoints of the sides AB, BC, CD and ADof a parallelogram ABCD, show thatar(EFGH) 1ar(ABCD)2=.

Answer 1

Answer:

▶ 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 .

\huge \green{ \mid \underline{ \overline{ \sf Solution :- } } \mid}∣

Solution:−

∣

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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