Question

If E, F G and H are respectively the midpoints of the sides AB, BC, CD and AD of a parallelogram ABCD, show that ar(EFGH) =1/2 ar(ABCD).

Answer 1

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▶ 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 .


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


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



#BeBrainly.

Answer 2

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$\bold {concept}$

If the question is “If P, Q, R and S are respectively the mid-points of the sides of a parallelogram ABCD taken in order then show that area (PQRS) is half of area (ABCD”, then the answer to this question is:
ABCD is a parallelogram
∴ AB = CD and AB||CD

⇒ ½ AB = ½ CD and AP||DR
⇒AP = DR and AP||DR
Thus, APRD is a parallelogram
ΔSPR and parallelogram APRD lie on the same base PR and between the same parallels PR and AD.
∴ area (ΔSPR) = ½ area (APRD) … (1)
Similarly, it can be proved that
area (ΔPQR) = ½area (PBCR) … (2)

Adding both sides of equations (1) and (2)

area (ΔSPR) + area (ΔPQR) = ½ [area (APRD) + area (PBCR)]
⇒ area (PQRS) = ½ area (ABCD)


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