if e,f,g and h are respectively the mid - points of the sides of a paralleogram abcd , show that ar(EFGH)=1/2ar(ABCD). draw a also figure
Answers
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)
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