efgh is a parallelogram with a and b mid point of sides gh and ef respectively then show that area of afph is half the area of total efgh
Answers
Answer:
Given: ABCD a parallelogram in which E,F,G,H, are mid points of AB,BC,CD,AD.
Then to prove: ar(EFGH) = ar(ABCD)
Construction: Draw HF parallel to AB and CD.
Proof: AB is parallel and equal to HF .
Therefore, ABFH is a parallelogram Since, triangle EFH and parallelogram ABFH lies on the same base HF and between same parallels AB and HF.
Therefore, ar(EFH) = 1/2 ar (ABFH) -(1)
Now, DC is parallel and equal to HF. Therefore,DCFH is a parallelogram Since, triangle GFH and parallelogram DCFH lies on the same base HF and between same parallels DC and HF.
Therefore, ar(GFH) = 1/2 ar (DCFH) -(2)
From 1 and 2 we get,
ar(EFH) + ar (GFH) = 1/2 ar (ABFH) + 1/2 ar (DCFH)
ar (EFGH) = 1/2 ( ar(ABFH) + ar(DCFH)) ar (EFGH) = 1/2 ( ar(ABCD)) ar (EFGH)
at (EFGH) = 1/2 ar (ABCD)
proved
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Step-by-step explanation:
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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