If e, f, g, h ar respectively the mid points of the sides of a parallelogram abcd, show that ar(efgh)= 1/2 ar(abcd)
Answers
Answer:
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).
Read more on Brainly.in - https://brainly.in/question/1797838#readmoreStep-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).
Read more on Brainly.in - https://brainly.in/question/1797838#readmoreStep-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 PROVED
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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