# Parallelogram ABCD and rectangle ABEF have same base and equal areas. prove that perimeter of parallelogram is greater than that of rectangle.

2
by Nicknameisnickname

2016-03-08T14:16:00+05:30
Since opposite sides of a|| gm and rectangle are equal.
Therefore AB = DC               [Since ABCD is a || gm]
and,           AB = EF               [Since ABEF is a rectangle]
Therefore  DC = EF                                                                     ... (1)
⇒          AB + DC = AB + EF  (Add AB in both sides)                   ... (2)
Since, of all the segments that can be drawn to a given line from a point not lying on it, the perpendicular segment is the shortest.
Therefore BE < BC and AF < AD
⇒ BC > BE and AD > AF
⇒ BC + AD > BE + AF                   ... (3)
Adding (2) and (3), we get
AB + DC + BC + AD > AB + EF + BE + AF
⇒ AB + BC + CD + DA > AB + BE + EF + FA
⇒  perimeter of || gm ABCD > perimeter of rectangle ABEF.
Hence,the perimeter of the parallelogram is greater than that of the rectangle.
2016-03-08T15:44:18+05:30
=> Opp sides of ||gm and rect are equal.
=> AB = DC
=> AB = EF
=> DC = EF (1)
=> AB + DC = AB + EF  (Adding AB in both sides) (2)
=> The perpendicular segment is the shortest
=> BE < BC & AF < AD
=> BC > BE & AD > AF
=> BC + AD > BE + AF (3)
=> Adding (2) and (3)
=> AB + DC + BC + AD > AB + EF + BE + AF
=> AB + BC + CD + DA > AB + BE + EF + FA
=> P of  ABCD > P of ABEF
Proved
it was correct
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