Question

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

Answer 1

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.

Answer 2

=> 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