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

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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.

=> 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