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.