Given,

|| gm ABCD and a rectangle ABEF have the same base AB and equal areas.

To prove,

Perimeter of || gm ABCD is greater than the perimeter of rectangle ABEF.

Proof:

We know that, the opposite sides of a|| gm and rectangle are equal.

AB = DC [As ABCD is a || gm]

and, AB = EF [As ABEF is a rectangle]

DC = EF … (i)

Adding AB on both sides, we get,

⇒ AB + DC = AB + EF … (ii)

We know that, the perpendicular segment is the shortest of all the segments that can be drawn to a given line from a point not lying on it.

BE < BC and AF < AD

⇒ BC > BE and AD > AF

⇒ BC + AD > BE + AF … (iii)

Adding (ii) and (iii), 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.

The perimeter of the parallelogram is greater than that of the rectangle.

Hence Proved.