In Fig., if parallelogram ABCD and rectangle ABEM are of equal area, then :

In rectangle ABEM,

AB = EM …(eq.i) .. [sides of a rectangle]

In parallelogram ABCD,

CD = AB …(eq.ii)

Adding, equations (i) and (ii),

We get, AB + CD = EM + AB

We know that The perpendicular distance between two parallel sides of a parallelogram is always less than the length of the other parallel sides.

BE < BC and AM < AD...[because, in a right-angled triangle, the hypotenuse is greater than the other side].

On adding both above inequalities, we get:

SE + AM < BC + AD or BC + AD > BE + AM

On adding AB + CD on both sides, we get:

⇒ AB + CD + BC + AD> AB + CD + BE + AM

⇒ AB + BC + CD + AD > AB + BE + EM + AM    [∴ CD = AB = EM]

Hence,

We get, The perimeter of parallelogram ABCD > perimeter of rectangle ABEM