In the given figure, a ‖gm ABCD and a rectangle ABEF are of equal area. Then,
Given: Area (‖gm ABCD) = Area (rectangle ABEF)
Consider ΔAFD
Clearly AD is the hypotenuse
∴ AD > AF
Perimeter of Rectangle ABEF = 2× (AB + AF) –1
Perimeter of Parallelogram ABCD = 2× (AB + AD) –2
On comparing –1 and –2, we can see that
Perimeter of ABCD > perimeter of ABEF (∵AD > AF)
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