Question

|| gm ABCD and rectangle ABEF have the same base AB and are equal in areas. Show that the perimeter of the || gm is greater than that of the rectangle.

Answer 1

We have:
A parallelogram ABCD and a rectangle ABEF are on the same base AB and in between the same parallel lines.
AB = CD and AB = FE
i.e., CD = FE
∴ AB + CD = AB + FE ...(i)
Now, we know that AD > AF and BC > BE. (∵ Hypotenuse is the longest side of a triangle)
∴​ AD + BC > AF + BE ...(ii)
We know that:
Perimeter of ABCD = AB + BC + CD + AD
Perimeter of ABEF = AB + BE+ FE+ AF
​From (i) and (ii), we get:
∴ AB+ CD + AD + BC > AB + FE+ AF + BE
Hence, the perimeter of ABCD is greater than that of ABEF.