Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

We have a parallelogram ABCD and rectangle ABEF such that

ar(||gm ABCD) = ar( rect. ABEF)

AB = CD [Opposite sides of parallelogram]
and AB = EF [Opposite sides of a rectangle]
$$\Rightarrow$$ CD = EF
$$\Rightarrow$$ AB + CD = AB + EF … (1)

BE < BC and AF < AD [In a right triangle, hypotenuse is the longest side]

$$\Rightarrow$$ (BC + AD) > (BE + AF) …(2)

From (1) and (2), we have

(AB + CD) + (BC+AD) > (AB + EF) +( BE + AF)
$$\Rightarrow$$ (AB + BC + CD + DA) > (AB + BE + EF + FA)
$$\Rightarrow$$ Perimeter of parallelogram ABCD > Perimeter of rectangle ABEF.