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.

Explanation

We have
AB = EF and CD = AB `rArr AB+CD = AB+EF." "…(i)`
We know that, of all the line segment to a given line from a point, outside it, the perpendicular is the least.
`therefore BEltBC and AF lt AD.`
`therefore BC+AD gt BE+AF." "...(ii)`
From (i) and (ii), we get
`AB+BC+CD+ADgtAB+BE+EF+AF.`
Hence, the perimeter of the ||gm is greater than the perimeter of the rectangle.