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

Parallelogram $ABCD$ and rectangle $ABEF$ on same base $AB$ have equal areas.           ....Given
$AB=CD$        ...opposite side of parallelogram
$AB=EF$         ...opposite side of rectangle

and, $\angle F=\angle E=90^{\circ }$
$AD>AF$ and $BC>BE$     ....(Hypotenuse of right angle is greater than other sides )

Now,
Perimeter of $\square ABCD=AB+BC+CD+DA$
Perimeter of $\square ABEF=AB+BE+EF+FA$