Question

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.

Answer 1

Given: Parallelogram $ABCD$ and rectangle $ABEF$ are on same base and have equal areas.
$AB=CD$ ...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 $\square ABEF=AB+BE+EF+FA$

Since, $CD=EF$ and $AD>AF,BC>BE$

$\therefore$ Perimeter of parallelogram $ABCD>$ Perimeter of rectangle $ABEF$