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.

Given: Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas.

To Prove: The perimeter of the parallelogram ABCD is greater than that of rectangle ABEF.
Proof: Let ABCD be the parallelogram and ABEF be the rectangle on the same base AB and between the same parallels AB and FC. Then, perimeter of the parallelogram ABCD = 2(AB + AD) and, perimeter of the rectangle ABEF = 2(AB + AF).

In ΔADF,
∵ ∠AFD = 90°
∴ ∠ADF is an acute angle. (< 90°)
| Angle sum property of a triangle
∴ ∠AFD > ∠ADF ∴ AD > AF
Side opposite to greater angle of a triangle is longer ∴ AB + AD > AB + AF
∴ 2(AB + AD) > 2(AB + AF)
∴ Perimeter of the parallelogram ABCD > Perimeter of the rectangle ABEF.