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.

Solution

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.

Some More Questions From Constructions Chapter

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Constructions

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.

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