Question

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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 the same base AB and have equal areas. To Prove: The perimeter of the parallelogram ABCD is greater than that of rectangle ABEF. ... Then, perimeter of the parallelogram ABCD = 2(AB + AD) and, perimeter of the rectangle ABEF = 2(AB + AF).

Answer 2

For , diagram refer to above picture

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Given :-

• Paralellogram ABCD & Rectangle ABEF are on same Base AB & have equal Areas.

To show :-

• Perimeter of the parallelogram is greater than that of the rectangle.

Proof :-

We know that, the opposite sides of a Parallelogram and rectangle are equal.

So,

AB = DC. [As ABCD is a || gm]

and, AB = EF. [As ABEF is a rectangle]

, DC = EF. ........................… (i)

Adding AB on both sides, we get,

⇒AB + DC = AB + EF …..... (ii)

We know that, the perpendicular segment is the shortest of all the segments that can be drawn to a given line from a point not lying on it.

BE < BC and AF < AD

⇒ BC > BE and AD > AF

⇒ BC+AD > BE+AF … (iii)

Adding (ii) and (iii), we get

AB+DC+BC+AD > AB+EF+BE+AF

⇒ AB+BC+CD+DA > AB+ BE+EF+FA

⇒ perimeter of || gm ABCD > perimeter of rectangle ABEF.

The perimeter of the parallelogram is greater than that of the rectangle.

Hence, Proved.