Given,
|| gm ABCD and a rectangleABEF have the same base AB and equal areas.
To prove,
Perimeter of || gmABCD is greater than the perimeter of rectangle ABEF.
Proof,
We know that, theopposite sides of a|| gm and rectangle are equal.
, AB = DC [As ABCD isa || gm]
and, AB = EF [As ABEFis a rectangle]
, DC = EF … (i)
Adding AB on bothsides, we get,
⇒AB + DC = AB + EF …(ii)
We know that, theperpendicular segment is the shortest of all the segments that can be drawn toa 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 || gmABCD > perimeter of rectangle ABEF.
, the perimeter of theparallelogram is greater than that of the rectangle.
Hence Proved.