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Hi there! Have questions about your homework? At Brainly, there are 60 million students who want to help each other learn. Questions are usually answered in less than 10 minutes. Try it for yourself by posting a question! :D

Since opposite sides of a|| gm and rectangle are equal. Therefore AB = DC [Since ABCD is a || gm] and, AB = EF [Since ABEF is a rectangle] Therefore DC = EF ... (1) ⇒ AB + DC = AB + EF (Add AB in both sides) ... (2) Since, of all the segments that can be drawn to a given line from a point not lying on it, the perpendicular segment is the shortest. Therefore BE < BC and AF < AD ⇒ BC > BE and AD > AF ⇒ BC + AD > BE + AF ... (3) Adding (2) and (3), 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. Hence,the perimeter of the parallelogram is greater than that of the rectangle.

=> Opp sides of ||gm and rect are equal. => AB = DC => AB = EF => DC = EF (1) => AB + DC = AB + EF (Adding AB in both sides) (2) => The perpendicular segment is the shortest => BE < BC & AF < AD => BC > BE & AD > AF => BC + AD > BE + AF (3) => Adding (2) and (3) => AB + DC + BC + AD > AB + EF + BE + AF => AB + BC + CD + DA > AB + BE + EF + FA => P of ABCD > P of ABEF Proved