Question

parallelogram ABCD and rectangle ABEF are on the same base and between the same parallels have equal areas . Show that perimeter of parallelogram is greater than perimeter of rectangle​

Answer 1

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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.

HOPES THIS HELPS YOU............☺☺

Answer 2

Step-by-step explanation:

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.

HOPES THIS HELPS YOU.☺☺