Parallelogrom ABCD and rectangle ABEF have the same base AB and also have equal areas.Show that the perimeter of the parallelogram is greater than that of the rectangle.

consider the above diagram

opposite sides of rectangle and parallelogram are equal so

AB= EF and AB= CD so EF=CD

therefore AB+CD=AB+EF  ...... (1)

Now since of all the segments that can be drawn to a given line from a point, the perpendicular segment is the shortest.so,

AF< AD and BE<BC

AD+BC > AF+ BE  ...... (2)

Adding (1) and (2)

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

⇒ Perimeter of parallelogram ABCD > perimeter of rectangle ABEF

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Since they are on the same base and have equal areas, they lie between the same parallels.

Triangles ADF and BCE are right triangles,

AD>AF and BC>BE (hypotenuse is the longest side)

AB=CD and AB=EF

So,CD=EF

Perimeter of //gm ABCD= AB+BC+CD+AD

Perimeter of rectangle ABEF= AB+BE+EF+AF

Since, AD>AF and BC>BE

So, Perimeter of //gm ABCD>Perimeter of rect. ABEF

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