Question

parallelogram ABCD and rectangle ABEF have same base AB and also 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. therefore parallelogram and rectangle are between the same set of parallel lines.

TPT: perimeter of ABCD > perimeter of ABEF

proof:

in the right angled triangle ADF , 

AD>AF [since hypotenuse is the longest side of the right angled triangle]........(1)

perimeter of parallelogram ABCD = AB+BC+CD+DA

=AB+AD+AB+AD  [since opposite sides of a parallelogram are equal]

=2(AB+AD)............(2)

perimeter of rectangle ABEF = AB+BE+EF+FA

=AB+AF+AB+AF [since opposite sides of a rectangle are equal]

= 2(AB+AF)

=2AB+2AF

<2AB+2AD  [using (1)]

<2(AB+AD)

< perimeter of parallelogram ABCD [using (2)]

thus the perimeter of the parallelogram is greater than the perimeter of rectangle.

hope this helps you. 

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