# Parallelogram ABCD and rectangle ABEF are on the same base and have equal areas. Show the perimeter of parallelogram is greater than rectangle. given: ABCD is a rectangle and EFCD is a parallelogram, with the same base CD and have equal areas.

TPT: perimeter of the parallelogram EFCD > perimeter of the rectangle

proof:

since parallelogram and rectangle have equal areas with same base CD, therefore it will be between same set of parallel lines.

CD=EF............(1) [opposite sides of the parallelogram]

CD=AB...........(2) [opposite sides of the rectangle]

from (1) and (2), EF=AB...........(3)

in the triangle DAE,

since ∠DAE=90 deg

ED>AD [since length of the hypotenuse is greater than other sides]..........(4)

perimeter of parallelogram EFCD

=EF+FC+CD+DE

=AB+FC+CD+DE  [using (3)]

which is the perimeter of the rectangle ABCD

therefore perimeter of parallelogram with the same base and with equal areas is greater than the perimeter of the rectangle.

• 73
In llgm ABCD and rectangle ABEF
AB=FE (opposite sides of llgm and res are equal)
& AB=CD(opposite sides of llgm and res are equal)
As in ∆AFC <F=90 and AC is hypotonus
Therefore, AF<AC
As AF=BE
AC=BD
=> AF< AC (i)
BE <BD (ii)
PERIMETER OF llgm ABCD= AB+BC+CD+DA (iii)
PERIMETER OF rectangle ABEF= AB+BE+EF+AF (iv)
FROM (i), (ii) ,(iii)& (iv) we get
AB+BC+CD+DA> AB+BE+EF+AF
=> PERIMETER OF llgm > PERIMETER OF rectangle ABEF

• 15
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