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


Answer :


We have a parallelogram ABCD and rectangle ABEF such that

ar(||gm ABCD) = ar( rect. ABEF)

AB = CD [Opposite sides of parallelogram]
and AB = EF [Opposite sides of a rectangle]
\(\Rightarrow \) CD = EF
\(\Rightarrow \) AB + CD = AB + EF … (1)

BE < BC and AF < AD [In a right triangle, hypotenuse is the longest side]

\(\Rightarrow \) (BC + AD) > (BE + AF) …(2)

From (1) and (2), we have

(AB + CD) + (BC+AD) > (AB + EF) +( BE + AF)
\(\Rightarrow \) (AB + BC + CD + DA) > (AB + BE + EF + FA)
\(\Rightarrow \) Perimeter of parallelogram ABCD > Perimeter of rectangle ABEF.

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