Parallelogram `A B C D` and rectangle `A B E F` have the same base `A B` and also have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle. GIVE : `A ^(gm)A B C D` and a rectangle `A B E F` with the same base `A B` and equal areas. TO PROVE : Perimeter of `^(gm)A B C D > P e r i m e t e rofr e c t a nge l` ABEF i.e. `A B+B C+C D+A D > A B+B E+E F+A F`

Explanation

AB=DC-(1)

AB=EF-(2)

DC=EF

adding equation 1 and 2

AB+DC=AB+EF-(3)

`BEltBC`and`AFltAD`

`BC>BE` and`ADltAF`

BC+AD>BE+AF-(4)

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