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` > Perimeter of rectange `ABEF` i.e. `A B+B C+C D+A D > A B+B E+E F+A F`

Explanation

Proof: `ABCD` is a parallelogram
`=>AB=CD`

`ABEF` is a rectangle
`=>AB=EF`

Perimeter of `ABCD= AB+BC+CD+AC`

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

Since, `CD=EF` and `AD>AF`, `BC>BE`
`therefore` Perimeter of a parallelogram is more than a rectangle.