Do the given perimeter and area corresponds to many shapes?

Question

I have a perimeter P and area A of a planar shape. How to prove that there are many shapes that corresponds to those perimeter and area values?

Answer 1

If $P^2=4A$, there is only one planar shape (up to translation) that matches: the disc of radius $\frac1{2\pi}P$.

If $P^2<4A$, there is no such shape.

If $P^2>4A$, you may start with a disc of area $A$, which has too short perimeter. If you deform a tiny part of the perimeter outside and another part inside, you can achieve that the aea remains unchanged, while the perimeter takes on as large a value as you want. As there is a lot of freedom of choice involved in this process, the claim follows.

Answer 2

If you start with a rectangle and have an indent on it, like in shape A, then you can always move that indent to any other place and the perimeter and the area will remain the same, as in shape B (starting with an "outdent" would work too of course.)

Perhaps this proof works for what you have in mind.