Is there any theorem about figures of equal area and perimeter being congruent? I had an idea, that all geometric objects, that are different, as they're not a translation, rotation, and a reflection of one another cannot have the same area AND perimeter, as compared to ONE ANOTHER. They can't be CONGRUENT. 
If the shapes are similar there is no similar shapes can contradict this "idea", or that is what I think. 
I know that there was some idea on this, is there any theorem, or specific idea, which this is expressed? 
 A: Counter-example without words (except for these):


A: This idea is wrong, however attractive.
Let's consider a polygon with $n$ unequal sides.  The side lengths determine the perimeter, so any rearrangement of the sides gives the same perimeter.
On the other hand any arrangement of the sides that inscribes the polygon in a circle maximizes the area contained by the polygon.  The maximum area is independent of the order of sides, and even given reflection and rotation, there will be $n!/(2n)$ such arrangements, i.e. $(n-1)!/2$ noncongruent polygons, having the same area and perimeter.
On the other hand, restricting ourselves to two figures that are similar is a very narrow restriction.  One has a single parameter of scale to vary between the figures.  In that setting either equal area OR equal perimeter is enough to compel congruence of the figures.
For an example of two polygons with different numbers of sides, say 4 sides vs. 6 sides, consider a perimeter of six and area of two.  We can achieve this with a rectangle of size $2 \times 1$.  To match the perimeter we start with a regular hexagon having unit length sides, which will have area $\frac{3\sqrt{3}}{2} \gt 2$.  We can maintain the perimeter at six but reduce the area down to two by "squeezing" a pair of opposing (parallel) sides closer together.  The exact distance between these two sides to get the area six can be found by solving a cubic equation, but clearly the motion of moving sides closer together could achieve any area between that of the regular hexagon and zero (as the two sides get arbitrarily close).
A: Another counterexample without words:

