I think you are operating under the mistaken and oversimplified belief that the perimeter of a shape provides more information than it does in determining the area of the shape.
In theory, you need to know every thing about the curve of the shape-- not just it's length/perimeter but every twist turn and angle.
Of course, if the shape has a general pattern or formula than one can probably determine a lot of this information for relatively little data. For example if you know the shape is a polygon, it's enough to know the lengths of all of the sides and the measurement of all the angles. And if you know all but one of the angles you can determine that final angle and if you know all the angles and all but one of the sides you can determine the final side.
But, in general you need to know a lot more than just the total perimeter and the number of sides.
But then again if you know more information about the shape you can often determine more information. If for example you know it is an equilateral $n$-gon and you know the perimeter is $P$ you know the sides are all $P/n$. (However if $n > 3$ you don't know the angles so you don't know the area.) If you know it is an equiangular $n$-gon you know the the total angle and there for every angle. If you know it is a rectangle then you know all the angles are right angles and it is a parallelagram so all you need for the area is the two side (width and height). If you know the perimeter you only need one side and you can determine the other side.
So, your question.
Perimeter = $P$.
Triangle: With sides $a,b,c: a+b+c = P$ and angles $A,B,C: A+B+C = 180$. Angle $A$ is opposite $a$, $B$ opposite $b$ and $C$ opposite $c$.
We don't have enough information. If we had two sides, we can get the third from the perimeter and from that we can use trig identities to get the angles. Or if we had two angles, we could get the third and then we could get the trig propotional lengths and from the perimeter we can get the sides. Or if we had one side and one angle using trig and the restriction of the perimeter we can the rest.
Basically: If we have $2$ out of the six pieces of information and the perimeter, we can get the remaining four pieces of information, and from the six pieces of information we can get the area.
The six pieces of information are related via:
$\frac {\sin A}a = \frac {\sin B}b = \frac{\sin C}c; A+B+C = 180; a+b+c = P$ and also $a^2 + b^2 + 2ab\cos C = c^2$, $b^2 + c^2 + 2bc\cos A = a^2$, or $a^2 + c^2 + 2ac\cos B$ . (Any two of the variables will get you the remaining four.)
The area of the triangle = $\frac 12 *base*height = \frac 12 *a* ( {\sin C}*b)$ or any of the other combinations of two sides and the angle between them.
Equilateral triangle:
In this case $s= a = b = c = P/3$ and $A = B = C = 60$ so the area is $\frac 12(\frac P3)^2\sin 60 = (\frac P3)^2\frac{\sqrt{3}}4= s^2 \frac{\sqrt{3}}4$.
Quadrilateral:
If you don't know whether it is a rectangle or a parallegram there is no formula in general. But you can know that if the sides are $a,b,c,d$ and the angles are $A,B,C,D$ (capitals opposite lower case) you know: $a + b + c + d = P$ and $A+B+C+D= 360$. And you can calculate the area by "cutting" it into two triangles and figuring the area of those and adding them.
Rectangle:
You know the angles are right angles and that it is a parallelagram so $a =c$ and $b = d$ so $2a + 2b = P$ and $b = \frac {P- 2a}2$ and the area is therefore $a*b = a*\frac{P - 2a}2 = \frac P{2a} - a^2$.
Square:
$s = P/4$ and area is $s^2 = \frac {P^2}{16}$
It's worth noting the area of a regular $n$-gon.
An regualar $n$-gon has $n$ sides all equal so $s = \frac Pn$.
And regular $n$-gon has $n$ angles. These angles add up to $({n-2})*180$ because we can can cut the $n$-gon into $n-2$ triangles whose angles are each $180$. So each angle of the $n$-gon is $\theta = \frac{n-2}n * 180$.
We can slice the $n$-gon into isoceles triangle wedges. Each isoceles tringle slice has a base of length $s = \frac Pn$. The base angle is $\frac {\theta}2 = \frac{n-2}n*90$. So the height of the triangle is $height = opposite = adjacent*\frac {opposite}{adjacent} = (\frac 12 s)*\tan(\frac{\theta} 2)$.
So the area of the triangle is $\frac 12*s* (\frac 12 s)*\tan(\frac{\theta} 2)=\frac 14 s^2 \tan(\frac{\theta} 2)$.
So the area of a regular $n$-gon is $n*\frac 14 s^2 \tan(\frac{\theta} 2)=\frac 1{4n}*P^2\tan(\frac{n-2}n*90)$.
For $n=3$ we get $\frac 1{12}*P^2 \tan 30 = \frac 1{12}*P^2*\frac {\sqrt 3}3=(\frac P3)^2\frac{\sqrt{3}}4 = s^2\frac{\sqrt{n}}4$ which is what we had for equilateral triangle above.
Likewise if $n=4$ we get $\frac 1{16}P^2 \tan 45 = (\frac {P}4)^2=s^2$ as we had for squares.
Believe it or not:
Area of circle = $\lim_{n\rightarrow \infty} \frac 1{4n}*P^2\tan(\frac{n-2}n*90) =P^2 \lim_{n\rightarrow \infty} \frac 1{4n}\tan(\frac{n-2}n*90) =P^2 *\frac 1{4\pi}= (\frac {P}{2\pi}) ^2 \pi = \pi r^2$
Best Answer
To count the turns, mark a fixed radius on the circle (in red below), then count how many times it turns around the center during the movement. While rolling along a side, the fixed radius turns at a constant (angular) rate, as expected. But watch what happens when you turn the corner. Right before the turn, the radius is perpendicular to the bottom side of the triangle. Right after the turn, the radius becomes perpendicular to the right side of the triangle. Therefore, the turn introduced an instantaneous additional rotation, and it's simple geometry to see that it's $120^\circ$. There is $3$ corners, so it adds up to one additional complete revolution by the time the circle rolls back to the starting position. Add that to the $3$ turns due to rolling, and it's the correct answer of $4$ turns total.
Once you visualize this right, it becomes obvious that there is nothing special about the triangle. If the circle rolled around a rectangle, instead, there would be $4$ turns of $90^\circ$ each, which would once again add up to one complete additional revolution. It doesn't even have to be a polygon, the same would happen around another circle, or around any simple closed curve, for that matter.
This additional revolution is the same reason why the moon actually rotates, even though we always see only one side of it from the earth (nicely illustrated in this animation on the NASA site). If your circle maintained the same point of contact with the triangle all along, so that there would be no rolling at all, it would still complete a full revolution just by sliding around the triangle.