A question regarding shapes and infinitesimals I have found myself quite confused about this.  It started when I was reading a book called Mathographics by Robert Dixon which explains different methods for deriving pi. One of the methods involves rectifying a circle which had me wondering about making shapes with the same perimeter measure.
At first, I assumed that if you take the perimeter of any object and manipulate it into any shape, it should always maintain the same area...but apparently this doesn't work with circles and squares..or even triangles.  Is it the case that any equilateral shape with same perimeter measure will never maintain the same area?  And why not?  I know mathematically it doesn't work.  I've done the math to prove it doesn't work with circles and triangles...but why??  Where is my confusion stemming from?   
 A: If $P_1P_2\dots P_n$ is a regular polygon. Let $O$ the center of the polygon. Let $\alpha=2\pi/n$, and $OP_1=r$.
The perimeter is $n\times 2r \sin(\alpha/2)=n \times 2r \sin(\pi/n)$.
The area is $n \times 2r\sin(\alpha/2)\times r \cos(\alpha/2)=n \times 2r\sin(\pi/n)\times r \cos(\pi/n)$.
If $Q_1Q_2 \dots Q_k$ is another regular polygon with same perimeter than $P_1P_2 \dots P_n$. The center is $O$ too. Let $OQ_1=r'$.Then $n \times 2r \sin(\pi/n)=k \times 2r'\sin(\pi/k)$.
If the area is the same, $n \times 2r\sin(\pi/n)\times r \cos(\pi/n)=k \times 2r'\sin(\pi/k)\times r' \cos(\pi/k)$.
So $r \cos (\pi/n)=r' \cos(\pi/k)$.
and $k \times 2r'\sin(\pi/k)=n \times 2r \sin(\pi/n)$.
So $k \sin(\pi/k)\cos(\pi/n)=n\sin(\pi/n)\cos(\pi/k)$.
So $k \tan(\pi/k)= n \tan(\pi/n)$.
But if $k, n \geq 3$, as $x \mapsto \tan(x)/x$ is strictly increasing on $]0, \pi/3]$, we can't have $k \tan(\pi/k)= n \tan(\pi/n)$, and so we can't have same area and same perimeter for $P_1\dots P_n$ and $Q_1 \dots Q_k$ if $n \neq k$.
A: You intuition is faulty.
Consider a square $12$ inches long and $12$ inches high.  It has perimeter $48$ inches and an area $144$ square inches.  
Squeeze it so that it is now $1$ inch high and $23$ inches long.  It still has perimeter $48$ inches.  But its area is now $23$ square inches.  Where did the extra $121$ square inches go?
In the comments you commented that by flatten it we are pushing one edge out by $11$ inches but shrinking the other by $11$ inches so the area should stay balanced.
But increasing $12$ inches by $11$ inches is proportionally much less significant that decreasing $12$ inches by $11$ inches.
We've increased the side by $\frac {12 + 11}{12} = \frac {23}{12} = 1.916666..... $ times longer.
We've decrease the other side to be only  $\frac {12 -11}{12} = \frac 1{12} = .083333....$ times as long.
So the area will now be $1.91666666..... * .0833333 = 0.159722222....$ as much.
Indeed the original area was $12*12 = 144$ and the new area is $12*12*\frac{12 +11}{12}*\frac{12 - 11}{12} = 12*12*\frac{23}{12}*\frac{1}{12} = 23*1$.
The "secret" is similar to the one of calulating wins vs. losses as percentages.  $a -x$ is a different percentage of $a$ then $a + x$ is of $a$.  So increasing one side by $x$ and decreasing the other $x$ are maintaining the same absolute value but not the same proportional values.  So the area will not maintain proportionality.  But the absolute perimeter will stay constant.
Or in other words.  $a*a = a^2$ but $(a+x)(a-x) = a^2 - x^2 \ne a^2$.  Area is not preserved when perimeter is.
(And vice-versa.  $a*b = Area$ but $(a+x)\frac {ab}{a+x} = ab = Area$ while  $\frac{ab}{a+x} \ne b - x$.  Area is preserved but perimeter is not.)
