Consider the regular hexagon
Imagine I want to stretch it uniformly in the horizontal direction ($x$). In order to keep the area constant, I naturally need to compress it in the vertical direction ($y$). In the end, all I'm doing is changing the aspect ratio of the hexagon. The perimeter, however, will change.
For a given stretching factor in the $x$ direction, $h_x$, how do I find the stretching factor $h_y$ such that the area $A$ is constant?
Also, how do I find the perimeter as a function of $h_x$, $P(h_x)$?
I understand my question is relatively broad, and no attempt at solving it is presented. I'm looking for hints and ideas.
For the area, the easy way to visualize this is to divide the figure into tiny squares. then the area is the sum of the area of the squares $$A=\sum_i A_i$$ When you stretch the figure, you will stretch each individual square into a rectangle. If the side of the individual square was $a$, the sides of the rectangle are now $h_x a$ and $h_y a$, so the new area of the figure will be $$A'=h_x h_y A$$ If you want $A=A'$, then $h_y=1/h_x$.
For the perimeter it gets more complicated, because individual sides change lengths in different ways. But you can still come up with a simple algorithm. Each side has originally some length $l_i$ that we can write as a combination of length along the $x$ axis and along $y$ axis using Pythagoras' theorem:$$l_i=\sqrt{l_{ix}^2+l_{iy}^2}$$You know now how each component stretches, so $$l_i'=\sqrt{h_x^2l_{ix}^2+\frac{1}{h_x^2}l_{iy}^2}$$