A Geometric Measure theoretic answer to your question could be the following.
Your example simply shows how things really go... There is nothing strange in it, nor in the "informal" notion of convergence you used: actually the trouble remains the same, even when all the underlying notions are well formalized.
The problem here is that the perimeter $\mathcal{P} (\cdot )$ is "only" a lower semicontinuous functional. This means: if a sequence of sets $E_n$ converges to another set $E$ in some sense (e.g. in the Hausdorff metric, which is a sort of uniform convergence for sets, or in some $L^p$ metric), then you have:
(*) $\displaystyle \mathcal{P} (E)\leq \liminf_{n\to \infty} \mathcal{P}(E_n)$;
moreover, in general the inequality is strict, even if the sequence at the RHside converges.
A less problematic example of this basic fact of GMT is the following. Let:
$E_n:=\{ (x,y)\in \mathbb{R}^2|\ x^2+y^2<1\text{ and } |y|\geq \frac{1}{n} |x|\}$,
so $E_n$ is the unit open circle with two symmetric slices (crossing the $x$ axis) removed and the slices become thinner and thinner as $n$ increases. Note that $E_n$ is a piecewise $C^1$ set with a finite number of corners, i.e. $5$, and this number does not increase with $n$ (on the contrary, in your example the number of corners becomes larger as $n\to \infty$).
Then $E_n$ converges to the unit open circle $D$ in the $L^1$ metric: in fact, the measure of the symmetric difference set $D\Delta E_n$ tends to $0$ as $n$ increases, i.e. $\lVert \chi_D -\chi_{E_n}\rVert_1\to 0$; on the other hand, one has:
$\displaystyle \mathcal{P} (E_n)=4+ 2\left( \pi -2\arctan \tfrac{1}{n}\right)$
(the summand $4$ appears because the boundary $\partial E_n$ contains four radii of lenght $=1$) therefore:
$\displaystyle \liminf_{n\to \infty} \mathcal{P}(E_n)=\lim_{n\to \infty} \mathcal{P} (E_n) =2\pi +4 >2\pi =\mathcal{P} (D)$.
For what is worth, $E_n$ converges to $D$ also in the strongest Hausdorff metric, because it is not hard to prove that the Haudorff distance:
$\displaystyle \text{dist}_H(E_n,D):=\inf \{ \epsilon >0| E_n\subseteq (1+\epsilon)D \text{ and } D\subseteq E_n+\epsilon D\}$
is given by:
$\displaystyle \text{dist}_H(E_n,D) =2-\frac{2}{\sqrt{1+\frac{1}{n^2}}}$,
and $\displaystyle \lim_{n\to \infty} \text{dist}_H(E_n,D) =0$.
If you choose $E_n$ as in your example, then you have the same situation: a sequence of sets which does converge to a triangle in some metrics (in particular, it converges in both $L^1$ and Hausdorff metric) and for which strict inequality holds in (*).
For a positive value $k$, the graph of $y = kf(x)$ is obtained from the graph of $y=f(x)$ by scaling in the vertical direction. In coordinates, if $(x_0, y_0)$ is on the graph of $y=f(x)$, then that point moves to $(x_0, ky_0)$ on the graph of $y=kf(x)$. Any point on the $x$-axis remains fixed (as then $y=0$). When $k > 1$, the graph is stretched vertically (away from the $x$-axis), while if $0 < k < 1$, the graph is compressed vertically (toward the $x$-axis).
Another way to think about this: The graph of $y=kf(x)$ is exactly the same as the graph of $y=f(x)$ if the $y$-axis is scaled so that $1$ is relabeled $k$, $2 \mapsto 2k$, $-1 \mapsto -k$, etc.
Hope this helps!
Best Answer
There's no physics here, but the polar graph $$ r = 1 + 0.375\operatorname{sech}(2.75(\theta + \tfrac{\pi}{2})) $$ is a good visual match for your image:
Edit: I hit on the parameters after a fortuitously small number of tries. Here are images showing the profiles if the "width" or "height" (respectively) of the sech hump is varied: