3dgeometry
The shape can be constructed by taking a cylinder with its height equal to its diameter and cutting a triangle out of it when viewing it from the side with the base of the triangle matching one end of the cylinder and the opposite point of the triangle in the center of where the opposite end of the cylinder was.
Is there a name for this shape?
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 (*).
$|AD|=H,|AE|=H/2$. We have $AE/AD=\cos BAD$, so $\angle BAD=60^o$.
$\angle ADB=90^o$. So $BD=\sqrt3\ AD$. Hence the area of the triangle (1/2 base x height) is $\sqrt3\ H^2$.
Best Answer
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It has three Forms depending from which side you look: