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 (*).
Place the vertices of the equilateral triangle at the points $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$, and denote the radius of the semicircles by $r$. The equilateral triangle formed by the diameters of the semicircles has side length $2\sqrt2$, so its area is $2\sqrt3$. To find the distance from a vertex where the diameter corresponding to that vertex intersects the semicircle corresponding to another vertex, take the diameter
$$
\pmatrix{1\\0\\0}+\lambda\pmatrix{0\\1\\-1}
$$
and find the point on it at distance $r$ from the vertex $\pmatrix{0\\0\\1}$:
$$
1+\lambda^2+(1+\lambda)^2=r^2\;,
\\
2\lambda^2+2\lambda+2=r^2\;,
\\
\lambda=-\frac12+\sqrt{\frac{r^2}2-\frac34}\;.
$$
Thus the side length of the three smaller equilateral triangles that are "missing" is
$$
\sqrt2\left(1-\left(-\frac12+\sqrt{\frac{r^2}2-\frac34}\right)\right)=\sqrt2\left(\frac32-\sqrt{\frac{r^2}2-\frac34}\right)\;,
$$
so their total area is
$$
3\cdot\frac{\sqrt3}4\cdot2\left(\frac94+\frac{r^2}2-\frac34-3\sqrt{\frac{r^2}2-\frac34}\right)=\frac34\sqrt3\left(3+r^2-3\sqrt{2r^2-3}\right)\;.
$$
Subtracting this from $2\sqrt3$ leaves
$$
\frac{\sqrt3}4\left(9\sqrt{2r^2-3}-3r^2-1\right)\;.
$$
Now we need to add back the three circular segments that extend into the three equilateral triangles we subtracted. Their radius is $r$, and their angle $\alpha$ is twice the angle between $\pmatrix{1\\\lambda\\-\lambda}-\pmatrix{0\\0\\1}$ and $\pmatrix{1\\1\\-1}-\pmatrix{0\\0\\1}$ and thus
$$
\alpha=2\arccos\frac{3(\lambda+1)}{\sqrt{6\cdot(2\lambda^2+2\lambda+2)}}=2\arccos\left(\sqrt{\frac38}\frac{1+\sqrt{2r^2-3}}r\right)\;.
$$
With the formula for the area of a circular segment, the total area is then
$$
\frac{\sqrt3}4\left(9\sqrt{2r^2-3}-3r^2-1\right)+\frac32r^2\left(\alpha-\sin\alpha\right)\;.
$$
Here's a plot of this total area for $r$ in the relevant range $[\sqrt2,\sqrt6]$, with the area ranging from $\pi-\sqrt3$ to $2\sqrt3$.
In this calculation the side length of the equilateral triangle was fixed at $\sqrt2$; for a different side length $a$, scale the radius by $\sqrt2/a$ and then scale the resulting area by $a^2/2$.
Best Answer
Here it is our right-angled Reuleaux triangle.
Assuming that the sides of the inner equilateral triangle have unit length, the blue arcs are arcs of circles with radius $\sqrt{3}$, associated to chords with unit length. It follows that the length of each arc is $\frac{\pi}{6}\sqrt{3}$. The enclosed area is the area of the equilateral triangle, $\frac{1}{4}\sqrt{3}$, plus three times the area of a circle segment, which can be seen as the difference between a circle sector with area $\frac{\pi}{4}$ and a triangle with area $\frac{abc}{4R}=\frac{3}{4}$.