Consider $P\_i$, which is a regular $i$-gon in $\mathbb{R}^2$ having diameter $1$.


Assume that a compact set $X$ in $[0,10]^2\subset \mathbb{R}^2$ is a union of convex hulls of $P\_i$. I want to prove that there is $0<r\_0$ s.t. $0<r<r\_0$ implies that area of $r$-tubular neighborhood $U\_r(X)$ of $X$ is less than ${\rm area}\ X + Cr$, where $C$ is independent of $X$.


(${\rm area}\ U\_r(P\_i)\leq {\rm area} \ P\_i + Cr$ is clear. But $X$ can have arbitrarilly many vertices.)



Area of tubular neighborhood of union of convex sets having diameter $1$ with vertices
$\DeclareMathOperator{\area}{area}$ The region $X$ is a non-convex polygon and for simplicity I will asume it is simply connected. Its vertices are of two types: convex vertices and non-convex vertices. Denote by $V\_+$ the set of convex vertices and by $V\_-$ the set of non-convex vertices. For a vertex $v$ denote by $\alpha\_v$ the exterior angle at $v$. Then there exists $r\_0=r\_0(X)>0$ such that, for $r<r\_0$,


$$\area\big(\; U\_r(X)\;\big) =\area(X)+\mathrm{perimeter}(X)r+\sum\_{v\in V\_+}r^2\alpha\_v/2-\sum\_{v\in V\_-}r^2\cot\alpha\_v/2. $$


The angles $\alpha\_v$ are constrained by


$$\sum\_{v\in V\_+}\alpha\_v-\sum\_{v\in V\_-}\alpha\_v=2\pi. $$


(To see this follow the outer normal to the boundary as you travel along the boundary counterclockwisely and see the jumps in the angle of the normal as it passes a vertex. ) Hence


$$\sum\_{v\in V\_+}r^2\alpha\_v/2=\pi r^2+\sum\_{v\in V\_-}\alpha\_v/2, $$


so


$$\area\big(\; U\_r(X)\;\big) =\area(X)+\mathrm{perimeter}(X)r+\pi r^2+r^2\sum\_{v\in V\_-} \big(\alpha\_v/2-\cot\alpha\_v/2\big). 
$$
The perimeter of $X$ will play a role. If $X$ has many vertices so its boundary is really jagged you can suspect that the perimeter if large. Your question would require that the perimeter of $X$ is bounded by some universal constant. I find this unlikely, but I do not yet have an argument.


**Comment** Let me point out a subtlety. Fix an equilateral triangle $T\_0$ of side $1$. For $\theta\in[0,2\pi]$ denote by $T\_\theta$ the triangle the equilateral triangle obtained from $T\_0$ by a counterclockwise rotation of angle $\theta$ about its center $O$. All the triangles $T\_\theta$ are contained in the disk $D$ of radius $\frac{\sqrt{3}}{2}$ and center $O$. For any natural number $n$ denote by $X\_n$ the region
$$X\_n=\bigcup\_{k=0}^{n-1} T\_{2k\pi/n}.$$


I believe that
$$
\lim\_{n\to\infty}\mathrm{perimeter}(X\_n)=\infty
$$
though I do not have an argument yet. On the other hand
$$
\lim\_{n\to \infty}\area(X\_n)=\area(D)=\frac{3\pi}{4}.
$$
This is telling me that $r\_0(X\_n)\to 0$ as $n\to \infty$. The reason is the presence of small angles at nonconvex vertices.


