Area of tubular neighborhood of union of convex sets having diameter $1$ with vertices 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.)
 A: $\DeclareMathOperator{\area}{area}$ 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.
A: Similar case - We will change the condition of diameter $1$ :

Assume that $P_i$ is a regular $i$-gon s.t. a unit circle $S$ is
inscribed
in $P_i$.

We consider $Y$ to be union of unit balls $B$ in ${\rm conv} \ P_i$ where $\partial B =S$. Hence ${\rm
perim}\ Y$ is bounded by const which is independent of $X$.
(cf
Length of a boundary of union of unit balls in $\mathbb{R}^2$)
Note that ${\rm perim}\ X\leq  4\cdot {\rm perim}\ Y$. Fix any $r_0\in
\mathbb{R}$. From Liviu Nicolaescu's answer, for any $X$, we have
$${\rm area}\ U_r(X) \leq {\rm area}\ X+ Cr+\pi r^2 $$ for any
$0<r<r_0$. Hence we have $ Cr+\pi r^2 \leq Dr$ where $D=D(r_0)$.
