[proofplan] We first prove the formula for convex polygons by decomposing the parallel body into the polygon itself, rectangular strips over its edges, and circular sectors at its vertices. The strip areas add to $t\,\operatorname{Per}(P)$, and the sector angles add to $2\pi$, producing the term $\pi t^2$. For a general convex body, we approximate it from inside by convex polygons whose Hausdorff distance, area, and perimeter converge to those of $K$. Applying the polygon formula to these approximants and using the inclusions forced by Hausdorff convergence passes the identity to the limit. [/proofplan]

[step:Prove the formula for a convex polygon by decomposing its parallel body]Let $P \subset \mathbb{R}^2$ be a convex polygon with vertices $v_1,\dots,v_m$ listed counterclockwise and edges \begin{align*} E_i := [v_i,v_{i+1}], \end{align*} where $v_{m+1}:=v_1$. Let $\ell_i := \mathcal{H}^1(E_i)$ be the length of $E_i$, and let $n_i \in \mathbb{R}^2$ be the outward unit normal to $E_i$. At the vertex $v_i$, let $\alpha_i \in (0,\pi]$ denote the exterior angle between the outward normals $n_{i-1}$ and $n_i$, measured in the exterior normal cone of $P$. For $t \geq 0$, define the rectangular strips \begin{align*} R_i := \{x+s n_i : x \in E_i,\ 0 < s \leq t\}, \end{align*} and the circular sectors \begin{align*} S_i := \{v_i + r u : 0 < r \leq t,\ u \in C_i \cap \partial B\}, \end{align*} where $C_i$ is the exterior normal cone of $P$ at $v_i$. We justify that these pieces exhaust $P+tB$ up to boundary sets. If $y\in (P+tB)\setminus P$, compactness of $P$ gives a point $x\in P$ with $|y-x|=\operatorname{dist}(y,P)\leq t$. The nearest-point condition for a closed convex set says $y-x$ belongs to the exterior normal cone of $P$ at $x$: indeed, if some $z\in P$ satisfied $(y-x)\cdot(z-x)>0$, then for sufficiently small $\lambda\in(0,1)$ the point $x+\lambda(z-x)\in P$ would have smaller distance to $y$, contradicting the choice of $x$. For a polygon, a boundary point $x$ lies either in the relative interior of a unique edge, where the exterior normal cone is the ray $\{s n_i:s\geq0\}$, or is a vertex $v_i$, where the exterior normal cone is $C_i$. Hence $P+tB$ is the disjoint union up to $\mathcal{L}^2$-null boundary pieces of $P$, the strips $R_i$, and the sectors $S_i$. The map \begin{align*} \Phi_i: E_i \times (0,t] &\to R_i \\ (x,s) &\mapsto x+s n_i \end{align*} is an isometry from the rectangle $E_i \times (0,t]$ onto $R_i$, so \begin{align*} \mathcal{L}^2(R_i)=t\ell_i. \end{align*} The sector $S_i$ has radius $t$ and angle $\alpha_i$, hence \begin{align*} \mathcal{L}^2(S_i)=\frac{\alpha_i}{2}t^2. \end{align*} Since the exterior angles of a convex polygon sum to $2\pi$, \begin{align*} \sum_{i=1}^m \alpha_i = 2\pi. \end{align*} Therefore \begin{align*} \mathcal{L}^2(P+tB) &= \mathcal{L}^2(P) + \sum_{i=1}^m \mathcal{L}^2(R_i) + \sum_{i=1}^m \mathcal{L}^2(S_i) \\ &= \mathcal{L}^2(P) + t\sum_{i=1}^m \ell_i + \frac{t^2}{2}\sum_{i=1}^m \alpha_i \\ &= \mathcal{L}^2(P) + t\,\operatorname{Per}(P) + \pi t^2. \end{align*}[/step]

[guided]The geometry of $P+tB$ is controlled by nearest points of $P$. We first justify that description. Let $y\in(P+tB)\setminus P$, and choose $x\in P$ with $|y-x|=\operatorname{dist}(y,P)\leq t$, which exists because $P$ is compact. The vector $y-x$ must be outward normal to $P$ at $x$: if there were $z\in P$ with $(y-x)\cdot(z-x)>0$, then for small $\lambda\in(0,1)$ the convexity of $P$ gives $x+\lambda(z-x)\in P$, and expanding the square shows this point is closer to $y$ than $x$ is. Thus $y-x$ belongs to the exterior normal cone of $P$ at $x$. Points whose nearest point lies in the relative interior of an edge therefore form a rectangle of width $t$ over that edge. Points whose nearest point is a vertex form a circular sector, because the allowed outward directions are exactly the directions in the exterior normal cone at that vertex. For each edge $E_i=[v_i,v_{i+1}]$, define its outward unit normal $n_i \in \mathbb{R}^2$ and its length $\ell_i=\mathcal{H}^1(E_i)$. The strip attached to $E_i$ is \begin{align*} R_i := \{x+s n_i : x \in E_i,\ 0 < s \leq t\}. \end{align*} The map \begin{align*} \Phi_i: E_i \times (0,t] &\to R_i \\ (x,s) &\mapsto x+s n_i \end{align*} preserves Euclidean lengths in the edge direction and in the normal direction, so it sends the rectangle $E_i \times (0,t]$ onto $R_i$ without area distortion. Hence \begin{align*} \mathcal{L}^2(R_i)=t\ell_i. \end{align*} At a vertex $v_i$, the exterior normal cone $C_i$ consists of the outward directions between the two adjacent outward normals. If $\alpha_i$ is the corresponding exterior angle, then the vertex contribution is the circular sector \begin{align*} S_i := \{v_i + r u : 0 < r \leq t,\ u \in C_i \cap \partial B\}. \end{align*} A sector of radius $t$ and angle $\alpha_i$ has area \begin{align*} \mathcal{L}^2(S_i)=\frac{\alpha_i}{2}t^2. \end{align*} These pieces overlap only along line segments and circular arcs, which have zero $\mathcal{L}^2$-measure. Therefore their areas add. Since the exterior angles of a convex polygon complete one full turn, \begin{align*} \sum_{i=1}^m \alpha_i=2\pi. \end{align*} Also, \begin{align*} \operatorname{Per}(P)=\sum_{i=1}^m \ell_i. \end{align*} Combining these identities gives \begin{align*} \mathcal{L}^2(P+tB) &= \mathcal{L}^2(P) + \sum_{i=1}^m t\ell_i + \sum_{i=1}^m \frac{\alpha_i}{2}t^2 \\ &= \mathcal{L}^2(P) + t\,\operatorname{Per}(P) + \pi t^2. \end{align*}[/guided]

[step:Approximate the convex body from inside by polygons with convergent area and perimeter]We use the following approximation fact for planar convex bodies. [claim:Inscribed convex polygons approximate a planar convex body in area and perimeter] There exists a sequence of convex polygons $(P_j)_{j=1}^{\infty}$ such that \begin{align*} P_j \subset K,\qquad d_H(P_j,K)\to 0,\qquad \mathcal{L}^2(P_j)\to \mathcal{L}^2(K),\qquad \operatorname{Per}(P_j)\to \operatorname{Per}(K), \end{align*} where $d_H$ denotes Hausdorff distance. [/claim] [proof] Because $\partial K$ is a compact rectifiable Jordan curve, its length is \begin{align*} \operatorname{Per}(K)=\mathcal{H}^1(\partial K). \end{align*} For each $j \in \mathbb{N}$, choose a finite subset $F_j \subset \partial K$ such that $F_j \subset F_{j+1}$, the set $F_j$ is a $1/j$-net in $\partial K$, and the polygonal curve through the points of $F_j$ in cyclic boundary order has length at least $\operatorname{Per}(K)-1/j$. Define \begin{align*} P_j := \operatorname{conv}(F_j). \end{align*} Since $K$ is convex and $F_j \subset K$, we have $P_j \subset K$. Let $\Gamma_j$ denote the polygonal curve obtained by joining the points of $F_j$ in cyclic boundary order. The curve $\Gamma_j$ is contained in $K$, and its image is the boundary of $P_j$ after deleting possible redundant collinear or non-extreme vertices. Deleting such vertices does not change the convex hull and cannot increase the polygonal length, while inserting points on the same boundary segment does not change the total length of that segment. Hence the length of $\Gamma_j$ equals $\operatorname{Per}(P_j)$. Since $\partial K$ is rectifiable and $\operatorname{Per}(K)$ is its curve length, every inscribed polygonal curve has length at most $\operatorname{Per}(K)$, so \begin{align*} \operatorname{Per}(P_j)\leq \operatorname{Per}(K). \end{align*} By the choice of $F_j$ and the equality between the length of $\Gamma_j$ and $\operatorname{Per}(P_j)$, \begin{align*} \operatorname{Per}(P_j)\geq \operatorname{Per}(K)-\frac{1}{j}. \end{align*} Therefore \begin{align*} \operatorname{Per}(P_j)\to \operatorname{Per}(K). \end{align*} We next verify Hausdorff convergence. Let $\varepsilon_j:=1/j$. Since $F_j$ is an $\varepsilon_j$-net in $\partial K$, every supporting point of $K$ is within Euclidean distance $\varepsilon_j$ of a point of $F_j$. Equivalently, for every unit vector $u \in \partial B$, the support functions \begin{align*} h_K(u)&:=\sup_{x\in K} u\cdot x,\\ h_{P_j}(u)&:=\sup_{x\in P_j} u\cdot x \end{align*} satisfy \begin{align*} h_K(u)-\varepsilon_j \leq h_{P_j}(u)\leq h_K(u). \end{align*} The [second inequality](/theorems/2899) follows from $P_j\subset K$. The first follows by choosing $x_u\in\partial K$ with $u\cdot x_u=h_K(u)$ and then choosing $p_u\in F_j$ with $|p_u-x_u|\leq \varepsilon_j$, which gives \begin{align*} h_{P_j}(u)\geq u\cdot p_u \geq u\cdot x_u-|p_u-x_u|=h_K(u)-\varepsilon_j. \end{align*} The support-function criterion for Hausdorff distance between compact convex sets gives \begin{align*} d_H(P_j,K)\leq \varepsilon_j\to 0. \end{align*} Finally we prove area convergence by showing that every interior point of $K$ eventually lies in $P_j$. Let $x\in \operatorname{int} K$, and choose $\rho>0$ such that $B(x,\rho)\subset K$. If $x\notin P_j$, the separating hyperplane theorem for the compact convex set $P_j$ gives a unit vector $u_j\in\partial B$ such that \begin{align*} u_j\cdot x>h_{P_j}(u_j). \end{align*} Because $x+\rho u_j\in K$, the support function of $K$ satisfies \begin{align*} h_K(u_j)\geq u_j\cdot(x+\rho u_j)=u_j\cdot x+\rho. \end{align*} The support-function estimate already proved gives \begin{align*} h_{P_j}(u_j)\geq h_K(u_j)-\varepsilon_j\geq u_j\cdot x+\rho-\varepsilon_j. \end{align*} For all sufficiently large $j$, $\varepsilon_j<\rho$, contradicting $u_j\cdot x>h_{P_j}(u_j)$. Hence $\operatorname{int}K\subset\bigcup_{j=1}^{\infty}P_j$. Since $P_j\subset K$, the only points of $K$ not covered by the union lie in $\partial K$. The boundary $\partial K$ has finite $\mathcal{H}^1$-measure and hence zero $\mathcal{L}^2$-measure, so \begin{align*} \mathcal{L}^2\left(K\setminus \bigcup_{j=1}^{\infty}P_j\right)=0. \end{align*} By continuity from below of the measure $\mathcal{L}^2$, \begin{align*} \mathcal{L}^2(P_j)\to \mathcal{L}^2(K). \end{align*} [/proof][/step]

[guided]The approximation must preserve three quantities at once: inclusion, area, and perimeter. We choose points on $\partial K$ that become dense and then take their convex hulls. Convexity guarantees these hulls stay inside $K$. For each $j\in\mathbb{N}$, choose a finite set $F_j\subset\partial K$ such that $F_j\subset F_{j+1}$, $F_j$ is a $1/j$-net in $\partial K$, and the polygonal curve through $F_j$ in cyclic order has length at least $\operatorname{Per}(K)-1/j$. Define \begin{align*} P_j:=\operatorname{conv}(F_j). \end{align*} Because $F_j\subset K$ and $K$ is convex, the convex hull $P_j$ is contained in $K$. The perimeter convergence follows from the definition of the length of a rectifiable curve as the supremum of lengths of inscribed polygonal curves. Let $\Gamma_j$ be the polygonal curve joining the points of $F_j$ in cyclic boundary order. Although the sides of $\Gamma_j$ are chords lying inside $K$, not arcs of $\partial K$, this is exactly the kind of inscribed polygonal curve used in the definition of curve length. Its image is the boundary of $P_j$ after deleting redundant collinear or non-extreme vertices; deleting such vertices does not change the convex hull and cannot increase polygonal length, while points inserted on the same side do not change the total length of that side. Therefore the length of $\Gamma_j$ equals $\operatorname{Per}(P_j)$. Since $\operatorname{Per}(K)$ is the supremum of the lengths of such inscribed polygonal curves, \begin{align*} \operatorname{Per}(P_j)\leq \operatorname{Per}(K). \end{align*} The choice of $F_j$ gives the reverse estimate up to $1/j$: \begin{align*} \operatorname{Per}(P_j)\geq \operatorname{Per}(K)-\frac{1}{j}. \end{align*} Hence \begin{align*} \operatorname{Per}(P_j)\to \operatorname{Per}(K). \end{align*} To prove Hausdorff convergence, use support functions. For a compact convex set $A\subset\mathbb{R}^2$, define \begin{align*} h_A:\partial B&\to\mathbb{R}\\ u&\mapsto \sup_{x\in A}u\cdot x. \end{align*} Since $P_j\subset K$, we have $h_{P_j}(u)\leq h_K(u)$ for every $u\in\partial B$. Conversely, choose $x_u\in\partial K$ with $u\cdot x_u=h_K(u)$. Since $F_j$ is a $1/j$-net in $\partial K$, there is $p_u\in F_j$ with $|p_u-x_u|\leq 1/j$. Then \begin{align*} h_{P_j}(u)\geq u\cdot p_u \geq u\cdot x_u-|p_u-x_u| \geq h_K(u)-\frac{1}{j}. \end{align*} Thus the support functions differ uniformly by at most $1/j$, and therefore \begin{align*} d_H(P_j,K)\leq \frac{1}{j}\to 0. \end{align*} For area, the inclusions $P_j\subset P_{j+1}\subset K$ imply that $\mathcal{L}^2(P_j)$ increases to the measure of $\bigcup_j P_j$. We must prove that no interior point of $K$ is permanently missed. Let $x\in\operatorname{int}K$, and choose $\rho>0$ with $B(x,\rho)\subset K$. Suppose, for contradiction, that $x\notin P_j$ for some large $j$. Since $P_j$ is compact and convex, the separating hyperplane theorem gives a unit vector $u_j\in\partial B$ such that \begin{align*} u_j\cdot x>h_{P_j}(u_j). \end{align*} But $x+\rho u_j\in K$, so \begin{align*} h_K(u_j)\geq u_j\cdot(x+\rho u_j)=u_j\cdot x+\rho. \end{align*} The support-function estimate gives \begin{align*} h_{P_j}(u_j)\geq h_K(u_j)-\frac{1}{j}\geq u_j\cdot x+\rho-\frac{1}{j}. \end{align*} For $j>1/\rho$, this contradicts $u_j\cdot x>h_{P_j}(u_j)$. Hence every interior point of $K$ belongs to $P_j$ for all sufficiently large $j$, and therefore \begin{align*} K\setminus\bigcup_{j=1}^{\infty}P_j\subset\partial K. \end{align*} Because $\partial K$ is rectifiable, it has zero two-dimensional Lebesgue measure. Therefore \begin{align*} \mathcal{L}^2\left(K\setminus \bigcup_{j=1}^{\infty}P_j\right)=0. \end{align*} Continuity from below for $\mathcal{L}^2$ gives \begin{align*} \mathcal{L}^2(P_j)\to\mathcal{L}^2(K). \end{align*}[/guided]

[step:Use Hausdorff convergence to trap the parallel body between polygonal parallel bodies] Let $(P_j)_{j=1}^{\infty}$ be the inscribed polygonal approximation from the previous step, and define \begin{align*} \varepsilon_j := d_H(P_j,K). \end{align*} Since $P_j\subset K$, we have \begin{align*} P_j+tB\subset K+tB. \end{align*} Since $d_H(P_j,K)=\varepsilon_j$ and $P_j\subset K$, every point of $K$ lies in $P_j+\varepsilon_j B$, so \begin{align*} K\subset P_j+\varepsilon_j B. \end{align*} Adding $tB$ to both sides and using $aB+bB=(a+b)B$ for $a,b\geq0$, we obtain \begin{align*} K+tB\subset P_j+(\varepsilon_j+t)B. \end{align*} Therefore \begin{align*} \mathcal{L}^2(P_j+tB) \leq \mathcal{L}^2(K+tB) \leq \mathcal{L}^2(P_j+(t+\varepsilon_j)B). \end{align*} [/step]

[step:Pass the polygon formula to the limit] Applying the polygon formula to $P_j$ with radii $t$ and $t+\varepsilon_j$ gives \begin{align*} \mathcal{L}^2(P_j)+t\,\operatorname{Per}(P_j)+\pi t^2 \leq \mathcal{L}^2(K+tB) \leq \mathcal{L}^2(P_j)+(t+\varepsilon_j)\operatorname{Per}(P_j)+\pi(t+\varepsilon_j)^2. \end{align*} By the approximation step, \begin{align*} \mathcal{L}^2(P_j)\to\mathcal{L}^2(K),\qquad \operatorname{Per}(P_j)\to\operatorname{Per}(K),\qquad \varepsilon_j\to0. \end{align*} Also $\operatorname{Per}(P_j)\leq \operatorname{Per}(K)$, so \begin{align*} \varepsilon_j\operatorname{Per}(P_j)\to0. \end{align*} Taking the limit in the lower and upper bounds yields \begin{align*} \mathcal{L}^2(K+tB) = \mathcal{L}^2(K)+t\,\operatorname{Per}(K)+\pi t^2. \end{align*} This is the desired identity for every $t\geq0$. [/step]