Vector area of a polygonal boundary ($\mathbf{A}=\frac{1}{2}{\sum }_i{B}_i\times B_{i+1}$)

For an oriented closed polygon, triangulate with a common vertex $P$. Its total vector area is $\frac{1}{2}{\sum }_i(B_i-P)\times (B_{i+1}-P)$; the terms involving $P$ telescope, giving the displayed boundary-only formula. Its signed projected area along a unit vector $V$ is $V\cdot \mathbf{A}$. The maximum is $|\mathbf{A}|$, in direction $\mathbf{A}/|\mathbf{A}|$ if nonzero, and zero projections have directions perpendicular to $\mathbf{A}$. If $\mathbf{A}=0$, every direction has zero signed projection despite possible nonzero unsigned surface area.