For a square lattice with lattice translation vector, $T=a\hat{x}+a\hat{y}$, we find that the reciprocal lattice vectors are given by, $$A=\frac{2\pi}{a}\hat{x},$$ and, $$B=\frac{2\pi}{a}\hat{y},$$ such that the Brillouin zone ranges from $-\frac{\pi}{a}\to\frac{\pi}{a}$. That is it ranges from $-\frac{|A|}{2}\to\frac{|A|}{2}$, where $|A|$ is the magnitude of $A$, and where $|A|=|B|$.
Now consider a hexagonal lattice with receiprocal lattice vectors, $$A=2\pi\hat{x}+\frac{2\pi}{\sqrt{3}}\hat{y},$$ and, $$B=\frac{4\pi}{\sqrt{3}},$$ both with magnitude, $|A|=|B|=\frac{4\pi}{\sqrt{3}}$. Does the first Brillouin zone run from, $=\frac{2\pi}{\sqrt{3}}\to\frac{2\pi}{\sqrt{3}}$? If this is the case, can we say that the first Brillouin zone is constructed similarly to the Wigner-Seitz unit cell in that it bisects lines to closest neighbors in reciprocal space to form a closed cell?