= Solution
For the <first two Brillouin zones of a rectangular lattice>, write $K=\pi/a$ and $L=\pi/b$, so $0<K<L$ and the <reciprocal lattice vectors> are $(2nK,2mL)$. The perpendicular bisectors to the four nearest axial reciprocal points enclose
$$
\boxed{\mathcal B_1=\{(k_x,k_y):|k_x|<K,\ |k_y|<L\}.}
$$
The <second Brillouin zone> consists of points beyond exactly one <Bragg plane>. In the first quadrant, it is the union of the following two regions:
$$
\begin{aligned}
K<k_x<2K,&\qquad 0<k_y<\frac{K^2+L^2-Kk_x}{L},\\
0<k_x<K,&\qquad L<k_y<\frac{K^2+L^2-Kk_x}{L}.
\end{aligned}
$$
Reflect these regions in both axes. The first is a trapezoidal side lobe; the second is half of a triangular top lobe. The sloping boundary is the bisector for $\mathbf G=(2K,2L)$. On the side lobe, $k_x<2K$ is the bisector condition for $\mathbf G=(4K,0)$: crossing that plane too would enter the third or a higher zone. These additional bisectors are why four rectangles pasted onto the first zone would be incorrect.
Equivalently, the full right lobe has vertices
$$
(K,-L),\quad(2K,-L+K^2/L),\quad(2K,L-K^2/L),\quad(K,L),
$$
and the top lobe has vertices $(-K,L),(K,L),(0,L+K^2/L)$. Add their left and bottom reflections. Their total area is
$$
2\left(2KL-\frac{K^3}{L}\right)+2\frac{K^3}{L}=4KL,
$$
equal to the first-zone area, as required by reciprocal-space state counting. These formulas remain valid for every $a>b$, including very elongated rectangles; boundaries themselves can be shared by convention.
\Image[/past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2012-rectangular-first-and-second-brillouin-zones.png]
{title=First and second Brillouin zones of rectangular lattices}
{description=Blue is the first zone and orange is the complete disconnected second zone; two aspect ratios show the effect of the higher axial Bragg plane. All coordinates are in units of pi divided by a.}
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