= Solution
The reciprocal primitive vectors are $(2\pi/a_x,0)$ and $(0,2\pi/a_y)$, so the first <Brillouin zone> is
$$
\boxed{|k_x|\leq\pi/a_x,\qquad |k_y|\leq\pi/a_y.}
$$
It is wider in the $k_x$ direction because $a_x<a_y$.
The geometry alone does not fix the hopping amplitudes or a unique <Fermi surface>. One consistent <tight-binding model> is
$$
E(\mathbf k)=\epsilon_0-2t_x\cos(k_xa_x)-2t_y\cos(k_ya_y),\qquad t_x>t_y>0,
$$
representing larger hopping between the closer atoms. It remains a nearest-neighbour bipartite band at half filling, so its <chemical potential> is $\epsilon_0$. Its <Fermi surface> consists of the two sheets
$$
\boxed{k_x=\pm\frac1{a_x}\arccos\!\left[-\frac{t_y}{t_x}\cos(k_ya_y)\right].}
$$
These meet the top and bottom zone edges and continue periodically into the neighbouring zones. The right panel of the figure gives an explicit example with $t_x=2t_y$ and $a_x=0.65a_y$. The central region between the sheets is occupied and has half the zone area. If the hopping amplitudes instead remained equal, a geometrically stretched diamond would also satisfy the stated information. The metric inequality $a_x<a_y$ alone does not determine which of these shapes occurs.
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