 Solution

In curvature-minus-one polar coordinates the hyperbolic metric is $d{s}^2=d{r}^2+{\sinh }^{2}rd{\theta }^2$. Thus the area element is $\sinh rdrd\theta$, givingIt is asymptotic to $\pi e^r$.

For a usual locally finite tessellation with compact congruent tiles of nonempty interior, let their diameter be at most $D$ and choose in each a congruently placed interior ball of radius $r_0>0$. These balls have disjoint interiors. The number of tiles meeting a geodesic segment of length one is uniformly bounded: their ball centres lie in a ball of radius $D+1$, and the disjoint $r_0$-balls lie in the larger ball of radius $D+1+r_0$. Comparing areas supplies a constant $K$. A path of length $R$, divided into at most $\lceil R\rceil$ such segments, therefore crosses at most $K(\lceil R\rceil +1)$ tiles. A small generic perturbation avoids vertices if steps are defined by crossing sides.

Consequently the union of tiles at most $n$ steps from the initial tile contains a metric ball of radius $cn-C$ for fixed $c>0,C$. If $A$ is tile area, their number $N(n)$ is at least $\mathrm{Area}B(cn-C)/A$, an exponential lower bound. Conversely any chain of $n$ adjacent tiles lies in $B(D(n+1))$ about a point in the initial tile; the interiors are disjoint, so $AN(n)\le \mathrm{Area}B(D(n+1))$. Thus $N(n)$ is bounded above and below by positive exponential functions of $n$. No precise common growth exponent is claimed.

An explicit example is the regular right-angled pentagon tiling $\{5,4\}$: reflect a regular hyperbolic pentagon with all angles $\pi /2$ in its sides. Four pentagons meet at each vertex, and their interiors tile the plane. Its tiles have area $3\pi -5\pi /2=\pi$ by the hyperbolic polygon area formula. The compactness and ordinary local finiteness assumptions exclude pathological decompositions with degenerate tiles.