In Euclidean space, $\pi$ is the constant value $3.14159\dots$ But I tried to measure the value of $\pi$ and found that $\pi$ is not constant!
So I wonder if there is a range of $\pi$. If so, is $\pi$ in hyperbolic geometry bigger than $3.14159\dots$ (in Euclidean geometry)?
As Blue noted in a comment, the range of perimeter/diameter ratio of hyperbolic circles is $(\pi,\infty)$. Blue's comment contains an exact formula; I'll instead give a qualitative analysis.
One way to realize the hyperbolic plane is the Euclidean disk $\{x\in\mathbb R^2:|x|<1\}$ on which the length of a curve $\gamma$ is computed is $$\int_{\gamma} \frac{|d\gamma(t)|}{ 1- |\gamma(t)|^2}$$ (Often there is a constant of $4$ here, to normalize the curvature to $-1$. This constant would cancel out in the computations below, anyway.)
Let's consider the circle of Euclidean radius $r\in (0,1)$ centered at $0$. Its hyperbolic radius is $$r_h = \int_0^r \frac{1}{1-t^2}\,dt$$ and its hyperbolic perimeter is $$p_h = \frac{2\pi r}{1-r^2}$$ The integral can be taken exactly, but the desired conclusion can be obtained without doing this. Indeed, $$r_h < \int_0^r \frac{1}{1-r^2}\,dt = \frac{r}{1-r^2} = \frac{p_h}{2\pi} $$ This tells you that the "hyperbolic $\pi$", $p_h/(2r_h)$, is strictly greater than the Euclidean $\pi$. To see that "hyperbolic $\pi$" can be arbitrarily large, note that $$r_h < \int_0^r \frac{1}{1-x}\,dx = -\log(1-r) $$ which implies $p_h/r_h\to \infty$ as $r\to 1$ (e.g., with L'Hospital's rule).