= Hyperbolic area bounds an inscribed disc
If a domain of finite <hyperbolic area> $A$ contains a hyperbolic disc of radius $r$ in curvature $-1$, then $2\pi(\cosh r-1)\leq A$, hence $r\leq\operatorname{arcosh}(1+A/(2\pi))$. For an interior point of a <hyperbolic triangle>, the open disc with radius equal to its distance from the boundary lies inside the triangle. Its area is at most $\pi$ by <Gauss-Bonnet theorem>, giving the elementary bound $r\leq\operatorname{arcosh}(3/2)$; this need not be the sharp triangle inradius bound.
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