Let $T$ be a triangle in $\mathbb{H}^2$. Its area is $\pi - \alpha - \beta - \gamma$, where $\alpha$, $\beta$, and $\gamma$ are the interior angles. You can find how slim this triangle is by considering an inscribed circle in $T$. The radius of this triangle, thus $\delta$, are bounded above by the area, so to find the $\delta$ that works for all triangles, you take the limit and consider an ideal triangle $T_\infty$. You can explicitly compute that the inscribed circle has radius $4 \log \phi$, where $\phi$ is the golden ratio. (See [here][1] and [here][2].)


  [1]: http://en.wikipedia.org/wiki/Ideal_triangle
  [2]: http://392c.wordpress.com/2009/02/12/more-quasi-isometry-and-hyperbolic-metric-spaces/