# In hyperbolic geometry, is the area of a circle approximately EQUAL to the circumference?

I know that the circumference of a circle of radius $R$ in hyperbolic geometry is $2π sinh R$. I believe this is approximately equal to $πe^R$.

I know that the area of a circle of radius $R$ in hyperbolic geometry is $2π(cosh R − 1)$. But this also appears to be approximately equal to $πe^R$.

I plotted all three of these functions on Wolframalpha, and the graphs do appear to be relatively identical when the range on the x-axis is large.

Have I made some mistake here? How can the circumference be equal to the area?

• Obviously both $\sinh(x)$ and $\cosh(x)$ are close to $e^x$ if $x$ is positive and large. Is it really surprising? I guess in hyperbolic geometry we just have a isoperimetric inequality with a different constant. – Jack D'Aurizio Jan 7 '17 at 23:31
• A quick reference: books.google.it/… – Jack D'Aurizio Jan 7 '17 at 23:33
• The way I like to think of this is that the area increases so quickly with radius that "most of" the area is very close to the circumference. – Chappers Jan 7 '17 at 23:41

What's likely happening is that you've discovered that: $$\sinh(x) = \frac{e^x-e^{-x}}{2},\qquad \cosh(x) = \frac{e^x+e^{-x}}{2}$$
So, we have that: $$2\pi\sinh R = \pi e^R-\pi e^{-R}$$ For large $R$ we get that $e^{-R}$ gets small, so we will get that $2\pi\sinh R\approx \pi e^R$.
Now, we have that: $$2\pi(\cosh R-1) = \pi e^R+\pi e^{-R}-2\pi$$ There's no reason that this should be approximately equal to $\pi e^R$, unless you're choosing $R$ large enough that $\pi(e^{-R}-2\pi)$ is negligable compared to $\pi e^R$. In this case, we would again have that it's approximately $\pi e^R$.
In both cases, they're aren't exactly equal. They can be equal to good approximation for large $R$ though.