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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?

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    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. – 2017-01-07
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    A quick reference: https://books.google.it/books?id=mjb0BwAAQBAJ&pg=PA139&lpg=PA139&dq=isoperimetric+inequality+hyperbolic+geometry&source=bl&ots=4jcGOdyhmx&sig=tBjn-J3JQUP0Poo4uQTpNZNr1oo&hl=it&sa=X&ved=0ahUKEwjzpP24mLHRAhVDxRQKHfR1AF8Q6AEIYzAH#v=onepage&q=isoperimetric%20inequality%20hyperbolic%20geometry&f=false – 2017-01-07
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    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. – 2017-01-07

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They aren't equal, but as you've noticed they're approximately equal.

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.