Find hyperbolic area of hyperbolic triangle Given $u=(0,0)$, $v=(\frac12,0)$, $w=(\frac12,\frac12)$. Find the hyperbolic area of the hyperbolic triangle $[u,v,w]$ in $\mathbb{H}^2$.
My approach:
One of the angles, say $\alpha$, is a right angle, so we have that $\alpha=\frac{\pi}2$, so that $[u,v,w]$ is a right triangle. Now, this triangle is also isosceles, so two of its sides, say $b$ and $c$, are equal.
Now, $$b=c=\cosh^{-1}\left(\frac53\right)$$
Also, the other side, say $a$, is determined as follows:
$$a=\cosh^{-1}\left(\frac73\right)$$
Now, we get
$$\frac73 = \cosh b=\frac{\cos\beta}{\sin\alpha}=\cot\beta$$
$$\implies \beta = \cot^{-1}\left(\frac73\right)=\alpha$$
$$\implies A=\frac{\pi}2-2\cot^{-1}\left(\frac73\right)$$
I'd appreciate if someone would let me know whether or not my reasoning is correct in this solution.

The area of a hyperbolic triangle is equal to the angle deficit, i.e. to the difference between the hyperbolic sum of interior angles and the Euclidean sum of $\pi$.
Poincaré half plane
Draw the triangle, and it will look like this:

This triangle has two ideal points. I'm not sure I'd call it isosceles, since two legs have infinite length, but you might as well. It has two angles of $0$ and one of $\frac\pi2$, so the sum is $\frac\pi2$, which is $\frac\pi2$ less than the Euclidean sum of a triangle. Thus the area is $\frac\pi2\approx1.5708$.
Poincaré disk
Draw the triangle, and it will look like this:

No right angle here, nor is it obviously isosceles. The segment $vw$ belongs to a circle with center at $\left(\frac54,\frac14\right)$. You can verify that by checking that inverting $v$ or $w$ results in a point on that circle, too. You can use that to compute the lines connecting that center to $v$ resp. $w$, then have the tangents in these points orthogonal to that, and finally compute the angle between tangent and the line to the origin. Sum the angles, subtract it from $\pi$ and you have the area $\pi-2\arctan3\approx0.6435$.
