How does one find the area and perimeter of the hyperbolic octagon with interior angles $\frac{\pi}2$? I'm completely stuck.
I have subdivided the octagon into eight hyperbolic triangles with two interior angles $\frac{\pi}4$. How can I now find the third angle?
Imagine we have your octagon we divide it in 8 equal sectors, (all meeting at the centre of the octagon) then every sector has a top angle at the centre of $ \frac {2 \pi} {8} = \frac {\pi} {4} $ and base angles are the internal angle / 2 $ = \frac {\pi / 2} {2} = \frac {\pi} {4} $
We have 8 of these sectortriangles.
(Only while writing this realised it were all equilateral triangles, but for being more general and following the reasoning in the comments, we just overlook this)
For trigonomic reasons we cut all sectortriangles in half from the top angle to the middle of the base
so we now have 16 triangles each having angles of
The area of this triangle is $ (\pi- \frac {\pi} {4} - \frac {\pi} {8} - \frac { \pi} {2} = \frac {\pi} {8} $
Following wikipedia.https://en.wikipedia.org/wiki/Hyperbolic_triangle#Relations_between_angles
$ \cos A = \cosh a \sin B $
$ \sin A = \frac{\cos B}{\cosh b} $
$ \tan A = \frac{\cot B}{\cosh c} $
Or rewritten
$ \cosh a = \frac{ \cos A} { \sin B} $
$ \cosh b = \frac{ \cos B} { \sin A } $
$ \cosh c = \frac{ \cot B} { \tan A} $
filled in
$ \cosh a = \frac{\cos \frac{\pi} {4}} {\sin \frac{\pi} {8}} $
$ \cosh b = \frac{\cos \frac {\pi} {8}} {\sin \frac{\pi} {4} } $
$ \cosh c = \frac{\cot \frac {\pi} {8}} {\tan \frac {\pi} {4}} $
caculated (not exact values)
$ \cosh a = 1.848 \to a = 1.224 $
$ \cosh b = 1.307 \to b = 0.764 $
$ \cosh c = 2.414 \to c = 1.529 $
Now we have all there is to say about a regular octagon with internal right angles:
the area is $ 16 * \frac {\pi} {8} = 2{\pi}$
the perimater is $ 16 * 0.764 = 12.229 $
the radius of the inscribed circle is $ 1.224$
the radius of the circumscribed circle is $ 1.529 $
Thats all :)