# Area of hyperbolic triangle definition

I found this question recently in my booklet on hyperbolic geometry asking a very simple question but I could not answer it:

Why can we not define the area of a hyperbolic triangle as in the plane as half the product of the perpendicular and the base?

I know the half plane model and the Poincare disk models but I cannot find a satisfactory explanation. I thought it might have something to do with that there are no rectangles in hyperbolic geometry but I cannot proceed. Help appreciated.

• I began to studied hyperbolic geometry three weeks ago, so I am not sure for the exact reason. But as far as I known, if we defined the area in hyperbolic plane as we did in Euclid plane, then at least the möbius transformations would not preserve the area as we except. I am really new at this topic, so I am not sure for my commenting. – KWSK Oct 2 '15 at 7:12
• Nice question, but above my knowledge, i made a relatd question of a (tiny) bit simpler problem see math.stackexchange.com/questions/1462778/… – Willemien Oct 3 '15 at 18:07
• Well, one reason is that in hyperbolic geometry $ah_a\neq bh_b$, so this 'area' would depend on the side we choose as the base — so in a sense it's not well-defined. – Grigory M Oct 3 '15 at 19:25
• @KWSK Well, (hyperbolic) lengths of both the base and the (hyperbolic) perpedicular are manifestly invariant under hyperbolic motions. – Grigory M Oct 3 '15 at 19:27

What is the area? Well, at least we want it to be (1) a non-negative function of a polygon that is (2) additive: $S(A\cup B)=S(A)+S(B)$ if $A\cap B$ has no interior points. It turns out, these two simple properties define $S$ almost uniquely — it's unique up to multiplication by a constant.
Now in Euclidean geometry one can prove that $S(\Delta)=ah_a$ by considering a triangle inside the rectangle — but because 'there are no hyperbolic rectangle' the same proof doesn't work in hyperbolic geometry.