User:Hans G. Oberlack/FS R



The rectangular isosceles triangle as base element.

Segments in the general case
The side of the rectangular isosceles triangle ABC $$|AB| =|AC| = s$$ The segment BC is the hypothenusis h of the triangle. h depends on the length of the side: Applying the Pythagorean theorem to the triangle ABC leads to: $$\quad |BC|^2 =|AB|^2 + |AC|^2=$$ $$\quad \Leftrightarrow |BC|^2= 2 \cdot s^2$$ $$\quad \Leftrightarrow |BC| = \sqrt 2 \cdot s$$

Perimeter in the general case
Perimeter of base rectangular isosceles triangle $$ P_0 = |AB|+|BC|+|AC| =s+ \sqrt 2 \cdot s +s =2 \cdot s + \sqrt 2 \cdot s =(2+\sqrt 2) \cdot s$$

Area in the general case
Area of the base rectangular isosceles triangle $$ A_0 = \frac {s^2}{2}$$

Centroids in the general case
By definition the centroid points of a base shape are $$x_0=0 \quad y_0=0$$. Relatively is the lower left point of the of the base of the triangle at: $$x_L= -\frac {s}{3} \quad   y_L= - \frac{s}{3} $$

Area in the normalised case
In the normalised case the area of the base isosceles triangle is set to $$A_0= 1$$.

Segment in the normalised case
With $$A_0 =\frac{s^2} {2} =1 \quad \Rightarrow s^2=2 \quad \Rightarrow s=\sqrt 2 = 1.41421...$$

Perimeter in the normalised case
Perimeter of base rectangular isosceles triangle $$ P_0 = (2+ \sqrt 2) \cdot s = (2+ \sqrt 2) \cdot \sqrt 2 = 2 \cdot (\sqrt 2+ 1) = 4.8284271... $$

Centroids in the normalised case
The positions of lower left point of the base rectangular isosceles triangle: $$\quad x_L= -\frac {s}{3}= -\frac {\sqrt 2} {3} = - 0.4714...$$ $$\quad  y_L=- \frac {s}{3}= -\frac {\sqrt 2} {3} = - 0.4714...$$

Identifying number
Apart of the base element there is no other shape allocated. Therefore the integer part of the identifying number is 0. The decimal part of the identifying number is the decimal part of the sum of the perimeters and the distances of the centroids in the normalised case. $$decimalpart (4.8284271...+0)=decimalpart(4.8284271...)=.8284271...$$ So the identifying number is: $$0.8284271$$