Shows the base element: Isoceles right triangle

Base is the isoceles right triangle of side length ${\displaystyle a}$ .

0) Side length of the triangle: ${\displaystyle a=|AB|=|AC|}$

0) Perimeter of the triangel: ${\displaystyle P_{0}=|AB|+|BC|+|AC|=a+{\sqrt {2}}\cdot a+a=(2+{\sqrt {2}})\cdot a}$

0) Area of the triangle: ${\displaystyle A_{0}={\frac {|AB|\cdot |AC|}{2}}={\frac {a^{2}}{2}}}$

By definition the centroid points of a base shape are ${\displaystyle x_{0}=0\quad y_{0}=0}$.
Relatively is the lower left point of the of the base of the triangle at: ${\displaystyle x_{L}=-{\frac {a}{3}}\quad y_{L}=-{\frac {a}{3}}}$

In the normalised case the area of the base is set to 1.
${\displaystyle ||ABC||=1\Rightarrow {\frac {a^{2}}{2}}=1\Leftrightarrow a^{2}=2\Leftrightarrow a={\sqrt {2}}}$

0) Side length of the triangle: ${\displaystyle a={\sqrt {2}}=1.4142...}$

0) Perimeter of the triangle: ${\displaystyle (2+{\sqrt {2}})\cdot a=(2+{\sqrt {2}})\cdot {\sqrt {2}}=2{\sqrt {2}}+2=4.8284271...}$
.

0) Area of the triangle: ${\displaystyle A_{0}={\frac {{\sqrt {2}}^{2}}{2}}=1}$

0) Centroid positions of the triangle: ${\displaystyle x_{0}=0\quad y_{0}=0}$

Apart of the base element there is no 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.

${\displaystyle decimalpart(4.8284271...+0.0)=decimalpart(4.8284271...)=.8284271}$

So the identifying number is: ${\displaystyle 0.8284271}$

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