The location of Steffen's Polyhedron's vertexes is listed as below:

$P_1 = (0, 0, 0)$

$P_2 = (-12, 0, 0)$

$P_3 = (\frac{1}{24}, -\frac{17*\sqrt[2]{287}}{24}, 0)$

$P_4 = (x, r\cos\theta ,r\sin\theta)$

$Note_1$: $x$ and $r$ in $P_4$ can be calculated with the following equation set :

$$ \begin{cases} \left\|P_4-P_1\right\|=5 \\ \left\|P_4-P_2\right\|=10 \end{cases} $$

$P_5$ to $P_7$ can be calculated with the following equation set :

$$ \begin{cases} \left\|P_7-P_6\right\|=10 \\ \left\|P_7-P_5\right\|=5 \\ \left\|P_7-P_3\right\|=12 \\ \left\|P_7-P_2\right\|=12 \\ \left\|P_6-P_5\right\|=12 \\ \left\|P_6-P_4\right\|=12 \\ \left\|P_6-P_1\right\|=10 \\ \left\|P_5-P_4\right\|=11 \\ \left\|P_5-P_2\right\|=10 \\ \end{cases} $$

$P_8$ and $P_9$ can be calculated with the following two equation sets :

$$ \begin{cases} \left\|P_8-P_7\right\|=5 \\ \left\|P_8-P_6\right\|=12 \\ \left\|P_8-P_3\right\|=10 \end{cases} $$

$$ \begin{cases} \left\|P_9-P_6\right\|=12 \\ \left\|P_9-P_3\right\|=10 \\ \left\|P_9-P_1\right\|=5 \end{cases} $$

Finally, the fourteens triangles that are used to make up this Polyhedron is $\triangle{p_1} {p_2} {p_3} , \triangle{p_7} {p_3} {p_2} , \triangle{p_1} {p_4} {p_2} , \triangle{p_2} {p_4} {p_5} , \triangle{p_2} {p_5} {p_7} , \triangle{p_1} {p_6} {p_4} , \triangle{p_4} {p_6} {p_5} , \triangle{p_5} {p_6} {p_7} , \triangle{p_6} {p_8} {p_7} , \triangle{p_6} {p_9} {p_8} , \triangle{p_1} {p_9} {p_6} , \triangle{p_3} {p_7} {p_8} , \triangle{p_3} {p_8} {p_9} , \triangle{p_1} {p_3} {p_9}$