# Volume of a tetrahedron

## Calculates the volume and surface area of a tetrahedron from six edge lengths.

edge length a1
a2
a3
a4
a5
a6

volume V
surface area S
 $\normal Tetrahedron\\\vspace{5}(1)\ volume:\\\hspace{40} V=sqrt{V^2}\\\vspace{5}\hspace{40} V^2=\frac{1}{144}[a_1^2a_5^2(a_2^2+a_3^2+a_4^2+a_6^2-a_1^2-a_5^2)\\\hspace{60}+a_2^2a_6^2(a_1^2+a_3^2+a_4^2+a_5^2-a_2^2-a_6^2)\\\hspace{60}+a_3^2a_4^2(a_1^2+a_2^2+a_5^2+a_6^2-a_3^2-a_4^2)\\\hspace{60}-a_1^2a_2^2a_4^2-a_2^2a_3^2a_5^2-a_1^2a_3^2a_6^2-a_4^2a_5^2a_6^2]\\\vspace{5}(2)\ surface\ area:\\\hspace{40} S=\sqrt{s_1(s_1-a_1)(s_1-a_2)(s_1-a_4)}\\\hspace{60}+\sqrt{s_2(s_2-a_2)(s_2-a_3)(s_2-a_5)}\\\hspace{60}+\sqrt{s_3(s_3-a_3)(s_3-a_6)(s_3-a_1)}\\\hspace{60}+\sqrt{s_4(s_4-a_4)(s_4-a_5)(s_4-a_6)}\\\vspace{5}\hspace{40}s1={\large \frac{a_1+a_2+a_4}{2}},\ s2={\large \frac{a_2+a_3+a_5}{2}}\\\hspace{40}s3={\large \frac{a_3+a_6+a_1}{2}},\ s4={\large \frac{a_4+a_5+a_6}{2}}\\$
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