Tetrahedron
Measurement

Given that the regular tetrahedron with edge length a {\displaystyle a}. The surface area of a regular tetrahedron A {\displaystyle A} is four times the area of an equilateral triangle: A = 4 ⋅ (3 4 a 2) = a 2 3 ≈ 1.732 a 2. {\displaystyle A=4\cdot \left({\frac {\sqrt {3}}{4}}a^{2}\right)=a^{2}{\sqrt {3}}\approx 1.732a^{2}.} The height of a regular tetrahedron is 6 3 a {\textstyle {\frac {\sqrt {6}}{3}}a}. The volume of a regular tetrahedron can be ascertained similarly as the other pyramids, one-third of the base and its height. Because the base is an equilateral, it is: V = 1 3 ⋅ (3 4 a 2) ⋅ 6 3 a = a 3 6 2 ≈ 0.118 a 3. {\displaystyle V={\frac {1}{3}}\cdot \left({\frac {\sqrt {3}}{4}}a^{2}\right)\cdot {\frac {\sqrt {6}}{3}}a={\frac {a^{3}}{6{\sqrt {2}}}}\approx 0.118a^{3}.} Its volume can also be obtained by dissecting a cube into three parts.

Its dihedral angle — the angle between two planar — and its angle between lines from the center of a regular tetrahedron between two vertices is respectively: arccos (1 3) = arctan (2 2) ≈ 70.529 ∘, arccos (− 1 3) = 2 arctan (2) ≈ 109.471 ∘. {\displaystyle {\begin{aligned}\arccos \left({\frac {1}{3}}\right)&=\arctan \left(2{\sqrt {2}}\right)\approx 70.529^{\circ },\\\arccos \left(-{\frac {1}{3}}\right)&=2\arctan \left({\sqrt {2}}\right)\approx 109.471^{\circ }.\end{aligned}}}

The radii of its circumsphere R {\displaystyle R}, insphere r {\displaystyle r}, midsphere r M {\displaystyle r_{\mathrm {M} }}, and exsphere r E {\displaystyle r_{\mathrm {E} }} are: R = 6 4 a, r = 1 3 R = a 24, r M = r R = a 8, r E = a 6. {\displaystyle {\begin{aligned}R={\frac {\sqrt {6}}{4}}a,&\qquad r={\frac {1}{3}}R={\frac {a}{\sqrt {24}}},\\r_{\mathrm {M} }={\sqrt {rR}}={\frac {a}{\sqrt {8}}},&\qquad r_{\mathrm {E} }={\frac {a}{\sqrt {6}}}.\end{aligned}}} For a regular tetrahedron with side length a {\displaystyle a}, the radius of its circumscribed sphere R {\displaystyle R}, and distances d i {\displaystyle d_{i}} from an arbitrary point in 3-space to its four vertices, it is: d 1 4 + d 2 4 + d 3 4 + d 4 4 4 + 16 R 4 9 = (d 1 2 + d 2 2 + d 3 2 + d 4 2 4 + 2 R 2 3) 2, 4 (a 4 + d 1 4 + d 2 4 + d 3 4 + d 4 4) = (a 2 + d 1 2 + d 2 2 + d 3 2 + d 4 2) 2. {\displaystyle {\begin{aligned}{\frac {d_{1}^{4}+d_{2}^{4}+d_{3}^{4}+d_{4}^{4}}{4}}+{\frac {16R^{4}}{9}}&=\left({\frac {d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}}{4}}+{\frac {2R^{2}}{3}}\right)^{2},\\4\left(a^{4}+d_{1}^{4}+d_{2}^{4}+d_{3}^{4}+d_{4}^{4}\right)&=\left(a^{2}+d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}\right)^{2}.\end{aligned}}}

With respect to the base plane the slope of a face (2 √ 2) is twice that of an edge (√ 2), corresponding to the fact that the horizontal distance covered from the base to the apex along an edge is twice that along the median of a face. In other words, if C is the centroid of the base, the distance from C to a vertex of the base is twice that from C to the midpoint of an edge of the base. This follows from the fact that the medians of a triangle intersect at its centroid, and this point divides each of them in two segments, one of which is twice as long as the other (see proof).

Its solid angle at a vertex subtended by a face is arccos (23 27) = π 2 − 3 arcsin (1 3) = 3 arccos (1 3) − π {\displaystyle {\begin{aligned}\arccos \left({\frac {23}{27}}\right)&={\frac {\pi }{2}}-3\arcsin \left({\frac {1}{3}}\right)\\&=3\arccos \left({\frac {1}{3}}\right)-\pi \end{aligned}}} This is approximately 0.55129 steradians, 1809.8 square degrees, or 0.04387 spats.
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