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# Prymid

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A regular tetrahedron is a triangular pyramid in which each face is an equilateral triangle. If the height of a regular tetrahedron is 20 inches then what is the length of each edge of the tetrahedron?

Sep 18, 2017

#1
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A regular tetrahedron is a triangular pyramid in which each face is an equilateral triangle.

If the height of a regular tetrahedron is 20 inches then what is the length of each edge of the tetrahedron?

For a regular tetrahedron of edge length a:
Height of pyramid  $${\displaystyle h={\frac {\sqrt {6}}{3}}a={\sqrt {\frac {2}{3}}}\,a\,}$$

$$\begin{array}{|rcll|} \hline h &=& \sqrt {\frac {2}{3}}\,a \\ a &=& \frac{h}{\sqrt {\frac {2}{3}}} \\ a &=& \sqrt {\frac {3}{2}}\,h \\ \mathbf {a} &\mathbf {=}& \mathbf {\sqrt {1.5}\,h } \\ \hline \end{array}$$

$$\begin{array}{rcll} a &=& \sqrt{1.5}\cdot 20 \\ a &=& 24.4948974278\, \text{inch} \\ \end{array}$$

The length of each edge of the tetrahedron is 24.5 inch

Sep 19, 2017
#2
+96275
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Let us find the edge length

Draw a segment from the apex of the tetrahedron perpendicular to the base.....this is the height....call the point where this segment intersects the base, A

Draw  a segment  from one of the bottom vertexes (call this point  B)  to  A

Find the midpoint  of one of the sides connecting to B....label this point  C

And BC  = (1/2) the side length  = (1/2)s

So.....we have the right triangle ABC lying on the base

And we have this relationship

cos 30°  =  ( BC) / (BA)

cos (30°) = (1/2)s / (BA)

BA  =  (1/2)s / [√ 3 / 2]  =  s / √ 3

So......we have another right triangle...one leg is the height, one leg is BA and the hypotenuse is the side length......and we have......

√ [ 20^2 + BA^2]  = s

√ [ 20^2  + (s/√3)^2  ]  = s

√ [(1200 + s^2) / 3 ]  = s

[1200 + s^2] / 3  = s^2

1200 + s^2  = 3s^2

1200  = 2s^2

600  = s^2

√ 600  = s

10√ 6   = s  ≈  24.49  inches

Sep 19, 2017
edited by CPhill  Sep 19, 2017
edited by CPhill  Sep 19, 2017