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?

AdminMod2
Sep 18, 2017

#1**0 **

**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**

heureka
Sep 19, 2017

#2**+1 **

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

CPhill
Sep 19, 2017