+168 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?
+26393 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
+129899
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
