+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?
+26396 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 h=√63a=√23a
h=√23aa=h√23a=√32ha=√1.5h
a=√1.5⋅20a=24.4948974278inch
The length of each edge of the tetrahedron is 24.5 inch
+130474
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
