The tetrahedron is a solid figure having four faces each of which is an equilateral triangle. It has four vertices and six edges.
Refer to Figure 6.1 A, B and C. In the figure the following notation is used:
a is the length of a side
h is the height of the figure
r is the radius of the inscribed sphere
R is the radius of the circumscribed sphere
is the angle that a line from a vertex to the centroid of the figure from a face
is the distance from a vertex to the centroid of the ‘base’. It is the radius of the circumscribing circle for the equilateral triangle that is the bottom of the figure. See Figure 6.1B.
From Equation 3.2) above, we know that
Figure 6.1.B shows the ‘bottom’ of the tetrahedron. The distance marked d is the distance from the midpoint of a face to the centroid of the face. That is, d is the radius of the circle inscribed in the equilateral triangle that is the bottom of the figure.
To find the height of the figure, h, consider Figure 6.1.C.
so we have that
h = a = a =
So finally we have
6.1) h =
To find the radius,R, of the circumscribed sphere, consider Figure 7.1.C
And from the figure we see that r = h – R so that we have
By collecting and simplifying, we have the following
So 2Rh =
R = = =
So finally we have that
6.2) R = where R is the radius of the circumscribing sphere
To find the radius of the inscribed sphere, note from Figure 6.1.C that r = h –R.
This gives the relationship
6.3) r = where r is the radius of the inscribed sphere.
To express r in terms of R, where R =
6.4) r =
To express a in terms of R, we have a = and we have finally
6.5) a =