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

**Q **

is the
angle that a line from a vertex to the centroid of the figure from a face

**x **

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 = _{}

or

R = _{} = _{} = _{}

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

r = _{}

**6.3)** r = _{} where **r** is
the radius of the inscribed sphere.

To express **r** in terms of **R**, where R = _{}

we have

**6.4)** r = _{}

To express **a** in terms of **R**, we have a = _{} and we have finally

**6.5)** a = _{}

Reference [5]

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