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### Finding the Height of a Tetrahedron

```
Date: 05/03/98 at 11:57:47
From: Andrew Jones
Subject: Volume of a Tetrahedron

Hi Dr. Math,

I am trying to find out the height of a tetrahedron, so I can work out
its volume using the formula 1/3*(base area)*(perpendicular height).
The tetrahedron's sides are all length x, and I want to know its
volume relative to x.

Please could you tell me how to work out the height of this
tetrahedron, and therefore its area?

Thank you very much.

Yours mathematically,
Andrew Jones
```

```
Date: 05/03/98 at 13:53:37
From: Doctor Sam
Subject: Re: Volume of a Tetrahedron

Andrew,

Great question! I wish I could draw you a picture . . . it is hard
enough to talk about two dimensions, but you have jumped into the
third.

Imagine your tetrahedron sitting on one of its equilateral triangles
as its base. You want to find its height, the perpendicular distance
from the vertex in space to the triangle base.

Try drawing this, and draw in the height h in the middle of the
tetrahedron. You want to find h and you know x, the side of the
tetrahedron. Since h is perpendicular to the base, there is a right
angle at the foot of this line. If you connect the foot of the
perpendicualar to one of the corners of the tetrahedron's base, you
will have a right triangle that stands up in space with only one of
its edges in the plane:

/|
/ |
x/  |h
/   |
/    |
-------

If we could find the length of the third side of this triangle (in
terms of x), we could use the Pythagorean Theorem to find h.

So now look closely at this third side. This is inside the equilateral
triangle that is the tetrahedron's base. In fact, if you join the foot
of h to all three of the vertices of the base, you will have a picture
of an equilateral triangle with three lines meeting in the center of
the triangle.

If you extend these lines, you will have a picture of an equilateral
triangle with three lines drawn from the vertices to the opposite
bases, all meeting in the center of the triangle. In an equilateral
triangle, these lines are medians (they divide the opposite side in
half) and also angle bisectors (they divide the angles in half) and
also altitudes (they are perpendicular to the sides of the
equilateral triangle).

Are you familiar with properties of these lines? There are two
properties that will let you figure out the length you want in terms
of x. The first is that medians divide the opposite side in half.
Therefore, in the base triangle, each median cuts the equilateral
triangle into two right triangles. The hypotenuse of these triangles
is x, the side of the tetrahedron. The edge of the triangle on the
base is x/2.

Together, these values will let you calculate the third side of the
right triangle, the median/altitude/angle bisector line.

The second important property is that the medians of any triangle
intersect at a point that divides the median into the ratio of 2 to 1.
That is, the length that you want (the base of the vertical triangle)
is two-thirds of the length of the median (in the base triangle).

I hope you were able to follow all that . . . it is difficult without
a figure to look at. So draw a good one for yourself!

Good luck, Andrew. I hope that helped.

-Doctor Sam, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Polyhedra

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