Height of a rectangular tetrahedron question?

Question

I have a problem that reads the following:

A rectangular tetrahedron has edges 36 units in length. What is the altitude of the tetrahedron? Give your answer in simplest radical form x sqrt y.

This is supposed to be a challenging problem, and though I'm in honors math, I haven't learned about these shapes. Please don't just give me the answer. Help to point in the right direction.

Thank you so much!

Answer 1

I think you meant a *regular* tetrahedron, not rectangular tetrahedron. A regular tetrahedron has four (tetra) faces that are all the same (equilateral triangles).

Place the tetrahedron on the table. One face is down, the other 3 faces form a 3-sided pyramid. What you want to calculate is the distance from the top of the pyramid straight down to the center of the bottom face.

Start with this bottom face. The center point of the equilateral triangle will be at a distance 2/3 from the vertex and 1/3 the way from a side. You can confirm this through some math or derive it another way, using the attached diagram. The angles are 30-60-90 so there are many ways to do this.
http://img141.imageshack.us/img141/7096/trianglecenterjq3.gif

Given that the side is 36, the length of b (distance from the vertex) will be 2/3(18√3) or 12√3.

Now look at the pyramid from the side. You have a right triangle with the altitude and this measurement (b) as the legs. And one of the sides (36) as the hypotenuse.

This is easy to solve using the Pythagorean theorem.
36² = a² +(12√3)²
36² = a² + 3*144
1296 = a² + 432
a² = 1296 - 432
a² = 864
a = sqrt(864)
a = sqrt(144 * 6)
a = 12√6
a ≈ 29.39 units

This matches with the formula for the altitude of a tetrahedron which is.
a = s(√6/3)

Answer 2

a tetrahedron has four congruent triangular faces.