Write the formula to solve for the edge of a tetrahedron.

Substitute the volume and solve.

Because a tetrahedron has four congruent equilateral triangles as its faces, we know the three equal angles of each face, as with all equilateral triangles, are each  .  We are given the height of a face, which is the length of a line that bisects one of the    angles and forms a right triangle with the side length as its hypotenuse. This means the angle between the height and the side length is  ,  whose cosine is equal to the adjacent side, the height, over the hypotenuse, the side length. This gives us:

Write the formula to solve for the edge of a tetrahedron.

Substitute the given volume to the equation and solve.

Write the tetrahedron formula to solve for the edge.

Substitute the volume and simplify.

A tetrahedron has four congruent equilateral triangle faces, so its surface area is four times the area of one of these equilateral triangles. This gives us:

Where    is the base of the triangle, or the edge length of the tetrahedron, and    is the height of each triangular face. We want to solve for  ,  but we are only given the surface area and not the height, so we need to express this value in terms of the base. One way of doing this is by recognizing that the height of each equilateral triangle face bisects an angle and forms two congruent right triangles for which it is the base and the edge length, or  ,  is the hypotenuse. Each angle in an equilateral triangle is  ,  so if one is bisected by the height then the angle between it and the hypotenuse is  .  The cosine of this angle is equal to the height over the hypotenuse, which gives us:

Now that we have an expression for the height in terms of the base, we can plug this value into the equation for surface area and be left with only one unknown, the base that is equivalent to the edge length for which we want to solve:

The base of the triangle has an area that can be found using the formula for the area of an equilateral triangle, substituting :

Now, in the formula for the volume of a pyramid, substitute :

The perimeter of the base is one yard, or 36 inches; its sidelength - and, sunsequently, its height - are one-fourth of that, or 9 inches, and the area of the base is  square inches. The volume of a pyramid is one-third the product of its height and the area of its base, so substitute  in the following:

 cubic inches.

A tetrahedron is a triangular pyramid and can be looked at as such.

Three of the vertices -  - are on the -plane, and can be seen as the vertices of the triangular base. This triangle, as seen below, is isosceles:

Its base is 60 and its height is 40, so its area is

The fourth vertex is off the -plane; its perpendicular distance to the aforementioned face is its -coordinate, 20, so this is the height of the pyramid. The volume of the pyramid is 

.

The tetrahedron looks like this:

 is the origin and  are the other three points, whose distances away from the origin on each of the three (perpendicular) axes are shown.

This is a triangular pyramid, and we can consider  the (right triangular) base; its area is half the product of its legs, or

.

The volume of the tetrahedron is one third the product of its base and its height, the latter of which is . Therefore,

.

The tetrahedron looks like this:

 is the origin and  are the other three points, which are 60 units away from the origin on each of the three (mutually perpendicular) axes.

This is a triangular pyramid, and we can consider  the (right triangular) base; its area is half the product of its legs, or

.

The volume of the tetrahedron is one third the product of its base and its height, the latter of which is 60. Therefore,

.