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

$\displaystyle e=\sqrt2 \sqrt[3]{3V}$

Substitute the volume and solve.

$\displaystyle e=\sqrt2 \sqrt[3]{3(\frac{1}{3})}= \sqrt2 \sqrt[3]{1}=\sqrt2$

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  $\displaystyle 60^{\circ}$.  We are given the height of a face, which is the length of a line that bisects one of the  $\displaystyle 60^{\circ}$  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  $\displaystyle 30^{\circ}$,  whose cosine is equal to the adjacent side, the height, over the hypotenuse, the side length. This gives us:

$\displaystyle \cos 30^{\circ}=\frac{h}{a}$

$\displaystyle \frac{\sqrt{3}}{2}=\frac{6\sqrt{3}}{a}$

$\displaystyle a=\frac{6\sqrt{3}}{\frac{\sqrt{3}}{2}}=6\sqrt{3}\cdot \frac{2}{\sqrt{3}}=12$

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

$\displaystyle e=\sqrt2 \sqrt[3]{3V}$

Substitute the given volume to the equation and solve.

$\displaystyle e=\sqrt2 \sqrt[3]{3(9)}= \sqrt2 \sqrt[3]{27}=3\sqrt2$

Write the tetrahedron formula to solve for the edge.

$\displaystyle e=\sqrt2 \sqrt[3]{3V}$

Substitute the volume and simplify.

$\displaystyle e=\sqrt2 \sqrt[3]{3(3)}= \sqrt2 \sqrt[3]{9}$

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:

$\displaystyle SA=4\left(\frac{1}{2}bh\right)=2bh$

Where  $\displaystyle b$  is the base of the triangle, or the edge length of the tetrahedron, and  $\displaystyle h$  is the height of each triangular face. We want to solve for  $\displaystyle b$,  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  $\displaystyle b$,  is the hypotenuse. Each angle in an equilateral triangle is  $\displaystyle 60^{\circ}$,  so if one is bisected by the height then the angle between it and the hypotenuse is  $\displaystyle 30^{\circ}$.  The cosine of this angle is equal to the height over the hypotenuse, which gives us:

$\displaystyle \cos30^{\circ}=\frac{h}{b}\rightarrow h=b\cos30^{\circ}\rightarrow h=\frac{\sqrt{3}}{2}b$

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:

$\displaystyle SA=2bh=2b\left(\frac{\sqrt{3}}{2}b\right)=\sqrt{3}b^2$

$\displaystyle 16\sqrt{3}=\sqrt{3}b^2$

$\displaystyle b^2=16\rightarrow b=4$

The base of the triangle has an area that can be found using the formula for the area of an equilateral triangle, substituting $\displaystyle s=10$:

$\displaystyle B = \frac{s^{2}\sqrt{3}}{4}$

$\displaystyle B = \frac{10^{2}\sqrt{3}}{4}$

$\displaystyle B= \frac{100\sqrt{3}}{4}$

$\displaystyle B= 25\sqrt{3}$

Now, in the formula for the volume of a pyramid, substitute $\displaystyle B = 25\sqrt{3}, h = 15$:

$\displaystyle V = \frac{1}{3} Bh$

$\displaystyle V = \frac{1}{3}\cdot 25\sqrt{3} \cdot 15$

$\displaystyle V = 125\sqrt{3}$

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 $\displaystyle 9 ^{2} = 81$ square inches. The volume of a pyramid is one-third the product of its height and the area of its base, so substitute $\displaystyle B = 81, h = 9$ in the following:

$\displaystyle V = \frac{1}{3} Bh = \frac{1}{3} \cdot 81 \cdot 9 = 243$ cubic inches.

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

Three of the vertices - $\displaystyle (0,0,0), (60,0,0), (30, 40, 0)$ - are on the $\displaystyle xy$-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

$\displaystyle B = \frac{1}{2} \cdot 60 \cdot 40 = 1,200$

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

$\displaystyle V = \frac{1}{3} \cdot 1,200 \cdot 20 = 8,000$.

The tetrahedron looks like this:

$\displaystyle O$ is the origin and $\displaystyle A,B,C$ 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 $\displaystyle \Delta OCB$ the (right triangular) base; its area is half the product of its legs, or

$\displaystyle B = \frac{1}{2} \cdot n \cdot 4n= 2n^{2}$.

The volume of the tetrahedron is one third the product of its base and its height, the latter of which is $\displaystyle 9n$. Therefore,

$\displaystyle V = \frac{1}{3} \cdot 9n \cdot 2n^{2} = 6 n^{3}$.

The tetrahedron looks like this:

$\displaystyle O$ is the origin and $\displaystyle A,B,C$ 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 $\displaystyle \Delta OAB$ the (right triangular) base; its area is half the product of its legs, or

$\displaystyle B = \frac{1}{2} \cdot 60 \cdot 60 = 1,800$.

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

$\displaystyle V = \frac{1}{3} \cdot 60 \cdot 1,800 = 36,000$.