Write the formula for finding the surface area of a tetrahedron.

\(\displaystyle A=\sqrt3 s^2\)

Substitute the edge and solve.

\(\displaystyle A=\sqrt3 (\sqrt3)^2 = \sqrt3 (3) = 3\sqrt3\)

The surface area is the area of all of the faces of the tetrahedron. To begin, we must find the area of one of the faces. Because a tetrahedron is made up of triangles, we simply plug the given values for base and height into the formula for the area of a triangle:

\(\displaystyle \small A=\frac{1}{2}bh\)

\(\displaystyle A=\frac{1}{2}(11cm)(9cm)\)

\(\displaystyle A=49.5cm^2\)

Therefore, the area of one of the faces of the tetrahedron is \(\displaystyle 49.5cm^2\). However, because a tetrahedron has 4 faces, in order to find the surface area, we must multiply this number by 4:

\(\displaystyle S.A.=4 \cdot 49.5cm^2\)

\(\displaystyle \small S.A.=198cm^2\)

Therefore, the surface area of the tetrahedron is \(\displaystyle 198cm^2\).

If this is a regular tetrahedron, then all four triangles are equilateral triangles. 

If the slant height is \(\displaystyle \frac{\sqrt3}{2}\), then that equates to the height of any of the triangles being \(\displaystyle \frac{\sqrt3}{2}\).

In order to solve for the surface area, we can use the formula

\(\displaystyle SA=\sqrt{3}\cdot a^2\)

where \(\displaystyle a\) in this case is the measure of the edge.

The problem has not given the edge; however, it has provided information that will allow us to solve for the edge and therefore the surface area. 

Picture an equilateral triangle with a height \(\displaystyle \frac{\sqrt3}{2}\).

Drawing in the height will divide the equilateral triangle into two 30/60/90 right triangles. Because this is an equilateral triangle, we can deduce that finding the measure of the hypotenuse will suffice to solve for the edge length (\(\displaystyle a\)). 

In order to solve for the hypotenuse of one of the right triangles, either trig functions or the rules of the special 30/60/90 triangle can be used. 

Using trig functions, one option is using \(\displaystyle cos(30^{\circ})=\frac{\frac{\sqrt3}{2}}{a}\).

Rearranging the equation to solve for \(\displaystyle a\), 

\(\displaystyle a \cdot cos(30^{\circ}) = \frac{\sqrt3}{2}\)

\(\displaystyle a = \frac{\frac{\sqrt3}{2}}{cos(30^{\circ})}\)

\(\displaystyle a = 1\)

Now that \(\displaystyle a\) has been solved for, it can be substituted into the surface area equation.

\(\displaystyle SA= \sqrt{3}\cdot (1)^2\)

\(\displaystyle SA= \sqrt{3} \cdot 1\)

\(\displaystyle SA = {\color{Blue} \sqrt{3}}\)

The problem is essentially asking us to go from a three-dimensional measurement to a two-dimensional one. In order to approach the problem, it's helpful to see how volume and surface area are related. 

This can be done by comparing the formulas for surface area and volume:

\(\displaystyle V= \frac{a^3}{6\sqrt{2}}\) 

\(\displaystyle SA = \sqrt{3}\cdot a^2\)

We can see that both calculation revolve around the edge length.

That means, if we can solve for \(\displaystyle a\) (edge length) using volume, we can solve for the surface area. 

\(\displaystyle 27=\frac{a^3}{6\sqrt{2}}\)

\(\displaystyle 27 \cdot 6\sqrt{2}= a^3\)

\(\displaystyle \sqrt[3]{27 \cdot 6\sqrt{2}}=a\)

\(\displaystyle {\color{Green} 6.12}=a\)

Now that we know \(\displaystyle a\), we can substitute this value in for the surface area formula:

\(\displaystyle SA=\sqrt{3} \cdot a^2\)

\(\displaystyle SA = \sqrt{3} \cdot (6.12)^2\)

\(\displaystyle SA={\color{Blue} 64.9}\)

Write the volume equation for a tetrahedron.

\(\displaystyle V=\frac{e^3}{6\sqrt2}\)

In this formula, \(\displaystyle V\) stands for the tetrahedron's volume and \(\displaystyle e\) stands for the length of one of its edges.

Substitute the given edge length and solve.

\(\displaystyle V=\frac{(\sqrt[3]6\:cm)^3}{6\sqrt2} = \frac{6\:cm^3}{6\sqrt2}= \frac{1}{\sqrt2}\:cm\)

Rationalize the denominator.

\(\displaystyle \frac{1}{\sqrt2}\:cm\cdot \frac{\sqrt2}{\sqrt2} = \frac{\sqrt2}{2}\:cm\)

Write the equation to find the volume of a tetrahedron.

\(\displaystyle V=\frac{a^3}{6\sqrt2}\)

Substitute the side length and solve for the volume.

\(\displaystyle V=\frac{(\frac{1}{6})^3}{6\sqrt2}= \frac{1}{216}\left(\frac{1}{6\sqrt2}\right)= \frac{1}{1296\sqrt2}\)

Rationalize the denominator.

\(\displaystyle V=\frac{1}{1296\sqrt2} \cdot \frac{\sqrt2}{\sqrt2}= \frac{\sqrt2}{1296\cdot 2} = \frac{\sqrt2}{2592}\)

The volume of a tetrahedron can be solved for by using the equation:

\(\displaystyle V= \frac{a^3}{6\sqrt{2}}\)

where \(\displaystyle a\) is the measurement of the edge of the tetrahedron. 

This problem can be quickly solved by substituting 6 in for \(\displaystyle a\). 

\(\displaystyle V=\frac{6^3}{6\sqrt{2}}=\frac{216}{6\sqrt{2}}=\frac{36}{\sqrt{2}}\)

\(\displaystyle {\color{Blue} V=25.5}\)

Solve the equaiton for the surface area of a sphere for the radius and plug in the values:

\(\displaystyle SA=4{\pi}r^2\rightarrow r=\sqrt{\frac{SA}{4\pi}}= \sqrt{\frac{64\pi}{4\pi}}=\sqrt16=4\)

Recall that the equation for the volume of a sphere is:

\(\displaystyle V=\frac{4}{3}\pi r^3\)

For our data, we know:

\(\displaystyle 2304\pi=\frac{4}{3}\pi r^3\)

Solve for \(\displaystyle r\). First, multiply both sides by \(\displaystyle \frac{3}{4}\):

\(\displaystyle \frac{3}{4}2304\pi=\pi r^3\)

\(\displaystyle 1728\pi=\pi r^3\)

Now, divide out the \(\displaystyle \pi\):

\(\displaystyle 1728= r^3\)

Using your calculator, you can solve for \(\displaystyle r\). Remember, if need be, you can raise \(\displaystyle 1728\) to the power of \(\displaystyle \frac{1}{3}\) if your calculator does not have a variable-root button.

This gives you:

\(\displaystyle r=12\)

If you get something like \(\displaystyle 11.999999....\), just round up. This is a rounding issue with some calculators.

Recall that the equation for the volume of a sphere is:

\(\displaystyle V=\frac{4}{3}\pi r^3\)

For our data, we know:

\(\displaystyle 972\pi = \frac{4}{3}\pi r^3\)

Solve for \(\displaystyle r\). Begin by dividing out the \(\displaystyle \pi\) from both sides:

\(\displaystyle 972 = \frac{4}{3} r^3\)

Next, multiply both sides by \(\displaystyle \frac{3}{4}\):

\(\displaystyle \frac{3}{4} * 972 =r^3\)

\(\displaystyle 729=r^3\)

Using your calculator, solve for \(\displaystyle r\). Recall that you can always use the \(\displaystyle \frac{1}{3}\) power if you don't have a variable-root button.  

You should get:

\(\displaystyle r=9\)  If you get \(\displaystyle r=8.99999999...\), just round up to \(\displaystyle 9\). This is a general rounding problem with calculators. Since you are looking for the diameter, you must double this to \(\displaystyle 18\).