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:

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

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

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

A tetrahedron comprises four triangular surfaces; if the tetrahedron is regular, then each surface is an equilateral triangle. The area of an equilateral triangle with sides of length \(\displaystyle s\) can be computed using the formula

\(\displaystyle A = \frac{s^{2}\sqrt{3}}{4}\);

The total surface area of the tetrahedron is four times this, or

\(\displaystyle S = 4A = 4 \cdot \frac{s^{2}\sqrt{3}}{4} = s^{2}\sqrt{3}\)

Set \(\displaystyle s = 60\) and substitute:

\(\displaystyle S = s^{2}\sqrt{3} = 60 ^{2}\sqrt{3} = 3,600 \sqrt{3}\).

A regular tetrahedron is composed of four equilateral triangles. The formula for the volume of a regular tetrahedron is:

\(\displaystyle V=\frac{a^3\sqrt{2}}{12}\), where \(\displaystyle a\) represents the length of the side.

Plugging in our values we get:

\(\displaystyle V=\frac{(2\: m)^3\sqrt{2}}{12}\)

\(\displaystyle V=\frac{8\: m^3\sqrt{2}}{12} = \frac{2\sqrt{2}}{3}m^3\)

Write the formula for the volume of a tetrahedron.

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

Substitute in the length of the edge provided in the problem.

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

Rationalize the denominator.

\(\displaystyle V= \frac{1}{6\sqrt2}\:cm\cdot \frac{\sqrt2}{\sqrt2} = \frac{\sqrt2}{12}\:cm^3\)

Write the formula for the volume of a tetrahedron.

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

Substitute in the length of the edge provided in the problem:

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

Cancel out the \(\displaystyle \sqrt2\) in the denominator with one in the numerator:

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

A square root is being raised to the power of two in the numerator; these two operations cancel each other out. After canceling those operations, reduce the remaining fraction to arrive at the correct answer:

\(\displaystyle V= \frac{(\sqrt2)^2}{6}\:in^3=\frac{2}{6}\:in^3=\frac{1}{3}\:in^3\)

Write the formula for finding the volume of a tetrahedron.

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

Substitute in the edge length provided in the problem. 

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

Cancel out the \(\displaystyle 6\) in the denominator with part of the \(\displaystyle 6^3\) in the numerator:

\(\displaystyle V= \frac{6^2}{\sqrt2}\:cm^3\)

Expand, rationalize the denominator, and reduce to arrive at the correct answer:

\(\displaystyle V= \frac{36}{\sqrt2}\:cm^3=\frac{36}{\sqrt2}\:cm^3 \cdot \frac{\sqrt2}{\sqrt2} =\frac{36\sqrt2}{2}\:cm^3 = 18\sqrt2\:cm^3\)

Write the formula the volume of a tetrahedron.

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

Substitute the edge length provided in the equation into the formula.

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

Cancel out the denominator with part of the numerator and solve the remaining part of the numerator to arrive at the correct answer.

\(\displaystyle V= (6\sqrt 2)^2 = 72\)

Write the formula the volume of a tetrahedron and substitute in the provided edge length.

\(\displaystyle V=\frac{s^3}{6\sqrt 2}= \frac{(\sqrt 5)^3}{6\sqrt 2} = \frac{25\sqrt 5}{6\sqrt2}\)

Rationalize the denominator to arrive at the correct answer.

\(\displaystyle \frac{25\sqrt5}{6\sqrt2} \cdot \frac{\sqrt2}{\sqrt2} = \frac{25\sqrt 10}{12}\)