A regular tetrahedron has 4 triangular faces. The area of one of these faces is given by:

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

Because the surface area is the area of all 4 faces combined, in order to find the area for one of the faces only, we must divide the surface area by 4. We know that the surface area is \(\displaystyle \small 684cm^2\), therefore:

\(\displaystyle \small A=\frac{S.A.}{4}\)

\(\displaystyle A=\frac{684cm^{2}}{4}=171cm^2\)

Since we now have the area of one face, and we know the height of one face is \(\displaystyle 18cm\), we can now plug these values into the original formula:

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

\(\displaystyle \small 171=\frac{1}{2}b(18)\)

\(\displaystyle \small 171=9b\)

\(\displaystyle \small b=19\)

Therefore, the length of the base of one face is \(\displaystyle \small 19cm\).

The only given information is the surface area of the regular tetrahedron.

This is a quick problem that can be easily solved for by using the formula for the surface area of a tetrahedron:

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

If we substitute in the given infomation, we are left with the edge being the only unknown. 

\(\displaystyle 156 = \sqrt{3} \cdot a^2\)

\(\displaystyle \frac{156}{\sqrt{3}}=a^2\)

\(\displaystyle \sqrt{\frac{156}{\sqrt{3}}}=a\)

\(\displaystyle a=9.49 \approx{\color{Blue} 9.5}\)

The problem provides the information for the slant height and the area of one of the equilateral triangle faces. 

The slant height merely refers to the height of this equilateral triangle. 

Therefore, if we're given the area of a triangle and it's height, we should be able to solve for it's base. The base in this case will equate to the measurement of the edge. It's helpful to remember that in this case, because all faces are equilateral triangles, the measure of one length will equate to the length of all other edges.

We can use the equation that will allow us to solve for the area of a triangle:

\(\displaystyle A=\frac{1}{2} \cdot b\cdot h\)

where \(\displaystyle b\) is base length and \(\displaystyle h\) is height.

Substituting in the information that's been provided, we get:

\(\displaystyle 43.3 = \frac{1}{2} \cdot b \cdot \frac{10\sqrt{3}}{2}\)

\(\displaystyle b \cdot \frac{10\sqrt{3}}{2}= 2 \cdot 43.3\)

\(\displaystyle b = 2 \cdot 43.3 \cdot \frac{2}{10\sqrt{3}}\)

\(\displaystyle b =9.99971\approx {\color{Blue} 10}\)

This becomes a quick problem by just utilizing the formula for the volume of a tetrahedron. 

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

Upon substituting the value for the volume into the formula, we are left with \(\displaystyle a\), which represents the edge length. 

\(\displaystyle 94.8= \frac{a^3}{6\sqrt{2}}\)

\(\displaystyle a^3=94.8 \cdot 6\sqrt{2}\)

\(\displaystyle a= \sqrt[3]{94.8 \cdot 6\sqrt{2}}\)

\(\displaystyle a= {\color{Blue} 9.3}\)

The problem states that the volume is:

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

The point of the problem is to solve for the length of the edge. Becasuse there are no numbers, the final answer will be an expression. 

In order to solve for it, we will have to rearrange the formula for volume in terms of \(\displaystyle a\). 

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

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

\(\displaystyle a= \sqrt[3]{\frac{V}{2\sqrt3}}\)

The diagonal of a shape is simply the length from a vertex to the center of the face or vertex opposite to it. With a regular tetrahedron, we have a face opposite to the vertex, and this basically amounts to calculating the height of our shape.

We know that the height of a tetrahedron is \(\displaystyle {\large }s\sqrt{\frac{2}{3}}\) where s is the side length, so we can put \(\displaystyle s = 9\) into this formula:

\(\displaystyle {\large} 9\sqrt{2/3} = 3(3/\sqrt3)(\sqrt2) = 3\sqrt{3\cdot2} = 3\sqrt{6}\)

which gives us the correct answer.

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:

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

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

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