The formula for the volume of a tetrahedron is:

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

When \(\displaystyle \small V= \frac{4}{3\sqrt{2}}\) we have \(\displaystyle \small \frac{4}{3\sqrt{2}}=\frac{a^3}{6\sqrt{2}}\) .  

Multiplying the left side by \(\displaystyle \small \frac{2}{2}\) gives us,

 \(\displaystyle \small \frac{8}{6\sqrt{2}}=\frac{a^3}{6\sqrt{2}}\) , or \(\displaystyle \small 8=a^3\). 

Finally taking the third root of both sides yields \(\displaystyle \small 2=a\)

A regular tetrahedron has four faces of equal area made of equilateral triangles.

Therefore, we know that one face will be equal to:

 \(\displaystyle \small \left(\frac{1}{4}\right)\left(16\sqrt{3}\right) ft^2\) , or \(\displaystyle \small 4\sqrt{3} ft^2\)

Since the surface of one face is an equilateral triangle, and we know that,

 \(\displaystyle \small Area=\left(\frac{1}{2}\right)b\times h\) , the problem can be expressed as:

\(\displaystyle \small 4\sqrt{3}=\left(\frac{1}{2}\right)b\times h\)

In an equilateral triangle, the height \(\displaystyle \small h\), is equal to \(\displaystyle \small \frac{1}{2}(b)\sqrt{3}\) so we can substitute for \(\displaystyle \small h\) like so:

\(\displaystyle \small 4\sqrt{3}=\left(\frac{1}{2}\right)b \times \left(\frac{1}{2}\right)b\sqrt{3}\)

Solving for \(\displaystyle \small b\) gives us the length of one edge.

\(\displaystyle \small 4\sqrt{3}=\frac{b^2\sqrt{3}}{4}\)

\(\displaystyle \small (4)(4\sqrt{3})=(4)\left(\frac{b^2\sqrt{3}}{4}\right)\)

\(\displaystyle \small 16\sqrt{3}=b^2\sqrt{3}\)

\(\displaystyle \small 16=b^2\)

\(\displaystyle \small b=\pm 4\) 

However, we know that the edge of the tetrahedron is a positive number so \(\displaystyle \small b=4\).

Since the base \(\displaystyle \small b\) is the same as one edge of the tetrahedron, and a tetrahedron has six edges we multiply \(\displaystyle \small 6 \times 4\) to arrive at \(\displaystyle \small 24 ft^2\)

A regular tetrahedron has four faces of equal area made of equilateral triangles.

Therefore, we know that one face will be equal to,

 \(\displaystyle \left(\frac{1}{4}\right)49\sqrt{3}\) cm , or \(\displaystyle \small \small \frac{49}{4}\sqrt{3}\) cm.

 Since the surface of one face is an equilateral triangle, and we know that,

 \(\displaystyle \small Area=\left(\frac{1}{2}\right)b\times h\) , the problem can be expressed as:

\(\displaystyle \small \small \frac{49}{4}\sqrt{3}=\left(\frac{1}{2}\right)b\times h\)

In an equilateral triangle, the height \(\displaystyle \small h\) is equal to \(\displaystyle \small \frac{1}{2}(b)\sqrt{3}\) so we can substitute for \(\displaystyle \small h\) like so:

\(\displaystyle \small \small \frac{49}{4}\sqrt{3}=\left(\frac{1}{2}\right)b \times \left(\frac{1}{2}\right)b\sqrt{3}\)

Solving for \(\displaystyle \small b\) gives us the length of one edge.

\(\displaystyle \small \small \frac{49}{4}\sqrt{3}=\frac{b^2\sqrt{3}}{4}\)

\(\displaystyle \small \small 49\sqrt{3}=b^2\sqrt{3}\)

\(\displaystyle \small \small 49=b^2\)

\(\displaystyle \small b=\pm7\)

However, we know that the edge of the tetrahedron is a positive number so

\(\displaystyle \small \small b=7\).

The formula for the volume of a tetrahedron is \(\displaystyle \small V=\frac{a^3}{6\sqrt{2}}\).  When \(\displaystyle \small \small V= 36\) we have \(\displaystyle \small 36=\frac{a^3}{6\sqrt{2}}\) .  

We simply solve for \(\displaystyle \small a\)...

\(\displaystyle \small 6\sqrt{2}\times36=a^3\)

\(\displaystyle \small 216\sqrt{2}=a^3\).

Take the cube root of both sides to find the answer for a.

\(\displaystyle \small \small 6\left ( \left ( 2\right^{1/2} ) \right )^{1/3}=a\)

\(\displaystyle \small \small a=6 \sqrt[6]{2}\)

The area of one side is given as \(\displaystyle \small 25\) \(\displaystyle \small m^2\).  The side of a regular tetrahedron is an equilateral triangle so area is determined by:

\(\displaystyle \small Area= \frac{1}{2}b\times h\).

In an equilateral triangle, \(\displaystyle \small h=\frac{1}{2}b\sqrt{3}\)  so we can substitute for \(\displaystyle \small h\) into the area formula:

\(\displaystyle \small Area = \frac{1}{2}b\times \frac{1}{2}b\sqrt{3}\).

Plugging in the value of the area which was given yields.

\(\displaystyle \small 25= \frac{1}{2}b\times \frac{1}{2}b\sqrt{3}=\frac{b^2\sqrt{3}}{4}\)

Solve for \(\displaystyle \small b\) will give us the length of an edge.

\(\displaystyle \small 100=b^2\sqrt{3}\)

\(\displaystyle \small b^2=\frac{100}{\sqrt{3}}\)

\(\displaystyle \small b=\frac{10}{\sqrt[4]{3}}\)

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 \(\displaystyle a\). 

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

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

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