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.

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

The formula for the volume of a regular tetrahedron is:

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

Where \(\displaystyle s\) is the length of side. Using this formula and the given values, we get:

\(\displaystyle V = \frac{5\:cm^3}{6\sqrt{2}} = 14.73\:cm^3\)

The volume of a tetrahedron is found with the formula:

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

where \(\displaystyle \small a\) is the length of the edges.

When \(\displaystyle \small a=6\), 

\(\displaystyle \small V=\frac{36}{\sqrt{2}}\).

The volume of a tetrahedron is found with the formula,

 \(\displaystyle \small V= \frac{a^3}{6\sqrt{2}}\) where \(\displaystyle \small a\) is the length of the edges.

When \(\displaystyle \small a= \frac{2}{3}\) the volume becomes,

\(\displaystyle \small V= \left ( \frac{2}{3} ^3\right )\div 6\sqrt{2}\)

\(\displaystyle \small V=\frac{8}{27}\div 6\sqrt{2}\)

\(\displaystyle \small V= \frac{8}{27}\times\frac{1}{6\sqrt{2}}=\frac{4}{81\sqrt{2}}\)

The answer is in volume, so it must be in a cubic measurement!

The volume of a tetrahedron is found with the formula \(\displaystyle \small V= \frac{a^3}{6\sqrt{2}}\) where \(\displaystyle \small a\) is the length of the edges.

When \(\displaystyle \small \small a= \frac{4}{5}\)

\(\displaystyle \small \small V= \left ( \frac{4}{5} ^3\right )\div 6\sqrt{2}\)

\(\displaystyle \small \small V=\frac{64}{125}\div 6\sqrt{2}\)

\(\displaystyle \small \small \small V= \frac{64}{125}\times\frac{1}{6\sqrt{2}}=\frac{64}{750\sqrt{2}}=\frac{32}{375\sqrt{2}}\)

This answer is not found, so it is "none of the above."

The volume of a regular tetrahedron is found with the formula \(\displaystyle \small \small V_{1}= \frac{a^3}{6\sqrt{2}}\) where \(\displaystyle \small a\) is the length of the edges. 

The volume of the same tetrahedron when the length of the edges are doubled would be \(\displaystyle \small \small \small V_{2}= \frac{(2a)^3}{6\sqrt{2}}\).

Therefore,

\(\displaystyle \small V_{2}=\frac{8a^3}{6\sqrt{2}}=8\times V_{1}\)

The volume of a tetrahedron is found with the formula \(\displaystyle \small V= \frac{a^3}{6\sqrt{2}}\) where \(\displaystyle \small a\) is the length of the edges.

When \(\displaystyle \small \small a= 22\),

 \(\displaystyle \small \small \small V=\frac{22^3}{6\sqrt{2}}\)

\(\displaystyle \small V=\frac{10648}{6\sqrt{2}}=\frac{5324}{3\sqrt{2}}\)

And, of course, volume should be in cubic measurements!

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.

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