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.

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

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 $\displaystyle a$. 

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

$\displaystyle {\color{Blue} V=25.5}$

The volume of a tetrahedron is $\displaystyle V=\frac{s^3}{6\sqrt2}$.

This tetrahedron has a side with a length of 8. 

$\displaystyle V=\frac{8^3}{6\sqrt2}$, which becomes $\displaystyle \frac{512}{6\sqrt2}$.

You can reduce that answer further, so that it becomes 

$\displaystyle \frac{256}{3\sqrt2}$.