How to find the volume of a tetrahedron - Geometry

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What is the volume of the following tetrahedron? Assume the figure is a regular tetrahedron.

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

, where represents the length of the side.

Plugging in our values we get:

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

$V=\frac{8: $m^3$\sqrt{2}$}{12} = $$\frac{2\sqrt{2}$}{3}m^3$$

Find the volume of a tetrahedron with an edge of .

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Write the formula for the volume of a tetrahedron.

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

Rationalize the denominator.

$V= $\frac{1}{6\sqrt2}$:cm\cdot $\frac{\sqrt2}{\sqrt2}$ = $$\frac{\sqrt2}{12}$:cm^3$$

Find the volume of a tetrahedron with an edge of $\sqrt2:in$ .

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Write the formula for the volume of a tetrahedron.

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

Cancel out the $\sqrt2$ in the denominator with one in the numerator:

$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:

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

Find the volume of a tetrahedron with an edge of .

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Write the formula for finding the volume of a tetrahedron.

Substitute in the edge length provided in the problem.

Cancel out the in the denominator with part of the in the numerator:

$V= $$\frac{6^2$$}{\sqrt2}$:cm^3$$

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

$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$$

Find the volume of a tetrahedron with an edge of $6\sqrt2$ .

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Write the formula the volume of a tetrahedron.

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

$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.

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

Find the volume of a tetrahedron with an edge of $\sqrt5$ .

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

Rationalize the denominator to arrive at the correct answer.

$\frac{25\sqrt5}{6\sqrt2}$ \cdot $\frac{\sqrt2}{\sqrt2}$ = $\frac{25\sqrt 10}{12}$

Find the volume of a regular tetrahedron if one of its edges is $\sqrt[3]{6}:cm$ long.

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Write the volume equation for a tetrahedron.

In this formula, stands for the tetrahedron's volume and stands for the length of one of its edges.

Substitute the given edge length and solve.

Rationalize the denominator.

$$\frac{1}{\sqrt2}$:cm\cdot $\frac{\sqrt2}{\sqrt2}$ = $\frac{\sqrt2}{2}$:cm$

Find the volume of the regular tetrahedron with side length .

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The formula for the volume of a regular tetrahedron is:

$V = $$\frac{s^3$}{6\sqrt{2}$}$

Where is the length of side. Using this formula and the given values, we get:

$V = $$\frac{5:cm^3$}{6\sqrt{2}$} = $14.73:cm^3$$

Find the volume of a tetrahedron if the side length is $\frac{1}{6}$ .

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Write the equation to find the volume of a tetrahedron.

Substitute the side length and solve for the volume.

Rationalize the denominator.

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

What is the volume of a regular tetrahedron with edges of ?

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The volume of a tetrahedron is found with the formula:

$\small V= $$\frac{a^3$}{6\sqrt{2}$}$ ,

where $\small a$ is the length of the edges.

When $\small a=6$ ,

$\small V=\frac{36}{\sqrt{2}$}$ .

What is the volume of a regular tetrahedron with edges of $\small $\frac{2}{3}$ cm$ ?

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The volume of a tetrahedron is found with the formula,

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

When $\small a= $\frac{2}{3}$$ the volume becomes,

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

$\small V=\frac{8}{27}$\div 6$\sqrt{2}$$

$\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!

What is the volume of a regular tetrahedron with edges of $\small \small $\frac{4}{5}$cm$ ?

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

When $\small \small a= $\frac{4}{5}$$

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

$\small \small V=\frac{64}{125}$\div 6$\sqrt{2}$$

$\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."

How is the volume of a regular tetrahedron effected when the length of each edge is doubled?

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

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

Therefore,

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

What is the volume of a regular tetrahedron with edges of $\small 22$ $\small cm$ ?

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

When $\small \small a= 22$ ,

$\small \small \small $V=\frac{22^3$}{6\sqrt{2}$}$

$\small V=\frac{10648}{6\sqrt{2}$}=\frac{5324}{3\sqrt{2}$}$

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

What is the volume of a regular tetrahedron with an edge length of 6?

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The volume of a tetrahedron can be solved for by using the equation:

$V= $$\frac{a^3$}{6\sqrt{2}$}$

where is the measurement of the edge of the tetrahedron.

This problem can be quickly solved by substituting 6 in for .

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

What is the volume of the tetrahedron shown below?

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The volume of a tetrahedron is .

This tetrahedron has a side with a length of 8.

, which becomes $\frac{512}{6\sqrt2}$ .

You can reduce that answer further, so that it becomes

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

What is the volume of the following tetrahedron? Assume the figure is a regular tetrahedron.

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

, where represents the length of the side.

Plugging in our values we get:

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

$V=\frac{8: $m^3$\sqrt{2}$}{12} = $$\frac{2\sqrt{2}$}{3}m^3$$

Find the volume of a tetrahedron with an edge of .

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Write the formula for the volume of a tetrahedron.

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

Rationalize the denominator.

$V= $\frac{1}{6\sqrt2}$:cm\cdot $\frac{\sqrt2}{\sqrt2}$ = $$\frac{\sqrt2}{12}$:cm^3$$

Find the volume of a tetrahedron with an edge of $\sqrt2:in$ .

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Write the formula for the volume of a tetrahedron.

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

Cancel out the $\sqrt2$ in the denominator with one in the numerator:

$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:

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

Find the volume of a tetrahedron with an edge of .

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Write the formula for finding the volume of a tetrahedron.

Substitute in the edge length provided in the problem.

Cancel out the in the denominator with part of the in the numerator:

$V= $$\frac{6^2$$}{\sqrt2}$:cm^3$$

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

$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$$