How to find the volume of a tetrahedron - Advanced Geometry

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Question

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

Answer

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

$V=\frac{a^3\sqrt{2}}{12}$ , where represents the length of the side.

Plugging in our values we get:

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

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

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Question

Find the volume of a tetrahedron with an edge of .

Answer

Write the formula for the volume of a tetrahedron.

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

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

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

Rationalize the denominator.

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

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Question

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

Answer

Write the formula for the volume of a tetrahedron.

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

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

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

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:

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

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Question

Find the volume of a tetrahedron with an edge of .

Answer

Write the formula for finding the volume of a tetrahedron.

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

Substitute in the edge length provided in the problem.

$V=\frac{(6:cm)^3}{6\sqrt 2}=\frac{6^3}{6\sqrt 2}:cm^3$

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:

$V= \frac{36}{\sqrt2}:cm^3=\frac{36}{\sqrt2}:cm^3 \cdot \frac{\sqrt2}{\sqrt2} =\frac{36\sqrt2}{2}:cm^3 = 18\sqrt2:cm^3$

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Question

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

Answer

Write the formula the volume of a tetrahedron.

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

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.

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

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Question

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

Answer

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

$V=\frac{s^3}{6\sqrt 2}= \frac{(\sqrt 5)^3}{6\sqrt 2} = \frac{25\sqrt 5}{6\sqrt2}$

Rationalize the denominator to arrive at the correct answer.

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

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Question

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

Answer

Write the volume equation for a tetrahedron.

$V=\frac{e^3}{6\sqrt2}$

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.

$V=\frac{(\sqrt[3]6:cm)^3}{6\sqrt2} = \frac{6:cm^3}{6\sqrt2}= \frac{1}{\sqrt2}:cm$

Rationalize the denominator.

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

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Question

Find the volume of the regular tetrahedron with side length .

Answer

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:

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

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Question

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

Answer

Write the equation to find the volume of a tetrahedron.

$V=\frac{a^3}{6\sqrt2}$

Substitute the side length and solve for the volume.

$V=\frac{(\frac{1}{6})^3}{6\sqrt2}= \frac{1}{216}\left(\frac{1}{6\sqrt2}\right)= \frac{1}{1296\sqrt2}$

Rationalize the denominator.

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

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Question

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

Answer

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 $\small a=6$ ,

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

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Question

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

Answer

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!

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Question

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

Answer

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."

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Question

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

Answer

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,

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

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Question

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

Answer

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!

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Question

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

Answer

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 .

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

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

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Question

What is the volume of the tetrahedron shown below?

Answer

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

This tetrahedron has a side with a length of 8.

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

You can reduce that answer further, so that it becomes

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

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