Geometry - How to find the volume of a tetrahedron

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

$\small \frac{36}{\sqrt{2}} cm^3$

$\small \frac{216}{\sqrt{2}} cm^3$

$\small \frac{36}{\sqrt{2}} cm^3$

$\small 6\sqrt{2} cm^3$

$\small \sqrt{2} cm^3$

Explanation

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 $\small a=6$ ,

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

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

$\frac{25\sqrt10}{12}$

$\frac{125}{6}$

$\frac{5\sqrt3}{9}$

$5\sqrt3$

$\frac{125\sqrt2}{12}$

Explanation

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.

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

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

$\frac{2\sqrt{2}}{3}m^3$

$\4\sqrt{3}m^3$

$\frac{\sqrt{2}}{3}m^3$

$\frac{8\sqrt{2}}{3}m^3$

$\frac{3\sqrt{2}}{2}m^3$

Explanation

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:

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

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

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

None of the above.

$\small \frac{32}{\sqrt{2}}$ $\small cm^3$

$\small \frac{125}{\sqrt{2}}$ $\small cm^3$

$\small 375\sqrt{2}$

$\small \frac{375}{32\sqrt{2}}$

Explanation

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."

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

Explanation

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 .

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

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

Find the volume of the regular tetrahedron with side length .

Explanation

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:

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

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

$\small \frac{5324}{3\sqrt{2}}$ $\small cm^3$

$\small \frac{5324}{3\sqrt{2}}$ $\small cm^2$

$\small \frac{2662}{3\sqrt{2}}$ $\small cm^3$

$\small \frac{2662}{3\sqrt{2}}$ $\small cm^2$

None of the above.

Explanation

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!

Find the volume of a tetrahedron with an edge of .

$18\sqrt2:cm^3$

$36\sqrt2:cm^3$

Explanation

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:

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

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

$\small \frac{4}{81\sqrt{2}} cm^3$

$\small \frac{4}{81\sqrt{2}} cm^2$

$\small \frac{2}{9\sqrt{2}}cm^3$

$\small \frac{2}{9\sqrt{2}}cm^2$

None of the above.

Explanation

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!

Find the volume of a tetrahedron with an edge of .

$\frac{\sqrt2}{12}:cm^3$

$6\sqrt{2}:cm^3$

$\frac{1}{6}:cm^3$

$\frac{\sqrt3}{4}:cm^3$

Explanation

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.

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

How to find the volume of a tetrahedron

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