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ACT Math : Tetrahedrons

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Example Questions

Example Question #1 : How To Find The Volume Of A Tetrahedron

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

Possible Answers:

$\frac{\sqrt2}{2}\;cm$ $\displaystyle \frac{\sqrt2}{2}\;cm$

$2\sqrt6\:cm$ $\displaystyle 2\sqrt6\:cm$

$2\sqrt3\:cm$ $\displaystyle 2\sqrt3\:cm$

$\sqrt6\:cm$ $\displaystyle \sqrt6\:cm$

$4\sqrt3\:cm$ $\displaystyle 4\sqrt3\:cm$

Correct answer:

$\frac{\sqrt2}{2}\;cm$ $\displaystyle \frac{\sqrt2}{2}\;cm$

Explanation:

Write the volume equation for a tetrahedron.

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

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.

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

Rationalize the denominator.

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

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Example Question #83 : Solid Geometry

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

Possible Answers:

$\frac{\sqrt2}{1296}$ $\displaystyle \frac{\sqrt2}{1296}$

$\frac{\sqrt2}{16}$ $\displaystyle \frac{\sqrt2}{16}$

$\frac{\sqrt2}{2592}$ $\displaystyle \frac{\sqrt2}{2592}$

$\frac{\sqrt2}{216}$ $\displaystyle \frac{\sqrt2}{216}$

$\frac{3\sqrt2}{2}$ $\displaystyle \frac{3\sqrt2}{2}$

Correct answer:

$\frac{\sqrt2}{2592}$ $\displaystyle \frac{\sqrt2}{2592}$

Explanation:

Write the equation to find the volume of a tetrahedron.

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

Substitute the side length and solve for the volume.

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

Rationalize the denominator.

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

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Example Question #1 : How To Find The Volume Of A Tetrahedron

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

Possible Answers:

24.2 $\displaystyle 24.2$

27.8 $\displaystyle 27.8$

26.6 $\displaystyle 26.6$

25.5 $\displaystyle 25.5$

32.1 $\displaystyle 32.1$

Correct answer:

25.5 $\displaystyle 25.5$

Explanation:

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

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

where $\displaystyle a$ is the measurement of the edge of the tetrahedron.

This problem can be quickly solved by substituting 6 in for $\displaystyle a$.

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

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

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ACT Math : Tetrahedrons

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #1 : How To Find The Volume Of A Tetrahedron

Example Question #83 : Solid Geometry

Example Question #1 : How To Find The Volume Of A Tetrahedron

All ACT Math Resources

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