PSAT Math : How to find the volume of a tetrahedron

Study concepts, example questions & explanations for PSAT Math

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Example Questions

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

Tetra_2

Note: Figure NOT drawn to scale.

The above triangular pyramid has volume 25. To the nearest tenth, evaluate \(\displaystyle h\).

Possible Answers:

\(\displaystyle h \approx 7.2\)

\(\displaystyle h \approx 10.8\)

\(\displaystyle h \approx 13.3\)

\(\displaystyle h \approx 8.8\)

Insufficient information is given to answer the problem.

Correct answer:

\(\displaystyle h \approx 10.8\)

Explanation:

We are looking for the height of the pyramid.

The base is an equilateral triangle with sidelength 4, so its area can be calculated as follows:

\(\displaystyle B = \frac{s^{2} \sqrt{3} }{4}\)

\(\displaystyle B = \frac{4^{2} \sqrt{3} }{4}\)

\(\displaystyle B = \frac{16 \sqrt{3} }{4}\)

\(\displaystyle B = 4\sqrt{3}\)

The height \(\displaystyle h\) of a pyramid can be calculated using the fomula

\(\displaystyle V = \frac{1}{3}Bh\)

We set \(\displaystyle V = 25\) and \(\displaystyle B = 4\sqrt{3}\) and solve for \(\displaystyle h\):

\(\displaystyle \frac{1}{3} \cdot 4\sqrt{3} \cdot h = 25\)

\(\displaystyle h = \frac{25 \cdot 3}{4 \sqrt{3}} = \frac{75}{4 \sqrt{3}} \approx \frac{75}{4 \cdot 1.7321} \approx 10.8\)

Example Question #251 : Psat Mathematics

Tetra_1

Note: Figure NOT drawn to scale.

Give the volume (nearest tenth) of the above triangular pyramid.

Possible Answers:

\(\displaystyle 27.7\)

\(\displaystyle 22.6\)

\(\displaystyle 55.4\)

\(\displaystyle 18.5\)

\(\displaystyle 15.1\)

Correct answer:

\(\displaystyle 18.5\)

Explanation:

The height of the pyramid is \(\displaystyle h = 8\). The base is an equilateral triangle with sidelength 4, so its area can be calculated as follows:

\(\displaystyle B = \frac{s^{2} \sqrt{3} }{4}\)

\(\displaystyle B = \frac{4^{2} \sqrt{3} }{4}\)

\(\displaystyle B = \frac{16 \sqrt{3} }{4}\)

\(\displaystyle B = 4\sqrt{3}\)

The volume of a pyramid can be calculated using the fomula

\(\displaystyle V = \frac{1}{3}Bh\)

\(\displaystyle V = \frac{1}{3} \cdot 4\sqrt{3} \cdot 8 = \frac{32\sqrt{3} }{3} \approx \frac{32 \cdot 1.7321}{3} \approx 18.5\)

Example Question #252 : Geometry

A regular tetrahedron has an edge length of \(\displaystyle 4\). What is its volume?

Possible Answers:

\(\displaystyle \frac{32}{\sqrt{2}}\)

\(\displaystyle \frac{32}{3}\)

\(\displaystyle \frac{64}{3\sqrt{2}}\)

\(\displaystyle \frac{32}{3\sqrt{2}}\)

\(\displaystyle 3\)

Correct answer:

\(\displaystyle \frac{32}{3\sqrt{2}}\)

Explanation:

The volume of a tetrahedron is found with the equation \(\displaystyle V=\frac{a^3}{6\sqrt{2}}\), where \(\displaystyle a\) represents the length of an edge of the tetrahedron.

Plug in 4 for the edge length and reduce as much as possible to find the answer:

 

\(\displaystyle V=\frac{4^3}{6\sqrt{2}}\)

\(\displaystyle V=\frac{64}{6\sqrt{2}}\)

\(\displaystyle V=\frac{32}{3\sqrt{2}}\)

The volume of the tetrahedron is \(\displaystyle \frac{32}{3\sqrt{2}}\).

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