How to find the volume of a tetrahedron

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PSAT Math › How to find the volume of a tetrahedron

Questions 1 - 3

1

A regular tetrahedron has an edge length of . What is its volume?

$\frac{32}{3\sqrt{2}}$

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

$\frac{64}{3\sqrt{2}}$

$\frac{32}{3}$

Explanation

The volume of a tetrahedron is found with the equation $V=\frac{a^3}{6\sqrt{2}}$ , where 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:

$V=\frac{4^3}{6\sqrt{2}}$

$V=\frac{64}{6\sqrt{2}}$

$V=\frac{32}{3\sqrt{2}}$

The volume of the tetrahedron is $\frac{32}{3\sqrt{2}}$ .

2

Tetra_2

Note: Figure NOT drawn to scale.

The above triangular pyramid has volume 25. To the nearest tenth, evaluate .

$h \approx 10.8$

$h \approx 13.3$

$h \approx 8.8$

$h \approx 7.2$

Insufficient information is given to answer the problem.

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:

$B = \frac{s^{2} \sqrt{3} }{4}$

$B = \frac{4^{2} \sqrt{3} }{4}$

$B = \frac{16 \sqrt{3} }{4}$

$B = 4\sqrt{3}$

The height of a pyramid can be calculated using the fomula

$V = \frac{1}{3}Bh$

We set and $B = 4\sqrt{3}$ and solve for :

$\frac{1}{3} \cdot 4\sqrt{3} \cdot h = 25$

$h = \frac{25 \cdot 3}{4 \sqrt{3}} = \frac{75}{4 \sqrt{3}} \approx \frac{75}{4 \cdot 1.7321} \approx 10.8$

3

Tetra_1

Note: Figure NOT drawn to scale.

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

Explanation

The height of the pyramid is . The base is an equilateral triangle with sidelength 4, so its area can be calculated as follows:

$B = \frac{s^{2} \sqrt{3} }{4}$

$B = \frac{4^{2} \sqrt{3} }{4}$

$B = \frac{16 \sqrt{3} }{4}$

$B = 4\sqrt{3}$

The volume of a pyramid can be calculated using the fomula

$V = \frac{1}{3}Bh$

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

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