How to find the volume of a tetrahedron - PSAT Math

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Question

Note: Figure NOT drawn to scale.

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

Answer

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 :

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

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

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Question

Note: Figure NOT drawn to scale.

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

Answer

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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Question

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

Answer

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

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