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PSAT Math : How to find the volume of a tetrahedron

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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:

$h \approx 7.2$ $\displaystyle h \approx 7.2$

$h \approx 10.8$ $\displaystyle h \approx 10.8$

$h \approx 8.8$ $\displaystyle h \approx 8.8$

$h \approx 13.3$ $\displaystyle h \approx 13.3$

Insufficient information is given to answer the problem.

Correct answer:

$h \approx 10.8$ $\displaystyle h \approx 10.8$

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:

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

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

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

$B = 4\sqrt{3}$ $\displaystyle B = 4\sqrt{3}$

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

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

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

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

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

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Example Question #2 : How To Find The Volume Of A Tetrahedron

Tetra_1

Note: Figure NOT drawn to scale.

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

Possible Answers:

55.4 $\displaystyle 55.4$

27.7 $\displaystyle 27.7$

18.5 $\displaystyle 18.5$

22.6 $\displaystyle 22.6$

15.1 $\displaystyle 15.1$

Correct answer:

18.5 $\displaystyle 18.5$

Explanation:

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

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

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

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

$B = 4\sqrt{3}$ $\displaystyle B = 4\sqrt{3}$

The volume of a pyramid can be calculated using the fomula

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

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

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

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

Possible Answers:

$\displaystyle 3$

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

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

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

$\frac{32}{3}$ $\displaystyle \frac{32}{3}$

Correct answer:

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

Explanation:

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

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

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

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

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

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PSAT Math : How to find the volume of a tetrahedron

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

Example Question #2 : How To Find The Volume Of A Tetrahedron

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

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