How to find the volume of a tetrahedron - PSAT Math

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Note: Figure NOT drawn to scale.

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

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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$

Note: Figure NOT drawn to scale.

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

Tap to see back →

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$

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

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The volume of a tetrahedron is found with the equation , 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:

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

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

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

Note: Figure NOT drawn to scale.

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

Tap to see back →

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$

Note: Figure NOT drawn to scale.

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

Tap to see back →

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$

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

Tap to see back →

The volume of a tetrahedron is found with the equation , 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:

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

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

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

Note: Figure NOT drawn to scale.

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

Tap to see back →

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$

Note: Figure NOT drawn to scale.

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

Tap to see back →

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$

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

Tap to see back →

The volume of a tetrahedron is found with the equation , 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:

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

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

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

Note: Figure NOT drawn to scale.

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

Tap to see back →

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$

Note: Figure NOT drawn to scale.

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

Tap to see back →

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$

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

Tap to see back →

The volume of a tetrahedron is found with the equation , 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:

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

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

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

Note: Figure NOT drawn to scale.

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

Tap to see back →

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$

Note: Figure NOT drawn to scale.

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

Tap to see back →

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$

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

Tap to see back →

The volume of a tetrahedron is found with the equation , 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:

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

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

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

Note: Figure NOT drawn to scale.

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

Tap to see back →

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$

Note: Figure NOT drawn to scale.

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

Tap to see back →

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$

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

Tap to see back →

The volume of a tetrahedron is found with the equation , 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:

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

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

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

Note: Figure NOT drawn to scale.

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

Tap to see back →

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$

Note: Figure NOT drawn to scale.

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

Tap to see back →

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$