How to find the volume of a tetrahedron - SSAT Upper Level Quantitative

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In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates .

What is the volume of this tetrahedron?

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The tetrahedron looks like this:

is the origin and are the other three points, which are fifteen units away from the origin on each of the three (perpendicular) axes.

This is a triangular pyramid, and we can consider $\Delta OCB$ the base; its area is half the product of its legs, or

$B = $\frac{1}{2}$ \cdot 15 \cdot 15 = 112$\frac{1}{2}$$ .

The volume of the tetrahedron is one third the product of its base and its height, the latter of which is 15. Therefore,

$V = $\frac{1}{3}$ \cdot 15 \cdot 112$\frac{1}{2}$ = 562 $\frac{1}{2}$$ .

Above is the base of a triangular pyramid, which is equilateral. The height of the pyramid is equal to the perimeter of its base. In terms of , give the volume of the pyramid.

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By the 30-60-90 Theorem, $AM = CM $\sqrt{3}$$ , or

$CM = $\frac{AM}{\sqrt{3}$} = $\frac{n}{\sqrt{3}$} = $\frac{n\sqrt{3}$}{3}$

is the midpoint of $\overline{BC}$ , so

$BC = 2 \cdot CM = 2 \cdot $\frac{n\sqrt{3}$}{3} = $\frac{2n\sqrt{3}$}{3}$

The area of the triangular base is half the product of its base and its height:

$B = $\frac{1}{2}$ \cdot n \cdot $\frac{2n\sqrt{3}$}{3}= $\frac{ $n^{2}$$$\sqrt{3}$}{3}$

The height of the pyramid is equal to the perimeter, so it will be three times , or

$p = 3 \cdot BC = 3 \cdot $\frac{2n\sqrt{3}$}{3} = 2n$\sqrt{3}$$

The volume of the pyramid is one third the product of this area and the height of the pyramid:

$V = $\frac{1}{3}$ \cdot 2n$\sqrt{3}$ \cdot $\frac{ $n^{2}$$$\sqrt{3}$}{3}$

$= $\frac{2 $n^{3}$$\cdot \sqrt {3} \cdot $\sqrt{3}$}{9}$

$= $\frac{6 $n^{3}$$ }{9}$

$= $\frac{2 $n^{3}$$ }{3}$

In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates

,

where

Give its volume in terms of .

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The tetrahedron looks like this:

is the origin and are the other three points.

This is a triangular pyramid, and we can consider $\Delta OCB$ the base; its area is half the product of its legs, or

$B = $\frac{1}{2}$ \cdot (n-1)\cdot n$ .

The volume of the tetrahedron is one third the product of its base and its height. Therefore,

$V = $\frac{1}{3}$ \cdot \left ( n+1\right ) \cdot $\frac{1}{2}$ \cdot (n-1)\cdot n$

After some rearrangement:

$V= $\frac{1}{6}$ ( n+1 ) (n-1) n$

$= $\frac{1}{6}$ $(n^{2}$-1) n$

$= $\frac{1}{6}$ $(n^{3}$-n)$

$= $\frac{1}{6}$ $n^{3}$-$\frac{1}{6}$n$

In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates .

Give its volume.

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A tetrahedron is a triangular pyramid and can be looked at as such.

Three of the vertices - - are on the -plane, and can be seen as the vertices of the triangular base. This triangle, as seen below, is isosceles:

Its base is 10 and its height is 18, so its area is

$B = $\frac{1}{2}$ \cdot 10 \cdot 18 = 90$

The fourth vertex is off the -plane; its perpendicular distance to the aforementioned face is its -coordinate, 8, so this is the height of the pyramid. The volume of the pyramid is

$V = $\frac{1}{3}$ \cdot 90 \cdot 8 = 240$

In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates

,

where

Give its volume in terms of .

Tap to see back →

The tetrahedron looks like this:

is the origin and are the other three points, each of which lies along one of the three (mutually perpendicular) axes.

This is a triangular pyramid, and we can consider $\Delta OAB$ the base; its area is half the product of its legs, or

$B = $\frac{1}{2}$ \cdot n \cdot 2n = $n^{2}$$ .

The volume of the tetrahedron is one third the product of its base area and its height . Therefore, the volume is

$V = $\frac{1}{3}$ \cdot3n \cdot $n^{2}$ = $n^{3}$$

Find the volume of a regular tetrahedron that has a side length of .

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Use the following formula to find the volume of a regular tetrahedron:

Now, plug in the given side length.

$\text{Volume}$=\frac{64\sqrt2}{12}$

$\text{Volume}$=\frac{16\sqrt2}{3}$

Find the volume of a regular tetrahedron that has a side length of .

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Use the following formula to find the volume of a regular tetrahedron:

Now, plug in the given side length.

$\text{Volume}$=\frac{125\sqrt2}{12}$

Find the volume of a regular tetrahedron with a side length of .

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Use the following formula to find the volume of a regular tetrahedron:

Now, plug in the given side length.

$\text{Volume}$=\frac{216\sqrt2}{12}$

$$\text{Volume}$=18\sqrt2$

Find the volume of a regular tetrahedron with side lengths of .

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Use the following formula to find the volume of a regular tetrahedron:

Now, plug in the given side length.

$\text{Volume}$=\frac{343\sqrt2}{12}$

Find the volume of a regular tetrahedron with side lengths of .

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Use the following formula to find the volume of a regular tetrahedron:

Now, plug in the given side length.

$\text{Volume}$=\frac{512\sqrt2}{12}$

$\text{Volume}$=\frac{128\sqrt2}{3}$

Find the volume of a tetrahedron with side lengths of .

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Use the following formula to find the volume of a regular tetrahedron:

Now, plug in the given side length.

$\text{Volume}$=\frac{1728\sqrt2}{12}$

$$\text{Volume}$=144\sqrt2$

Find the volume of a regular tetrahedron with side lengths of .

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Use the following formula to find the volume of a regular tetrahedron:

Now, plug in the given side length.

$\text{Volume}$=\frac{1000\sqrt2}{12}$

$\text{Volume}$=\frac{250\sqrt2}{3}$

Find the volume of a regular tetrahedron with side lengths of .

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Use the following formula to find the volume of a regular tetrahedron:

Now, plug in the given side length.

Find the volume of a regular tetrahedron with side lengths of .

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Use the following formula to find the volume of a regular tetrahedron:

Now, plug in the given side length.

Find the volume of a regular tetrahedron with side lengths of .

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Use the following formula to find the volume of a regular tetrahedron:

Now, plug in the given side length.

Find the volume of a regular tetrahedron with side lengths of $\frac{1}{2}$ .

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Use the following formula to find the volume of a regular tetrahedron:

Now, plug in the given side length.

$$\text{Volume}$=\frac{\sqrt2}{12}$($\frac{1}{8}$)$

$\text{Volume}$=\frac{\sqrt2}{96}$

Find the volume of a regular tetrahedron with side lengths of $\frac{1}{3}$ .

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Use the following formula to find the volume of a regular tetrahedron:

Now, plug in the given side length.

$$\text{Volume}$=\frac{\sqrt2}{12}$($\frac{1}{27}$)$

$\text{Volume}$=\frac{\sqrt2}{324}$

In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates .

What is the volume of this tetrahedron?

Tap to see back →

The tetrahedron looks like this:

is the origin and are the other three points, which are fifteen units away from the origin on each of the three (perpendicular) axes.

This is a triangular pyramid, and we can consider $\Delta OCB$ the base; its area is half the product of its legs, or

$B = $\frac{1}{2}$ \cdot 15 \cdot 15 = 112$\frac{1}{2}$$ .

The volume of the tetrahedron is one third the product of its base and its height, the latter of which is 15. Therefore,

$V = $\frac{1}{3}$ \cdot 15 \cdot 112$\frac{1}{2}$ = 562 $\frac{1}{2}$$ .

Above is the base of a triangular pyramid, which is equilateral. The height of the pyramid is equal to the perimeter of its base. In terms of , give the volume of the pyramid.

Tap to see back →

By the 30-60-90 Theorem, $AM = CM $\sqrt{3}$$ , or

$CM = $\frac{AM}{\sqrt{3}$} = $\frac{n}{\sqrt{3}$} = $\frac{n\sqrt{3}$}{3}$

is the midpoint of $\overline{BC}$ , so

$BC = 2 \cdot CM = 2 \cdot $\frac{n\sqrt{3}$}{3} = $\frac{2n\sqrt{3}$}{3}$

The area of the triangular base is half the product of its base and its height:

$B = $\frac{1}{2}$ \cdot n \cdot $\frac{2n\sqrt{3}$}{3}= $\frac{ $n^{2}$$$\sqrt{3}$}{3}$

The height of the pyramid is equal to the perimeter, so it will be three times , or

$p = 3 \cdot BC = 3 \cdot $\frac{2n\sqrt{3}$}{3} = 2n$\sqrt{3}$$

The volume of the pyramid is one third the product of this area and the height of the pyramid:

$V = $\frac{1}{3}$ \cdot 2n$\sqrt{3}$ \cdot $\frac{ $n^{2}$$$\sqrt{3}$}{3}$

$= $\frac{2 $n^{3}$$\cdot \sqrt {3} \cdot $\sqrt{3}$}{9}$

$= $\frac{6 $n^{3}$$ }{9}$

$= $\frac{2 $n^{3}$$ }{3}$

In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates

,

where

Give its volume in terms of .

Tap to see back →

The tetrahedron looks like this:

is the origin and are the other three points.

This is a triangular pyramid, and we can consider $\Delta OCB$ the base; its area is half the product of its legs, or

$B = $\frac{1}{2}$ \cdot (n-1)\cdot n$ .

The volume of the tetrahedron is one third the product of its base and its height. Therefore,

$V = $\frac{1}{3}$ \cdot \left ( n+1\right ) \cdot $\frac{1}{2}$ \cdot (n-1)\cdot n$

After some rearrangement:

$V= $\frac{1}{6}$ ( n+1 ) (n-1) n$

$= $\frac{1}{6}$ $(n^{2}$-1) n$

$= $\frac{1}{6}$ $(n^{3}$-n)$

$= $\frac{1}{6}$ $n^{3}$-$\frac{1}{6}$n$