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

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Question

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?

Answer

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,

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

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Question

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.

Answer

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:

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

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

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

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

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Question

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 .

Answer

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:

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

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

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

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

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Question

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

Give its volume.

Answer

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

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

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Question

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 .

Answer

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 $n^{2}$ and its height . Therefore, the volume is

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

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Question

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

Answer

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

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}4^3$

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

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

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Question

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

Answer

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

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}5^3$

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

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Question

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

Answer

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

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}6^3$

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

$\text{Volume}=18\sqrt2$

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Question

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

Answer

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

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}7^3$

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

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Question

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

Answer

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

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}8^3$

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

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

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Question

Find the volume of a tetrahedron with side lengths of .

Answer

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

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}12^3$

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

$\text{Volume}=144\sqrt2$

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Question

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

Answer

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

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}10^3$

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

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

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Question

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

Answer

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

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}(2x)^3$

$\text{Volume}=\frac{8x^3\sqrt2}{12}$

$\text{Volume}=\frac{2x^3\sqrt2}{3}$

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Question

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

Answer

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

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}(4a)^3$

$\text{Volume}=\frac{64a^3\sqrt2}{12}$

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

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Question

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

Answer

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

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}(12h)^3$

$\text{Volume}=\frac{1728h^3\sqrt2}{12}$

$\text{Volume}=144h^3\sqrt2$

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Question

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

Answer

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

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}(\frac{1}{2})^3$

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

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

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Question

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

Answer

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

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}(\frac{1}{3})^3$

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

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

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Question

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?

Answer

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,

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

Compare your answer with the correct one above

Question

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.

Answer

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:

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

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

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

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

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Question

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 .

Answer

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:

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

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

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

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

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