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SSAT Upper Level Math : Volume of a Three-Dimensional Figure

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Example Questions

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

Tetrahedron

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.

Possible Answers:

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

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

$\frac{2 n^{3} }{9}$

$2n^{3}$

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

Correct answer:

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

Explanation:

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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Example Question #3 : How To Find The Volume Of A Tetrahedron

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

(0,0,0), (n-1,0,0), (0,n,0), (0,0, n+1) ,

where n > 1

Give its volume in terms of .

Possible Answers:

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

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

$\frac{1}{6} n^{3}- \frac{1}{3}n^{2} +\frac{1}{6}n$

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

$\frac{1}{6} n^{3}- \frac{1}{3}n^{2} -\frac{1}{6}n$

Correct answer:

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

Explanation:

The tetrahedron looks like this:

Tetrahedron

is the origin and A,B,C 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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Example Question #5 : How To Find The Volume Of A Tetrahedron

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

(0,0,0), (n ,0,0), (0,2n,0), (0,0, 3n) ,

where n > 1

Give its volume in terms of .

Possible Answers:

$3 n^{3}$

$2 n^{3}$

$n^{3}$

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

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

Correct answer:

$n^{3}$

Explanation:

The tetrahedron looks like this:

Tetrahedron

is the origin and A,B,C 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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Example Question #41 : Volume Of A Three Dimensional Figure

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

Possible Answers:

$\frac{5\sqrt2}{6}$

$\frac{25\sqrt2}{12}$

$\frac{125\sqrt2}{12}$

$20\sqrt2$

Correct answer:

$\frac{125\sqrt2}{12}$

Explanation:

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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Example Question #7 : How To Find The Volume Of A Tetrahedron

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

Possible Answers:

$\frac{16\sqrt2}{3}$

$\frac{8\sqrt2}{3}$

$32\sqrt2$

$\frac{64}{3\sqrt2}$

Correct answer:

$\frac{16\sqrt2}{3}$

Explanation:

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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Example Question #8 : How To Find The Volume Of A Tetrahedron

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

Possible Answers:

$24\sqrt2$

$6\sqrt2$

$18\sqrt2$

$36\sqrt2$

Correct answer:

$18\sqrt2$

Explanation:

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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Example Question #42 : Volume Of A Three Dimensional Figure

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

Possible Answers:

$\frac{343\sqrt2}{12}$

$18\sqrt2$

$\frac{7\sqrt2}{6}$

$\frac{49\sqrt2}{12}$

Correct answer:

$\frac{343\sqrt2}{12}$

Explanation:

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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Example Question #43 : Volume Of A Three Dimensional Figure

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

Possible Answers:

$\frac{128\sqrt2}{3}$

$\frac{16\sqrt2}{3}$

$10\sqrt2$

$2\sqrt2$

Correct answer:

$\frac{128\sqrt2}{3}$

Explanation:

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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Example Question #11 : How To Find The Volume Of A Tetrahedron

Find the volume of a tetrahedron with side lengths of .

Possible Answers:

$1728\sqrt2$

$966\sqrt2$

$144\sqrt2$

$168\sqrt2$

Correct answer:

$144\sqrt2$

Explanation:

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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Example Question #12 : How To Find The Volume Of A Tetrahedron

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

Possible Answers:

$\frac{250\sqrt2}{3}$

$125\sqrt2$

$\frac{500\sqrt2}{3}$

$\frac{1000\sqrt2}{3}$

Correct answer:

$\frac{250\sqrt2}{3}$

Explanation:

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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SSAT Upper Level Math : Volume of a Three-Dimensional Figure

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

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

Example Question #3 : How To Find The Volume Of A Tetrahedron

Example Question #5 : How To Find The Volume Of A Tetrahedron

Example Question #41 : Volume Of A Three Dimensional Figure

Example Question #7 : How To Find The Volume Of A Tetrahedron

Example Question #8 : How To Find The Volume Of A Tetrahedron

Example Question #42 : Volume Of A Three Dimensional Figure

Example Question #43 : Volume Of A Three Dimensional Figure

Example Question #11 : How To Find The Volume Of A Tetrahedron

Example Question #12 : How To Find The Volume Of A Tetrahedron

All SSAT Upper Level Math Resources

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