Advanced Geometry : Solid Geometry

Study concepts, example questions & explanations for Advanced Geometry

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Example Questions

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

Find the volume of the regular tetrahedron with side length \displaystyle 5\:cm.

Possible Answers:

\displaystyle 125.00\:cm^3

\displaystyle 14.73\:cm^3

\displaystyle 20.83\:cm^3

\displaystyle 12.03\:cm^3

\displaystyle 2.95\:cm^3

Correct answer:

\displaystyle 14.73\:cm^3

Explanation:

The formula for the volume of a regular tetrahedron is:

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

Where \displaystyle s is the length of side. Using this formula and the given values, we get:

\displaystyle V = \frac{5\:cm^3}{6\sqrt{2}} = 14.73\:cm^3

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

What is the volume of a regular tetrahedron with edges of \displaystyle 6cm?

Possible Answers:

\displaystyle \small \frac{36}{\sqrt{2}} cm^3

\displaystyle \small \sqrt{2} cm^3

\displaystyle \small 6\sqrt{2} cm^3

\displaystyle \small \frac{36}{\sqrt{2}} cm^3

\displaystyle \small \frac{216}{\sqrt{2}} cm^3

Correct answer:

\displaystyle \small \frac{36}{\sqrt{2}} cm^3

Explanation:

The volume of a tetrahedron is found with the formula:

 \displaystyle \small V= \frac{a^3}{6\sqrt{2}} ,

where \displaystyle \small a is the length of the edges.

When \displaystyle \small a=6

\displaystyle \small V=\frac{36}{\sqrt{2}}.

Example Question #91 : Geometry

What is the volume of a regular tetrahedron with edges of \displaystyle \small \frac{2}{3} cm?

Possible Answers:

\displaystyle \small \frac{4}{81\sqrt{2}} cm^3

\displaystyle \small \frac{4}{81\sqrt{2}} cm^2

None of the above.

\displaystyle \small \frac{2}{9\sqrt{2}}cm^2

\displaystyle \small \frac{2}{9\sqrt{2}}cm^3

Correct answer:

\displaystyle \small \frac{4}{81\sqrt{2}} cm^3

Explanation:

The volume of a tetrahedron is found with the formula,

 \displaystyle \small V= \frac{a^3}{6\sqrt{2}} where \displaystyle \small a is the length of the edges.

When \displaystyle \small a= \frac{2}{3} the volume becomes,

\displaystyle \small V= \left ( \frac{2}{3} ^3\right )\div 6\sqrt{2}

\displaystyle \small V=\frac{8}{27}\div 6\sqrt{2}

\displaystyle \small V= \frac{8}{27}\times\frac{1}{6\sqrt{2}}=\frac{4}{81\sqrt{2}}

The answer is in volume, so it must be in a cubic measurement!

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

What is the volume of a regular tetrahedron with edges of \displaystyle \small \small \frac{4}{5}cm?

Possible Answers:

\displaystyle \small \frac{32}{\sqrt{2}} \displaystyle \small cm^3

None of the above.

\displaystyle \small \frac{375}{32\sqrt{2}}

\displaystyle \small 375\sqrt{2}

\displaystyle \small \frac{125}{\sqrt{2}} \displaystyle \small cm^3

Correct answer:

None of the above.

Explanation:

The volume of a tetrahedron is found with the formula \displaystyle \small V= \frac{a^3}{6\sqrt{2}} where \displaystyle \small a is the length of the edges.

When \displaystyle \small \small a= \frac{4}{5}

\displaystyle \small \small V= \left ( \frac{4}{5} ^3\right )\div 6\sqrt{2}

\displaystyle \small \small V=\frac{64}{125}\div 6\sqrt{2}

\displaystyle \small \small \small V= \frac{64}{125}\times\frac{1}{6\sqrt{2}}=\frac{64}{750\sqrt{2}}=\frac{32}{375\sqrt{2}}

 

This answer is not found, so it is "none of the above."

Example Question #11 : Volume

How is the volume of a regular tetrahedron effected when the length of each edge is doubled?

Possible Answers:

It is 4 times greater.

It is doubled.

It increases by 50%.

It cannot be determined by the information given.

It is 8 times greater.

Correct answer:

It is 8 times greater.

Explanation:

The volume of a regular tetrahedron is found with the formula \displaystyle \small \small V_{1}= \frac{a^3}{6\sqrt{2}} where \displaystyle \small a is the length of the edges. 

The volume of the same tetrahedron when the length of the edges are doubled would be \displaystyle \small \small \small V_{2}= \frac{(2a)^3}{6\sqrt{2}}.

Therefore,

\displaystyle \small V_{2}=\frac{8a^3}{6\sqrt{2}}=8\times V_{1}

 

Example Question #81 : Solid Geometry

What is the volume of a regular tetrahedron with edges of \displaystyle \small 22 \displaystyle \small cm?

Possible Answers:

\displaystyle \small \frac{5324}{3\sqrt{2}} \displaystyle \small cm^2

\displaystyle \small \frac{2662}{3\sqrt{2}} \displaystyle \small cm^3

\displaystyle \small \frac{2662}{3\sqrt{2}} \displaystyle \small cm^2

None of the above.

\displaystyle \small \frac{5324}{3\sqrt{2}} \displaystyle \small cm^3

Correct answer:

\displaystyle \small \frac{5324}{3\sqrt{2}} \displaystyle \small cm^3

Explanation:

The volume of a tetrahedron is found with the formula \displaystyle \small V= \frac{a^3}{6\sqrt{2}} where \displaystyle \small a is the length of the edges.

When \displaystyle \small \small a= 22,

 \displaystyle \small \small \small V=\frac{22^3}{6\sqrt{2}}

\displaystyle \small V=\frac{10648}{6\sqrt{2}}=\frac{5324}{3\sqrt{2}}

And, of course, volume should be in cubic measurements!

Example Question #42 : Tetrahedrons

Find the volume of a regular tetrahedron if one of its edges is \displaystyle \sqrt[3]{6}\:cm long.

Possible Answers:

\displaystyle 2\sqrt6\:cm

\displaystyle 4\sqrt3\:cm

\displaystyle 2\sqrt3\:cm

\displaystyle \frac{\sqrt2}{2}\;cm

\displaystyle \sqrt6\:cm

Correct answer:

\displaystyle \frac{\sqrt2}{2}\;cm

Explanation:

Write the volume equation for a tetrahedron.

\displaystyle V=\frac{e^3}{6\sqrt2}

In this formula, \displaystyle V stands for the tetrahedron's volume and \displaystyle e stands for the length of one of its edges.

Substitute the given edge length and solve.

\displaystyle V=\frac{(\sqrt[3]6\:cm)^3}{6\sqrt2} = \frac{6\:cm^3}{6\sqrt2}= \frac{1}{\sqrt2}\:cm

Rationalize the denominator.

\displaystyle \frac{1}{\sqrt2}\:cm\cdot \frac{\sqrt2}{\sqrt2} = \frac{\sqrt2}{2}\:cm

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

Find the volume of a tetrahedron if the side length is \displaystyle \frac{1}{6}.

Possible Answers:

\displaystyle \frac{\sqrt2}{16}

\displaystyle \frac{3\sqrt2}{2}

\displaystyle \frac{\sqrt2}{2592}

\displaystyle \frac{\sqrt2}{216}

\displaystyle \frac{\sqrt2}{1296}

Correct answer:

\displaystyle \frac{\sqrt2}{2592}

Explanation:

Write the equation to find the volume of a tetrahedron.

\displaystyle V=\frac{a^3}{6\sqrt2}

Substitute the side length and solve for the volume.

\displaystyle V=\frac{(\frac{1}{6})^3}{6\sqrt2}= \frac{1}{216}\left(\frac{1}{6\sqrt2}\right)= \frac{1}{1296\sqrt2}

Rationalize the denominator.

\displaystyle V=\frac{1}{1296\sqrt2} \cdot \frac{\sqrt2}{\sqrt2}= \frac{\sqrt2}{1296\cdot 2} = \frac{\sqrt2}{2592}

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

What is the volume of a regular tetrahedron with an edge length of 6?

Possible Answers:

\displaystyle 24.2

\displaystyle 27.8

\displaystyle 32.1

\displaystyle 26.6

\displaystyle 25.5

Correct answer:

\displaystyle 25.5

Explanation:

The volume of a tetrahedron can be solved for by using the equation:

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

where \displaystyle a is the measurement of the edge of the tetrahedron. 

This problem can be quickly solved by substituting 6 in for \displaystyle a

\displaystyle V=\frac{6^3}{6\sqrt{2}}=\frac{216}{6\sqrt{2}}=\frac{36}{\sqrt{2}}

\displaystyle {\color{Blue} V=25.5}

Example Question #81 : Advanced Geometry

What is the volume of the tetrahedron shown below? 


Screen shot 2015 10 21 at 7.16.10 pm

Possible Answers:

\displaystyle 512

\displaystyle \frac{36}{\sqrt2}

\displaystyle \frac{256}{6\sqrt2}

\displaystyle \frac{256}{3\sqrt2}

\displaystyle 25.63

Correct answer:

\displaystyle \frac{256}{3\sqrt2}

Explanation:

The volume of a tetrahedron is \displaystyle V=\frac{s^3}{6\sqrt2}.

This tetrahedron has a side with a length of 8. 

\displaystyle V=\frac{8^3}{6\sqrt2}, which becomes \displaystyle \frac{512}{6\sqrt2}.

You can reduce that answer further, so that it becomes 

\displaystyle \frac{256}{3\sqrt2}

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