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SAT II Math I : Geometry

Study concepts, example questions & explanations for SAT II Math I

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All SAT II Math I Resources

6 Diagnostic Tests 113 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

The volume of a tetrahedron is found with the formula:

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

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

When $\small a=6$ ,

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

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

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

Possible Answers:

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

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

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

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

None of the above.

Correct answer:

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

Explanation:

The volume of a tetrahedron is found with the formula,

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

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

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

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

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

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

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Example Question #6 : How To Find The Volume Of A Tetrahedron

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

Possible Answers:

None of the above.

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

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

$\small 375\sqrt{2}$

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

Correct answer:

None of the above.

Explanation:

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

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

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

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

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

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

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Example Question #91 : Geometry

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 8 times greater.

It is doubled.

It cannot be determined by the information given.

It increases by 50%.

Correct answer:

It is 8 times greater.

Explanation:

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

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

Therefore,

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

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Example Question #42 : Tetrahedrons

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

Possible Answers:

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

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

None of the above.

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

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

Correct answer:

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

Explanation:

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

When $\small \small a= 22$ ,

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

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

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

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Example Question #81 : Solid Geometry

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

Possible Answers:

$2\sqrt3\:cm$

$2\sqrt6\:cm$

$4\sqrt3\:cm$

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

$\sqrt6\:cm$

Correct answer:

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

Explanation:

Write the volume equation for a tetrahedron.

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

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

Substitute the given edge length and solve.

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

Rationalize the denominator.

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

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Example Question #82 : Solid Geometry

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

Possible Answers:

$\frac{3\sqrt2}{2}$

$\frac{\sqrt2}{2592}$

$\frac{\sqrt2}{16}$

$\frac{\sqrt2}{216}$

$\frac{\sqrt2}{1296}$

Correct answer:

$\frac{\sqrt2}{2592}$

Explanation:

Write the equation to find the volume of a tetrahedron.

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

Substitute the side length and solve for the volume.

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

Rationalize the denominator.

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

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Example Question #82 : Solid Geometry

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

Possible Answers:

25.5

32.1

26.6

24.2

27.8

Correct answer:

25.5

Explanation:

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

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

where is the measurement of the edge of the tetrahedron.

This problem can be quickly solved by substituting 6 in for .

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

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

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Example Question #21 : Volume

What is the volume of the tetrahedron shown below?

Screen shot 2015 10 21 at 7.16.10 pm

Possible Answers:

$\frac{256}{6\sqrt2}$

$\frac{256}{3\sqrt2}$

$\frac{36}{\sqrt2}$

25.63

512

Correct answer:

$\frac{256}{3\sqrt2}$

Explanation:

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

This tetrahedron has a side with a length of 8.

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

You can reduce that answer further, so that it becomes

$\frac{256}{3\sqrt2}$ .

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Example Question #92 : Geometry

Find the volume of sphere whose radius is 15.

Possible Answers:

$300\pi$

$4500\pi$

$562.5\pi$

$1200\pi$

$36000\pi$

Correct answer:

$4500\pi$

Explanation:

The volume of a sphere is given by the equation:

$V=\frac{4}{3} \pi r^3$

The problem says that the radius is 15, so plug in 15 for r and simplify.

$V=\frac{4}{3} \pi r^3=\frac{4}{3} \pi 15^3=4500\pi$

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All SAT II Math I Resources

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SAT II Math I : Geometry

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

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

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

Example Question #6 : How To Find The Volume Of A Tetrahedron

Example Question #91 : Geometry

Example Question #42 : Tetrahedrons

Example Question #81 : Solid Geometry

Example Question #82 : Solid Geometry

Example Question #82 : Solid Geometry

Example Question #21 : Volume

Example Question #92 : Geometry

All SAT II Math I Resources

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