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SAT II Math I : SAT Subject Test in Math I

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 #2 : How To Find The Volume Of A Tetrahedron

Find the volume of a tetrahedron with an edge of $6\:cm$ .

Possible Answers:

$36\sqrt2\:cm^3$

$18\sqrt2\:cm^3$

Correct answer:

$18\sqrt2\:cm^3$

Explanation:

Write the formula for finding the volume of a tetrahedron.

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

Substitute in the edge length provided in the problem.

$V=\frac{(6\:cm)^3}{6\sqrt 2}=\frac{6^3}{6\sqrt 2}\:cm^3$

Cancel out the in the denominator with part of the 6^3 in the numerator:

$V= \frac{6^2}{\sqrt2}\:cm^3$

Expand, rationalize the denominator, and reduce to arrive at the correct answer:

$V= \frac{36}{\sqrt2}\:cm^3=\frac{36}{\sqrt2}\:cm^3 \cdot \frac{\sqrt2}{\sqrt2} =\frac{36\sqrt2}{2}\:cm^3 = 18\sqrt2\:cm^3$

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

Find the volume of a tetrahedron with an edge of $6\sqrt2$ .

Possible Answers:

Correct answer:

Explanation:

Write the formula the volume of a tetrahedron.

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

Substitute the edge length provided in the equation into the formula.

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

Cancel out the denominator with part of the numerator and solve the remaining part of the numerator to arrive at the correct answer.

$V= (6\sqrt 2)^2 = 72$

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Example Question #4 : How To Find The Volume Of A Tetrahedron

Find the volume of a tetrahedron with an edge of $\sqrt5$ .

Possible Answers:

$\frac{25\sqrt10}{12}$

$5\sqrt3$

$\frac{125}{6}$

$\frac{5\sqrt3}{9}$

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

Correct answer:

$\frac{25\sqrt10}{12}$

Explanation:

Write the formula the volume of a tetrahedron and substitute in the provided edge length.

$V=\frac{s^3}{6\sqrt 2}= \frac{(\sqrt 5)^3}{6\sqrt 2} = \frac{25\sqrt 5}{6\sqrt2}$

Rationalize the denominator to arrive at the correct answer.

$\frac{25\sqrt5}{6\sqrt2} \cdot \frac{\sqrt2}{\sqrt2} = \frac{25\sqrt 10}{12}$

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

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

Possible Answers:

$14.73\:cm^3$

$125.00\:cm^3$

$2.95\:cm^3$

$12.03\:cm^3$

$20.83\:cm^3$

Correct answer:

$14.73\:cm^3$

Explanation:

The formula for the volume of a regular tetrahedron is:

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

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

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

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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 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 #11 : Volume

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

Possible Answers:

It cannot be determined by the information given.

It is 4 times greater.

It is 8 times greater.

It is doubled.

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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SAT II Math I : SAT Subject Test in Math I

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

All SAT II Math I 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 #4 : How To Find The Volume Of A Tetrahedron

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

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 #11 : Volume

Example Question #42 : Tetrahedrons

Example Question #81 : Solid Geometry

All SAT II Math I Resources

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