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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 : 3 Dimensional Geometry

Example cylinder

Figure not drawn to scale.

What is the volume of the cylinder above?

Possible Answers:

52.36 in³

48.79 in³

66.13 in³

56.55 in³

94.25 in³

Correct answer:

56.55 in³

Explanation:

In order to find the volume of a cylinder, you find the area of the circular top and multiply it by the height.

Example cylinder

$\small \small Volume=\pi r{^{2}}h$

$\small \pi (3{^{2}})(2)=\pi(9)(2)=18\pi=56.55$

The volume of the cylinder is 56.55 in³

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Example Question #1 : Volume

Cone example

Figure not drawn to scale

If the volume of the cone above is 47.12 ft³, what is the radius of the base?

Possible Answers:

5 ft

2 ft

3 ft

4 ft

3.5ft

Correct answer:

3 ft

Explanation:

Because we have been given the volume of the cone and have been asked to find the radius of the base of the cone, we must work backwards using the volume formula.

Cone example

$\small Volume= {\frac{1}{3}}\pi r{^{2}}h$

$\small \small 47.12={\frac{1}{3}}\pi r{^{2}}(5)={\frac{5}{3}}\pi r{^{2}}$

$\small \left({\frac{3}{5}} \right)47.12=\left({\frac{3}{5}} \right )\left({\frac{5}{3}} \right )\pi r{^{2}}$

$\small \frac{28.27}{\pi}=r{^{2}}$

$\small 9.00=r{^{2}}$

$\small r=3$

The radius of the base of the cone is 3 ft.

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Example Question #1 : Volume

Box example

Figure not drawn to scale.

What is the volume of the above image?

Possible Answers:

64 yd³

42 yd²

42 yd³

32 yd³

84 yd²

Correct answer:

42 yd³

Explanation:

You can find the volume of a box by following the equation below:

$\small Volume=(length)(height)(depth)$

$\small (7)(3)(2)=42$

The surface area of the box is 42 yd³(remember that volume measurements are cubic units NOT square units)

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Example Question #1 : Volume

What is the volume of the following tetrahedron? Assume the figure is a regular tetrahedron.

Possible Answers:

$\4\sqrt{3}m^3$

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

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

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

$\frac{8\sqrt{2}}{3}m^3$

Correct answer:

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

Explanation:

A regular tetrahedron is composed of four equilateral triangles. The formula for the volume of a regular tetrahedron is:

$V=\frac{a^3\sqrt{2}}{12}$ , where represents the length of the side.

Plugging in our values we get:

$V=\frac{(2\: m)^3\sqrt{2}}{12}$

$V=\frac{8\: m^3\sqrt{2}}{12} = \frac{2\sqrt{2}}{3}m^3$

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Example Question #2 : Volume

Find the volume of a tetrahedron with an edge of $1\:cm$ .

Possible Answers:

$3\:cm^3$

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

$\frac{\sqrt2}{12}\:cm^3$

$\frac{1}{6}\:cm^3$

$\frac{\sqrt3}{4}\:cm^3$

Correct answer:

$\frac{\sqrt2}{12}\:cm^3$

Explanation:

Write the formula for the volume of a tetrahedron.

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

Substitute in the length of the edge provided in the problem.

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

Rationalize the denominator.

$V= \frac{1}{6\sqrt2}\:cm\cdot \frac{\sqrt2}{\sqrt2} = \frac{\sqrt2}{12}\:cm^3$

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

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

Possible Answers:

$\frac{1}{6}\:in^3$

$\sqrt3\:in^3$

$\frac{1}{2}\:in^3$

$\frac{2}{3}\:in^3$

$\frac{1}{3}\:in^3$

Correct answer:

$\frac{1}{3}\:in^3$

Explanation:

Write the formula for the volume of a tetrahedron.

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

Substitute in the length of the edge provided in the problem:

$V=\frac{(\sqrt2\:in)^3}{6\sqrt2}$

Cancel out the $\sqrt2$ in the denominator with one in the numerator:

$V= \frac{(\sqrt2)^3}{6\sqrt2}\:in^3$

A square root is being raised to the power of two in the numerator; these two operations cancel each other out. After canceling those operations, reduce the remaining fraction to arrive at the correct answer:

$V= \frac{(\sqrt2)^2}{6}\:in^3=\frac{2}{6}\:in^3=\frac{1}{3}\:in^3$

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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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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 : 3 Dimensional Geometry

The volume of the cylinder is 56.55 in³

Example Question #1 : Volume

If the volume of the cone above is 47.12 ft³, what is the radius of the base?

The radius of the base of the cone is 3 ft.

Example Question #1 : Volume

The surface area of the box is 42 yd³(remember that volume measurements are cubic units NOT square units)

Example Question #1 : Volume

Example Question #2 : Volume

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

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

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 : 3 Dimensional Geometry

The volume of the cylinder is 56.55 in3

Example Question #1 : Volume

If the volume of the cone above is 47.12 ft3, what is the radius of the base?

The radius of the base of the cone is 3 ft.

Example Question #1 : Volume

The surface area of the box is 42 yd3 (remember that volume measurements are cubic units NOT square units)

Example Question #1 : Volume

Example Question #2 : Volume

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

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

All SAT II Math I Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

The volume of the cylinder is 56.55 in³

If the volume of the cone above is 47.12 ft³, what is the radius of the base?

The surface area of the box is 42 yd³(remember that volume measurements are cubic units NOT square units)