SAT II Math I : Volume

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

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Example Questions

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 

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

Example Question #21 : Volume

Find the volume of sphere whose radius is 15.

Possible Answers:

\(\displaystyle 562.5\pi\)

\(\displaystyle 300\pi\)

\(\displaystyle 1200\pi\)

\(\displaystyle 4500\pi\)

\(\displaystyle 36000\pi\)

Correct answer:

\(\displaystyle 4500\pi\)

Explanation:

The volume of a sphere is given by the equation:

\(\displaystyle V=\frac{4}{3} \pi r^3\)

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

\(\displaystyle V=\frac{4}{3} \pi r^3=\frac{4}{3} \pi 15^3=4500\pi\)

Example Question #532 : Sat Subject Test In Math I

Find the volume of a sphere with the diameter of \(\displaystyle 3\).

Possible Answers:

\(\displaystyle \frac{9}{4}\pi\)

\(\displaystyle 9\pi\)

\(\displaystyle 12\pi\)

\(\displaystyle 27\pi\)

\(\displaystyle \frac{9}{2}\pi\)

Correct answer:

\(\displaystyle \frac{9}{2}\pi\)

Explanation:

Write the equation for the volume of a sphere.

\(\displaystyle V=\frac{4}{3}\pi r^3\)

The radius is half the diameter, or \(\displaystyle \frac{3}{2}\).

Substitute the radius into the equation.

\(\displaystyle V=\frac{4}{3}\pi( \frac{3}{2})^3\)

\(\displaystyle V=\frac{4}{3}\pi( \frac{27}{8})\)

Reduce the fractions.

The answer is:  \(\displaystyle \frac{9}{2}\pi\)

Example Question #22 : Volume

Determine the volume of a rectangular prism if the length is \(\displaystyle \frac{1}{2}\), width is \(\displaystyle \frac{2}{3}\), and the height is \(\displaystyle \frac{5}{6}\).

Possible Answers:

\(\displaystyle \frac{1}{3}\)

\(\displaystyle \frac{5}{9}\)

\(\displaystyle \frac{2}{3}\)

\(\displaystyle \frac{5}{18}\) 

\(\displaystyle \frac{9}{5}\)

Correct answer:

\(\displaystyle \frac{5}{18}\) 

Explanation:

Write the formula for the volume of a rectangular prism.

\(\displaystyle A=\textup{ (Length) (Width) (Height)}\)

Substitute the dimensions into the formula.

\(\displaystyle A = \frac{1}{2}\cdot \frac{2}{3}\cdot \frac{5}{6} = \frac{10}{36}\)

Reduce this fraction.

\(\displaystyle \frac{5\times 2}{18\times 2}\)

The answer is:  \(\displaystyle \frac{5}{18}\)

Example Question #532 : Sat Subject Test In Math I

The length of a box is half its height and two-thirds its width. The volume of the box is four cubic meters. Give the length of the box to the nearest centimeter.

Possible Answers:

\(\displaystyle 175 \textrm{ cm}\)

\(\displaystyle 110 \textrm{ cm}\)

\(\displaystyle 229\textrm{ cm}\)

\(\displaystyle 144 \textrm{ cm}\)

None of these

Correct answer:

\(\displaystyle 110 \textrm{ cm}\)

Explanation:

Call \(\displaystyle L\)\(\displaystyle H\), and \(\displaystyle W\) the length, width, and height of the crate. 

The length of the crate is half its height, so 

\(\displaystyle L = \frac{1}{2} H\)

\(\displaystyle 2 \cdot L = 2 \cdot \frac{1}{2} H\)

\(\displaystyle 2 L = H\)

The length of the crate is two-thirds its width, so

\(\displaystyle L = \frac{2}{3} W\)

\(\displaystyle \frac{3} {2} \cdot L = \frac{3} {2} \cdot \frac{2}{3} W\)

\(\displaystyle \frac{3} {2} L = W\)

The dimensions of the crate in terms of \(\displaystyle L\) are \(\displaystyle L\)\(\displaystyle \frac{3} {2} L\), and \(\displaystyle 2L\). The volume is their product:

\(\displaystyle V= LWH\)

Substitute:

\(\displaystyle LWH = V\)

\(\displaystyle L \cdot \frac{3} {2} L \cdot 2L = 4\)

\(\displaystyle 3 L ^{3} = 4\)

Solve for \(\displaystyle L\):

\(\displaystyle 3 L ^{3} = 4\)

\(\displaystyle \frac{3 L ^{3} }{3} = \frac{4 }{3}\)

\(\displaystyle L ^{3} = \frac{4 }{3}\)

\(\displaystyle L =\sqrt[3]{ \frac{4 }{3}}\)

\(\displaystyle L \approx 1.101\) meters.

Since one meter comprises 100 centimeters, multiply by 100 to convert to centimeters:

\(\displaystyle 1.101 \cdot 100 = 110.1\),

which rounds to 110 centimeters.

 

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