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SSAT Upper Level Math : Geometry

Study concepts, example questions & explanations for SSAT Upper Level Math

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All SSAT Upper Level Math Resources

6 Diagnostic Tests 311 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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

Find the volume of a regular tetrahedron with side lengths of $\frac{1}{2}$ $\displaystyle \frac{1}{2}$.

Possible Answers:

$\frac{4\sqrt2}{3}$ $\displaystyle \frac{4\sqrt2}{3}$

$\frac{\sqrt2}{24}$ $\displaystyle \frac{\sqrt2}{24}$

$\frac{\sqrt2}{96}$ $\displaystyle \frac{\sqrt2}{96}$

$\frac{\sqrt2}{48}$ $\displaystyle \frac{\sqrt2}{48}$

Correct answer:

$\frac{\sqrt2}{96}$ $\displaystyle \frac{\sqrt2}{96}$

Explanation:

Use the following formula to find the volume of a regular tetrahedron:

$\text{Volume}=\frac{\sqrt2}{12}side^3$ $\displaystyle \text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}(\frac{1}{2})^3$ $\displaystyle \text{Volume}=\frac{\sqrt2}{12}(\frac{1}{2})^3$

$\text{Volume}=\frac{\sqrt2}{12}(\frac{1}{8})$ $\displaystyle \text{Volume}=\frac{\sqrt2}{12}(\frac{1}{8})$

$\text{Volume}=\frac{\sqrt2}{96}$ $\displaystyle \text{Volume}=\frac{\sqrt2}{96}$

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

Find the volume of a regular tetrahedron with side lengths of $\frac{1}{3}$ $\displaystyle \frac{1}{3}$.

Possible Answers:

$\frac{\sqrt2}{27}$ $\displaystyle \frac{\sqrt2}{27}$

$\frac{\sqrt2}{12}$ $\displaystyle \frac{\sqrt2}{12}$

$\frac{\sqrt2}{324}$ $\displaystyle \frac{\sqrt2}{324}$

$\frac{\sqrt2}{54}$ $\displaystyle \frac{\sqrt2}{54}$

Correct answer:

$\frac{\sqrt2}{324}$ $\displaystyle \frac{\sqrt2}{324}$

Explanation:

Use the following formula to find the volume of a regular tetrahedron:

$\text{Volume}=\frac{\sqrt2}{12}side^3$ $\displaystyle \text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}(\frac{1}{3})^3$ $\displaystyle \text{Volume}=\frac{\sqrt2}{12}(\frac{1}{3})^3$

$\text{Volume}=\frac{\sqrt2}{12}(\frac{1}{27})$ $\displaystyle \text{Volume}=\frac{\sqrt2}{12}(\frac{1}{27})$

$\text{Volume}=\frac{\sqrt2}{324}$ $\displaystyle \text{Volume}=\frac{\sqrt2}{324}$

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Example Question #671 : Geometry

A given rectangular prism has a length of $9\:cm$ $\displaystyle 9\:cm$, a width of $5\:cm$ $\displaystyle 5\:cm$, and a height of $7\:cm$ $\displaystyle 7\:cm$. What is the volume of the prism?

Possible Answers:

$45\:cm^{3}$ $\displaystyle 45\:cm^{3}$

$63\:cm^{3}$ $\displaystyle 63\:cm^{3}$

$315\:cm^{2}$ $\displaystyle 315\:cm^{2}$

$63\:cm^{2}$ $\displaystyle 63\:cm^{2}$

$315\:cm^{3}$ $\displaystyle 315\:cm^{3}$

Correct answer:

$315\:cm^{3}$ $\displaystyle 315\:cm^{3}$

Explanation:

The volume of a given prism V=Bh $\displaystyle V=Bh$, where $\displaystyle B$ is the base area and $\displaystyle h$ is the height. For a rectangular prism, the base area $B=l \times w$ $\displaystyle B=l \times w$, or length times width. Therefore:

V=Bh $\displaystyle V=Bh$

$V=(l\times w)h$ $\displaystyle V=(l\times w)h$

Substituting in our known values:

$V=(9\:cm\times 5\:cm)7\:cm$ $\displaystyle V=(9\:cm\times 5\:cm)7\:cm$

$V=45\:cm^{2}\times 7\:cm$ $\displaystyle V=45\:cm^{2}\times 7\:cm$

$V=315\:cm^{3}$ $\displaystyle V=315\:cm^{3}$

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

Find the volume of a circular prism with a base radius of $11\:cm$ $\displaystyle 11\:cm$ and a height of $20\:cm$ $\displaystyle 20\:cm$.

Possible Answers:

$440\pi\:cm^{2}$ $\displaystyle 440\pi\:cm^{2}$

$2420\pi\:cm^{3}$ $\displaystyle 2420\pi\:cm^{3}$

$2420\pi\:cm^{2}$ $\displaystyle 2420\pi\:cm^{2}$

$440\pi\:cm^{3}$ $\displaystyle 440\pi\:cm^{3}$

$220\pi\:cm^{3}$ $\displaystyle 220\pi\:cm^{3}$

Correct answer:

$2420\pi\:cm^{3}$ $\displaystyle 2420\pi\:cm^{3}$

Explanation:

The volume of a given prism V=Bh $\displaystyle V=Bh$, where $\displaystyle B$ is the base area and $\displaystyle h$ is the height. For a circular prism, the base area $B=\pi r^{2}$ $\displaystyle B=\pi r^{2}$, or pi times the square of the radius. Therefore:

$V=(\pi r^{2})h$ $\displaystyle V=(\pi r^{2})h$

Substituting in our known values:

$V=(\pi (11\:cm)^{2})(20\:cm)$ $\displaystyle V=(\pi (11\:cm)^{2})(20\:cm)$

$V=(121\pi\:cm^{2})(20\:cm)$ $\displaystyle V=(121\pi\:cm^{2})(20\:cm)$

$V=2420\pi\:cm^{3}$ $\displaystyle V=2420\pi\:cm^{3}$

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

A triangular prism has a triangle base length of $4\:cm$ $\displaystyle 4\:cm$, a triangle height length of $8\:cm$ $\displaystyle 8\:cm$, and a prism height of $12\:cm$ $\displaystyle 12\:cm$. What is the volume of the prism?

Possible Answers:

$192\:cm^{3}$ $\displaystyle 192\:cm^{3}$

$192\:cm^{2}$ $\displaystyle 192\:cm^{2}$

Not enough information provided

$384\:cm^{2}$ $\displaystyle 384\:cm^{2}$

$384\:cm^{3}$ $\displaystyle 384\:cm^{3}$

Correct answer:

$192\:cm^{3}$ $\displaystyle 192\:cm^{3}$

Explanation:

The volume of a given prism V=Bh $\displaystyle V=Bh$, where $\displaystyle B$ is the base area and $\displaystyle h$ is the prism height. For a triangular prism, the base area $B=\frac{1}{2}bh_{2}$ $\displaystyle B=\frac{1}{2}bh_{2}$, where $\displaystyle b$ is the base of the triangle and $h_{2}$ $\displaystyle h_{2}$ is the height of the triangle (not of the prism).

Therefore, we can substitute the base area equation into the equation for the volume of a prism:

V=Bh $\displaystyle V=Bh$

$V=(\frac{1}{2}bh_{2})h$ $\displaystyle V=(\frac{1}{2}bh_{2})h$

Substituting in our known values:

$V=(\frac{1}{2}(4\:cm)(8\:cm))(12\:cm)$ $\displaystyle V=(\frac{1}{2}(4\:cm)(8\:cm))(12\:cm)$

$V=(16\:cm^{2})(12\:cm)$ $\displaystyle V=(16\:cm^{2})(12\:cm)$

$V=192\:cm^{3}$ $\displaystyle V=192\:cm^{3}$

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

A given rectangular prism has a length of $\displaystyle 12$, a width of $\displaystyle 6$, and a height of $\displaystyle 8$. What is the volume of the prism?

Possible Answers:

572 $\displaystyle 572$

$\displaystyle 60$

574 $\displaystyle 574$

576 $\displaystyle 576$

$\displaystyle 26$

Correct answer:

576 $\displaystyle 576$

Explanation:

The volume $\displaystyle V$ of a rectangular prism is the product of its base area $\displaystyle A$ and its height $\displaystyle h$: V=Ah $\displaystyle V=Ah$. Since we can determine the base area of the rectangular prism from its length and width, we can rewrite the equation and solve:

V=Ah $\displaystyle V=Ah$

V=lwh $\displaystyle V=lwh$

$\displaystyle V=(12)(6)(8)$

$V=12 \times48$ $\displaystyle V=12 \times48$

V=576 $\displaystyle V=576$

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Example Question #2 : How To Find The Volume Of A Prism

A given rectangular prism has a length of 10cm $\displaystyle 10cm$, a width of 7cm $\displaystyle 7cm$, and a height of 15cm $\displaystyle 15cm$, what is its volume?

Possible Answers:

$85cm^{3}$ $\displaystyle 85cm^{3}$

$525cm^{3}$ $\displaystyle 525cm^{3}$

$32cm^{3}$ $\displaystyle 32cm^{3}$

$1050cm^{3}$ $\displaystyle 1050cm^{3}$

$1050cm^{2}$ $\displaystyle 1050cm^{2}$

Correct answer:

$1050cm^{3}$ $\displaystyle 1050cm^{3}$

Explanation:

V=Ah $\displaystyle V=Ah$

V=lwh $\displaystyle V=lwh$

$\displaystyle V=(10)(7)(25)$

V=1050 $\displaystyle V=1050$

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Example Question #1743 : Common Core Math: Grade 5

What is the volume of the shape below?

Screen shot 2015 07 28 at 3.39.34 pm

Possible Answers:

$\small 5cm^3$ $\displaystyle \small 5cm^3$

$\small 3cm^3$ $\displaystyle \small 3cm^3$

$\small 6cm^3$ $\displaystyle \small 6cm^3$

$\small 7cm^3$ $\displaystyle \small 7cm^3$

$\small 8cm^3$ $\displaystyle \small 8cm^3$

Correct answer:

$\small 6cm^3$ $\displaystyle \small 6cm^3$

Explanation:

The formula for volume of a rectangular prism is $\small v=l\times w\times h$ $\displaystyle \small v=l\times w\times h$

$\small v=3\times1\times2$ $\displaystyle \small v=3\times1\times2$

$\small v=6cm^3$ $\displaystyle \small v=6cm^3$

Remember, volume is always labeled as units to the third power.

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Example Question #1 : Apply The Volume Formula: Ccss.Math.Content.5.Md.C.5b

What is the volume of the shape below?

Screen shot 2015 07 28 at 3.47.24 pm

Possible Answers:

$\small 10cm^3$ $\displaystyle \small 10cm^3$

$\small 36cm^3$ $\displaystyle \small 36cm^3$

$\small 29cm^3$ $\displaystyle \small 29cm^3$

$\small 26m^3$ $\displaystyle \small 26m^3$

$\small 15cam^3$ $\displaystyle \small 15cam^3$

Correct answer:

$\small 36cm^3$ $\displaystyle \small 36cm^3$

Explanation:

The formula for volume of a rectangular prism is $\small v=l\times w\times h$ $\displaystyle \small v=l\times w\times h$

$\small v=6\times3\times2$ $\displaystyle \small v=6\times3\times2$

$\small v=36cm^3$ $\displaystyle \small v=36cm^3$

Remember, volume is always labeled as units to the third power.

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Example Question #2 : Apply The Volume Formula: Ccss.Math.Content.5.Md.C.5b

What is the volume of the shape below?

Screen shot 2015 07 28 at 3.56.22 pm

Possible Answers:

$\small 6cm^3$ $\displaystyle \small 6cm^3$

$\small 12cm^3$ $\displaystyle \small 12cm^3$

$\small 14cm^3$ $\displaystyle \small 14cm^3$

$\small 42cm^3$ $\displaystyle \small 42cm^3$

$\small 21cm^3$ $\displaystyle \small 21cm^3$

Correct answer:

$\small 42cm^3$ $\displaystyle \small 42cm^3$

Explanation:

The formula for volume of a rectangular prism is $\small v=l\times w\times h$ $\displaystyle \small v=l\times w\times h$

$\small v=2\times3\times7$ $\displaystyle \small v=2\times3\times7$

$\small v=42cm^3$ $\displaystyle \small v=42cm^3$

Remember, volume is always labeled as units to the third power.

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SSAT Upper Level Math : Geometry

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

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

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

Example Question #671 : Geometry

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

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

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

Example Question #2 : How To Find The Volume Of A Prism

Example Question #1743 : Common Core Math: Grade 5

Example Question #1 : Apply The Volume Formula: Ccss.Math.Content.5.Md.C.5b

Example Question #2 : Apply The Volume Formula: Ccss.Math.Content.5.Md.C.5b

All SSAT Upper Level Math Resources

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