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GED Math : 3-Dimensional Geometry

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Example Questions

Example Question #91 : 3 Dimensional Geometry

A cylindrical soft drink can has a radius of $\displaystyle 2$cm and a height of $\displaystyle 14$cm. In cubic centimeters, what volume of soft drink does the can hold? Round your answer to the nearest hundredths place.

Possible Answers:

193.40 $\displaystyle 193.40$

175.93 $\displaystyle 175.93$

142.59 $\displaystyle 142.59$

133.58 $\displaystyle 133.58$

Correct answer:

175.93 $\displaystyle 175.93$

Explanation:

Recall how to find the volume of a cylinder:

$\text{Volume}=\pi r^2 h$ $\displaystyle \text{Volume}=\pi r^2 h$

Plug in the given radius and height to find the volume.

$\text{Volume}=\pi (2)^2(14)=56\pi=175.93$ $\displaystyle \text{Volume}=\pi (2)^2(14)=56\pi=175.93$

The soft drink can is able to hold a volume of 173.93 $\displaystyle 173.93$ cubic centimeters of liquid.

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Example Question #92 : 3 Dimensional Geometry

If a cylinder has a height of $\displaystyle 3$ inches and a diameter of $\displaystyle 8$ inches, what is its volume?

Possible Answers:

$48\pi$ $\displaystyle 48\pi$ cubed inches

$12\pi$ $\displaystyle 12\pi$ cubed inches

81.3 $\displaystyle 81.3$ squared inches

$53.2\pi$ $\displaystyle 53.2\pi$ cubed inches

$53.2\pi$ $\displaystyle 53.2\pi$ squared inches

Correct answer:

$48\pi$ $\displaystyle 48\pi$ cubed inches

Explanation:

The volume of a cylinder can be solved for very simply. Just make sure you have the formula for volume handy!

The formula for the volume of a cylinder is: $V = \pi r^2h$ $\displaystyle V = \pi r^2h$, where r is radius and h is the height. It's helpful to remember that volume will provide a value in units cubed, where area or surface area will provide a value in units squared. This minor detail confuses many students.

Looking at the problem, we have been provided with the height and the diameter. This problem provides a slight curve ball in that we indirectly have been given the radius, so we need to do some detective work. What is the relationship between radius and diameter? The diameter is twice the radius. Or in math speak: d = 2r $\displaystyle d = 2r$. This means we can solve for our radius by taking half of the diameter. Therefore, the radius is $\displaystyle 4$ inches.

Now we are ready to "plug and chug" to get our final answer.

$V = \pi \times 4^{2} \times 3$ $\displaystyle V = \pi \times 4^{2} \times 3$

$V = \pi \times 16 \times 3$ $\displaystyle V = \pi \times 16 \times 3$

$V= \pi \times 48$ $\displaystyle V= \pi \times 48$

$V = 48\pi$ $\displaystyle V = 48\pi$ cubed inches

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Example Question #93 : 3 Dimensional Geometry

If a right cylinder has a diameter of $\displaystyle 12$ and a height that is three times the radius, what is its volume?

Possible Answers:

$18\pi$ $\displaystyle 18\pi$

$\displaystyle 94$

$648\pi$ $\displaystyle 648\pi$

$\displaystyle 27$

$36\pi$ $\displaystyle 36\pi$

Correct answer:

$648\pi$ $\displaystyle 648\pi$

Explanation:

The volume of a cylinder can be solved for very simply. Just make sure you have the formula for volume handy!

The formula for the volume of a cylinder is: $V = \pi r^2h$ $\displaystyle V = \pi r^2h$, where r is radius and h is the height. It's helpful to remember that volume will provide a value in units cubed, where area or surface area will provide a value in units squared. This minor detail confuses many students.

Looking at the problem, we have been provided with the height kind of) and the diameter. This problem provides a slight curve ball in that we indirectly have been given the radius and height, so we need to do some detective work. What is the relationship between radius and diameter? The diameter is twice the radius. Or in math speak: d = 2r $\displaystyle d = 2r$. This means we can solve for our radius by taking half of the diameter. Therefore, the radius is $\displaystyle 6$. Now we can figure out the value of the height. The problem says it is three times the radius. Therefore, $6 \times 3 = 18$ $\displaystyle 6 \times 3 = 18$, so the height is 18.

Now we are ready to "plug and chug" to get our final answer.

$V = \pi \times 6^{2} \times 18$ $\displaystyle V = \pi \times 6^{2} \times 18$

$V = \pi \times 36 \times 18$ $\displaystyle V = \pi \times 36 \times 18$

$V= \pi \times 648$ $\displaystyle V= \pi \times 648$

$V = 648\pi$ $\displaystyle V = 648\pi$ cubed units

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Example Question #94 : 3 Dimensional Geometry

A cylinder has a height of $\displaystyle 2x$ and a circumference of $z\pi$ $\displaystyle z\pi$. What is its volume?

Possible Answers:

$z^2x^2\pi$ $\displaystyle z^2x^2\pi$

$\frac{1}{2}zx\pi$ $\displaystyle \frac{1}{2}zx\pi$

$\frac{1}{8}z^2x^3\pi$ $\displaystyle \frac{1}{8}z^2x^3\pi$

$\frac{1}{2}z^2x\pi$ $\displaystyle \frac{1}{2}z^2x\pi$

$\frac{1}{4}z^3x\pi$ $\displaystyle \frac{1}{4}z^3x\pi$

Correct answer:

$\frac{1}{2}z^2x\pi$ $\displaystyle \frac{1}{2}z^2x\pi$

Explanation:

The volume of a cylinder can be solved for very simply. Just make sure you have the formula for volume handy!

The formula for the volume of a cylinder is: $V = \pi r^2h$ $\displaystyle V = \pi r^2h$, where r is radius and h is the height. It's helpful to remember that volume will provide a value in units cubed, where area or surface area will provide a value in units squared. This minor detail confuses many students.

Looking at the problem, we have been provided with the height and the circumference. This problem provides a slight curve ball in that we indirectly have been given the radius, so we need to do some detective work. What is the relationship between radius and circumference? The circumference is $\pi$ $\displaystyle \pi$ times the diameter. This can be mathematically defined as: $C=\pi d$ $\displaystyle C=\pi d$.

We may solve for the diameter by using this formula. Keep in mind we are solving for d.

$z\pi = \pi d$ $\displaystyle z\pi = \pi d$

Now divide both sides by $\pi$ $\displaystyle \pi$ to solve for d.

$\frac{z\pi}{\pi} = \frac{d{\color{Red} \pi}}{{\color{Red} \pi}}$ $\displaystyle \frac{z\pi}{\pi} = \frac{d{\color{Red} \pi}}{{\color{Red} \pi}}$

d=z $\displaystyle d=z$

We're almost there. Now we need to solve for the radius. The diameter is twice the radius. Or in math speak: d = 2r $\displaystyle d = 2r$. This means we can solve for our radius by taking half of the diameter. Therefore, the radius is $\frac{1}{2}z$ $\displaystyle \frac{1}{2}z$.

Now we are ready to "plug and chug" to get our final answer.

$V = \pi \times (\frac{1}{2}z)^{2} \times 2x$ $\displaystyle V = \pi \times (\frac{1}{2}z)^{2} \times 2x$

$V = \pi \times \frac{1}{4}z^2 \times 2x$ $\displaystyle V = \pi \times \frac{1}{4}z^2 \times 2x$

$V= \pi \times \frac{2}{4}z^2x$ $\displaystyle V= \pi \times \frac{2}{4}z^2x$

$V=\frac{1}{2}z^2x\pi$ $\displaystyle V=\frac{1}{2}z^2x\pi$

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Example Question #1 : Volume Of A Pyramid

Pyramid_1

The above square pyramid has volume 100. Evaluate $\displaystyle x$ to the nearest tenth.

Possible Answers:

14.1 $\displaystyle 14.1$

17.3 $\displaystyle 17.3$

6.7 $\displaystyle 6.7$

5.8 $\displaystyle 5.8$

Correct answer:

6.7 $\displaystyle 6.7$

Explanation:

The volume of a square pyramid with base of area $\displaystyle B$ and with height $\displaystyle h$ is

$V = \frac{1}{3} Bh$ $\displaystyle V = \frac{1}{3} Bh$.

The base, being a square of sidelength $\displaystyle x$, has area $x^{2}$ $\displaystyle x^{2}$. The height is $\displaystyle x$. Therefore, setting $V = 100, B = x^{2}, h = x$ $\displaystyle V = 100, B = x^{2}, h = x$, we solve for $\displaystyle x$ in the equation

$\frac{1}{3} x^{2} \cdot x = 100$ $\displaystyle \frac{1}{3} x^{2} \cdot x = 100$

$\frac{1}{3} x^{3} = 100$ $\displaystyle \frac{1}{3} x^{3} = 100$

$x^{3} = 300$ $\displaystyle x^{3} = 300$

$x = \sqrt[3]{300} \approx 6.7$ $\displaystyle x = \sqrt[3]{300} \approx 6.7$

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Example Question #2 : Volume Of A Pyramid

A square pyramid whose base has sidelength $\displaystyle A$ has volume $2A^{3}$ $\displaystyle 2A^{3}$. What is the ratio of the height of the pyramid to the sidelength of its base?

Possible Answers:

4:1 $\displaystyle 4:1$

3:1 $\displaystyle 3:1$

8:1 $\displaystyle 8:1$

6:1 $\displaystyle 6:1$

Correct answer:

6:1 $\displaystyle 6:1$

Explanation:

Since the correct answer is independent of the value of $\displaystyle A$, for simplicity's sake, assume that A = 1 $\displaystyle A = 1$.

The volume of a square pyramid with base of area $\displaystyle B$ and with height $\displaystyle h$ is

$V = \frac{1}{3} Bh$ $\displaystyle V = \frac{1}{3} Bh$.

The base, being a square of sidelength 1, has area 1. In the volume formula, we set B = 1 $\displaystyle B = 1$ and $V = 2 A^{3} = 2 \cdot 1^{3} = 2$ $\displaystyle V = 2 A^{3} = 2 \cdot 1^{3} = 2$, and solve for $\displaystyle h$:

$\frac{1}{3} \cdot 1 \cdot h = 2$ $\displaystyle \frac{1}{3} \cdot 1 \cdot h = 2$

$\frac{1}{3} h = 2$ $\displaystyle \frac{1}{3} h = 2$

$3 \cdot \frac{1}{3} h =3 \cdot 2$ $\displaystyle 3 \cdot \frac{1}{3} h =3 \cdot 2$

h = 6 $\displaystyle h = 6$

This means that the height-to-sidelength ratio h:A $\displaystyle h:A$ is equal to 6:1 $\displaystyle 6:1$, which is the correct response.

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Example Question #3 : Volume Of A Pyramid

Suppose a triangular pyramid has base area of 10, and a height of 6. What is the volume?

Possible Answers:

$\displaystyle 60$

$\displaystyle 40$

$\displaystyle 80$

$\displaystyle 20$

120 $\displaystyle 120$

Correct answer:

$\displaystyle 20$

Explanation:

Write the formula for the volume of a pyramid.

$V= \frac{B\times h}{3}$ $\displaystyle V= \frac{B\times h}{3}$

Substitute the known base area and the height into the formula.

$V= \frac{10\times 6}{3} = \frac{60}{3} = 20$ $\displaystyle V= \frac{10\times 6}{3} = \frac{60}{3} = 20$

The answer is: $\displaystyle 20$

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Example Question #4 : Volume Of A Pyramid

If the base of a pyramid is a square, with a length of 5, and the height of the pyramid is 9, what must be the volume?

Possible Answers:

$\displaystyle 45$

225 $\displaystyle 225$

150 $\displaystyle 150$

$\displaystyle 75$

$\displaystyle 15$

Correct answer:

$\displaystyle 75$

Explanation:

Write the formula for the volume of pyramid.

$V=\frac{Bh}{3}$ $\displaystyle V=\frac{Bh}{3}$

The base area of a square is s^2 $\displaystyle s^2$.

Substituting the side length:

$\displaystyle B=s^2 = 5^2 = 25$

Substitute the base area and the height.

$V=\frac{25\times 9}{3} = 25\times 3 = 75$ $\displaystyle V=\frac{25\times 9}{3} = 25\times 3 = 75$

The answer is: $\displaystyle 75$

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Example Question #5 : Volume Of A Pyramid

Suppose the base area of a pyramid is 24, and the height is 10. What must the volume be?

Possible Answers:

240 $\displaystyle 240$

160 $\displaystyle 160$

$\textup{There is not enough information.}$ $\displaystyle \textup{There is not enough information.}$

720 $\displaystyle 720$

$\displaystyle 80$

Correct answer:

$\displaystyle 80$

Explanation:

Write formula for a pyramid.

$V=\frac{(LW)H}{3} = \frac{BH}{3}$ $\displaystyle V=\frac{(LW)H}{3} = \frac{BH}{3}$

Substitute the base and height.

$V = \frac{24\times 10}{3} = 8\times 10 = 80$ $\displaystyle V = \frac{24\times 10}{3} = 8\times 10 = 80$

The answer is: $\displaystyle 80$

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Example Question #6 : Volume Of A Pyramid

Find the volume of a pyramid with a square base area of $\displaystyle 6$ and a height of $\displaystyle 10$.

Possible Answers:

$\displaystyle 30$

$\displaystyle 90$

$\displaystyle 15$

$\displaystyle 20$

$\displaystyle 60$

Correct answer:

$\displaystyle 20$

Explanation:

Write the formula of the volume of a pyramid.

$V = \frac{LWH}{3}$ $\displaystyle V = \frac{LWH}{3}$

The base area is represented by $\displaystyle LW$. This means the volume is:

$V = \frac{LWH}{3} =\frac{BH}{3}$ $\displaystyle V = \frac{LWH}{3} =\frac{BH}{3}$

Substitute the base and height.

$V=\frac{6(10)}{3} = 20$ $\displaystyle V=\frac{6(10)}{3} = 20$

The answer is: $\displaystyle 20$

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GED Math : 3-Dimensional Geometry

Study concepts, example questions & explanations for GED Math

All GED Math Resources

Example Questions

Example Question #91 : 3 Dimensional Geometry

Example Question #92 : 3 Dimensional Geometry

Example Question #93 : 3 Dimensional Geometry

Example Question #94 : 3 Dimensional Geometry

Example Question #1 : Volume Of A Pyramid

Example Question #2 : Volume Of A Pyramid

Example Question #3 : Volume Of A Pyramid

Example Question #4 : Volume Of A Pyramid

Example Question #5 : Volume Of A Pyramid

Example Question #6 : Volume Of A Pyramid

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