GED Math : 3-Dimensional Geometry

Study concepts, example questions & explanations for GED Math

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Example Questions

Example Question #5 : Volume Of A Pyramid

Find the volume of a square pyramid with a base area of 12 and a height of 4.

Possible Answers:

\displaystyle 192

\displaystyle 16

\displaystyle 576

\displaystyle 48

Correct answer:

\displaystyle 16

Explanation:

Write the formula for the volume of a pyramid.

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

The base area constitutes \displaystyle LW of the given equation.

Substitute values into the formula.

\displaystyle V = \frac{12(4)}{3} = 4(4) = 16

The answer is:  \displaystyle 16

Example Question #1614 : Ged Math

Find the volume of a square pyramid if the base area is 2, and the height is 5.

Possible Answers:

\displaystyle 10

\displaystyle \frac{10}{3}

\displaystyle 50

\displaystyle 20

Correct answer:

\displaystyle \frac{10}{3}

Explanation:

Write the formula for the volume of a pyramid.

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

The base area constitutes \displaystyle LW, and can be replaced with the numerical area.

\displaystyle V =\frac{LWH}{3} =\frac{BH}{3} = \frac{2\times 5}{3} =\frac{10}{3}

The answer is:  \displaystyle \frac{10}{3}

Example Question #5 : Volume Of A Pyramid

If the length, width, and height of a pyramid is 2, 7, and 9, respectively, what must be the volume?

Possible Answers:

\displaystyle 42

\displaystyle 32

\displaystyle 180

\displaystyle 81

\displaystyle 126

Correct answer:

\displaystyle 42

Explanation:

Write the formula for the volume of a pyramid.

\displaystyle V= \frac{LWH}{3} = \frac{2(7)(9)}{3} = 2(7)(3) = 42

The answer is:  \displaystyle 42

Example Question #6 : Volume Of A Pyramid

Find the volume of a pyramid with a length, width, and height of \displaystyle 2,5,12, respectively.

Possible Answers:

\displaystyle 120

\displaystyle 20

\displaystyle 160

\displaystyle 80

\displaystyle 40

Correct answer:

\displaystyle 40

Explanation:

Write the volume formula for the pyramid.

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

Substitute the dimensions.

\displaystyle V=\frac{2\times 5 \times 12}{3} = (2)(5)(4) = 40

The answer is:  \displaystyle 40

Example Question #1 : Volume Of Other Solids

Swimming_pool

The above depicts a rectangular swimming pool for an apartment. 

On the right edge, the pool is three feet deep; on the left edge, it is eight feet deep. Going from the right to the left, its depth increases uniformly. In cubic feet, how much water does the pool hold?

Possible Answers:

\displaystyle 1,870 \textrm{ ft}^{3}

\displaystyle 2,070 \textrm{ ft}^{3}

\displaystyle 9,625 \textrm{ ft}^{3}

\displaystyle 9,875 \textrm{ ft}^{3}

Correct answer:

\displaystyle 9,625 \textrm{ ft}^{3}

Explanation:

The pool can be looked at as a trapezoidal prism with "height" 35 feet and its bases the following shape (depth exaggerated):

Pool

The area of this trapezoidal base, which has height 50 feet and bases 3 feet and 8 feet, is

\displaystyle B = \frac{1}{2} (3+8) \cdot 50 = 275 square feet;

The volume of the pool is the base multiplied by the "height", or 

\displaystyle 275 \cdot 35 = 9,625 cubic feet, the capacity of the pool.

Example Question #2 : Volume Of Other Solids

Cone_1

Refer to the right circular cone in the above diagram. What is its volume, to the nearest whole cubic centimeter?

Possible Answers:

\displaystyle 314,159 \textrm{ cm}^{3}

\displaystyle 510,509 \textrm{ cm}^{3}

\displaystyle 340,339 \textrm{ cm}^{3}

\displaystyle 471,239 \textrm{ cm}^{3}

Correct answer:

\displaystyle 314,159 \textrm{ cm}^{3}

Explanation:

The volume of a right circular cone of height \displaystyle h and base with radius \displaystyle r is 

\displaystyle V = \frac{1}{3} \pi r^{2}h.

The radius is 50. To find the height, we need to use the Pythagorean Theorem with the radius 50 as one leg and the slant height 130 as the hypotenuse of a right triangle, and the height \displaystyle h as the other leg:

\displaystyle h^{2} + 50^{2} = 130 ^{2}

\displaystyle h^{2} + 2,500= 16,900

\displaystyle h^{2} + 2,500- 2,500= 16,9 00 - 2,500

\displaystyle h^{2} = 14,4 00

\displaystyle h = \sqrt{14,400} = 120

Substitute 120 for \displaystyle h and 50 for \displaystyle r in the volume formula:

\displaystyle V = \frac{1}{3} \pi r^{2}h

\displaystyle V = \frac{1}{3} \times 3.14 \times 50^{2} \times 120

\displaystyle V = \frac{1}{3} \times 3.14 \times 2,500 \times 120 \approx 314,159 cubic centimeters.

Example Question #1 : Volume Of Other Solids

Find the volume of a cube with a height of 12cm.

Possible Answers:

\displaystyle 2532\text{cm}^3

\displaystyle 1728\text{cm}^3

\displaystyle 2028\text{cm}^3

\displaystyle 1584\text{cm}^3

\displaystyle 1331\text{cm}^3

Correct answer:

\displaystyle 1728\text{cm}^3

Explanation:

To find the volume of a cube, we will use the following formula:

\displaystyle V = l \cdot w \cdot h

where l is the length, w is the width, and h is the height of the cube. 

Now, we know the height of the cube is 12cm. Because it is a cube, all sides (lengths, widths, heights) are equal. Therefore, the length and the width are also 12cm. So, we can substitute. We get

\displaystyle V = 12\text{cm} \cdot12\text{cm} \cdot12\text{cm}

\displaystyle V = 144\text{cm}^2 \cdot 12\text{cm}

\displaystyle V = 1728\text{cm}^3

Example Question #3 : Volume Of Other Solids

A cone has a diameter of 10in and a height of 6in.  Find the volume.

Possible Answers:

\displaystyle 30\pi \text{ in}^3

\displaystyle 20\pi \text{ in}^3

\displaystyle 80\pi \text{ in}^3

\displaystyle 50\pi \text{ in}^3

\displaystyle 60\pi \text{ in}^3

Correct answer:

\displaystyle 50\pi \text{ in}^3

Explanation:

To find the volume of a cone, we will use the following formula:

\displaystyle V = \pi r^2 \frac{h}{3}

where r is the radius and h is the height of the cone.

Now, we know the diameter of the cone is 10in. We also know the diameter is two times the radius. Therefore, the radius is 5in.

We know the height of the cone is 6in.

Knowing all of this, we can substitute into the formula. We get

\displaystyle V = \pi \cdot (5\text{in})^2 \cdot \frac{6\text{in}}{3}

\displaystyle V = \pi \cdot 25\text{in}^2 \cdot 2\text{in}

\displaystyle V = \pi \cdot 50\text{in}^3

\displaystyle V = 50\pi \text{ in}^3

Example Question #661 : Geometry And Graphs

Find the volume of a cube with a height of 11in.

Possible Answers:

\displaystyle 726\text{in}^3

\displaystyle 864\text{in}^3

\displaystyle 1210\text{in}^3

\displaystyle 1440\text{in}^3

\displaystyle 1331\text{in}^3

Correct answer:

\displaystyle 1331\text{in}^3

Explanation:

To find the volume of a cube, we will use the following formula:

\displaystyle V = l \cdot w \cdot h

where l is the length, w is the width, and h is the height of the cube.

Now, we know the height of the cube is 11in. Because it is a cube, all sides are equal. Therefore, the width and the length are also 11in. So, we can substitute. We get

\displaystyle V = 11\text{in} \cdot11\text{in} \cdot11\text{in}

\displaystyle V = 121\text{in}^2 \cdot 11\text{in}

\displaystyle V = 1331\text{in}^3

Example Question #662 : Geometry And Graphs

Find the volume of a cube with a height of 10cm.

Possible Answers:

\displaystyle 1000\text{cm}^3

\displaystyle 1500\text{cm}^3

\displaystyle 300\text{cm}^3

\displaystyle 2500\text{cm}^3

\displaystyle 3000\text{cm}^3

Correct answer:

\displaystyle 1000\text{cm}^3

Explanation:

To find the volume of a cube, we will use the following formula:

\displaystyle V = l \cdot w \cdot h

where l is the length, w is the width, and h is the height of the cube.

Now, we know the height of the cube is 10cm. Because it is a cube, all sides (lengths, widths, heights) are equal. Therefore, the length and the width are also 10cm. So, we can substitute. We get

\displaystyle V = 10\text{cm} \cdot 10\text{cm} \cdot 10\text{cm}

\displaystyle V = 100\text{cm}^2 \cdot 10\text{cm}

\displaystyle V = 1000\text{cm}^3

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