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GED Math : GED Math

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Example Questions

Example Question #102 : 3 Dimensional Geometry

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

Possible Answers:

120

160

Correct answer:

Explanation:

Write the volume formula for the pyramid.

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

Substitute the dimensions.

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

The answer is:

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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:

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

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

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

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

Correct answer:

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

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

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

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

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

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Example Question #102 : 3 Dimensional Geometry

Cone_1

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The volume of a right circular cone of height and base with radius is

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

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 as the other leg:

$h^{2} + 50^{2} = 130 ^{2}$

$h^{2} + 2,500= 16,900$

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

$h^{2} = 14,4 00$

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

Substitute 120 for and 50 for in the volume formula:

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

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

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

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Example Question #1 : Volume Of Other Solids

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

Possible Answers:

$2532\text{cm}^3$

$1728\text{cm}^3$

$2028\text{cm}^3$

$1584\text{cm}^3$

$1331\text{cm}^3$

Correct answer:

$1728\text{cm}^3$

Explanation:

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

$V = l \cdot w \cdot h$

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

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

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

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

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Example Question #2 : Volume Of Other Solids

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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

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

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Example Question #1 : Volume Of Other Solids

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

Possible Answers:

$1331\text{in}^3$

$726\text{in}^3$

$1440\text{in}^3$

$864\text{in}^3$

$1210\text{in}^3$

Correct answer:

$1331\text{in}^3$

Explanation:

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

$V = l \cdot w \cdot h$

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

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

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

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

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Example Question #3 : Volume Of Other Solids

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

Possible Answers:

$300\text{cm}^3$

$2500\text{cm}^3$

$1000\text{cm}^3$

$3000\text{cm}^3$

$1500\text{cm}^3$

Correct answer:

$1000\text{cm}^3$

Explanation:

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

$V = l \cdot w \cdot h$

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

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

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

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

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Example Question #4 : Volume Of Other Solids

A cone has a diameter of 8in and a height of 9in. Find the volume.

Possible Answers:

$24\pi \text{ in}^3$

$48\pi \text{ in}^3$

$36\pi \text{ in}^3$

$72\pi \text{ in}^3$

$45\pi \text{ in}^3$

Correct answer:

$48\pi \text{ in}^3$

Explanation:

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

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

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

Now, we know the diameter of the cone is 8in. We also know the diameter is two times the radius. Therefore, the radius is 4in.

We know the height of the cone is 9in.

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

$V = \pi \cdot (4\text{in})^2 \cdot \frac{9\text{in}}{3}$

$V = \pi \cdot 16\text{in}^2 \cdot 3\text{in}$

$V = \pi \cdot 48\text{in}^3$

$V = 48\pi \text{ in}^3$

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Example Question #3 : Volume Of Other Solids

Find the volume of a cube with a height of 7in.

Possible Answers:

$294\text{in}^3$

$196\text{in}^3$

$216\text{in}^3$

$343\text{in}^3$

$245\text{in}^3$

Correct answer:

$343\text{in}^3$

Explanation:

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

$V = l \cdot w \cdot h$

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 7in. Because it is a cube, all sides are equal. Therefore, the width and the length are also 7in. So, we can substitute. We get

$V = 7\text{in} \cdot 7\text{in} \cdot 7\text{in}$

$V = 49\text{in}^2 \cdot 7\text{in}$

$V = 343\text{in}^3$

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Example Question #1 : Volume Of Other Solids

Let $\pi = 3.14$ .

A cone has a height of 5in and a diameter of 12in. Find the volume.

Possible Answers:

$452.16\text{in}^3$

$188.4\text{in}^3$

$376.80\text{in}^3$

$471\text{in}^3$

$94.20\text{in}^3$

Correct answer:

$188.4\text{in}^3$

Explanation:

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

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

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

We know $\pi=3.14$ .

We know the diameter of the cone is 12in. We know the diameter is two times the radius. So, the radius is 6in.

We know the height is 5in.

Now, we can substitute. We get

$V = 3.14 \cdot (6\text{in})^2 \cdot \frac{5\text{in}}{3}$

$V = 3.14 \cdot 36\text{in}^2 \cdot \frac{5\text{in}}{3}$

Now, we can simplify to make things easier. The 36 and the 3 can both be divided by 3. So, we get

$V = 3.14 \cdot 12\text{in}^2 \cdot \frac{5\text{in}}{1}$

$V = 3.14 \cdot 12\text{in}^2 \cdot 5\text{in}$

$V = 3.14 \cdot 60\text{in}^3$

$V = 188.4\text{in}^3$

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GED Math : GED Math

Study concepts, example questions & explanations for GED Math

All GED Math Resources

Example Questions

Example Question #102 : 3 Dimensional Geometry

Example Question #1 : Volume Of Other Solids

Example Question #102 : 3 Dimensional Geometry

Example Question #1 : Volume Of Other Solids

Example Question #2 : Volume Of Other Solids

Example Question #1 : Volume Of Other Solids

Example Question #3 : Volume Of Other Solids

Example Question #4 : Volume Of Other Solids

Example Question #3 : Volume Of Other Solids

Example Question #1 : Volume Of Other Solids

All GED Math Resources

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