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GED Math : 3-Dimensional Geometry

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Example Questions

Example Question #673 : Geometry And Graphs

A cube has a length of 9cm. Find the volume.

Possible Answers:

$486\text{cm}^3$

$567\text{cm}^3$

$81\text{cm}^3$

$729\text{cm}^3$

$381\text{cm}^3$

Correct answer:

$729\text{cm}^3$

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 length of the cube is 9cm. Because it is a cube, all sides/lengths are equal. Therefore, the width and height are also 9cm.

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

$V = 9\text{cm} \cdot 9\text{cm} \cdot 9\text{cm}$

$V = 81\text{cm}^2 \cdot 9\text{cm}$

$V = 729\text{cm}^3$

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Example Question #673 : Geometry And Graphs

Find the volume of a cube with a height of 6cm.

Possible Answers:

$180\text{cm}^3$

$216\text{cm}^3$

$36\text{cm}^3$

$252\text{cm}^3$

$27\text{cm}^3$

Correct answer:

$216\text{cm}^3$

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 6cm. Because it is a cube, all lengths/sides/etc are equal. Therefore, the length and the width are also 6cm.

So, we substitute. We get

$V = 6\text{cm} \cdot 6\text{cm} \cdot 6\text{cm}$

$V = 36\text{cm}^2 \cdot 6\text{cm}$

$V = 216\text{cm}^3$

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Example Question #674 : Geometry And Graphs

Find the volume of a cone with a radius of 8in and a height of 6in.

Possible Answers:

$384\pi \text{ in}^3$

$128\pi \text{ in}^3$

$144\pi \text{ in}^3$

$96\pi \text{ in}^3$

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

Correct answer:

$128\pi \text{ in}^3$

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 radius is 8in. We know the height is 6in. So, we can substitute. We get

$V = \pi \cdot (8\text{in})^2 \cdot \frac{6\text{in}}{3}$

$V = \pi \cdot 64\text{in}^2 \cdot 2\text{in}$

$V = \pi \cdot 128\text{in}^3$

$V = 128\pi \text{ in}^3$

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Example Question #122 : 3 Dimensional Geometry

An office uses cone-shaped paper cups for water in their water cooler. The cups have a radius of inches and a height of 4.25 inches. If the water cooler can hold 1155 cubic inches of water, how many complete cups of water can the water cooler fill?

Possible Answers:

Correct answer:

Explanation:

Start by finding the volume of a cup.

Recall how to find the volume of a cone:

$\text{Volume of Cone}=\frac{1}{3}\pi r^2 h$

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

$\text{Volume}=\frac{1}{3}\pi(3)^2(4.25)=40.055$

Now divide the total volume of the water in the water cooler by the volume of one cup in order to find how many complete cups the water cooler can fill.

$\text{Number of cups filled}=\frac{1155}{40.055}=28.84$

Since the question asks for the number of complete cups that can be filled, we must round down to .

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Example Question #121 : 3 Dimensional Geometry

How many edges and vertices are found on a square pyramid?

Possible Answers:

$8\ edges,\ 8\ vertices$

$8\ edges,\ 5\ vertices$

$6\ edges,\ 4\ vertices$

$6\ edges,\ 6\ vertices$

Correct answer:

$8\ edges,\ 5\ vertices$

Explanation:

The base of a square pyramid is, as the name suggests, a square which has edges and vertices. The vertices of the square each have edges that meet at a single point, adding an additional vertex and additional edges. Together, a square pyramid has edges and vertices.

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Example Question #122 : 3 Dimensional Geometry

How many vertices does an octagonal pyramid have?

Possible Answers:

Seven

Eight

Six

Nine

Correct answer:

Nine

Explanation:

An octagonal pyramid has a base with eight vertices, each of which is a vertex of the pyramid. There is one more vertex, or the apex, which is connected to each of the vertices of the base by an edge. Nine is the correct choice.

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Example Question #123 : 3 Dimensional Geometry

A circular swimming pool at an apartment complex has diameter 18 meters and depth 2.5 meters throughout.

The apartment manager needs to get the interior of the swimming pool painted. The paint she wants to use covers 40 square meters per can. How many cans of paint will she need to purchase?

You may use 3.14 for $\pi$ .

Possible Answers:

Correct answer:

Explanation:

The pool can be seen as a cylinder with depth (or height) 2.5 meters and a base with diameter 18 meters - and radius half this, or 9 meters.

The bottom of the pool - the base of the cylinder - is a circle with radius 9 meters, so its area is

$B = \pi r ^{2} = 3.14 \cdot 9 \cdot 9 = 254.34$ square meters.

Its side - the lateral face of the cylinder - has area

$LA = 2 \pi r h = 2 \cdot 3.14 \cdot 18 \cdot 2.5 = 141.3$ square meters.

Their sum - the total area to be painted - is 254.34 + 141.3 = 395.64 square feet. Since one can of paint covers 40 square meters, divide:

$395.64 \div 40 \approx 9.891$

Nine cans of paint and part of a tenth will be required, so the correct response is ten.

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Example Question #1 : Faces And Surface Area

A water tank takes the shape of a sphere whose exterior has radius 24 feet; the tank is six inches thick throughout. To the nearest hundred, give the surface area of the interior of the tank in square feet.

Use 3.14 for $\pi$ .

Possible Answers:

$54,300 \textrm{ ft}^{2}$

$6,600 \textrm{ ft}^{2}$

$6,900 \textrm{ ft}^{2}$

$50,900 \textrm{ ft}^{2}$

Correct answer:

$6,900 \textrm{ ft}^{2}$

Explanation:

Six inches is equal to 0.5 feet, so the radius of the interior of the tank is

24-0.5= 23.5 feet.

The surface area of the interior of the tank can be calculated using the formula

$SA = 4 \pi r^{2}$

$SA \approx 4 \cdot 3.14 \cdot 23.5 ^{2} \approx 6,936$ ,

which rounds to 6,900 square feet.

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Example Question #124 : 3 Dimensional Geometry

Cone_1

Give the total surface area of the above cone to the nearest square meter.

Possible Answers:

$176\textrm{ m}^{2}$

$186\textrm{ m}^{2}$

$126\textrm{ m}^{2}$

$135\textrm{ m}^{2}$

Correct answer:

$186\textrm{ m}^{2}$

Explanation:

The base is a circle with radius r = 4 , and its area can be calculated using the area formula for a circle:

$B= \pi r^{2} = \pi \times 4^{2} \approx 3.14 \times 16 = 50.24$ square meters.

To find the lateral area, we need the slant height of the cone. This can be found by way of the Pythagorean Theorem. Treating the height h = 10 and the radius r = 4 as the legs and slant height as the hypotenuse, calculate:

$l^{2} = h^{2} + r^{2}$

$l^{2} = 10^{2} + 4^{2} = 100 + 16 = 116$

$l = \sqrt{116} \approx 10.77$ meters.

The formula for the lateral area can be applied now:

$LA = \pi r l \approx 3.14 \cdot 4 \cdot 10.77 \approx 135.34$

Add the base and the lateral area to obtain the total surface area:

$SA = B + LA \approx 50.24 + 135.34 \approx 185.58$ .

This rounds to 186 square meters.

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Example Question #125 : 3 Dimensional Geometry

A water tank takes the shape of a closed cylinder whose exterior has a height of 40 feet and a base with radius 15 feet; the tank is three inches thick throughout. To the nearest hundred, give the surface area of the interior of the tank in square feet.

Use 3.14 for $\pi$ .

Possible Answers:

$3,600\textrm{ ft}^{2}$

$3,700\textrm{ ft}^{2}$

$4,900\textrm{ ft}^{2}$

$5,000\textrm{ ft}^{2}$

Correct answer:

$5,000\textrm{ ft}^{2}$

Explanation:

Three inches is equal to 0.25 feet, so the height of the interior of the tank is

$40 - 2 \times 0.25 = 39.5$ feet.

The radius of the interior of the tank is

15 - 0.25 = 14.75 feet.

The surface area of the interior of the tank can be determined by using this formula:

$SA = 2 \pi r (r + h)$

$SA = 2 \cdot 3.14 \cdot 14.75 (14.75 + 39.5)$

$SA \approx 2 \cdot 3.14 \cdot 14.75 \cdot 54.25\approx 5,025$ ,

which rounds to 5,000 square feet.

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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 #673 : Geometry And Graphs

Example Question #673 : Geometry And Graphs

Example Question #674 : Geometry And Graphs

Example Question #122 : 3 Dimensional Geometry

Example Question #121 : 3 Dimensional Geometry

Example Question #122 : 3 Dimensional Geometry

Example Question #123 : 3 Dimensional Geometry

Example Question #1 : Faces And Surface Area

Example Question #124 : 3 Dimensional Geometry

Example Question #125 : 3 Dimensional Geometry

All GED Math Resources

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