GED Math Quiz: Geometry 3d

15 questions · exam conditions

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Geometry 3dQuestion 1 of 15

Refer to the figure. A solid metal sphere of radius $6$ cm is melted and recast into small cylinders, each with radius $1$ cm and height $2$ cm. Assuming no metal is lost, what is the maximum number of complete small cylinders that can be made?

$Question graphic$

48

144

216

288

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GED Math Quiz

GED Math Quiz: Geometry 3d

Practice Geometry 3d in GED Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Geometry 3d, giving you a quick way to practice the rules, question types, and explanations that matter most for GED Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$48$
$144$ (correct answer)
$216$
$288$

Explanation: Sphere volume:

\tfrac{4}{3}\pi(6)^3=288\pi

. Cylinder volume:

\pi(1)^2(2)=2\pi

. Number:

288\pi/2\pi=144

. A forgets the

\tfrac{4}{3}

factor. C uses sphere volume formula

\tfrac{4}{3}\pi r^3

but divides wrongly. D doubles the answer.

Question 2

A cube has a total surface area of $486\text{ cm}^2$ . What is the volume of the cube?

$729\text{ cm}^3$ (correct answer)
$216\text{ cm}^3$
$125\text{ cm}^3$
$64\text{ cm}^3$

Explanation: This problem tests your understanding of surface area and volume formulas for cubes, and how to work backwards from given information to find what you need. A cube has 6 identical square faces. If each face has side length

s

, then each face has area

s^2

, making the total surface area

6s^2

. Since the surface area is

486\text{ cm}^2

, you can set up the equation:

6s^2 = 486

. Dividing both sides by 6 gives

s^2 = 81

, so

s = 9\text{ cm}

. Now that you know the side length, you can find the volume using

V = s^3 = 9^3 = 729\text{ cm}^3

. This confirms that A is correct. Looking at the wrong answers: B (

216\text{ cm}^3

) equals

6^3

, which suggests someone incorrectly used

s = 6

as the side length—this comes from mistakenly thinking

s^2 = 36

instead of

s^2 = 81

. C (

125\text{ cm}^3

) equals

5^3

, indicating the error of using

s = 5

. D (

64\text{ cm}^3

) equals

4^3

, showing the mistake of using

s = 4

. These wrong answers likely result from computational errors when solving

6s^2 = 486

or from incorrectly taking the square root of 81. Remember: when working with cube problems, always identify what information you're given and what formulas connect that to what you need to find. Surface area problems often require you to find the side length first, then use it to calculate volume.

Question 3

A shipping box is a right rectangular prism that measures $18 \text{ in}$ long, $12 \text{ in}$ wide, and $10 \text{ in}$ high. What is the volume of the box?

$2\,160\text{ in}^3$ (correct answer)
$1\,440\text{ in}^3$
$600\text{ in}^3$
$360\text{ in}^3$

Explanation: When you encounter a problem asking for the volume of a rectangular prism (or box), you're working with one of the most fundamental 3D geometry formulas. Volume measures how much space is inside a three-dimensional object, and for rectangular prisms, you simply multiply length × width × height. For this shipping box, you have all three dimensions: 18 inches long, 12 inches wide, and 10 inches high. Multiply these together:

18 × 12 × 10 = 2{,}160 \text{ cubic inches}

. Notice that volume is always expressed in cubic units (

\text{in}^3

\text{ft}^3

, etc.) because you're multiplying three linear measurements. Let's examine why the other answers are incorrect. Answer B (

1{,}440\text{ in}^3

) results from multiplying only two dimensions correctly and making an error with the third—perhaps calculating

18 × 12 × 8

instead of

18 × 12 × 10

. Answer C (

600\text{ in}^3

) comes from adding the dimensions instead of multiplying them (

18 + 12 + 10 = 40

, then possibly multiplying by something incorrectly). Answer D (

360\text{ in}^3

) appears to come from multiplying only some of the dimensions, like

18 × 10 = 180

, then doubling it. Remember: for any rectangular prism volume problem, always multiply all three dimensions—length × width × height. Double-check that you're multiplying (not adding) and that your final answer includes cubic units. This formula appears frequently on the GED, so master it completely.

Question 4

A grain silo is shown in the figure. It consists of a cylinder with a hemisphere on top. The cylindrical portion has a radius of $4$ feet and a height of $15$ feet. Refer to the figure. What is the total volume of the silo, in cubic feet, rounded to the nearest cubic foot?

$754$
$888$ (correct answer)
$1,021$
$1,155$

Explanation: Cylinder volume:

\pi(4)^2(15)=240\pi

. Hemisphere volume:

\frac{1}{2}\cdot\frac{4}{3}\pi(4)^3=\frac{128\pi}{3}

. Total:

240\pi+\frac{128\pi}{3}=\frac{848\pi}{3}\approx 888

. Choice A uses only the cylinder. Choice C adds a full sphere instead of a hemisphere. Choice D adds a cone with height 4 instead of a hemisphere.

Question 5

A cone has the dimensions shown. Refer to the figure. If the radius is doubled and the height is halved, by what factor does the volume change?

The volume is unchanged.
The volume is multiplied by $2$ . (correct answer)
The volume is multiplied by $4$ .
The volume is multiplied by $\tfrac{1}{2}$ .

Explanation: Original volume

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

. New volume:

\tfrac{1}{3}\pi(2r)^2(h/2)=\tfrac{1}{3}\pi\cdot 4r^2\cdot h/2=2V

. A assumes effects cancel (forgetting radius is squared). C forgets to halve the height. D reverses the logic.

Question 6

Refer to the figure. A cylindrical water tank has an inner radius of $3$ feet and a height of $8$ feet. The tank is currently filled with water to $\tfrac{3}{4}$ of its capacity. How many cubic feet of water are in the tank? (Use $\pi\approx 3.14$ .)

$56.52$
$169.56$ (correct answer)
$226.08$
$678.24$

Explanation: Full volume:

\pi(3)^2(8)=72\pi\approx 226.08

. Three-quarters:

0.75(226.08)=169.56

. A takes

\tfrac{1}{4}

instead of

\tfrac{3}{4}

. C gives full tank volume. D uses diameter 6 as radius.

Question 7

Refer to the figure. A sphere is inscribed in a cube of edge length $10$ cm so that the sphere touches all six faces. What is the volume of the empty space inside the cube but outside the sphere, in cubic cm, to the nearest whole number?

$477$ (correct answer)
$523$
$816$
$1,000$

Explanation: Cube volume:

1000

. Sphere radius is half the edge =

5

. Sphere volume:

\tfrac{4}{3}\pi(5)^3\approx 523.6

. Empty space:

1000-523.6\approx 476.4\approx 477

. B gives the sphere volume. C uses radius 10. D gives cube volume only.

Question 8

Refer to the figure. A cone has a radius of $5$ cm and a height of $12$ cm. What is the lateral (side) surface area of the cone, in square centimeters? Leave the answer in terms of $\pi$ .

$60\pi$
$65\pi$ (correct answer)
$90\pi$
$25\pi$

Explanation: Slant height:

\ell=\sqrt{5^2+12^2}=\sqrt{169}=13

. Lateral area:

\pi r\ell=\pi(5)(13)=65\pi

. A uses height 12 as slant. C uses diameter. D is base area only.

Question 9

Refer to the figure. A cylindrical candle has a diameter of $8$ cm and a height of $15$ cm. A wax manufacturer wants to wrap the curved (lateral) surface and the top circular face of each candle with a decorative film. How many square centimeters of film are needed per candle? (Use $\pi\approx 3.14$ .)

$376.8$
$427.04$ (correct answer)
$477.28$
$527.52$

Explanation: Radius =

4

cm. Lateral area:

2\pi rh=2\pi(4)(15)=120\pi\approx 376.8

. Top circle:

\pi(4)^2=16\pi\approx 50.24

. Total:

376.8+50.24=427.04

. A omits the top. C includes both top and bottom. D uses diameter as radius for one term.

Question 10

Refer to the figure. A hemispherical bowl has an inner radius of $6$ inches. Water is poured into the bowl until it is filled to a depth equal to the full radius (completely full). The water is then poured into a cylindrical container with an inner radius of $4$ inches. To what height, in inches, will the water rise in the cylinder?

$6$
$9$ (correct answer)
$12$
$18$

Explanation: Hemisphere volume:

\tfrac{1}{2}\cdot\tfrac{4}{3}\pi(6)^3=\tfrac{2}{3}\pi(216)=144\pi

. Set equal to cylinder volume:

\pi(4)^2 h=16\pi h=144\pi

, so

h=9

. A uses ratio

6/...

incorrectly. C uses full sphere formula. D forgets to square the cylinder radius.

Question 11

Refer to the figure. A regular hexagonal prism has a base edge length of $4$ cm and a height of $10$ cm. What is the volume of the prism, in cubic centimeters? (The area of a regular hexagon with side $s$ is $\tfrac{3\sqrt{3}}{2}s^2$ .)

$120\sqrt{3}$
$240\sqrt{3}$ (correct answer)
$480\sqrt{3}$
$960\sqrt{3}$

Explanation: Base area:

\tfrac{3\sqrt{3}}{2}(4)^2=24\sqrt{3}

. Volume:

24\sqrt{3}\cdot 10=240\sqrt{3}

. A uses half the base area. C doubles the answer (perhaps confused with surface area). D uses

3\sqrt{3}s^2

(forgot the

\tfrac{1}{2}

Question 12

A cylindrical water tank has a radius of 4 feet and a height of 12 feet. If the tank is currently filled to 75% of its capacity, how many cubic feet of water need to be added to fill it completely?

$48\pi$ cubic feet (correct answer)
$192\pi$ cubic feet
$144\pi$ cubic feet
$64\pi$ cubic feet

Explanation: First, find the total volume:

V = \pi r^2 h = \pi(4)^2(12) = 192\pi

cubic feet. Currently filled to 75%, so current volume is

0.75 \times 192\pi = 144\pi

cubic feet. Water needed:

192\pi - 144\pi = 48\pi

cubic feet. Choice B is the total volume, choice C is the current volume, and choice D incorrectly uses radius instead of radius squared in the calculation.

Question 13

A cylindrical can has a radius of $4\text{ cm}$ and a height of $15\text{ cm}$ . What is the surface area of the can to the nearest square centimeter? (Use $\pi=3.14$ .)

$1\,510\text{ cm}^2$
$477\text{ cm}^2$ (correct answer)
$380\text{ cm}^2$
$301\text{ cm}^2$

Explanation: When you encounter a surface area problem for a cylinder, you need to visualize what surfaces make up the complete shape: two circular bases (top and bottom) plus the curved side that wraps around. The surface area formula for a cylinder is

SA = 2\pi r^2 + 2\pi rh

, where the first term represents the two circular bases and the second term represents the curved lateral surface. With

r = 4

cm and

h = 15

cm:

Area of both bases: $2\pi r^2 = 2(3.14)(4^2) = 2(3.14)(16) = 100.48$ cm²
Lateral surface area: $2\pi rh = 2(3.14)(4)(15) = 376.8$ cm²
Total surface area: $100.48 + 376.8 = 477.28$ cm²

Rounded to the nearest square centimeter, this gives us

477

cm², which is answer choice B. Looking at the wrong answers: A)

1{,}510

cm² is far too large and likely comes from a major calculation error or using the wrong formula entirely. C)

380

cm² is close to just the lateral surface area (

376.8

), suggesting someone forgot to include the circular bases. D)

301

cm² is too small and might result from various computational mistakes or only calculating one base plus lateral area. Remember that cylinder surface area problems always involve three components: top circle, bottom circle, and the rectangular "wrapper" that forms the curved side. Missing any of these parts will lead you to an incorrect answer choice.

Question 14

A cone-shaped paper cup has a radius of $5\text{ cm}$ and a height of $9\text{ cm}$ . How many cubic centimeters of water can the cup hold when filled to the brim? (Use $\pi=3.14$ .)

$235\text{ cm}^3$ (correct answer)
$471\text{ cm}^3$
$706\text{ cm}^3$
$848\text{ cm}^3$

Explanation: When you encounter volume problems involving three-dimensional shapes, you need to identify the shape and apply the correct formula. This cone problem requires the volume formula for a cone:

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

. Let's substitute the given values: radius = 5 cm, height = 9 cm, and

\pi = 3.14

V = \frac{1}{3} \times 3.14 \times 5^2 \times 9

V = \frac{1}{3} \times 3.14 \times 25 \times 9

V = \frac{1}{3} \times 706.5

V = 235.5 \text{ cm}^3

Rounding to the nearest whole number gives us

235 \text{ cm}^3

, which is answer A. Now let's examine why the other answers are wrong. Answer B (

471 \text{ cm}^3

) represents a common error where students forget to divide by 3 in the cone formula, essentially calculating

\frac{2}{3}\pi r^2 h

instead. Answer C (

706 \text{ cm}^3

) occurs when students completely omit the

\frac{1}{3}

factor and use the cylinder formula

\pi r^2 h

instead. Answer D (

848 \text{ cm}^3

) suggests using an incorrect formula altogether, possibly confusing cone volume with surface area calculations. The key strategy for cone volume problems is remembering that a cone's volume is exactly one-third of a cylinder with the same base and height. Always double-check that you've included the

\frac{1}{3}

factor in your calculation, as this is the most frequent mistake students make on these problems.

Question 15

The ice‐cream scoop forms a perfect sphere with a diameter of $6\text{ cm}$ . What is the volume of one scoop? (Use $\pi=3.14$ .)

$113\text{ cm}^3$ (correct answer)
$226\text{ cm}^3$
$452\text{ cm}^3$
$904\text{ cm}^3$

Explanation: When you encounter a sphere volume problem, you're working with three-dimensional geometry. The key is identifying what information you have and applying the correct volume formula. For any sphere, the volume formula is

V = \frac{4}{3}\pi r^3

, where

r

is the radius. Since the problem gives you a diameter of 6 cm, you first need to find the radius:

r = \frac{6}{2} = 3

cm. Now substitute into the formula:

V = \frac{4}{3} \times 3.14 \times 3^3 = \frac{4}{3} \times 3.14 \times 27

. Calculate step by step:

3.14 \times 27 = 84.78

, then

\frac{4}{3} \times 84.78 = \frac{339.12}{3} = 113.04

cm³. Rounding gives us 113 cm³. Looking at the wrong answers: Answer B (226 cm³) appears to result from forgetting the

\frac{4}{3}

coefficient and just calculating

\pi r^3

, then doubling it. Answer C (452 cm³) likely comes from using the diameter instead of radius in the formula, calculating

\frac{4}{3}\pi \times 6^3

but making arithmetic errors. Answer D (904 cm³) represents using the full diameter cubed without the proper fractional coefficient. Study tip: Always write down the sphere volume formula first, then carefully identify whether you're given radius or diameter. Most errors on sphere problems come from confusing these two measurements or forgetting the

\frac{4}{3}

coefficient that makes spheres different from cylinders.