Grade 8

Grade 8

Geometry 8.G.C.9

9. Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems.

If your students start to find these geometry topics a bit two-dimensional—well, they might be onto something. Most of these geometry concepts are in two dimensions. It's simpler, clearer—but, alas!—boring-er. You know what'll really get their adrenaline pumping? Let's go 3D.

Yeah, we're talking about volume.

Students should understand that volume is a measure of three-dimensional space. Like area, but with an extra dimension added in. While you could always find the volume by counting how many little cubes you can fit into a figure, there's an easier way. Plus, those little cubes get to be a drag when you have to carry them around everywhere.

Instead, your students can make use of the volume formulas. They should already know how to calculate the volumes of simpler three-dimensional figures, like prisms and pyramids. Might be time to round off the corners and get to know cones, cylinders, and spheres.

Students should know not only the volume formulas of cylinders, cones, and spheres (V = πr2h, V = ⅓πr2h, and V = 4⁄3πr3, where r is the radius and h is the height), but also have a basic understanding of where they come from. It might help to compare the volume formulas of prisms and cylinders, looking for similarities and differences. (Spoiler alert: they're really the same formula!)

Once students have a solid understanding of how to use the formulas and which dimension to plug in where, they can work on applying these formulas to real-life scenarios where the dimensions aren't as explicitly stated. Also, combine these formulas with other geometric concepts and formulas that the students should already know.

Don't forget to tell your students about the importance of units and how to convert between them! Volume is always in units cubed because we're dealing with three dimensions—so the conversions are also cubed, too. It might seem like a minor thing, but it's likely to trip them up from time to time, especially if they're too caught up in playing the "Which Formula Do I Use?" game (which is rarely as fun as it sounds).

 

Drills

  1. What is the volume of a sphere that has a radius of 6 cm?

    Correct Answer:

    288π cm3

    Answer Explanation:

    To find the volume of the sphere, we just have to use our handy dandy formula: V = 4⁄3πr3. All we have to do is plug in 6 cm for r, and that gives us 288π cm3 as our answer. Squaring the radius rather than cubing it would give (A), using three-fourths instead of four-thirds would give (B), and (D) results from taking the radius to the fourth power and multiplying by ¾.


  2. What is the volume of a cylinder with a radius of 4 m and a height of 5 m?

    Correct Answer:

    80π m3

    Answer Explanation:

    The formula for the volume of a cylinder is V = πr2h, where r is the radius of the circular base and h is the height of the cylinder. Plugging in r = 4 and h = 5, we have V = π(4 m)2(5 m) = (16 m2)(5 m)π = 80π m3. If you got (A), you probably accidentally replaced 42 with 8 instead of 16, (C) may have resulted from switching the height and the radius, and (D) comes from squaring both the height and the radius.


  3. What is the volume of a cone with radius 9 in and height 3 in?

    Correct Answer:

    81π in3

    Answer Explanation:

    The volume formula for a cone is given by , or one-third the volume of a cylinder with the same dimensions. Plugging in all the values in the right places should give a volume of . Forgetting to square the radius would give (A), while (B) comes from switching the height and the radius. Also, remember not to multiply anything by  before squaring! Lots of multiples of 3 out there, and while Three's Company, it's not always welcome.


  4. If a sphere has a volume 2304π m3, what is the length of the radius?

    Correct Answer:

    12 m

    Answer Explanation:

    Time to apply the volume formula for spheres: . This time, we know the volume and not the radius, so after rearranging the formula to solve for the radius, we an substitute V with 2304π m3. That should give a radius of 12 m. If you're really gung-ho about it, you can double-check it by plugging 12 m back in for r and getting . Yep, that's our answer. Make sure not to add the π into the length of the radius!


  5. If a cylinder has a volume of 320π km3 and a height of 5 km, what is the length of its radius?

    Correct Answer:

    8 km

    Answer Explanation:

    Since this is a cylinder we're dealing with, we need to be sure we're using the right formula: V = πr2h. We're looking for the radius, so some rearranging might be useful before plugging in the values for V and h to solve for r. If we do all our math correctly, we should get (B) as our answer. We can check it, too: V = πr2h = π(8 km)2(5 km) = π(64 km2)(5 km) = 320π km3. Bingo.


  6. What is the height of a cone with a radius of 6 feet and a volume of 60π cubic feet?

    Correct Answer:

    4 feet

    Answer Explanation:

    First off, it's best to take a gander at the volume formula we need: . We know that r = 6 feet and V = 60π cubic feet, so all we need to do is solve for h. Rearranging the formula and substituting in those numbers should give us h = 5 feet. Remember your order of operations (simplify exponents before multiplying!) and make sure to plug the numbers in for the right variables, and you should be just fine.


  7. Which of the following has the greatest volume?

    Correct Answer:

    A sphere with a radius of 4 units

    Answer Explanation:

    As long as you remember which formula applies to which solid, you should be fine. In fact, we don't even have to calculate the volumes all the way; we can just look at the differences in the factors. In (A), our formula turns out to be . For (B), it's  = π(4 units2)(4 units), and (C) has a volume of V = π(4 units)2(4 units) = π(4 units)3. We have 4 to the third power in (A) and (C), but (A) has an additional four-thirds multiplied into it, making it a bit larger than (C) and definitely more than (B). You want real numbers? Fine. Compare them yourself: 85.3π, 64π, and 16π.


  8. A basketball has a circumference of 29.5 inches. Using 3.14 as an approximation for π, what is the basketball's volume to the nearest cubic inch?

    Correct Answer:

    434 in3

    Answer Explanation:

    It helps to know that the circumference is given by C = 2πr, where r is the radius, and that the basketball is essentially a sphere with volume . Performing all the calculations correctly gives us a radius of about 4.7 inches and a volume of about 434 in3. Dropping the 2 in 2πr, forgetting to multiply by π in the volume formula, or switching four-thirds with three-fourths would yield the other wrong answers.


  9. Kenny's Canning Company sells cans in a variety of sizes. With the economic downturn, it plans to discontinue a can size that has a radius of 1 foot and a height of 6 inches. Which of the following will have the same volume as one of the discontinued cans?

    Correct Answer:

    Two cans, each with a radius of 4 and a height of 27 inches

    Answer Explanation:

    This problem is all about the volume formula for cylinders, V = πr2h. The discontinued can has a volume of V = π(12 in)2(6 in) = 864π in3. If we calculate the volume of the individual cans in our answer options and multiply them by the number of cans, we should end up with 864π in3, 600π in3, 720π in3, and 288π in3, respectively. The only volume that matches is that in (A). Make sure to multiply volumes by the right number of cans!


  10. A giant cylindrical container of jawbreakers has a radius of 1 foot and a height of 3 feet. If the diameter of a single jawbreaker is 2 inches, how many jawbreakers fit into the container?

    Correct Answer:

    3888 jawbreakers

    Answer Explanation:

    Before we do anything else, we should really convert all units into inches. The container has a 12-inch radius and is 36 inches tall. Since it's a cylinder, we can apply the volume formula for cylinders: V = πr2h = π(12 in)2(36 in) = 5184π in3. Then, we can calculate the volume of a single jawbreaker and see how many of them fit inside the volume of the container. Jawbreakers are spheres with a diameter of 2 inches (meaning a radius of 1 inch), so  for a single jawbreaker. Dividing the total volume of the container (5184π in3) by the volume of a single jawbreaker (1.33π in3) will give us the maximum number of jawbreakers that will fit into the container, and 5184π ÷ 1.33π = 3888 jawbreakers.


More standards from Grade 8 - Geometry

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