- A cube has a volume of 6 cubic centimeters. If all side lengths are doubled,what is the volume of the new cube?
- A rectangular prism has a volume of 32 cubic feet. What is the new volume if the length and width are doubled?
- The dimensions of a large cube are three times the dimensions of a small cube. What could be the volumes of the two cubes?
- The volume of a cone is 4π square units. If the radius is doubled and the height is tripled, what is the new volume?
- Explain the relationship between the volume of a cylinder and the volume of a cone with the same base & height?
- A cylinder has a volume of 81π cubic cm. What is the volume of a cone with the same base and height as this cylinder?
- Find the exact volume of a cone with radius 2cm. and height 15 cm.
- A cylinder has a volume of 100π and a radius of 2. What is the height of the cylinder?
- If the radius of a cylinder is doubled, what happens to the volume of the cylinder?
- The volume of a rectangular prism is 640 cubic cm. What is the volume of a similar rectangular prism that has edge lengths one fourth as long as the edge lengths of the original prism?
- A cone with a radius of 3 cm has a volume of 300π cubic cm. What is the height of the cone?
- A cone and a sphere both have a radius of 9 cm. What must the height of the cone be in order for the cone and sphere to have the same volume?

Wow, I'm guessing this is the entire worksheet? I'm not going to fill out the actual answers but want you to see how the questions are framed by multiplying the dimensions you are already given by a number to simply increase (or decrease) the size of the shape.

To start remember your formulas, I see cones, cylinders, spheres, cubes, and rectangular prisms.

So for our volumes:

V_{cube} = length*length*length = l^{3}

V_{rprism} = length *width*height = l*w*h

V_{sphere} = (4/3)πr^{3}

V_{cylinder} = πr^{2}h

V_{cone} = (1/3)πr^{2}h

To begin:

1. l^{3} = 6 cm^{3} if you doubled the lengths you would have (2l)*(2l)*(2l) = 2^{3}l^{3} = 8l^{3}

2. l*w*h = 32 ft^{3} if you doubled the length and width you'd get (2l)*(2w)*h = 4(l*w*h)

3. Make it easy on yourself and use the l =1 for the small cube so volume is l^{3} = 1*1*1 = 1 units^{3}

if the large one has sides of 3l then its (3l)*(3l)*(3l) = 27l^{3}

4. There's something wrong with the way you typed the question here because all volumes should have cubic units not square units. Area has square units, but assuming its a typo and you meant volume:

V = (1/3)πr^{2}h = 4π units^{3} new volume would simply be (1/3)π(2r)^{2}(3h) = 12[(1/3)πr^{2}h] units^{3}

5. The base being the same implies that the radii are equal as well. This means that the two dimensional circle that they are derived from is exactly the same (notice how they both have the formula for the area of a circle in them --> πr^{2} ). The difference is that the cone's radius diminishes to zero as the height increases (it comes to a point) where the radius of the cylinder is constant the entirety of the height.

6. This relates the two formulas together, you know that: (1/3)V_{cyl} = V_{cone} just from looking at them.

7. Again, plug in the values for r and h into (1/3)πr^{2}h to get V = (1/3)π(2)^{2}(15)

8. V_{cyl} = 100π = π(2)^{2}h, just solve for h here.

9. V_{cyl} = πr^{2}h if r is doubled it would yield V = π(2r)^{2}h very similar to #4 above.

10. Another dimensional change but this time a reduction to each dimension. If you have l*w*h = 640 cm^{3} then (1/4)l*(1/4)w*(1/4)h would be the way to your answer.

11. V = 300π = (1/3)π(3)^{2}h and solve for h just like #8 but using the formula for a cone instead of a cylinder.

12. Here the volumes have to be the same so use both formulas and equate them.

(4/3)π(9)^{3} = (1/3)π(9)^{2}h now just solve for h like #s 8 and 11

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