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Can anyone help me with this Measuring Area & Volume study guide?

  1. A cube has a volume of 6 cubic centimeters. If all side lengths are doubled,what is the volume of the new cube?
  2. A rectangular prism has a volume of 32 cubic feet. What is the new volume if the length and width are doubled?
  3. The dimensions of a large cube are three times the dimensions of a small cube. What could be the volumes of the two cubes?
  4. The volume of a cone is 4π square units. If the radius is doubled and the height is tripled, what is the new volume?
  5. Explain the relationship between the volume of a cylinder and the volume of a cone with the same base & height?
  6. 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?
  7. Find the exact volume of a cone with radius 2cm. and height 15 cm.
  8. A cylinder has a volume of 100π and a radius of 2. What is the height of the cylinder?
  9. If the radius of a cylinder is doubled, what happens to the volume of the cylinder?
  10. 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?
  11. A cone with a radius of 3 cm has a volume of 300π cubic cm. What is the height of the cone?
  12. 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?
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2 Answers

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:

Vcube = length*length*length = l3

Vrprism = length *width*height = l*w*h

Vsphere = (4/3)πr3

Vcylinder = πr2h

Vcone = (1/3)πr2h

To begin:

1. l3 = 6 cm3 if you doubled the lengths you would have (2l)*(2l)*(2l) = 23l3 = 8l3

2.  l*w*h = 32 ft3 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 l3 = 1*1*1 = 1 units3

if the large one has sides of 3l then its (3l)*(3l)*(3l) = 27l3

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)πr2h = 4π units3  new volume would simply be (1/3)π(2r)2(3h) = 12[(1/3)πr2h] units3

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 --> πr2 ). 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)Vcyl = Vcone just from looking at them.

7. Again, plug in the values for r and h into (1/3)πr2h to get V = (1/3)π(2)2(15)

8. Vcyl = 100π = π(2)2h, just solve for h here.

9. Vcyl = πr2h if r is doubled it would yield V = π(2r)2h very similar to #4 above.

10. Another dimensional change but this time a reduction to each dimension. If you have l*w*h = 640 cm3 then (1/4)l*(1/4)w*(1/4)h would be the way to your answer.

11. V = 300π = (1/3)π(3)2h 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)2h now just solve for h like #s 8 and 11

 

 

Courtney - 

Some of these problems have a similar approach, so I'll start you on some and leave others for you

1 - For a cube, we know all sides are equal, so if we let s = length of a side, then the volume V=s3. If we double each side, then the new cube would have sides of length 2s, and we cube this to get the volume:

 V2 = (2s)3 = 8s3

So the volume of the new cube would be 8 times the volume of the original cube. 

2 - Similar approach as #1. We know the volume of a rectangular prism is V = l*w*h. So now, if the length and width are doubled, the volume
V2 = (2l)*(2w)*h = 4lwh

So the volume of the new prism is 4 times the volume of the original. 

3 - Similar to #1, left for the student. 

4 - Volume of a Cone V = 1/3 ∏ r2h. If the radius is doubled and the height tripled, we plug these new values into our formula:
V2 = 1/3 ∏ (2r)2(3h) = 1/3 ∏ 4r2*3h. 

Rearranging our factors by pulling the 4 and 3 out front, we get 
V2 = (4*3)(1/3 ∏ r2h), =  12 (1/3 ∏ r2h)

So the new volume is 12 times the original volume. Since they ask for the new volume, and the original volume is 4∏, I will leave that final calculation to the student. 

5 - Volume of a cone is Vcone=1/3 ∏ r2h.   Vcylinder =  ∏ r2h. So for a cone with the same base and height of a cylinder, the cone's Volume is 1/3 the Cylinder's volume. 

6 - Use the relationship in #5 to solve this one. 

7 - Use the formula for volume of a cone for this one. 

8 - Vcylinder = ∏ r2h . Plugging numbers in, we get 100∏ = ∏ (2)2h. I'll leave the calculations to the student. 

9 - This is similar to #2. Vcylinder = ∏ r2h . If we double the radius, our new volume is 

Vcylinder = ∏ (2r)2h . The rest is left to the student. 

10 - Similar to #2, only instead of multiplying 2 sides by 2, we multiply 3 sides by 1/4:
V = (1/4)l * (1/4)w * (1/4)h. 

11 - Straight application of the formula for volume of a cone. Plug in the numbers where appropriate and solve for h.

12 - Volume of a Sphere is Vsphere = 4/3 ∏r3 . Volume of a cone is Vcone=1/3 ∏ r2h.

If we want the two volumes to be equal, let's set them to each other and solve for h:

4/3 ∏r3 = 1/3 ∏ r2h  

Multiply each side by 3 to remove denominators:  
4 ∏r3 =  ∏ r2h

Divide by pi to remove it:
4 r3 =  r2h

Now divide both sides by r2:
4r = h

So the height of the cone must be 4 times the radius for the volume of the cone and cylinder to be equal. 

 

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