# PSAT Math : How to find the volume of a cylinder

## Example Questions

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### Example Question #1 : How To Find The Volume Of A Cylinder

The volume of a cylinder is 36π. If the cylinder’s height is 4, what is the cylinder’s diameter?

6

3

12

9

4

6

Explanation:

Volume of a cylinder? V = πr2h. Rewritten as a diameter equation, this is:

V = π(d/2)2h = πd2h/4

Sub in h and V: 36p = πd2(4)/4 so 36p = πd2

Thus d = 6

### Example Question #1 : How To Find The Volume Of A Cylinder

A cylinder has a height of 5 inches and a radius of 3 inches.  Find the lateral area of the cylinder.

15π

8π

30π

45π

24π

30π

Explanation:

LA = 2π(r)(h) = 2π(3)(5) = 30π

### Example Question #3 : Cylinders

A cylinder has a volume of 20. If the radius doubles, what is the new volume?

80

100

40

20

60

80

Explanation:

The equation for the volume of the cylinder is πr2h. When the radius doubles (r becomes 2r) you get π(2r)2h = 4πr2h. So when the radius doubles, the volume quadruples, giving a new volume of 80.

### Example Question #4 : Cylinders

A cylinder has a height that is three times as long as its radius. If the lateral surface area of the cylinder is 54π square units, then what is its volume in cubic units?

27π

54π

243π

81π

81π

Explanation:

Let us call r the radius and h the height of the cylinder. We are told that the height is three times the radius, which we can represent as h = 3r.

We are also told that the lateral surface area is equal to 54π. The lateral surface area is the surface area that does not include the bases. The formula for the lateral surface area is equal to the circumference of the cylinder times its height, or 2πrh. We set this equal to 54π,

2πrh = 54π

Now we substitute 3r in for h.

2πr(3r) = 54π

6πr2 = 54π

Divide by 6π

r2 = 9.

Take the square root.

r = 3.

h = 3r = 3(3) = 9.

Now that we have the radius and the height of the cylinder, we can find its volume, which is given by πr2h.

V = πr2h

V = π(3)2(9) = 81π

### Example Question #2 : How To Find The Volume Of A Cylinder

What is the volume of a hollow cylinder whose inner radius is 2 cm and outer radius is 4 cm, with a height of 5 cm?

60π cm3

80π cm3

50π cm3

20π cm3

100π cm3

60π cm3

Explanation:

The volume is found by subtracting the inner cylinder from the outer cylinder as given by V = πrout2 h – πrin2 h. The area of the cylinder using the outer radius is 80π cm3, and resulting hole is given by the volume from the inner radius, 20π cm3. The difference between the two gives the volume of the resulting hollow cylinder, 60π cm3.

### Example Question #6 : Cylinders

What is the volume of a right cylinder with a circumference of 25π in and a height of 41.3 in?

4813.33π in3

6453.125π in3

1032.5π in3

25812.5π in3

3831.34π in3

6453.125π in3

Explanation:

The formula for the volume of a right cylinder is: V = A * h, where A is the area of the base, or πr2.  Therefore, the total formula for the volume of the cylinder is: V = πr2h.

First, we must solve for r by using the formula for a circumference (c = 2πr): 25π = 2πr; r = 12.5.

Based on this, we know that the volume of our cylinder must be: π*12.52*41.3 = 6453.125π in3

### Example Question #1 : Cylinders

An 8-inch cube has a cylinder drilled out of it. The cylinder has a radius of 2.5 inches. To the nearest hundredth, approximately what is the remaining volume of the cube?

203.34 in3

391.33 in3

354.92 in3

157.08 in3

462 in3

354.92 in3

Explanation:

We must calculate our two volumes and subtract them. The volume of the cube is very simple: 8 * 8 * 8, or 512 in3.

The volume of the cylinder is calculated by multiplying the area of its base by its height. The height of the cylinder is 8 inches (the height of the cube through which it is being drilled). Therefore, its volume is πr2h = π * 2.5* 8 = 50π in3

The volume remaining in the cube after the drilling is: 512 – 50π, or approximately 512 – 157.0795 = 354.9205, or 354.92 in3.

### Example Question #2 : Cylinders

An 12-inch cube of wood has a cylinder drilled out of it. The cylinder has a radius of 3.75 inches. If the density of the wood is 4 g/in3, what is the mass of the remaining wood after the cylinder is drilled out?

3813.3 g

4921.4 g

2594.11 g

4791.43 g

3193.33 g

4791.43 g

Explanation:

We must calculate our two volumes and subtract them. Following this, we will multiply by the density.

The volume of the cube is very simple: 12 * 12 * 12, or 1728 in3.

The volume of the cylinder is calculated by multiplying the area of its base by its height. The height of the cylinder is 8 inches (the height of the cube through which it is being drilled). Therefore, its volume is πr2h = π * 3.75* 12 = 168.75π in3.

The volume remaining in the cube after the drilling is: 1728 – 168.75π, or approximately 1728 – 530.1433125 = 1197.8566875 in3. Now, multiply this by 4 to get the mass: (approx.) 4791.43 g.

### Example Question #9 : Cylinders

A hollow prism has a base 5 in x 6 in and a height of 10 in. A closed, cylindrical can is placed in the prism. The remainder of the prism is then filled with gel around the cylinder. The thickness of the can is negligible. Its diameter is 4 in and its height is half that of the prism. What is the approximate volume of gel needed to fill the prism?

203.44 in3

103.33 in3

249.73 in3

187.73 in3

237.17 in3

237.17 in3

Explanation:

The general form of our problem is:

Gel volume = Prism volume – Can volume

The prism volume is simple: 5 * 6 * 10 = 300 in3

The volume of the can is found by multiplying the area of the circular base by the height of the can. The height is half the prism height, or 10/2 = 5 in. The area of the base is equal to πr2. Note that the prompt has given the diameter. Therefore, the radius is 2, not 4. The base's area is: 22π = 4π. The total volume is therefore: 4π * 5 = 20π in3.

The gel volume is therefore: 300 – 20π or (approx.) 237.17 in3.

### Example Question #10 : Cylinders

A hollow prism has a base 12 in x 13 in and a height of 42 in. A closed, cylindrical can is placed in the prism. The remainder of the prism is then filled with gel, surrounding the can. The thickness of the can is negligible. Its diameter is 9 in and its height is one-fourth that of the prism. The can has a mass of 1.5 g per in3, and the gel has a mass of 2.2 g per in3. What is the approximate overall mass of the contents of the prism?

973.44 g

139.44 g

15.22 kg

13.95 kg

11.48 kg

13.95 kg

Explanation:

We must find both the can volume and the gel volume. The formula for the gel volume is:

Gel volume = Prism volume – Can volume

The prism volume is simple: 12 * 13 * 42 = 6552 in3

The volume of the can is found by multiplying the area of the circular base by the height of the can. The height is one-fourth the prism height, or 42/4 = 10.5 in. The area of the base is equal to πr2. Note that the prompt has given the diameter. Therefore, the radius is 4.5, not 9. The base's area is: 4.52π = 20.25π. The total volume is therefore: 20.25π * 10.5 = 212.625π in3.

The gel volume is therefore: 6552 – 212.625π or (approx.) 5884.02 in3.

The approximate volume for the can is: 667.98 in3

From this, we can calculate the approximate mass of the contents:

Gel Mass = Gel Volume * 2.2 = 5884.02 * 2.2 = 12944.844 g

Can Mass = Can Volume * 1.5 = 667.98 * 1.5 = 1001.97 g

The total mass is therefore 12944.844 + 1001.97 = 13946.814 g, or approximately 13.95 kg.

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