Cylinders - SAT Math

The height of the submerged part of the cylinder is 27cm. 2πrh + πr2 is equal to 270π + 25π = 295π

The lateral area (not including its bases) is equal to the circumference of the base times the height of the cylinder. Think of it like a label that is wrapped around a soup can. Therefore, we can write this area as:

A = h * π * d or A = h * π * 2r = 2πrh

Now, substituting in our values, we get:

512.5π = 2 * 41*rπ; 512.5π = 82rπ

Solve for r by dividing both sides by 82π:

6.25 = r

From here, we can calculate the area of a base:

A = 6.252π = 39.0625π

NOTE: The question asks for the area of the bases. Therefore, the answer is 2 * 39.0625π or 78.125π in2.

We have the following categories to consider:

= ( + ) * 0.0015

= () * 0.0001

= 2 * 0.00125 = $0.0025

The area of ends of the can are each equal to π*52 or 25π. For two ends, that is 50π.

The lateral area of the can is equal to the circumference of the top times the height, or 2 * π * r * h = 2 * 5 * 12 * π = 120π.

Therefore, the total surface area of the aluminum can is 120π + 50π = 170π. The cost is 170π * 0.0015 = 0.255π, or approximately $0.80.

The area of the label is NOT the same as the lateral area of the can. (Recall that it must be 2 inches longer than the circumference of the can.) Therefore, the area of the label is (2 + 2 * π * 5) * 12 = (2 + 10π) * 12 = 24 + 120π. Multiply this by 0.0001 to get 0.0024 + 0.012π = (approximately) $0.04.

Therefore, the total cost is approximately 0.80 + 0.04 + 0.0025 = $0.8425, or $0.84.

We need to find expressions for the surface area and the volume of a cylinder. The surface area of the cylinder consists of the sum of the surface areas of the two bases plus the lateral surface area.

surface area of cylinder = surface area of bases + lateral surface area

The bases of the cylinder will be two circles with radius r. Thus, the area of each will be _πr_2, and their combined surface area will be 2_πr_2.

The lateral surface area of the cylinder is equal to the circumference of the circular base multiplied by the height. The circumferece of a circle is 2_πr_, and the height is h, so the lateral area is 2_πrh_.

surface area of cylinder = 2_πr_2 + 2_πrh_

Next, we need to find an expression for the volume. The volume of a cylinder is equal to the product of the height and the area of one of the bases. The area of the base is _πr_2, and the height is h, so the volume of the cylinder is πr_2_h.

volume = πr_2_h

Then, we must set the volume and surface area expressions equal to one another and solve for r in terms of h.

2_πr_2 + 2_πrh_ = πr_2_h

First, let's factor out 2_πr_ from the left side.

2_πr_(r + h) = πr_2_h

We can divide both sides by π.

2_r_(r + h) = r_2_h

We can also divide both sides by r, because the radius cannot equal zero.

2(r + h) = rh

Let's now distribute the 2 on the left side.

2_r_ + 2_h_ = rh

Subtract 2_r_ from both sides to get all the r's on one side.

2_h_ = rh – 2_r_

r_h – 2_r = 2_h_

Factor out an r from the left side.

r(h – 2) = 2_h_

Divide both sides by h – 2

r = 2_h_/(h – 2)

The answer is r = 2_h_/(h – 2).

When you're calculating the surface area of a cylinder, note that the cylinder will have two circles, one for the top and one for the bottom, and one rectangle that wraps around the "side" of the cylinder (it's helpful to picture peeling the label off a can of soup - it's curved when it's on the can, but really it's a rectangle that has been wrapped around). You know the area of the circle formula; for the rectangle, note that the height is given to you but the width of the rectangle is one you have to intuit: it's the circumference of the circle, because the entire distance around the circle from one point around and back again is the horizontal distance that the area must cover.

Therefore the surface area of a cylinder =

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

The equation for the volume of a cylinder is V = Ah, where A is the area of the base and h is the height.

Thus, the volume can also be expressed as V = πr2h.

The diameter is 13 inches, so the radius is 13/2 = 6.5 inches.

Now we can easily calculate the volume:

V = 6.52π * 27.5 = 1161.88π in3

Volume of cylinder = π * (base radius)2 x height = π * (base diameter / 2 )2 x height

Volume Cylinder 1 = π * (3 / 2 )2 x 9 = π * (1.5 )2 x 9 = π * 20.25

Volume Cylinder 2 = π * (4 / 2 )2 x 9 = π * (2 )2 x 9 = π * 36

Total Volume = π * 20.25 + π * 36

Volume of Cylinder 3 = π * (6 / 2 )2 x H = π * (3 )2 x H = π * 9 x H

Set Total Volume equal to the Volume of Cylinder 3 and solve for H

π * 20.25 + π * 36 = π * 9 x H

20.25 + 36 = 9 x H

H = (20.25 + 36) / 9 = 6.25”

LA = 2π(r)(h) = 2π(3)(5) = 30π

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.

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

The answer is 81π.

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.

To find the final volume, we will need to subtract the volume of the hole from the total initial volume of the cylinder.

The volume of a cylinder is given by the product of the base area times the height: .

Find the initial volume using the given base area and height.

Next, find the volume of the hole that was drilled. The base area of this cylinder can be calculated from the radius of the hole. Remember that the height of the hole is only half the height of the block.

Finally, subtract the volume of the hole from the total initial volume.

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

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 * 2.52 * 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.

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 * 3.752 * 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.

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 πr_2. Note that the prompt has given the diameter. Therefore, the radius is 2, not 4. The base's area is: 22_π = 4_π_. The total volume is therefore: 4_π_ * 5 = 20_π_ in3.

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

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 πr_2. Note that the prompt has given the diameter. Therefore, the radius is 4.5, not 9. The base's area is: 4.52_π = 20.25_π_. The total volume is therefore: 20.25_π_ * 10.5 = 212.625_π_ in3.

The gel volume is therefore: 6552 – 212.625_π_ 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.

Jessica needs to fill up gallons at the effective rate of . divided by is equal to . Notice how the units work out.