All SAT Math Resources
Example Questions
Example Question #31 : How To Find The Volume Of A Cylinder
Jessica wishes to fill up a cylinder with water at a rate of gallons per minute. The volume of the cylinder is gallons. The hole at the bottom of the cylinder leaks out gallons per minute. If there are gallons in the cylinder when Jessica starts filling it, how long does it take to fill?
Jessica needs to fill up gallons at the effective rate of . divided by is equal to . Notice how the units work out.
Example Question #161 : Solid Geometry
The figure below represents a cylinder with a smaller cylinder removed from its middle.
Find the volume of the figure.
In order to find the volume of the figure, we will first need to find the volume of both cylinders.
Recall how to find the volume of the cylinder:
Now, use the given radius and height to find the volume of the larger cylinder.
Next, use the given radius and height to find the volume of the smaller cylinder.
Subtract the volume of the smaller cylinder from the volume of the larger one to find the volume of the figure.
Make sure to round to places after the decimal.
Example Question #1 : How To Find The Surface Area Of A Cylinder
An upright cylinder with a height of 30 and a radius of 5 is in a big tub being filled with oil. If only the top 10% of the cylinder is visible, what is the surface area of the submerged cylinder?
325π
295π
270π
300π
345π
295π
The height of the submerged part of the cylinder is 27cm. 2πrh + πr2 is equal to 270π + 25π = 295π
Example Question #1 : Cylinders
A right circular cylinder has a height of 41 in. and a lateral area (excluding top and bottom) 512.5π in2. What is the area of its bases?
None of the other answers
312.5 in2
78.125π in2
156.25 in2
39.0625π in2
78.125π in2
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.
Example Question #162 : Solid Geometry
The diameter of the lid of a right cylindrical soup can is 5 in. If the can is 12 inches tall and the label costs $0.00125 per square inch to print, what is the cost to produce a label for a can? (Round to the nearest cent.)
$0.29
$0.08
$1.18
$0.24
$0.16
$0.24
The general mechanics of this problem are simple. The lateral area of a right cylinder (excluding its top and bottom) is equal to the circumference of the top times the height of the cylinder. Therefore, the area of this can's surface is: 5π * 12 or 60π. If the cost per square inch is $0.00125, a single label will cost 0.00125 * 60π or $0.075π or approximately $0.24.
Example Question #163 : Solid Geometry
Aluminum is sold to a soup manufacturer at a rate of $0.0015 per square inch. The cans are made so that the ends perfectly fit on the cylindrical body of the can. It costs $0.00125 to attach the ends to the can. The outer label (not covering the top / bottom) costs $0.0001 per in2 to print and stick to the can. The label must be 2 inches longer than circumference of the can. Ignoring any potential waste, what is the manufacturing cost (to the nearest cent) for a can with a radius of 5 inches and a height of 12 inches?
$0.45
$0.57
$0.77
$0.91
$0.84
$0.84
We have the following categories to consider:
<Aluminum Cost> = (<Area of the top and bottom of the can> + <Lateral area of the can>) * 0.0015
<Label Cost> = (<Area of Label>) * 0.0001
<Attachment cost> = 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.
Example Question #3 : How To Find The Surface Area Of A Cylinder
The number of square units in the surface area of a right circular cylinder is equal to the number of cubic units in its volume. If r and h represent the length in units of the cylinder's radius and height, respectively, which of the following is equivalent to r in terms of h?
r = 2h/(h – 2)
r = h/(2h – 2)
r = 2h2 + 2
r = h2/(h + 2)
r = h2 + 2h
r = 2h/(h – 2)
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 πr2, and their combined surface area will be 2πr2.
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 2πr, and the height is h, so the lateral area is 2πrh.
surface area of cylinder = 2πr2 + 2πrh
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 πr2, and the height is h, so the volume of the cylinder is πr2h.
volume = πr2h
Then, we must set the volume and surface area expressions equal to one another and solve for r in terms of h.
2πr2 + 2πrh = πr2h
First, let's factor out 2πr from the left side.
2πr(r + h) = πr2h
We can divide both sides by π.
2r(r + h) = r2h
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.
2r + 2h = rh
Subtract 2r from both sides to get all the r's on one side.
2h = rh – 2r
rh – 2r = 2h
Factor out an r from the left side.
r(h – 2) = 2h
Divide both sides by h – 2
r = 2h/(h – 2)
The answer is r = 2h/(h – 2).
Example Question #4 : How To Find The Surface Area Of A Cylinder
What is the surface area of a cylinder with a radius of and a height of ?
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 =
Example Question #41 : 3 Dimensional Geometry
What is the surface area of a cone with a radius of 4 and a height of 3?
Here we simply need to remember the formula for the surface area of a cone and plug in our values for the radius and height.
Example Question #211 : Geometry
The lateral area is twice as big as the base area of a cone. If the height of the cone is 9, what is the entire surface area (base area plus lateral area)?
27π
9π
81π
54π
90π
81π
Lateral Area = LA = π(r)(l) where r = radius of the base and l = slant height
LA = 2B
π(r)(l) = 2π(r2)
rl = 2r2
l = 2r
From the diagram, we can see that r2 + h2 = l2. Since h = 9 and l = 2r, some substitution yields
r2 + 92 = (2r)2
r2 + 81 = 4r2
81 = 3r2
27 = r2
B = π(r2) = 27π
LA = 2B = 2(27π) = 54π
SA = B + LA = 81π