All PSAT Math Resources
Example Questions
Example Question #1 : How To Find The Surface Area Of A Prism
What is the total surface area of a box without a lid, if the dimensions of the base of the box are , and the box is tall?
The surface area of a rectangular prism is . Our prism, however, is missing its top, so it will be . This gives us .
Example Question #1 : Pyramids
The volume of a 6-foot-tall square pyramid is 8 cubic feet. How long are the sides of the base?
Volume of a pyramid is
Thus:
Area of the base is .
Therefore, each side is .
Example Question #2 : Pyramids
A right pyramid with a square base has a height that is twice the length of one edge of the base. If the height of the pyramid is 6 meters, find the volume of the pyramid.
12
6
18
30
24
18
If the height, which is twice the length of the base edges, measures 6 meters, then each base edge must measure 3 meters.
Since the base is a square, the area of the base is 3 x 3 = 9.
Therefore the volume of the right pyramid is V = (1/3) x area of the base x height = 1/3(9)(6) = 18.
Example Question #1 : How To Find The Volume Of A Pyramid
What is the approximate volume of a square pyramid with one edge of the base measuring and a height equal to the diagonal of its base?
If one edge of the base is 2.7 inches, and the height of the pyramid is equal to the diagonal of the base, we can find the height using the edge. The height is going to be equal to . The area of the base is . Volume of a pyramid is equal to
This leaves us with
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?
300π
325π
295π
270π
345π
295π
The height of the submerged part of the cylinder is 27cm. 2πrh + πr2 is equal to 270π + 25π = 295π
Example Question #1 : How To Find The Surface Area Of A Cylinder
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?
39.0625π in2
156.25 in2
None of the other answers
78.125π in2
312.5 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 #1 : How To Find The Surface Area Of A Cylinder
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
$0.16
$0.24
$0.08
$1.18
$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 #2 : How To Find The Surface Area Of A Cylinder
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?
$0.84
$0.45
$0.77
$0.91
$0.57
$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 #1 : How To Find Surface Area
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 = h2/(h + 2)
r = 2h2 + 2
r = 2h/(h – 2)
r = h/(2h – 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 #1 : How To Find Surface Area
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 =
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