All HSPT Math Resources
Example Questions
Example Question #2121 : Hspt Mathematics
Find the surface area of a cube with a side length of .
Write the formula for the surface area of a cube.
Substitute the length.
Example Question #2122 : Hspt Mathematics
What is the surface area of a cube with a side length of three?
Write the formula for the surface area of a cube.
Substitute the side length into the equation.
Simplify the square inside the parentheses and multiply.
Example Question #2123 : Hspt Mathematics
Find the surface area of a cube with side length .
To solve, simply use the following formula for the surface area of a cube.
Thus,
Example Question #2124 : Hspt Mathematics
If a cube has an area of on one of its sides, what is the total surface area?
A cube has sides that have equal length edges and also equal side areas.
To find the total surface area, you just need to multiple the side area () by which is,
.
Example Question #26 : How To Find Surface Area
A sphere has diameter 12. What is 75% of its surface area?
The radius of a sphere is half its diameter, which here is 12, so the radius is 6. The surface area of the sphere can be calculated by setting in the formula:
75% of this is
Example Question #27 : How To Find Surface Area
Which expression is equal to 40% of the surface area of the above cylinder?
The surface area of the cylinder can be calculated by setting and in the formula
40% of this surface area is
Example Question #231 : Geometry
According to regulations, the maximum radius of a (spherical) bowling ball is 7.11 centimeters. What does that make the maximum surface area, to the nearest square centimeter?
(Note: disregard the holes)
Set and use the formula for the surface area of a sphere:
Example Question #1 : Cylinders
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?
345π
295π
270π
300π
325π
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
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.)
$0.16
$0.24
$1.18
$0.08
$0.29
$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
$0.77
$0.45
$0.57
$0.91
$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.