How to find surface area - HSPT Math
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A room has dimensions of 18ft by 15ft by 9ft. The last dimension is the height of the room. It has one door that is 3ft by 7ft and two windows, each 2ft by 5ft. There is no trim to the floor, wall, doors, or windows. What is the total exposed wall space?
A room has dimensions of 18ft by 15ft by 9ft. The last dimension is the height of the room. It has one door that is 3ft by 7ft and two windows, each 2ft by 5ft. There is no trim to the floor, wall, doors, or windows. What is the total exposed wall space?
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If broken down into parts, this is an easy problem. It is first necessary to isolate the dimensions of the walls. If the room is 9 ft high, we know 18 x 15 designates the area of the floor and ceiling. Based on this, we know that the room has the following dimensions for the walls: 18 x 9 and 15 x 9. Since there are two of each, we can calculate the total area of walls - ignoring doors and windows - by doubling the sum of these two areas:
2 * (18 * 9 + 15 * 9) = 2 * (162 + 135) = 2 * 297 = 594 ft2
Now, we merely need to calculate the area "taken out" of the walls:
For the door: 3 * 7 = 21 ft2
For the windows: 2 * (2 * 5) = 20 ft2
The total wall space is therefore: 594 – 21 – 20 = 553 ft2
If broken down into parts, this is an easy problem. It is first necessary to isolate the dimensions of the walls. If the room is 9 ft high, we know 18 x 15 designates the area of the floor and ceiling. Based on this, we know that the room has the following dimensions for the walls: 18 x 9 and 15 x 9. Since there are two of each, we can calculate the total area of walls - ignoring doors and windows - by doubling the sum of these two areas:
2 * (18 * 9 + 15 * 9) = 2 * (162 + 135) = 2 * 297 = 594 ft2
Now, we merely need to calculate the area "taken out" of the walls:
For the door: 3 * 7 = 21 ft2
For the windows: 2 * (2 * 5) = 20 ft2
The total wall space is therefore: 594 – 21 – 20 = 553 ft2
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A spherical orange fits snugly inside a small cubical box such that each of the six walls of the box just barely touches the surface of the orange. If the volume of the box is 64 cubic inches, what is the surface area of the orange in square inches?
A spherical orange fits snugly inside a small cubical box such that each of the six walls of the box just barely touches the surface of the orange. If the volume of the box is 64 cubic inches, what is the surface area of the orange in square inches?
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The volume of a cube is found by V = s3. Since V = 64, s = 4. The side of the cube is the same as the diameter of the sphere. Since d = 4, r = 2. The surface area of a sphere is found by SA = 4π(r2) = 4π(22) = 16π.
The volume of a cube is found by V = s3. Since V = 64, s = 4. The side of the cube is the same as the diameter of the sphere. Since d = 4, r = 2. The surface area of a sphere is found by SA = 4π(r2) = 4π(22) = 16π.
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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?
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?
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The height of the submerged part of the cylinder is 27cm. 2πrh + πr2 is equal to 270π + 25π = 295π
The height of the submerged part of the cylinder is 27cm. 2πrh + πr2 is equal to 270π + 25π = 295π
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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.)
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.)
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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.
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.
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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?
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?
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We have the following categories to consider:
= ( + ) * 0.0015
= 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 have the following categories to consider:
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.
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A room has dimensions of 23ft by 17ft by 10ft. The last dimension is the height of the room. It has one door that is 2.5ft by 8ft and one window, 3ft by 6ft. There is no trim to the floor, wall, doors, or windows. If one can of paint covers 57 ft2 of surface area. How many cans of paint must be bought to paint the walls of the room.
A room has dimensions of 23ft by 17ft by 10ft. The last dimension is the height of the room. It has one door that is 2.5ft by 8ft and one window, 3ft by 6ft. There is no trim to the floor, wall, doors, or windows. If one can of paint covers 57 ft2 of surface area. How many cans of paint must be bought to paint the walls of the room.
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If broken down into parts, this is an easy problem. It is first necessary to isolate the dimensions of the walls. If the room is 10ft high, we know 23 x 17 designates the area of the floor and ceiling. Based on this, we know that the room has the following dimensions for the walls: 23 x 10 and 17 x 10. Since there are two of each, we can calculate the total area of walls - ignoring doors and windows - by doubling the sum of these two areas:
2 * (23 * 10 + 17 * 10) = 2 * (230 + 170) = 2 * 400 = 800 ft2
Now, we merely need to calculate the area "taken out" of the walls:
For the door: 2.5 * 8 = 20 ft2
For the windows: 3 * 6 = 18 ft2
The total wall space is therefore: 800 – 20 – 18 = 762 ft2
Now, if one can of paint covers 57 ft2, we calculate the number of cans necessary by dividing the total exposed area by 57: 762/57 = (approx.) 13.37.
Since we cannot buy partial cans, we must purchase 14 cans.
If broken down into parts, this is an easy problem. It is first necessary to isolate the dimensions of the walls. If the room is 10ft high, we know 23 x 17 designates the area of the floor and ceiling. Based on this, we know that the room has the following dimensions for the walls: 23 x 10 and 17 x 10. Since there are two of each, we can calculate the total area of walls - ignoring doors and windows - by doubling the sum of these two areas:
2 * (23 * 10 + 17 * 10) = 2 * (230 + 170) = 2 * 400 = 800 ft2
Now, we merely need to calculate the area "taken out" of the walls:
For the door: 2.5 * 8 = 20 ft2
For the windows: 3 * 6 = 18 ft2
The total wall space is therefore: 800 – 20 – 18 = 762 ft2
Now, if one can of paint covers 57 ft2, we calculate the number of cans necessary by dividing the total exposed area by 57: 762/57 = (approx.) 13.37.
Since we cannot buy partial cans, we must purchase 14 cans.
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A certain cube has a side length of 25 m. How many square tiles, each with an area of 5 m2, are needed to fully cover the surface of the cube?
A certain cube has a side length of 25 m. How many square tiles, each with an area of 5 m2, are needed to fully cover the surface of the cube?
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A cube with a side length of 25m has a surface area of:
25m * 25m * 6 = 3,750 m2
(The surface area of a cube is equal to the area of one face of the cube multiplied by 6 sides. In other words, if the side of a cube is s, then the surface area of the cube is 6_s_2.)
Each square tile has an area of 5 m2.
Therefore, the total number of square tiles needed to fully cover the surface of the cube is:
3,750m2/5m2 = 750
Note: the volume of a cube with side length s is equal to _s_3. Therefore, if asked how many mini-cubes with side length n are needed to fill the original cube, the answer would be:
s3/n3
A cube with a side length of 25m has a surface area of:
25m * 25m * 6 = 3,750 m2
(The surface area of a cube is equal to the area of one face of the cube multiplied by 6 sides. In other words, if the side of a cube is s, then the surface area of the cube is 6_s_2.)
Each square tile has an area of 5 m2.
Therefore, the total number of square tiles needed to fully cover the surface of the cube is:
3,750m2/5m2 = 750
Note: the volume of a cube with side length s is equal to _s_3. Therefore, if asked how many mini-cubes with side length n are needed to fill the original cube, the answer would be:
s3/n3
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A company wants to build a cubical room around a cone so that the cone's height and diameter are 3 inch less than the dimensions of the cube. If the volume of the cone is 486π ft3, what is the surface area of the cube?
A company wants to build a cubical room around a cone so that the cone's height and diameter are 3 inch less than the dimensions of the cube. If the volume of the cone is 486π ft3, what is the surface area of the cube?
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To begin, we need to solve for the dimensions of the cone.
The basic form for the volume of a cone is: V = (1/3)πr_2_h
Using our data, we know that h = 2r because the height of the cone matches its diameter (based on the prompt).
486_π_ = (1/3)πr_2 * 2_r = (2/3)_πr_3
Multiply both sides by (3/2_π_): 729 = _r_3
Take the cube root of both sides: r = 9
Note that this is in feet. The answers are in square inches. Therefore, convert your units to inches: 9 * 12 = 108, then add 3 inches to this: 108 + 3 = 111 inches.
The surface area of the cube is defined by: A = 6 * _s_2, or for our data, A = 6 * 1112 = 73,926 in2
To begin, we need to solve for the dimensions of the cone.
The basic form for the volume of a cone is: V = (1/3)πr_2_h
Using our data, we know that h = 2r because the height of the cone matches its diameter (based on the prompt).
486_π_ = (1/3)πr_2 * 2_r = (2/3)_πr_3
Multiply both sides by (3/2_π_): 729 = _r_3
Take the cube root of both sides: r = 9
Note that this is in feet. The answers are in square inches. Therefore, convert your units to inches: 9 * 12 = 108, then add 3 inches to this: 108 + 3 = 111 inches.
The surface area of the cube is defined by: A = 6 * _s_2, or for our data, A = 6 * 1112 = 73,926 in2
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Angie is painting a 2 foot cube for a play she is in. She needs 
of paint for every square foot she paints. How much paint does she need?
Angie is painting a 2 foot cube for a play she is in. She needs
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First we must calculate the surface area of the cube. We know that there are six surfaces and each surface has the same area:
Now we will determine the amount of paint needed
First we must calculate the surface area of the cube. We know that there are six surfaces and each surface has the same area:
Now we will determine the amount of paint needed
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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)
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)
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Set 
and use the formula for the surface area of a sphere:
Set
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Steve's bedroom measures 20' by 18' by 8' high. He wants to paint the ceiling and all four walls using a paint that gets 360 square feet of coverage per gallon. A one-gallon can of the paint Steve wants costs $36.00; a one-quart can costs $13.00. What is the least amount of money that Steve can expect to spend on paint in order to paint his room?
Steve's bedroom measures 20' by 18' by 8' high. He wants to paint the ceiling and all four walls using a paint that gets 360 square feet of coverage per gallon. A one-gallon can of the paint Steve wants costs $36.00; a one-quart can costs $13.00. What is the least amount of money that Steve can expect to spend on paint in order to paint his room?
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Two of the walls have area 
; two have area 
; the ceiling has area 
.
Therefore, the total area Steve wants to cover is
Divide 968 by 360 to get the number of gallons Steve needs to paint his bedroom:
Since 
, Steve needs to purchase either two gallon cans and three quart cans, or three gallon cans.
The first option will cost him 
; the second option will cost him 
. The latter is the more economical option.
Two of the walls have area
Therefore, the total area Steve wants to cover is
Divide 968 by 360 to get the number of gallons Steve needs to paint his bedroom:
Since
The first option will cost him
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Give the surface area of the above box in square centimeters.
Give the surface area of the above box in square centimeters.
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100 centimeters make one meter, so convert each of the dimensions of the box by multiplying by 100.
centimeters
centimeters
Use the surface area formula, substituting 
:
square centimeters
100 centimeters make one meter, so convert each of the dimensions of the box by multiplying by 100.
Use the surface area formula, substituting
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Note: Figure NOT drawn to scale.
Refer to the above diagram, which shows a square. Give the ratio of the area of the yellow region to that of the white region.
Note: Figure NOT drawn to scale.
Refer to the above diagram, which shows a square. Give the ratio of the area of the yellow region to that of the white region.
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The area of the entire square is the square of the length of a side, or
.
The area of the right triangle is half the product of its legs, or
.
The area of the yellow region is therefore the difference of the two, or
.
The ratio of the area of the yellow region to that of the white region is
; that is, 55 to 9.
The area of the entire square is the square of the length of a side, or
The area of the right triangle is half the product of its legs, or
The area of the yellow region is therefore the difference of the two, or
The ratio of the area of the yellow region to that of the white region is
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The above depicts a rectangular swimming pool for an apartment. The pool is five feet deep everywhere.
An apartment manager wants to paint the four sides and the bottom of the swimming pool. One one-gallon can of the paint he wants to use covers 
square feet. How many cans of the paint will the manager need to buy?
The above depicts a rectangular swimming pool for an apartment. The pool is five feet deep everywhere.
An apartment manager wants to paint the four sides and the bottom of the swimming pool. One one-gallon can of the paint he wants to use covers
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The bottom of the swimming pool has area
square feet.
There are two sides whose area is
square feet,
and two sides whose area is
square feet.
Add the areas:
square feet.
One one-gallon can of paint covers 350 square feet, so divide:
Seven full gallons and part of another are required, so eight is the correct answer.
The bottom of the swimming pool has area
There are two sides whose area is
and two sides whose area is
Add the areas:
One one-gallon can of paint covers 350 square feet, so divide:
Seven full gallons and part of another are required, so eight is the correct answer.
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The surface area of the above cylinder is 80% of what number?
The surface area of the above cylinder is 80% of what number?
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The surface area of the cylinder can be calculated by setting 
and 
in the formula
This is 80% of the number
The surface area of the cylinder can be calculated by setting
This is 80% of the number
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What is the surface area of a cube with a side edge of 7?
What is the surface area of a cube with a side edge of 7?
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First you must find the area of one side of the cube which is side squared.
.
Then you multiply that area by six since that is how many sides a cube has.
Your answer is 
.
First you must find the area of one side of the cube which is side squared.
Then you multiply that area by six since that is how many sides a cube has.
Your answer is
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The volume of a sphere is 2304_π_ in3. What is the surface area of this sphere in square inches?
The volume of a sphere is 2304_π_ in3. What is the surface area of this sphere in square inches?
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To solve this, we must first begin by finding the radius of the sphere. To do this, recall that the volume of a sphere is:
V = (4/3)_πr_3
For our data, we can say:
2304_π_ = (4/3)_πr_3; 2304 = (4/3)_r_3; 6912 = 4_r_3; 1728 = _r_3; 12 * 12 * 12 = _r_3; r = 12
Now, based on this, we can ascertain the surface area using the equation:
A = 4_πr_2
For our data, this is:
A = 4_π_*122 = 576_π_
To solve this, we must first begin by finding the radius of the sphere. To do this, recall that the volume of a sphere is:
V = (4/3)_πr_3
For our data, we can say:
2304_π_ = (4/3)_πr_3; 2304 = (4/3)_r_3; 6912 = 4_r_3; 1728 = _r_3; 12 * 12 * 12 = _r_3; r = 12
Now, based on this, we can ascertain the surface area using the equation:
A = 4_πr_2
For our data, this is:
A = 4_π_*122 = 576_π_
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A sphere has its center at the origin. A point on its surface is found on the x-y axis at (6,8). In square units, what is the surface area of this sphere?
A sphere has its center at the origin. A point on its surface is found on the x-y axis at (6,8). In square units, what is the surface area of this sphere?
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To find the surface area, we must first find the radius. Based on our description, this passes from (0,0) to (6,8). This can be found using the distance formula:
62 + 82 = _r_2; _r_2 = 36 + 64; _r_2 = 100; r = 10
It should be noted that you could have quickly figured this out by seeing that (6,8) is the hypotenuse of a 6-8-10 triangle (which is a multiple of the "easy" 3-4-5).
The rest is easy. The surface area of the sphere is defined by:
A = 4_πr_2 = 4 * 100 * π = 400_π_
To find the surface area, we must first find the radius. Based on our description, this passes from (0,0) to (6,8). This can be found using the distance formula:
62 + 82 = _r_2; _r_2 = 36 + 64; _r_2 = 100; r = 10
It should be noted that you could have quickly figured this out by seeing that (6,8) is the hypotenuse of a 6-8-10 triangle (which is a multiple of the "easy" 3-4-5).
The rest is easy. The surface area of the sphere is defined by:
A = 4_πr_2 = 4 * 100 * π = 400_π_
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A rectangular prism has a volume of 70 m3. If the length, width, and height of the prism are integers measured in meters, which of the following is NOT a possible measure of the surface area of the prism measured in square meters?
A rectangular prism has a volume of 70 m3. If the length, width, and height of the prism are integers measured in meters, which of the following is NOT a possible measure of the surface area of the prism measured in square meters?
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Since the volume is the product of length, width, and height, and each of these three dimensions are integers, it is important to know the factors of the volume. 70 = (2)(5)(7). This implies that each of these factors (and only these factors with the exception of 1) will show up in the three dimensions exactly once. This creates precisely the following five possibilities:
2, 5, 7
SA = 2((2)(5)+(2)(7)+(5)(7)) = 118
1, 7, 10
SA = 2((1)(7)+(1)(10)+(7)(10)) = 174
1, 5, 14
SA = 2((1)(5)+(1)(14)+(5)(14)) = 178
1, 2, 35
SA = 2((1)(2)+(1)(35)+(2)(35)) = 214
1, 1, 70
SA = 2((1)(1)+(1)(70)+(1)(70)) = 282
Since the volume is the product of length, width, and height, and each of these three dimensions are integers, it is important to know the factors of the volume. 70 = (2)(5)(7). This implies that each of these factors (and only these factors with the exception of 1) will show up in the three dimensions exactly once. This creates precisely the following five possibilities:
2, 5, 7
SA = 2((2)(5)+(2)(7)+(5)(7)) = 118
1, 7, 10
SA = 2((1)(7)+(1)(10)+(7)(10)) = 174
1, 5, 14
SA = 2((1)(5)+(1)(14)+(5)(14)) = 178
1, 2, 35
SA = 2((1)(2)+(1)(35)+(2)(35)) = 214
1, 1, 70
SA = 2((1)(1)+(1)(70)+(1)(70)) = 282
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A cube has a surface area of 10m2. If a cube's sides all double in length, what is the new surface area?
A cube has a surface area of 10m2. If a cube's sides all double in length, what is the new surface area?
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The equation for surface area of the original cube is 6s2. If the sides all double in length the new equation is 6(2s)2 or 6 * 4s2. This makes the new surface area 4x that of the old. 4x10 = 40m2
The equation for surface area of the original cube is 6s2. If the sides all double in length the new equation is 6(2s)2 or 6 * 4s2. This makes the new surface area 4x that of the old. 4x10 = 40m2
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