# Law parallel the size of the rectangles ## Contents

• 1 parallel size rectangles
• 2 Examples of calculating the size of cuboid
• 3 video about the size and space cuboid
• 4 References

## Parallel size rectangles

Known size as a vacuum quantity or material that exist within a three-shape dimensions, measured by the size unit per cubic meter in accordance with the international system of units,  and can be the size of cuboid account, which is a form three-dimensional through the law as follows: 

• Cuboid size = length × width × height, and symbols: h = a × b × c; Where: H: parallel the size of the rectangles. A: parallel length of rectangles. B: View cuboid. A: high cuboid.
• H: parallel the size of the rectangles.
• A: parallel length of rectangles.
• B: View cuboid.
• A: high cuboid.

## Examples of calculating the size of cuboid

• The first example: What is the size of the parallel rectangles that length of 14 cm, and width 12 cm, and a height of 8 cm?  Solution: parallel size rectangles = length × width × height, thus: Parallel size rectangles = 14 × 12 × 8 = 1344 cm 3.
• Solution: The size of the cuboid = length × width × height, thus: Parallel size rectangles = 14 × 12 × 8 = 1344 cm 3.
• Parallel size rectangles = 14 × 12 × 8 = 1344 cm 3.
• The second example: What is the size of the parallel length of 14 cm, and width of 50 mm, and a height of 10 cm?  Solution: parallel size rectangles = length × width × height As the length and height in centimeters, it must convert the display to become the unity of a centimeter, so as to become all dimensions of the same unit, it is known that 10 mm = 1 cm, so the width is equal to: 50 mm / 10 cm = 5 cm. After becoming the same dimensions of the unit, the size can be found as follows: Parallel size rectangles = 14 × 5 × 10 = 700 cm 3.
• Solution: The size of the cuboid = length × width × height
• As the length and height in centimeters, it must convert the display to become the unity of a centimeter, so as to become all dimensions of the same unit, it is known that 10 mm = 1 cm, so the width is equal to: 50 mm / 10 cm = 5 cm.
• After becoming the same dimensions of the unit, the size can be found as follows: Parallel size rectangles = 14 × 5 × 10 = 700 cm 3.
• Parallel size rectangles = 14 × 5 × 10 = 700 cm 3.
• Third example: What is the cost of buying bricks that should be used to build a wall on a parallel form of rectangles length of 20 m and a height of 2 m, and width of 0.75 m, note that each height of 7.5 cm brick mold, and a length of 25 cm, and width 10 cm, and that every 1000 block of brick coin worth 900 cash?  solution: Wall size represents the size of the parallel rectangles, and can be calculated as follows: Wall = length size × width × height = 20 m × 2 m × 0.75 m = 30 m³. Bricks size also represents the size of the parallel rectangles, and can be calculated as follows: Size brick mold = 25 cm × 10 cm × 7.5 cm = 1875 Sm³. Number of bricks required size = wall / size bricks, but the size of bricks size cubic centimeters, the size of the wall Fmkas cubic meter; So you must unify the units by turning the wall to the size of the cubic centimeter in size by dividing the value of (1,000,000); Because every 1 = 1,000,000 cubic meters Sm³, and it: the size of the brick cubic meters = 1875 / 1,000,000 = 0.001875 cubic meters. Number of bricks = 30 / 0.001875 = 16,000 block of brick. An operation ratio, and proportionality between the number of templates, and their cost is as follows: Template 1000 ← cost 900 cash currency Each template 16.000 ← ?? Perform a multiplication cross, the cost of templates = 900 × 16,000 / 1,000, and is equal to 14,400 cash currency.
• The solution:
• Wall size represents the size of the parallel rectangles, and can be calculated as follows: Wall = length size × width × height = 20 m × 2 m × 0.75 m = 30 m³.
• Wall = length size × width × height = 20 m × 2 m × 0.75 m = 30 m³.
• Bricks size also represents the size of the parallel rectangles, and can be calculated as follows: Size brick mold = 25 cm × 10 cm × 7.5 cm = 1875 Sm³.
• Size brick mold = 25 cm × 10 cm × 7.5 cm = 1875 Sm³.
• Number of bricks required size = wall / size bricks, but the size of bricks size cubic centimeters, the size of the wall Fmkas cubic meter; So you must unify the units by turning the wall to the size of the cubic centimeter in size by dividing the value of (1,000,000); Because every 1 = 1,000,000 cubic meters Sm³, and it: the size of the brick cubic meters = 1875 / 1,000,000 = 0.001875 cubic meters. Number of bricks = 30 / 0.001875 = 16,000 block of brick.
• Number of bricks = 30 / 0.001875 = 16,000 block of brick.
• An operation ratio, and proportionality between the number of templates, and their cost is as follows: Template 1000 ← cost 900 cash currency Each template 16.000 ← ?? Perform a multiplication cross, the cost of templates = 900 × 16,000 / 1,000, and is equal to 14,400 cash currency.
• Template 1000 ← cost 900 cash currency
• Each template 16.000 ← ??
• Perform a multiplication cross, the cost of templates = 900 × 16,000 / 1,000, and is equal to 14,400 cash currency.
• Fourth example: a swimming pool for the Olympic Games in length 50 m, width 25 m, and the depth of water in which 14 o'clock, what is the amount of water that can accommodate her this blessing?  Solution: Can be expressed as the amount of water in the pond by using this size, and the size of the water is equal to the size of the parallel rectangles, and can be found as follows: Parallel size rectangles = length × width × height = 50 × 25 × 2 = 2500 m 3, which is the amount of water in this pond.
• The solution: Can be expressed as the amount of water in the pond by using this size, and the size of the water is equal to the size of the parallel rectangles, and can be found as follows: Parallel size rectangles = length × width × height = 50 × 25 × 2 = 2500 m 3, which is the amount of water in this pond.
• Can be expressed as the amount of water in the pond by using this size, and the size of the water is equal to the size of the parallel rectangles, and can be found as follows:
• Parallel size rectangles = length × width × height = 50 × 25 × 2 = 2500 m 3, which is the amount of water in this pond.
• Fifth example: If the length of parallel rectangles 8 cm, and a height of 3 cm, what is currently note that the size of 120 cm 3?  Solution: Size cuboid = length × width × height, it: 120 = 8 × width × 3 Solving this equation, the width = 5 cm.
• The solution: Size cuboid = length × width × height, it: 120 = 8 × width × 3 Solving this equation, the width = 5 cm.
• Size cuboid = length × width × height, it:
• 120 = 8×العرض×3
• Solving this equation, the width = 5 cm.
• Sixth example: designed Fouad fund a parallel form Rectangles size 2500 cm 3, and a height of 25 cm, square base shape, and then realized that it needs a smaller fund size hatching from its height to become a size of 1000 cm 3, and remained an area of ​​its base as it is, how much has become a height, and whether became a cubic fund form ?  solution: The base area is calculated using the law of parallel size rectangles = length × width × height. As the size = 2500 cm 3, height = 25 cm, and compensate these values ​​in the law of size can be obtained on the base area of ​​a square shape as follows :. 2500 = (length × width) × height = (length × width) × 25, and by dividing the parties (25) that produces: 100 cm 2 = length × width, which represents the base area. Calculate the length, the base width and square shape as follows: The base area = (rib) 2, the length of it: the length of the rib = 100√ = 10 cm, and as the base square shape, the display is equal to 10 cm as well. Calculate the height of the fund after cutting part of its height by law parallel the size of the rectangles, the result that: the size of the fund after the shear = length × width × height, it: 1000 = 10 × 10 × height, and dividing the parties (100) that produces: new height = 10 cm. Since the height = width = height, the output format is a cube.
• The solution:
• The base area is calculated using the law of parallel size rectangles = length × width × height. As the size = 2500 cm 3, height = 25 cm, and compensate these values ​​in the law of size can be obtained on the base area of ​​a square shape as follows :. 2500 = (length × width) × height = (length × width) × 25, and by dividing the parties (25) that produces: 100 cm 2 = length × width, which represents the base area.
• As the size = 2500 cm 3, height = 25 cm, and compensate these values ​​in the law of size can be obtained on the base area of ​​a square shape as follows :.
• 2500 = (length × width) × height = (length × width) × 25, and by dividing the parties (25) that produces: 100 cm 2 = length × width, which represents the base area.
• Calculate the length, the base width and square shape as follows: The base area = (rib) 2, the length of it: the length of the rib = 100√ = 10 cm, and as the base square shape, the display is equal to 10 cm as well.
• The base area = (rib) 2, the length of it: the length of the rib = 100√ = 10 cm, and as the base square shape, the display is equal to 10 cm as well.
• Calculate the height of the fund after cutting part of its height by law parallel the size of the rectangles, the result that: the size of the fund after the shear = length × width × height, it: 1000 = 10 × 10 × height, and dividing the parties (100) that produces: new height = 10 cm.
• 1000 = 10 × 10 × height, and dividing the parties (100) that produces: new height = 10 cm.
• Since the height = width = height, the output format is a cube.
• Seventh example: What is the amount of air that there is room inside a parallel form of rectangles is equal to the length of 5 m, width 6 p.m. and 10 p.m. height?  Solution: the amount of air inside the room = capacity = room size cuboid. Cuboid size = length × width × height Parallel size rectangles = 5 × 6 × 10 = 300 m 3, so the amount of air that exists inside the room 300 m 3.
• Solution: the amount of air inside the room = capacity of the room size = parallel rectangles.
• Cuboid size = length × width × height
• Parallel size rectangles = 5 × 6 × 10 = 300 m 3, so the amount of air that exists inside the room 300 m 3.
• Eighth example: a metal rod in the form of a parallelogram length of 10 m and width of 60 cm and a thickness of 25 cm, what is the price if the price per cubic meter per 250 dollars?  Solution: To calculate the penis the price of the metal must first account of its size; Because the price = cost per cubic meter size × cuboid, and it: Parallel size rectangles = length × width × height = 10 × (60/100) × (25/10), and it should be noted that it was divisible by 100 to convert from cm to meters. Parallel size rectangles = 1.5 m 3. The price of the metal rod = 1.5 × 250 = 375 dollars.
• Solution: To calculate the price of the metal penis must first calculate the size; Because the price = cost per cubic meter size × cuboid, and it:
• Parallel size rectangles = length × width × height = 10 × (60/100) × (25/10), and it should be noted that it was divisible by 100 to convert from cm to meters.
• Parallel size rectangles = 1.5 m 3.
• The price of the metal rod = 1.5 × 250 = 375 dollars.
• Ninth example: What is the height of parallel rectangles note that the size of 300 cm 3, and the base area of ​​30 cm?  The solution: the size of cuboid = length × width × height, and the height can be found as follows: Al-Qaeda is a rectangular shape, and therefore the area = length × width, equal to 30 cm. Can be found in the rise of the Code size as follows: 300 = 30 × height, it: Height = 300/30 = 10 cm.
• Solution: The size of the cuboid = length × width × height, and the height can be found as follows: Al-Qaeda is a rectangular shape, and therefore the area = length × width, equal to 30 cm. Can be found in the rise of the Code size as follows: 300 = 30 × height, it: Height = 300/30 = 10 cm.
• Al-Qaeda is a rectangular shape, and therefore the area = length × width, equal to 30 cm.
• Can be found in the rise of the Code size as follows: 300 = 30 × height, it:
• Height = 300/30 = 10 cm.
• X example: pool empty swimming pool on a parallel form of rectangles length of 25 m, width 10 m and depth of 2 p.m., it can be refilled with water at a rate of 800 liters per minute how much of the time in minutes, hours required for full packaged note that each cubic meter = 1000 liters?  Solution: Calculate the amount of water needed to fill the pond, which can be calculated using the size of the law cuboid = length × width × height, it: Parallel size rectangles = 25 × 10 × 2 = 500 m 3, the amount of water needed to fill the pond with water. The time that is required to complete packaged size = / fill rate, but it must first convert packing rate per liter to cubic meter, and so on Baksmth (1000); Because each cubic meter = 1000 liters; Ie 800 l / min = 800/1000 = 0.8 m³ / min, and thus: The time required to fill the pond full = 500 m³ / ((0.8) cubic meters / min), and from time in minutes = 625 minutes, and the time in hours = 625/60 = 10 hours and almost half
• The solution:
• Calculate the amount of water needed to fill the pond, which can be calculated using the size of the law cuboid = length × width × height, it: Parallel size rectangles = 25 × 10 × 2 = 500 m 3, the amount of water needed to fill the pond with water.
• Parallel size rectangles = 25 × 10 × 2 = 500 m 3, the amount of water needed to fill the pond with water.
• The time that is required to complete packaged size = / fill rate, but it must first convert packing rate per liter to cubic meter, and so on Baksmth (1000); Because each cubic meter = 1000 liters; Ie 800 l / min = 800/1000 = 0.8 m³ / min, and thus: The time required to fill the pond full = 500 m³ / ((0.8) cubic meters / min), and from time in minutes = 625 minutes, and the time in hours = 625/60 = 10 hours and almost half
• The time required to fill the pond full = 500 m³ / ((0.8) cubic meters / min), and from time in minutes = 625 minutes, and the time in hours = 625/60 = 10 hours and almost half
• Example atheist ten: two boxes A, B on the parallel form of rectangles If the dimensions (ie, length, width) Fund Base: a 10 cm × 8 cm, and the dimensions of the Fund Base B: 15 cm × 10 cm, if the mobilization of the fund a water reaching a height of 15 cm, then this water was poured in the fund (b) to what high water will rise in this fund?  solution: The quantity (size) of water in a box = quantity (size) Water Fund B, and compensation in the law of parallel size rectangles = length × width × height, results that: 10 × 8 × 15 = 15 × 10 × height, and solving the equation that produces: Height = 8 cm.
• The solution: The quantity (size) of water in a box = quantity (size) Water Fund B, and compensation in the law of parallel size rectangles = length × width × height, results that: 10 × 8 × 15 = 15 × 10 × height, and solving the equation that produces: Height = 8 cm.
• The quantity (size) of water in a box = quantity (size) Water Fund B, and compensation in the law of parallel size rectangles = length × width × height, results that:
• 10 × 8 × 15 = 15 × 10 × height, and solving the equation that produces: Height = 8 cm.
• Twelfth example: If the size of the fund on a parallel form of rectangles 1440 m 3, and a length of 15 m and a height of 8 meter. What is the height?  Solution: Parallel size rectangle = length × width × height, it: 1440 = 15 × 8 × height, and solving the equation that produces: Height = 1440/120 = 12 Eng.
• The solution:
• Parallel size rectangle = length × width × height, it:
• 1440 = 15 × 8 × height, and solving the equation that produces: Height = 1440/120 = 12 Eng.
• XIII example: If the dimensions of the fund base parallelogram 80 form cm × 40 cm, the size of 160 liters, and wanted Ahmed coating all aspects of the fund with the exception of the lower base, and the cost of paint 6000 coin / m², very expensive coating this fund.  The solution: Fund account by using the height of the size of the law cuboid, but it must first be converted to cubic centimeter per liter to unify the units by multiplying the volume value (1,000); Because 1 liter = 1,000 Sm³ to produce that: the size of the rectangle parallel = 160 liters = 160,000 Sm³, the compensation value in the law of parallel size rectangles: length × width × height, the result that: 160,000 = 80 × 40 × height, and from: Height = 50 cm. Fund calculate the area with the exception of the lower base to calculate the cost of painted: A parallel area of ​​rectangles except the lower base side = space + space upper the base 2 × height = × (length + PowerPoint) + length × width = 2 × 50 × (80 + 40) + 80 × 40 = 15,200 cm² = 1.52 m ²; Because each 1 m² = 1000 cm². Calculate the cost of paint = fund space × cost of paint = 1.52 × 6000 m² coin cash / m² = 9.120 currency cash.
• The solution:
• Fund account by using the height of the size of the law cuboid, but it must first be converted to cubic centimeter per liter to unify the units by multiplying the volume value (1,000); Because 1 liter = 1,000 Sm³ to produce that: the size of the rectangle parallel = 160 liters = 160,000 Sm³, the compensation value in the law of parallel size rectangles: length × width × height, the result that: 160,000 = 80 × 40 × height, and from: Height = 50 cm.
• 160,000 = 80 × 40 × height, and from: Height = 50 cm.
• Fund calculate the area with the exception of the lower base to calculate the cost of painted: A parallel area of ​​rectangles except the lower base side = space + space upper the base 2 × height = × (length + PowerPoint) + length × width = 2 × 50 × (80 + 40) + 80 × 40 = 15,200 cm² = 1.52 m ²; Because each 1 m² = 1000 cm².
• A parallel area of ​​rectangles except the lower base side = space + space upper the base 2 × height = × (length + PowerPoint) + length × width = 2 × 50 × (80 + 40) + 80 × 40 = 15,200 cm² = 1.52 m ²; Because each 1 m² = 1000 cm².
• Calculate the cost of paint = fund space × cost of paint = 1.52 × 6000 m² coin cash / m² = 9.120 currency cash.

## Video about the size and space cuboid

To learn this geometry continued video: 

## References

• ^ أ ب "Cube and Cuboid", www.toppr.com, Retrieved 5-4-2020. Edited.
• ↑ "Volume of a Cuboid", www.mathsisfun.com, Retrieved 5-4-2020. Edited.
• ^ أ ب ت "Cuboid", www.math-only-math.com, Retrieved 5-4-2020. Edited.
• ^ أ ب "Volume of a Cuboid ", www.mathopolis.com, Retrieved 5-4-2020. Edited.
• ↑ "Volume Of Cuboid", byjus.com, Retrieved 5-4-2020. Edited.
• ^ أ ب "Volume of a Cuboid", www.web-formulas.com, Retrieved 5-4-2020. Edited.
• ↑ "Volume of a Cuboid", corbettmaths.com, Retrieved 5-4-2020 (page 3). Edited.
• ^ أ ب "Volume of a cuboid", www.cuemath.com, Retrieved 5-4-2020. Edited.
• ↑ "VOLUME OF CUBOID WORD PROBLEMS", www.onlinemath4all.com, Retrieved 5-4-2020. Edited.
• ↑ Video about the size and space cuboid.

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