# Rectangles parallel space law

## Contents

• 1 parallel area of ​​rectangles
• 2 space side of the cuboid
• 3 Examples of rectangles parallel space account
• 4 video about the size and space cuboid
• 5 References

## Parallel space Rectangles

Rectangles parallel contains six aspects, it can be an area of ​​account by finding the sum of these areas facets, but as the opposite sides in the cuboid are identical, we need only three ways to express space, using dimensional triangular to express them, namely: length, width, The rise, as follows: [1]

• Parallel total area of ​​rectangles = (2 × height × width) + (2 × width × height) + (2 × height × height), and symbols: a parallel area of ​​rectangles = (2 × a × b) + (2 × b × p) + (2 × a × p); Where: A: parallel length of rectangles. B: View cuboid. P: high cuboid.
• A: parallel length of rectangles. B: View cuboid. P: high cuboid.
• A: parallel length of rectangles.
• B: View cuboid.
• P: high cuboid.

It should be noted here that it has been beaten number 2; Because each of the two sides opposite cuboid identical; Any of the same area, and the area is measured in units of longitudinal square. [1]

For more information and examples of cuboid you can read the following articles: parallel definition of rectangles.

## Side area of ​​the cuboid

Know the side area of ​​the cuboid (in English: Cuboid Lateral Surface Area) as the sum of all aspects of spaces except upper-sided and lower, and the following formula can be used to find a side space: [2]

• Side area of ​​the cuboid = four aspects of the cuboid area = 2 × (width × height) + 2 × (height × length), and remove the height common factor, and the order of the border, the side space is equal to: side area of ​​the cuboid = 2 × height × (length + PowerPoint), and symbols: side area of ​​the cuboid p = 2 × × (a + b); Where: A: parallel length of rectangles. B: View cuboid. P: high cuboid.
• A: parallel length of rectangles. B: View cuboid. P: high cuboid.
• A: parallel length of rectangles.
• B: View cuboid.
• P: high cuboid.

## Examples of rectangles parallel space account

• The first example: What is the total area of ​​the Fund on the cuboid shape of a length of 6 cm, and width of 5 cm, and a height of 4 cm? [3] Solution: You can find the total area by following the following steps: Parallel area of ​​rectangles = 2 × (length × display + width × height + height × length) = 2 × (6 × 5 + 5 × 4 + 4 × 6) = 2 × (30 + 20 + 24) = 2 × 74 = 148 cm 2.
• Solution: You can find the total area by following the following steps: Parallel area of ​​rectangles = 2 × (length × display + width × height + height × length) = 2 × (6 × 5 + 5 × 4 + 4 × 6) = 2 × (30 + 20 + 24) = 2 × 74 = 148 cm 2.
• Parallel area of ​​rectangles = 2 × (length × display + width × height + height × length) = 2 × (6 × 5 + 5 × 4 + 4 × 6) = 2 × (30 + 20 + 24) = 2 × 74 = 148 cm 2.
• The second example: What is the total area of ​​rectangles, which parallel the length of 20 cm, and width 12 cm, and a height of 9 cm? [4] Solution: You can find the total area by following the following steps: Parallel area of ​​rectangles = 2 × (length × display + width × height + length × height) = 2 × ((20 × 12) + (12 × 9) + (20 × 9)) = 2 × (240 + 108 + 180 ) = 2 × 528 = 1056 cm 2.
• Solution: You can find the total area by following the following steps: Parallel area of ​​rectangles = 2 × (length × display + width × height + length × height) = 2 × ((20 × 12) + (12 × 9) + (20 × 9)) = 2 × (240 + 108 + 180 ) = 2 × 528 = 1056 cm 2.
• Parallel area of ​​rectangles = 2 × (length × display + width × height + length × height) = 2 × ((20 × 12) + (12 × 9) + (20 × 9)) = 2 × (240 + 108 + 180 ) = 2 × 528 = 1056 cm 2.
• Third example: What is the space side of the parallel rectangles that length 3 p.m. and 5 p.m. display, and a height of 4 m? [4] Solution: You can find a side area by following the following steps: Side area of ​​the cuboid = 2 × height × (length + PowerPoint) = 2 × 4 × (3 + 5) Side Area = 8 × 8 Side = 64 m 2 space.
• Solution: You can find a side area by following the following steps: Side area of ​​the cuboid = 2 × height × (length + PowerPoint) = 2 × 4 × (3 + 5) Side Area = 8 × 8 Side = 64 m 2 space.
• Side area of ​​the cuboid = 2 × height × (length + PowerPoint) = 2 × 4 × (3 + 5)
• Side Area = 8 × 8
• Side = 64 m 2 space.
• Fourth example: What is the space side of the cuboid if the length of 12 cm, and width 13 cm, and a height of 15 cm? [5] Solution: You can find a side area by following the following steps: Side area of ​​the cuboid = 2 × height × (length + PowerPoint) = 2 × 15 × (12 + 13) = 750 cm 2.
• Solution: You can find a side area by following the following steps: Side area of ​​the cuboid = 2 × height × (length + PowerPoint) = 2 × 15 × (12 + 13) = 750 cm 2.
• Side area of ​​the cuboid = 2 × height × (length + PowerPoint) = 2 × 15 × (12 + 13) = 750 cm 2.
• Fifth example: parallelogram 40 m2, side 26 m 2 area, what is the base area? [6] This question can be resolved by following the following steps: Total area = 2 × area of ​​the base + space side, and it: 40 = 2 × Al Qaeda Area + 26, and the order of the equation by subtracting (26) of the parties, and then divided by (2), that produces: 2 × 14 = area of ​​the base, and it: Rule 7 = m 2.
• This question can be resolved by following the following steps: Total area = 2 × area of ​​the base + space side, and it: 40 = 2 × Al Qaeda Area + 26, and the order of the equation by subtracting (26) of the parties, and then divided by (2), that produces: 2 × 14 = area of ​​the base, and it: Rule 7 = m 2.
• Total area = 2 × area of ​​the base + space side, and it:
• 40 = 2 × Al Qaeda Area + 26, and the order of the equation by subtracting (26) of the parties, and then divided by (2), that produces:
• 2 × 14 = area of ​​the base, and it: Rule 7 = m 2.
• Sixth example: parallelogram length of 16 cm, and width 14 cm, height 10 cm, what is the college? [7 area] Solution: The space can be found by following the following steps: Total = 2 space × (length × display + width × height + length × height) = 2 × (16 × 14+ 14 × 10 + 10 × 16) = 2 × (224 + 140 + 160) = 2 × 524 = 1048 cm 2 .
• Solution: The space can be found by following the following steps: Total = 2 space × (length × display + width × height + length × height) = 2 × (16 × 14+ 14 × 10 + 10 × 16) = 2 × (224 + 140 + 160) = 2 × 524 = 1048 cm 2 .
• Total = 2 space × (length × display + width × height + length × height) = 2 × (16 × 14+ 14 × 10 + 10 × 16) = 2 × (224 + 140 + 160) = 2 × 524 = 1048 cm 2 .
• Seventh: parallelogram example, an area of ​​base 20 cm 2, and around 20 cm, if the height of 6 cm, what is the college? [8 area] Solution: cuboid consists of two bases, and four faces, so the parallel area of ​​rectangles = 2 × (the base area) + space The four aspects of space or side, and it: Space rules = 2 × area of ​​the base, and thus: space rules = 2 × 20 = 40 cm 2. The four faces of space or side = 2 space × height × (length + PowerPoint), and because the base perimeter of the rectangular = 2 × (length + PowerPoint), then your side area of ​​the cuboid = perimeter of the base × height = 20 × 6 = 120 cm 2. Which: parallel area of ​​rectangles = 120 40 + = 160 cm 2.
• Solution: cuboid consists of two bases, and four faces, so the parallel area of ​​rectangles = 2 × (the base area) + four aspects of an area or side space, it:
• Space rules = 2 × area of ​​the base, and thus: space rules = 2 × 20 = 40 cm 2.
• The four faces of space or side = 2 space × height × (length + PowerPoint), and because the base perimeter of the rectangular = 2 × (length + PowerPoint), then your side area of ​​the cuboid = perimeter of the base × height = 20 × 6 = 120 cm 2.
• Which: parallel area of ​​rectangles = 120 40 + = 160 cm 2.
• Example VIII: lounge on a parallel form of rectangles to be the four walls of paint coating If the cost per square meter 8 coins, what is the cost of painting the walls of this hall paint note that the perimeter of its land is 250 m, and height is 6 m? [9] Solution: the cost of paint = wall space × cost per square meter Walls can calculate the area by calculating the side area of ​​the cuboid, as follows: Area side 2 × height = × (length + PowerPoint) Can be found 2 × (length + PowerPoint) through the ocean; Where the ocean is equal to 2 × (length + PowerPoint), and equal to 250 m; This is because all aspects of cuboid be a rectangular shape, and the perimeter of the rectangle = 2 × (length + PowerPoint). Compensation in the side area of ​​the law, the side area = 6 × 250 = 1500 m 2. The cost of paint = 1500 × 8 = 12,000 coin cash.
• Solution: the cost of paint = space walls × the cost per square meter
• Walls can calculate the area by calculating the side area of ​​the cuboid, as follows: Area side 2 × height = × (length + PowerPoint) Can be found 2 × (length + PowerPoint) through the ocean; Where the ocean is equal to 2 × (length + PowerPoint), and equal to 250 m; This is because all aspects of cuboid be a rectangular shape, and the perimeter of the rectangle = 2 × (length + PowerPoint). Compensation in the side area of ​​the law, the side area = 6 × 250 = 1500 m 2.
• Area side 2 × height = × (length + PowerPoint)
• Can be found 2 × (length + PowerPoint) through the ocean; Where the ocean is equal to 2 × (length + PowerPoint), and equal to 250 m; This is because all aspects of cuboid be a rectangular shape, and the perimeter of the rectangle = 2 × (length + PowerPoint).
• Compensation in the side area of ​​the law, the side area = 6 × 250 = 1500 m 2.
• The cost of paint = 1500 × 8 = 12,000 coin cash.
• Ninth example: parallelogram total area of ​​214 cm 2, size 210 cm 3, and its base area 42 cm 2, what are the three dimensions of length, width, and height? [10] To resolve this question is the following steps: This question can be resolved by using the following laws: The total area of ​​the cuboid = 2 × (length × display + width × height + height × length) Cuboid size = length × width × height The base area = length × width, because the base rectangular shape. Through a legal size, and space height calculation, as follows: The base area = 42 = length × width, and compensate this value in the law that size produces: Parallel size rectangles = 42 × height = 210, and dividing the parties (42) produces the height = 5 cm. Compensate for the rise in the area of ​​law cuboid as follows: 2 × (42 + width × 5 + 5 × length) = 214; This is because the value (length × width) represents space, equal to 42, and by dividing the parties (2), then subtract (42) of the parties produce that: width × 5 + 5 × height = 65, and dividing the parties (5) results that: length + PowerPoint = 13. We became our equations Alatetan: The first equation: Length + Delivery = 13 The second equation: length × width = 42 After solving these equations for compensation, it produces the values ​​of length and width are: length = 6 cm, width = 7 cm.
• To resolve this question is the following steps:
• This question can be resolved by using the following laws: The total area of ​​the cuboid = 2 × (length × display + width × height + height × length) Cuboid size = length × width × height The base area = length × width, because the base rectangular shape.
• The total area of ​​the cuboid = 2 × (length × display + width × height + height × length)
• Cuboid size = length × width × height
• The base area = length × width, because the base rectangular shape.
• Through a legal size, and space height calculation, as follows: The base area = 42 = length × width, and compensate this value in the law that size produces: Parallel size rectangles = 42 × height = 210, and dividing the parties (42) produces the height = 5 cm.
• The base area = 42 = length × width, and compensate this value in the law that size produces:
• Parallel size rectangles = 42 × height = 210, and dividing the parties (42) produces the height = 5 cm.
• Compensate for the rise in the area of ​​law cuboid as follows: 2 × (42 + width × 5 + 5 × length) = 214; This is because the value (length × width) represents space, equal to 42, and by dividing the parties (2), then subtract (42) of the parties produce that: width × 5 + 5 × height = 65, and dividing the parties (5) results that: length + PowerPoint = 13.
• We became our equations Alatetan: The first equation: Length + Delivery = 13 The second equation: length × width = 42 After solving these equations for compensation, it produces the values ​​of length and width are: length = 6 cm, width = 7 cm.
• The first equation: Length + Delivery = 13
• The second equation: length × width = 42
• After solving these equations for compensation, it produces the values ​​of length and width are: length = 6 cm, width = 7 cm.

For more information and examples about the size of the cuboid you can read the following articles: Law parallel the size of the rectangles.

• Example X: swimming pool, a parallel form of rectangles in length 20 m, width 15 m ,, and depth is 4 m, a very cost paintjob if it is equal to 20 coin cash per square meter? [9] Solution: the cost of paint = side walls space × cost per square meter Walls can calculate the area by calculating the side area of ​​the cuboid, as follows: Side Area = 2 × height × (length + PowerPoint) = 2 × 4 × (20 +15) = 280 m 2. Calculate the cost of paint = 280 × 20 = 5,600 coin cash.
• Solution: the cost of paint = area side walls × cost per square meter
• Walls can calculate the area by calculating the side area of ​​the cuboid, as follows: Side Area = 2 × height × (length + PowerPoint) = 2 × 4 × (20 +15) = 280 m 2.
• Side Area = 2 × height × (length + PowerPoint) = 2 × 4 × (20 +15) = 280 m 2.
• Calculate the cost of paint = 280 × 20 = 5,600 coin cash.

## Video about the size and space cuboid

To learn this geometry continued video: [11]

## References

• ^ أ ب "What is a Cuboid? - Definition, Shape, Area & Properties", www.tutors.com, Retrieved 3-4-2020. Edited.
• ↑ "https://www.vedantu.com/maths/cuboid-and-cube", www.vedantu.com, Retrieved 3-4-2020. Edited.
• ↑ "Total Surface Area of a Cuboid", www.mathsteacher.com.au, Retrieved 3-4-2020. Edited.
• ^ أ ب "cuboids", www.onlinemath4all.com, Retrieved 3-4-2020. Edited.
• ↑ "Cube and Cuboid", byjus.com, Retrieved 3-4-2020. Edited.
• ↑ "A cuboid", www.toppr.com, Retrieved 3-4-2020. Edited.
• ↑ "Surface Area of a Cuboid", www.web-formulas.com, Retrieved 3-4-2020. Edited.
• ↑ "Surface Area of a Cuboid", brilliant.org, Retrieved 3-4-2020. Edited.
• ^ أ ب "A cuboide ", www.toppr.com, Retrieved 3-4-2020. Edited.
• ↑ "surface area of a cuboid", www.topperlearning.com, Retrieved 3-4-2020. Edited.
• ↑ Video about the size and space cuboid.

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