# Semi-cube law ## Contents

• 1 Overview of the cube and the semi-cube
• 2 laws cuboid
• 3 examples of a variety of semi-cube
• 4 References

## Overview of the cube and the semi-cube

The cube (known in English: Cube) as one of the engineering polyhedra, and its dimensions are (length, width, height), and consists of the cube of the six aspects of square shape and equal in the area, and since all sides of the box are identical, the cube length = display = height, it also consists of 12 characters, and 8 heads resulting from the confluence of three characters with each other, and all corners of the list. 

For more information about the cube you can read the following articles: cube space law, how to calculate the size of the cube.

The semi-cube It is a cuboid, which is considered one of polyhedra, and a list of angles, the three dimensions are: length, width, height, and consists of semi-cube of six aspects of rectangular shape, each of two sides of which are identical, and converge these faces when the characters, which are a straight lines meet all three of them in the points known as headers, and differs from the cube of hand that the rectangular faces and are not identical, and different dimensions and unequal.  

## Semi-cube laws

• Law semi-cube space: The semi-cube three-dimensional multi-faceted, and calculate the area must calculate the total areas of each side of the six faces rectangular, as follows:   The total area of ​​the semi-cube = area of ​​preferential treatment side + rules = 2 space × ( The first face space) + 2 × (second face area) + 2 × (the third face area) = 2 × (length × width) + 2 × (length × height) + 2 × (width × height); Note that the law of the rectangle area = rectangle length × width of the rectangle. Semi-cube side space = space side facets = 2 × (length × height) + 2 × (width × height) = 2 × height × (length + PowerPoint) = perimeter of the base (rectangle) × height; Note that: the perimeter of the rectangle = 2 × (the length of the rectangle × width of the rectangle).
• The total area of ​​the semi-cube = area of ​​preferential treatment side + rules = 2 space × (the first face area) + 2 × (second face area) + 2 × (the third face area) = 2 × (length × width) + 2 × (length × height ) + 2 × (width × height); Note that the law of the rectangle area = rectangle length × width of the rectangle.
• Semi-cube side space = space side facets = 2 × (length × height) + 2 × (width × height) = 2 × height × (length + PowerPoint) = perimeter of the base (rectangle) × height; Note that: the perimeter of the rectangle = 2 × (the length of the rectangle × width of the rectangle).

• Semi-cube size law: defines the size in general as the amount of vacuum that fills the shape three-dimensional, and can by calculating quasi-size cube calculate the amount of water needed to fill the water tank on the semi-cube-shaped, for example, and other things, and can size is calculated using the following relationship: [ 4] semi-cube size = length × width × height.
• Semi-cube size = length × width × height.

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

• Diameter law cuboid: can calculate the length of diameters trapezoidal using the following relationship:  diameters semi-cube = (Ataiwl² + Aerd² + Alaratvaa²) √.
• Diameters semi-cube = (Ataiwl² + Aerd² + Alaratvaa²) √.

## Examples of a variety of semi-cube

• The first example: grandfather size and area of ​​semi-cube if the length of 10 cm, and width of 4 cm, and a height of 5 cm.  Solution: Compensation in the law: cuboid size = length × width × height = 10 × 4 × 5 = 200 cm 3. Compensation in the law: cuboid total area = 2 × (length × width) + 2 × (length × height) + 2 × (width × height) = 2 × (10 × 4) + 2 × (10 × 5) +2 × (4 × 5) = 80 + 100 + 40 = 220 cm 2.
• The solution:
• Compensation in the law: cuboid size = length × width × height = 10 × 4 × 5 = 200 cm 3.
• Compensation in the law: cuboid total area = 2 × (length × width) + 2 × (length × height) + 2 × (width × height) = 2 × (10 × 4) + 2 × (10 × 5) +2 × (4 × 5) = 80 + 100 + 40 = 220 cm 2.
• The second example: grandfather size and area of ​​semi-cube if the length of 8 cm, and width of 5 cm, and a height of 4 cm.  Solution: Compensation in the law: cuboid size = length × width × height = 8 × 5 × 4 = 160 cm 3. Compensation in the law: cuboid total area = 2 × (length × width) + 2 × (length × height) + 2 × (width × height) = 2 × (8 × 5) + 2 × (8 × 4) +2 × (5 × 4) = 80 + 64 + 40 = 184 cm 2.
• The solution:
• Compensation in the law: cuboid size = length × width × height = 8 × 5 × 4 = 160 cm 3.
• Compensation in the law: cuboid total area = 2 × (length × width) + 2 × (length × height) + 2 × (width × height) = 2 × (8 × 5) + 2 × (8 × 4) +2 × (5 × 4) = 80 + 64 + 40 = 184 cm 2.
• Third example: very cuboid and the total area of ​​the side if the length of 10 cm, and width of 8 cm, and a height of 7 cm.  Solution: Compensation in the law: cuboid total area = 2 × (length × width) + 2 × (length × height) + 2 × (width × height) = 2 × (10 × 8) + 2 × (10 × 7) +2 × (8 × 7) = 160 + 140 + 112 = 412 cm 2. Compensation in the law: semi-cube space side = 2 × (length × height) + 2 × (width × height) = 2 × (10 × 7) + 2 × (8 × 7) = 140 + 112 = 262 cm 2.
• The solution:
• Compensation in the law: cuboid total area = 2 × (length × width) + 2 × (length × height) + 2 × (width × height) = 2 × (10 × 8) + 2 × (10 × 7) +2 × (8 × 7) = 160 + 140 + 112 = 412 cm 2.
• Compensation in the law: semi-cube space side = 2 × (length × height) + 2 × (width × height) = 2 × (10 × 7) + 2 × (8 × 7) = 140 + 112 = 262 cm 2.
• Fourth example: very length of the diameters semi-cube if the length of 15 cm, and width 9 cm, and a height of 6 cm.  Solution: Compensation in the law: the countries of sub-cube = (Ataiwl² + Aerd² + Alaratvaa²) √ = (15² + 9² + 6²) √ = 18.49 cm.
• The solution:
• Compensation in the law: the countries of sub-cube = (Ataiwl² + Aerd² + Alaratvaa²) √ = (15² + 9² + 6²) √ = 18.49 cm.
• V example: If the paste four cubes, each side length of 4 cm next to each other, to be sub-cube, a very expensive if the price of painted coating per square centimeter than 100 coin.  The solution: To calculate the cost of paint must first calculate the total area of ​​the semi-cube, and account must first calculate the length of the rib dimensions of the semi-cube, and are as follows: Cuboid length = 4 × side length of the cube = 4 × 4 = 16 cm. Semi-cube height = width = cube semi-side length of the cube = 4 cm. Application of the law: cuboid total area = 2 × (length × width) + 2 × (length × height) + 2 × (width × height) = 2 × (16 × 4) + 2 × (16 × 4) + 2 × (4 × 4) = 128 + 128 + 32 = 288 cm 2. Calculate the total cost to paint semi-cube: total cost = cuboid space × cost of paint per square centimeter = 288 × 100 = 28,800 currency cash.
• The solution:
• To calculate the cost of paint must first calculate the total area of ​​the semi-cube, and account must first calculate the length of the rib dimensions of the semi-cube, and are as follows: Cuboid length = 4 × side length of the cube = 4 × 4 = 16 cm. Semi-cube height = width = cube semi-side length of the cube = 4 cm.
• Cuboid length = 4 × side length of the cube = 4 × 4 = 16 cm.
• Semi-cube height = width = cube semi-side length of the cube = 4 cm.
• Application of the law: cuboid total area = 2 × (length × width) + 2 × (length × height) + 2 × (width × height) = 2 × (16 × 4) + 2 × (16 × 4) + 2 × (4 × 4) = 128 + 128 + 32 = 288 cm 2.
• Calculate the total cost to paint semi-cube: total cost = cuboid space × cost of paint per square centimeter = 288 × 100 = 28,800 currency cash.
• Sixth example: grandfather total mass of the four iron bars on semi-cube-shaped if the length of each 0.2 m, and width of 0.1 m and a height of 0.6 m, note that the mass per cubic centimeter is 8 grams.  Solution: To calculate the mass of iron bars must first calculate the size, using the law: semi-cube size = length × width × height = 0.2 × 0.1 × 0.6 = 0.012 m 3. Size conversion from cubic meter to cubic centimeter, so beat at 1,000,000 to become: 0.012 × 1,000,000 = 12,000 cm 3. Hit the mass of each cubic centimeter-sized semi-cube to calculate the mass of the iron penis per: penis per block = 12,000 × 8 = 96,000 g, block eight bars = 4 × 96,000 = 384,000 g = 384 kg.
• The solution:
• To calculate the mass of iron bars must first calculate the size, using the law: semi-cube size = length × width × height = 0.2 × 0.1 × 0.6 = 0.012 m 3.
• Size conversion from cubic meter to cubic centimeter, so beat at 1,000,000 to become: 0.012 × 1,000,000 = 12,000 cm 3.
• Hit the mass of each cubic centimeter-sized semi-cube to calculate the mass of the iron penis per: penis per block = 12,000 × 8 = 96,000 g, block eight bars = 4 × 96,000 = 384,000 g = 384 kg.

## References

• ↑ "Cube Hexahedron", www.mathsisfun.com, Retrieved 29-11-2017. Edited.
• ↑ "Cuboid", www.mathworld.wolfram.com, Retrieved 9-12-2017. Edited.
• ^ A b T. Rajai Samih al-Assar, Jawad Younis Abu Halil, Mohammed Zuhair Abu Sabih (2013), to the Olympics and mathematics competitions (first edition), Riyadh: King Fahd University of Petroleum and Minerals Deanship Search Alalma_mketbh Obeikan, page 85-90, part the first. Adapted.
• ^ أ ب "What is a Cuboid Shape? - Definition, Area & Properties", www.study.com, Retrieved 9-12-2017. Edited.
• ↑ "Cuboids, Rectangular Prisms and Cubes", www.mathsisfun.com, Retrieved 9-12-2017. Edited.
• ↑ "Spinning Cube (Hexahedron) ", www.mathopolis.com, Retrieved 21-4-2020. Edited.
• ↑ "Surface Area of a Cuboid", www.web-formulas.com, Retrieved 21-4-2020. Edited.
• ^ أ ب ت Lata Wishram, UNDERSTANDING MATHEMATICS, Page 323. Edited.

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