Mathematics Volume of Solids Formulae for Volume of Solids Cube | Cuboid | Triangular Prism | Cylinder | Cone | Pyramid | Sphere | AnyPrism | s3 | lwh | ½bhl | Πr2h | 1/3πr2h | 1/3Ah | 4/3πr3 | Ah | A = area of the base of the figure s = length of a side of the figure l = length of the figure w = width of the figure h = height of the figure π = 22/7 or 3.14 1. Compute the volume of a cube with side 7cm. Volume of cube: s3 s = 7cm s3 = (7cm x 7cm x 7cm) = 343cm3 2. Compute
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AREA (i) The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude are 24.8 cm and 16.5 cm respectively. If one of the diagonal of the rhombus is 22 cm‚ find the length of the other diagonal. (ii) The floor of a rectangular hall has a perimeter 250m. If the cost of paining the four walls at the rate of Rs 10 per m2 is Rs 1500. Find the height of the hall. (iii) A room is half as long again as it is broad. The cost of carpeting the room at Rs
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In Parts A and C‚ the relationship between surface area and volume was investigated. Plasticine was formed into a cube and a sphere; both shapes were cut in half. It was found that plasticine volume should not vary‚ two halves have a greater surface area than a whole‚ and cubes have a greater surface area than spheres of the same volume. In Part B‚ the relationship between diffusion and surface area to volume ratio was investigated. Three agar-phenolphthalein-sodium hydroxide cubes of different sizes
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Investigating Ratios of Areas and Volumes In this portfolio‚ I will be investigating the ratios of the areas and volumes formed from a curve in the form y = xn between two arbitrary parameters x = a and x = b‚ such that a < b. This will be done by using integration to find the area under the curve or volume of revolution about an axis. The two areas that will be compared will be labeled ‘A’ and ‘B’ (see figure A). In order to prove or disprove my conjectures‚ several different values for n will
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diffusion. Diffusion is the process that cells use to obtain oxygen‚ water and food. Also‚ how they lose waste substances‚ for example‚ urea and carbon dioxide. Basically‚ Diffusion is when particles move from an area of high concentration to an area of low concentration. The surface area to volume ratio of the cell is an important factor in diffusion. It is the effect of this factor that will be investigated in this practical. Materials used in the practical consist of blocks of agar jelly containing
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or be found in multicellular organisms. The unicellular and multicellular organisms are linked to cell size and surface area to volume ratio. The experiment for cell size and diffusion was set to see how and how much water can go to the cells. This movement of water is called Osmosis. Osmosis is the movement of water molecules from an area of low concentration (lots of water) to an area of high concentration (little water) through a semi permeable membrane‚ demonstrated in ‘figure 1’. A semi permeable
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Surface area Surface area is the measure of how much exposed area a solid object has‚ expressed in square units. Mathematical description of the surface area is considerably more involved than the definition of arc length of a curve. For polyhedra (objects with flat polygonal faces) the surface area is the sum of the areas of its faces. Smooth surfaces‚ such as a sphere‚ are assigned surface area using their representation as parametric surfaces. This definition of the surface area is based on methods
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Risk Assessment 5 Method 6 Results 6 Analysis/Discussion 7 Conclusion 8 Acknowledgments/Bibliography 8 Table of Contents Background Research As an object becomes bigger its surface area and volume increases but the surface area to volume ratio decreases‚ this is because volume increases quicker than the surface area; as volume is three dimensional. This concept applies to cells and reaction rate because cells need to absorb their food‚ water and oxygen through their cell membrane and depending on
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Intro: Surface Area and Volume Multiple Choice Identify the choice that best completes the statement or answers the question. Find the surface area of the space figure represented by the net. ____ 1. 12 in. 4 in. 6 in. 4 in. 4 in. 6 in. a. 288 in.2 ____ 2. b. 144 in.2 c. 240 in.2 d. 288 in.2 5 cm 5 cm 7 cm 8 cm 4 cm ____ a. 124 cm2 b. 110 cm2 c. 150 cm2 d. 164 cm2 3. Find the surface area of the cylinder. Use a calculator. Round to the nearest tenth. 4m 3m a
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Surface Area Formulas In general‚ the surface area is the sum of all the areas of all the shapes that cover the surface of the object. Cube | Rectangular Prism | Prism | Sphere | Cylinder | Units Note: "ab" means "a" multiplied by "b". "a2" means "a squared"‚ which is the same as "a" times "a". Be careful!! Units count. Use the same units for all measurements. Examples |Surface Area of a Cube = 6 a 2
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