Amsco Integrated Algebra I: Surface Areas of Solids
 This is a free lesson from our course in Amsco's Integrated Algebra

 This lesson explains the basics of and how to find the surface area of solids such as rectangular solid, cube and right circular cylinder. The surface area of a solid is the sum of the areas of the surfaces that makes up the solid. The surface area of a right prism is the sum of the areas of the bases and the faces of the solid. Here a right prism is a solid with congruent bases and with a height that is perpendicular to these bases. The number of faces of a prism is equal to the number of sides of a base. E.g. if the dimensions of a rectangular solid are represented by l, w & h, then surface Area of Rectangular solid = 2lw+2wh+2lh. Similarly in case of a cube, if s represents the length of each side, then surface area of the cube is 6 s2. (More text below video...)
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 (Continued from above) Further you’ll learn: A right circular cylinder is a solid with two bases that are circles of the same size, and with a height that is perpendicular to these bases. The surface area of circular cylinder is: 2πrh + 2πr2. E.g. to get the surface area of a right circular cylinder with the radius = 5mm and height of 10mm, you take two times pi times the radius, times the height. Then add that to two times pi times, the radius squared i.e. Surface Area = 2πrh + 2πr2 =2π*5*10 + 2π (5)2=100π + 50π= 150πmm2 Similarly if you work out the surface area of a square pyramid where the length of each side of the base is 10 meters and the slant height is also 10 meters, it answers 300m2.The video above explains in detail with the help of several examples.

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