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Prisms Mathematical Definition- A polyhedron with two congruent faces that lie in a parallel plane Prism is composed of a base which is on both sides of the object and then lateral faces Lateral faces are every face of the prism that are not bases.

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Examples of Prisms

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Faces of a Prism The highlighted faces of this triangular prism are the bases, while the clear ones are the lateral faces. In this rectangular prism, the highlighted faces are again the base.

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Surface Area of Prism Using the net method is one way to find the area. You break the 3-D shape down into easy 2 dimensional shapes. The triangular prism to the right when breaks down into two triangles as the base and three rectangles as the lateral faces. ent/mejhm/html/object_interacti ves/surfaceArea/use_it.htmlhttp://www.learnalberta.ca/cont ent/mejhm/html/object_interacti ves/surfaceArea/use_it.html

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Formula for Surface Area Surface Area = 2(area of base) + (Perimeter of Base)(height) SA= 2(B)+ Ph

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Examples Ex SA= 2B+ Ph B= (2)(5) = 10 P= = 14 H= 10 =2(10)+ (14)10 =160 units²

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Examples SA= 2B+Ph B= ½ (4)(6)= 12 cm P= 4+6+7= 17 cm H= 12 cm SA= 2(12)+ 17(12) = =228 cm squared

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Surface Area of a Cylinder A cylinder has a total of three surfaces: a top, bottom, and middle. The top and bottom, which are circles To find the surface area we must separate the cylinder into these parts and find the area of each separately. The first parts are circles (top and bottom of cylinder). The area of a circle is πr^2. So, the area of two circles would be πr^2 + πr2 = 2πr^2.

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The third surface, the lateral surface area, is the curved wall of the cylinder. We can manipulate the curved wall of a cylinder to produce a recognizable shape. First start with the cylinder and make a cut down the middle as shown in Step One. Next the cut edge will be cut open as shown in Step Two. When the cylinder's wall is completely open, it makes the shape of a rectangle.

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To calculate the rectangles area, we need to go back to the top of the original cylinder which is a circle. The distance around a circle is called its circumference, C = 2πr. The circumference has been marked in red in the diagram. When the cylinder wall is completely open, we see that the circumference of the circle becomes the length of the final rectangle. The dimensions of the rectangle are the circumference, C = 2πr, and the height of the cylinder, h. So, the area of the rectangle is A = l x w = C x h = 2πr x h = 2πrh.

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The next step is to put all the surfaces together which include the top the bottom and the walls (middle). Those values are πr^2 + πr^2 + 2πrh = 2πr^2 + 2πrh. This can also be written as SA = 2B + Ph where B=area of base, P=perimeter of base, and h=height

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Example SA = 2πr^2 + 2πrh = 2(3.14)(6)^2 + 2(3.14)(6)(4) = = If r = 6 ft and h = 4 ft then what is the surface area?

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Surface Area of a Pyramid Pyramids that have a square base have a total of five surfaces: one base and four sides.

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The first step in finding the surface area is to find the the slant height value, l. The height of the pyramid is a leg of the right triangle. The base of the right triangle is half the length of the base edge of the pyramid, s. At this point, we would use The Pythagorean Theorem to calculate the slant height h^2 + (1/2s)^2 = l^2

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Now find the area of each surface The base is a square with dimensions, s, so its area is s x s = s^2 The other four surfaces are triangles with an area of ½(base)(height) or ½(s)(I) ½(s)(I) x 4 = 2(s)(l) The total surface area equation is s^2 + 2sl

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Example Given: s = 8 m and h = 3 m, find the surface area First we calculate the slant height: 3^2 + (½x8)^2 = (I)^ = (l)^2 25 = (I)^2 5 = I Next use the formula to calculate the surface area: SA = (s)^2 + 2(s)(l) = (7 in)^2 + 2(7 in)(6.95 in) = 49 in^ in^2 = in^2.

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Surface Area of Cones Separates into 2 parts: Base and Side Area of Base = r^2 Area of Side = (radius of base)(slant height) = r^2 + rS

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Example First identify your radius of base and slant height Use Pythagorean Theorem to find S 12^2 + 5^2 = 13^2 r^2 + rS = (12)^2 + (12)(13) = S 12

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Example 2 Find surface area of the collar Large cone – small cone rs - rs (6)(10) - (3)(5) =

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