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Surface Area ofPyramids (and Cones) Section 10.5 P. 548 - 552
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• In this section, you will practice finding the “surface” area of pyramids. These pyramids will include both triangular and square bases. However, the base will be “regular” polygon – so either squares or equilateral triangles.• This is not true of all pyramids, however.
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Is the height of a lateral face (any face that isNOT the base)
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GUIDED PRACTICE for Example 1 Find the surface area. Round to the nearest tenth. 1. Area of the B = 8 x 8 = 64 ft2 Area of one lateral side is (8 x 12) / 2 = 48 4 Triangles x 48 = 192 Add 192 + 64 = 256 ft 2 is the surface area
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Area of the B = 6 x 6 = 36Area of one lateral side triangle = (6x11) /2 = 33 33 x 4 = 132Add 132 + 36 = 168 cm2 for the surface area
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GUIDED PRACTICE for Example 1 Area of the base + Area of the three lateral sides B = 62.4 One lateral Triangle (12 x 10) /2 = 60 60 x 3 = 180 (for the 3 Tri. Sides) SA = 180 + 62.4 = 242.4 in2
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Area of B is given = 84.9One lateral Triangle – (14 x 16) / 2 = 112 Time 3 (for the three triangle) 3 x 112 = 336Add 336 + 84.9 = 420.9 yd2 for the surface area
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Look at the sphere formula for surface area S = 4πr2You will NOT need to memorize this, butyou should be able to substitute in thevalues correctly and solve the problem. diameter= 9.4 S = 4πr2 S = 4 x 3.14 x 4.72 S = 277.5 in2
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• Assignment: P 550 #2-4, 17-19• Show all your steps – Draw the nets only if you want to• Remember – this is still area and labels will be square units.
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