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anonymous
 5 years ago
Please help.. I need to find the area of the shaded regions
anonymous
 5 years ago
Please help.. I need to find the area of the shaded regions

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how would it be 67 for #11

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Fr the first image notice that half of the annulus is shaded so ½(36ππ16) = Sum of area of shaded parts

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For #10. (49ππ) = area of non shaded part

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I take that back. ans for #10 is wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Here's a helpful diagram: http://www.gmathacks.com/pdfs/equiTriCirc.jpg OP = 6 since OPB is a special triangle, PB = 6√3 OB = OC = OA = 2*6=12 (because they are all equidistant) Area of triangle = ½*base*height base = 2*PB height = OP+OC base = 2*6√3 = 12√3 height = 6+12 = 18 Therefore area of shaded part = 108√3  36π

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There are a couple of ways to do #12 area of quadrant with shaded part = 1/4*area of circle =1/4*π*36 = 9π Area of triangle within quadrant = ½(6*6) = 18 Area of shaded part in first quadrant = 9π18 We have to shaded parts in the circle so total are of shaded part = 2(9π18) = 18π36 or 18(π2) Or area of half circle = 1/2*π*36 = 18π area of triangle = 1/2*12*6 = 36 Area of shaded parts = 18π36 or 18(π2)
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