anonymous
  • anonymous
Please help.. I need to find the area of the shaded regions
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
#11= 67
anonymous
  • anonymous
how would it be 67 for #11

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anonymous
  • anonymous
Fr the first image notice that half of the annulus is shaded so ½(36π-π16) = Sum of area of shaded parts
anonymous
  • anonymous
For #10. (49π-π) = area of non shaded part
anonymous
  • anonymous
I take that back. ans for #10 is wrong
anonymous
  • anonymous
ok
anonymous
  • anonymous
Here'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
  • anonymous
That was #11
anonymous
  • anonymous
so #11 is 865.92
anonymous
  • anonymous
There 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 = 9π-18 We have to shaded parts in the circle so total are of shaded part = 2(9π-18) = 18π-36 or 18(π-2) Or area of half circle = 1/2*π*36 = 18π area of triangle = 1/2*12*6 = 36 Area of shaded parts = 18π-36 or 18(π-2)
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