Question

Please help.. I need to find the area of the shaded regions

Answer 1

#11= 67

Answer 2

how would it be 67 for #11

Answer 3

Fr the first image notice that half of the annulus is shaded so
½(36π-π16) = Sum of area of shaded parts

Answer 4

For #10. (49π-π) = area of non shaded part

Answer 5

I take that back.
ans for #10 is wrong

Answer 6

ok

Answer 7

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

Answer 8

That was #11

Answer 9

so #11 is 865.92

Answer 10

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)