Asquare is inscribed in a circle. a point in the figure is selected at random. find the probability that the point will be in the shaded region.

Let the side of the square = x . therefore, the diameter of the square is half the side = x/2, and radius = x/4 

P( selecting point on the not shaded ) = not shaded area/whole area 
P( selecting point on the not shaded ) = ( the whole square - the four shaded circles ) / the whole square 

P( selecting point on the not shaded ) = ( x^2 - 4 * ( π * r^2 ) )/ x^2 
P( selecting point on the not shaded ) = ( x^2 - 4 * ( π * (x/4)^2 ) )/ x^2 
P( selecting point on the not shaded ) = ( x^2 - 4 * ( π * x^2/16 ) )/ x^2 
P( selecting point on the not shaded ) = ( x^2 - π * x^2/4 )/ x^2 
P( selecting point on the not shaded ) = x^2( 1 - π/4 )/ x^2 
P( selecting point on the not shaded ) = ( 1 - π/4 ) 
P( selecting point on the not shaded ) = (4 - π)/4 ≈ 0.21 ≈ 21% < close to C. 20% 

 
2) 
P( selecting point on the shaded ) = shaded area/whole area 
P( selecting point on the shaded ) = ( the four shaded circles ) / the whole square 
P( selecting point on the shaded ) = ( 4 * ( π * r^2 ) )/ x^2 
P( selecting point on the shaded ) = ( 4 * ( π * (x/4)^2 ) )/ x^2 
P( selecting point on the shaded ) = ( 4 * ( π * x^2/16 ) )/ x^2 
P( selecting point on the shaded ) = ( π * x^2/4 )/ x^2 
P( selecting point on the shaded ) = x^2( π/4 )/ x^2 
P( selecting point on the shaded ) = π/4 ≈ 0.7854 ≈ 79% < close to A. 80% 

 
3) 
I need to show a picture of what we are going to refer to. the picture is: 
http://imageshack.us/photo/my-images/59/... 

reading about equilateral triangle, "In an equilateral triangle, the internal angle bisectors,..." : 
http://www.algebra.com/algebra/homework/... 

let the side of the equilateral triangle = x 

radius of circle: 
tan(30) = r/(x/2) 
1/√(3) = 2r/x 
x/2√(3) = r 
x/(2√(3)) * (√(3)/√(3)) = r > rationalize 
√(3)x/(2 * 3) = r > rationalize 
√(3)x/6 = r 

P( selecting point on the shaded ) = shaded area/whole area 
P( selecting point on the shaded ) = ( the shaded circle ) / the whole triangle 
P( selecting point on the shaded ) = ( π * r^2 )/ ( √(3) * s^2/4 ) 
P( selecting point on the shaded ) = ( π * (√(3)x/6)^2 )/ ( √(3) * x^2/4 ) 
P( selecting point on the shaded ) = ( π * 3x^2/36 )/ ( √(3) * x^2/4 ) 
P( selecting point on the shaded ) = ( π * x^2/12 )/ ( √(3) * x^2/4 ) 
P( selecting point on the shaded ) = ( π/12 )/ ( √(3)/4 ) 
P( selecting point on the shaded ) = ( π/12 ) * 4/√(3) 
P( selecting point on the shaded ) = ( π/3 ) * 1/√(3) ≈ 0.605 ≈ 61% < close to C. 60%       
I hope this helps :)