geometry

I don't have enough tiem to do it properly but here are some ideas.

The area of sector is 60/360 * area of the whole circle.

the quad is a rhombus which means both are equilateral triangles.

Find the area of one triangle. Might want to find height first though not strictly necessary.

subtract and then double to get shaded area.

Sorry do not have time to write instructions well.

We can use the same idea here, tertre, as we did the other day on this type of problem

The line of symmetry divides the shaded area into equal halves

If we only consider the left half.....we will have the sector of a circle QRS  which will be (60° /360°) =  1/6 the area of a circle with a radius of 12  = (1/6) pi (12)^2   =  144/6  pi     =  24pi   units^2   (1)

Now....cosider triangle QRS...its area is  (1/2) (12)^2  sin (60°)  = 72√3 / 2   =   36√3  units^2   (2)

So   1/2   of the shaded area  = (1)  - (2)  =

[24 pi - 36√3]  unts^2  =    12  [ 2pi - 3√3 ] units^2

So....twice this is the shaded region's area  =   24 [ 2pi - 3√3 ] units^2

Oh, ok! Thank you both CPhill and Melody! 

Area of Segment = 1/2R^2[x - sin(x)]  = 60 degrees =pi/3(Radians)
                                  =1/2*144[pi/3 - sqrt(3)/2]
                                  =72 (π/3 - sqrt(3)/2) - units^2 - area of one segment
                                  =144 (π/3 - sqrt(3)/2) - units^2 

                                  =24[2π - 3sqrt(3)] - units^2 - Area of shaded segments.