please help

The circle has area 64pi and that means angle AOB is 60 degrees since it's equalateral so we have 32pi/3 for the sector and for the triangle is 8*4sqrt3=32sqrt3 we then divide by 2 to get 16sqrt3 we subtract to get 32pi/3-16sqrt3 so we have \(\boxed{(B)\dfrac{32\pi}3-16\sqrt3}\)

Angle o is 60 degrees (equilateral triangle)

60 / 360    x    pi  x 8^2 -   8^2 sqrt3/4                 red is area of equilateral triangle  

AO = OB = AB =>  a

Area = 1/6 [ a² pi - a² (3√3) / 2]