Question 1186471


The area of the sector is {{{A[s] = ((pi/6)/(2*pi))*pi*r^2 = (1/12)*pi*4^2 = (4*pi)/3}}} sq.ft.


The area of the triangle defined by the radius and the half-radius is given by {{{A[t] = (1/2)*r*(r/2)*sin(theta) = (1/2)*4*(4/2)*sin(pi/6)  = 4*(1/2) = 2}}}sq.ft.


Therefore the area of the desired region is {{{A[s] - A[t] = (4*pi)/3 - 2}}} ≈ 2.18879 sq.ft, to 5 d.p.