I wonder if I can use the area of trapezium for the shape CAOB? I've abit of doubt for the arc slope as often, the slope of the trapezium appears to be a straight line. For part a) I'm just checking if I'm right, by taking the area of trapezium CAOB minus the area of the quadrant AOB to find the area of shaded region?
Area of Quadrant = 0.25*5*5*pi = 6.25pi
AC = AB = sqrt(5*5 + 5*5) = 5*sqrt(2)
Area of Trapezium = 0.5*(5+5*sqrt(2))*(5) = 30.1777
Area shaded = Area Trapezium - Area of Quadrant = 30.177 - 6.25pi = 10.5421 cm^2
Since $$S_{sector OAB}=S_{sector ACB},$$ it's just $$S_{sector OAB}-S_{segment AB}=S_{\Delta OAB}=\frac{5\cdot5}{2}=12.5.$$