Answer: Connect CA, CA', and the area of shaded part = the total area of irregular figure Abd' a '- the area of rectangular ABCD- sector DCD''

AB=？ √3，BC= 1，AC=2

S total = S△ABC+S sector ACA '+S△A ' CD '

=? √3/2+ 1/4π×22+√3/2=√3+π

S rectangle =√3

S sector DCD'= 1/4π×(√3)2=3/4π.

S shadow =(√3+π)-√3-3/4π= 1/4π.