find the area of the shaded region

Let C and D be the centre of the circle where AD = 12cm and BC = 4 cm

In triangle CDE,

CD² = DE² + CE²

CE² = CD² - DE²

CE² = 16² - 8² = 192

So, CE = 83

i.e Height of the trapezium, h = 83

Area of the trapezium = 1/2 x (BC + AD) x h

                                       = 1/2 x 16 x 83

                                       = 643 cm²

In triangle CDE, 

tan x = 83  / 8

x = 60^o

So, FCE = 30^o          

i.e BCF = 90^o + 30^o = 120^o   

Now, Area of sector ADF =   x (12)² x (60/360) = 75.36 cm²

Now, Area of sector BFC =   x (4)² x (120/360) = 16.74 cm²

Area of shaded region =Area of Trapezium ABCD - (Area of Sector ADF + Area of Sector BFC)

                                     = 643 - (75.36 + 16.74)  = 18.75 cm²