Solution :

Given :ABCD is a TRAPEZIUM with AE and BF are perpendicular to CD side of trapezium and,

AD= BC = 25cm

Inside the trapezium there is a circle of radius 3.5cm and if we clear notice the diagram diameter of the circle will give us the LENGTH of EF,AB,AE,BF.

Diameter of circle (2r) = length of EF 

So, EF = 2× 3.5

EF = AB = AE = BF = 7cm

Length of DE =√(AD² - AE²) 

Length of DE =√(25² - 7²) 

Length of DE =√(625 - 49) 

Length of DE =√576 = 24cm

Since BC = AD and BF = AE 

DE = FC = 24cm

Area of trapezium ABCD = (AE/2)(AB + DC)

AE =7cm , AB = 7cm , DC = DE + EF + FC = 24 + 7 + 24 = 55cm 

Area of trapezium ABCD = (7/2)(7 + 55)

Area of trapezium ABCD = (3.5)(62)

Area of trapezium ABCD =217cm²

Area of trapezium ABCE = (BF/2)(AB + EC)

BF = 7cm , AB = 7cm , EC = EF + FC = 7 + 24 = 31cm

Area of trapezium ABCE = (7/2)(7 + 31)

Area of trapezium ABCE = (3.5)(38)

Area of trapezium ABCE = 133cm²

Required ratio:

Area of trapezium ABCD : Area of trapezium ABCE 

= 217 : 133

= 31 : 19

So, the correct option is 3).31 : 19