Let DF be x and FC be y and the height of the rectangle (BC) be h. Asshown in figure below:



Now area AECF =2 and writing it in terms of our variable we get

area AECF =2= $y \times h$.

Similarly area EBFD =3 and writing it in terms of our variable we get

area EBFD =3= $x \times h$.

Now area of the rectangle is clearly =$(x+y) \times h$.

Clearly if we add the area of EBFD and AECF we get the area of rectangle = $(x+y) \times h= x \times h + y \times h = 3+2= 5$

Hence option C