Let x=OD, OA=OD+AD=x+2. Let y=angle AOB=DOC.
xy=2cm the length of arc DC; (x+2)y=2.8cm the length of arc AB.
xy+2y=2.8 so 2+2y=2.8 and 2y=0.8 therefore y=0.4 radians=angle AOB.
(x+2)0.4=2.8; x+2=7 so x=5cm. The radius r=x+2=7cm.
If A=area of sector AOB, then A/(πr^2)=y/(2π), so A=yπr^2/2π=r^2y/2=49*0.4/2=9.8 sq cm.
B=area of sector AOC, r=5, B=25*0.4/2=5 sq cm.
Area of ABCD=9.8-5=4.8 sq cm.