(i) arc AB/20(pi)=120/360=1/3, so AB=20(pi)/3 (one third of the circumference)=20.94cm.
(ii) draw perpendicular from O to chord AB at C to make two back-to-back triangles of equal area with OC as the common side. Angle AOC=BOC=60 degrees. AC/OA=sin60=sqrt(3)/2, AC=5sqrt(3), because OA=10. OC=OAcos60=10/2=5. Triangle OCB is congruent to OCA, so they have the same area=(1/2)AC*OC=(1/2)25sqrt(3). Therefore the area of AOB=25sqrt(3)=43.3 sq cm.
(iii) area of major segment AB/area of circle=(360-120)/360=240/360=2/3, therefore area=200(pi)/3=209.44 sq cm.