The correct option is A 5082
Given sector is OAPB,
Area of the minor sector
=θ360∘×πr2
=30∘360∘×π×r2
=112×πr2
=πr212
Area of the major sector=Area of the circle−Area of the minor sector
Area of the major sector=(πr2 −πr212) cm2
Area of the major sector=(12πr2−πr212) cm2
Area of the major sector=(11πr212) cm2
Now, using π=227
Area of the major sector=(11×22×42×4212×7) cm2
Area of the major sector=11×11×42 cm2
Area of the major sector=121×42 cm2
Area of the major sector=5082 cm2