∫AB is a chord in the circle O.
Perpendicular bisector with OD of AB
∴∠AOD=∠BOD=∠AOB/2=60
∴OA=2OE
AB = 6
∴AE=BE=AB/2=3
Let OE be x and OA be 2x.
In RT△AOE, OA 2-OE 2 = AE 2.
Solution: x=√3
∴S△OAB=AB*OE/2=6√3/2=3√3
Hope to adopt! thank you