∫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