an equilateral triangle of side 9 cm is inscribed in a circle.Find the radius of the circle
∆ABC is an equilateral triangle.
AB = BC = CA = 9 cm
O is the circumcentre of ∆ABC.
∴ OD is the perpendicular bisector of the side BC. (O is the point of intersection of the perpendicular bisectors of the sides of the triangle)
In ∆OBD and ∆OCD,
OB = OC (Radius of the circle)
BD = DC (D is the mid point of BC)
OD = OD (Common)
∴ ∆OBD ≅ ∆OCD (SSS congruence criterion)
⇒ ∠BOD = ∠COD (CPCT)
∠BOC = 2 ∠BAC = 2 × 60° = 120° ( The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle)
Thus, the radius of the circle is .