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)
In ∆BOD,
Sin ∠BOD
Thus, the radius of the circle is .