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 .

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