Experts please answer it fast.

Experts please answer it fast. An equilateral triangle of side 6 cm is inscribed in a circle. Find

Dear student,

∆ABC is an equilateral triangle.

AB = BC = CA = 6cm

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, Radius of the circle=2√3 cm
Regards