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An equilateral triangle of side 9cm inscribed a circle,find the radius of the circle. |
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Answer» ∆ABC is an equilateral triangle.AB = BC = CA = 9 cmO 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 3√3 awesome another method plzzzz |
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