Question

An equilateral triangle of side 9cm is inscribed in a circle find height of equilateral triangle

Answer 1



∆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 3√3cm