In case of an equilateral triangle inscribed in a circle-

Radius of circle = 2/3 of height of the equilateral triangle

Why is this true?

In case of an equilateral triangle inscribed in a circle-

Radius of circle = 2/3 of height of the equilateral triangle

Why is this true?

let C be the centre of the circle, and A and D be points on the circle.

∴ CD and CA will be the radius of the circle.

let CD = CA = r.

since ∠DAB = 60° ( angle of equilateral triangle )

∠CAB = 30° since line CA bisects ∠DAB.

CA = r, ∠CAB = 30°

sin(x) = opposite/ hypotenuse

∴sin(30) = CB / CA

∴0.5 = CB/r

∴CB = 0.5r.

height of triangle is DB

DB = DC + CB

DB = r + 0.5r

DB = 1.5r

∴ height of triangle is 1.5 times radius

so radius is height divided by 1.5

radius is 2/3 height!