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!