If for a triangle, the radius of the circumcircle is double the radius of the inscribed circle (incircle), then which one of the following is correct?

Concept:

Let the side of the triangle are a, b & c

Semi perimeter S =\(\frac{a\ +\ b\ +\ c}{2}\) 

Circumradius R = \(\frac{abc}{{4}{\times Area \ of \ triangle}}\ =\ \frac{abc}{4Δ}\)

Inradius r = \(\frac {area \ of \ triangle}{semi \ perimeter \ of \ triangle}\ =\ \frac{Δ}{S}\)

In an equilateral triangle, the ratio of the radius of the circumcircle to that of the incircle Is 2:1. i.e. R = 2r

Calculation:

Let a, b, c be the sides of a triangle,

Δ = area of the triangle,

S = semi-perimeter \(= \frac{a\ +\ b\ +\ c}{2}\)

\(R\ =\ \frac{abc}{4Δ}\) and \(r\ =\ \frac{Δ}{S}\)

According to the question, the radius of the circumcircle is double the radius of the incircle

\(⇒\ \ \frac{abc}{4Δ}\ =\ \frac{2\Delta}{S}\)

\(⇒\ \ {abc}\ =\ \frac{8\Delta^2}{S}\)

\(⇒\ \ {abc}\ =\ \frac{8[S(S\ -\ a)(S\ -\ b)(S\ -\ c)]}{S}\)

⇒ abc = (b + c − a)(c + a − b)(a + b − c)

This will be true only if, a = b = c

Hence, the triangle is an equilateral.