ABC is an isosceles triangle inscribed in a circle. If AB = AC = 12$$\sqrt 5 $$ and BC = 24 cm then radius of circle is:

Solution(By Examveda Team)

$$\eqalign{ & {R_2} = \frac{{abc}}{{4\vartriangle }} \cr & \vartriangle = \sqrt {S\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \cr} $$
$$ = \sqrt {12\left( {\sqrt 5 + 1} \right)\left( {12} \right) \times 12 \times 12\left( {\sqrt 5 - 1} \right)} $$
where a = 12$$\sqrt 5 $$ , b = 12$$\sqrt 5 $$  & c = 24
$$\eqalign{ & S = \frac{{a + b + c}}{2} \cr & S = \frac{{24\sqrt 5 + 24}}{2} \cr & S = 12\left( {\sqrt 5 + 1} \right) \cr & {R_2} = \frac{{12\sqrt 5 \times 12\sqrt 5 \times 24}}{{4 \times 12 \times 12 \times 2}} \cr & {R_2} = \frac{{30}}{2} \cr & {R_2} = 15\,{\text{cm}} \cr} $$