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

Solution (By Examveda Team)

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