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

Answer: Option B

Solution

Answer: Option B

Solution:

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