Question
ABC is an isosceles triangle inscribed in a circle. If AB = AC = 12$$\sqrt 5 $$ and BC = 24 cm then radius of circle is:
Answer: Option B
$$\eqalign{
& {R_2} = \frac{{abc}}{{4\vartriangle }} \cr
& \vartriangle = \sqrt {S\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \cr} $$
$$ = \sqrt {12\left( {\sqrt 5 + 1} \right)\left( {12} \right) \times 12 \times 12\left( {\sqrt 5 - 1} \right)} $$
where a = 12$$\sqrt 5 $$ , b = 12$$\sqrt 5 $$Â & c = 24
$$\eqalign{
& S = \frac{{a + b + c}}{2} \cr
& S = \frac{{24\sqrt 5 + 24}}{2} \cr
& S = 12\left( {\sqrt 5 + 1} \right) \cr
& {R_2} = \frac{{12\sqrt 5 \times 12\sqrt 5 \times 24}}{{4 \times 12 \times 12 \times 2}} \cr
& {R_2} = \frac{{30}}{2} \cr
& {R_2} = 15\,{\text{cm}} \cr} $$
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$$\eqalign{
& {R_2} = \frac{{abc}}{{4\vartriangle }} \cr
& \vartriangle = \sqrt {S\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \cr} $$
$$ = \sqrt {12\left( {\sqrt 5 + 1} \right)\left( {12} \right) \times 12 \times 12\left( {\sqrt 5 - 1} \right)} $$
where a = 12$$\sqrt 5 $$ , b = 12$$\sqrt 5 $$Â & c = 24
$$\eqalign{
& S = \frac{{a + b + c}}{2} \cr
& S = \frac{{24\sqrt 5 + 24}}{2} \cr
& S = 12\left( {\sqrt 5 + 1} \right) \cr
& {R_2} = \frac{{12\sqrt 5 \times 12\sqrt 5 \times 24}}{{4 \times 12 \times 12 \times 2}} \cr
& {R_2} = \frac{{30}}{2} \cr
& {R_2} = 15\,{\text{cm}} \cr} $$
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