The radius of the circle passing through the vertices of the triangle $A B C$ , is

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Solution

given $A B = 12, B C = 6, C A = 12$ $⟹ a = 6, b = 12, c = 12$
$s = \frac{a + b + c}{2} = \frac{6 + 12 + 12}{2} = 15 ⟹ s - a = 15 - 6 = 9$
$s - b = 15 - 12 = 3$
$s - c = 15 - 12 = 3$
$Δ = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{15 (9) (3) (3)} = 9 \sqrt{15}$
$r = \frac{Δ}{s} = \frac{9 \sqrt{15}}{15} = \frac{3 \sqrt{15}}{5}, r_{1} = \frac{Δ}{s - a} = \frac{9 \sqrt{15}}{9} = \sqrt{15}, r_{2} = r_{3} = \frac{Δ}{s - b} = \frac{9 \sqrt{15}}{3} = 3 \sqrt{15}$
$r_{1} + r_{2} + r_{3} - r = 4 R ⟹ \sqrt{15} + 3 \sqrt{15} + 3 \sqrt{15} - \frac{3 \sqrt{15}}{5} = 4 R ⟹ R = \frac{8 \sqrt{15}}{5}$

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