Consider a triangle ABC and let a, b and c denote the lengths of the sides opposite to vertices A, B and C respectively. Suppose a = 6, b = 10 and the area of the triangle is 15 \sqrt{3} . \textrm{ }\textrm{ }\textrm{ } If \textrm{ }\textrm{ }\textrm{ } \angle ACB is obtuse and if r denotes the radius of the incircle of the triangle, then r2 is equal to

Engineering

Mathematics

Circumradius Inradius and exradius

Question

Consider a triangle ABC and let a, b and c denote the lengths of the sides opposite to vertices A, B and C respectively. Suppose a = 6, b = 10 and the area of the triangle is $15 \sqrt{3} . If ∠ ACB$ is obtuse and if r denotes the radius of the incircle of the triangle, then r² is equal to

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$\frac{1}{2} absinC = 15 \sqrt{3}; 6 \times 10 \times sinC = 30$

$sinC = \frac{\sqrt{3}}{2} \Rightarrow C = 120 º {∵ C is obtuse}$

$\cos 120 º = \frac{36 + 100 - c^{2}}{2 \times 6 \times 10} = \frac{- 1}{2} \Rightarrow 136 - c^{2} = - 60 \Rightarrow c^{2} = 196 \Rightarrow c = 14$

$∴ s = \frac{a + b + c}{2} = \frac{6 + 10 + 14}{2} = 15$

$\Rightarrow r = \frac{Δ}{s} = \frac{15 \sqrt{3}}{15} = \sqrt{3} \Rightarrow r^{2} = 3$

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