Consider a triangle ABC and let a,b,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 $A CB$ is obtuse and if r denotes the radius of the in circle of the triangle then r² is equal to

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answer is 3.

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Detailed Solution

$Δ = 15 \sqrt{3}$
Area of triangle $= \frac{1}{2} ab \sin c$
$\begin{array}{l} 15 \sqrt{3} = \frac{1}{2} (6) (10) \sin c \\ \sin c = \frac{\sqrt{3}}{2} \Rightarrow c = 120^{\circ} (c is obtuse) \\ cosc = \frac{a^{2} + b^{2} - c^{2}}{2 ab} \\ \cos 120 = \frac{36 + 100 - c^{2}}{120} \\ c^{2} = 196 \Rightarrow c = 14 \\ s = \frac{a + b + c}{2} = 15 \end{array}$
$\begin{array}{l} r = \frac{Δ}{s} = \frac{15 \sqrt{3}}{15} = \sqrt{3} \\ r^{2} = 3 \end{array}$