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 $\angle ACB$ is obtuse and if $r$ denotes the radius of the incircle of the triangle, then $r^2$ is equal to ______.

Answer 1

Answer: 3

$\Delta =\frac{1}{2}ab\sin C$
$\Rightarrow \sin C=\frac{2\Delta }{ab}=\frac{2\times 15\sqrt{3}}{6\times 10}=\frac{\sqrt{3}}{2}$
$\Rightarrow C=120^{\circ }$
$\Rightarrow c=\sqrt{a^2+b^2-2ab\cos C}$
$=\sqrt{6^2+10^2-2\times 6\times 10\times \cos 120^{\circ }}=14$
$\therefore r=\frac{\Delta }{s}$
$\Rightarrow r^2=\frac{225\times 3}{{\left(\frac{6+10+14}{2}\right)}^2}=3$