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

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is A $3$

$\Delta =\frac{1}{2}ab\sin C$

$\Rightarrow \sin C=\frac{2\Delta }{ab}=\frac{2\times 15\sqrt{3}}{6\times 10}=\frac{\sqrt{3}}{2}$

$\Rightarrow C={120}^o$

$\Rightarrow c=\sqrt{a^2+b^2-2ab\cos C}$

$=\sqrt{6^2+{10}^2-2\times 6\times 10\times \cos {120}^o}=14$

$\therefore r=\frac{\Delta }{s}$

$\Rightarrow r^2=\frac{225\times 3}{{\left(\frac{6+10+14}{2}\right)}^2}=3$.