Question

In $△ABC,\angle C>\frac{\pi }{2},a=6,b=10$ and Area of the triangle is $15\sqrt{3}$. What is the value of $r^2$?

• A. 1
• B. 3
• C. 5
• D. 2

Answer 1

The correct option is B 3

$\Delta =\frac{1}{2}ab\sin C⟹15\sqrt{3}=\frac{1}{2}\times 6\times 10\times \sin C$
$⟹\sin C\frac{\sqrt{3}}{2}⟹\angle C={120}^{\circ }$

$c^2=a^2+b^2–2ab\cos C=36+100-120\times \frac{-1}{2}=196⟹c=14$

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

$r=\frac{\Delta }{s}=\frac{15\sqrt{3}}{15}=\sqrt{3}⟹r^2=3$