Question

If in a triangle $ABC$, $r_1=2,r_2=3$ and $r_3=6$, then $a=$

Answer 1

We know that
$r_1=\frac{\Delta }{s-a}=2\Rightarrow a=s-\frac{\Delta }{2}...\left(i\right)$
$r_2=\frac{\Delta }{s-b}=3\Rightarrow b=s-\frac{\Delta }{3}...\left(ii\right)$
$r_3=\frac{\Delta }{s-c}=6\Rightarrow c=s-\frac{\Delta }{6}...\left(iii\right)$
Adding equation $\left(i\right),\left(ii\right)$ and $\left(iii\right)$, we get
$a+b+c=3s-\Delta$
$\Rightarrow s=\Delta$
Now, Area of triangle $ABC$
$\Rightarrow \Delta =\sqrt{\Delta \left(\frac{\Delta }{2}\right)\left(\frac{\Delta }{3}\right)\left(\frac{\Delta }{6}\right)}$
$\Rightarrow \Delta =\frac{{\Delta }^2}{6}$
Now, $\Delta =6=s$
$r_1=\frac{\Delta }{s-a}$
$2=\frac{6}{6-a}$
$\therefore a=3$