Question

In $△ABC$, with usual notation, if $r=r_1-r_2-r_3$, then $2R=$

• A. $a$
• B. $b+c$
• C. $C$
• D. $c+a$

Answer 1

Answer: (A)

In a $△ABC$, it is given that,
$r=r_1-r_2-r_3$
$\Rightarrow \frac{\Delta }{s}=\frac{\Delta }{s-a}-\frac{\Delta }{s-b}-\frac{\Delta }{s-c}$
$\Rightarrow \frac{1}{s-b}+\frac{1}{s-c}=\frac{1}{s-a}-\frac{1}{s}$
$\Rightarrow \frac{2s-b-c}{(s-b)(s-c)}=\frac{a}{s(s-a)}$
$\Rightarrow s(s-a)=s^2-(b+c)s+bc$
$\Rightarrow s^2-sa=s^2-(b+c)s+bc$
$\Rightarrow (b+c-a)s=bc$
$\Rightarrow (b+c{)}^2-a^2=2bc$
$\Rightarrow b^2+c^2+2bc-a^2=2bc$
$\Rightarrow b^2+c^2=a^2$
$\therefore △ABC$ is the right angled triangle with angle $A=90^{\circ }$,
so $2R=a$.