Q67 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Mathematics

In a $\triangle A B C$, with usual notation, if $r=r_1-r_2-r_3$, then $2 R=$

  • Correct answer
    A
    $\mathrm{a}$
  • B
    $b+c$
  • C
    $\mathrm{C}$
  • D
    $c+a$

Solution

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

Hence, the correct answer is option: 'A'.

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