Q67 / 160 · AP EAMCET 2024 (23 May Shift 1) · Mathematics

$\text { In } \triangle \mathrm{ABC}, \frac{r_2\left(r_1+r_3\right)}{\sqrt{r_1 r_2+r_2 r_3+r_3 r_1}}=$

  • A
    $a$
  • Correct answer
    B
    $b$
  • C
    $c$
  • D
    $s$

Solution

$\begin{aligned} & \frac{r_2\left(r_1+r_3\right)}{\sqrt{r_1 r_2+r_2 r_3+r_3 r_1}} \\ & =\frac{\frac{\Delta}{s-b}\left(\frac{\Delta}{s-a}+\frac{\Delta}{s-c}\right)}{\sqrt{\frac{\Delta^2}{(s-a)(s-b)}+\frac{\Delta^2}{(s-b)(s-c)}+\frac{\Delta^2}{(s-c)(s-a)}}} \\ & =\frac{\frac{\Delta^2}{s-b}\left(\frac{s-c-s-a}{(s-a)(s-c)}\right)}{\Delta \sqrt{\frac{s-c+s-a+s-b}{(s-a)(s-b)(s-c)}}}=\frac{\frac{\Delta(2 s-c-a)}{(s-a)(s-b)(s-c)}}{\sqrt{\frac{3 s-(a+b+c)}{(s-a)(s-b)(s-c)}}} \\ & =\frac{\Delta b}{(s-a)(s-b)(s-c)} \times \frac{\sqrt{(s-a)(s-b)(s-c)}}{\sqrt{s}} \\ & =\frac{\Delta b}{\sqrt{s(s-a)(s-b)(s-c)}}=b \end{aligned}$

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

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