Q69 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics

In a triangle $A B C, \frac{a\left(m_1+r_2 m_3\right)}{n_1-r+r_2+r_3}=$

  • Correct answer
    A
    $\sqrt{r_1 r_2 r_3}$
  • B
    $r_1 r_2+r_2 r_3+r_3 r_1$
  • C
    $2(R+r)$
  • D
    $2+\frac{r}{2 R}$

Solution

$r_1-r+r_2+r_3=\frac{\Delta}{s-a}-\frac{\Delta}{s}+\frac{\Delta}{s-b}+\frac{\Delta}{s-c}$ $\begin{aligned} & =\Delta\left[\begin{array}{l}s(s-b)(s-c)-(s-a)(s-b)(s-c) \\ +s(s-a)(s-c)+s(s-a)(s-b) \\ s(s-a)(s-b)(s-c)\end{array}\right] \\ & =\Delta\left[\frac{(s-b)(s-c)[s-s+a]+s(s-a)[s-c+s-b]}{\Delta^2}\right] \\ & =\Delta a\left[\frac{(s-b)}{\Delta} \frac{(s-c)}{\Delta}+\frac{s}{\Delta} \frac{(s-a)}{\Delta}\right] \\ & =\Delta a\left[\frac{1}{r_2} \times \frac{1}{r_3}+\frac{1}{r} \cdot \frac{1}{r_1}\right]=\Delta a\left[\frac{rr_1 r_2 r_3}{rr_1 r_2 r_3}\right] \\ & \Rightarrow \frac{a\left(r r_1+r_2 r_3\right)}{r_1-r+r_2+r_3}=\frac{rr_1 r_2 r_3}{\Delta}=\frac{rr_1 r_2 r_3}{\sqrt{rr_1 r_2 r_3}}=\sqrt{rr_1 r_2 r_3} .\end{aligned}$

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

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