Q68 / 160 · TS EAMCET 2018 (05 May Shift 2) · Mathematics

In $\triangle A B C_1 r_1+r_2+r_3=$

  • Correct answer
    A
    $4 R$
  • B
    $4 R+r$
  • C
    $4 R-r$
  • D
    $4 R+s^2$

Solution

We know that $$ \begin{aligned} r_1 & =\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b} \text { and } r_3=\frac{\Delta}{s-c} . \\ \therefore \quad & r_1+r_2+r_3=\frac{\Delta}{s-a}+\frac{\Delta}{s-b}+\frac{\Delta}{s-c} \\ & =\Delta\left[\frac{1}{s-a}+\frac{1}{s-b}\right]+\frac{\Delta}{s-c}-\frac{\Delta}{s}+\frac{\Delta}{s} \\ & =\Delta\left[\frac{s-b+s-a}{(s-a)(s-b)}\right]+\frac{\Delta(s-s+c)}{s(s-c)}+\frac{\Delta}{s} \\ & =\frac{\Delta c}{(s-a)(s-b)}+\frac{\Delta c}{s(s-c)}+\frac{\Delta}{s} \\ & =\Delta c\left[\frac{s^2-c s+s^2-a s-b s+a b}{s(s-a)(s-b)(s-c)}\right]+\frac{\Delta}{s} \\ & =\left[\frac{\Delta c\left[2 s^2-s(a+b+c)+a b\right.}{s(s-a)(s-b)(s-c)}\right]+\frac{\Delta}{s} \\ & =\frac{\Delta c\left[2 s^2-2 s^2+a b\right]}{\Delta^2}+\frac{\Delta}{s} \end{aligned} $$ $$ \begin{aligned} & =\frac{a b c}{\Delta}+\frac{\Delta}{s} \\ & =4 R+r \quad\left[\because r=\frac{\Delta}{s} \text { and } R=\frac{a b c}{4 \Delta}\right] \end{aligned} $$

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

Relevant Questions