Q70 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics

In $\triangle \mathrm{ABC}$, if $\mathrm{a}=7, \mathrm{~b}=8$ and $\mathrm{c}=9$ then $\frac{1}{\mathrm{r}_1^2}+\frac{1}{\mathrm{r}_2^2}+\frac{1}{\mathrm{r}_3^2}=$

  • A
    $\frac{97}{360}$
  • Correct answer
    B
    $\frac{5}{72}$
  • C
    $\frac{169}{360}$
  • D
    $\frac{67}{72}$

Solution

$\begin{aligned} & \text {} a=7, b=8, c=9 \\ & \frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}=\left(\frac{s-a}{\Delta}\right)^2+\left(\frac{s-b}{\Delta}\right)^2+\left(\frac{s-c}{\Delta}\right)^2 \\ & =\frac{(s-a)^2+(s-b)^2+(s-c)^2}{\Delta^2} \\ & =\frac{5^2+4^2+3^2}{12(5 \cdot 4 \cdot 3)}=\frac{50}{720}=\frac{5}{72} .\end{aligned}$

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

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