Q70 / 160 · TS EAMCET 2023 (13 May Shift 1) · Mathematics
If the roots of the equation $x^3-11 x^2+.36 x-36=0$ are the ex-radii of a triangle $A B C$ then the perimeter of the triangle $\mathrm{ABC}$ is
- A$24$
- B$18$
- Correct answerC$12$
- D$9$
Solution
$x^3-11 x^2+36 x-36=0$
$\begin{aligned} & (x-2)(x-3)(x-6)=0 \\ & \therefore \quad r_1=2, r_2=3, r_3=6 \\ & \Rightarrow \quad \frac{1}{r}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6} \\ & \therefore \quad r=1 \Rightarrow r=\frac{\Delta}{s}=1 \\ & \therefore \quad \Delta=s \\ & \Rightarrow \quad r_1=\frac{\Delta}{s-a}=\frac{s}{s-a} \\ & \therefore \quad r_2=\frac{s}{s-b} \text { and } r_3=\frac{s}{s-c} \\ & \quad r_1=2 \Rightarrow \frac{s}{s-a}=2\end{aligned}$
$\Rightarrow, s=2 s-2 a \Rightarrow 2 a=s$ ...(i)
$r_2=3 \Rightarrow \frac{s}{s-b}=3$
$\Rightarrow s=3 s-3 b \Rightarrow 3 b=2 s$ ...(ii)
$r_3-6 \Rightarrow \frac{s}{s-c}=6$
$\Rightarrow s=6 s-6 c \Rightarrow 6 c=5 s$ ...(iii)
$\begin{aligned} & \text { (i) }+ \text { (ii) }+ \text { (iii), } \\ & \quad 2 a+3 b+6 c=8 s=4 a+4 b+4 c\end{aligned}$
$\therefore \quad 2 a+b=2 c$ ...(iv)
From (i) and (ii),
$$
a+b=\frac{s}{2}+\frac{2 s}{3}=\frac{7 s}{6}
$$
From (i), (ii), (iii)
$$
\begin{aligned}
& a=3, b=4, c=5 \\
\therefore \quad & \text { Perimeter }=3+4+5=12 \mathrm{~cm} .
\end{aligned}
$$
Hence, the correct answer is option: 'C'.