If $\triangle \mathrm{ABC}$ is a right angled isosceles triangle and $\angle \mathrm{C}=90^{\circ}$, then $\mathrm{r}: \mathrm{r}_3=$

• A

  $\sqrt{2}+1: \sqrt{2}-1$
• Correct answer
  B

  $\sqrt{2}-1: \sqrt{2}+1$
• C

  $\sqrt{2}: 1$
• D

  $1: \sqrt{2}$

Solution

$\because A B C$ is isosceles triangle. $\therefore A C=B C$ and $\angle C=90^{\circ}$ Let $A C=B C \Rightarrow a=b$
In $\triangle A B C$, $\begin{aligned} & c^2=a^2+b^2=a^2+a^2=2 a^2 \\ & \Rightarrow \quad c=a \sqrt{2} \end{aligned}$ Now, $\frac{r}{r_3}=\frac{\frac{\Delta}{s}}{\frac{\Delta}{s-c}}=\frac{s-c}{s}$ $\begin{aligned} & =\frac{\frac{a+b+c}{2}-c}{\frac{a+b+c}{2}}=\frac{a+b-c}{a+b+c} \\ & =\frac{a+a-a \sqrt{2}}{a+a+a \sqrt{2}}=\frac{2-\sqrt{2}}{2+\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}+1} \\ & \Rightarrow r: r_3=(\sqrt{2}-1):(\sqrt{2}+1) . \end{aligned}$

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