Question

The ratio between the circumradius and inradius of a right angled isosceles triangle is

• A. 2:1
• B. $(\sqrt{2}+1):1$
• C. $1:(\sqrt{2}-1)$
• D. $(\sqrt{2}+1):\sqrt{2}$

Answer 1

Answer: (B), (C)

The correct options are
B $(\sqrt{2}+1):1$
C $1:(\sqrt{2}-1)$

$R=\frac{abc}{4\Delta }$

$=\frac{a\sqrt{2}\times a^2}{4\times \frac{1}{2}\times a^2}=\frac{a}{\sqrt{2}}$

$r=\frac{\frac{1}{2}{a}^2}{\frac{(a+a+a\sqrt{2})}{2}}$

$r=\frac{1}{\sqrt{2}(\sqrt{2}+1)}$

$\frac{R}{r}=(\sqrt{2}+1)$.