Question

The angles of a right-angled triangle are in $A.P.$ The ratio of the inradius and the primeter is

• A. $(2-\sqrt{3}):2\sqrt{3}$
• B. $1:8\sqrt{3}(2+\sqrt{3})$
• C. $(2+\sqrt{3}):4\sqrt{3}$
• D. none of these

Answer 1

The correct option is A $(2-\sqrt{3}):2\sqrt{3}$

$(\alpha -\beta )+\alpha +(\alpha +\beta )={180}^0$ and $\alpha +\beta ={90}^0$

$\therefore$ the angles are ${30}^0,{60}^0,{90}^0$

$\therefore \frac{a}{\sin {30}^0}+\frac{b}{\sin {60}^0}+\frac{c}{\sin {90}^0}=2R\Rightarrow a=R,b=\sqrt{3}R,c=2R$

$r=\frac{\Delta }{s}=\frac{\frac{1}{2}.R.\sqrt{3}R}{\frac{1}{2}(R+\sqrt{3}R+2R)}=\frac{\sqrt{3}R}{3+\sqrt{3}}$ and $2s=R+\sqrt{3}R+2R=(3+\sqrt{3})R$

$\therefore \frac{r}{2s}=\frac{\sqrt{3}R}{3+\sqrt{3}}\times \frac{1}{(3+\sqrt{3})R}=\frac{1}{(3+\sqrt{3})(\sqrt{3}+1)}$

$=\frac{1}{6+4\sqrt{3}}=\frac{1}{2\sqrt{3}(2+\sqrt{3})}=\frac{2-\sqrt{3}}{2\sqrt{3}}$