Question

The angles of a right - angled triangle are in AP. The ratio of the inradius and the perimeter is

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is A

$(\alpha -\beta )+\alpha +(\alpha +\beta )={180}^{0}and\alpha +\beta ={90}^0\therefore \text{The angles are}{30}^0,{60}^0,{90}^0\therefore \frac{a}{sin{30}^0}=\frac{b}{sin{60}^0}=\frac{c}{sin{90}^0}=2Rora=R,b=\sqrt{3}R,c=2Rr=\frac{\Delta }{s}=\frac{\frac{1}{2}R.\sqrt{3}R}{\frac{1}{2}(R+\sqrt{3}R+2R)}=\frac{\sqrt{3}R}{3+\sqrt{3}}and2s=R+\sqrt{3}R+2R=(3+\sqrt{3})R\therefore \frac{r}{2s}=\frac{\sqrt{3}R}{3+\sqrt{3}}\times \frac{1}{3+\sqrt{3}}R=\frac{1}{(3+\sqrt{3})(\sqrt{3}+1)}=\frac{1}{6+4\sqrt{3}}=\frac{1}{2\sqrt{3}(2+\sqrt{3})}=\frac{2-\sqrt{3}}{2\sqrt{3}}$