Question
The angles of a right-angled triangle are in A.P. The ratio of the inradius and the perimeter is
A.
$$\left( {2 - \sqrt 3 } \right):2\sqrt 3 $$
B.
$$1:8\sqrt 3 \left( {2 + \sqrt 3 } \right)$$
C.
$$\left( {2 + \sqrt 3 } \right):4\sqrt 3 $$
D.
None of these
Answer :
$$\left( {2 - \sqrt 3 } \right):2\sqrt 3 $$
Solution :
$$\left( {\alpha - \beta } \right) + \alpha + \left( {\alpha + \beta } \right) = {180^ \circ }\,\,{\text{and }}\alpha + \beta = {90^ \circ }.$$
∴ the angles are $${30^ \circ },{60^ \circ },{90^ \circ }.$$
$$\eqalign{
& \therefore \,\,\frac{a}{{\sin {{30}^ \circ }}} = \frac{b}{{\sin {{60}^ \circ }}} = \frac{c}{{\sin {{90}^ \circ }}} = 2R\,\,\,{\text{or, }}a = R,b = \sqrt 3 R,c = 2R. \cr
& r = \frac{\vartriangle }{s} = \frac{{\frac{1}{2}R \cdot \sqrt 3 R}}{{\frac{1}{2}\left( {R + \sqrt 3 R + 2R} \right)}} = \frac{{\sqrt 3 R}}{{3 + \sqrt 3 }} \cr
& {\text{and, }}2s = R + \sqrt 3 R + 2R = \left( {3 + \sqrt 3 } \right)R. \cr
& \therefore \,\,\frac{r}{{2s}} = \frac{{\sqrt 3 R}}{{3 + \sqrt 3 }} \times \frac{1}{{\left( {3 + \sqrt 3 } \right)R}} = \frac{1}{{\left( {3 + \sqrt 3 } \right)\left( {\sqrt 3 + 1} \right)}} \cr
& \frac{r}{{2s}} = \frac{1}{{6 + 4\sqrt 3 }} = \frac{1}{{2\sqrt 3 \left( {2 + \sqrt 3 } \right)}} = \frac{{2 - \sqrt 3 }}{{2\sqrt 3 }}. \cr} $$