Question
In a $$\vartriangle ABC,$$ the sides are in the ratio 4 : 5 : 6. The ratio of the circumradius and the inradius is
A.
8 : 7
B.
3 : 2
C.
7 : 3
D.
16 : 7
Answer :
16 : 7
Solution :
Here, $$a = 4k, b = 5k, c = 6k.$$
$$\eqalign{ & \therefore \,\,s = \frac{{15k}}{2}. \cr & \therefore \,\,\vartriangle = \sqrt {\frac{{15k}}{2}\left( {\frac{{15k}}{2} - 4k} \right)\left( {\frac{{15k}}{2} - 5k} \right)\left( {\frac{{15k}}{2} - 6k} \right)} = \frac{{15\sqrt 7 }}{4}{k^2}. \cr & {\text{But }}\,R = \frac{{abc}}{{4\vartriangle }} = \frac{{4k \cdot 5k \cdot 6k}}{{15\sqrt 7 {k^2}}} = \frac{8}{{\sqrt 7 }}k \cr & {\text{and }}\,r = \frac{\vartriangle }{s} = \frac{{15\sqrt 7 }}{4}{k^2} \cdot \frac{2}{{15k}} = \frac{{\sqrt 7 }}{2}k. \cr & \therefore \,\,\frac{R}{r} = \frac{{\frac{{8k}}{{\sqrt 7 }}}}{{\frac{{\sqrt 7 k}}{2}}} = \frac{{16}}{7}. \cr} $$
Here, $$a = 4k, b = 5k, c = 6k.$$
$$\eqalign{ & \therefore \,\,s = \frac{{15k}}{2}. \cr & \therefore \,\,\vartriangle = \sqrt {\frac{{15k}}{2}\left( {\frac{{15k}}{2} - 4k} \right)\left( {\frac{{15k}}{2} - 5k} \right)\left( {\frac{{15k}}{2} - 6k} \right)} = \frac{{15\sqrt 7 }}{4}{k^2}. \cr & {\text{But }}\,R = \frac{{abc}}{{4\vartriangle }} = \frac{{4k \cdot 5k \cdot 6k}}{{15\sqrt 7 {k^2}}} = \frac{8}{{\sqrt 7 }}k \cr & {\text{and }}\,r = \frac{\vartriangle }{s} = \frac{{15\sqrt 7 }}{4}{k^2} \cdot \frac{2}{{15k}} = \frac{{\sqrt 7 }}{2}k. \cr & \therefore \,\,\frac{R}{r} = \frac{{\frac{{8k}}{{\sqrt 7 }}}}{{\frac{{\sqrt 7 k}}{2}}} = \frac{{16}}{7}. \cr} $$