Let a, b and c be the length of sides of a triangle ABC such that [a + b 7 = b + c 8 = c +…

Question

Let a, b and c be the length of sides of a triangle ABC such that $${{a + b} \over 7} = {{b + c} \over 8} = {{c + a} \over 9}$$. If r and R are the radius of incircle and radius of circumcircle of the triangle ABC, respectively, then the value of $${R \over r}$$ is equal to :

Accepted Answer

$${{a + b} \over 7} = {{b + c} \over 8} = {{c + a} \over 9} = \lambda $$

$$a + b = 7\lambda $$

$$b + c = 8\lambda $$

$$c + a = 9\lambda $$

$$a + b + c = 12\lambda $$

$$\therefore$$ $$a = 4\lambda ,\,b = 3\lambda ,\,c = 5\lambda $$

$$S = {{4\lambda  + 3\lambda  + 5\lambda } \over 2} = 6\lambda $$

$$\Delta  = \sqrt {S(s - a)(s - b)(s - c)}  = \sqrt {(6\lambda )(2\lambda )(3\lambda )(\lambda )}  = 6{\lambda ^2}$$

$$R = {{abc} \over {4\Delta }} = {{(4\lambda )(3\lambda )(5\lambda )} \over {4(6{\lambda ^2})}} = {5 \over 2}\lambda $$

$$r = {\Delta  \over s} = {{6{\lambda ^2}} \over {6\lambda }} = \lambda $$

$${R \over r} = {{{5 \over 2}\lambda } \over \lambda } = {5 \over 2}$$

Solutions · 1