Let a, b and c be the length of sides of a triangle ABC such that...

Let $\dfrac{a + b}{7} = \dfrac{b + c}{8} = \dfrac{c + a}{9} = k$

so $a + b = 7k, b + c = 8k, c + a = 9k$

Solving this we get $a = 4k, b = 3k, c = 5k$ and calculating semi-perimeter

$S = \dfrac{a + b + c}{2}$ $⇒ S = 6k$

We know that $\dfrac{r}{R} = 4sin\dfrac{A}{2}sin\dfrac{B}{2}sin\dfrac{C}{2}$

$⇒ \dfrac{r}{R} = 4\sqrt{\dfrac{\left(s - b\right)\left(s - c\right)}{bc}} \times \sqrt{\dfrac{\left(s - a\right)\left(s - c\right)}{ac}} \times \sqrt{\dfrac{\left(s - a\right)\left(s - b\right)}{ab}}$

$⇒ \dfrac{r}{R} = 4 \cdot \dfrac{\left(s - a\right)\left(s - b\right)\left(s - c\right)}{abc}$

$⇒ \dfrac{r}{R} = 4 \cdot \dfrac{\left(2k\right)\left(3k\right)\left(k\right)}{\left(4k\right)\left(3k\right)\left(5k\right)}$ $⇒ \dfrac{R}{r} = \dfrac{5}{2}$