Q33 / 90 · JEE Main 2022 (25 Jun Shift 1) · Mathematics

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

Solution

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

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

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

$S = \frac{a + b + c}{2}$ $\Rightarrow S = 6 k$

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

$\Rightarrow \frac{r}{R} = 4 \sqrt{\frac{(s - b) (s - c)}{b c}} \times \sqrt{\frac{(s - a) (s - c)}{a c}} \times \sqrt{\frac{(s - a) (s - b)}{a b}}$

$\Rightarrow \frac{r}{R} = 4 \cdot \frac{(s - a) (s - b) (s - c)}{a b c}$

$\Rightarrow \frac{r}{R} = 4 \cdot \frac{(2 k) (3 k) (k)}{(4 k) (3 k) (5 k)}$ $\Rightarrow \frac{R}{r} = \frac{5}{2}$

Hence, the correct answer is option: 'C'.