QJEEM22Madfb6: JEE Main 2022 Mathematics - Class 11 JEEM22Madfb6 Solution: Let a, b and c be the length of sides of a triangle ABC such…

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

$a + b = 7\lambda$

$b + c = 8\lambda$

$c + a = 9\lambda$

$a + b + c = 12\lambda$

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

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

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

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

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

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