a + b 7 = b + c 8 = c + a 9 = λ {{a + b} \over 7} = {{b + c} \over 8} = {{c + a} \over 9} = \lambda 7 a + b = 8 b + c = 9 c + a = λ a + b = 7 λ a + b = 7\lambda a + b = 7 λ b + c = 8 λ b + c = 8\lambda b + c = 8 λ c + a = 9 λ c + a = 9\lambda c + a = 9 λ a + b + c = 12 λ a + b + c = 12\lambda a + b + c = 12 λ ∴ \therefore ∴ a = 4 λ , b = 3 λ , c = 5 λ a = 4\lambda ,\,b = 3\lambda ,\,c = 5\lambda a = 4 λ , b = 3 λ , c = 5 λ S = 4 λ + 3 λ + 5 λ 2 = 6 λ S = {{4\lambda + 3\lambda + 5\lambda } \over 2} = 6\lambda S = 2 4 λ + 3 λ + 5 λ = 6 λ Δ = S ( s − a ) ( s − b ) ( s − c ) = ( 6 λ ) ( 2 λ ) ( 3 λ ) ( λ ) = 6 λ 2 \Delta = \sqrt {S(s - a)(s - b)(s - c)} = \sqrt {(6\lambda )(2\lambda )(3\lambda )(\lambda )} = 6{\lambda ^2} Δ = S ( s − a ) ( s − b ) ( s − c ) = ( 6 λ ) ( 2 λ ) ( 3 λ ) ( λ ) = 6 λ 2 R = a b c 4 Δ = ( 4 λ ) ( 3 λ ) ( 5 λ ) 4 ( 6 λ 2 ) = 5 2 λ R = {{abc} \over {4\Delta }} = {{(4\lambda )(3\lambda )(5\lambda )} \over {4(6{\lambda ^2})}} = {5 \over 2}\lambda R = 4Δ ab c = 4 ( 6 λ 2 ) ( 4 λ ) ( 3 λ ) ( 5 λ ) = 2 5 λ r = Δ s = 6 λ 2 6 λ = λ r = {\Delta \over s} = {{6{\lambda ^2}} \over {6\lambda }} = \lambda r = s Δ = 6 λ 6 λ 2 = λ R r = 5 2 λ λ = 5 2 {R \over r} = {{{5 \over 2}\lambda } \over \lambda } = {5 \over 2} r R = λ 2 5 λ = 2 5