In a triangle ABC,if r_1 r_2 + r_3 r = 35,r_2 | vedclass

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(C) Given equations: $r_1 r_2 + r_3 r = 35$,$r_2 r_3 + r r_1 = 63$,and $r_3 r_1 + r r_2 = 45$. Using the identities $r_1 = \frac{\Delta}{s-a}$,$r_2 =

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$21$

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(C) Given equations: $r_1 r_2 + r_3 r = 35$,$r_2 r_3 + r r_1 = 63$,and $r_3 r_1 + r r_2 = 45$. Using the identities $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$,and $r = \frac{\Delta}{s}$,we know that $r_1 r_2 + r_3 r = ab$,$r_2 r_3 + r r_1 = bc$,and $r_3 r_1 + r r_2 = ac$. Thus,$ab = 35$,$bc = 63$,and $ac = 45$. Multiplying these gives $(abc)^2 = 35 	imes 63 	imes 45 = 99225$,so $abc = 315$. Then $c = \frac{abc}{ab} = \frac{315}{35} = 9$,$a = \frac{abc}{bc} = \frac{315}{63} = 5$,and $b = \frac{abc}{ac} = \frac{315}{45} = 7$. Therefore,$2s = a + b + c = 5 + 7 + 9 = 21$.

In a triangle $ABC$,if $r_1 r_2 + r_3 r = 35$,$r_2 r_3 + r r_1 = 63$,and $r_3 r_1 + r r_2 = 45$,then $2s =$

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