In triangle ABC,suppose the exradii opposite | vedclass

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(A) For $	riangle ABC$,the exradii are given by $r_1=\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b}, r_3=\frac{\Delta}{s-c}$ and the inradius is $r=\frac

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$4R$

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(A) For $	riangle ABC$,the exradii are given by $r_1=\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b}, r_3=\frac{\Delta}{s-c}$ and the inradius is $r=\frac{\Delta}{s}$,where $\Delta$ is the area of the triangle and $s$ is the semi-perimeter.We know the identity $r_1+r_2+r_3 = 4R+r$.Rearranging this,we get $r_1+r_2+r_3-r = 4R$.Given $r_1=2, r_2=3, r_3=6$,we can find $r$ using the relation $\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}$.$\frac{1}{r} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1$.Thus,$r=1$.Substituting the values,$r_1+r_2+r_3-r = 2+3+6-1 = 10$.Since $r_1+r_2+r_3-r = 4R$,we have $10 = 4R$,so $R = 2.5$.The expression $r_1+r_2+r_3-r$ is equal to $4R$.

In $\triangle ABC$,suppose the exradii opposite to angles $A, B$ and $C$ are denoted by $r_1, r_2$ and $r_3$ respectively. If $r_1=2, r_2=3, r_3=6$ and $R$ is the radius of the circumcircle,then the value of $r_1+r_2+r_3-r$ is:

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