In triangle ABC,if B=90^{circ},then 2(r+R)= | vedclass

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(C) Given,in $	riangle ABC$,$B=90^{\circ}$.Using the sine rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = \frac{1}{2R}$.Since $B=90^{\

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$a+c$

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(C) Given,in $	riangle ABC$,$B=90^{\circ}$.Using the sine rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = \frac{1}{2R}$.Since $B=90^{\circ}$,$\frac{\sin 90^{\circ}}{b} = \frac{1}{2R} \implies b = 2R$.The inradius $r$ is given by $r = (s-b) 	an(\frac{B}{2})$.Substituting $B=90^{\circ}$,$r = (s-b) 	an(45^{\circ}) = s-b$.Since $s = \frac{a+b+c}{2}$,we have $r = \frac{a+b+c}{2} - b = \frac{a-b+c}{2}$.Thus,$2r = a-b+c$.Finally,$2(r+R) = 2r + 2R = (a-b+c) + b = a+c$.

In $\triangle ABC$,if $B=90^{\circ}$,then $2(r+R)=$

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