A) \[a+b\]
B) \[b+c\]
C) \[c+a\]
D) \[a+b+c\]
Correct Answer: A
Solution :
Here, \[R=OA=OB=OC=\frac{1}{2}AB=\frac{c}{2}\] \[r=\frac{\Delta }{s}=\frac{\frac{1}{2}ab}{\frac{1}{2}(a+b+c)}=\frac{ab}{a+b+c}\] \[\therefore r+R=\frac{ab}{a+b+c}+\frac{c}{2}=\frac{2ab+c(a+b+c)}{2(a+b+c)}\] \[=\frac{2ab+ca+bc+{{a}^{2}}+{{b}^{2}}}{2(a+b+c)},\,\,(\because \,\,\,{{c}^{2}}={{a}^{2}}+{{b}^{2}})\] \[=\frac{{{(a+b)}^{2}}+c(a+b)}{2(a+b+c)}\,=\frac{(a+b)(a+b+c)}{2(a+b+c)}\] \[\therefore \,\,2(r+R)=a+b.\]You need to login to perform this action.
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