question_answer

In a \[\Delta ABC,\]let \[\angle C=\frac{\pi }{2}.\]If \[r\]and \[R\]are in radius and the circum-radius respectively of the triangle, then \[2(r+R)\] is equal to                           [IIT Screening 2000; AIEEE 2005]

A) \[a+b\]

B) \[b+c\]

C) \[c+a\]

D) \[a+b+c\]

Correct Answer: A

Solution :

Here, \[R=OA=OB=OC=\frac{1}{2}AB=\frac{c}{2}\] \[r=\frac{\Delta }{s}=\frac{\frac{1}{2}ab}{\frac{1}{2}(a+b+c)}=\frac{ab}{a+b+c}\] \[\therefore r+R=\frac{ab}{a+b+c}+\frac{c}{2}=\frac{2ab+c(a+b+c)}{2(a+b+c)}\]            \[=\frac{2ab+ca+bc+{{a}^{2}}+{{b}^{2}}}{2(a+b+c)},\,\,(\because \,\,\,{{c}^{2}}={{a}^{2}}+{{b}^{2}})\]           \[=\frac{{{(a+b)}^{2}}+c(a+b)}{2(a+b+c)}\,=\frac{(a+b)(a+b+c)}{2(a+b+c)}\] \[\therefore \,\,2(r+R)=a+b.\]