A) \[\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{2R}\]
B) \[\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{R}\]
C) \[\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{4R}\]
D) \[\frac{2({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}{R}\]
Correct Answer: A
Solution :
Let area of triangle be\[\Delta \], then according to question, \[\Delta =\frac{1}{2}ax=\frac{1}{2}by=\frac{1}{2}cz\] \[\therefore \] \[\,\frac{bx}{c}+\frac{cy}{a}+\frac{az}{b}=\frac{b}{c}\left( \frac{2\Delta }{a} \right)+\frac{c}{a}\left( \frac{2\Delta }{b} \right)+\frac{a}{b}\left( \frac{2\Delta }{c} \right)\] \[=\frac{2\Delta ({{b}^{2}}+{{c}^{2}}+{{a}^{2}})}{abc}=\frac{2({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}{abc}\text{ }\text{. }\frac{abc}{4R}=\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{2R}\].
You need to login to perform this action.
You will be redirected in
3 sec
