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 :
| [a] Let area of triangle be \[\Delta \], then according to question, |
| \[\Delta =\frac{1}{2}ax=\frac{1}{2}by=\frac{1}{2}cz\therefore \frac{bx}{c}+\frac{cy}{a}+\frac{az}{b}\] |
| \[=\frac{b}{c}\left( \frac{2\Delta }{a} \right)+\frac{c}{a}\left( \frac{2\Delta }{b} \right)+\frac{a}{b}\left( \frac{2\Delta }{c} \right)\] |
| \[=\,\,\frac{2\Delta ({{b}^{2}}+{{c}^{2}}+{{a}^{2}})}{abc}\] |
| \[=\frac{2({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}{abc}.\frac{abc}{4R}=\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{2R}\] |
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