A) equilateral
B) right angled and isosceles
C) right angled and not isosceles
D) None of the above
Correct Answer: C
Solution :
We have\[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=ac+\sqrt{3}ab\] \[\Rightarrow \]\[\frac{{{a}^{2}}}{4}+{{c}^{2}}-ac+\frac{3{{a}^{2}}}{4}+{{b}^{2}}-\sqrt{3}ab=0\] \[\Rightarrow \]\[{{\left( \frac{a}{2}-c \right)}^{2}}+{{\left( \frac{\sqrt{3}a}{2}-b \right)}^{2}}=0\] \[\Rightarrow \]\[\frac{a}{2}-c=0\]and\[\frac{\sqrt{3}}{2}a-b=0\] \[\Rightarrow \]\[a=2c\]and \[\sqrt{3}a=2b\] \[\Rightarrow \]\[a=\frac{2b}{\sqrt{3}}=2c=\lambda \] [say] \[\Rightarrow \]\[a=\lambda ,b=\frac{\sqrt{3}}{2}\lambda \]and \[c=\frac{\lambda }{2}\] Now,\[{{b}^{2}}+{{c}^{2}}={{\left( \frac{\sqrt{3}\lambda }{2} \right)}^{2}}+{{\left( \frac{\lambda }{2} \right)}^{2}}\] \[=\frac{3{{\lambda }^{2}}}{2}+\frac{{{\lambda }^{2}}}{4}={{\lambda }^{2}}\] \[\therefore \]\[{{b}^{2}}+{{c}^{2}}={{a}^{2}}\]
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