A) \[\frac{y}{\sqrt{3}}\]
B) \[\frac{c}{3}\]
C) \[\frac{c}{\sqrt{3}}\]
D) \[\frac{3}{2}y\]
Correct Answer: C
Solution :
Let a and b be the two sides of the triangle. Then, \[a+b=x\] and \[ab=y\] Now,\[{{x}^{2}}-{{c}^{2}}=y\Rightarrow {{(a+b)}^{2}}-{{c}^{2}}=ab\] \[\Rightarrow \]\[{{a}^{2}}+{{b}^{2}}-{{c}^{2}}=-ab\] \[\Rightarrow \]\[\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}=-\frac{1}{2}\] \[\Rightarrow \]\[\cos C=-\frac{1}{2}\] \[\Rightarrow \]\[\angle C=\frac{2\pi }{3}\] Now, let R be the circumradius of the triangle. \[\therefore \]\[R=\frac{c}{2\sin C}=\frac{c}{\sqrt{3}}\]
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