A) \[\frac{155}{8}\sqrt{3}\,sq\,unit\]
B) \[\frac{165}{8}\sqrt{3}\,sq\,unit\]
C) \[\frac{175}{8}\sqrt{3}\,sq\,unit\]
D) \[\frac{185}{8}\sqrt{3}\,sq\,unit\]
E) \[\frac{195}{8}\sqrt{3}\,sq\,unit\]
Correct Answer: B
Solution :
Given, \[{{x}^{2}}-{{y}^{2}}-7x+9y+5=0\] \[\therefore \] \[R=\sqrt{{{\left( \frac{-7}{2} \right)}^{2}}+{{\left( \frac{9}{2} \right)}^{2}}-5}\] \[=\sqrt{\frac{49}{4}+\frac{81}{4}-5}=\frac{\sqrt{110}}{2}\] In \[\Delta OAB,\] \[\cos 30{}^\circ =\frac{AB}{R}\] \[\Rightarrow \] \[\frac{\sqrt{3}}{2}=\frac{AB}{\sqrt{110/2}}\] \[\Rightarrow \] \[\frac{\sqrt{330}}{4}=AB\] \[\therefore \]Length \[AC=\frac{\sqrt{330}}{2}\] \[\therefore \]Area of equilateral\[\Delta =\frac{\sqrt{3}}{4}{{(a)}^{2}}\] \[=\frac{\sqrt{3}}{4}\times \frac{330}{4}=\frac{165\sqrt{3}}{8}\]sq unitYou need to login to perform this action.
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