question_answer

Area of the equilateral triangle inscribed in the circle\[{{x}^{2}}+{{y}^{2}}-7x+9y+5=0\]is

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 unit