question_answer

The area of an equilateral triangle that can be inscribed in\[{{x}^{2}}+{{y}^{2}}-4x-6y-12=0,\]is

A)  \[\frac{25\sqrt{3}}{4}\,sq\] unit               

B)  \[\frac{35\sqrt{3}}{4}sq\,\] unit

C)  \[\frac{55\sqrt{3}}{4}sq\] unit

D)         \[\frac{75\sqrt{3}}{4}sq\] unit

E)  \[\frac{25}{4}sq\] unit

Correct Answer: D

Solution :

The equation of circle is \[{{x}^{2}}+{{y}^{2}}-4x-6y-12=0\] Radius of this circle\[=\sqrt{4+9+12}\]                                 \[=\sqrt{25}=5\,unit\] In\[\Delta BOD,\] \[\cos 30{}^\circ \frac{BD}{OB}\] \[\Rightarrow \]               \[BD=\frac{\sqrt{3}}{2}\times 5=\frac{5\sqrt{3}}{2}\]unit \[\therefore \]  \[BC=2BD=5\sqrt{3}\] Hence, area of\[\Delta ABC=\frac{\sqrt{3}}{4}{{(5\sqrt{3})}^{2}}\] \[=\frac{75}{4}\sqrt{3}\]sq unit