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 unitYou need to login to perform this action.
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