A) \[\frac{155}{8}\sqrt{3}\] square units
B) \[\frac{165}{8}\sqrt{3}\] square units
C) \[\frac{175}{8}\sqrt{3}\] square units
D) \[\frac{165}{8}\sqrt{3}\] square units
Correct Answer: D
Solution :
[d] Given circle : \[{{x}^{2}}+{{y}^{2}}-7x+9y+5=0\] |
\[\therefore \] Centre \[=\left( \frac{7}{2},\frac{-9}{2} \right)\] |
Radius \[=\sqrt{\frac{49}{4}+\frac{81}{4}-5}=\frac{\sqrt{110}}{2}\] |
Since \[\Delta ABC\] is an equilateral |
\[\therefore \angle MAL=30{}^\circ ,\angle MLA=90{}^\circ \] |
Also \[MA=\frac{\sqrt{110}}{2}\] |
\[\therefore AL=MA\cos 30{}^\circ =\frac{\sqrt{110}}{2}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{330}}{4}\] |
\[\therefore \] Side of \[\Delta =2.AL=\frac{\sqrt{330}}{2}\] |
Area of equilateral \[\Delta =\frac{\sqrt{3}}{4}{{a}^{2}}=\frac{\sqrt{3}}{4}\times \frac{330}{4}\] |
\[=\frac{165}{8}\sqrt{3}\] sq. units |
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