Conic Sections Question 238
Question: Area of the equilateral triangle inscribed in the circle $ x^{2}+y^{2}-7x+9y+5=0 $ is
Options:
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
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Answer:
Correct Answer: D
Solution:
[d] Given circle : $ x^{2}+y^{2}-7x+9y+5=0 $
$ \therefore $ Centre $ =( \frac{7}{2},\frac{-9}{2} ) $ 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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