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प्रश्न
The area of an equilateral triangle inscribed in the circle x2 + y2 − 6x − 8y − 25 = 0 is
पर्याय
\[\frac{225\sqrt{3}}{6}\]
25π
50π − 100
none of these
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उत्तर
\[\frac{225\sqrt{3}}{6}\]

Let ABC be the required equilateral triangle.
The equation of the circle is x2 + y2 − 6x − 8y − 25 = 0.
Therefore, coordinates of the centre O is \[\left( 3, 4 \right)\].
Radius of the circle = OA = OB = OC = \[\sqrt{9 + 16 + 25} = 5\sqrt{2}\]
In \[∆\] BOD, we have:
\[\sin60° = \frac{DB}{BO}\]
\[ \Rightarrow DB = \frac{\sqrt{3}}{2}\left( 5\sqrt{2} \right)\]
\[ \Rightarrow BC = 2BD = \sqrt{3}\left( 5\sqrt{2} \right) = 5\sqrt{6}\]
Now, area of \[\bigtriangleup ABC\] = \[\frac{\sqrt{3}}{4}B C^2 = \frac{\sqrt{3}}{4} \left( 5\sqrt{6} \right)^2 = \frac{\sqrt{3}\left( 150 \right)}{4} = \frac{\sqrt{3}\left( 75 \right)}{2} = \frac{\sqrt{3}\left( 225 \right)}{6}\]
square units
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