The area of an equilateral triangle inscribed | vedclass

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(D) The equation of the given circle is $x^2+y^2-6x+2y-28=0$.Completing the square,we get $(x-3)^2+(y+1)^2 = 28+9+1 = 38$.Thus,the radius $r$ of the c

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$\frac{57 \sqrt{3}}{2}$

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(D) The equation of the given circle is $x^2+y^2-6x+2y-28=0$.Completing the square,we get $(x-3)^2+(y+1)^2 = 28+9+1 = 38$.Thus,the radius $r$ of the circle is $\sqrt{38}$.An equilateral triangle inscribed in a circle of radius $r$ has an area given by $A = \frac{3\sqrt{3}}{4} R^2$,where $R$ is the circumradius.Here,$R = r = \sqrt{38}$.Therefore,the area is $\frac{3\sqrt{3}}{4} 	imes (\sqrt{38})^2 = \frac{3\sqrt{3}}{4} 	imes 38 = \frac{3\sqrt{3} 	imes 19}{2} = \frac{57\sqrt{3}}{2}$ sq. units.

The area of an equilateral triangle inscribed in the circle $x^2+y^2-6x+2y-28=0$ is . . . . . . sq. units.

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