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If the area of an equilateral triangle inscribed in the circle, is3 x^2+3 y^2+10 x+15 y-c=0sq27 \sqrt{2} units, then c is equal to ?

 

Option: 1

187.28


Option: 2

82.05


Option: 3

125


Option: 4

100


Answers (1)

best_answer

The equation of the circle in the form:
\begin{aligned} & 3 x^2+3 y^2+10 x+15 y-c=0 \\ & \quad 3\left(x^2+(10 / 3) x\right)+3\left(y^2+5 y\right)=c \\ & 3\left(x^2+(10 / 3) x+(10 / 6)^2\right)+3\left(y^2+5 y+(5 / 2)^2\right)=c+3(10 / 6)^2+3(5 / 2)^2 \\ & 3(x+5 / 3)^2+3(y+5 / 2)^2=c+125 / 2 \end{aligned}

Comparing this with the standard equation of a circle:
(x-h)^2+(y-k)^2=r^2

The center of the circle is at (-5 / 3,-5 / 2)

and the radius of the circle isr=\sqrt{(} c+125 / 6)

Let A, B, and C be the vertices of the equilateral triangle. Then, the sides of the triangle are AB, BC, and CA, and they are all equal in length to the radius of the circle.

The length of each side s
s=r=\sqrt{(} c+125 / 6)

The area of an equilateral triangle with side length s is given by:


\begin{aligned} & c=216 \sqrt{(2)} / \sqrt{(3)}-125 c=187.28 \\ & \end{aligned}

Therefore, c is approximately 187.28.

 

 

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SANGALDEEP SINGH

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