The area of equilateral triangle inscribed in the circle x^2+y^2-4x+8y+11=0 is of the form a*sqrt(b)/c. Find a+b+c.
+2666 To find the circumcenter, we first need to know the radius of the circle.
We can rewrite the circle equation to: \(x^2-4x+y^2+8y=-11\)
We can now complete the square, and we get: \((x-2)^2+(y+4)^2=4+16-11\)
The left hand is equal to 9, meaning the radius is 3.
We now know the circumcenter is 3.
The equation for the circumcenter is: \(\large{{abc} \over {4[ABC]}}\)
But, we have an equilateral triangle, so \(a = b= c\). This means we can substitute both b and c for a.
This gives us the equation: \(\large{{a^3} \over 4[ABC] } = 3\)
The formula for the area of an equilateral triangle is \({\sqrt 3 \over 4} s^2\)
We can substitute this in, and we get: \({{a^3} \over {{4 \sqrt3 \over 4} a^2}} = 3\)
We can simplify and divide by \(a^2\) to get: \(\large{a \over \sqrt3} = 3\)
This means that the side of the triangle is \(3 \sqrt 3 \).
Can you find the area from here?
Hint: Use the formula for the area of the triangle
