# A triangle has corners at #(1 ,1 )#, #(7 ,9 )#, and #(4 ,2 )#. What is the area of the triangle's circumscribed circle?

Area of triangle's circumscribed circle is

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To find the area of the circumscribed circle of a triangle, we first need to determine the radius of the circumscribed circle. The radius ( R ) of the circumscribed circle is given by the formula:

[ R = \frac{abc}{4A} ]

where ( a, b, c ) are the lengths of the sides of the triangle, and ( A ) is the area of the triangle.

First, we calculate the lengths of the sides of the triangle using the given coordinates:

- Side ( a ) is between (1, 1) and (7, 9).
- Side ( b ) is between (7, 9) and (4, 2).
- Side ( c ) is between (4, 2) and (1, 1).

Using the distance formula ( d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} ), we find:

- Side ( a ): ( d_a = \sqrt{(7 - 1)^2 + (9 - 1)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 )
- Side ( b ): ( d_b = \sqrt{(4 - 7)^2 + (2 - 9)^2} = \sqrt{9 + 49} = \sqrt{58} )
- Side ( c ): ( d_c = \sqrt{(4 - 1)^2 + (2 - 1)^2} = \sqrt{9 + 1} = \sqrt{10} )

Next, we calculate the area ( A ) of the triangle using Heron's formula:

[ A = \sqrt{s(s - a)(s - b)(s - c)} ]

where ( s ) is the semiperimeter given by ( s = \frac{a + b + c}{2} ).

Plugging in the values:

[ s = \frac{10 + \sqrt{58} + \sqrt{10}}{2} ]

[ A = \sqrt{\frac{10 + \sqrt{58} + \sqrt{10}}{2} \cdot \left(\frac{10 + \sqrt{58} + \sqrt{10}}{2} - 10\right) \cdot \left(\frac{10 + \sqrt{58} + \sqrt{10}}{2} - \sqrt{58}\right) \cdot \left(\frac{10 + \sqrt{58} + \sqrt{10}}{2} - \sqrt{10}\right)} ]

After calculating ( A ), we can substitute the values of ( a, b, c, ) and ( A ) into the formula for the radius of the circumscribed circle:

[ R = \frac{abc}{4A} ]

Finally, once we have the radius ( R ), we can find the area of the circumscribed circle using the formula ( \text{Area} = \pi R^2 ).

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