A triangle has corners at #(2 ,4 )#, #(6 ,5 )#, and #(4 ,2 )#. What is the area of the triangle's circumscribed circle?

Set the right side of equation 1 equal to right side of equation 2 and equation 3:

Combine like terms with h and k terms on the left and constants on the right:

To find the area of the circumscribed circle of a triangle, you can use the formula:

$A = \frac{{abc}}{{4R}}$

Where:

• $A$ is the area of the triangle.
• $a, b,$ and $c$ are the lengths of the sides of the triangle.
• $R$ is the radius of the circumscribed circle.

First, calculate the lengths of the sides of the triangle using the distance formula:

$a = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ $b = \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2}$ $c = \sqrt{(x_1 - x_3)^2 + (y_1 - y_3)^2}$

Given the coordinates: $(2, 4), (6, 5),$ and $(4, 2)$

$a = \sqrt{(6 - 2)^2 + (5 - 4)^2} = \sqrt{16 + 1} = \sqrt{17}$ $b = \sqrt{(4 - 6)^2 + (2 - 5)^2} = \sqrt{4 + 9} = \sqrt{13}$ $c = \sqrt{(2 - 4)^2 + (4 - 2)^2} = \sqrt{4 + 4} = \sqrt{8}$

Now, find the area of the triangle using Heron's formula:

$s = \frac{{a + b + c}}{2}$

$s = \frac{{\sqrt{17} + \sqrt{13} + \sqrt{8}}}{2}$

$s \approx \frac{{4.123 + 3.606 + 2.828}}{2} \approx 5.278$

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

$A \approx \sqrt{5.278(5.278 - \sqrt{17})(5.278 - \sqrt{13})(5.278 - \sqrt{8})}$ $A \approx \sqrt{5.278(0.278)(1.672)(2.45)} \approx \sqrt{2.763} \approx 1.662$

Now, to find the radius of the circumscribed circle, use the formula:

$R = \frac{{abc}}{{4A}}$

$R = \frac{{\sqrt{17} \cdot \sqrt{13} \cdot \sqrt{8}}}{{4 \cdot 1.662}}$ $R = \frac{{\sqrt{176}}}{{6.648}} \approx \frac{{13.267}}{{6.648}} \approx 1.994$

Finally, the area of the circumscribed circle is given by:

$Area = \pi R^2$

$Area \approx \pi (1.994)^2$ $Area \approx \pi \cdot 3.976$ $Area \approx 12.51$

Therefore, the area of the triangle's circumscribed circle is approximately $12.51$ square units.