To find the perimeter of the triangle with vertices at the points \(A(1, 5)\), \(B(4, 1)\), and \(C(-4, -5)\), we first need to calculate the lengths of each side of the triangle using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
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Length \(AB\): \[ AB = \sqrt{(4 - 1)^2 + (1 - 5)^2} = \sqrt{(3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
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Length \(BC\): \[ BC = \sqrt{(-4 - 4)^2 + (-5 - 1)^2} = \sqrt{(-8)^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \]
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Length \(CA\): \[ CA = \sqrt{(1 - (-4))^2 + (5 - (-5))^2} = \sqrt{(1 + 4)^2 + (5 + 5)^2} = \sqrt{(5)^2 + (10)^2} = \sqrt{25 + 100} = \sqrt{125} \approx 11.2 \]
Now, we can calculate the perimeter of the triangle by adding the lengths of all sides:
\[ \text{Perimeter} = AB + BC + CA = 5 + 10 + 11.2 = 26.2 \]
Therefore, the perimeter of the triangle, rounded to the nearest tenth of a unit, is 26.2 units.