To find the perimeter of the triangle formed by the points \( A(1, 5) \), \( B(4, 1) \), and \( C(-4, -5) \), Nathan needs to calculate the lengths of each of the sides of the triangle and then add those lengths together. The perimeter \( P \) of the triangle is given by the formula:
\[ P = AB + BC + CA \]
Step 1: Calculate the Lengths of the Sides
We will use the distance formula to find the lengths of the sides. The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
a. Length of Side \( AB \)
For points \( A(1, 5) \) and \( B(4, 1) \):
\[ AB = \sqrt{(4 - 1)^2 + (1 - 5)^2} = \sqrt{(3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
b. Length of Side \( BC \)
For points \( B(4, 1) \) and \( C(-4, -5) \):
\[ BC = \sqrt{(-4 - 4)^2 + (-5 - 1)^2} = \sqrt{(-8)^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \]
c. Length of Side \( CA \)
For points \( C(-4, -5) \) and \( A(1, 5) \):
\[ CA = \sqrt{(1 - (-4))^2 + (5 - (-5))^2} = \sqrt{(1 + 4)^2 + (5 + 5)^2} = \sqrt{(5)^2 + (10)^2} = \sqrt{25 + 100} = \sqrt{125} = 5\sqrt{5} \]
Calculating \( \sqrt{125} \):
\[ 5\sqrt{5} \approx 5 \times 2.236 = 11.18 \text{ (using } \sqrt{5} \approx 2.236\text{)} \]
Step 2: Calculate the Perimeter
Now add the lengths of all three sides:
\[ P = AB + BC + CA = 5 + 10 + 5\sqrt{5} \] \[ P \approx 5 + 10 + 11.18 = 26.18 \]
Step 3: Round to the Nearest Tenth
Rounding \( 26.18 \) to the nearest tenth gives:
\[ \text{Perimeter} \approx 26.2 \]
Final Answer
The perimeter of the triangle, rounded to the nearest tenth of a unit, is \( \boxed{26.2} \).