To calculate the perimeter of the triangle with vertices $J$, $K$, and $L$, we will use the distance formula:

$\text{Distance between two points } (x_1, y_1) \text{ and } (x_2, y_2) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$

Step 1: Identify the coordinates of $J$, $K$, and $L$ from the graph:

• $J(-10, 8)$
• $K(-2, 2)$
• $L(6, 6)$

Step 2: Calculate the lengths of each side.

1. Distance $JK$: $JK = \sqrt{(-2 - (-10))^2 + (2 - 8)^2} = \sqrt{(8)^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$

2. Distance $KL$: $KL = \sqrt{(6 - (-2))^2 + (6 - 2)^2} = \sqrt{(8)^2 + (4)^2} = \sqrt{64 + 16} = \sqrt{80} \approx 8.94$

3. Distance $LJ$: $LJ = \sqrt{(6 - (-10))^2 + (6 - 8)^2} = \sqrt{(16)^2 + (-2)^2} = \sqrt{256 + 4} = \sqrt{260} \approx 16.12$

Step 3: Add the side lengths to find the perimeter.

$\text{Perimeter} = JK + KL + LJ = 10 + 8.94 + 16.12 = 35.06$

Final Answer:

The perimeter of the triangle is 35.06 units.

Would you like further clarification on any steps or related concepts? Here are some related questions you might consider:

1. How do we generalize the distance formula to three-dimensional coordinates?
2. What are the properties of triangles in coordinate geometry?
3. How can the midpoint formula complement calculations for the triangle’s sides?
4. Can the triangle's area be determined using these same coordinates?
5. How does rounding affect precision in geometric problems