Math Problem Statement

Find the perimeter of the triangle in units. Round to two decimal places as necessary.

Solution

To find the perimeter of the triangle △JKL\triangle JKL, I will calculate the distances between the points JJ, KK, and LL using the distance formula:

d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

From the graph, the coordinates of the points are approximately:

  • J(−6,8)J(-6, 8)
  • K(0,2)K(0, 2)
  • L(6,6)L(6, 6)

Step 1: Calculate JKJK

= \sqrt{(6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} \approx 8.49$$ ### Step 2: Calculate $$KL$$ $$d_{KL} = \sqrt{(6 - 0)^2 + (6 - 2)^2} = \sqrt{(6)^2 + (4)^2} = \sqrt{36 + 16} = \sqrt{52} \approx 7.21$$ ### Step 3: Calculate $$JL$$ $$d_{JL} = \sqrt{(6 - (-6))^2 + (6 - 8)^2} = \sqrt{(12)^2 + (-2)^2} = \sqrt{144 + 4} = \sqrt{148} \approx 12.17$$ ### Step 4: Add distances to find the perimeter $$\text{Perimeter} = d_{JK} + d_{KL} + d_{JL} = 8.49 + 7.21 + 12.17 \approx 27.87$$ ### Final Answer: The perimeter of the triangle is approximately **27.87 units**. Would you like a detailed explanation of any step? ### Follow-up Questions: 1. How is the distance formula derived from the Pythagorean theorem? 2. Can you calculate the area of $$\triangle JKL$$? 3. How can the perimeter be verified graphically? 4. What are the implications of rounding during such calculations? 5. How would this process change if the coordinates included decimals? ### Tip: Always double-check graph points for accuracy when solving geometry problems based on plotted coordinates.

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Math Problem Analysis

Mathematical Concepts

Distance calculation
Perimeter
Analytic geometry

Formulas

Distance formula: d = √((x2 - x1)^2 + (y2 - y1)^2)

Theorems

Pythagorean Theorem (used indirectly in the distance formula)

Suitable Grade Level

Grades 8-10

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