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To calculate the perimeter of quadrilateral \( GHJI \), we need to find the lengths of all its sides and sum them up. We can determine these lengths by applying the distance formula to each pair of adjacent vertices.

The distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

### Coordinates of the vertices (as seen from the graph):
- \( G(-8, 4) \)
- \( H(-4, -8) \)
- \( I(4, -6) \)
- \( J(6, 8) \)

We will now calculate the distances:

1. **Distance \( GH \):**
\[
d(GH) = \sqrt{(-4 - (-8))^2 + (-8 - 4)^2} = \sqrt{(4)^2 + (-12)^2} = \sqrt{16 + 144} = \sqrt{160} \approx 12.6
\]

2. **Distance \( HI \):**
\[
d(HI) = \sqrt{(4 - (-4))^2 + (-6 - (-8))^2} = \sqrt{(8)^2 + (2)^2} = \sqrt{64 + 4} = \sqrt{68} \approx 8.2
\]

3. **Distance \( IJ \):**
\[
d(IJ) = \sqrt{(6 - 4)^2 + (8 - (-6))^2} = \sqrt{(2)^2 + (14)^2} = \sqrt{4 + 196} = \sqrt{200} \approx 14.1
\]

4. **Distance \( JG \):**
\[
d(JG) = \sqrt{(6 - (-8))^2 + (8 - 4)^2} = \sqrt{(14)^2 + (4)^2} = \sqrt{196 + 16} = \sqrt{212} \approx 14.6
\]

### Perimeter:
Now, summing these distances to get the perimeter:
\[
\text{Perimeter} \approx 12.6 + 8.2 + 14.1 + 14.6 = 49.5 \, \text{units}
\]

The perimeter of quadrilateral \( GHJI \) is approximately **49.5 units**.

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Would you like a detailed breakdown of any step? Here are 5 related questions to deepen your understanding:

1. How does the distance formula apply to other polygons?
2. How would the perimeter change if the vertices of the quadrilateral were shifted?
3. What are the possible ways to calculate the area of quadrilateral \( GHJI \)?
4. Can this method be applied to a three-dimensional shape? How would it differ?
5. How does rounding affect the accuracy of the perimeter calculation?

**Tip:** Always double-check the coordinates of each point to ensure accurate distance calculation!

What is the perimeter of quadrilateral GHJI?

Learn how to calculate the perimeter of quadrilateral GHJI by applying the distance formula between points on a coordinate plane. This problem demonstrates key concepts in geometry and coordinate geometry, suitable for students in Grades 8-10.

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