Identify the polygon with vertices $$ P(2,-2), Q(4,2) $$, $$ R(6,2) $$, and $$ S(4,-2) $$, and then find the perimeter and area of the polygon.

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We are given the vertices  
$$
P(2,-2),\quad Q(4,2),\quad R(6,2),\quad S(4,-2)
$$  
and we wish to identify the polygon, then compute its perimeter and its area.

---

**Step 1. Identify the Polygon**

Connect the vertices in order: $$ P \to Q \to R \to S \to P $$.

1. Compute vectors corresponding to the sides:
   - $$ \overrightarrow{PQ} = (4-2,\; 2-(-2)) = (2,4) $$
   - $$ \overrightarrow{QR} = (6-4,\; 2-2) = (2,0) $$
   - $$ \overrightarrow{RS} = (4-6,\; -2-2) = (-2,-4) $$
   - $$ \overrightarrow{SP} = (2-4,\; -2-(-2)) = (-2,0) $$

2. Notice that:
   - $$\overrightarrow{PQ} = (2,4)$$ and $$\overrightarrow{RS} = (-2,-4)$$ have the same magnitude and are parallel (opposite in direction),
   - $$\overrightarrow{QR} = (2,0)$$ and $$\overrightarrow{SP} = (-2,0)$$ are horizontal, equal in magnitude, and parallel (again oppositely directed).

3. In a quadrilateral, if both pairs of opposite sides are equal and parallel, the shape is a **parallelogram**.

Thus, the polygon is a parallelogram.

---

**Step 2. Find the Perimeter**

1. Find the lengths of each side:
   - Length of $$PQ$$:
     $$
     PQ = \sqrt{(2)^2 + (4)^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}.
     $$
     
   - Length of $$QR$$:
     $$
     QR = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2.
     $$
     
   - Length of $$RS$$:
     $$
     RS = \sqrt{(-2)^2 + (-4)^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}.
     $$
     
   - Length of $$SP$$:
     $$
     SP = \sqrt{(-2)^2 + (0)^2} = \sqrt{4} = 2.
     $$

2. The perimeter $$P$$ is the sum of the side lengths:
   $$
   P = PQ+QR+RS+SP = 2\sqrt{5} + 2 + 2\sqrt{5} + 2 = 4\sqrt{5} + 4.
   $$

---

**Step 3. Find the Area**

One way to find the area of a parallelogram is by using the magnitude of the cross product of two adjacent side vectors. Choose vertex $$P$$ with adjacent sides $$\overrightarrow{PQ}$$ and $$\overrightarrow{PS}$$.

1. We have:
   - $$\overrightarrow{PQ} = (2,4)$$,
   - $$\overrightarrow{PS} = S - P = (4-2,\; -2-(-2)) = (2,0)$$.

2. The area $$A$$ is given by the absolute value of the determinant:
   $$
   A = \left| \det\begin{pmatrix} 2 & 4 \ 2 & 0 \end{pmatrix} \right| = \left| (2)(0) - (4)(2) \right| = |0-8| = 8.
   $$

---

**Final Answers:**

- The polygon is a **parallelogram**.
- The perimeter is  
  $$
  4\sqrt{5}+4.
  $$
- The area is  
  $$
  8.
  $$
The polygon is a parallelogram. Its perimeter is $$4\sqrt{5} + 4$$ and its area is 8.

Identify the polygon with vertices \( P(2,-2), Q(4,2) \), \( R(6,2) \), and \( S(4,-2) \), and then find the perimeter and area of the polygon.

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