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.

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 \[ 8. \]