Identify the polygon with vertices \( H(-6,-2), A(-10,4) \), \( T(-6,6) \), and \( C(-2,4) \) and then find the perimeter and area of the polygon.

To identify the polygon with the given vertices \( H(-6,-2) \), \( A(-10,4) \), \( T(-6,6) \), and \( C(-2,4) \), we can start by plotting these points on a Cartesian coordinate system and connecting them in the order they are given. The vertices form a quadrilateral since there are four points. To find the perimeter, we need to calculate the lengths of each side: 1. Distance \( HA \): \[ d = \sqrt{(-10 - (-6))^2 + (4 - (-2))^2} = \sqrt{(-4)^2 + (6)^2} = \sqrt{16 + 36} = \sqrt{52} \approx 7.21 \] 2. Distance \( AT \): \[ d = \sqrt{(-6 - (-10))^2 + (6 - 4)^2} = \sqrt{(4)^2 + (2)^2} = \sqrt{16 + 4} = \sqrt{20} \approx 4.47 \] 3. Distance \( TC \): \[ d = \sqrt{(-2 - (-6))^2 + (4 - 6)^2} = \sqrt{(4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} \approx 4.47 \] 4. Distance \( CH \): \[ d = \sqrt{(-6 - (-2))^2 + (-2 - 4)^2} = \sqrt{(-4)^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} \approx 7.21 \] Now, summing all these lengths gives us the perimeter: \[ \text{Perimeter} = HA + AT + TC + CH \approx 7.21 + 4.47 + 4.47 + 7.21 \approx 23.36 \] Next, to find the area of the quadrilateral formed by the vertices \( H \), \( A \), \( T \), and \( C \), we can use the shoelace formula: \[ \text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right| \] Assigning the points as follows: - \( H(-6, -2) \) (point 1) - \( A(-10, 4) \) (point 2) - \( T(-6, 6) \) (point 3) - \( C(-2, 4) \) (point 4) The calculation becomes: \[ \text{Area} = \frac{1}{2} \left| (-6)(4) + (-10)(6) + (-6)(4) + (-2)(-2) - ((-2)(-10) + 4(-6) + 6(-2) + 4(-6)) \right| \] Calculating the right-hand side: \[ \text{Area} = \frac{1}{2} \left| -24 -60 -24 + 4 - (20 -24 -12 -24) \right| = \frac{1}{2} \left| -104 + 8 \right| = \frac{1}{2} \left| -96 \right| = 48 \] Thus, the area of the polygon is \( 48 \) square units. Final Results: - **Identified Polygon**: Quadrilateral - **Perimeter**: Approximately \( 23.36 \) - **Area**: \( 48 \) square units.