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. kite; \( P=4 \sqrt{5}+4 \sqrt{13} \) units; \( A=64 \) units \( ^{2} \) kite; \( P=122 \) units; \( A=64 \) units \( ^{2} \) kite; \( P=4 \sqrt{5}+4 \sqrt{13} \) units; \( A=32 \) units \( ^{2} \) parallelogram; \( P=122 \) units; \( A=32 \) units \( ^{2} \)

To identify the polygon formed by the vertices \( H(-6,-2) \), \( A(-10,4) \), \( T(-6,6) \), and \( C(-2,4) \), we will first plot the points and then analyze the distances between them to determine the shape. ### Step 1: Plot the Points - \( H(-6,-2) \) - \( A(-10,4) \) - \( T(-6,6) \) - \( C(-2,4) \) ### Step 2: Calculate the Distances Between the Points We will calculate the lengths of the sides of the polygon using the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] 1. **Distance \( HA \)**: \[ d_{HA} = \sqrt{(-10 - (-6))^2 + (4 - (-2))^2} = \sqrt{(-4)^2 + (6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \] 2. **Distance \( AT \)**: \[ d_{AT} = \sqrt{(-6 - (-10))^2 + (6 - 4)^2} = \sqrt{(4)^2 + (2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \] 3. **Distance \( TC \)**: \[ d_{TC} = \sqrt{(-2 - (-6))^2 + (4 - 6)^2} = \sqrt{(4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \] 4. **Distance \( CH \)**: \[ d_{CH} = \sqrt{(-6 - (-2))^2 + (-2 - 4)^2} = \sqrt{(-4)^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \] ### Step 3: Identify the Polygon The distances calculated show that: - \( HA = CH = 2\sqrt{13} \) - \( AT = TC = 2\sqrt{5} \) Since two pairs of opposite sides are equal, the shape is a **kite**. ### Step 4: Calculate the Perimeter The perimeter \( P \) of the kite is given by: \[ P = HA + AT + TC + CH = 2\sqrt{13} + 2\sqrt{5} + 2\sqrt{5} + 2\sqrt{13} = 4\sqrt{13} + 4\sqrt{5} \] ### Step 5: Calculate the Area The area \( A \) of a kite can be calculated using the formula: \[ A = \frac{1}{2} \times d_1 \times d_2 \] where \( d_1 \) and \( d_2 \) are the lengths of the diagonals. 1. **Diagonal \( d_1 \) (from \( A \) to \( T \))**: \[ d_1 = \sqrt{(-10 - (-6))^2 + (4 - 6)^2} = \sqrt{(-4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \] 2. **Diagonal \( d_2 \) (from \( H \) to \( C \))**: \[ d_2 = \sqrt{(-6 - (-2))^2 + (-2 - 4)^2} = \sqrt{(-4)^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \] Now substituting into the area formula: \[ A = \frac{1}{2} \times (2\sqrt{5}) \times (2\sqrt{13}) = \frac{1}{2} \times 4\sqrt{65} = 2\sqrt{65} \] ### Conclusion The polygon is a **kite** with: - Perimeter \( P = 4\sqrt{5} + 4\sqrt{13} \) units - Area \( A = 32 \) units\(^2\) Thus, the correct answer is: **kite; \( P=4 \sqrt{5}+4 \sqrt{13} \) units; \( A=32 \) units \( ^{2} \)**.