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 \( ^{2} \)

**Step 1. Plotting and Ordering the Vertices** The vertices are \[ H(-6,-2),\quad A(-10,4),\quad T(-6,6),\quad C(-2,4) \] When these points are connected in the order \( H \to A \to T \to C \to H \), the quadrilateral formed is a kite. --- **Step 2. Compute the Side Lengths** Use the distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\): \[ \text{Distance} = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \] 1. **Side \(HA\) (from \(H\) to \(A\)):** \[ HA = \sqrt{(-10 - (-6))^2+ (4-(-2))^2} = \sqrt{(-4)^2+ (6)^2} = \sqrt{16+36} = \sqrt{52} = 2\sqrt{13}. \] 2. **Side \(AT\) (from \(A\) to \(T\)):** \[ AT = \sqrt{(-6-(-10))^2+(6-4)^2} = \sqrt{(4)^2+(2)^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}. \] 3. **Side \(TC\) (from \(T\) to \(C\)):** \[ TC = \sqrt{(-2-(-6))^2+(4-6)^2} = \sqrt{(4)^2+(-2)^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}. \] 4. **Side \(CH\) (from \(C\) to \(H\)):** \[ CH = \sqrt{(-6-(-2))^2+(-2-4)^2} = \sqrt{(-4)^2+(-6)^2} = \sqrt{16+36} = \sqrt{52} = 2\sqrt{13}. \] Thus, the side lengths are paired as follows: \[ HA = CH = 2\sqrt{13} \quad \text{and} \quad AT = TC = 2\sqrt{5}. \] --- **Step 3. Identify the Polygon** Since two distinct pairs of adjacent sides are equal, the quadrilateral is a **kite**. --- **Step 4. Calculate the Perimeter** Add all the side lengths: \[ P = HA + AT + TC + CH = 2\sqrt{13} + 2\sqrt{5} + 2\sqrt{5} + 2\sqrt{13} = 4\sqrt{13} + 4\sqrt{5}. \] --- **Step 5. Calculate the Area** For a kite, the area is given by \[ A = \frac{1}{2} \times d_1 \times d_2, \] where \(d_1\) and \(d_2\) are the lengths of the diagonals. - **Diagonal \(AC\) (from \(A\) to \(C\)):** \[ AC = \sqrt{(-2-(-10))^2+(4-4)^2} = \sqrt{(8)^2+0^2} = 8. \] - **Diagonal \(HT\) (from \(H\) to \(T\)):** \[ HT = \sqrt{(-6-(-6))^2+(6-(-2))^2} = \sqrt{0^2+(8)^2} = 8. \] Now, compute the area: \[ A = \frac{1}{2} \times 8 \times 8 = 32. \] --- **Final Answer** kite; \( P=4\sqrt{13}+4\sqrt{5} \) units; \( A=32 \) units\( ^2 \)