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.
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To identify the polygon with the given vertices \( H(-6,-2) \), \( A(-10,4) \), \( T(-6,6) \), and \( C(-2,4) \), we can plot the points on a coordinate plane to see their arrangement. 1. **Identifying the Polygon**: Plot the points on a graph. When connected in the order \( H \), \( A \), \( T \), and \( C \) (returning to \( H \)), we find that the shape formed is a quadrilateral. 2. **Calculating the perimeter**: To calculate the perimeter, we need to find the lengths of each side: - \( HA = \sqrt{(-10 + 6)^2 + (4 + 2)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \) - \( AT = \sqrt{(-6 + 10)^2 + (6 - 4)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \) - \( TC = \sqrt{(-2 + 6)^2 + (4 - 6)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \) - \( CH = \sqrt{(-6 + 2)^2 + (-2 - 4)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \) Now, summing the lengths gives: \[ \text{Perimeter} = 2\sqrt{13} + 2\sqrt{5} + 2\sqrt{5} + 2\sqrt{13} = 4\sqrt{13} + 4\sqrt{5} \] 3. **Calculating the area**: We can use the shoelace theorem to calculate the area of the quadrilateral. With the vertices \( H(-6,-2) \), \( A(-10,4) \), \( T(-6,6) \), and \( C(-2,4) \), we arrange them as follows: \[ \text{Area} = \frac{1}{2} \left| -6\cdot4 + -10\cdot6 + -6\cdot4 + -2\cdot(-2) - (-2\cdot-10 + 4\cdot-6 + 6\cdot-2 + 4\cdot-6) \right| \] \[ = \frac{1}{2} \left| -24 - 60 - 24 + 4 - (20 + 24 + 12 + 24) \right| \] \[ = \frac{1}{2} \left| -104 + 80 \right| = \frac{1}{2} \left| -24 \right| = 12 \] So, the polygon is a quadrilateral with a perimeter of \( 4\sqrt{13} + 4\sqrt{5} \) and an area of \( 12 \) square units.
