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 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 determine the shape by connecting the vertices in order.

1. **Plotting the Points:**
   - $$ H(-6,-2) $$
   - $$ A(-10,4) $$
   - $$ T(-6,6) $$
   - $$ C(-2,4) $$

2. **Connecting the Points:**
   - Connect $$ H $$ to $$ A $$
   - Connect $$ A $$ to $$ T $$
   - Connect $$ T $$ to $$ C $$
   - Connect $$ C $$ back to $$ H $$

3. **Identifying the Shape:**
   - The vertices are connected in the order $$ H \to A \to T \to C \to H $$.
   - By examining the coordinates, we can see that the shape has four sides, which indicates it is a quadrilateral.

Next, we will calculate the lengths of the sides to find the perimeter and use the coordinates to find the area.

### Step 1: Calculate the Lengths of the Sides

Using the distance formula between two points $$ (x_1, y_1) $$ and $$ (x_2, y_2) $$:
$$
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
$$

- **Length $$ HA $$**:
$$
d_{HA} = \sqrt{(-10 - (-6))^2 + (4 - (-2))^2} = \sqrt{(-4)^2 + (6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}
$$

- **Length $$ AT $$**:
$$
d_{AT} = \sqrt{(-6 - (-10))^2 + (6 - 4)^2} = \sqrt{(4)^2 + (2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}
$$

- **Length $$ TC $$**:
$$
d_{TC} = \sqrt{(-2 - (-6))^2 + (4 - 6)^2} = \sqrt{(4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}
$$

- **Length $$ CH $$**:
$$
d_{CH} = \sqrt{(-6 - (-2))^2 + (-2 - 4)^2} = \sqrt{(-4)^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}
$$

### Step 2: Calculate the Perimeter

The perimeter $$ P $$ of the polygon is the sum of the lengths of all sides:
$$
P = d_{HA} + d_{AT} + d_{TC} + d_{CH} = 2\sqrt{13} + 2\sqrt{5} + 2\sqrt{5} + 2\sqrt{13} = 4\sqrt{13} + 4\sqrt{5}
$$

### Step 3: Calculate the Area

To find the area of the quadrilateral, we can use the shoelace formula:
$$
\text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right|
$$
Substituting the coordinates $$ H(-6,-2) $$, $$ A(-10,4) $$, $$ T(-6,6) $$, $$ C(-2,4) $$:
$$
\text{Area} = \frac{1}{2} \left| (-6)(4) + (-10)(6) + (-6)(4) + (-2)(-2) - ((-2)(-10) + (4)(-6) + (6)(-2) + (4)(-6)) \right|
$$
Calculating each term:
$$
= \frac{1}{2} \left| -24 - 60 - 24 + 4 - (20 - 24 - 12 - 24) \right|
$$
$$
= \frac{1}{2} \left| -84 + 20 \right| = \frac{1}{2} \left| -64 \right| = 32
$$

### Final Results

- **Polygon Type**: Quadrilateral
- **Perimeter**: $$ 4\sqrt{13} + 4\sqrt{5} $$
- **Area**: $$ 32 $$

Thus, the polygon is a quadrilateral with a perimeter of $$ 4\sqrt{13} + 4\sqrt{5} $$ and an area of $$ 32 $$.
The polygon is a quadrilateral with a perimeter of $$ 4\sqrt{13} + 4\sqrt{5} $$ and an area of $$ 32 $$.

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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