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 $$ ^{2} $$

Question

UpStudy AI · Accepted Answer

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 $$ ^{2} $$**.
The polygon is a kite with a perimeter of $$ 4\sqrt{5} + 4\sqrt{13} $$ units and an area of 32 square units.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Mind Expander

preguntas relacionadas

Latest Geometry Questions