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 $$ ^{2} $$

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UpStudy AI · Accepted Answer

**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$$ ^2 $$
kite; perimeter $$ = 4\sqrt{13} + 4\sqrt{5} $$ units; area $$ = 32 $$ square units

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