Identify the polygon with vertices $$ K(0,1), L(2,-4) $$, $$ M(-3,-2) $$, and $$ N(-5,3) $$, and then find the perimeter and area of the polygon. rhombus; $$ P=4 \sqrt{29} $$ units; $$ A=21 $$ units $$ ^{2} $$ parallelogram; $$ P=20 $$ units; $$ \mathrm{A}=21 $$ units $$ ^{2} $$ parallelogram; $$ P=4 \sqrt{29} $$ units; $$ ^{2} A=29 $$ units $$ ^{2} $$ rhombus; $$ P=12 $$ units; $$ \mathrm{A}=29 $$ units $$ ^{2} $$

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UpStudy AI · Accepted Answer

To identify the polygon formed by the vertices $$ K(0,1) $$, $$ L(2,-4) $$, $$ M(-3,-2) $$, and $$ N(-5,3) $$, we will follow these steps:

1. **Calculate the lengths of the sides** to determine the type of polygon.
2. **Calculate the perimeter** of the polygon.
3. **Calculate the area** of the polygon.

### Step 1: Calculate the lengths of the sides

We will use the distance formula to find the lengths of the sides:

$$
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
$$

Calculating the lengths of each side:

- Length $$ KL $$:
$$
d_{KL} = \sqrt{(2 - 0)^2 + (-4 - 1)^2} = \sqrt{2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29}
$$

- Length $$ LM $$:
$$
d_{LM} = \sqrt{(-3 - 2)^2 + (-2 + 4)^2} = \sqrt{(-5)^2 + 2^2} = \sqrt{25 + 4} = \sqrt{29}
$$

- Length $$ MN $$:
$$
d_{MN} = \sqrt{(-5 + 3)^2 + (3 + 2)^2} = \sqrt{(-2)^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29}
$$

- Length $$ NK $$:
$$
d_{NK} = \sqrt{(0 + 5)^2 + (1 - 3)^2} = \sqrt{5^2 + (-2)^2} = \sqrt{25 + 4} = \sqrt{29}
$$

### Step 2: Determine the type of polygon

Since all sides $$ KL, LM, MN, NK $$ are equal in length ($$ \sqrt{29} $$), the polygon is a rhombus.

### Step 3: Calculate the perimeter

The perimeter $$ P $$ of a rhombus is given by:

$$
P = 4 \times \text{length of one side}
$$

Thus,

$$
P = 4 \times \sqrt{29}
$$

### Step 4: Calculate the area

The area $$ A $$ of a rhombus 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. We will first find the diagonals $$ KM $$ and $$ LN $$.

- Length $$ KM $$:
$$
d_{KM} = \sqrt{(-3 - 0)^2 + (-2 - 1)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}
$$

- Length $$ LN $$:
$$
d_{LN} = \sqrt{(-5 - 2)^2 + (3 + 4)^2} = \sqrt{(-7)^2 + 7^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2}
$$

Now, substituting into the area formula:

$$
A = \frac{1}{2} \times (3\sqrt{2}) \times (7\sqrt{2}) = \frac{1}{2} \times 21 \times 2 = 21
$$

### Conclusion

The polygon is a rhombus with:

- Perimeter $$ P = 4\sqrt{29} $$ units
- Area $$ A = 21 $$ units$$^2$$

Thus, the correct answer is:

**rhombus; $$ P=4 \sqrt{29} $$ units; $$ A=21 $$ units $$ ^{2} $$**.
The polygon is a rhombus with a perimeter of $$ 4\sqrt{29} $$ units and an area of 21 square units.

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