Question

What is the area of a rectangle with vertices at (−3, −1), (1, 3), (3, 1), and (−1, −3)?

Enter your answer in the box. Do not round any side lengths.

Answer 1

To find the area of the rectangle with vertices at \((-3, -1)\), \( (1, 3) \), \( (3, 1) \), and \((-1, -3)\), we first need to identify the lengths of the sides of the rectangle.

1. **Identifying the coordinates:**
- \( A = (-3, -1) \)
- \( B = (1, 3) \)
- \( C = (3, 1) \)
- \( D = (-1, -3) \)

2. **Calculating the side lengths:**
To find the lengths of the sides, we can compute the distances between the points that are supposed to be adjacent vertices.

- **Distance between \( A (-3, -1) \) and \( B (1, 3) \):**

\[
AB = \sqrt{(1 - (-3))^2 + (3 - (-1))^2} = \sqrt{(1 + 3)^2 + (3 + 1)^2} = \sqrt{(4)^2 + (4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}
\]

- **Distance between \( B (1, 3) \) and \( C (3, 1) \):**

\[
BC = \sqrt{(3 - 1)^2 + (1 - 3)^2} = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]

- **Distance between \( C (3, 1) \) and \( D (-1, -3) \):**

\[
CD = \sqrt{(-1 - 3)^2 + (-3 - 1)^2} = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}
\]

- **Distance between \( D (-1, -3) \) and \( A (-3, -1) \):**

\[
DA = \sqrt{(-3 - (-1))^2 + (-1 - (-3))^2} = \sqrt{(-2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]

3. **Width and height of the rectangle:**
From our calculations, we see that the length of one side (width) is \( AB = 4\sqrt{2} \) and the length of the adjacent side (height) is \( BC = 2\sqrt{2} \).

4. **Calculating the area:**
Finally, the area of the rectangle is given by the product of the lengths of two adjacent sides:

\[
\text{Area} = AB \times BC = (4\sqrt{2}) \times (2\sqrt{2}) = 8 \times 2 = 16
\]

Thus, the area of the rectangle is:

\[
\boxed{16}
\]