Question

Suppose a rectangle is drawn on a coordinate plane. What is the perimeter of the rectangle if its verticies are at coordinates (0,0), (4,4), (5,3), and (1,-1)? Round your answer to the nearest tenth. (1 point) The perimeter is approximately units.

Answer 1

To find the perimeter of the rectangle with the given vertices at coordinates \((0,0)\), \((4,4)\), \((5,3)\), and \((1,-1)\), we first need to verify if these points indeed form a rectangle and then calculate the lengths of the sides.

1. **Calculate the lengths of the sides**:
- We can find the distance between pairs of points using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

2. **Calculate distance for each pair of points**:
- Between \((0,0)\) and \((4,4)\):
\[
d_1 = \sqrt{(4 - 0)^2 + (4 - 0)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}
\]

- Between \((0,0)\) and \((1,-1)\):
\[
d_2 = \sqrt{(1 - 0)^2 + (-1 - 0)^2} = \sqrt{1 + 1} = \sqrt{2}
\]

- Between \((1,-1)\) and \((5,3)\):
\[
d_3 = \sqrt{(5 - 1)^2 + (3 - (-1))^2} = \sqrt{(4)^2 + (4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}
\]

- Between \((4,4)\) and \((5,3)\):
\[
d_4 = \sqrt{(5 - 4)^2 + (3 - 4)^2} = \sqrt{1 + 1} = \sqrt{2}
\]

3. **Identify the lengths of the sides**:
- The distances \(d_1\) and \(d_3\) are both \(4\sqrt{2}\), and the distances \(d_2\) and \(d_4\) are both \(\sqrt{2}\).
- It seems we have two pairs of equal lengths which indicates that we could have a rectangle.

4. **Calculating Perimeter**:
The perimeter \(P\) of a rectangle is given by:
\[
P = 2 \times (\text{length} + \text{width})
\]
Here, we can take \(\text{length} = 4\sqrt{2}\) and \(\text{width} = \sqrt{2}\):
\[
P = 2 \times (4\sqrt{2} + \sqrt{2}) = 2 \times (5\sqrt{2}) = 10\sqrt{2}
\]

5. **Approximate \(10\sqrt{2}\)**:
Using \(\sqrt{2} \approx 1.414\):
\[
10\sqrt{2} \approx 10 \times 1.414 = 14.14
\]
Rounding to the nearest tenth, we get \(14.1\).

Thus, the perimeter of the rectangle is approximately **14.1 units**.