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 formed by the given vertices \((0,0)\), \((4,4)\), \((5,3)\), and \((1,-1)\), we first need to determine if these points truly form a rectangle.

Next, we'll calculate the distances between each pair of adjacent vertices.

1. **Distance between \((0,0)\) and \((4,4)\)**:
\[
d = \sqrt{(4 - 0)^2 + (4 - 0)^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \approx 5.7
\]

2. **Distance between \((4,4)\) and \((5,3)\)**:
\[
d = \sqrt{(5 - 4)^2 + (3 - 4)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \approx 1.4
\]

3. **Distance between \((5,3)\) and \((1,-1)\)**:
\[
d = \sqrt{(1 - 5)^2 + (-1 - 3)^2} = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \approx 5.7
\]

4. **Distance between \((1,-1)\) and \((0,0)\)**:
\[
d = \sqrt{(0 - 1)^2 + (0 - (-1))^2} = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \approx 1.4
\]

Now that we have calculated the lengths of all four sides, we can compute the perimeter \(P\):
\[
P = 2 \times (5.7 + 1.4) = 2 \times 7.1 = 14.2
\]

Therefore, the perimeter of the rectangle is approximately \( \boxed{14.2} \) units.