Question

Suppose a parallelogram is drawn on a coordinate plane. Using vertices with the coordinates (0,0) , (2,3) , (5,2) , and (3,−1) , compute the perimeter. Round your answer to the nearest tenth.

Answer 1

To find the perimeter of the parallelogram formed by the given vertices, we first need to calculate the lengths of its sides using the distance formula between points.

The given vertices of the parallelogram are:
- A (0, 0)
- B (2, 3)
- C (5, 2)
- D (3, -1)

The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is computed using the formula:

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

Let's calculate the lengths of the sides \( AB \) and \( BC \).

1. **Calculating \( AB \)**:
\[
AB = \sqrt{(2 - 0)^2 + (3 - 0)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}
\]

2. **Calculating \( BC \)**:
\[
BC = \sqrt{(5 - 2)^2 + (2 - 3)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}
\]

Noting that \( AD \) and \( CD \) will have the same lengths as \( AB \) and \( BC \) respectively, we will find:

3. **Calculating \( AD \)**:
\[
AD = \sqrt{(3 - 0)^2 + (-1 - 0)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}
\]

4. **Calculating \( CD \)**:
\[
CD = \sqrt{(5 - 3)^2 + (2 - (-1))^2} = \sqrt{2^2 + (2 + 1)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}
\]

Now, we have:

- Length of \( AB = \sqrt{13} \)
- Length of \( BC = \sqrt{10} \)

The perimeter \( P \) of the parallelogram is given by:

\[
P = 2(AB + BC) = 2(\sqrt{13} + \sqrt{10})
\]

Calculating the approximate values:

\[
\sqrt{13} \approx 3.60555
\]
\[
\sqrt{10} \approx 3.16228
\]

Thus, we have:

\[
AB + BC \approx 3.60555 + 3.16228 \approx 6.76783
\]

Now calculating the perimeter:

\[
P \approx 2 \times 6.76783 \approx 13.53566
\]

Finally, rounding to the nearest tenth:

\[
\text{Perimeter} \approx 13.5
\]

Therefore, the perimeter of the parallelogram is approximately \( \boxed{13.5} \).