Question

Suppose a parallelogram is drawn on a coordinate plane. Using vertices with the coordinates  (2,1) ,  (2,3) ,  (4,4) , and  (4,2) , compute the perimeter. Round your answer to the nearest tenth.(1 point)The perimeter is approximately  units.

Answer 1

To find the perimeter of the parallelogram defined by the vertices \( (2,1) \), \( (2,3) \), \( (4,4) \), and \( (4,2) \), we need to first find the lengths of its sides.

1. **Identify the vertices:**
- \( A(2, 1) \)
- \( B(2, 3) \)
- \( C(4, 4) \)
- \( D(4, 2) \)

2. **Calculate the lengths of the sides:**
- **Length of side AB:**
\[
AB = \sqrt{(2 - 2)^2 + (3 - 1)^2} = \sqrt{0 + 4} = \sqrt{4} = 2
\]

- **Length of side BC:**
\[
BC = \sqrt{(4 - 2)^2 + (4 - 3)^2} = \sqrt{(2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5}
\]

- **Length of side CD:**
\[
CD = \sqrt{(4 - 4)^2 + (2 - 4)^2} = \sqrt{0 + 4} = \sqrt{4} = 2
\]

- **Length of side DA:**
\[
DA = \sqrt{(2 - 4)^2 + (1 - 2)^2} = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}
\]

3. **Calculate the perimeter:**
The perimeter \( P \) of the parallelogram is the sum of the lengths of its sides:
\[
P = AB + BC + CD + DA = 2 + \sqrt{5} + 2 + \sqrt{5} = 4 + 2\sqrt{5}
\]

4. **Approximate the value of \( \sqrt{5} \):**
We know that \( \sqrt{5} \approx 2.236 \).
Therefore,
\[
2\sqrt{5} \approx 2 \times 2.236 \approx 4.472
\]

5. **Calculate the perimeter numerically:**
\[
P \approx 4 + 4.472 \approx 8.472
\]

6. **Round to the nearest tenth:**
\[
P \approx 8.5
\]

Thus, the perimeter of the parallelogram is approximately **8.5 units**.