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.

The perimeter is approximately (answer here) units.

Answer 1

To find the perimeter of the parallelogram given its vertices at \((0,0)\), \((2,3)\), \((5,2)\), and \((3,-1)\), we can calculate the lengths of the sides using the distance formula. The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

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

### Step 1: Calculate the lengths of the sides

**1. Distance between \((0,0)\) and \((2,3)\):**

\[
d_1 = \sqrt{(2 - 0)^2 + (3 - 0)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}
\]

**2. Distance between \((2,3)\) and \((5,2)\):**

\[
d_2 = \sqrt{(5 - 2)^2 + (2 - 3)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}
\]

**3. Distance between \((5,2)\) and \((3,-1)\):**

\[
d_3 = \sqrt{(3 - 5)^2 + (-1 - 2)^2} = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}
\]

**4. Distance between \((3,-1)\) and \((0,0)\):**

\[
d_4 = \sqrt{(0 - 3)^2 + (0 - (-1))^2} = \sqrt{(-3)^2 + (1)^2} = \sqrt{9 + 1} = \sqrt{10}
\]

### Step 2: Calculate the perimeter

The perimeter \(P\) of the parallelogram is the sum of the lengths of all four sides:

\[
P = d_1 + d_2 + d_3 + d_4
\]

Substituting the values we calculated:

\[
P = \sqrt{13} + \sqrt{10} + \sqrt{13} + \sqrt{10} = 2\sqrt{13} + 2\sqrt{10}
\]

### Step 3: Approximate the values

Next, we need numeric approximations of \(\sqrt{13}\) and \(\sqrt{10}\):

\[
\sqrt{13} \approx 3.60555 \quad \text{and} \quad \sqrt{10} \approx 3.16228
\]

Calculating the perimeter:

\[
P \approx 2(3.60555) + 2(3.16228) = 7.2111 + 6.32456 \approx 13.53566
\]

### Step 4: Round to the nearest tenth

Rounding to the nearest tenth:

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

Thus, the perimeter is approximately \(\boxed{13.5}\) units.