Question

Suppose a parallelogram is drawn on a coordinate plane. Using the vertices with the coordinates H=(2,1)
, A=(2,3)
, W=(4,4)
, and K=(4,2)
, compute the perimeter. Round your answer to the nearest tenth.

Answer 1

To find the perimeter of the parallelogram formed by the vertices \( H(2,1) \), \( A(2,3) \), \( W(4,4) \), and \( K(4,2) \), we first need to calculate the lengths of the sides.

1. **Calculate the length of side \( HA \)**:
\[
HA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(2 - 2)^2 + (3 - 1)^2} = \sqrt{0 + 4} = \sqrt{4} = 2
\]

2. **Calculate the length of side \( AW \)**:
\[
AW = \sqrt{(4 - 2)^2 + (4 - 3)^2} = \sqrt{(2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5}
\]

3. **Calculate the length of side \( WK \)**:
\[
WK = \sqrt{(4 - 4)^2 + (2 - 4)^2} = \sqrt{0 + (-2)^2} = \sqrt{4} = 2
\]

4. **Calculate the length of side \( KH \)**:
\[
KH = \sqrt{(4 - 2)^2 + (1 - 2)^2} = \sqrt{(2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}
\]

Now, we have the lengths of the sides:
- \( HA = 2 \)
- \( AW = \sqrt{5} \)
- \( WK = 2 \)
- \( KH = \sqrt{5} \)

Since the opposite sides of a parallelogram are equal, we can calculate the perimeter \( P \):
\[
P = HA + AW + WK + KH = 2 + \sqrt{5} + 2 + \sqrt{5} = 4 + 2\sqrt{5}
\]

Now, we evaluate \( \sqrt{5} \) for a numerical approximation:
\[
\sqrt{5} \approx 2.236
\]
Thus,
\[
P \approx 4 + 2 \cdot 2.236 \approx 4 + 4.472 \approx 8.472
\]

Rounding to the nearest tenth:
\[
P \approx 8.5
\]

Thus, the perimeter of the parallelogram is approximately \(\boxed{8.5}\).