Question

Use the coordinates to compute the approximate area of the rectangle.
The vertices coordinates are (3,1),(4,-2), (-2,-4), and (-3,-1)

Answer 1

To find the area of the rectangle formed by the given vertices, we can follow these steps:

1. **Identify the vertices**: We have the vertices at points A(3, 1), B(4, -2), C(-2, -4), and D(-3, -1).

2. **Calculate the lengths of the sides**: To do this, we need to find the distances between the vertices that are adjacent to each other.

Let's pair the points with those that are adjacent to each other in a potential rectangle configuration.

- The distance between A(3, 1) and B(4, -2):
\[
d_{AB} = \sqrt{(4 - 3)^2 + (-2 - 1)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}
\]

- The distance between B(4, -2) and C(-2, -4):
\[
d_{BC} = \sqrt{(-2 - 4)^2 + (-4 - (-2))^2} = \sqrt{(-6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}
\]

- The distance between C(-2, -4) and D(-3, -1):
\[
d_{CD} = \sqrt{(-3 - (-2))^2 + (-1 - (-4))^2} = \sqrt{(-1)^2 + (3)^2} = \sqrt{1 + 9} = \sqrt{10}
\]

- The distance between D(-3, -1) and A(3, 1):
\[
d_{DA} = \sqrt{(3 - (-3))^2 + (1 - (-1))^2} = \sqrt{(6)^2 + (2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}
\]

3. **Compute the area**: The area of a rectangle is given by \( \text{length} \times \text{width} \). From our distances calculated:

- Let’s define the width as \( \sqrt{10} \) and the length as \( 2\sqrt{10} \). Thus, the area \( A \) can be computed as:
\[
A = \text{length} \times \text{width} = 2\sqrt{10} \times \sqrt{10} = 2 \times 10 = 20
\]

Thus, the approximate area of the rectangle is \( \boxed{20} \).