Question

A rectangle is drawn on a coordinate plane with vertices (0,0), (2,2), (9,−4), and (7,−6). Using the coordinates, what is the approximate area of the rectangle?

Answer 1

To determine whether the given vertices form a rectangle and to find its area, we need to analyze the coordinates provided: (0,0), (2,2), (9,−4), and (7,−6).

First, let's organize the coordinates:
- \( A(0, 0) \)
- \( B(2, 2) \)
- \( C(9, -4) \)
- \( D(7, -6) \)

Next, we can calculate the distances between the points to check for the sides of a rectangle:

1. **Distance \( AB \)**:
\[
d_{AB} = \sqrt{(2-0)^2 + (2-0)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]

2. **Distance \( BC \)**:
\[
d_{BC} = \sqrt{(9-2)^2 + (-4-2)^2} = \sqrt{(7)^2 + (-6)^2} = \sqrt{49 + 36} = \sqrt{85}
\]

3. **Distance \( CD \)**:
\[
d_{CD} = \sqrt{(7-9)^2 + (-6 - (-4))^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]

4. **Distance \( DA \)**:
\[
d_{DA} = \sqrt{(0-7)^2 + (0-(-6))^2} = \sqrt{(-7)^2 + (6)^2} = \sqrt{49 + 36} = \sqrt{85}
\]

Now, we can group the distances:
- \( AB \) and \( CD \) are both \( 2\sqrt{2} \).
- \( BC \) and \( DA \) are both \( \sqrt{85} \).

Since opposite sides are equal and all angles would be right angles (which we can check further), the four points (0,0), (2,2), (9,-4), and (7,-6) represent vertices of a rectangle.

Next, we will compute the area of the rectangle, which is given by the formula:
\[
\text{Area} = \text{length} \times \text{width}
\]
Here, we can take:
- Length = \( \sqrt{85} \)
- Width = \( 2\sqrt{2} \)

Now we calculate the area:
\[
\text{Area} = \sqrt{85} \times 2\sqrt{2} = 2\sqrt{170}
\]

To approximate the numerical value of the area, let's compute:
- Approximating \( \sqrt{170} \):
\[
\sqrt{170} \approx 13.0384 \quad (\text{using a calculator})
\]
Therefore,
\[
\text{Area} \approx 2 \times 13.0384 \approx 26.0768
\]

Thus, the approximate area of the rectangle is:
\[
\boxed{26.08}
\]