Question

A rectangle is drawn on a coordinate plane with vertices (0, 0), (2, 2), (9, -4), and (7, -6). Using the coordinates, which is the approximate area of the rectangle?
A. 52.2 units^2
B. 25.76 units^2
C. 24.1 units^2
D. 12.05 units^2

Answer 1

To determine the area of the rectangle, we will first identify two pairs of opposite vertices.

The vertices given are:
- A (0, 0)
- B (2, 2)
- C (9, -4)
- D (7, -6)

We can check if these points form the corners of a rectangle by calculating the lengths of its sides, which can be determined using the distance formula.

1. **Calculating distances between the vertices:**

Let's find the distance between A (0, 0) and B (2, 2):

\[
d_{AB} = \sqrt{(2 - 0)^2 + (2 - 0)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \approx 2.83
\]

Now, let’s calculate the distance between C (9, -4) and D (7, -6):

\[
d_{CD} = \sqrt{(7 - 9)^2 + (-6 + 4)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \approx 2.83
\]

Next, we calculate the distance between A (0, 0) and D (7, -6):

\[
d_{AD} = \sqrt{(7 - 0)^2 + (-6 - 0)^2} = \sqrt{7^2 + (-6)^2} = \sqrt{49 + 36} = \sqrt{85} \approx 9.22
\]

And then the distance between B (2, 2) and C (9, -4):

\[
d_{BC} = \sqrt{(9 - 2)^2 + (-4 - 2)^2} = \sqrt{7^2 + (-6)^2} = \sqrt{49 + 36} = \sqrt{85} \approx 9.22
\]

2. **Finding the area of the rectangle:**

Now, from the calculated lengths:
- One pair of sides has lengths \(d_{AB} = d_{CD} \approx 2.83\)
- The other pair of sides has lengths \(d_{AD} = d_{BC} \approx 9.22\)

The area \(A\) of the rectangle can be computed using:
\[
A = \text{length} \times \text{width} \approx 9.22 \times 2.83 \approx 26.1
\]

Now, we compare this area to the provided options:
A. 52.2 units²
B. 25.76 units²
C. 24.1 units²
D. 12.05 units²

The closest option to our calculated area of approximately \(26.1\) units² is **B. 25.76 units²**.

Thus, the correct answer is **B. 25.76 units²**.