To determine the most precise name for the quadrilateral ABCD, we first need to analyze the coordinates and calculate the distances between the vertices to identify the shape.

The coordinates are given as:

• $A(0, 2)$
• $B(5, 2)$
• $C(-2, -5)$
• $D(-7, -5)$

Step 1: Analyze the sides

1. Distance between $A(0, 2)$ and $B(5, 2)$:

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(5 - 0)^2 + (2 - 2)^2} = \sqrt{25} = 5$

2. Distance between $B(5, 2)$ and $C(-2, -5)$:

$BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(-2 - 5)^2 + (-5 - 2)^2} = \sqrt{49 + 49} = \sqrt{98} \approx 7$

3. Distance between $C(-2, -5)$ and $D(-7, -5)$:

$CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(-7 - (-2))^2 + (-5 - (-5))^2} = \sqrt{25 + 0} = 5$

4. Distance between $D(-7, -5)$ and $A(0, 2)$:

$DA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(0 - (-7))^2 + (2 - (-5))^2} = \sqrt{49 + 49} = \sqrt{98} \approx 7$

Step 2: Analyze the parallel sides

• $AB = 5$ and $CD = 5$ → These are parallel sides because they are both horizontal (same $y$-coordinate values for both points).
• $BC = 7$ and $DA = 7$ → These are also parallel sides because they both have the same slope. We can check this by calculating the slopes.

Slope of $AB$ (horizontal line, so slope = 0):

$\text{Slope of } AB = \frac{2 - 2}{5 - 0} = 0$

Slope of $CD$ (horizontal line, so slope = 0):

$\text{Slope of } CD = \frac{-5 - (-5)}{-7 - (-2)} = 0$

Slope of $BC$:

$\text{Slope of } BC = \frac{-5 - 2}{-2 - 5} = \frac{-7}{-7} = 1$

Slope of $DA$:

$\text{Slope of } DA = \frac{2 - (-5)}{0 - (-7)} = \frac{7}{7} = 1$

Since the slopes of $BC$ and $DA$ are the same, they are parallel lines.

Step 3: Conclusion on the type of quadrilateral

• The quadrilateral has two pairs of parallel sides: $AB \parallel CD$ and $BC \parallel DA$.
• It has opposite sides of equal length: $AB = CD = 5$ and $BC = DA = 7$.

This makes the quadrilateral a parallelogram, and specifically, since all sides are not equal, we can classify it as a rectangle.

Step 4: Area of the parallelogram

The area of a parallelogram is given by the formula:

$\text{Area} = \text{base} \times \text{height}$

Using $AB = 5$ as the base, and the distance between the two parallel lines (height), which is the vertical distance between line $AB$ and $CD$, we find the height to be $7$ (from the calculation above).

Thus, the area of quadrilateral ABCD is:

$\text{Area} = 5 \times 7 = 35 \, \text{square units}$

Final Answer:

• Most precise name for the quadrilateral: Parallelogram (specifically a rectangle)
• Area: 35 square units

Would you like to see more details on how the area or classification was determined? Or do you have any other questions?

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Here are 5 related questions you might find interesting:

1. How do we calculate the area of other types of quadrilaterals like rhombuses or trapezoids?
2. What would happen if the quadrilateral was a square instead of a rectangle?
3. How do you find the area of a triangle given three vertices in a coordinate plane?
4. Can a quadrilateral with only one pair of parallel sides still be considered a parallelogram?
5. How does the Pythagorean theorem apply to calculating side lengths or distances in the coordinate plane?

Tip: When working with coordinate geometry, always double-check if the sides are parallel (by comparing slopes) to identify the shape correctly!