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The problem is asking to find the area of a trapezoid given four vertices with coordinates \( A(0, 0), B(3, 3), C(7, 3), D(5, 1) \).

### Steps to solve:

1. **Identify if the shape is a trapezoid**:
   A trapezoid has two parallel sides. We need to check if any pair of sides are parallel.

   - Slope of line \( AB \): 
     \[
     m_{AB} = \frac{3 - 0}{3 - 0} = 1
     \]
   - Slope of line \( CD \):
     \[
     m_{CD} = \frac{3 - 1}{7 - 5} = 1
     \]
   Since the slopes of \( AB \) and \( CD \) are equal, these two sides are parallel, confirming the shape is a trapezoid.

2. **Use the trapezoid area formula**:
   \[
   \text{Area} = \frac{1}{2} \times (b_1 + b_2) \times h
   \]
   where \( b_1 \) and \( b_2 \) are the lengths of the parallel sides, and \( h \) is the height (the perpendicular distance between the parallel sides).

3. **Find the lengths of the parallel sides**:
   - \( AB \) (using the distance formula):
     \[
     AB = \sqrt{(3 - 0)^2 + (3 - 0)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}
     \]
   - \( CD \):
     \[
     CD = \sqrt{(7 - 5)^2 + (3 - 1)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
     \]

4. **Find the height (perpendicular distance between the parallel lines)**:
   The parallel lines have the same slope. To find the height, we compute the perpendicular distance from any point on one line to the other. For simplicity, let's find the perpendicular distance from point \( D(5, 1) \) to line \( AB \).

   The formula for the perpendicular distance from a point \( (x_1, y_1) \) to a line \( ax + by + c = 0 \) is:
   \[
   d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}
   \]
   The equation of line \( AB \) is \( y = x \) or equivalently \( x - y = 0 \).

   Substituting \( D(5, 1) \) into the distance formula:
   \[
   d = \frac{|5 - 1|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}
   \]
   So, the height \( h = 2\sqrt{2} \).

5. **Calculate the area**:
   \[
   \text{Area} = \frac{1}{2} \times (3\sqrt{2} + 2\sqrt{2}) \times 2\sqrt{2}
   \]
   \[
   = \frac{1}{2} \times 5\sqrt{2} \times 2\sqrt{2} = \frac{1}{2} \times 20 = 10
   \]

### Final Answer:
The area of the trapezoid is **10 square units**.

Would you like more details or have any further questions?

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Here are 5 related questions:
1. How do you find the slopes of different line segments?
2. Can the area of a trapezoid be calculated if it's not on a grid?
3. How does the distance formula apply in 3D coordinates?
4. What makes a trapezoid different from other quadrilaterals?
5. Can you use trigonometry to find the height of a trapezoid?

**Tip:** When checking if two sides are parallel, comparing slopes is the quickest method.

Find the area of the trapezoid with vertices A(0, 0), B(3, 3), C(7, 3), and D(5, 1).

Solve for the area of a trapezoid with vertices A(0, 0), B(3, 3), C(7, 3), and D(5, 1) using coordinate geometry. This solution involves calculating slopes, distances, and applying the area formula for trapezoids. Suitable for students in Grades 9-12.

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