1. **State the problem:** Find the area of the quadrilateral with vertices at points $A(2,2)$, $B(2,8)$, $C(7,7)$, and $D(7,3)$. 2. **Formula used:** To find the area of a polygon given vertices, we can use the Shoelace formula: $$\text{Area} = \frac{1}{2} \left| \sum_{i=1}^{n-1} (x_i y_{i+1} - y_i x_{i+1}) + (x_n y_1 - y_n x_1) \right|$$ where $(x_i, y_i)$ are the coordinates of the vertices in order. 3. **Apply the formula:** Label vertices in order: $A(2,2)$, $B(2,8)$, $C(7,7)$, $D(7,3)$. Calculate the sum: $$S_1 = x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_1 = 2 \times 8 + 2 \times 7 + 7 \times 3 + 7 \times 2 = 16 + 14 + 21 + 14 = 65$$ $$S_2 = y_1 x_2 + y_2 x_3 + y_3 x_4 + y_4 x_1 = 2 \times 2 + 8 \times 7 + 7 \times 7 + 3 \times 2 = 4 + 56 + 49 + 6 = 115$$ 4. **Calculate area:** $$\text{Area} = \frac{1}{2} |S_1 - S_2| = \frac{1}{2} |65 - 115| = \frac{1}{2} \times 50 = 25$$ 5. **Interpretation:** The area of the quadrilateral is 25 square units. **Final answer:** $$\boxed{25}$$