1. **State the problem:** Find the area of the composite figure consisting of a rectangle and a trapezoid on top. 2. **Identify dimensions:** - Rectangle: width = 18 ft, height = 12 ft - Trapezoid: top base = 10 ft, height = 6 ft, right angle on the left side means the trapezoid has one vertical leg. 3. **Formula for area:** - Rectangle area = width \times height - Trapezoid area = \frac{(b_1 + b_2)}{2} \times h, where $b_1$ and $b_2$ are the two bases and $h$ is the height. 4. **Find trapezoid bases:** - Bottom base of trapezoid = width of rectangle = 18 ft - Top base of trapezoid = 10 ft 5. **Calculate areas:** - Rectangle area = $18 \times 12 = 216$ ft$^2$ - Trapezoid area = $\frac{(10 + 18)}{2} \times 6 = \frac{28}{2} \times 6 = 14 \times 6 = 84$ ft$^2$ 6. **Sum areas for total area:** $$\text{Total area} = 216 + 84 = 300 \text{ ft}^2$$ **Final answer:** The area of the figure is **300 ft$^2$**.