1. **Problem 1: Area of the trapezoid with a right triangle inside** Given dimensions: height $h=6$ m, bases $b_1=7.4$ m and $b_2=10.4$ m. Formula for trapezoid area: $$\text{Area} = \frac{1}{2} (b_1 + b_2) h$$ Calculate: $$\text{Area} = \frac{1}{2} (7.4 + 10.4) \times 6$$ $$= \frac{1}{2} \times 17.8 \times 6$$ $$= \cancel{\frac{1}{2}} \times 17.8 \times \cancel{6}$$ (showing cancellation of 2 and 6 as $6=2\times3$) $$= 17.8 \times 3 = 53.4 \text{ m}^2$$ 2. **Problem 2: Area of the triangle with sides 12 yd, 22.3 yd, and 10 yd, doubled** The problem states to use $\frac{1}{2} \times \text{base} \times \text{height}$ and multiply by 2. Assuming base $b=7.3$ yd and height $h=10$ yd (from the note $\frac{1}{2} 7.3 \times 10$), area of one triangle: $$\text{Area} = \frac{1}{2} \times 7.3 \times 10 = \frac{1}{2} \times 73 = 36.5 \text{ yd}^2$$ Multiply by 2: $$2 \times 36.5 = 73 \text{ yd}^2$$ 3. **Problem 3: Area of composite shape (triangle + rectangle)** Triangle base $b=20.6$ in, height $h=22$ in. Triangle area: $$\text{Area}_{\triangle} = \frac{1}{2} \times 20.6 \times 22 = \frac{1}{2} \times 453.2 = 226.6 \text{ in}^2$$ Rectangle dimensions: 13 in by 6 in. Rectangle area: $$\text{Area}_{\square} = 13 \times 6 = 78 \text{ in}^2$$ Total area: $$226.6 + 78 = 304.6 \text{ in}^2$$ **Final answers:** - Trapezoid area = $53.4$ m$^2$ - Triangle area (doubled) = $73$ yd$^2$ - Composite shape area = $304.6$ in$^2$