1. **Problem statement:** Find the area of the trapezoid at bottom-left. 2. **Formula for trapezoid area:** $$\text{Area} = \frac{1}{2} \times h \times (b_1 + b_2)$$ where $h$ is the height, $b_1$ and $b_2$ are the lengths of the two bases. 3. **Given values:** Height $h = 6$ m, bases $b_1 = 10.4$ m and $b_2 = 7.4$ m. 4. **Calculate the sum of the bases:** $$10.4 + 7.4 = 17.8$$ 5. **Calculate the area:** $$\text{Area} = \frac{1}{2} \times 6 \times 17.8 = 3 \times 17.8 = 53.4 \text{ m}^2$$ --- 1. **Problem statement:** Find the area of the triangle at bottom-center-left. 2. **Formula for triangle area:** $$\text{Area} = \frac{1}{2} \times b \times h$$ 3. **Given values:** Base $b = 7.3$ yd, height $h = 10$ yd. 4. **Calculate the area of one triangle:** $$\frac{1}{2} \times 7.3 \times 10 = 36.5 \text{ yd}^2$$ 5. **Since the figure is two such triangles, multiply by 2:** $$36.5 \times 2 = 73 \text{ yd}^2$$ --- 1. **Problem statement:** Find the area of the triangle at bottom-center-right. 2. **Formula for triangle area:** $$\text{Area} = \frac{1}{2} \times b \times h$$ 3. **Given values:** Base $b = 13$ in, height $h = 6$ in. 4. **Calculate the area:** $$\frac{1}{2} \times 13 \times 6 = \frac{1}{2} \times 78 = 39 \text{ in}^2$$ **Note:** The other calculation $(\frac{1}{2} \times 16 \times 20.6 = 164.8)$ seems unrelated to this triangle's base and height. --- **Final answers:** - Trapezoid area = $53.4$ m$^2$ - Bottom-center-left triangle area = $73$ yd$^2$ - Bottom-center-right triangle area = $39$ in$^2$