1. **State the problem:** We need to find the area of Field 1 and then use that to find the area of Field 2. The fields form two triangles within a larger triangle ABCD. 2. **Identify given information:** - Side AC = 35 m - Side AB = 30 m - Side BD = 10 m - Angle at A = 50° - Angle ABC is not a right angle 3. **Find the area of Field 1 (triangle ABC):** We use the formula for the area of a triangle given two sides and the included angle: $$\text{Area} = \frac{1}{2}ab\sin(C)$$ Here, sides AB and AC with included angle at A (50°): $$\text{Area}_{Field1} = \frac{1}{2} \times 30 \times 35 \times \sin(50^\circ)$$ 4. **Calculate the sine value:** $$\sin(50^\circ) \approx 0.7660$$ 5. **Calculate the area of Field 1:** $$\text{Area}_{Field1} = \frac{1}{2} \times 30 \times 35 \times 0.7660 = 15 \times 35 \times 0.7660 = 525 \times 0.7660 = 402.15$$ Rounded to nearest m²: $$\text{Area}_{Field1} \approx 402\,m^2$$ 6. **Find the area of Field 2 (triangle BCD):** Since BD = 10 m and BC is vertical line dividing the fields, the height of Field 2 is the same as the height of Field 1 from point C to AB. 7. **Calculate height from C to AB:** Area of Field 1 = $\frac{1}{2} \times base \times height$ Base = AB = 30 m $$402.15 = \frac{1}{2} \times 30 \times h \Rightarrow 402.15 = 15h \Rightarrow h = \frac{402.15}{15} = 26.81\,m$$ 8. **Calculate area of Field 2:** Base = BD = 10 m Height = 26.81 m (same height as Field 1) $$\text{Area}_{Field2} = \frac{1}{2} \times 10 \times 26.81 = 5 \times 26.81 = 134.05$$ Rounded to nearest m²: $$\text{Area}_{Field2} \approx 134\,m^2$$ **Final answers:** - Area of Field 1 = 402 m² - Area of Field 2 = 134 m²