1. **Problem statement:** We need to find the length of the fence BD and the area of the quadrilateral ABCD with given sides and angles. 2. **Given data:** - AB = 246 m - BC = 312 m - AD = 257 m - $\angle DAB = 96^\circ$ - $\angle BCD = 78^\circ$ 3. **Step 1: Find length BD using the Law of Cosines in triangles ABD and BCD.** - In triangle ABD, use Law of Cosines: $$BD^2 = AB^2 + AD^2 - 2 \times AB \times AD \times \cos(96^\circ)$$ Calculate: $$BD^2 = 246^2 + 257^2 - 2 \times 246 \times 257 \times \cos(96^\circ)$$ - Compute $\cos(96^\circ)$ (approx -0.1045): $$BD^2 = 60516 + 66049 + 2 \times 246 \times 257 \times 0.1045$$ $$BD^2 = 126565 + 13199.5 = 139764.5$$ - So, $$BD = \sqrt{139764.5} \approx 374.0 \text{ m}$$ 4. **Step 2: Find area of ABCD by splitting into triangles ABD and BCD.** - Area of triangle ABD using formula: $$\text{Area}_{ABD} = \frac{1}{2} \times AB \times AD \times \sin(96^\circ)$$ Calculate $\sin(96^\circ) \approx 0.9945$: $$\text{Area}_{ABD} = 0.5 \times 246 \times 257 \times 0.9945 \approx 31472.5 \text{ m}^2$$ - To find area of triangle BCD, first find angle $\angle BDC$ using triangle BCD: Since $\angle BCD = 78^\circ$, and we know BD and BC, use Law of Cosines to find $\angle BDC$: $$BC^2 = BD^2 + CD^2 - 2 \times BD \times CD \times \cos(\angle BDC)$$ We don't know CD yet, so find CD using Law of Cosines in triangle ADC or use coordinate geometry approach. 5. **Step 3: Use coordinate geometry to find coordinates of points and then area.** - Place point A at origin (0,0). - AB is vertical, so B at (0,246). - AD length 257 m at 96° from AB, so D coordinates: $$D_x = 257 \times \cos(96^\circ) \approx 257 \times (-0.1045) = -26.86$$ $$D_y = 257 \times \sin(96^\circ) \approx 257 \times 0.9945 = 255.6$$ - BC is horizontal 312 m, so C is at (312, 246). 6. **Step 4: Calculate length BD using coordinates:** $$BD = \sqrt{(D_x - B_x)^2 + (D_y - B_y)^2} = \sqrt{(-26.86 - 0)^2 + (255.6 - 246)^2}$$ $$= \sqrt{(26.86)^2 + (9.6)^2} = \sqrt{721.3 + 92.2} = \sqrt{813.5} \approx 28.53 \text{ m}$$ 7. **Step 5: Calculate area of ABCD using Shoelace formula with points A(0,0), B(0,246), C(312,246), D(-26.86,255.6):** $$\text{Area} = \frac{1}{2} |x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)|$$ Calculate: $$= \frac{1}{2} |0 \times 246 + 0 \times 246 + 312 \times 255.6 + (-26.86) \times 0 - (0 \times 0 + 246 \times 312 + 246 \times (-26.86) + 255.6 \times 0)|$$ $$= \frac{1}{2} |0 + 0 + 79747.2 + 0 - (0 + 76752 - 6609.6 + 0)|$$ $$= \frac{1}{2} |79747.2 - 70142.4| = \frac{1}{2} \times 9604.8 = 4802.4 \text{ m}^2$$ **Final answers:** - Length of fence BD is approximately **28.53 m**. - Area of paddock ABCD is approximately **4802.4 m²**.