Circle Segment Area 7Bfe0E

1. **State the problem:** We need to find the area of the shaded segment of a circle where the central angle $\angle WNB = 120^\circ$ and the radius $BN = 8$ ft. 2. **Identify the formula:** The area of a segment of a circle is given by: $$\text{Area of segment} = \text{Area of sector} - \text{Area of triangle}$$ 3. **Calculate the area of the sector:** The area of a sector with central angle $\theta$ (in degrees) and radius $r$ is: $$\text{Area of sector} = \frac{\theta}{360} \times \pi r^2$$ Given $\theta = 120^\circ$ and $r = 8$ ft, $$\text{Area of sector} = \frac{120}{360} \times \pi \times 8^2 = \frac{1}{3} \times \pi \times 64 = \frac{64\pi}{3}$$ 4. **Calculate the area of the triangle formed by the two radii and the chord:** The triangle $WNB$ is isosceles with sides $BN = NW = 8$ ft and included angle $120^\circ$. Area of triangle using formula: $$\text{Area} = \frac{1}{2} r^2 \sin(\theta) = \frac{1}{2} \times 8^2 \times \sin(120^\circ) = \frac{1}{2} \times 64 \times \sin(120^\circ)$$ Since $\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}$, $$\text{Area} = 32 \times \frac{\sqrt{3}}{2} = 16\sqrt{3}$$ 5. **Calculate the area of the segment:** $$\text{Area of segment} = \frac{64\pi}{3} - 16\sqrt{3}$$ 6. **Approximate the numerical value:** Using $\pi \approx 3.1416$ and $\sqrt{3} \approx 1.732$, $$\frac{64\pi}{3} \approx \frac{64 \times 3.1416}{3} = 67.02$$ $$16\sqrt{3} \approx 16 \times 1.732 = 27.71$$ So, $$\text{Area of segment} \approx 67.02 - 27.71 = 39.31$$ 7. **Final answer:** The area of the shaded segment is approximately **39.3** square feet to the nearest tenth.