To the nearest tenth, what is the area of the shaded segment when BN = 8 ft?


Answer:
Step-by-step explanation:
Answer is
39.3 ft²
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We need to find the area of that sector of the circle.
Then we subtract that by the area of the ΔBNW.
Area of a sector of a circle where \theta is the cent. ang of the sector (in degrees) and r is the radius:
A_{\text{sector}} = \pi r^2 \frac{\theta}{360}
We know that the radius is 8 ft; it is given to us. The central angle is also given as 120°
Number crunching:
\begin{aligned} A_{\text{sector}} &= \pi(8)^2 \frac{120}{360} \\ &= 64 \pi \cdot \frac{1}{3} \\ &= \frac{64}{3}\pi \end{aligned}
Area of a triangle is given by
A_{\text{triangle}} = \frac{1}{2}ab\sin C
where a and b are sides that are not opposite to angle C.
Well we know two sides of the triangle; they're identical sides because they're both the same length of the radius. We also know an angle in the triangle that is not opposite to either of our known sides.
C = 120° and a = b = 8 ft, therefore
\begin{aligned} A_{\text{triangle}} &= \tfrac{1}{2}(8)(8) \sin120 \\ &= 32\sin 120 \end{aligned}
Area of the shaded segment (use a calculator, degree mode)
\begin{aligned} A_\text{shaded} &= A_{\text{sector}} - A_{\text{triangle}} \\ &= \tfrac{64}{3}\pi - 32\sin 120 \\ &= 39.3\text{ ft}^2 \end{aligned}
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