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

Step-by-step explanation:

Answer is

  39.3 ft²

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}

Read more on -

start with a line segment pq that we will copy. mark a point r that will be one endpoint of the new line segment. set the compasses' point on the point p of the line segment to be copied. adjust the compasses' width to the point q. the compasses' width is now equal to the length of the line segment pq.

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