Chapter 11.3, Problem 42PPS

To determine

The area of the shaded region.

Answer to Problem 42PPS

The value of area is $\text{A}\left(\text{shaded}\right)=22.1{\text{mm}}^2$

Explanation of Solution

Given information:

The area of each shaded region

  

Formula used:

  $\text{A}\left(\text{sector}\right)=\frac{x}{360}\times \pi r^2$

Calculation:

The area of the shaded region is the difference between the area covered by the minor arc and the area of the triangle. 

The central angle of the minor arc is 360 − 240 = 120. 

  $\begin{array}{l}\text{A}\left(\text{sector}\right)=\frac{x}{360}\times \pi r^2\\ =\frac{120}{360}\pi {\left(6\right)}^2\\ =\frac{1}{3}\pi \left(36\right)\\ \text{A}\left(\text{sector}\right)=12\pi \end{array}$

The central angle is 60°, so the triangle is equilateral. Use 36-60-90 triangles to find the height

  

The area of the triangle is calculated as,

  $\begin{array}{l}A=\frac{1}{2}bh\\ =\frac{1}{2}\left(6\right)\left(3\sqrt{3}\right)\\ A=9\sqrt{3}\end{array}$

  $\begin{array}{l}\text{A}\left(\text{shaded}\right)=\text{A}\left(\text{sector}\right)\text{-A}\left(\text{triangle}\right)\\ =12\pi -9\sqrt{3}\\ \text{A}\left(\text{shaded}\right)=22.1{\text{mm}}^2\end{array}$

Conclusion:

The value of area is $\text{A}\left(\text{shaded}\right)=22.1{\text{mm}}^2$