Chapter 11.6, Problem 16WE

To determine

To Find: The area of shaded region.

Answer to Problem 16WE

  $18\pi -18\sqrt{3}$

Explanation of Solution

Given:

  

  $\begin{array}{l}r=6\\ d=2(6)=12\end{array}$

In triangle using trigonometric ratios.

  $\begin{array}{l}\cos 60=\frac{a}{12}=\frac{1}{2}\\ a=6\\ \sin 60=\frac{b}{12}=\frac{\sqrt{3}}{2}\\ b=6\sqrt{3}\end{array}$

Area of shaded region

  $\begin{array}{l}=\frac{1}{2}\pi r^2-\frac{1}{2}ab\\ $ =\frac{1}{2}\pi {(6)}^2-\frac{1}{2}(6)(6\sqrt{3})$=18\pi -18\sqrt{3}\end{array}$