Chapter 11.6, Problem 17WE

To determine

Find the area of the shaded region.

Answer to Problem 17WE

7.653

Explanation of Solution

Given:

  

Calculation:

The given diagram can be sectioned as :

  

In the right triangle OGC,

  $\begin{array}{l}\sin 30°=\frac{\text{length of opposite side to 30}°}{\text{length of hypotenuse}}=\frac{1}{r}\\ \Rightarrow \frac{1}{2}=\frac{1}{r}\\ \Rightarrow r=2\end{array}$

  $\begin{array}{l}\cos 30°=\frac{\text{length of adjacent side to 30}°}{\text{length of hypotenuse}}=\frac{h}{r}\\ \Rightarrow \frac{\sqrt{3}}{2}=\frac{h}{2}\\ \Rightarrow h=\sqrt{3}\end{array}$

Area of sector COA :

  $\begin{array}{l}\frac{60°}{360°}\left(\pi r^2\right)\\ =\frac{1}{6}\left(4\pi \right)\end{array}$

Area of triangle COA:

  $\frac{1}{2}(2)h=h=\sqrt{3}$

So, the area of CGAD:

  $\begin{array}{l}\frac{1}{6}(4\pi )-\sqrt{3}\\ =\frac{2}{3}\pi -\sqrt{3}\\ =\frac{2\pi -3\sqrt{3}}{3}\\ =0.362\end{array}$

Area of the other side of the figure is equal to this .

Area of rectangle EFAC is :

  $2(2h)=4h=4\sqrt{3}$

So, the total area of the shaded region is :

  $0.362+4\sqrt{3}+0.362=7.653$