Chapter 11.6, Problem 15WE

To determine

To Find: The area of shaded region.

Answer to Problem 15WE

  $A=12\pi +8sq.units$

Explanation of Solution

Given:

  

  

  $d=8$

  $r=\frac{8}{2}=4$

Area of the sector.

  $\begin{array}{l}A=\frac{x}{360}(\pi r^2)\\ A=\frac{90}{360}(\pi {\left(4\right)}^2)\\ A_s=4\pi \end{array}$

Now area of triangle

  $\begin{array}{l}A=\frac{1}{2}bh\\ A=\frac{1}{2}(4)(4)\\ A_t=8\end{array}$

Area of unshaded region.

  $\begin{array}{l}A=A_s-A_t\\ A=4\pi -8sq.units\end{array}$

Area of circle.

  $\begin{array}{l}{A}_c=\pi r^2\\ $ A_c=\pi {(4)}^2$A_c=16\pi \end{array}$

Area of shaded region = area of circle − area of unshaded region

  $\begin{array}{l}A=16\pi -(4\pi -8)\\ A=12\pi +8sq.units\end{array}$

Hence, $A=12\pi +8sq.units$