Chapter 12.6, Problem 59SPR

To determine

Calculate the area of the shaded region.

Answer to Problem 59SPR

The area of the shaded region is $80.17unit{s}^2$ .

Explanation of Solution

Given:

It is given in the question that there is a figure given below:

  

Concept Used:

In this, use the concept of trigonometric ratios and area of a circle and area of a octagon i.e. $A_{circle}=\pi r^2\&A_{octagon}=\frac{1}{2}aP$ .

Calculation: Let the central angle be $\alpha$ . Now, use the trigonometric ratios,

  

  $\alpha =360÷16={22.5}^{\circ }$

  $\begin{array}{l}\sin {22.5}^{\circ }=\frac{x}{16}\\ x=16\sin {22.5}^{\circ }\end{array}$

Now,

  $s=2x=2\cdot 16\sin {22.5}^{\circ }=32\sin {22.5}^{\circ }$

  $P=n\cdot s=8\cdot 32\sin {22.5}^{\circ }=25632\sin {22.5}^{\circ }$

Now,

  $\begin{array}{l}{A}_{circle}=\pi r^2\\ A_{circle}=\pi {(}^{16}\\ A_{circle}=256\pi \end{array}$

  $\begin{array}{l}{A}_{octagon}=\frac{1}{2}aP\\ A_{octagon}=\frac{1}{2}(16\cos {22.5}^{\circ })(256\sin {22.5}^{\circ })\\ A_{octagon}=2048\cos {22.5}^{\circ }\cdot \sin {22.5}^{\circ }\end{array}$

  $\begin{array}{l}{A}_{shaded}=A_{circle}-A_{octagon}\\ A_{shaded}=256\pi -2048\cos {22.5}^{\circ }\cdot \sin {22.5}^{\circ }\\ A_{shaded}=80.17unit{s}^2\end{array}$

Conclusion:

The area is $80.17unit{s}^2$ .