Chapter 11.5, Problem 32WE

To determine

To Find: The area enclosed between the circle and the octagon in terms of $r$ .

Answer to Problem 32WE

  $A=0.31r^2$

Explanation of Solution

Given: A regular octagon is inscribed in a circle with radius $r$

  

Using central angle property.

  $\begin{array}{l}n=8\\ \angle AOX=\frac{180}{8}={55}^{°}\\ \cos \theta =\cos {45}^{°}=\frac{h}{r}\\ h=r\cos \frac{{45}^{°}}{2}\\ similarly\\ \sin {15}^{°}=\frac{s}{r}\\ s=2r\sin \left(\frac{{45}^{°}}{2}\right)\end{array}$

Now area of shaded region = Area of sector − area of triangle

Then

  $\begin{array}{l}A=\frac{2\theta }{360}\pi r^2-\frac{1}{2}bh\\ A=\frac{45}{360}\pi r^2-\frac{1}{2}sh\\ A=\frac{45}{360}\pi r^2-\frac{1}{2}(2r\sin \frac{{45}^{°}}{2})(r\cos \frac{{45}^{°}}{2})\\ A=\frac{\pi r^2}{8}-\frac{r^2}{2\sqrt{2}}\\ A=\frac{r^2}{8}(\pi -2\sqrt{2})\end{array}$

Now area of shaded region is eight times this area.

Thus

  $\begin{array}{l}A=(8)\frac{r^2}{8}(\pi -2\sqrt{2})\\ A=r^2(\pi -2\sqrt{2})\\ using\pi =3.14,\sqrt{2}=1.414\\ A=0.31r^2\end{array}$

Hence, $A=0.31r^2$