Chapter 11, Problem 26SGR

To determine

To find: The area of the octagon.

Answer to Problem 26SGR

The area of the octagon is $100.68{\mathrm{cm}}^2$ .

Explanation of Solution

Given Information: The side of regular octagon is $6cm$ .

Calculation: The side of regular octagon is $6cm$ .

  

Central angle $=\frac{{360}^{\circ }}{8}={45}^{\circ }$

Apothem $\mathrm{co}$ is height of the $\mathrm{AOB}$ it bisect the angle $\mathrm{AOB}$ and the side $\mathrm{AB}$ .

So angle $\mathrm{AOC}=\frac{{45}^{\circ }}{2}={22.5}^{\circ }$ .

  $\begin{array}{l}\frac{AC}{OA}=\sin {22.5}^{\circ }\\ \frac{AC}{6}=0.38\\ AC=6\times 0.38\\ AC=2.28\mathrm{cm}\\ AB=2\times AC=2\times 2.28=4.56\mathrm{cm}\\ \frac{OC}{OA}=\cos {22.5}^{\circ }\\ \frac{OC}{6}=0.92\\ OC=6\times 0.92\\ OC=5.52\mathrm{cm}\end{array}$

So $a=5.52\mathrm{cm}$ .

Perimeter :

  $\begin{array}{l}=8\times \mathrm{side}\\ =8\times 4.56\\ =36.48\mathrm{cm}\end{array}$

Area of octagon:

  $\begin{array}{l}=\frac{1}{2}aP\\ =\frac{1}{2}(5.52)(36.48)\\ =(5.52)(18.24)\\ =100.68{\mathrm{cm}}^2\end{array}$

Thus, the area of octagon is $100.68{\mathrm{cm}}^2$ .