Chapter 11.4, Problem 3CCYP

To determine

To calculate: The area of the octagon.

Answer to Problem 3CCYP

The area of the octagon is $106.016{\text{ in}}^2$.

Explanation of Solution

Given information: The length of apothem is $8cm$ .

Formula Used: The area of the octagon is equal to 8 times the area of a triangle since octagon contains 8 equal triangles in it. Therefore,

  $A_{\text{Octagon}}=8\times A_{Triangle}$ .......(1)

• Here, $A_{\text{Octagon}}$is the area of the octagon and $A_{Triangle}$ is the area of the triangle.

Calculation:

The length of the apothem is $8cm$ .

The central angle of the triangle is,

  $\begin{array}{c}\theta =\frac{{360}^{\circ }}{8}\\ ={45}^{\circ }\end{array}$

The measure of the angle $\angle BAC$ is,

  $\begin{array}{c}\angle BAC=\frac{{45}^{\circ }}{2}\\ ={22.5}^{\circ }\end{array}$

Consider the figure shown below:

  

The area of the octagon is,

  $\begin{array}{c}{A}_{\text{Octagon}}=8\times A_{Triangle}\\ =8\times \frac{1}{2}\times \text{Base}\times \text{ Height}\\ =\text{4}\times \text{Base}\times \text{ 8}\\ =\text{32}\times \text{Base}\end{array}$

Therefore, the base of the triangle is,

  $\text{Base}=2\times BC$

Calculate the length of the BC.

  $\begin{array}{c}\tan {22.5}^{\circ }=\frac{BC}{AB}\\ \tan {22.5}^{\circ }=\frac{BC}{8}\\ BC=8\tan {22.5}^{\circ }\\ =3.313\end{array}$

Thus, the area of the octagon is,

  $\begin{array}{c}{A}_{\text{Octagon}}=8\times A_{Triangle}\\ =8\times \frac{1}{2}\times \text{Base}\times \text{ Height}\\ =\text{4}\times \text{Base}\times \text{ 8}\\ =\text{32}\times \text{Base}\\ \text{=32}\times \text{3}\text{.313}\\ \text{=106}\text{.016}\end{array}$

Hence, the area of the octagon is $106.016{\text{ in}}^2$.