Chapter 11.4, Problem 21WE

To determine

To Find: Perimeter and area.

Answer to Problem 21WE

  $\begin{array}{l}\\ p=6.21\end{array}$

  $A\approx 3sq.unit$

Explanation of Solution

Given: Regular dodecagon having twelve sides

  

Given: $r=1$

Solving

  $\begin{array}{l}m\angle AOB=\frac{{360}^{°}}{n}=\frac{360}{12}\\ m\angle AOB={30}^{°}\\ m\angle AOX=\frac{1}{2}m\angle AOB=\frac{1}{2}(30)={15}^{°}\end{array}$

Now,

  $\begin{array}{l}In\Delta AOX\\ \sin {15}^{°}=\frac{AX}{AO}=\frac{s}{2(r)}=\frac{s}{2}\\ s=2\sin {15}^{°}\\ \cos 15=\frac{OX}{OA}=\frac{a}{r}=\frac{a}{1}\\ a=\cos {15}^{°}\end{array}$

Now

Perimeter

  $\begin{array}{l}p=ns\\ p=12s\\ p=12(2\sin {15}^{°})\\ p=24\sin {15}^{°}\\ p=24(0.2588)\\ p=6.21\end{array}$

i.e.

  $\begin{array}{l}a=0.9659\\ A=\frac{1}{2}pa\\ A=\frac{1}{2}(6.211)(0.9659)\\ A\approx 3sq.unit\end{array}$