Chapter 11.4, Problem 15WE

To determine

To find: The area $A$ .

Answer to Problem 15WE

  $A=216\sqrt{3}sq.unit$

Explanation of Solution

Given:

  

  

Here Given: $p=72$

Solving

  $\begin{array}{l}m\angle AOB=\frac{{360}^{°}}{n}=\frac{360}{6}\\ m\angle AOB={60}^{°}\\ m\angle AOX=\frac{1}{2}m\angle AOB=\frac{1}{2}(60)={30}^{°}\end{array}$

Now,

  $\begin{array}{l}In\Delta AOX\\ \tan {30}^{°}=\frac{AX}{OX}=\frac{s}{2(a)}=\frac{1}{\sqrt{3}}\\ a=\frac{\sqrt{3}}{2}s\mathrm{..}(i)\\ \cos 30=\frac{OX}{OA}=\frac{a}{r}=\frac{\sqrt{3}}{2}\\ a=\frac{\sqrt{3}}{2}r\mathrm{..}(ii)\\ from(i)and(ii)\\ r=s\end{array}$

Now

  $\begin{array}{l}P=72\\ P=6ss\\ 72=6\\ s=12\\ a=\frac{\sqrt{3}}{2}s=\frac{\sqrt{3}}{2}\left(12\right)\\ a=6\sqrt{3}\\ A=\frac{1}{2}pa\\ A=\frac{1}{2}\left(72\right)(6\sqrt{3})\\ A=216\sqrt{3}sq.unit\end{array}$