Chapter 11.4, Problem 8WE

To determine

To find: Radius $r$ , apothem $a$ and area $A$ .

Answer to Problem 8WE

  $r=3,a=\frac{3}{2},A=\frac{27\sqrt{3}}{4}sq.unit$

Explanation of Solution

Given:

  

  

Here Given: $P=9\sqrt{3}$

Solving

  $\begin{array}{l}m\angle AOB=\frac{{360}^{°}}{n}=\frac{360}{3}\\ m\angle AOB={120}^{°}\\ m\angle AOX=\frac{1}{2}m\angle AOB=\frac{1}{2}(120)={60}^{°}\end{array}$

Now,

  $\begin{array}{l}In\Delta AOX\\ \tan {60}^{°}=\frac{AX}{OX}=\frac{s}{2(a)}=\sqrt{3}\\ s=2a\sqrt{3}\\ \cos 60=\frac{OX}{OA}=\frac{a}{r}=\frac{1}{2}\\ 2a=r\end{array}$

Now

Perimeter

  $\begin{array}{l}p=ns\\ p=3s\end{array}$

i.e.

  $\begin{array}{l}P=9\sqrt{3}\\ P=3s\\ 9\sqrt{3}=3s\\ s=3\sqrt{3}\\ r=2(a)=2\left(\frac{s}{2\sqrt{3}}\right)=2\left(\frac{3\sqrt{3}}{2\sqrt{3}}\right)\\ r=3\\ a=\frac{r}{2}=\frac{3}{2}\\ a=\frac{3}{2}\\ A=\frac{1}{2}ap\\ A=\frac{1}{2}\left(\frac{3}{2}\right)(9\sqrt{3})\\ A=\frac{27\sqrt{3}}{4}sq.unit\end{array}$