Chapter 10, Problem 26RP

a.

To determine

To find: the length of a side of each inner polygon.

Answer to Problem 26RP

The length of the side of each inner polygon is $6\sqrt{3},6\sqrt{2},6$ respectively.

Explanation of Solution

Given information: Each circle below is inscribed in a regular polygon and is

circumscribed about another regular polygon.

The length of each outer polygon is 12.

  

Calculation:

Below figure shows the first circle.

  

If the radius is x, then inner polygon side is x and outer polygon side is $\frac{2x}{\sqrt{3}}$ .

  $\begin{array}{c}12=\frac{2x}{\sqrt{3}}\\ x=6\sqrt{3}\end{array}$

Below figure shows the second circle.

  

If the radius is r, then inner polygon side is $\frac{2r}{\sqrt{2}}=\sqrt{2}r$ and outer polygon side is 2r .

  $\begin{array}{c}12=2r\\ r=6\end{array}$

Inner polygon side is $6\sqrt{2}$

Below figure shows the third circle.

  

If the radius is r, then inner polygon side is $\frac{2\times 3r}{2\sqrt{3}}=r\sqrt{3}$ and outer polygon side is $\frac{2\times 3r}{\sqrt{3}}=2r\sqrt{3}$ .

  $\begin{array}{c}12=2r\sqrt{3}\\ r=2\sqrt{3}\end{array}$

Inner side is 6.

b.

To determine

To find: the ratio of the sides of the smaller polygon to the sides of the larger polygon, in each case.

Answer to Problem 26RP

The ratio of the sides of the smaller polygon to the sides of the larger polygon, in each case is $\frac{\sqrt{3}}{2},\frac{\sqrt{2}}{2},\frac{1}{2}$ respectively.

Explanation of Solution

Given information:

Calculation:

Below figure shows the first circle.

  

If the radius is x, then inner polygon side is x and outer polygon side is $\frac{2x}{\sqrt{3}}$ .

  $\begin{array}{c}12=\frac{2x}{\sqrt{3}}\\ x=6\sqrt{3}\\ \text{The}\text{ratio}\text{is}\frac{x}{\frac{2x}{\sqrt{3}}}=\frac{\sqrt{3}}{2}\end{array}$

Below figure shows the second circle.

  

If the radius is r, then inner polygon side is $\frac{2r}{\sqrt{2}}=\sqrt{2}r$ and outer polygon side is 2r .

  $\begin{array}{c}12=2r\\ r=6\end{array}$

Inner polygon side is $6\sqrt{2}$ .

  $\text{The}\text{ratio}\text{is}\frac{r\sqrt{2}}{2r}=\frac{\sqrt{2}}{2}$

Below figure shows the third circle.

  

If the radius is r, then inner polygon side is $\frac{2\times 3r}{2\sqrt{3}}=r\sqrt{3}$ and outer polygon side is $\frac{2\times 3r}{\sqrt{3}}=2r\sqrt{3}$ .

  $\begin{array}{c}12=2r\sqrt{3}\\ r=2\sqrt{3}\end{array}$

Inner side is 6.

  $\text{The}\text{ratio}\text{is}\frac{r\sqrt{3}}{2r\sqrt{3}}=\frac{1}{2}$