Chapter 12.3, Problem 28E

To determine

To find : the polygon where themeasure of an interior angle and an exterior angle ratio is $1\text{ to }1$ , $2\text{ to }1$ and $2\text{ to }3$ .

Answer to Problem 28E

the polygon where the measure of an interior angle and an exterior angle ratio is $1\text{ to }1$ , $2\text{ to }1$ and $2\text{ to }3$ is tetrahedral , hexagon and triangle.

Explanation of Solution

Given : the ratioprovided in the question,

  $1\text{ to }1$

  $2\text{ to }1$

  $2\text{ to }3$

Formula used: the measure of an interior angle of a regular n -gon is $\frac{\left(n-2\right)·{180}^{\circ }}{n}$ .

Each exterior angle for a regular n-gon has measure equal to $\frac{{360}^{\circ }}{n}$ .

Calculation: for ratios we find the number of sides.

For ratio $1\text{ to }1$ ,

  $\begin{array}{l}\frac{\left(n-2\right)·{180}^{\circ }}{n}:\frac{{360}^{\circ }}{n}\\ \frac{\left(n-2\right)·{180}^{\circ }}{n}·n:\frac{{360}^{\circ }}{n}·n\text{ [multiply n by each side]}\\ \left(n-2\right)·{180}^{\circ }:{360}^{\circ }\\ \frac{\left(n-2\right)·{180}^{\circ }}{{180}^{\circ }}:\frac{{360}^{\circ }}{{180}^{\circ }}\text{ [divide each side by }{180}^{\circ }]\end{array}$

  $\begin{array}{l}\left(n-2\right):2\\ n-2+2:2+2\text{ [add 2 both side]}\\ n:4\end{array}$

So for ratio $1\text{ to }1$ the polygon is tetrahedral.

For ratio $2\text{ to }1$

  $\begin{array}{l}\frac{\left(n-2\right)·{180}^{\circ }}{n}:\frac{{360}^{\circ }}{n}·2\\ \frac{\left(n-2\right)·{180}^{\circ }}{n}·n:\frac{{720}^{\circ }}{n}·n\text{ [substitute each side by n]}\\ \left(n-2\right)·{180}^{\circ }:{720}^{\circ }\\ \frac{\left(n-2\right)·{180}^{\circ }}{{180}^{\circ }}:\frac{{720}^{\circ }}{{180}^{\circ }}\text{ [divide each side by }{180}^{\circ }]\end{array}$

  $\begin{array}{l}\left(n-2\right):4\\ n-2+2:4+2\text{ [add 2 both side]}\\ n:6\end{array}$

So for ratio $2\text{ to }1$ , the polygon is hexagon.

For ratio $2\text{ to }3$ ,

  $\begin{array}{l}\frac{\left(n-2\right)·{180}^{\circ }}{n}·3:\frac{{360}^{\circ }}{n}·2\text{ [multiply]}\\ \frac{\left(n-2\right)·{540}^{\circ }}{n}·n:\frac{{720}^{\circ }}{n}·n\text{ [multiply both side by n]}\\ \left(n-2\right)·{540}^{\circ }:{720}^{\circ }\\ \frac{\left(n-2\right)·{540}^{\circ }}{{540}^{\circ }}:\frac{{720}^{\circ }}{{540}^{\circ }}\text{ [divide each side by 54}{0}^{\circ }]\end{array}$

  $\begin{array}{l}\left(n-2\right):1.33\\ n-2+2:1.33+2\\ n:3\end{array}$

So for ratio $2\text{ to }3$ the polygon is triangle.

Hence,

Thus,the polygon where the measure of an interior angle and an exterior angle ratio is $1\text{ to }1$ , $2\text{ to }1$ and $2\text{ to }3$ is tetrahedral , hexagon and triangle.