Chapter 12.6, Problem 21E

i.

To determine

To complete the table.

Answer to Problem 21E

The complete table is:

Regular polygon	Quadrilateral	Hexagon	Octagon	$10$ -gon
Sides	$4$	$6$	$8$	$10$
Angles of rotation (in either direction)	${90}^{\circ },{180}^{\circ }$	${60}^{\circ },{120}^{\circ },{180}^{\circ }$	${45}^{\circ },{90}^{\circ },{135}^{\circ },{180}^{\circ }$	${36}^{\circ },{72}^{\circ },{108}^{\circ },{144}^{\circ },{180}^{\circ }$

Explanation of Solution

Given:

The table provided in the question is:

Regular polygon	Quadrilateral	Hexagon	Octagon	$10$ -gon
Sides	$4$	$?$	$?$	$?$
Angles of rotation (in either direction)	${90}^{\circ },{180}^{\circ }$	$?$	$?$	$?$

Formula Used:

  $\text{Number of sides of regular polygon }\left(n\right)=\frac{{360}^{\circ }}{\text{Interior angle of regular polygon}}$

The angle of rotation of a regular polygon is the result of dividing $360$ by number of sides of regular polygon, which keep rotating itself at a same angle up till ${180}^{\circ }$ .

Calculation:

  $\begin{array}{l}\text{Interior angle of Hexagon}={60}^{\circ }\\ \text{Number of sides of Hexagon}=\frac{{360}^{\circ }}{{60}^{\circ }}=6\end{array}$

The angle of rotation of Hexagon

  $={60}^{\circ },{120}^{\circ },{180}^{\circ }$ .

  $\begin{array}{l}\text{Interior angle of Octagon}={45}^{\circ }\\ \text{Number of sides of Octagon}=\frac{{360}^{\circ }}{{45}^{\circ }}=8\end{array}$

The angle of rotation of Octagon

  $={45}^{\circ },{90}^{\circ },{135}^{\circ },{180}^{\circ }$

  $\begin{array}{l}\text{Interior angle of }10\text{-gon}={36}^{\circ }\\ \text{Number of sides of }10\text{-gon}=\frac{{360}^{\circ }}{{36}^{\circ }}=10\end{array}$

The angle of rotation of $10$ -gon

  $={36}^{\circ },{72}^{\circ },{108}^{\circ },{144}^{\circ },{180}^{\circ }$

Hence,

The complete table is:

Regular polygon	Quadrilateral	Hexagon	Octagon	$10$ -gon
Sides	$4$	$6$	$8$	$10$
Angles of rotation (in either direction)	${90}^{\circ },{180}^{\circ }$	${60}^{\circ },{120}^{\circ },{180}^{\circ }$	$\begin{array}{l}{45}^{\circ },{90}^{\circ },\\ {135}^{\circ },{180}^{\circ }\end{array}$	$\begin{array}{l}{36}^{\circ },{72}^{\circ },{108}^{\circ },\\ {144}^{\circ },{180}^{\circ }\end{array}$

ii.

To determine

To explain the statement.

Answer to Problem 21E

The number of sides of a regular polygon is twice the number of angle of rotation of a regular polygon.

Explanation of Solution

Given:

Number of sides related to angle of rotation.

The number of sides of a regular polygon is twice the number of angle of rotation of a regular polygon.

For example,

A regular hexagon has $6$ sides and its number of angle of rotation is $3$ , which is ${60}^{\circ },{120}^{\circ },\text{ and }{180}^{\circ }$ .

iii.

To determine

To complete the table.

Answer to Problem 21E

The complete table is:

Regular polygon	Quadrilateral	Hexagon	Octagon	$10$ -gon	$16$ -gon
Sides	$4$	$6$	$8$	$10$	16
Angles of rotation (in either direction)	${90}^{\circ },{180}^{\circ }$	${60}^{\circ },{120}^{\circ },{180}^{\circ }$	$\begin{array}{l}{45}^{\circ },{90}^{\circ },\\ {135}^{\circ },{180}^{\circ }\end{array}$	$\begin{array}{l}{36}^{\circ },{72}^{\circ },{108}^{\circ },\\ {144}^{\circ },{180}^{\circ }\end{array}$	$\begin{array}{l}{22.5}^{\circ },{45}^{\circ },{67.5}^{\circ },\\ {90}^{\circ },{112.5}^{\circ },{135}^{\circ },\\ {157.5}^{\circ },{180}^{\circ }\end{array}$

Explanation of Solution

Given:

The table provided in the question is:

Regular polygon	Quadrilateral	Hexagon	Octagon	$10$ -gon	$16$ -gon
Sides	$4$	$?$	$?$	$?$	$?$
Angles of rotation (in either direction)	${90}^{\circ },{180}^{\circ }$	$?$	$?$	$?$	$?$

Formula used:

  $\text{Number of sides of regular polygon }\left(n\right)=\frac{{360}^{\circ }}{\text{Interior angle of regular polygon}}$

The angle of rotation of a regular polygon is the result of dividing $360$ by number of sides of regular polygon, which keep rotating itself at a same angle up till ${180}^{\circ }$ .

Calculation:

  $\begin{array}{l}\text{Interior angle of }16\text{-gon}={22.5}^{\circ }\\ \text{Number of sides of }16\text{-gon}=\frac{{360}^{\circ }}{{22.5}^{\circ }}=16\end{array}$

The angle of rotation of $16$ -gon

  $={22.5}^{\circ },{45}^{\circ },{67.5}^{\circ },{90}^{\circ },{112.5}^{\circ },{135}^{\circ },{157.5}^{\circ },{180}^{\circ }$ .

Hence,

The complete table is:

Regular polygon	Quadrilateral	Hexagon	Octagon	$10$ -gon	$16$ -gon
Sides	$4$	$6$	$8$	$10$	16
Angles of rotation (in either direction)	${90}^{\circ },{180}^{\circ }$	${60}^{\circ },{120}^{\circ },{180}^{\circ }$	$\begin{array}{l}{45}^{\circ },{90}^{\circ },\\ {135}^{\circ },{180}^{\circ }\end{array}$	$\begin{array}{l}{36}^{\circ },{72}^{\circ },{108}^{\circ },\\ {144}^{\circ },{180}^{\circ }\end{array}$	$\begin{array}{l}{22.5}^{\circ },{45}^{\circ },{67.5}^{\circ },\\ {90}^{\circ },{112.5}^{\circ },{135}^{\circ },\\ {157.5}^{\circ },{180}^{\circ }\end{array}$