Chapter 12.1, Problem 54E

a.

To determine

To calculate: The sum of interior angles of polygons that have 4, 5, 6 and 7 as their sides.

Answer to Problem 54E

The sum of interior angles of polygons with 4 sides is $360°$ , 5 sides is $540°$ , 6 sides is $720°$ and 7 sides is $900°$ .

Explanation of Solution

Given information:

The interior angles sum of a triangle is $180°$ .

Formula used:

Sum of angles of a convex polygon with n sides is $\left(n-2\right)\cdot 180$ .

Calculation:

Consider the provided information that interior angles sum of a triangle is $180°$ .

Recall that Sum of angles of a convex polygon with n sides is $\left(n-2\right)\cdot 180$ .

Substitute n as 4 to compute the sum of interior angle of 4 sided polygon.

  $\begin{array}{c}\left(4-2\right)\cdot 180°=2\cdot 180°\\ =360°\end{array}$

Substitute n as 5 to compute the sum of interior angle of 5 sided polygon.

  $\begin{array}{c}\left(5-2\right)\cdot 180°=3\cdot 180°\\ =540°\end{array}$

Substitute n as 6 to compute the sum of interior angle of 6 sided polygon.

  $\begin{array}{c}\left(6-2\right)\cdot 180°=4\cdot 180°\\ =720°\end{array}$

Substitute n as 7 to compute the sum of interior angle of 7 sided polygon.

  $\begin{array}{c}\left(7-2\right)\cdot 180°=5\cdot 180°\\ =900°\end{array}$

Thus, the sum of interior angles of polygons with 4 sides is $360°$ , 5 sides is $540°$ , 6 sides is $720°$ and 7 sides is $900°$ .

b.

To determine

To show: The sum of interior angles of polygons that have 3, 4, 5, 6 and 7 as their sides form an arithmetic sequence.

Explanation of Solution

Given information:

The interior angles sum of a triangle is $180°$ .

Formula used:

Sum of angles of a convex polygon with n sides is $\left(n-2\right)\cdot 180$ .

A sequence with first term as $a_1$ and next term written as sum of preceding term and common difference then it is called an arithmetic sequence. It is denoted as $a_1,a_1+d,a_1+2d+\dots$ .

Calculation:

Consider the provided information that interior angles sum of a triangle is $180°$ .

Recall that Sum of angles of a convex polygon with n sides is $\left(n-2\right)\cdot 180$ .

Substitute n as 4 to compute the sum of interior angle of 4 sided polygon.

  $\begin{array}{c}\left(4-2\right)\cdot 180°=2\cdot 180°\\ =360°\end{array}$

Substitute n as 5 to compute the sum of interior angle of 5 sided polygon.

  $\begin{array}{c}\left(5-2\right)\cdot 180°=3\cdot 180°\\ =540°\end{array}$

Substitute n as 6 to compute the sum of interior angle of 6 sided polygon.

  $\begin{array}{c}\left(6-2\right)\cdot 180°=4\cdot 180°\\ =720°\end{array}$

Substitute n as 7 to compute the sum of interior angle of 7 sided polygon.

  $\begin{array}{c}\left(7-2\right)\cdot 180°=5\cdot 180°\\ =900°\end{array}$

The sequence is $180°,360°,540°,720°,900°$ .

Compute the difference between the consecutive terms,

  $\begin{array}{c}{a}_2-a_1=360°-180°\\ =180°\end{array}$

And

  $\begin{array}{c}{a}_3-a_2=540°-360°\\ =180°\end{array}$

Therefore, common difference is $180°$ .

Since, the common difference is same between consecutive terms so the sequence $180°,360°,540°,720°,900°$ forms an arithmetic sequence.

c.

To determine

To calculate: The sum of interior angles of polygons that has 35 sides.

Answer to Problem 54E

The sum of interior angles of polygons that has 35 sidesis $5940°$ .

Explanation of Solution

Given information:

The interior angles sum of a triangle is $180°$ .

Formula used:

A sequence with first term as $a_1$ and next term written as sum of preceding term and common difference then it is called an arithmetic sequence. It is denoted as $a_1,a_1+d,a_1+2d+\dots$ .

The $n\text{th}$ term of the arithmetic sequence is $a_n=a_1+\left(n-1\right)d$ .

Calculation:

Consider the sequence $180°,360°,540°,720°,900°,\dots$ ., it denote sum of interior angles of polygons that have 3, 4, 5, 6, 7, and so on as their sides and it form an arithmetic sequence.

Recall that $n\text{th}$ term of the arithmetic sequence is $a_n=a_1+\left(n-1\right)d$ .

Compute the difference between the consecutive terms,

  $\begin{array}{c}{a}_2-a_1=360°-180°\\ =180°\end{array}$

And

  $\begin{array}{c}{a}_3-a_2=540°-360°\\ =180°\end{array}$

Therefore, common difference is $180°$ .

Here $a_1=180°,n=35\text{ and }d=180°$ .

Therefore, sum of interior angles of polygons that has 35 sides is,

  $\begin{array}{c}{a}_{35}=180°+\left(35-1\right)180°\\ a_{35}=180°+\left(34\right)180°\\ a_{35}=5940°\end{array}$

Thus, the sum of interior angles of polygons that has 35 sides is $5940°$ .