Chapter 9.5, Problem 53E

Solution

To Prove: The sum of the interior angles of an n -sided polygon is $180°\left(n-2\right)$ for $n\ge 3$ .

The statement has proved.

Given Information:

The figure is show as:

  

Explanation:

Consider the given information,

  $P_n:{\displaystyle \sum _{i=1}^n{A}_i}=180°\left(n-2\right)$

Where, n is a positive integers, the condition is true for $n\ge 3$ , and $A_i$ is the interior angle of n -sided polygon.

Use the mathematical induction to prove the condition.

First of all, check the conditions is true for $n=3$ .

  $\begin{array}{c}{P}_3:{\displaystyle \sum _{i=1}^3{A}_i}=180°\left(3-2\right)\\ A_1+A_2+A_3=180°\left(1\right)\\ 180°=180°\end{array}$

So, the condition is true for 3.

Assume that the condition is true for $k$ .

  $P_k:{\displaystyle \sum _{i=1}^k{A}_i}=180°\left(k-2\right)$

Now, prove the condition for $n=k+1$ .

  $\begin{array}{l}{P}_{k+1}:{\displaystyle \sum _{i=1}^{k+1}{A}_i}=180°\left(k+1-2\right)\\ P_{k+1}:{\displaystyle \sum _{i=1}^{k+1}{A}_i}=180°\left(k-1\right)\end{array}$

So, start from $P_k$ , that is assumed to be true for polygon drawn in blue colour.

  ${\displaystyle \sum _{i=1}^k{B}_i}=180°\left(k-2\right)$

Where, $B_i$ is representing the angles of blue polygons.

  

To get the sum of the interior angles of a polygon with  $k+1$  angles, add the sum of the interior angles of the polygon with  $k$ angles to the sum of the grey triangle's angles:

  $\begin{array}{c}{\displaystyle \sum _{i=1}^k{A}_i}=180°\left(k-2\right)+180°\\ =180°k-2\left(180°\right)+180°\\ =180°k-\left(180°\right)\\ =180°\left(k-1\right)\end{array}$

Where, $A_i$ is representing the angles of big polygons.

This is exactly  $P_{k+1}$ ?. So it has proved that if  $P_k$  is true, then  $P_{k+1}$ ? is true.

As  $P_1$ ? is also true.

Hence, the statement has proved.