Chapter 12, Problem 12PTTS

To determine

To find: The sum of the interior angles of a regular $10\text{-gon}$ .

Answer to Problem 12PTTS

The sum of the interior angles of a regular $10\text{-gon}$ is ${1440}^{\circ }$ .

Explanation of Solution

Given information: A regular $10\text{-gon}$ .

Calculation: A regular $10\text{-gon}$ is a ten-sided polygon.

The sum of the interior angles of a polygon $=(\text{n}-2)\times {180}^{\circ }$

Where, for the $10\text{-gon}$ , $\text{n}=10$

Let’s find the sum of the interior angles of the $10\text{-gon}$ is,

  $\begin{array}{l}=(\text{n}-2)\times {180}^{\circ }\\ =(10-2)\times {180}^{\circ }\\ =8\times {180}^{\circ }\\ ={1440}^{\circ }\end{array}$

Hence, The sum of the interior angles of a regular $10\text{-gon}$ is ${1440}^{\circ }$ .