Chapter 12.3, Problem 6E

To determine

The sum of the measures of the exterior angles of a convex $12$ -gon.

Answer to Problem 6E

The sum of the measures of the exterior angles of a convex $12$ -gon is ${360}^{\circ }$ .

Explanation of Solution

Given information : Here, the polygon is convex $12$ -gon.

Formula used :

The sum of the measure of the all interior angles of a convex n-gon is $\left(n-2\right)\times {180}^{\circ }$

Measure of exterior angle $={180}^{\circ }-$ interior angle

Calculation :

Here, polygon is convex $12$ -gon i.e. $n=12$

Let, the measures of the interior angles of the $12$ -gon are $m\angle 1,m\angle 2,m\angle 3,m\angle 4,m\angle 5,m\angle 6,m\angle 7,m\angle 8,m\angle 9,m\angle 10,m\angle 11,m\angle 12$

The sum of the measure of the all interior angles of a convex $12$ -gon is

  $\begin{array}{l}\left(12-2\right)\times {180}^{\circ }\\ =10\times {180}^{\circ }\\ ={1800}^{\circ }\end{array}$

i.e.

  $\begin{array}{l}(m\angle 1+m\angle 2+m\angle 3+m\angle 4+m\angle 5+m\angle 6+m\angle 7+m\angle 8+m\angle 9+m\angle 10+m\angle 11+m\angle 12)\\ ={1800}^{\circ }\end{array}$

Therefore, the sum of the measures of the exterior angles of a convex $12$ -gon is

  $\begin{array}{l}\left({180}^{\circ }-m\angle 1\right)+\left({180}^{\circ }-m\angle 2\right)+\left({180}^{\circ }-m\angle 3\right)+\left({180}^{\circ }-m\angle 4\right)+\left({180}^{\circ }-m\angle 5\right)+\\ \left({180}^{\circ }-m\angle 6\right)+\left({180}^{\circ }-m\angle 7\right)+\left({180}^{\circ }-m\angle 8\right)+\left({180}^{\circ }-m\angle 9\right)+\left({180}^{\circ }-m\angle 10\right)+\\ \left({180}^{\circ }-m\angle 11\right)+\left({180}^{\circ }-m\angle 12\right)\\ =\left({180}^{\circ }+{180}^{\circ }+{180}^{\circ }+{180}^{\circ }+{180}^{\circ }+{180}^{\circ }+{180}^{\circ }+{180}^{\circ }+{180}^{\circ }+{180}^{\circ }+{180}^{\circ }+{180}^{\circ }\right)\\ -(m\angle 1+m\angle 2+m\angle 3+m\angle 4+m\angle 5+m\angle 6+m\angle 7+m\angle 8+m\angle 9+m\angle 10+m\angle 11+m\angle 12)\\ =\left(12\times {180}^{\circ }\right)-\left({1800}^{\circ }\right)\\ ={2160}^{\circ }-{1800}^{\circ }\\ ={360}^{\circ }\end{array}$

The sum of the measures of the exterior angles of a convex

  $12$ -gon isw ${360}^{\circ }$ .