Qk6pvf28jkidn
Answered

Connect with knowledgeable experts and enthusiasts on IDNLearn.com. Find reliable solutions to your questions quickly and easily with help from our experienced experts.

Given: An [tex]$n$[/tex]-gon

Prove: The sum of the measures of the interior angles is [tex]$180(n-2)^{\circ}$[/tex].

Complete the missing parts of the paragraph proof.

We are given an [tex]$n$[/tex]-gon, which has [tex]$n$[/tex] sides and [tex]$n$[/tex] vertices. If we choose one of the vertices, we can draw [tex]$(n-3)$[/tex] diagonals. These diagonals form [tex]$(n-2)$[/tex] triangles. The sum of the interior angle measures of a triangle is [tex]$180$[/tex] degrees. Therefore, [tex]$n-2$[/tex] triangles would have an interior angle measure sum of [tex]$180(n-2)$[/tex] degrees. Therefore, the sum of the measures of the interior angles of an [tex]$n$[/tex]-gon is [tex]$180(n-2)^{\circ}$[/tex].

Sagot :

GinnyAnsweridn
We are given an [tex]$n$[/tex]-gon, which has [tex]$n$[/tex] sides and [tex]$n$[/tex] vertices. If we choose one of the vertices, we can draw [tex]$n - 3$[/tex] diagonals. These diagonals form [tex]$n - 2$[/tex] triangles. The sum of the interior angle measures of a triangle is 180 degrees. [tex]$n - 2$[/tex] triangles would have an interior angle measure sum of [tex]$180(n-2)$[/tex]. Therefore, the sum of the measures of the interior angles of an [tex]$n$[/tex]-gon is [tex]$180(n-2)^{\circ}$[/tex].
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Your questions deserve precise answers. Thank you for visiting IDNLearn.com, and see you again soon for more helpful information.