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

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]$180^\circ$[/tex].

Therefore, [tex]$n-2$[/tex] triangles would have an interior angle measure sum of [tex]$180(n-2)^\circ$[/tex].

Thus, the sum of the measures of the interior angles of an [tex]$n$[/tex]-gon is [tex]$180(n-2)^\circ$[/tex].

Sagot :

GinnyAnsweridn
Here's the detailed, step-by-step solution:

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 from that vertex. 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].
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