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Proving the Polygon Interior Angle Sum Theorem

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

Sagot :

GinnyAnswer
Let's complete the missing parts of the paragraph proof step-by-step for proving the Polygon Interior Angle Sum Theorem.

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. [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].

So, the complete proof is:

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. [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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