To prove the Polygon Interior-Angle Sum Theorem, which states that the sum of the measures of the interior angles of a polygon with \( n \) sides is \( (n - 2) \times 180^\circ \), we can use the concept of triangulation.
Proof:
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Triangulation of the Polygon: We can divide a polygon with \( n \) sides into triangles. This can be done by drawing diagonals from one vertex to the other non-adjacent vertices.
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Counting the Triangles: Starting from one vertex, we can draw diagonals to the other \( n - 3 \) vertices that are not adjacent to it (since two vertices are adjacent to the chosen vertex, and one vertex is itself). Thus, we can create \( n - 2 \) triangles in total.
To visualize this, let’s consider a simple polygon. For example:
- A triangle (3 sides) has no diagonals, so we have \( 1 \) triangle.
- A quadrilateral (4 sides) can be divided into \( 2 \) triangles by drawing one diagonal.
- A pentagon (5 sides) can be divided into \( 3 \) triangles by drawing two diagonals.
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Angle Sum of the Triangles: Each triangle has an angle sum of \( 180^\circ \). Therefore, if we have \( n - 2 \) triangles, the total sum of all angles in those triangles is: \[ (n - 2) \times 180^\circ \]
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Conclusion: Since the interior angles of the polygon consist of the angles from all of these triangles, the sum of the measures of the interior angles of the polygon is also \( (n - 2) \times 180^\circ \).
Answer to the Second Part:
Each triangle formed has an angle sum of \( 180 \) degrees.
Summary:
The sum of the measures of the interior angles of a polygon with \( n \) sides is given by the formula: \[ (n - 2) \times 180^\circ \] This is derived from the fact that each triangle has an angle sum of \( 180^\circ \) and that a polygon can be divided into \( n - 2 \) triangles through triangulation.