Question
Prove the Polygon Interior-Angle Sum Theorem.
Given - A polygon with n sides.
Prove - The sum of the measures of the interior angles is (n - 2) 180 degrees.
6. The sum of the measures of the interior angles can be rewritten as
Given - A polygon with n sides.
Prove - The sum of the measures of the interior angles is (n - 2) 180 degrees.
6. The sum of the measures of the interior angles can be rewritten as
Answers
Answered by
GPT-4o mini
To prove the Polygon Interior-Angle Sum Theorem, we will use a polygon with \(n\) sides and demonstrate that the sum of the measures of the interior angles of this polygon is \((n - 2) \times 180^\circ\).
### Steps of the Proof:
1. **Understanding the Polygon**:
Let's begin with a polygon that has \(n\) sides. This polygon can be a triangle, quadrilateral, pentagon, etc., up to an \(n\)-gon.
2. **Dividing the Polygon into Triangles**:
One effective method for finding the sum of the interior angles of a polygon is to divide it into triangles. We can do this by drawing diagonals from one vertex to all other non-adjacent vertices.
3. **Counting the Number of Triangles**:
In a polygon with \(n\) sides, if we select one vertex and connect it to the other vertices (excluding the adjacent vertices), we create \(n - 3\) triangles. This is because:
- There are \(n - 1\) vertices excluding the selected vertex.
- Out of these, \(2\) are adjacent to the selected vertex and cannot be connected to form a triangle.
- Hence, the number of diagonals that can be drawn from a single vertex is \(n - 3\).
4. **Total Number of Triangles Formed**:
The original polygon can now be seen as composed of these \(n - 2\) triangles (the original polygon plus the triangles formed by diagonals equals \(n - 2\) triangles in total).
5. **Sum of Interior Angles of the Triangles**:
Each triangle has a sum of interior angles equal to \(180^\circ\). Therefore, if there are \(n - 2\) triangles, the sum of the interior angles of all these triangles is:
\[
(n - 2) \times 180^\circ
\]
6. **Conclusion**:
Since the interior angles of the polygon are made up of the angles of these triangles, the interior angle sum of the polygon is also:
\[
\text{Sum of interior angles of the polygon} = (n - 2) \times 180^\circ
\]
Thus, we have proven that for a polygon with \(n\) sides, the sum of the measures of the interior angles is \((n - 2) \times 180^\circ\).
\[
\text{Hence, } \text{the sum of the measures of the interior angles is } (n - 2) \times 180^\circ.
\]
### Steps of the Proof:
1. **Understanding the Polygon**:
Let's begin with a polygon that has \(n\) sides. This polygon can be a triangle, quadrilateral, pentagon, etc., up to an \(n\)-gon.
2. **Dividing the Polygon into Triangles**:
One effective method for finding the sum of the interior angles of a polygon is to divide it into triangles. We can do this by drawing diagonals from one vertex to all other non-adjacent vertices.
3. **Counting the Number of Triangles**:
In a polygon with \(n\) sides, if we select one vertex and connect it to the other vertices (excluding the adjacent vertices), we create \(n - 3\) triangles. This is because:
- There are \(n - 1\) vertices excluding the selected vertex.
- Out of these, \(2\) are adjacent to the selected vertex and cannot be connected to form a triangle.
- Hence, the number of diagonals that can be drawn from a single vertex is \(n - 3\).
4. **Total Number of Triangles Formed**:
The original polygon can now be seen as composed of these \(n - 2\) triangles (the original polygon plus the triangles formed by diagonals equals \(n - 2\) triangles in total).
5. **Sum of Interior Angles of the Triangles**:
Each triangle has a sum of interior angles equal to \(180^\circ\). Therefore, if there are \(n - 2\) triangles, the sum of the interior angles of all these triangles is:
\[
(n - 2) \times 180^\circ
\]
6. **Conclusion**:
Since the interior angles of the polygon are made up of the angles of these triangles, the interior angle sum of the polygon is also:
\[
\text{Sum of interior angles of the polygon} = (n - 2) \times 180^\circ
\]
Thus, we have proven that for a polygon with \(n\) sides, the sum of the measures of the interior angles is \((n - 2) \times 180^\circ\).
\[
\text{Hence, } \text{the sum of the measures of the interior angles is } (n - 2) \times 180^\circ.
\]