Sum of the interior angles of a polygon

We know that the sum of the interior angles of a triangle is $\pi$. And for a quadrilateral? And for a pentagon? There is a general formula that holds for every convex polygon.

Theorem — Sum of the interior angles of a polygon

In a convex polygon of $n$ sides the sum of the interior angles is: $(n-2)\cdot\pi \qquad\text{that is}\qquad (n-2)\cdot 180^\circ.$

Proof

The idea is to subdivide the polygon into triangles and then subtract the “excess” angles.

1. Construction. I choose a point $O$ interior to the polygon and join it to all the vertices, obtaining $n$ triangles.
2. I observe. The sum of all the angles of the triangles thus formed is $n\cdot\pi$.
3. I observe. However, the angles that face onto the point $O$ form a full angle: $2\pi$. These angles do not belong to the interior angles of the polygon.
4. I deduce. Subtracting, the sum of the interior angles of the polygon is $n\pi - 2\pi = (n-2)\pi$.

$\blacksquare$

Links

Topics: Euclidean geometry
Concepts: Proof · Polygon · Sum of the angles of a polygon · Sum of the angles of a triangle
Skills: Proving · Using formulae