The angle sum of any polygon

One new idea — splitting a shape into triangles — applied to a theorem this course already has.

A triangle’s angles sum to 180. A quadrilateral has four angles instead of three, and no single triangle to appeal to directly. This lesson is about the one idea that closes that gap: cut the quadrilateral into triangles, and let the last lesson do the rest.

A quadrilateral, A, B, C and D, all draggable, with the diagonal from A to C splitting it into two triangles.

That total holds because the diagonal from A to C splits the quadrilateral into two triangles, each independently guaranteed 180 by the last lesson — so together, 360. The quadrilateral’s own four angles are made up of exactly those two triangles’ six angles, with nothing left over and nothing shared twice.

Naming the cut

Cutting a shape into triangles that share only edges and vertices, covering it exactly once, is called triangulation. A quadrilateral needs only one diagonal to triangulate it into two triangles. A pentagon, cut the same way from a single vertex, triangulates into three. In general, an $n$-sided polygon triangulates into $n - 2$ triangles from one vertex, because the two sides touching that vertex are not diagonals, and everything else fans out from it in one triangle at a time.

From one quadrilateral to every polygon

The diagonal from A to C does two things at once: it cuts angle A into two pieces and angle C into two pieces, while leaving angle B and angle D whole, each entirely inside one of the two triangles. Add up all six angles across both triangles — three from triangle ABC, three from triangle ACD — and every one of the quadrilateral’s four original angles is accounted for exactly once, split or not. Two triangles, each summing to 180 by the last lesson, means those six angles sum to 360 together, and since they are the same six angles the quadrilateral’s four angles are made of, the quadrilateral’s angles sum to 360 too.

The same idea, run again, gives every polygon a formula. Triangulate an $n$-gon from one vertex into $n - 2$ triangles, and its interior angles sum to $(n - 2) \times 180$: 180 for a triangle, 360 for a quadrilateral, 540 for a pentagon, and onward by 180 for every side added.

This is the second time in a row this course has built a theorem entirely out of an earlier one (the exterior angle theorem did the same with a straight line plus an angle sum), and it will not be the last: most of what makes geometry feel large is a small number of theorems, reused in combination, rather than a large number of separate ones.

A hexagon has six sides. How many triangles does triangulating it from one vertex produce, and what do its interior angles sum to?

$6 - 2 = 4$ triangles, so its interior angles sum to $4 \times 180 = 720$ degrees.

Why can a diagonal not be drawn from a vertex to one of its own two neighbours?

Because a line to an adjacent vertex is already a side of the polygon, not a new cut across it. A diagonal has to reach a vertex that is not already directly connected, which is exactly why a triangulating vertex produces two fewer triangles than it has sides: its own two sides are excluded.

Does this triangulation method work for a polygon with a dent in it, not just a convex one?

Not from every vertex — a diagonal drawn from a vertex next to the dent can leave the polygon’s interior and cross an edge, which is not a triangulation any more. Triangulating a non-convex polygon still works, but it may need to start from a different vertex, or use more than one diagonal from more than one point.

Do the exercise

This lesson counts as done once you get its exercise right. Nothing to tick off by hand.