MAT 2800: Selected Solutions to Assignment 1
•Show that the sum of the interior (corner) angles of a polygon with nsides is 180◦(n−2).
Solution 1: verbal. Fix one vertex of the polygon and introduce the diagonals from this vertex
to each of the vertices not adjacent to it. Since two vertices are adjacent to our chosen vertex,
there will be n−3 diagonals, dividing the polygon into n−2 triangles.
We have shown that the sum of the degree measures of the interior angles of each of these triangles is
180◦; thus, the sum of the degree measures of all of the interior angles of the triangles is (n−2)180◦.
Since the sum of all of the angles of the triangles may be regrouped, using the commutative and
associative properties of addition, so that each group adds up to the degree measure of one of the
interior angles of the polygon, it follows that the sum of the degree measures of the interior angles
of the polygon is (n−2)180◦.
Advantage: informal; no need for complicated notation.
Disadvantage: somewhat unclear and hard to read.
Solution 2: open formula (implicit induction). Lab el the vertices of the polygon v0, v1, v2, . . . , vn−1.
The degree measure of ∠v1v0vn−1is given by the sum
(∠v1v0vn−1)◦=
n−2
X
i=1
(∠viv0vi+1)◦.
•Use the result of the first exercise to prove that the angle measure of each corner angle of a regular
n-gon is n−2
n180◦.