Is it possible to prove by induction that an n-sided polygon's angles add up to 180(n-2)?

Question

I'd love to see a proof of it by induction (if it's even possible). I gave it a try and got nowhere.

Answer 1

First step of the mathematical induction:
Test to see if it's true for n = 3.

n = 3 is the case of the triangle.

(3-1)*180 = 180.
The interior angles of the triangle add up to 180 degrees.
The formula is true for n = 3.

Now assume it's true for some k, k greater than or equal to 3.

We will prove that if it's true for n = k, then it must also be true for n = k+1.

Construct a polygon of (k+1) sides.
Label the points p1,p2,p3,... p_(k+1).

Draw a line from p1 to p_k.

Note that p1 to p_k to p_(k+1) back to p1 form a triangle.
Note that p1 to p2 to p3 to . . . to p_k back to p1 for a k sided polygon.

Note that the sum of the interior angles of the (k+1) sided polygon
is the sum of the interior angles of the k sided polygon we made
plus the sum of the interior angles of the triangle we made.
That is

sum of the interior angles of the (k+1) sided polygon is

(k-2)*180 + 180 = ( k - 1) * 180 = ( [ k + 1] - 2) * 180.

We have proven that if the formula is true for n = k,
then it must also be true for n = k+1.

The mathematical induction is complete.
The formula true for n = 3 implies it's true for n = 4
implies it's true for n = 5 implies it's true for n = 6 implies . . .

Answer 2

You should be able to do it. It's true for n = 3. The angles of a triangle add to 180 degrees. Assume it is true for (n-1)-sided polygons. Consider an arbitrary n-gon. Take two vertices two vertices apart from each other and connect them, throwing out the edges that connect these vertices to an intermediate point. The result is an n-1 - sided polygon. Using induction, we conclude that the sum of the angles of the (n-1)-sided polygon is 180 ((n-1)-2). Can you use this construction we did to conclude that the sum of the angles of the n-gon is 180(n-2)?

Answer 3

Sum Of Angles In Polygon

Answer 4

It is possible.

When n = 3, it is a triangle. Therefore, the sum of angles is 180 = 180 (3-2). It works!

Assume when n = k, it also works: sum = 180(k-2).

When n = k+1, we added one more side to the k-th polygon, which is equivalent to adding a triangle to the k-th polygon. Therefore, we added 180 more degrees to the sum of the k-th polygon.

sum = 180 + 180(k-2). = 180[(k+1)-2]

End of proof.

Answer 5

Yes, by drawing a line from one vertex to each other vertex of an n-sided convex polygon, the number of triangles formed is n-2 and because there are 180 in a triangle, 180(n-2) represnts the formula you are trying to prove.

Answer 6

Another way to do the proof is to realize when completing a loop you have to turn a sum of 360 degrees. The average angle turned is = 360/n.
The internal angle at each vertex is complementary. So each inside angle is an average of 180-a. So the the total is n*(180-360/n) =
180*(n-2)

Answer 7

that the polygon is irregular for one thing.

Answer 8

It's probably possible, but it's much easier by deduction.

Answer 9

Yes. Just assume it does work.

Answer 10

Does it work for a triangle? yes!
Does it work for a square? yes!
Does it work for a hexagon? yes!
Does it work for a decagon? yes!