Sum of Interior Angles of a Polygon

Showing a generalized way to find the sum of the interior angles of any polygon

We already know that the sum of the interior angles of a triangle

add up to 180 degrees

If the measure of this angle is A,

the measure of this angle over here is B, and

the measure of this angle is C, we know that A + B + C = 180 degrees

What happens when we have polygons with more than 3 sides?

a quadrilateral

probably applies to any quadrilateral with four sides,

not just things that have right angles and parallel lines

and all the rest

Actually, that looks a little too close to being parallel

so let me draw it like this

The way you can think about it, with the 4-sided quadrilateral is,

we already know about this:

the measures of the interior angles of a triangle add up to 180

so maybe we can divide this into 2 triangles

From this point right over here, if we draw a line like this,

Then if measure of this angle is A, measure of this is B,

measure of that is C

We know that A + B + C = 180 degrees

Then if we call this over here, X, this over here, Y and that, Z

Those are the measures of those angles

We know that X + Y + Z = 180 degrees

So, if we want the measure of the sum of all of the interior angles,

all of the interior angles are going to be:

plus this angle, which is going to be A + X

A + X is that whole angle for the quadrilateral

Plus this whole angle, which is going to be C + Y

And where you know A + B + C is 180 degrees

And we know that Z + X + Y = 180 degrees

So plus 180 degrees which is equal to 360 degrees

I think you see the general idea here

We just have to figure out how many triangles

we can divide something into

Then we just multiply it by 180 degrees,

since each of those triangles will have 180 degrees

can we fit into that thing

Let me draw an irregular pentagon

1, 2, 3, 4, 5

Looks more like a bit of a side-ways house there

Once again, we can draw our triangles inside of this pentagon

That would be one triangle there

That would be another triangle

that perfectly cover this pentagon

This is one triangle, the other triangle and the other one

We know each of those have 180 degrees,

if we take the sum of their angles

We also know the sum of all those interior angles are equal

to the sum of the interior angles of the polygon as a whole

To see that, clearly this interior angle is

one of the angles of the polygon

This is, as well

When you take the sum of this one and this one,

We take the sum of that one and that one, you get that entire one

Then when you take the sum of that one, plus that one,

plus that one, you get that entire interior angle

So if you take the sum of all the interior angles of

all of the interior angles of the polygon

In this case you have 1, 2, 3 triangles

3 times 180 degrees is equal to what?

300 +240 = 540 degrees

To generalize it,

we have up to use up four sides

We have to use up all the four sides of this quadrilateral

We had to use up four of the five sides right here in this pentagon

1, 2, and then 3, 4

So four sides give you two triangles

It seems like maybe every incremental side you have after that,

you can get another triangle out of it

1, 2, 3, 4, 5, 6 sides

and I can get one triangle out of these 2 sides

1, 2 sides of the actual hexagon

I can get another triangle out of these 2 sides of the actual hexagon

And it looks like I can get another triangle

out of each of the remaining sides

So one out of that one, and then,

one out of that one right over there

S-sided polygon

or 6 sides

So we can assume that S is greater than 4 sides

I want to figure out how many non-overlapping triangles

that perfectly cover that polygon

How many can I fit inside of it

Then I just have to multiply the number of triangles

times 180 degrees

to figure out what are the sum of the interior angles of that polygon

as a function of the number of sides

Once again, four of the sides are going to be used

to make two triangles

and we have two sides right over there

I can have, I can draw one triangle

what happens to the rest of the sides of the polygon

You can imagine putting a big black piece of construction paper

There might be other sides here

So out of these 2 sides I can draw one triangle just like that

Out of these two sides, I can draw another triangle right over there

So 4 sides used for two triangles

Then, no matter how many sides I have left over,

if I have all sorts of craziness here

Let me draw a little bit neater than that

So I can have all sorts of craziness right over here

It looks like every other incremental side

I can get another triangle out of it

one triangle out of that side, one triangle out of that side

and then one triangle out of this side

For example, this figure that I have drawn is a very irregular

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, is that right?

1, 2, 3, 4, 5, 6, 7 ,8 ,9 10 It is a decagon

In this decagon, four of the sides were used for two triangles

Then the other 6 sides I was able to get a triangle each

I have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Did I count, am I just not seeing something?

Oh I see, I have to draw another line right over here

These are two different sides

I can get another triangle out of that right over there

There you have it

I have these two triangles out of 4 sides

Then out of the other 6 remaining sides I get a triangle each

Plus 6 triangles, I got a total of 8 triangles

So we can generally think about it

Let me write this down

Our number of triangles is going to be equal to 2,

The remaining sides, I get a triangle each

The remaining sides are going to be S minus 4

The number of triangles are going to be 2 plus S minus 4

So, if I have an S-sided polygon, I can get S minus two triangles

that perfectly cover that polygon

Which tells us that an S-sided polygon if it has S minus 2 triangles,

that the interior angles in it are going to be

S minus 2 times 180 degrees, which is a pretty cool result

So someone told you that they had a 102-sided polygon

So, S is equal to 102 sides

You can say, okay the number of interior angles are going to be

Which is equal to 180, with two more zeros behind it

of a 102-sided polygon