Суми внутрішніх та зовнішніх кутів многокутника

The sum of the three angles in a triangle is 180 degrees.

You can find the proof anywhere, but let's go through it here:

Draw line ECD going through C, parallel to AB.

Then angle ECA equals angle CAB and angle DCB equals angle CBA (because ECD and AB are parallel lines; explanation is here). Since ECD is a straight angle, it has measure 180.

Now, we can attempt to show that the sum of the interior angles of a n-gon is $180(n-2)$.

We'll go with the "simple proof".

Consider a polygon like the pentagon below:

We have a pentagon, and we split it into 3 triangles. Each triangle has angle sum 180, so the total sum is $180 \cdot 3 = 540$.

We can split an n-gon into n-2 triangles like this, and therefore the sum of angles is $180(n-2)$.

Explanation:

There are actually other ways to directly calculate the sum of exterior angles, but we can talk through how to do it with what we've just learned here. If an exterior angle is the angle between one side of a polygon and the extension of an adjacent side, then for each vertex of a polygon, its exterior and interior angles sum to 180 degrees (the straight angle). Therefore, for one vertex, the exterior angle equals 180 minus the interior angle. Since the sum of all interior angles equals $180(n-2)$, the sum of all exterior angles is therefore $180 n - 180(n-2) = 360$ degrees.

A regular polygon is a polygon that has all equal angles and all equal sides. They have many useful properties.

A regular hexagon is shown below:

Notice that the center, O, divides the hexagon into six regions that are congruent to triangle OAB, if we were to draw OA, OB, OC, ... OF.

We proved that there were $180(n-2)$ degrees in the interior angles of a polygon, so a hexagon would have $180(6-2)=720$ degrees in its interior angles. Since all angles are equal, each angle of a hexagon has $\frac{1}{6} \cdot 720 = 120$ degrees.

In general, the degree measure of a regular n-gon is $\frac{180(n-2)}{n}$

Let's go back to our hexagon example:

Notice that $OG$ is perpendicular to $AB$. ($G$ is the midpoint of $AB$). This segment $OG$ is called the apothem of the hexagon.

The most useful property of the apothem is that in a regular n-gon: $area = apothem * semiperimeter$.

We can derive this formula by splitting the n-gon into $n$ triangles. Each one has area $0.5 * base * height$ (where the height is the apothem, and the length of one side is the base). We take $n$ of these for the full area, or $n * 0.5 * base * height$ which is equivalent to $apothem * semiperimeter$ (where semiperimeter equals $\frac{1}{2}$ of the perimeter).

More terms: the apothem is also the inradius of the n-gon, and $OA$ is the circumradius of the hexagon.

Explanation:

We know that regular polygons have equal interior angles, and we know that each interior angle in a regular hexagon is 120 degrees. The center O divides the hexagon into six congruent triangles, which are necessarily equilateral: we know that angle BOA is 60 degrees (360 / 6), and we know that angle ABO is 60 degrees (120 / 2). Triangle AOB is therefore an equilateral triangle with side = 1.

Remembering that the area of a regular n-gon is apothem * semiperimeter, we now know that the semiperimeter is 3 and the apothem is $\frac{\sqrt{3}}{2}$. The area therefore equals $3 \cdot \frac{\sqrt{3}}{2}$

An exterior angle is the angle between one side of a polygon and the extension of an adjacent side. You can see in the diagram below that all of the exterior angles have been marked and labeled:

(For obvious reasons, this concept will only apply to convex polygons!)

In our arbitrary pentagon ABCDE, we can we can see that the exterior angles add to 360 because the rays of the extended sides are basically "pivoting" around different fulcrums, namely each of the vertices. After it has pivoted around all five vertices and covered each of the exterior angles, it returns to its original position, implying that it has turned a full circle, or 360 degrees.

At least, that's how I think about it...

Another way to think about this is "zooming out": click for animation.

Since $n=10$ in the formula, we have $180^\circ \cdot \left( 10 - 2 \right) = 1440^\circ$.

Note that the problem statement implies that $$180 \cdot \left( n - 2 \right) > 2014 \implies n > \frac {1197}{90}.$$The smallest integer value possible is $14$.

Theorem. The sum of the interior angles of an $n$-sided polygon is $180^\circ \cdot \left( n - 2 \right)$.

Proof 1. Pick a point on the polygon and draw every diagonal of the polygon from that point. You have now divided the polygon into $n-2$ disjoint triangles, each inside the polygon, and whose union make the polygon, with sum of angles $180^\circ$. Therefore, the total sum is $180^\circ \cdot \left( n - 2 \right)$. $\Box$

Proof 2. Consider a point $P$ inside the polygon. Call the polygon $A_1A_2\cdots A_n$. Now, draw $PA_i$, for all $1 \le i \le n$. You have now divided the polygon into $n$ disjoing triangles as in the previous proof. If we add all the interior angles of the triangle, we have overcounted the sum of the interior angles of the polygon by the $360^\circ$ formed around the point $P$. Thus, the desired sum is $180^\circ \cdot n - 360^\circ = 180^\circ \cdot \left( n - 2 \right)$. $\Box$

Find more proofs? Edit them into this post.

There are several ways to think about the sum of interior angles in a regular polygon. We can use any of the following:

Exterior angles

Diagonals

Triangles

First, let us see how we can use exterior angles to come up with a simple formula for the sum of the interior angles in a regular polygon.

As you know, in regular polygons, all of the sides are congruent, all of the angles have the same degree measure, and there are the same amount of sides as interior angles. For example, a regular square has 4 equal sides, 4 congruent angles, and there are the same amount of angles as sides.

We know from the earlier lesson that the degree of an exterior angle in a regular n-sided polygon is $\frac{360}{n}$.

Since an interior angle and an exterior angle in a regular polygon add up to $180$, we can write the interior angle measure as $180$ minus the exterior angle measure, or $180-\frac{360}{n}$.

Since there are $n$ interior angles in an n-sided regular polygon with measure $180-\frac{360}{n}$, the sum of all of them is just $n(180-\frac{360}{n})=\boxed{180n-360}$

Note: n>2 because 2 points determine a line, not a polygon.

Since you know the sum of the interior angles in a polygon, you can find the measure of one interior angle of a regular polygon by dividing the sum by the amount of edges. So, have fun doing so on this worksheet!