Regular polygon. Number of sides of a regular polygon

broken line

Definition

broken line, or shorter, broken line, is called a finite sequence of segments, such that one of the ends of the first segment serves as the end of the second, the other end of the second segment serves as the end of the third, and so on. In this case, adjacent segments do not lie on the same straight line. These segments are called polyline links.

Types of broken line

The broken line is called closed if the beginning of the first segment coincides with the end of the last one.

The broken line can cross itself, touch itself, lean on itself. If there are no such singularities, then such a broken line is called simple.

Polygons

Definition

A simple closed polyline, together with a part of the plane bounded by it, is called polygon.

Comment

At each vertex of a polygon, its sides define some angle of the polygon. It can be either less than deployed, or more than deployed.

Property

Each polygon has an angle less than $180^\circ$.

Proof

Let a polygon $P$ be given.

Let's draw some straight line that does not intersect it. We will move it parallel to the side of the polygon. At some point, for the first time we obtain a line $a$ that has at least one common point with the polygon $P$. The polygon lies on one side of this line (moreover, some of its points lie on the line $a$).

The line $a$ contains at least one vertex of the polygon. Its two sides converge in it, located on the same side of the line $a$ (including the case when one of them lies on this line). So, at this vertex, the angle is less than the developed one.

Definition

The polygon is called convex if it lies on one side of each line containing its side. If the polygon is not convex, it is called non-convex.

Comment

A convex polygon is the intersection of half-planes bounded by lines that contain the sides of the polygon.

Properties of a convex polygon

A convex polygon has all angles less than $180^\circ$.

A line segment connecting any two points of a convex polygon (in particular, any of its diagonals) is contained in this polygon.

Proof

Let's prove the first property

Take any corner $A$ of a convex polygon $P$ and its side $a$ coming from the vertex $A$. Let $l$ be a line containing side $a$. Since the polygon $P$ is convex, it lies on one side of the line $l$. Therefore, its angle $A$ also lies on the same side of this line. Hence the angle $A$ is less than the straightened angle, that is, less than $180^\circ$.

Let's prove the second property

Take any two points $A$ and $B$ of a convex polygon $P$. The polygon $P$ is the intersection of several half-planes. The segment $AB$ is contained in each of these half-planes. Therefore, it is also contained in the polygon $P$.

Definition

Diagonal polygon is called a segment connecting its non-neighboring vertices.

Theorem (on the number of diagonals of an n-gon)

The number of diagonals of a convex $n$-gon is calculated by the formula $\dfrac(n(n-3))(2)$.

Proof

From each vertex of an n-gon one can draw $n-3$ diagonals (one cannot draw a diagonal to neighboring vertices and to this vertex itself). If we count all such possible segments, then there will be $n\cdot(n-3)$, since there are $n$ vertices. But each diagonal will be counted twice. Thus, the number of diagonals of an n-gon is $\dfrac(n(n-3))(2)$.

Theorem (on the sum of the angles of an n-gon)

The sum of the angles of a convex $n$-gon is $180^\circ(n-2)$.

Proof

Consider an $n$-gon $A_1A_2A_3\ldots A_n$.

Take an arbitrary point $O$ inside this polygon.

The sum of the angles of all triangles $A_1OA_2$, $A_2OA_3$, $A_3OA_4$, \ldots, $A_(n-1)OA_n$ is $180^\circ\cdot n$.

On the other hand, this sum is the sum of all interior angles of the polygon and the total angle $\angle O=\angle 1+\angle 2+\angle 3+\ldots=30^\circ$.

Then the sum of the angles of the considered $n$-gon is equal to $180^\circ\cdot n-360^\circ=180^\circ\cdot(n-2)$.

Consequence

The sum of the angles of a non-convex $n$-gon is $180^\circ(n-2)$.

Proof

Consider a polygon $A_1A_2\ldots A_n$ whose only angle $\angle A_2$ is non-convex, that is, $\angle A_2>180^\circ$.

Let's denote the sum of his catch $S$.

Connect the points $A_1A_3$ and consider the polygon $A_1A_3\ldots A_n$.

The sum of the angles of this polygon is:

$180^\circ\cdot(n-1-2)=S-\angle A_2+\angle 1+\angle 2=S-\angle A_2+180^\circ-\angle A_1A_2A_3=S+180^\circ-( \angle A_1A_2A_3+\angle A_2)=S+180^\circ-360^\circ$.

Therefore, $S=180^\circ\cdot(n-1-2)+180^\circ=180^\circ\cdot(n-2)$.

If the original polygon has more than one non-convex corner, then the operation described above can be done with each such corner, which will lead to the assertion being proved.

Theorem (on the sum of the exterior angles of a convex n-gon)

The sum of the exterior angles of a convex $n$-gon is $360^\circ$.

Proof

The exterior angle at vertex $A_1$ is $180^\circ-\angle A_1$.

The sum of all external angles is:

$\sum\limits_(n)(180^\circ-\angle A_n)=n\cdot180^\circ - \sum\limits_(n)A_n=n\cdot180^\circ - 180^\circ\cdot(n -2)=360^\circ$.

In the basic geometry course, it is proved that the sum of the angles of a convex n-gon is 180° (n-2). It turns out that this statement is also true for non-convex polygons.

Theorem 3. The sum of the angles of an arbitrary n-gon is 180° (n - 2).

Proof. Let's divide the polygon into triangles by drawing diagonals (Fig. 11). The number of such triangles is n-2, and in each triangle the sum of the angles is 180°. Since the angles of the triangles are the angles of the polygon, the sum of the angles of the polygon is 180° (n - 2).

Let us now consider arbitrary closed broken lines, possibly with self-intersections A1A2…AnA1 (Fig. 12, a). Such self-intersecting broken lines will be called star-shaped polygons (Fig. 12, b-d).

Let us fix the direction of counting the angles counterclockwise. Note that the angles formed by a closed polyline depend on the direction in which it is traversed. If the direction of the polyline bypass is reversed, then the angles of the polygon will be the angles that complement the angles of the original polygon up to 360°.

If M is a polygon formed by a simple closed broken line passing in a clockwise direction (Fig. 13, a), then the sum of the angles of this polygon will be equal to 180 ° (n - 2). If the broken line is passed in the counterclockwise direction (Fig. 13, b), then the sum of the angles will be equal to 180 ° (n + 2).

Thus, the general formula for the sum of the angles of a polygon formed by a simple closed polyline has the form = 180 ° (n 2), where is the sum of the angles, n is the number of angles of the polygon, "+" or "-" is taken depending on the direction of bypassing the polyline.

Our task is to derive a formula for the sum of the angles of an arbitrary polygon formed by a closed (possibly self-intersecting) polyline. To do this, we introduce the concept of the degree of a polygon.

The degree of a polygon is the number of revolutions made by a point during a complete sequential bypass of its sides. Moreover, the turns made in the counterclockwise direction are considered with the “+” sign, and the turns in the clockwise direction - with the “-” sign.

It is clear that the degree of a polygon formed by a simple closed broken line is +1 or -1, depending on the direction of the traversal. The degree of the broken line in Figure 12, a is equal to two. The degree of star heptagons (Fig. 12, c, d) is equal to two and three, respectively.

The notion of degree is defined similarly for closed curves in the plane. For example, the degree of the curve shown in Figure 14 is two.

To find the degree of a polygon or curve, you can proceed as follows. Suppose that, moving along the curve (Fig. 15, a), we, starting from some place A1, made a full turn, and ended up at the same point A1. Let's remove the corresponding section from the curve and continue moving along the remaining curve (Fig. 15b). If, starting from some place A2, we again made a full turn and got to the same point, then we delete the corresponding section of the curve and continue moving (Fig. 15, c). Counting the number of remote sections with the signs "+" or "-", depending on their direction of bypass, we obtain the desired degree of the curve.

Theorem 4. For an arbitrary polygon, the formula

180° (n+2m),

where is the sum of the angles, n is the number of angles, m is the degree of the polygon.

Proof. Let the polygon M have degree m and is conventionally shown in Figure 16. M1, …, Mk are simple closed broken lines, passing through which the point makes full turns. A1, …, Ak are the corresponding self-intersection points of the polyline, which are not its vertices. Let us denote the number of vertices of the polygon M that are included in the polygons M1, …, Mk by n1, …, nk, respectively. Since, in addition to the vertices of the polygon M, vertices A1, …, Ak are added to these polygons, the number of vertices of the polygons M1, …, Mk will be equal to n1+1, …, nk+1, respectively. Then the sum of their angles will be equal to 180° (n1+12), …, 180° (nk+12). Plus or minus is taken depending on the direction of bypassing broken lines. The sum of the angles of the polygon M0, remaining from the polygon M after the removal of the polygons M1, ..., Mk, is equal to 180° (n-n1- ...-nk+k2). The sums of the angles of the polygons M0, M1, …, Mk give the sum of the angles of the polygon M, and at each vertex A1, …, Ak we additionally obtain 360°. Therefore, we have the equality

180° (n1+12)+…+180° (nk+12)+180° (n-n1-…-nk+k2)=+360°k.

180° (n2…2) = 180° (n+2m),

where m is the degree of the polygon M.

As an example, consider the calculation of the sum of the angles of a five-pointed asterisk (Fig. 17, a). The degree of the corresponding closed polyline is -2. Therefore, the desired sum of the angles is 180.

Triangle, square, hexagon - these figures are known to almost everyone. But not everyone knows what a regular polygon is. But this is all the same Regular polygon is called the one that has equal angles and sides. There are a lot of such figures, but they all have the same properties, and the same formulas apply to them.

Properties of regular polygons

Any regular polygon, be it a square or an octagon, can be inscribed in a circle. This basic property is often used when constructing a figure. In addition, a circle can also be inscribed in a polygon. In this case, the number of points of contact will be equal to the number of its sides. It is important that a circle inscribed in a regular polygon will have a common center with it. These geometric figures subject to the same theorems. Any side of a regular n-gon is associated with the radius R of the circumscribed circle around it. Therefore, it can be calculated using the following formula: a = 2R ∙ sin180°. Through you can find not only the sides, but also the perimeter of the polygon.

How to find the number of sides of a regular polygon

Any one consists of a certain number of segments equal to each other, which, when connected, form a closed line. In this case, all the corners of the formed figure have the same value. Polygons are divided into simple and complex. The first group includes a triangle and a square. Complex polygons have more sides. They also include star-shaped figures. For complex regular polygons, the sides are found by inscribing them in a circle. Let's give a proof. Draw a regular polygon with an arbitrary number of sides n. Describe a circle around it. Specify the radius R. Now imagine that some n-gon is given. If the points of its angles lie on a circle and are equal to each other, then the sides can be found by the formula: a = 2R ∙ sinα: 2.

Finding the number of sides of an inscribed right triangle

An equilateral triangle is a regular polygon. The same formulas apply to it as to the square and the n-gon. A triangle will be considered correct if it has the same length sides. In this case, the angles are 60⁰. Construct a triangle with given side length a. Knowing its median and height, you can find the value of its sides. To do this, we will use the method of finding through the formula a \u003d x: cosα, where x is the median or height. Since all sides of the triangle are equal, we get a = b = c. Then the following statement is true: a = b = c = x: cosα. Similarly, you can find the value of the sides in an isosceles triangle, but x will be the given height. At the same time, it should be projected strictly on the base of the figure. So, knowing the height x, we find the side a isosceles triangle according to the formula a \u003d b \u003d x: cosα. After finding the value of a, you can calculate the length of the base c. Let's apply the Pythagorean theorem. We will look for the value of half the base c: 2=√(x: cosα)^2 - (x^2) = √x^2 (1 - cos^2α) : cos^2α = x ∙ tgα. Then c = 2xtanα. In such a simple way, you can find the number of sides of any inscribed polygon.

Calculating the sides of a square inscribed in a circle

Like any other inscribed regular polygon, a square has equal sides and angles. The same formulas apply to it as to the triangle. You can calculate the sides of a square using the value of the diagonal. Let's consider this method in more detail. It is known that the diagonal bisects the angle. Initially, its value was 90 degrees. Thus, after division, two are formed. Their angles at the base will be equal to 45 degrees. Accordingly, each side of the square will be equal, that is: a \u003d b \u003d c \u003d d \u003d e ∙ cosα \u003d e √ 2: 2, where e is the diagonal of the square, or the base of the right triangle formed after division. This is not the only way to find the sides of a square. Let's inscribe this figure in a circle. Knowing the radius of this circle R, we find the side of the square. We will calculate it as follows a4 = R√2. The radii of regular polygons are calculated by the formula R \u003d a: 2tg (360 o: 2n), where a is the length of the side.

How to calculate the perimeter of an n-gon

The perimeter of an n-gon is the sum of all its sides. It is easy to calculate it. To do this, you need to know the values ​​of all sides. For some types of polygons, there are special formulas. They allow you to find the perimeter much faster. It is known that any regular polygon has equal sides. Therefore, in order to calculate its perimeter, it is enough to know at least one of them. The formula will depend on the number of sides of the figure. In general, it looks like this: P \u003d an, where a is the value of the side, and n is the number of angles. For example, to find the perimeter of a regular octagon with a side of 3 cm, you need to multiply it by 8, that is, P = 3 ∙ 8 = 24 cm. For a hexagon with a side of 5 cm, we calculate as follows: P = 5 ∙ 6 = 30 cm. And so for each polygon.

Finding the perimeter of a parallelogram, square and rhombus

Depending on how many sides a regular polygon has, its perimeter is calculated. This makes the task much easier. Indeed, unlike other figures, in this case it is not necessary to look for all its sides, just one is enough. By the same principle, we find the perimeter of quadrangles, that is, a square and a rhombus. Despite the fact that these are different figures, the formula for them is the same P = 4a, where a is the side. Let's take an example. If the side of a rhombus or square is 6 cm, then we find the perimeter as follows: P \u003d 4 ∙ 6 \u003d 24 cm. A parallelogram has only opposite sides. Therefore, its perimeter is found using a different method. So, we need to know the length a and the width b of the figure. Then we apply the formula P \u003d (a + c) ∙ 2. A parallelogram, in which all sides and angles between them are equal, is called a rhombus.

Finding the perimeter of an equilateral and right triangle

The perimeter of the correct one can be found by the formula P \u003d 3a, where a is the length of the side. If it is unknown, it can be found through the median. AT right triangle only two sides are equal. The basis can be found through the Pythagorean theorem. After the values ​​​​of all three sides become known, we calculate the perimeter. It can be found by applying the formula P \u003d a + b + c, where a and b are equal sides, and c is the base. Recall that in an isosceles triangle a \u003d b \u003d a, therefore, a + b \u003d 2a, then P \u003d 2a + c. For example, the side of an isosceles triangle is 4 cm, find its base and perimeter. We calculate the value of the hypotenuse according to the Pythagorean theorem c \u003d √a 2 + in 2 \u003d √16 + 16 \u003d √32 \u003d 5.65 cm. Now we calculate the perimeter P \u003d 2 ∙ 4 + 5.65 \u003d 13.65 cm.

How to find the angles of a regular polygon

A regular polygon occurs in our lives every day, for example, an ordinary square, triangle, octagon. It would seem that there is nothing easier than building this figure yourself. But this is just at first glance. In order to construct any n-gon, you need to know the value of its angles. But how do you find them? Even scientists of antiquity tried to build regular polygons. They guessed to fit them into circles. And then the necessary points were marked on it, connected by straight lines. For simple figures, the construction problem has been solved. Formulas and theorems have been obtained. For example, Euclid in his famous work "The Beginning" was engaged in solving problems for 3-, 4-, 5-, 6- and 15-gons. He found ways to construct them and find angles. Let's see how to do this for a 15-gon. First you need to calculate the sum of its internal angles. It is necessary to use the formula S = 180⁰(n-2). So, we are given a 15-gon, which means that the number n is 15. We substitute the data we know into the formula and get S = 180⁰ (15 - 2) = 180⁰ x 13 = 2340⁰. We have found the sum of all interior angles of a 15-gon. Now we need to get the value of each of them. There are 15 angles in total. We do the calculation of 2340⁰: 15 = 156⁰. This means that each internal angle is 156⁰, now using a ruler and a compass, you can build a regular 15-gon. But what about more complex n-gons? For centuries, scientists have struggled to solve this problem. It was only found in the 18th century by Carl Friedrich Gauss. He was able to build a 65537-gon. Since then, the problem has officially been considered completely solved.

Calculation of angles of n-gons in radians

Of course, there are several ways to find the corners of polygons. Most often they are calculated in degrees. But you can also express them in radians. How to do it? It is necessary to proceed as follows. First, we find out the number of sides of a regular polygon, then subtract 2 from it. So, we get the value: n - 2. Multiply the found difference by the number n ("pi" \u003d 3.14). Now it remains only to divide the resulting product by the number of angles in the n-gon. Consider these calculations using the example of the same fifteen-sided. So, the number n is 15. Let's apply the formula S = p(n - 2) : n = 3.14(15 - 2) : 15 = 3.14 ∙ 13: 15 = 2.72. This is of course not the only way to calculate an angle in radians. You can simply divide the size of the angle in degrees by the number 57.3. After all, that many degrees is equivalent to one radian.

Calculation of the value of angles in degrees

In addition to degrees and radians, you can try to find the value of the angles of a regular polygon in grads. This is done in the following way. From total subtract 2 corners, divide the resulting difference by the number of sides of a regular polygon. We multiply the result found by 200. By the way, such a unit of measurement of angles as degrees is practically not used.

Calculation of external corners of n-gons

For any regular polygon, in addition to the internal one, you can also calculate the external angle. Its value is found in the same way as for other figures. So, to find the outer corner of a regular polygon, you need to know the value of the inner one. Further, we know that the sum of these two angles is always 180 degrees. Therefore, we do the calculations as follows: 180⁰ minus the value of the internal angle. We find the difference. It will be equal to the value of the angle adjacent to it. For example, the inner corner of a square is 90 degrees, so the outer angle will be 180⁰ - 90⁰ = 90⁰. As we can see, it is not difficult to find it. The external angle can take a value from +180⁰ to, respectively, -180⁰.

In the 8th grade, in geometry lessons at school, students for the first time get acquainted with the concept of a convex polygon. Very soon they will learn that this figure has a very interesting property. No matter how complex it may be, the sum of all the internal and external angles of a convex polygon takes on a strictly defined value. In this article, a tutor in mathematics and physics talks about what the sum of the angles of a convex polygon is.

The sum of the interior angles of a convex polygon

How to prove this formula?

Before proceeding to the proof of this statement, we recall which polygon is called convex. A polygon is called convex if it lies entirely on one side of the line containing any of its sides. For example, the one shown in this picture:

If the polygon does not satisfy the specified condition, then it is called non-convex. For example, like this:

The sum of the interior angles of a convex polygon is , where is the number of sides of the polygon.

The proof of this fact is based on the theorem on the sum of angles in a triangle, well known to all schoolchildren. I am sure that you are familiar with this theorem. The sum of the interior angles of a triangle is .

The idea is to split a convex polygon into multiple triangles. It can be done different ways. Depending on which method we choose, the evidence will be slightly different.

1. Divide a convex polygon into triangles by all possible diagonals drawn from some vertex. It is easy to understand that then our n-gon will be divided into triangles:

Moreover, the sum of all the angles of all the resulting triangles is equal to the sum of the angles of our n-gon. After all, each angle in the resulting triangles is a partial angle in our convex polygon. That is, the required amount is equal to .

2. You can also select a point inside the convex polygon and connect it to all vertices. Then our n-gon will be divided into triangles:

Moreover, the sum of the angles of our polygon in this case will be equal to the sum of all the angles of all these triangles minus the central angle, which is equal to . That is, the desired amount is again equal to .

The sum of the exterior angles of a convex polygon

Let us now ask ourselves the question: “What is the sum of the external angles of a convex polygon?” This question can be answered in the following way. Each outer corner is adjacent to the corresponding inner corner. Therefore it is equal to:

Then the sum of all external angles is . That is, it is equal to .

That is a very funny result. If we lay aside sequentially one after another all the external corners of any convex n-gon, then as a result exactly the entire plane will be filled.

This interesting fact can be illustrated as follows. Let's proportionally reduce all sides of some convex polygon until it merges into a point. After this happens, all the outer corners will be set aside one from the other and thus fill the entire plane.

Interesting fact, isn't it? And there are a lot of such facts in geometry. So learn geometry, dear students!

The material on what the sum of the angles of a convex polygon is equal to was prepared by Sergey Valerievich