Polygons. Types of polygons. Inner and outer corners of a convex polygon. The sum of the interior angles of a convex n-gon (theorem). The sum of the external angles of a convex n-gon (theorem). Regular polygons. Circle circumscribed about a regular polygon (theorem, corollary 1.2)

The interior angle of a convex polygon at a given vertex is the angle formed by its sides converging at that vertex. The exterior angle of a convex polygon at a given vertex is the angle adjacent to the interior angle at that vertex. inner corner outer corner

Theorem. The sum of the interior angles of a convex polygon is (n - 2) · 180 o, where n is the number of sides of the polygon. Given: a convex n-gon. Prove: α = (n – 2) 180 o Proof Inside the n-gon, take an arbitrary point O and connect it to all vertices. The polygon will be divided into n triangles with a common vertex O. The sum of the angles of each triangle is 180 o, therefore, the sum of the angles of all triangles is 180 o n. This sum, in addition to the sum of all the internal angles of the polygon, includes the sum of the angles of the triangles at the vertex O, equal to 360 o. Thus, the sum of all internal angles of the polygon is 180 o n - 360 o \u003d (n - 2) 180 o. So, n \u003d (n - 2) 180 o. Ch.t.d. about

Theorem. The sum of the external angles of a convex polygon, taken one at each vertex, does not depend on n and is equal to 360, where n is the number of sides of the n-gon. Proof. Since the external corner of the polygon is adjacent to the corresponding internal angle, and the sum of the adjacent angles is 180, then the sum of the external angles of the polygon is: . External and internal internal So, the sum of the external angles of a convex polygon, taken one at each vertex, does not depend on n and is equal to 360 o, where n is the number of sides of the n-gon. Ch.t.d.

Theorem. Any regular polygon can be inscribed with a circle, and moreover, only one. Proof. Let А1,А2,…,А n be a regular polygon, О be the center of the circumscribed circle. ОА1А2 =ОА2А3= ОАnА1, therefore the heights of these triangles, drawn from the vertex О, are also equal to ОН1=ОН2=…=ОНn. Therefore, the circle with therefore the circle with center O and radius OH1 passes through the points H1, H2, ..., Hn and touches the sides of the polygon at these points, i.e. the circle is inscribed in the given polygon. Hn H1 H2 H3 A1 A2 A3 An

Let us prove that there is only one inscribed circle. Suppose there is another inscribed circle with center O and radius OA. Then its center is equidistant from the sides of the polygon, i.e., the point O1 lies on each of the bisectors of the angles of the polygon, and therefore coincides with the point O of the intersection of these bisectors. The radius of this circle is equal to the distance from the point O to the sides of the polygon, i.e. is equal to OH1. The theorem is proved. Corollary 1 A circle inscribed in a regular polygon touches the sides of the polygon at their midpoints. Corollary 2 The center of a circle circumscribed about a regular polygon coincides with the center of a circle inscribed in the same polygon.

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 are 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 of an isosceles triangle using 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. In a 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. Subtract 2 from the total number of angles, 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.

Purpose: Derive a formula for finding the sum of the angles of a convex polygon;

• investigate the question of the sum of the external angles of a polygon, taken one at each vertex;
• to form a positive motivation for cognitive activity;
• develop logical thinking;
• develop attention, observation, the ability to analyze the drawing;
• to form the ability to apply the acquired knowledge to solve problems;
• to develop the communicative culture of students.

During the classes

Great Russian scientist, pride of the Russian Land,

Mikhailo Vasilyevich Lomonosov, said: "Violent work overcomes obstacles." I hope that today in the lesson our work with you will help us overcome all obstacles.

1. Actualization of basic knowledge. (Front poll.)

Presentation. (Slides 2-4)

- Formulate the definition of a polygon, name its main elements.
– Definition of a convex polygon.
- Give examples of quadrilaterals known to you, which are convex polygons.
Can a triangle be considered a convex polygon?
What is an exterior angle of a convex polygon?

2. Statement of the problem (output on the topic of the lesson).

Oral front work.

Find the sum of the angles of the given polygons (Slides 5-6)

- a triangle; rectangle:
- trapezoid; arbitrary heptagon.

In case of difficulty, the teacher asks questions:

- Formulate the definition of a trapezoid.
Name the bases of a trapezoid.
- What can be said about a pair of angles A and D, what property do they have?
- Can you still name a pair of internal one-sided catches on the drawing?
Can you find the sum of the angles of a heptagon? What is the question? (Is there a formula for finding the sum of the angles of an arbitrary polygon?)

So, it is clear that our knowledge today is not enough to solve this problem.

How can we formulate the topic of our lesson? - Sum of angles convex polygon.

3. Solution Problems. To answer this question, let's do a little research.

We already know the triangle sum theorem. Can we apply it in any way?

– What should be done for this? (Break the polygon into triangles.)

How can a polygon be divided into triangles? Think about it, discuss it and offer your best options.

There is work in groups, each group works on a separate computer on which the program "Geo Gebra" is installed.

At the end of the work, the teacher displays the results of the work of the groups on the screen. (Slide 7)

- Let's analyze the proposed options and try to choose the most optimal for our study.

Let's define the selection criteria: what do we want to get as a result of splitting? (The sum of all the angles of the constructed triangles must be equal to the sum of the angles of the polygon.)

- What options can be immediately discarded? Why?

(Option 1, since the sum of the angles of all triangles is not equal to the sum of the angles of the polygon.)

- Which option is the most suitable? Why? (Option 3.)

How did you get this option? (We drew diagonals from one vertex of the polygon

drawing	n is the number of polygon vertices	Number of diagonals drawn from one vertex	Number of received triangles
	4
	5
	6
	7
	n

- Let's try to establish a relationship between the number of polygon vertices, the number of diagonals that can be drawn from one vertex and the number of triangles obtained.

Each group receives a table that they must complete during the research process.

After discussion in groups, children formulate their conclusions:
from one vertex of an n-gon, n - 3 diagonals can be drawn (since a diagonal cannot be drawn to the chosen vertex itself and to two neighboring ones). In this case, we get n - 2 triangles.

Therefore, the sum of the angles of a convex polygon is 180 0 (n-2).

- Let's return to the proposed options for splitting a polygon into triangles.

Is it possible to use the variant proposed in Figure 4 to prove this theorem?

How many triangles are obtained with such a partition? ( P things)
What is the difference between the sum of the angles of all triangles and the sum of the angles of a polygon? (On 360 0)
- How can you calculate the sum of the angles of a polygon in this case?

(180P– 360 = 180n - 180x2 \u003d 180 (n -2)) (Clay 8)

– Does the variant proposed in Figure 2 satisfy the main requirement that we made for partitioning? (Yes.)

- Why is it not advisable to use it to find the sum of the angles of a polygon? (Harder to count the number of resulting triangles.)

Well, now let's return to the problem that we could not solve at the beginning of the lesson.

(Children verbally count the sum of the angles of the heptagon and two more similar exercises.) (Slide 9 and 10)

4. Application of acquired knowledge .

We have derived a formula for finding the sum of the interior angles of a convex polygon. Now let's talk about the sum of the external angles of the polygon, taken one at each vertex.

So, the task is: which is greater: the sum of the external angles, taken one at each vertex, for a convex hexagon or for a triangle? (Slide 11)

The children make their guesses. The teacher suggests conducting research to resolve this issue.

Each group is given a task to solve independently.

Group 1.

1) Find the sum of the external angles, taken one at each vertex, of a regular triangle.
2) - At a triangle, the degree values ​​of the angles of which are respectively 70 0 , 80 0 and 30 0 .

Group 2

1) Find the sum of the outer corners, taken one at each vertex, of the rectangle.
2) - At a quadrilateral, the interior angles of which are respectively 70 0 , 80 0 and 120 0 and 90 0 .

Group 3.

1) Find the sum of the external angles, taken one at each vertex, of a regular hexagon.
2) - At a hexagon, the internal angles of which are respectively 170 0 , 80 0 and 130 0 , 100 0 , 70 0 , 170 0.

After the end of the work, the children report their results, the teacher enters them in a table and displays them on the screen. (Slide 12)

So, what conclusion can be drawn from the obtained results? (The sum of the external angles, taken one at each vertex, for any polygon is 360 0.)

Now let's try to prove this fact for any n-gon.

If difficulties arise, the proof plan is discussed collectively:

1. Designate the interior angles of the polygon as α, β, γ, etc.
2. Express through the introduced notation the degree measures of external angles
3. Write an expression for finding the sum of the external angles of a polygon
4. Transform the resulting expression, use the previously obtained formula for the sum of the interior angles of the polygon.

The proof is written on the board:

(180 - α) + (180 - β) + (180 - γ) + ... = 180 p - (α + β + γ + ...) = 180 p - 180 (p - 2) = 360

5. Consolidation of the studied material. Problem solving.

Problem 1. Does there exist a convex polygon with such interior angles: 45 0 , 68 0 , 73 0 and 56 0 ? Explain your answer.

Let us prove by contradiction. If a convex polygon has four acute interior angles, then there are four obtuse exterior angles, which means that the sum of all exterior angles of the polygon is greater than 4*90 0 = 360 0 . We have a contradiction. The assertion has been proven.

A convex polygon has three angles of 80 degrees and the rest are 150 degrees. How many corners are in a convex polygon?

Because: for a convex n-gon, the sum of the angles is 180°(n – 2) , then 180(n - 2)=3*80 + x*150, where 3 angles of 80 degrees are given to us according to the condition of the problem, and the number of other angles is still unknown to us, which means we denote their number by x.

However, from the entry on the left side, we determined the number of corners of the polygon as n, since we know the values ​​of three of them from the condition of the problem, it is obvious that x=n-3.

So the equation will look like this: 180(n - 2) = 240 + 150(n - 3)

We solve the resulting equation

180n - 360 = 240 + 150n - 450

180n - 150n = 240 + 360 - 450

Answer: 5 peaks.

6. Summing up the lesson.

So, let's sum it up. Formulate your questions for the guys from another group based on the materials of today's lesson.

What do you think is the best question?

Discuss the degree of participation of each member of the group in collective work, name the most active.

Whose work in the group was the most productive?

7. Homework:

1. Task.

A polygon has three angles of 113 degrees, and the rest are equal to each other and their degree measure is an integer. Find the number of vertices of the polygon.

2. item 114 pp. 169–171, Pogorelov A.V. "Geometry 7–9".

Video lesson 2: Polygons. Problem solving

Lecture: Polygon. Sum of angles of a convex polygon

Polygons- these are the figures that surround us everywhere - this is also the form of honeycombs in which bees store their honey, architectural structures, and much more.

As mentioned earlier, polygons are shapes that have more than two corners. They consist of a closed broken line.

Moreover, the corners of the polygons can be external and internal. For example, a star is a figure that has 10 corners, some of which are convex and others concave:

Examples of convex polygons:

Please note that the figure shows regular polygons - these are the ones that are studied in detail in the school mathematics course.

Any polygon has the same number of vertices as the number of sides. Also note that neighboring vertices are those that have one common side. For example, a triangle has all adjacent vertices.

The more angles a regular polygon has, the greater their degree measure. However, the degree measure of an angle of a convex polygon cannot be greater than or equal to 180 degrees.

To determine the general degree measure of a polygon, you must use the formula.

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 indicated 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. This can be done in 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