Do you know what a regular hexagon looks like?
This question was not asked by chance. Most students in grade 11 do not know the answer to it.
A regular hexagon is one in which all sides are equal and all angles are also equal..
Iron nut. Snowflake. A cell of honeycombs in which bees live. Benzene molecule. What do these objects have in common? - The fact that they all have a regular hexagonal shape.
Many schoolchildren are lost when they see tasks for a regular hexagon, and they believe that some special formulas are needed to solve them. Is it so?
Draw the diagonals of a regular hexagon. We got six equilateral triangles. 
We know that the area right triangle: .
Then the area of a regular hexagon is six times larger.
Where is the side of a regular hexagon.
Please note that in a regular hexagon, the distance from its center to any of the vertices is the same and equal to the side of the regular hexagon.
This means that the radius of a circle circumscribed around a regular hexagon is equal to its side.
The radius of a circle inscribed in a regular hexagon is easy to find.
He is equal.
Now you can easily solve any USE problems in which a regular hexagon appears.
Find the radius of a circle inscribed in a regular hexagon with side .
The radius of such a circle is .
Answer: .
What is the side of a regular hexagon inscribed in a circle with a radius of 6?
We know that the side of a regular hexagon is equal to the radius of the circle circumscribed around it.
The topic of polygons is held in school curriculum but do not pay enough attention to it. Meanwhile, it is interesting, and this is especially true for a regular hexagon or hexagon - after all, many natural objects. These include honeycombs and more. This form is very well applied in practice.
Definition and construction
A regular hexagon is a plane figure that has six sides equal in length and the same number of equal angles.
If we recall the formula for the sum of the angles of a polygon
it turns out that in this figure it is equal to 720 °. Well, since all the angles of the figure are equal, it is easy to calculate that each of them is equal to 120 °.
Drawing a hexagon is very simple, all you need is a compass and a ruler.
The step by step instructions will look like this:
If desired, you can do without a line by drawing five circles of equal radius.
The figure thus obtained will be a regular hexagon, and this can be proved below.
Properties are simple and interesting
To understand the properties of a regular hexagon, it makes sense to break it into six triangles:
This will help in the future to more clearly display its properties, the main of which are:
- circumscribed circle diameter;
- diameter of the inscribed circle;
- square;
- perimeter.
The circumscribed circle and the possibility of construction
It is possible to describe a circle around a hexagon, and moreover, only one. Since this figure is correct, you can do it quite simply: draw a bisector from two adjacent angles inside. They intersect at point O, and together with the side between them form a triangle.
The angles between the side of the hexagon and the bisectors will be 60° each, so we can definitely say that a triangle, for example, AOB, is isosceles. And since the third angle will also be equal to 60 °, it is also equilateral. It follows that the segments OA and OB are equal, which means that they can serve as the radius of the circle.
After that, you can go to the next side, and also draw a bisector from the angle at point C. It will turn out another equilateral triangle, and side AB will be common to two at once, and OS will be the next radius through which the same circle goes. There will be six such triangles in total, and they will have a common vertex at point O. It turns out that it will be possible to describe the circle, and it is only one, and its radius is equal to the side of the hexagon:
That is why it is possible to construct this figure with the help of a compass and a ruler.
Well, the area of \u200b\u200bthis circle will be standard:
Inscribed circle
The center of the circumscribed circle coincides with the center of the inscribed one. To verify this, we can draw perpendiculars from the point O to the sides of the hexagon. They will be the heights of those triangles that make up the hexagon. And in an isosceles triangle, the height is the median with respect to the side on which it rests. Thus, this height is nothing but the perpendicular bisector, which is the radius of the inscribed circle.
The height of an equilateral triangle is calculated simply:
h²=a²-(a/2)²= a²3/4, h=a(√3)/2
And since R=a and r=h, it turns out that
r=R(√3)/2.
Thus, the inscribed circle passes through the centers of the sides of a regular hexagon.
Its area will be:
S=3πa²/4,
that is, three-quarters of that described.
Perimeter and area
Everything is clear with the perimeter, this is the sum of the lengths of the sides:
P=6a, or P=6R
But the area will be equal to the sum of all six triangles into which the hexagon can be divided. Since the area of a triangle is calculated as half the product of the base and the height, then:
S \u003d 6 (a / 2) (a (√3) / 2) \u003d 6a² (√3) / 4 \u003d 3a² (√3) / 2 or
S=3R²(√3)/2
Those who wish to calculate this area through the radius of the inscribed circle can be done like this:
S=3(2r/√3)²(√3)/2=r²(2√3)
Entertaining constructions
A triangle can be inscribed in a hexagon, the sides of which will connect the vertices through one:
There will be two of them in total, and their imposition on each other will give the Star of David. Each of these triangles is equilateral. This is easy to verify. If you look at the AC side, then it belongs to two triangles at once - BAC and AEC. If in the first of them AB \u003d BC, and the angle between them is 120 °, then each of the remaining ones will be 30 °. From this we can draw logical conclusions:
- The height of ABC from vertex B will be equal to half the side of the hexagon, since sin30°=1/2. Those who wish to verify this can be advised to recalculate according to the Pythagorean theorem, it fits here perfectly.
- The AC side will be equal to two radii of the inscribed circle, which is again calculated using the same theorem. That is, AC=2(a(√3)/2)=а(√3).
- Triangles ABC, CDE and AEF are equal in two sides and the angle between them, and hence the equality of sides AC, CE and EA follows.
Intersecting with each other, the triangles form a new hexagon, and it is also regular. It's easy to prove:
Thus, the figure meets the signs of a regular hexagon - it has six equal sides and corners. From the equality of triangles at the vertices, it is easy to deduce the length of the side of the new hexagon:
d=а(√3)/3
It will also be the radius of the circle described around it. The radius of the inscribed will be half the side of the large hexagon, which was proved when considering the triangle ABC. Its height is exactly half of the side, therefore, the second half is the radius of the circle inscribed in the small hexagon:
r₂=а/2
S=(3(√3)/2)(а(√3)/3)²=а(√3)/2
It turns out that the area of the hexagon inside the star of David is three times smaller than that of the large one in which the star is inscribed.
From theory to practice
The properties of the hexagon are very actively used both in nature and in various fields of human activity. First of all, this applies to bolts and nuts - the hats of the first and second are nothing more than a regular hexagon, if you do not take into account the chamfers. The size of wrenches corresponds to the diameter of the inscribed circle - that is, the distance between opposite faces.
Has found its application and hexagonal tiles. It is much less common than a quadrangular one, but it is more convenient to lay it: three tiles meet at one point, not four. Compositions can be very interesting:
Concrete paving slabs are also produced.
The prevalence of the hexagon in nature is explained simply. Thus, it is easiest to fit circles and balls tightly on a plane if they have the same diameter. Because of this, honeycombs have such a shape.
Mathematical properties
A feature of a regular hexagon is the equality of its side and the radius of the circumscribed circle, since
All angles are 120°.
The radius of the inscribed circle is:
The perimeter of a regular hexagon is:
The area of a regular hexagon is calculated by the formulas:
Hexagons tiling the plane, that is, they can fill the plane without gaps and overlaps, forming the so-called parquet.
Hexagonal parquet (hexagonal parquet)- tessellation of the plane with equal regular hexagons located side to side.
Hexagonal parquet is dual to triangular parquet: if you connect the centers of adjacent hexagons, then the segments drawn will give a triangular parquet. The Schläfli symbol of a hexagonal parquet is (6,3), which means that three hexagons converge at each vertex of the parquet.
Hexagonal parquet is the most dense packing of circles on the plane. In two-dimensional Euclidean space, the best filling is to place the centers of the circles at the vertices of a parquet formed by regular hexagons, in which each circle is surrounded by six others. The density of this packing is . In 1940, it was proved that this packing is the densest.
A regular hexagon with a side is a universal cover, that is, any set of diameter can be covered by a regular hexagon with a side (Pal's lemma).
A regular hexagon can be constructed using a compass and straightedge. Below is the construction method proposed by Euclid in the Elements, Book IV, Theorem 15.
Regular hexagon in nature, technology and culture
show the partition of the plane into regular hexagons. The hexagonal shape more than the others allows you to save on the walls, that is, less wax will be spent on honeycombs with such cells.
Some complex crystals and molecules, such as graphite, have a hexagonal crystal lattice.
Formed when microscopic water droplets in clouds are attracted to dust particles and freeze. The ice crystals that appear in this case, which at first do not exceed 0.1 mm in diameter, fall down and grow as a result of condensation of moisture from the air on them. In this case, six-pointed crystalline forms are formed. Due to the structure of water molecules, only 60° and 120° angles are possible between the rays of the crystal. The main water crystal has the shape of a regular hexagon in the plane. New crystals are then deposited on the tops of such a hexagon, new ones are deposited on them, and thus various forms of snowflake stars are obtained.
Scientists from Oxford University were able to simulate the emergence of such a hexagon in the laboratory. To find out how such a formation occurs, the researchers placed a 30-liter bottle of water on a turntable. She modeled the atmosphere of Saturn and its usual rotation. Inside, scientists placed small rings that rotate faster than the container. This generated miniature eddies and jets, which the experimenters visualized with green paint. The faster the ring rotated, the larger the eddies became, causing the nearby stream to deviate from a circular shape. Thus, the authors of the experiment managed to obtain various shapes - ovals, triangles, squares and, of course, the desired hexagon.
A natural monument of about 40,000 interconnected basalt (rarely andesitic) columns, formed as a result of an ancient volcanic eruption. Located in the north-east of Northern Ireland, 3 km north of the city of Bushmills.
The tops of the columns form a kind of springboard, which starts at the foot of the cliff and disappears under the surface of the sea. Most of the columns are hexagonal, although some have four, five, seven or eight corners. The tallest column is about 12 meters high.
About 50-60 million years ago, during the Paleogene period, the Antrim site was subject to intense volcanic activity when molten basalt permeated through the deposits, forming extensive lava plateaus. With rapid cooling, the volume of the substance decreased (this is observed when the mud dries). Horizontal compression resulted in the characteristic structure of hexagonal pillars.
The cross section of the nut has the form of a regular hexagon.
Construction of a regular hexagon inscribed in a circle. The construction of a hexagon is based on the fact that its side is equal to the radius of the circumscribed circle. Therefore, to build, it is enough to divide the circle into six equal parts and connect the found points to each other (Fig. 60, a).
A regular hexagon can be constructed using a T-square and a 30X60° square. To perform this construction, we take the horizontal diameter of the circle as the bisector of angles 1 and 4 (Fig. 60, b), build sides 1-6, 4-3, 4-5 and 7-2, after which we draw sides 5-6 and 3- 2.
Construction of an equilateral triangle inscribed in a circle. The vertices of such a triangle can be constructed using a compass and a square with angles of 30 and 60 °, or only one compass.
Consider two ways to construct an equilateral triangle inscribed in a circle.
First way(Fig. 61, a) is based on the fact that all three angles of the triangle 7, 2, 3 each contain 60 °, and the vertical line drawn through point 7 is both the height and the bisector of angle 1. Since the angle 0-1- 2 is equal to 30°, then to find the side
1-2, it is enough to build an angle of 30 ° at point 1 and side 0-1. To do this, set the T-square and square as shown in the figure, draw a line 1-2, which will be one of the sides of the desired triangle. To build side 2-3, set the T-square to the position shown by the dashed lines, and draw a straight line through point 2, which will define the third vertex of the triangle.
Second way is based on the fact that if you build a regular hexagon inscribed in a circle, and then connect its vertices through one, you get an equilateral triangle.
To construct a triangle (Fig. 61, b), we mark a vertex-point 1 on the diameter and draw a diametrical line 1-4. Further, from point 4 with a radius equal to D / 2, we describe the arc until it intersects with the circle at points 3 and 2. The resulting points will be two other vertices of the desired triangle.
Construction of a square inscribed in a circle. This construction can be done using a square and a compass.
The first method is based on the fact that the diagonals of the square intersect in the center of the circumscribed circle and are inclined to its axes at an angle of 45°. Based on this, we install a T-square and a square with angles of 45 ° as shown in Fig. 62, a, and mark points 1 and 3. Further, through these points, we draw the horizontal sides of the square 4-1 and 3-2 with the help of a T-square. Then, using a T-square along the leg of the square, we draw the vertical sides of the square 1-2 and 4-3.
The second method is based on the fact that the vertices of the square bisect the arcs of the circle enclosed between the ends of the diameter (Fig. 62, b). We mark points A, B and C at the ends of two mutually perpendicular diameters, and from them with a radius y we describe the arcs until they intersect.
Further, through the points of intersection of the arcs, we draw auxiliary lines, marked on the figure with solid lines. Their points of intersection with the circle will define vertices 1 and 3; 4 and 2. The vertices of the desired square obtained in this way are connected in series with each other.
Construction of a regular pentagon inscribed in a circle.
To inscribe a regular pentagon in a circle (Fig. 63), we make the following constructions.
We mark point 1 on the circle and take it as one of the vertices of the pentagon. Divide segment AO in half. To do this, with the radius AO from point A, we describe the arc until it intersects with the circle at points M and B. Connecting these points with a straight line, we get point K, which we then connect with point 1. With a radius equal to segment A7, we describe the arc from point K to the intersection with the diametrical line AO at point H. Connecting point 1 with point H, we get the side of the pentagon. Then, with a compass opening equal to the segment 1H, having described the arc from vertex 1 to the intersection with the circle, we find vertices 2 and 5. Having made serifs from vertices 2 and 5 with the same compass opening, we obtain the remaining vertices 3 and 4. We connect the found points sequentially with each other.
Construction of a regular pentagon given its side.
To construct a regular pentagon along its given side (Fig. 64), we divide the segment AB into six equal parts. From points A and B with radius AB we describe arcs, the intersection of which will give point K. Through this point and division 3 on the line AB we draw a vertical line.
We get the point 1-vertex of the pentagon. Then, with a radius equal to AB, from point 1 we describe the arc to the intersection with the arcs previously drawn from points A and B. The intersection points of the arcs determine the vertices of the pentagon 2 and 5. We connect the found vertices in series with each other.
Construction of a regular heptagon inscribed in a circle.
Let a circle of diameter D be given; you need to inscribe a regular heptagon into it (Fig. 65). Divide the vertical diameter of the circle into seven equal parts. From point 7 with a radius equal to the diameter of the circle D, we describe the arc until it intersects with the continuation of the horizontal diameter at point F. Point F is called the pole of the polygon. Taking point VII as one of the vertices of the heptagon, we draw rays from the pole F through even divisions of the vertical diameter, the intersection of which with the circle will determine the vertices VI, V and IV of the heptagon. To obtain vertices / - // - /// from points IV, V and VI, we draw horizontal lines until they intersect with the circle. We connect the found vertices in series with each other. The heptagon can be constructed by drawing rays from the F pole and through odd divisions of the vertical diameter.
The above method is suitable for constructing regular polygons with any number of sides.
The division of a circle into any number of equal parts can also be done using the data in Table. 2, which shows the coefficients that make it possible to determine the dimensions of the sides of regular inscribed polygons.
The most famous figure with more than four corners is the regular hexagon. In geometry, it is often used in problems. And in life, this is exactly what honeycombs have on the cut.
How is it different from wrong?
First, a hexagon is a figure with 6 vertices. Secondly, it can be convex or concave. The first one differs in that four vertices lie on one side of a straight line drawn through the other two.
Thirdly, a regular hexagon is characterized by the fact that all its sides are equal. Moreover, each corner of the figure also has the same value. To determine the sum of all its angles, you will need to use the formula: 180º * (n - 2). Here n is the number of vertices of the figure, that is, 6. A simple calculation gives a value of 720º. So each angle is 120 degrees.
In everyday activities, a regular hexagon is found in a snowflake and a nut. Chemists see it even in the benzene molecule.
What properties do you need to know when solving problems?
To what is stated above should be added:
- the diagonals of the figure, drawn through the center, divide it into six triangles, which are equilateral;
- the side of a regular hexagon has a value that coincides with the radius of the circumscribed circle around it;
- using such a figure, it is possible to fill the plane, and between them there will be no gaps and no overlaps.
Introduced notation
Traditionally, the side of a regular geometric figure is denoted by the Latin letter "a". To solve problems, area and perimeter are also required, these are S and P, respectively. A circle is inscribed in a regular hexagon or circumscribed about it. Then values for their radii are entered. They are denoted respectively by the letters r and R.
In some formulas, an internal angle, a semi-perimeter and an apothem (which is a perpendicular to the middle of any side from the center of the polygon) appear. Letters are used for them: α, p, m.
Formulas that describe a figure
To calculate the radius of an inscribed circle, you need this: r= (a * √3) / 2, and r = m. That is, the same formula will be for the apothem.
Since the perimeter of a hexagon is the sum of all sides, it will be determined as follows: P = 6 * a. Given that the side is equal to the radius of the circumscribed circle, for the perimeter there is such a formula for a regular hexagon: P \u003d 6 * R. From the one given for the radius of the inscribed circle, the relationship between a and r is derived. Then the formula takes the following form: Р = 4 r * √3.
For the area of a regular hexagon, this might come in handy: S = p * r = (a 2 * 3 √3) / 2.
Tasks
No. 1. Condition. There is a regular hexagonal prism, each edge of which is equal to 4 cm. A cylinder is inscribed in it, the volume of which must be determined.
Solution. The volume of a cylinder is defined as the product of the area of the base and the height. The latter coincides with the edge of the prism. And it is equal to the side of a regular hexagon. That is, the height of the cylinder is also 4 cm.
To find out the area of its base, you need to calculate the radius of the circle inscribed in the hexagon. The formula for this is shown above. So r = 2√3 (cm). Then the area of the circle: S \u003d π * r 2 \u003d 3.14 * (2√3) 2 \u003d 37.68 (cm 2).
Answer. V \u003d 150.72 cm 3.
No. 2. Condition. Calculate the radius of a circle that is inscribed in a regular hexagon. It is known that its side is √3 cm. What will be its perimeter?
Solution. This task requires the use of two of the above formulas. Moreover, they must be applied without even modifying, just substitute the value of the side and calculate.
Thus, the radius of the inscribed circle turns out to be 1.5 cm. For the perimeter, the following value turns out to be correct: 6√3 cm.
Answer. r = 1.5 cm, Р = 6√3 cm.
No. 3. Condition. The radius of the circumscribed circle is 6 cm. What value will the side of a regular hexagon have in this case?
Solution. From the formula for the radius of a circle inscribed in a hexagon, one easily obtains the one by which the side must be calculated. It is clear that the radius is multiplied by two and divided by the root of three. It is necessary to get rid of the irrationality in the denominator. Therefore, the result of actions takes the following form: (12 √3) / (√3 * √3), that is, 4√3.
Answer. a = 4√3 cm.
