Building inscribed in the circumference of the correct hexagon. The construction of the hexagon is based on the fact that its side is equal to the radius of the circle described. Therefore, for the construction, it is enough to separate the circle to six equal parts and connect the points found among themselves (Fig. 60, a).

The right hexagon can be built using a 30x60 ° coal. To perform this construct, we adopt the horizontal diameter of the circle for bisector of the angles 1 and 4 (Fig. 60, b), build the parties 1 -6, 4-3, 4-5 and 7-2, after which we carry out the parties 5-6 and 3- 2.

Building inscribed in the circumference of the equilateral triangle. The vertices of such a triangle can be constructed using a circulation and angle with angles of 30 and 60 ° or only one circular.

Consider two ways to build inscribed in the circumference of the equilateral triangle.

First method (FIG. 61, a) is based on the fact that all three angle of the triangle 7, 2, 3 contains 60 °, and the vertical straight line, carried out through point 7, is simultaneously height and bisector angle 1. Since angle 0-1- 2 is 30 °, then to find a side

1-2 sufficiently construct at point 1 and side 0-1 an angle of 30 °. To do this, we establish a flight and a square as shown in the figure, carry out a line 1-2, which will be one of the sides of the desired triangle. To build a side 2-3, we set the flight to the position shown by the dashing lines, and through point 2 we carry out a straight line that will determine the third vertex of the triangle.

Second way It is based on the fact that, if you build a regular hexagon, inscribed in a circle, and then combine its vertices through one, then the equilateral triangle will be.

To construct a triangle (FIG. 61, b) we plan on the diameter of the vertex point 1 and carry out the diametrical line 1-4. Next, from point 4 by a radius equal to D / 2, describe an arc before intersection with a circle at points 3 and 2. The obtained points will be two other vertices of the desired triangle.

Building a square inscribed in a circle. This construction can be performed using a square and a circular.

The first method is based on the fact that the diagonal of the square intersect in the center of the described circle and tilted to its axes at an angle of 45 °. Based on this, we establish a reynchin and angle with angle of 45 ° as shown in FIG. 62, and, and we mark points 1 and 3. Next through these points, we carry out the horizontal sides of the square 4-1 and 3-2 by these points. Then, with the help of a coal cathet, we spend 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 are divided by in half an arc of the circle concluded between the ends of the diameter (Fig. 62, b). We plan at the ends of two mutually perpendicular diameters of the point A, B and C and from them radius, we describe the arc to the mutual intersection of them.

Next, through the points of intersection of arcs, we carry out auxiliary straight lines marked on the figure with solid lines. The points of their intersection with the circle will determine the vertices 1 and 3; 4 and 2. Thus, the vertices of the desired square are combined consistently between them.

Building inscribed in the circumference of the right pentagon.

To enter into a circle, the right pentagon (FIG. 63), we produce the following constructions.

We look at the circumference point 1 and accept it for one of the tops of the pentagon. We divide the segment of JSC in half. For this, the radius of AO from the point A is described by the arc before intersection with the circle at the points M and B. By connecting these points directly, we obtain the point K, which can then connect with a point 1. Radius equal to the segment A7, describe from a point to an arc before intersection from diametral AO line at the point H. By connecting point 1 with a point H, we get the side of the pentagon. Then, with a solution of a circulation, equal to a segment of 1H, describing the arc from the vertex 1 to the intersection with a circle, we find the vertices 2 and 5. When making a sort of vertex 2 and 5, we obtain the remaining vertices 3 and 4. The points found sequentially connect each other.

Building the right pentagon on this side.

To build a right pentagon on this side (FIG. 64), we divide the AB segment to six equal parts. From the points A and in the AB radius, we describe the arc, the intersection of which will give the point K. through this point and division 3 on a straight AB carry out a vertical direct.

We get a 1-pin pentagon point. Then, with a radius equal to AB, from point 1, describe an arc before intersection with arcs previously carried out from the points A and B. The point of intersection of the arcs determine the vertices of the pentagon 2 and 5. The vertices found connect each other.

Building inscribed in the circumference of the right sevenginous.

Let the circumference of the diameter D; It is necessary to enter in it the right sevenfone (Fig. 65). We divide the vertical diameter of the circle to seven equal parts. From a point 7 by a radius equal to the diameter of the circumference D, describe the arc to the intersection with the continuation of the horizontal diameter at the point F. The point f will call the pole of the polygon. Taking a point VII for one of the vertices of the semi-angry, we carry out from the pole of F through a clear division of the vertical diameter of the rays, the intersection of which with a circle will determine the vertices VI, V and IV of the sevendent. To obtain vertices / - // - /// From points IV, V and VI, we spend the horizontal straight lines to cross the circle. The vertices found are connecting between themselves. The sevenfoon can be built by conducting rays from the pole f and through the odd divisions of the vertical diameter.

The given method is suitable for the construction of the right polygons with any number of parties.

The division of the circle on any number of equal parts can also be made using the data of the table. 2, in which the coefficients give the ability to determine the size of the sides of the right inscribed polygons.

Kuklin Aleksey

Work is a referractable character with elements of research. It addresses various ways to build correct N-koms. The paper contains a detailed answer to the question that you can always build an N-carbon with a circulation and a ruler. The presentation is attached to work, which can be found on this mini-site.

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Signatures for slides:

Building the right polygon work performed: student of grade 9 "in" MBOU SOSH No. 10 Kuklin Alexei

The right polygons are called a convex polygon, in which all sides and corners are equal. Go to Examples The convex polygon is called a polygon, all points of which lie on one side from any direct passing through the two of its neighboring vertices.

Back Proper polygons

The founders of the section of mathematics on the right polygons were ancient Greek scientists. Some of them were Archimed and Euclide.

Proof of the existence of the correct N-parliament if N (number of corners of the polygon) is greater than 2, then such a polygon exists. Let's try to build an 8th square and prove it. Evidence

Take the circle of an arbitrary radius with the center at the point O. We divide it for a certain number of equal arcs, in our case 8. To do this, we will carry out the radii so that it turns out 8 arcs, and the angle between the two nearest radii was equal to 360 °: the number of parties (in our Case 8), respectively, each angle will be 45 °.

3. We obtain points A1, A2, A3, A4, A5, A6, A7, A8. Alternately connect them and get the right octagon. Back

Building the right polygon on the side using turning the correct polygon can be built, knowing its corners. We know that the sum of the angles of the convex N-parliament is 180 ° (N - 2). From this, you can calculate the angle of the polygon, separating the amount on n. Corners Building

The corner of the correct: 3-square is equal to 60 ° 4-° C. equal to 90 ° 5-° C 108 ° 6-Caller is 120 ° 8-° C. 135 ° 9-° 90 ° 10 ° C is 144 ° 12-° C. 150 ° Conduous measure of the corners of the right triangles ago

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In 1796, one of the greatest mathematicians of all times by Karl Friedrich Gauss showed the possibility of constructing the correct N-koms, if equality is performed, where n is the number of angles, and the K-any natural number. Thus, it turned out that within 30 it is possible to divide the circle at 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30 equal parts. In 1836, Vantzel proved that the right polygons that do not satisfy this equality with the help of a ruler and the circulation cannot be built. Theorem Gaussa

Building a triangle construct a circle with a center at point O. We will construct another circumference of the same radius passing through the point O.

3. Connect the centers of circles and one of their intersection points by receiving the right polygon. Back construction triangles

Building a hexagon 1. We construct a circle with the center at the point O. 2. We will spend a straight line through the center of the circle. 3. We carry out an arc of the circumference of the same radius with the center at the point of intersection of the straight line with the circle to the intersection with the circle.

4. We will conduct direct across the center of the initial circle and the intersection point of the arc with this circle. 5. We connect the intersection points of all direct with the source circle and get the right hexagon. Building hexagon

Building a quadrilateral construct a circle with a center at point O. We carry out 2 mutually perpendicular diameter. From points in which the diameters relate to the circle carry out other circumference of this radius to their intersection (circles).

Building a quadrilateer 4. We carry out straight through the counting points of the circles. 5. Connect the intersection points of direct and circles and get the right quadrangle.

Building an octagon can be built any correct polygon in which 2 times more corners than this. Build an octagon with a quadrangle. Connect the opposite vertices of the quadrilateral. We carry out bisector of the angles of the corners of the intersecting diagonals.

4. Connect the points lying on the circle, while getting the right octagon. Building an octagon

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Building a decidagon construct a circle with a center at point O. We carry out 2 mutually perpendicular diameter. We divide the circle radius in half and from the points obtained on it will carry out a circle passing through the point O.

Building a decidagon 4. We will spend a segment from the center of a small circle to a point in which a large circle relates to its radius. 5. From the point of contacting the large circumference and its radius will carry out a circle so that it will come into contact with the small one.

Building a decidagon 6. From the points of intersection of a large and obtained circles, we carry out the circle built last time and will be carried out until the neighboring circles come in touch. 7. Connect points and get a decidagon.

Building a pentagon to build a correct pentagon needed during the construction of the right decagon to connect alternately not all points, but through one.

Approximate construction of the right pentagon by Dürer by constructing 2 circles passing through each other's center. Connect the direct centers by getting one of the sides of the pentagon. Connect the crossing points of the circles.

Approximate construction of the correct pentagon by Dürer 4. We will carry out another circumference of the same radius with the center at the intersection point of the other two circles. 5. We will spend 2 segments as indicated in the figure.

Approximate construction of the right pentagon by Dürer 6. Connect the points of contact with these segments with circles with the ends of the constructed pentagon. 7. Mount to pentagon.

Approximate construction of the right pentagon methods of cowarzhik, biona

In drawing, it is often necessary to build positive polygons. So let's say positive octagons Apply on road signboards.

You will need

• - Circul
• - ruler
• - Pencil

Instruction

1. Let the segment set, equal to the length of the side of the welcome octagon. Requires a faithful octagon. The first step will build an equifiable triangle on a given segment, applying a segment as a base. To do this, at first build the square with a side, equal to the segment, swipe the diagonally in it. Now build the bisector of the corners during the diagonals (in the figure of bisector is shown blue), the vertex of the bisector is formed on the crossing of the bisector, the sides of which are equal to the circle radius described around the faithful octagon.

2. Build a circle with a center at the top of the triangle. The radius of the circle is equal to the side of the triangle. Now imach the circus per distance equal to the value of a given segment. Set aside this distance on the circle, ranging from every end of the segment. Combine all the received points in the octagon.

3. If the circle is asked in which the octagon should be entered, the construction will be even easier. Build two perpendicular axial lines passing through the center of the circle. At the intersection of axial and circle, four vertices of the coming octagon will turn out. It remains to divide the distance between these points on the arc of the circumference of Hunt, in order to receive more four vertices.

Right triangle - One that all parties own an identical length. Based on this definition, building similar varieties triangle And is an easy task.

You will need

• Rule, sheet of voluntary paper, pencil

Instruction

1. Take a sheet of clean paper that is rapid into the cell, a ruler and notice three points on paper so that they were on an identical distance (Fig. 1)

2. With the profile of the line, combine the speed-fitted point of the point step, each other, so, as shown in Figure 2.

Note!
In the right (equilateral) triangle, all corners are equal to 60 degrees.

Helpful advice
The equilateral triangle is also equally chaled. If the triangle is a preceded, it means that 2 of the 3rd sides are equal, and the third party is considered the basis. Any positive triangle is a challenged, while the back statement is not correct.

Octagon - This is, in essence, two squares displaced on each other 45 ° and combined on top of a solid line. And therefore, in order to positively depict such a geometric shape, you need a solid pencil in a dirty, according to the rules, draw the square or a circle with which follow-up. The presentation is focused on the length of the side of 20 cm. And therefore, when the drawing is location, we consider the vertical and horizontal line 20 cm long fit on a sheet of paper.

You will need

• Rule, rectangular triangle, transport, pencil, circulation, sheet of paper

Instruction

1. Method 1. Inscribe at the bottom of the horizontal line 20 cm long. After that, on the one hand, notice the forward angle with the transporter, the one that is 90 °. The same is allowed to do with the support of the direct triangle. Swipe a vertical line and check 20 cm. Do the same manipulations from the other side. Combine the two points obtained by the horizontal line. As a result, it turned out a geometric figure - a square.

2. In order to build the 2nd (shifted) square, the center of the figure will be required. To do this, share any side of the square into 2 parts. Combine 2 points of parallel top and bottom of the parallel top and bottom, and then the pixels of the side. Spend via the center of the square 2 straight lines perpendicular to each other. Starting from the center, measure on new direct length of 10 cm, which as a result will give 4 straight lines. Combine 4 outdoor points among themselves, as a result, it will turn out the 2nd square. Now every point of 8 of the angles received is combined with each other. Thus, an octagon will be drawn.

3. Method 2. This will require a circulation, ruler and transportation. From the center of the sheet with the support of the circulation, draw a circle with a diameter of 20 cm (10 cm radius). Through the center point, swipe the line. After that, draw the second line perpendicular to it. The same is allowed to fulfill with the imaginary transportation or direct triangle. As a result, the circle will be divided into 4 equal parts. Further, any of the sections will extend another 2 parts. To do this, it is also allowed to use the transporter, measuring 45 ° or a rectangular triangle, the one that attach to a sharp angle of 45 ° and swipes. From the center on any straight line, measure 10 cm. In the end, it will turn out 8 "races" that combine each other. As a result, it turns out an octagon.

4. Method 3. For this, also draw the circle, spend through the middle line. After that, take the shipping, put it on the center and measure the angles, considering that every section of the octagon has an angle of 45 ° in the center. Later, at the resulting rays, measure the length of 10 cm. And combine them with each other. Octagon Ready.

Helpful advice
Make a hard pencil drawing, sidelines on which after that easily allowed will delete

The faithful octagon is a geometric shape, which is 135?, All sides between them are equal. This figure is often used in architecture, for example, when building a column, as well as in the manufacture of the Stop road sign. How to draw a positive octagon?

You will need

• - album sheet;
• - pencil;
• - line;
• - Circle;
• - Eraser.

Instruction

1. Draw at the beginning square. After that, spend the circle so so that the square was inside the circle. Now spend two axial middle lines of the square - horizontal and vertical to intersection with a circle. Combine the direct sections of the point of intersection of axes with a circle and the point of touch of the described circle with a square. Thus, get the sides of the right octagon.

2. Draw a faithful octagon with a different method. Initially draw a circle. After that, spend a horizontal line through its center. Stick the point of intersection of the extreme right border of the circle with horizontal. This point will be the center of another circumference, the radius of equal to the previous figure.

3. Spend a vertical line through the intersection points of the 2nd circle from the first. Put the feet of the circulation to the point of intersection of the vertical with horizontal and draw a small range with a radius equal to the distance from the center of the tiny circle to the center of the initial circle.

4. Draw a straight line in two points - the center of the initial circle and the point of intersection of the vertical and tiny circle. Continue it before intersection with the frontier of the original figure. It will be the top of the vertex of the octagon. Circle stuck another point, having a circle with a center at the intersection point of the initial circle with a horizontal and a radius, equal to the distance from the center to the existing vertex of the octagon.

5. Spend a straight line in two points - the center of the initial circle and the last new formed point. Continue the straight line to the intersection with the boundaries of the initial figure.

6. Combine the straight segments of stepped: the point of intersection of the horizontal with the right front of the initial figure, after that clockwise all formed points, including axes intersection points with an initial circle.

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