To construct any drawing or perform planar markings of a workpiece before processing it, it is necessary to carry out a number of graphic operations - geometric constructions.
In Fig. Figure 2.1 shows a flat part - a plate. To draw its drawing or mark a contour on a steel strip for subsequent manufacturing, you need to do it on the construction plane, the main ones are numbered with numbers written on the pointer arrows. In numbers 1 indicates the construction of mutually perpendicular lines, which must be performed in several places, with the number 2 – drawing parallel lines, in numbers 3 – pairing these parallel lines with an arc of a certain radius, a number 4 – conjugation of an arc and a straight arc given radius, which in in this case equal to 10 mm, number 5 – pairing of two arcs with an arc of a certain radius.
As a result of performing these and other geometric constructions, the contour of the part will be drawn.
Geometric construction is a method of solving a problem in which the answer is obtained graphically without any calculations. Constructions are carried out with drawing (or marking) tools as carefully as possible, because the accuracy of the solution depends on this.
Lines, given by conditions tasks, as well as constructions, are performed as solid subtle ones, and the results of construction are performed as solid basic ones.
When starting to make a drawing or marking, you must first determine which of the geometric constructions need to be applied in this case, i.e. analyze the graphic composition of the image.
Rice. 2.1.
Analysis of the graphic composition of the image called the process of dividing the execution of a drawing into separate graphic operations.
Identifying the operations required to construct a drawing makes it easier to choose how to execute it. If you need to draw, for example, the plate shown in Fig. 2.1, then analysis of the contour of its image leads us to the conclusion that we must apply the following geometric constructions: in five cases, draw mutually perpendicular center lines (figure 1 in a circle), in four cases draw parallel lines(number 2 ), draw two concentric circles (0 50 and 70 mm), in six cases construct mates of two parallel straight lines with arcs of a given radius (figure 3 ), and in four - the pairing of an arc and a straight arc of radius 10 mm (figure 4 ), in four cases, construct a pairing of two arcs with an arc of radius 5 mm (number 5 in a circle).
To carry out these constructions, you need to remember or repeat from the textbook the rules for drawing them.
In this case, it is advisable to choose a rational way to complete the drawing. Choice rational way solving a problem reduces the time spent on work. For example, when building equilateral triangle, inscribed in a circle, a more rational method is to construct it using a crossbar and a square with an angle of 60° without first determining the vertices of the triangle (see Fig. 2.2, a, b). A less rational way to solve the same problem is using a compass and a crossbar with preliminary determination of the vertices of the triangle (see Fig. 2.2, V).
Dividing segments and constructing angles
Constructing right angles
It is rational to construct a 90° angle using a crossbar and a square (Fig. 2.2). To do this, it is enough to draw a straight line and restore a perpendicular to it using a square (Fig. 2.2, A). It is rational to build a perpendicular to the inclined segment by moving (Fig. 2.2, b) or turning (Fig. 2.2, V) square.
Rice. 2.2.
Construction of obtuse and acute angles
Rational methods for constructing angles of 120, 30 and 150, 60 and 120, 15 and 165, 75 and 105.45 and 135° are shown in Fig. 2.3, which shows the positions of the squares for constructing these angles.
Rice. 2.3.
Dividing an angle into two equal parts
From the vertex of the corner, describe an arc of a circle of arbitrary radius (Fig. 2.4).
Rice. 2.4.
From points ΜηΝ intersection of the arc with the sides of the angle with a compass solution, more than half arcs ΜΝ, make two intersecting at a point A serifs.
Through the received point A and the vertex of the angle draw a straight line (the bisector of the angle).
Dividing a right angle into three equal parts
From the top right angle describe an arc of a circle of arbitrary radius (Fig. 2.5). Without changing the angle of the compass, make notches from the points of intersection of the arc with the sides of the angle. Through the received points M And Ν and the vertex of the angle are drawn by straight lines.
Rice. 2.5.
In this way, only right angles can be divided into three equal parts.
Constructing an angle equal to a given one. From the top ABOUT given an angle, draw an arc of arbitrary radius R, intersecting the sides of the angle at points M And N(Fig. 2.6, A). Then draw a straight segment, which will serve as one of the sides of the new angle. From point ABOUT 1 on this straight line with the same radius R draw an arc, getting a point Ν 1 (Fig. 2.6, b). From this point describe an arc of radius R 1, equal to the chord MN. The intersection of arcs gives a point Μ 1, which is connected by a straight line to the vertex of the new angle (Fig. 2.6, b).
Rice. 2.6.
Dividing a line segment into two equal parts. From the ends given segment with a compass opening greater than half its length, describe the arcs (Fig. 2.7). Straight line connecting the obtained points M And Ν, divides a segment into two equal parts and is perpendicular to it.
Rice. 2.7.
Constructing a perpendicular at the end of a straight line segment. From an arbitrary point O taken above the segment AB, describe a circle passing through a point A(end of a line segment) and intersecting the line at the point M(Fig. 2.8).
Rice. 2.8.
Through the received point M and center ABOUT circles draw a straight line until they meet opposite side circle at a point N. Full stop N connect a straight line to a point A.
Dividing a line segment by any number equal parts. From any end of a segment, for example from a point A, draw a straight line at an acute angle to it. On it, with a measuring compass, they lay down the right number equal segments of arbitrary size (Fig. 2.9). The last point is connected to the second end of the given segment (to the point IN). From all division points, using a ruler and a square, draw straight lines parallel to the straight line 9V, which will divide the segment AB into given number equal parts.
Rice. 2.9.
In Fig. Figure 2.10 shows how to apply this construction to mark the centers of holes evenly spaced on a straight line.
When building or developing home design projects, it is often necessary to build an angle equal to an existing one. Templates come to the rescue school knowledge geometry.
Instructions
- An angle is formed by two straight lines emanating from one point. This point will be called the vertex of the angle, and the lines will be the sides of the angle.
- Use three letters to represent corners: one at the top, two at the sides. The angle is named starting with the letter that stands on one side, then the letter that stands at the apex is named, and then the letter on the other side. Use other ways to indicate angles if you prefer otherwise. Sometimes only one letter is named, which is at the top. Can you mark the angles? Greek letters, for example, α, β, γ.
- There are situations when it is necessary to draw an angle so that it is equal to an already given angle. If it is not possible to use a protractor when constructing a drawing, you can only get by with a ruler and compass. Let's say on a straight line marked in the drawing with the letters MN, you need to construct an angle at point K, so that it is equal to angle B. That is, from point K it is necessary to draw a straight line forming an angle with the line MN, which will be equal to angle B.
- First, mark a point on each side of a given angle, for example, points A and C, then connect points C and A with a straight line. Get triangle ABC.
- Now construct the same triangle on line MN so that its vertex B is on the line at point K. Use the rule for constructing a triangle on three sides. Lay off the segment KL from point K. It must be equal to the segment BC. Get the L point.
- From point K, draw a circle with a radius equal to segment BA. From L, draw a circle with radius CA. Connect the resulting point (P) of intersection of two circles with K. Obtain triangle KPL, which will be equal to triangle ABC. This way you will get angle K. It will be equal to angle B. To make this construction more convenient and faster, set aside from vertex B equal segments, using one compass opening, without moving the legs, describe a circle with the same radius from point K.
Lesson objectives:
- Formation of the ability to analyze the studied material and the skills of applying it to solve problems;
- Show the significance of the concepts being studied;
- Development cognitive activity and independence in acquiring knowledge;
- Cultivating interest in the subject and a sense of beauty.
Lesson objectives:
- Develop skills in constructing an angle equal to a given one using a scale ruler, compass, protractor and drawing triangle.
- Test students' problem-solving skills.
Lesson plan:
- Repetition.
- Constructing an angle equal to a given one.
- Analysis.
- Construction example first.
- Construction example two.
Repetition.
Corner.
Flat angle- an unlimited geometric figure formed by two rays (sides of an angle) emerging from one point (vertex of the angle).
An angle is also called a figure formed by all points of the plane enclosed between these rays (Generally speaking, two such rays correspond to two angles, since they divide the plane into two parts. One of these angles is conventionally called internal, and the other - external.
Sometimes, for brevity, the angle is called the angular measure.
There is a generally accepted symbol to denote an angle: , proposed in 1634 by the French mathematician Pierre Erigon.
Corner is a geometric figure (Fig. 1), formed by two rays OA and OB (sides of the angle), emanating from one point O (vertex of the angle).
An angle is denoted by a symbol and three letters indicating the ends of the rays and the vertex of the angle: AOB (and the letter of the vertex is the middle one). Angles are measured by the amount of rotation of ray OA around vertex O until ray OA moves to position OB. There are two widely used units for measuring angles: radians and degrees. For radian measurement of angles, see below in the paragraph “Arc Length”, as well as in the chapter “Trigonometry”.
Degree system for measuring angles.
Here the unit of measurement is a degree (its designation is °) - this is a rotation of the beam by 1/360 of a full revolution. Thus, full turn beam is equal to 360 o. One degree is divided into 60 minutes (symbol ‘); one minute – respectively for 60 seconds (designation “). An angle of 90° (Fig. 2) is called right; an angle less than 90° (Fig. 3) is called acute; an angle greater than 90° (Fig. 4) is called obtuse.
Straight lines forming a right angle are called mutually perpendicular. If the lines AB and MK are perpendicular, then this is denoted: AB MK.
Constructing an angle equal to a given one.
Before starting construction or solving any problem, regardless of the subject, you need to carry out analysis. Understand what the assignment says, read it thoughtfully and slowly. If after the first time you have doubts or something was not clear or clear but not completely, it is recommended to read it again. If you are doing an assignment in class, you can ask the teacher. IN otherwise your task, which you misunderstood, may not be solved correctly, or you may find something that is not what was required of you, and it will be considered incorrect and you will have to redo it. As for me - It’s better to spend a little more time studying the task than to redo the task all over again.
Analysis.
Let a be the given ray with vertex A, and the angle (ab) be the desired one. Let's choose points B and C on rays a and b, respectively. By connecting points B and C, we get triangle ABC. IN equal triangles the corresponding angles are equal, and hence the method of construction follows. If on the sides of a given angle we select points C and B in some convenient way, and from a given ray into a given half-plane we construct a triangle AB 1 C 1 equal to ABC (and this can be done if we know all the sides of the triangle), then the problem will be solved.
When carrying out any constructions Be extremely careful and try to carry out all constructions carefully. Since any inconsistencies can result in some kind of errors, deviations, which can lead to an incorrect answer. And if the task of this type is performed for the first time, the error will be very difficult to find and fix.
Construction example first.
Let's draw a circle with its center at the vertex of this angle. Let B and C be the points of intersection of the circle with the sides of the angle. With radius AB we draw a circle with the center at point A 1 – the starting point of this ray. Let us denote the point of intersection of this circle with this ray as B 1 . Let us describe a circle with center at B 1 and radius BC. The intersection point C 1 of the constructed circles in the indicated half-plane lies on the side of the desired angle.
Triangles ABC and A 1 B 1 C 1 are equal on three sides. Angles A and A 1 are the corresponding angles of these triangles. Therefore, ∠CAB = ∠C 1 A 1 B 1
For greater clarity, you can consider the same constructions in more detail.
Construction example two.
The task remains to also set aside an angle equal to a given angle from a given half-line into a given half-plane.
Construction.
Step 1. Let's draw a circle with arbitrary radius and centers at vertex A of a given angle. Let B and C be the points of intersection of the circle with the sides of the angle. And let's draw segment BC.
Step 2. Let's draw a circle of radius AB with the center at point O - the starting point of this half-line. Let us denote the point of intersection of the circle with the ray as B 1 .
Step 3. Now we describe a circle with center B 1 and radius BC. Let point C 1 be the intersection of the constructed circles in the indicated half-plane. 
Step 4. Let's draw a ray from point O through point C 1. Angle C 1 OB 1 will be the desired one.
Proof.
Triangles ABC and OB 1 C 1 are congruent triangles with corresponding sides. And therefore angles CAB and C 1 OB 1 are equal.
Interesting fact:
In numbers.
In the objects of the surrounding world, you first of all notice them individual properties that distinguish one object from another.
Abundance of private individual properties obscures the general properties inherent in absolutely all objects, and it is therefore always more difficult to detect such properties.
One of the most important general properties of objects is that all objects can be counted and measured. We reflect this general property objects in the concept of number.
People mastered the process of counting, that is, the concept of number, very slowly, over centuries, in a persistent struggle for their existence.
In order to count, one must not only have objects that can be counted, but also already have the ability to abstract when considering these objects from all their other properties except number, and this ability is the result of a long historical development based on experience.
Every person now learns to count with the help of numbers imperceptibly in childhood, almost simultaneously with the time he begins to speak, but this counting, which is familiar to us, has gone through a long path of development and has taken different forms.
There was a time when only two numerals were used to count objects: one and two. In the process of further expansion of the number system, parts were involved human body and first of all the fingers, and if this kind of “numbers” was not enough, then also sticks, stones and other things.
N. N. Miklouho-Maclay in his book "Trips" talks about a funny method of counting used by the natives of New Guinea:
Questions:
- Define angle?
- What types of angles are there?
- What is the difference between diameter and radius?
List of sources used:
- Mazur K. I. “Solving the main competition problems in mathematics of the collection edited by M. I. Skanavi”
- Mathematical savvy. B.A. Kordemsky. Moscow.
- L. S. Atanasyan, V. F. Butuzov, S. B. Kadomtsev, E. G. Poznyak, I. I. Yudina “Geometry, 7 – 9: textbook for educational institutions”
Worked on the lesson:
Levchenko V.S.
Poturnak S.A.
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Subjects > Mathematics > Mathematics 7th gradeIn construction problems we will consider the construction geometric figure which can be done using a ruler and compass.
Using a ruler you can:
arbitrary straight line;
an arbitrary straight line passing through a given point;
a straight line passing through two given points.
Using a compass, you can describe a circle of a given radius from a given center.
Using a compass you can plot a segment on a given line from a given point.
Let's consider the main construction tasks.
Task 1. Construct a triangle with given sides a, b, c (Fig. 1).
Solution. Using a ruler, draw an arbitrary straight line and take on it arbitrary point B. Using a compass opening equal to a, we describe a circle with center B and radius a. Let C be the point of its intersection with the line. With a compass opening equal to c, we describe a circle from center B, and with a compass opening equal to b, we describe a circle from center C. Let A be the intersection point of these circles. Triangle ABC has sides equal to a, b, c.
Comment. In order for three straight segments to serve as sides of a triangle, it is necessary that the largest of them be less than the sum of the other two (and< b + с).
Task 2.
Solution. This angle with vertex A and the ray OM are shown in Figure 2.
Let us draw an arbitrary circle with its center at vertex A of the given angle. Let B and C be the points of intersection of the circle with the sides of the angle (Fig. 3, a). With radius AB we draw a circle with the center at point O - the starting point of this ray (Fig. 3, b). Let us denote the point of intersection of this circle with this ray as C 1 . Let us describe a circle with center C 1 and radius BC. Point B 1 of the intersection of two circles lies on the side of the desired angle. This follows from the equality Δ ABC = Δ OB 1 C 1 (the third sign of equality of triangles).
Task 3. Construct the bisector of this angle (Fig. 4).
Solution. From vertex A of a given angle, as from the center, we draw a circle of arbitrary radius. Let B and C be the points of its intersection with the sides of the angle. From points B and C we describe circles with the same radius. Let D be their intersection point, different from A. Ray AD bisects angle A. This follows from the equality Δ ABD = Δ ACD (the third criterion for the equality of triangles).
Task 4. Draw a perpendicular bisector to this segment (Fig. 5).
Solution. Using an arbitrary but identical compass opening (larger than 1/2 AB), we describe two arcs with centers at points A and B, which will intersect each other at some points C and D. The straight line CD will be the desired perpendicular. Indeed, as can be seen from the construction, each of the points C and D is equally distant from A and B; therefore, these points must lie on the perpendicular bisector to segment AB.
Task 5. Divide this segment in half. It is solved in the same way as problem 4 (see Fig. 5).
Task 6. Through a given point draw a line perpendicular to the given line.
Solution. There are two possible cases:
1) given point O lies on a given straight line a (Fig. 6).
From point O we draw a circle of arbitrary radius intersecting line a at points A and B. From points A and B we draw circles with the same radius. Let O 1 be the point of their intersection, different from O. We obtain OO 1 ⊥ AB. In fact, points O and O 1 are equidistant from the ends of the segment AB and, therefore, lie on the perpendicular bisector to this segment.