Image of spatial figures. Visual Guide (2019). Examples of constructing axonometric projections of geometric shapes

Lecture on the topic "The subject of stereometry"

Geometry is the science that studies the properties geometric shapes.

The school geometry course is divided intotwo sections:planimetry andstereometry.

Planimetry - a branch of geometry that studies the properties of geometric figures on a plane.

We studied planimetry in grades 7-9.

This year we are starting to study the second section of geometry - solid geometry

Stereometry is a branch of geometry that studies the properties of geometric shapes in space.

The word "stereometry" comes from the Greek words "stereos" volumetric, spatial and "metrio" to measure.

Solid geometry considers mathematical models of those material objects that architects, designers, builders and other specialists deal with.

In addition, the school course of stereometry serves as the basis for drawing and descriptive geometry - the most important disciplines of any technical university.

The main figures of stereometry

So, stereometry studies the properties of geometric shapes in space.

Geometric shapes in space.

are called bodies.

In stereometry, we will study the properties of geometric bodies, calculate their areas and volumes.

When studying spatial figures, their image in the drawing is used.

The image of a spatial figure is its projection onto a particular plane. The same figure admits different representations.

Usually, one of them is chosen, which is most convenient for studying its properties.

On the screen you see polyhedra - a cube, a parallelepiped and a pyramid, bodies of revolution - a sphere, a cone and a cylinder.

When depicting spatial figures, the invisible parts of these figures are shown with dashed lines.

Where does stereometry begin?

Just like planimetry.

We began to study planimetry with the basic concepts, figures and axioms.

Basic concepts of stereometry

First, it is a point and a line, as in planimetry. And the plane is added.

So, the basic concepts of stereometry are: melancholy, straight line, plane. They are accepted without prejudice.

A new concept for us is the plane.

A plane, like a straight line in planimetry, is infinite. It extends in all directions for an unlimited distance.

The geometric models of a part of the plane are, for example, the surface of a table, boards, etc.

Depict the plane in the form of a parallelogram, or in the form of an arbitrary area.

The notation we will use.

Dots. As before, the points will be denoted by capital Latin lettersA, B, C ….

There are 4 dots on the screen. They are marked with lettersA, B, C and D

Direct. Straight lines are denoted by lowercase Latin lettersa, b, c..., or two capital Latin lettersAB, CD, …

In the second case, we use the notation

two points through which the line passes.

On the screen you see a directa. There are dots on it.MandN.

Straightacan also be labeled asMN.

Planes. Planes are usually denoted by lowercase Greek letters (alpha, beta, gamma, delta, ...)

Planes can also be named after the three points through which the planes pass.

For example, on the screen the plane of blue color denoted as α, it can also be calledABC.

Plane beige colour denoted by β, it can also be denoted asKLN or KLM. Any three points through which the plane passes are taken.

Just as in planimetry, in stereometry we will use the sign for points: (belongs to the plane), and for straight lines the sign is: (lies in the plane).

Crossed out signs mean negation - does not belong to the plane, does not lie in the plane.

In the figure, you can see that two pointsA and Bbelong to the planeα (the plane passes through these points), and the points M, N,Kdo not belong to this plane (the plane does not pass through these points).

Briefly, it is written like this:

Point A belongs to the plane α, pointBbelongs to the plane α.

Dot MNdoes not belong to the plane α, the pointKdoes not belong to the plane α.

In this lesson, we got acquainted with a new section of geometry - stereometry.

We learned that the basic concepts of stereometry are a point, a line, a plane. Remember how points and lines are drawn. We learned how the plane is depicted and denoted.

Let's move on to problem solving.

Task 1.

Given:

points A, B, C and Dnot lying in the same plane.

Specify the planes to which it belongs:

a) straight AB;

b) point F;

c) point C.

Solution.

a) Direct ABlies in two planes:ABC and ABD;

b) Point Fbelongs to the planes:ABC and BCD;

c) Point Cbelongs to three planes:ABC, BCD, ACD.

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Introduction

§one. The role and place of geometric constructions in the school course

§2. Technique for solving problems in stereometry

§3. Fundamentals of the theory of geometric constructions

3.1 General axioms of constructive geometry

3.2 Construction task

§four. Methodology for solving construction problems in stereometry

4.1 Analysis

4.2 Construction

4.3 Proof

4.4 Research

Conclusion

Literature

Introduction

The entire history of geometry and some other branches of mathematics is closely connected with the development of the theory of geometric constructions. The most important axioms of geometry, formulated by the founder of the scientific geometric system Euclid around 300 BC, clearly show what role geometric constructions played in the formation of geometry. “A straight line can be drawn from any point to any point”, “A limited straight line can be continuously extended”, “A circle can be described from any center and any solution” - these postulates of Euclid clearly indicate the basic position of constructive methods in the geometry of the ancients.

Ancient Greek mathematicians considered "truly geometric" only constructions produced only by a compass and a ruler, not recognizing the use of other means for solving constructive problems as "legitimate". At the same time, in accordance with the postulates of Euclid, they considered the ruler as unlimited and one-sided, and the ability to draw circles of any size was attributed to the compass. Problems for building with a compass and a ruler are still considered very interesting today, and for more than a hundred years this has been the traditional material for a school geometry course.

One of the most valuable aspects of such tasks is that they develop search skills for solving practical problems, introduce them to feasible independent research, contribute to the development of specific geometric representations, as well as a more thorough processing of skills and abilities. And this, in turn, enhances the applied and polytechnical orientation of teaching geometry. Construction tasks do not allow a formal approach to them, they are a qualitatively new situation for applying the studied theorems and, thus, make it possible to carry out problematic repetition. Such tasks can be successfully connected with the new ideas of the school geometry course (transformations, vectors).

Geometric constructions can play a serious role in the mathematical preparation of a student. None of the tasks provides as much material for the development of the mathematical initiative and logical skills of the student as geometric construction tasks. These tasks usually do not allow a standard approach to them and formal perception by students. Construction tasks are convenient for consolidating the theoretical knowledge of students in any section of the school geometry course. By solving geometric construction problems, the student acquires many useful drawing skills.

In this term paper the methodology for solving problems for constructions in stereometry will be considered, as well as the role and place of geometric constructions in the school course.

§one. The role and place of geometric constructions in the school course

Construction tasks are tasks in which it is required to build some geometric figure according to predetermined data using a limited set of drawing tools (most often, rulers and compasses).

The role of construction tasks in the school course:

1. It contributes to the development of the imagination of schoolchildren, since even before solving this problem, it is necessary to clearly present the desired image.
2. Develops students' constructive abilities and reinforces relevant drawing skills.
3. Analysis and study of the obtained solution, consideration of the relationship between the data and the desired elements contributes to the development of the logical thinking of schoolchildren, in particular - mental operations: analysis, synthesis, abstraction; awaken their initiative.
4. Contributes to the solid consolidation of the theoretical material of the course.

Thematic planning of material related to geometric constructions assumes the following distribution by stages:

1. Introductory stage (grades 1-4). Here, for the first time, schoolchildren get acquainted with drawing tools - a ruler, a compass, a triangle and solve the simplest tasks for constructing a straight line, segment, circle, angle.
2. Propaedeutic stage (5-6 cells). more attention to geometric constructions prepares students to solve more challenging tasks systematic course. Ruler, compass, protractor, triangle are used. The construction of parallel and perpendicular lines with the help of a square and a ruler is considered; triangle with a ruler, compass and protractor; circle, square, rectangle.
3. Systematic course of geometry (7-11 cells).

7th grade. Here, for the first time, students meet the basic requirement for geometric drawings - all constructions must be carried out only with the help of a compass and ruler. This requirement follows from two postulates of Euclid in the Elements: a) a straight line can be drawn from any point to any point; b) from any center, any solution of a compass can describe a circle. In this case, it becomes necessary to prove that the constructed figure satisfies the requirements of the problem. In the 7th grade, students get acquainted with elementary tasks for constructing, constructing a circle, inscribed and described around a triangle; in addition, students learn the first general method solving construction problems - the method of geometric places (method of intersections).

Class. In the topic "Quadangles" the corresponding tasks are solved for the construction of geometric places by the method; in the topic "Movement" - all types of movement are used to solve problems for the construction; in the topic "Cartesian coordinates on the plane" constructions on the coordinate plane (construction of a straight line, circle, intersection points) are considered.

Class. In the topic "Similar Figures" - construction tasks using homothety and similarity transformation; in the topic " Regular polygons» - tasks for constructing inscribed and circumscribed regular polygons.

(grades 10-11). In stereometry, two types of geometric constructions are considered: a) imaginary constructions based only on the axioms of stereometry (often used in solving constructive problems such as “Prove that it is possible to draw through a point outside the plane ...”; b) constructions on a projection drawing, when other than points are indicated figures of their projection on the projection plane.

The process of solving problems consists of four stages, which students get acquainted with in the 7th grade:

1)analysis;

2)construction (synthesis);

3) proof;

)study.

Not all of these stages from the very beginning must be explicitly present when solving construction problems. In the simplest constructive problems, where the construction algorithm is obvious, it is permissible not to analyze the problem in an explicit form; if the proof follows directly from the construction, it can also be omitted (for example, when constructing in grades 7-8, it is usually either absent or limited to checking the feasibility of each operation and conducting a study to find the number of solutions (if possible)).

§2. Technique for solving problems in stereometry

I. We can single out the following main tasks to be solved in the study of stereometry:

1)development and consolidation of content lines begun in an incomplete high school; generalization of basic mathematical methods to the case of space;

2)study of the basic properties of spatial figures;

3)mastering the skills of depicting spatial figures on a plane based on the properties of parallel projection;

4)development of logical thinking, spatial representations of students in solving problems and proving theorems of the course of stereometry.

In the study of stereometry at school, two main stages can be distinguished:

) Formation of initial ideas about spatial figures (grades 1-9);

) Systematic course of stereometry (grades 10-11).

The systematic course in stereometry, which is taught for approximately 70 hours in the tenth and eleventh grades, provides for the following topics:

1.Axioms of stereometry and their simplest consequences.

2.Parallelism of lines and planes in space.

.Perpendicularity of lines and planes in space.

.Coordinates, vectors, geometric transformations in space.

.Polyhedra.

.Rotation theme.

.Surface area and volume of geometric bodies.

.The image of spatial figures on the plane.

Current textbooks put different substantive accents in the study of stereometry.

Atanasyan's textbook: material of different content questions is often included in one chapter (fusionism). At the same time, there is a frequent repetition of the material, an appeal to already familiar issues. Greater attention than Pogorelov's is given to vectors, movement to coordinates.

Pogorelov's textbook: has a clear logical structure, less attention vectors and geometric transformations. This implicitly carries the danger of obscuring the natural connections between the topics.

Let us highlight some methodological features of the study of stereometry.

1.The course of solid geometry is completely based on the course of planimetry.

most of the tasks of the course are reduced to solving planimetric problems, respectively, all the shortcomings that occurred in the study of planimetry are felt in the study of solid geometry.

Consequently, for a successful study of stereometry, the teacher must constantly return to planimetric material; before studying one or another theorem, it is necessary to repeat the necessary planimetric information.

2. In stereometry, there is a fundamentally different approach to geometric constructions.

If, when studying planimetry, students use drawings that give clear ideas about the object being studied, then in stereometry there are no drawing tools that allow you to depict spatial figures. Here we are not dealing with the object itself, but only with its image.

Each stereometric task is also a task for constructing an image of a figure using the properties of a parallel projection. This requires much more effort from students than they are required to solve planimetric problems.

3. In the course of stereometry, much attention is paid to the logical side of the conclusions being drawn; you have to justify each of your conclusions, clearly establishing the prerequisites.

The program for stereometry assumes a faster pace of material passing than for planimetry. At the same time, much more time is required to solve problems; accordingly, more significant space is occupied by independent work schoolchildren. Careful selection of tasks in the lesson is necessary - include only the most necessary.

5. The course of stereometry is built axiomatically. When studying the axiomatics of stereometry, it is necessary to solve two main methodological problems:

) the axioms of planimetry for space are reformulated (some should be clarified).

Here, in fact, under the guise of an agreement between the teacher and the student, a new axiom is introduced, as it were:

In any plane of space, all the axioms of planimetry are satisfied.

) new specific axioms of space are added, which at the first stages of study are illustrated with the help of models, a stereometric box, drawing, classroom geometry.

II. The formation of spatial representations takes place in several stages and includes:

the ability to present according to the drawing a holistic image of a geometric figure, the relative position of its elements;

the ability to mentally change the position of the figure - look from the other side;

the ability to mentally dismember a figure, to make a new object out of it;

the ability to depict a figure in a drawing, adequately reflecting the existing relationships;

the ability to imagine a figure based on its verbal description, etc.

At stage I, on a visual basis, the prerequisites are formed for creating a holistic image of a figure with highlighting its essential features. At this stage, the teacher should make extensive use of models, real objects of the surrounding world. After that, a drawing is built, which fixes the consideration of the corresponding geometric configuration.

At the end of stage I and stage II, schoolchildren form images of figures and their combinations, which they can imagine under almost unchanged conditions.

The scheme for the formation of spatial representations at stages I and II is as follows:

model drawing representation

At stage II, the role of models is somewhat reduced, because otherwise the development of the abilities of students to mentally imagine the features of the location of the figure and its elements will be inhibited.

When constructing a drawing at these stages, the teacher should not immediately demonstrate finished drawing, but try to perform it gradually with students for the purpose of gradual perception or spatial images.

Stage III: - mastering the ability to operate with images in changed conditions. Schoolchildren first work with the main drawing, which, however, often does not make it possible to see the features of the location of the figure from different positions. Therefore, the drawing, as a rule, must be supported by consideration of the corresponding model. The demonstration is accompanied by specially selected questions.

For example: What figures can be obtained by crossing a tetrahedron of a plane? Show different cases on the model and drawing. Justify the answer.

The scheme for the formation of spatial representations at stage III:

drawing model representation.

Stage IV: Students must construct stereometric objects on their own based on previously formulated ideas. In this case, neither a drawing nor a pre-prepared model is used, but you can only ask the teacher questions to clarify the location of the figure.

Scheme at stage IV: presentation drawing.

Imaginary constructions (Vp) - a formal-logical method of construction in space with the rejection of real constructions with the help of drawing tools, are carried out as if mentally; the figure accompanying them is purely illustrative.

From a mathematical point of view, V.p. are considered as problems of proving the existence of figures defined by certain known conditions. The proof itself consists in reducing the process of constructing figures (or their combinations) to a finite number of basic constructions, which are determined axiomatically. In this case, the solution (proof) may or may not be accompanied by a drawing.

The teacher draws students' attention to a number of difficulties that arise when constructing in space (you cannot build a plane, a polyhedron, etc.). Therefore, it is necessary to agree exactly: what it means to carry out this or that construction.

Based on the axioms of stereometry, we can assume the possibility of the following basic constructions in space:

) A plane can be built if the following elements are specified that determine its position in space:

a) a line and a point not lying on it,

b) two intersecting lines,

c) two parallel lines

d) three points that do not lie on the same line.

) A straight line in space can be constructed as a line of intersection of two planes.

) All planimetric constructions are feasible in space only on some given plane.

) A sphere can be constructed if the position of its center and radius R are given.

The execution of all other constructions is reduced to a finite number of basic ones.

On the projection drawing, points and lines are set together with their projections onto a certain plane, which is called the main one.

Projection drawings allow constructive means to build points and lines of intersection of the figures depicted on it. They have very importance for the development of spatial imagination of schoolchildren.

It is recommended to familiarize schoolchildren in the 10th grade with projection drawings when studying a parallel projection of its properties. Here the teacher leads the students to the conclusion that the figures in the drawing can be set by its projection on the projection plane.

Moreover, if a point or figure coincides with its projection, then given point or the figure lies on the projection plane.

The projection drawing can be illustrated with a parallelepiped model, where the projection plane is the plane of the bottom base, the projection direction is determined by the side ribs, and the projection of the top base is the bottom base.

The main type of stereometric tasks for constructing on a projection drawing are tasks for constructing sections of polyhedra. The school considers two methods for constructing sections:

1)trace method; 2) method interior design

(Sometimes a combination is used.)

In accordance with the method of traces, the trace of the secant plane is first constructed on the projection plane, and then the lines of intersection of the secant plane with the faces of the polyhedron are successively found.

The main disadvantage of this method is that the trace of the cutting plane may be removed from the main part of the drawing, therefore, it is necessary to reduce the drawing, which is undesirable.

The internal design method is based on the correspondence between the points of the section and the base points of the polyhedron. All constructions are inside it, but it is more difficult to explain the logic of construction, and the drawing is cluttered.

§3. Fundamentals of the theory of geometric constructions

1 General axioms of constructive geometry

A figure in geometry is any collection of points (containing at least one point).

We will assume that a certain plane is given in space, which we will call the main plane. We confine ourselves to considering only such figures that belong to this plane.

One figure is called a part of another figure if every point of the first figure belongs to the second figure. So, for example, the parts of a straight line will be: any segment lying on it, a ray lying on this straight line, a point on this straight line, the straight line itself.

The connection of two or more figures is the collection of all points belonging to at least one of these figures.

The intersection or common part of two or more figures is the collection of all points that are common to these figures.

The difference of two figures F and F is the set of all such points of the figure F that do not belong to the figure F.

It may turn out that the intersection (or difference) of two figures does not contain a single point. In this case, we say that the intersection (or, respectively, the difference) of these figures is an empty set of points.

The branch of geometry in which geometric constructions are studied is called constructive geometry. The basic concept of constructive geometry is the concept of constructing a geometric figure.

If it is said about any figure that it is given, then it is naturally understood that it has already been depicted, drawn, i.e. built. Thus, the first basic requirement of constructive geometry is as follows:

1. Each given figure is constructed.

Note that one should not confuse the concepts of "a given figure" and "a figure given (or determined) by such and such given elements of it."

1. If two (or more) figures are constructed, then the connection of these figures is also constructed.

3. If two figures are constructed, then it is possible to establish whether their difference is an empty set or not.

If the difference of two constructed figures is not an empty set, then this difference is constructed.

If two figures are constructed, then it is possible to establish whether their intersection is an empty set or not.

If the intersection of two constructed figures is not empty, then it is constructed.

The next three basic requirements talk about the ability to plot individual points.

It is possible to construct any finite number of common points of the two constructed figures, if such points exist.

You can construct a point that obviously belongs to the constructed figure.

You can construct a point that obviously belongs to the constructed figure.

2 Construction task

A construction task is a sentence that indicates by what data, with what tools, what geometric figure is required to be built (drawn on a plane) so that this figure satisfies certain conditions.

To solve a construction problem with the help of a compass and straightedge means to reduce it to a set of five elementary constructions, which are considered feasible in advance. Let's list them.

If two points A and B are constructed, then a straight line AB connecting them is constructed, as well as a segment AB and any of the rays AB and BA (the ruler axiom).

If point O and segment AB are constructed, then a circle is constructed with center at point O and radius AB, as well as any of the arcs of this circle.

If two lines are constructed, then their intersection point is constructed (if it exists).

If a line and a circle are constructed, then any of their intersection points (if it exists) is constructed.

If two circles are constructed, then any of their intersection points (if it exists) is constructed.

To solve a construction problem means to find all its solutions.

The last definition needs some clarification.

Figures that satisfy the conditions of the problem can differ both in shape and size, as well as in position on the plane. Differences in position on the plane are taken into account or not taken into account depending on the formulation of the construction problem itself, namely, depending on whether the condition of the problem provides or does not provide for a certain position of the desired figure relative to any given figures. Let's explain this with examples.

Consider the following the simplest task: construct a triangle given three sides and an angle between them. The exact meaning of this problem is as follows: construct a triangle so that its two sides are respectively equal to two given segments, and the angle between them is equal to the given angle. Here, the desired figure (triangle) is connected with these figures (two segments and an angle) only by equality relations, while the location of the desired triangle relative to these figures is indifferent. In this case, it is easy to construct a triangle that satisfies the condition of the problem. All triangles equal to this triangle also satisfy the condition of the problem. However, there is no point in considering these triangles as different solutions to this problem, because they differ from one another only in their position on the plane, about which nothing is said in the condition of the problem. Therefore, we assume that the problem has a unique solution.

So, if the condition of the problem does not provide for a certain location of the desired figure relative to these figures, then we agree to search only for all unequal figures that satisfy the condition of the problem. We can say that problems of this kind are solved "up to equality." This means that the problem is considered solved if:

) A certain number of unequal figures F1, F2, ... Fn are constructed, satisfying the conditions of the problem

) it is proved that any figure that satisfies the conditions of the problem is equal to one of these figures. It is assumed that the problem has n different solutions.

If the condition of the problem provides for a certain location of the desired figure relative to any given figure, then the complete solution consists in constructing all the figures that satisfy the condition of the problem (if such figures exist) in a finite number.

§four. Methodology for solving construction problems in stereometry

The essence of solving the construction problem lies in the fact that it is required to construct a certain figure with the tools indicated in advance, if a certain figure is given and certain relationships between the elements of the desired figure and the elements of this figure are indicated.

Each figure that satisfies the conditions of the problem is called a solution to this problem.

To find a solution to the problem of construction means to reduce it to a finite number of basic constructions, that is, to indicate a finite sequence of basic constructions, after which, the desired figure will already be considered built due to the accepted axioms of constructive geometry.

One of the main problems of the teaching methodology for solving construction problems is the method of introducing and studying the stages of solving constructive problems. Even in the IV century. BC e. ancient Greek geometers developed general scheme solving construction problems, which we use now. The process of solving the problem is divided into 4 stages: analysis, construction, proof and research. Let's consider each stage in more detail on the task.

Given points A (A ), B (B ), C (C ) and D (D ). Construct a plane passing through point D (D ), parallel to plane ABC.

4.1 Analysis

Analysis is milestone problem solving, which we understand as finding a way to solve a construction problem. At this stage, such dependencies between these figures and the desired figure should be noticed, which would allow us to construct this desired figure in the future (if we know how to build the desired figure, then no analysis is needed anymore).

To facilitate the search for connections between the desired figure and these figures, it usually turns out to be advantageous to have an auxiliary drawing in front of your eyes, a sketch drawing depicting the data and the desired figures approximately in the arrangement provided for by the condition of the problem. The drawing can be done by hand, by eye - this is a draft drawing, which should be formed when the problem has already been solved.

In the auxiliary drawing, these elements and the most important elements of interest should be highlighted. In practice, it is often more convenient to start building an auxiliary drawing not from a given figure, but from an approximate image of the original figure, attaching data to it so that they are in the relationships indicated in the problem statement.

If the auxiliary drawing does not suggest a way to construct the desired figure, then they try to find some part of the desired figure or, in general, some figure that can be built, and which can then be used to construct the desired figure.

The following points are taken into account:

) if in the auxiliary drawing it is not possible to directly notice the connections between the data and the desired elements necessary for solving, then it is advisable to introduce auxiliary figures into the drawing: connect the existing points with straight lines, mark the intersection points of the existing lines, continue some segments, etc. Sometimes it is useful draw parallels or perpendiculars to existing lines;

) if, according to the condition of the problem, the sum or difference of the segments or angles is given, then these quantities should be entered into the drawing, that is, they should be depicted on the sketch drawing, if they are not already on it;

) in the process of analysis, it is useful to recall theorems and previously solved problems in which there are dependencies between the elements that are mentioned in the condition of the problem under consideration.

Appendix 3 provides an analysis of the construction task: Construct a triangle given the base, the smaller angle at the base, and the difference between the other two sides .

From this example it can be seen that when finding a solution to the problem of construction, as well as for arithmetic problems, the analytic-synthetic method is used. Following from the question of the problem, we take into account what elements are known to us, and, conversely, we combine the initial data in such a way as to build the desired figure.

Stage name analysis does not mean that only analytical method, just as in the proof, which is sometimes called synthesis , the synthetic method of reasoning is not always applied. When analyzing a problem, when finding ways to solve it, analysis and synthesis are in constant interaction, complement and test each other.

Let's return to our problem and analyze it.

1.Let's find the point S 1, in which the AAs lying in the projecting plane intersect B straight AB and A B , point S 2 where lines AC and A intersect C , and point S 3, where lines AD and A intersect D .

2.In the AS plane 1S 3 construct a line passing through the point D, parallel to the line AS1 and in the plane AS 2S 3 passing through point D, parallel to line AS 2.

.Through the lines obtained, we construct the required plane.

2 Construction

The second stage of solving construction problems consists of two parts:

) enumeration in a certain order of all elementary constructions that need to be performed, according to the analysis, to solve the problem;

) direct execution these constructions on the drawing using drawing tools. Indeed, to solve a problem with the help of certain tools means to indicate a finite set of elementary constructions that are acceptable for given tools, the execution of which in a certain sequence allows us to answer the question of the problem.

This stage is introduced when solving the very first construction problem, which is usually the problem of constructing a segment equal to a given one on a given ray with an end at the beginning of this ray. In the conversation accompanying the introduction of the stage, it should be noted what the solution of any construction problem consists of and indicate that the implementation of this stage consists precisely in listing a finite number of operations for constructing the desired figure.

Let's return to our problem and consider its construction.

Building:

1.AB∩A B =S 1

2.AC∩A C = S 2

3.AD∩A D =S 3

4.D.S. 4║AS 1

5.D.S. 5║AS 2

6.D.S. 4S 5

4.3 Proof

After the figure is built, it is necessary to establish whether it satisfies the conditions of the problem, that is, to show that the figure obtained from these elements by a certain construction satisfies all the conditions of the problem. Hence, the proof essentially depends on the method of construction. The same problem can be solved different ways, depending on the construction plan outlined in the analysis, and therefore, the proof in each case will be different. The proof is a part of the solution of the problem, in its logical content, the opposite of the analysis. If it is established in analysis that any figure that satisfies the conditions set can be found in such and such a way, then in this third part of the solution the opposite is proved. This is the reverse position general view can be formulated as follows: if a certain figure is obtained from the given elements by such and such a construction, then it really satisfies the conditions set.

When solving the simplest problems, when all the conditions of the problem are directly reflected in the construction plan, there is no need to prove that the figure obtained from these elements by such a construction is the desired one. For example: Construct a triangle given two sides and an angle between them . Here the proof is reduced to a simple check whether the sides are the same as given, and whether the constructed angle is equal to the given one. In such problems, the proof is redundant, because the correctness of the solution is ensured by the correspondence of the construction to the analysis and the given conditions of the problem.

The proof does not just depend on analysis and construction, there is an interrelation and interdependence between them. The construction is carried out according to the plan drawn up during the analysis. There are several such plans. Construction and proof are a kind of criterion for the correctness and rationality of the plan. If the plan is not feasible with the available tools, or if the construction turns out to be irrational, we are forced to look for a new solution plan. In a similar way, both proof and research influence the analysis, often predetermining the choice of the solution plan.

Although the proof in solving construction problems is carried out similarly to the proof of theorems, using axioms, theorems and properties of geometric figures, there is also some difference between them. When proving theorems, in most cases it is easy to single out a condition and a conclusion. When solving construction problems, it is already more difficult to find data on the basis of which it is possible to prove that the constructed figure is the desired one. Therefore, when solving constructive problems in a class, it is sometimes expedient to specifically highlight what is given and what needs to be proved. For example, when solving a problem: Construct a rhombus along its two diagonals we invite the student to write down what is given (the diagonals are mutually perpendicular and, intersecting, are divided in half) and what needs to be proved (the sides are equal). In turn, when solving problems at home and in control work it is possible not to require the execution of evidence with the allocation of a separate condition and conclusion. There is no need to require a special proof in problems where the correctness of the solution is obvious.

Let us return to our problem and consider its proof.

Proof: the lines DS4 and DS5 pass through the point D and are parallel to the plane ABC by construction.

4 Research

When constructing, one usually confines oneself to finding one of some solutions, and it is assumed that all steps of the construction are actually feasible. For a complete solution of the problem, it is still necessary to clarify the following questions: 1) is it always possible (that is, for any choice of data) to perform the construction in the chosen way; 2) is it possible and how to construct the desired figure if the chosen method cannot be applied; 3) how many solutions does the problem have for each possible choice data? Consideration of all these questions constitutes the content of the study.

Thus, the study aims to establish the solvability conditions and determine the number of solutions. It is not uncommon for schoolchildren and even teachers to conduct research randomly choosing one or another case, and it is not clear why these and not some other cases are considered. It also remains unclear whether all possible cases have been considered. In practice, in most cases, it is possible to achieve the necessary completeness of the study if this study is carried out in the course of construction, which is the most accessible and expedient way. The essence of this technique is to enumerate successively all the steps that make up the construction, and for each step to establish whether the construction indicated at this step is always feasible, and if feasible, whether it is unique.

Consider the study of our problem.

Study: given task has a solution, and moreover, only one, since only one plane can be drawn parallel to a given plane and a straight line not lying on it.

Tasks

Task number 1.

Given: SABCD pyramid, PSB, KSC, MSA.

Construct: Section of SABCD by MKR plane

Solution: Since the points M, K and P lie on the side edges of the pyramid, it is immediately possible to construct two sides of the MP section

R To

M AT FROM

O H

BUT D

and RK. After that, you need to find the point H of intersection of the secant plane with the edge SD.

Now in the plane (BSD) we have two points of the secant plane: O1 and P. Hence, the point H sought on the edge SD will be the intersection point of the edge SD and the line PO1.

The point is found, the last two sides of the section MN and NK are easy to construct. Thus, MKRN is the desired cross section.

Task #2

Given: Construct a section of the prism ABCDA1B1C1D1 - prism, PAA1, QBB1, RCC1

Find: section ABCDA1B1C1D1 by a plane passing through points P, Q, R

Solution: Let's construct a trace of the secant plane on the plane of the lower base of the prism. Consider the face AA1B1B. This face contains the section points P and Q. Draw the line PQ. We continue the line PQ, which belongs to the section, to the intersection with the line AB. We obtain the point S1 belonging to the trace. Similarly, we obtain the point S2 by the intersection of the lines QR and BC. Straight line S1S2 - trace of the secant plane on the plane of the lower base of the prism. The line S1S2 intersects the side AD at the point U, the side CD at the point T. Let's connect the points P and U, since they lie in the same plane of the face AA1D1D. Similarly, we get TU and RT. PQRTU - desired section.

Task #3

Given: Construct a section of the prism ABCDA1B1C1D1 - prism, MA1B1, NAD, PDC

Find: Section ABCDA1B1C1D1 by a plane passing through points M, N, P

Solution: Points N and P lie in the plane of the section and in the plane of the lower base of the parallelepiped. Let's construct a line passing through these points. This line is the trace of the secant plane on the plane of the base of the parallelepiped. Let us continue the line on which the side AB of the parallelepiped lies. The lines AB and NP intersect at some point S. This point belongs to the section plane. Since the point M also belongs to the section plane and intersects the line AA1 at some point X. The points X and N lie in the same plane of the face AA1D1D, we connect them and get the line XN. Since the planes of the faces of the parallelepiped are parallel, it is possible to draw a line through the point M in the face A1B1C1D1 parallel to the line NP. This line will intersect the B1C1 side at the point Y. Similarly, we draw the line YZ, parallel to the line XN. We connect Z with P and get the required section - MYZPNX.

Also, tasks for constructing sections can be solved in the program "Live Geometry".

Given: ABCDA1B1C1D1-parallepiped, P CC1D1D, Q AA1D1D, R BB1. Construct: section ABCDA1B1C1D1 by plane PQR.

Solution:

Given: Points P, Q and R are taken on the surface of the parallelepiped ABCDA1B1C1D1 as follows: point P lies on face CC1D1D, point Q lies on edge B1C1, and point R lies on edge AA1.

Construct: section of a parallelepiped by a plane (PQR).

Solution:

Given: Points P and S are taken on the edges A1B1 and DD1 of the parallelepiped ABCDA1B1C1D1, respectively, and points Q and R are taken on the faces DD1C1C and AA1D1D, respectively.

Construct: a section of a parallelepiped by a plane passing through the point S parallel to the plane PQR.

Solution:

3.Independent problem solving

Each student receives a card with a task. On the same sheet, the construction of the section and the description of this construction are performed. Checking assignments can be done at the lesson in the TMC “Mathematics, grades 5-11. Practicum»

Task 1-7: construct a section passing through the points M, K, L.

Task 8: construct a section passing through point P and line KL.

Task 9: construct a section passing through the point K and the line PQ.

Exercise 1

Task 2

Task 3

Task 4 Task 5

Task 6

Task 7

Task 8 Task 9

Solutions of tasks in the EMC “Mathematics, grades 5-11. Practicum»

Conclusion

A systematic study of geometric constructions is necessary in the school course, since in the process of studying problems they concentrate knowledge from other areas of mathematics, develop skills in practical graphics, form search skills for solving practical problems, introduce them to feasible independent research, contribute to the development of specific geometric representations, and also to a more thorough processing of skills and abilities.

In this term paper, the role and place of constructions in the school course were considered, as well as the methodology for solving problems for construction in stereometry and basic geometric constructions was considered.

Literature

stereometry

1. Aleksandrov, I.I. Collection of geometric construction problems with solutions / II Aleksandrov. - M.: Uchpedgiz, 1954.

2. Argunov, B.I. Elementary geometry: textbook. allowance for ped. in-tov / B.I. Argunov, M.B. Balk. - M.: Enlightenment, 1966.

3. Konovalova, V.S. Solving construction problems in the course of geometry as a means of developing logical thinking / V.S. Konovalova, Z.V. Shilova // Knowledge of the processes of teaching physics: a collection of articles. Issue 9. - Kirov: Publishing house of VyatGGU, 2008. - S. 59-69.

4.Misyurkeev, I.V. Geometric constructions. Manual for teachers / I.V.Misyurkeev. - M: Uchpedgiz, 1950.

5. Ponarin, Ya.P. Elementary geometry: In 2 volumes - V.2: Stereometry, transformations of space / Ya.P. Ponarin - M.: MTsNMO, 2006.

6. Prasolov, V.V. Problems in stereometry. Part 1 / V.V. Prasolov. - M.: Nauka, 1991.

7. Sarantsev, G.I. Teaching mathematical proofs and rebuttals at school / G.I. Sarantsev. - M.: VLADOS, 2005.

8. Sharygin, I.F. Problems in Geometry (Stereometry) / I.F. Sharygin. - M.: Nauka, 2009.

The features of constructing images of figures, primarily flat ones, are considered, as well as tasks for constructing images on images.

When studying the question of the representation of figures in stereometry, we will focus on the representation of plane figures. And this is understandable, since looking at a real physical object (a house, a dice, a book, etc.), we see a surface, in many cases consisting of flat parts (Fig. 201 - 203). In the drawings and technical drawings first of all, they try to depict the surface of an object, and our life experience makes it possible to see the object as a whole behind the details of the surface.

Since the main geometric figure is a triangle, let's find out which figure can be an image of a triangle. And then we will be able to discuss the question of the image of other polygons known from planimetry. In addition, we will talk about the image of the simplest spatial figures.

We take parallel projection as the geometric basis of the image. First of all, it is necessary to clarify the content of the concept of “image”, because it is enough to understand the image of a figure as its direct parallel projection.

uncomfortable. figure large sizes it is simply impossible to project onto a sheet of paper - in order for the image to fit, the parallel projection of the figure must be proportionally reduced (or increased in other situations).

An image of a spatial figure is a figure similar to a parallel projection of a given figure onto a certain plane.

This definition needs to be supplemented. It is clear that the image should contain as much information about the figure as possible. It is unlikely that the parallel projection of the cube in fig. 204, a) fully reflects the features of this figure. That's why

on the image of polyhedra, their vertices and edges are depicted, visible and invisible. As already noted, invisible lines are depicted with dashed lines. Thus, the image of the cube in Fig. 204b)

gives more complete information about the cube. On the image of the spatial fi-

gurus also highlight images of its important elements (for example, diagonals, sections, etc.).

Note that neither the projection plane nor the projection direction is fixed in the definition. This is understandable, since a position convenient for consideration can be chosen arbitrarily.

Now let's answer the question: what figure can be the image of a triangle? The case when the triangle lies in the projecting plane

ty, we will not consider. In this case, it is projected onto a segment (Fig. 205).

Since the parallel projection of the triangle is a triangle (except for the case noted above), then the image of the triangle must also be

square. At the same time, the question arises: “And what triangle can be considered the image of this triangle?” As is known, with a parallel project

rovaniya change the length of the segments, measures of angles. It is clear that the parallel projection isosceles triangle is, generally speaking, a scalene triangle, the projection of an obtuse triangle can be an acute one, etc.

Carrying out simple experiments with cardboard models of triangles when receiving their shadow from the Sun or from a distant lamp shows that the shape of parallel projections of a triangle can be different. Moreover, one can make sure that due to the appropriate placement of the model, a triangle of a given shape can be obtained as a projection. Thus, considering the different shadows of one triangle, we can come to the following conclusion.

The image of this triangle can be an arbitrary triangle.

The mathematical substantiation of this fact will be done later. Using it, one can draw certain conclusions regarding the image of some four

rectangles. From the properties of parallel design

it follows that the image of a parallelogram is an arbitrary

parallelogram. Indeed, a diagonal parallelogram is divided into two equal triangles (Fig. 206, a). Picture-

An arbitrary triangle A 1 B 1 D 1 can be an expression of the triangle ABD. Dostro-

Yves triangle A 1 B 1 D 1 to parallelogram-

ma (Fig. 206, b), which is uniquely determined by this triangle, we get the following conclusion.

The images of this parallelogram can be an arbitrary parallelogram.

With regard to trapezoids, a similar conclusion cannot be drawn in the images, since the ratio of the lengths of the parallel bases must be preserved in parallel design. If, for example, one of the bases is half the second, then this ratio should be preserved in the image. Although, of course, the image of a trapezoid should be a trapezoid (but not arbitrary!).

As for the image of other polygons, it is possible to choose three of their points that do not lie on one straight line (for example, three vertices). These points define a triangle, which can be represented by an arbitrary triangle. Further, using the properties of parallel projection (they are also properties of images), in some cases it is possible to build an image of the entire polygon.

Having learned to depict some flat figures placed in space, we can begin to depict the simplest spatial figures.

Images of a rectangular parallelepiped, a cube are no different from images of an arbitrary parallelepiped, since arbitrary parallelograms can be images of squares and rectangles. Most often, the cube is depicted as it is done in Fig. 207a). On fig. 207, b)–d) images of a cube are also given. However, unlike Fig. 207, a), it is difficult to get an idea about the properties of the cube from these images. On fig. 207, b), c) the images are simple and correct, that is, they are made according to the laws of parallel design. However, they are not visual. This does not mean that in some cases we will not need each of the above images.

Let us consider in more detail the construction of the image of a parallelepiped. In §7, the parallelepiped was considered as a polyhedron whose faces are six parallelograms. In §8, an approach to constructing figures from segments was considered. Let's use

them. In a given plane α, we construct a parallelogram ABCD and through all its vertices we draw parallel lines intersecting the plane α (Fig. 208). On these lines, on one side of the plane α, we lay out the segments AA 1, BB 1, CC 1, DD 1 of the same length. It is easy to prove that the points A 1 , B 1 , C 1 , D 1 lie in the same plane and are the vertices of the parallelogram A 1 B 1 C 1 D 1 . Action-

Therefore, since AA 1 D 1 D, ABCD and BB 1 C 1 C are parallelograms, then A 1 D 1 ||AD ,AD ||BC, BC ||B 1 C 1 and, according to the criterion of parallelism of lines (Theorem 2 § 8) ,A 1 D 1 ||B 1 C 1 . This, in particular, gives us the opportunity to assert that the points A 1 , B 1 , C 1 , D 1 lie in the same plane.

Similarly, we have that A 1 B 1 ||D 1 C 1, that is, the quadrilateral A 1 B 1 C 1 D 1 is a parallelogram.

The set of all points of the segments connecting the points of the parallelograms ABCD and A 1 B 1 C 1 D 1 form a figure that is parallelepiped(Fig. 209). It is clear that when constructing parallelepipeds, one can get by with parallel segments connecting the corresponding points of the parallelograms. The image is made, as in Fig. 208, only taking into account the fact that the box is "filled" with points, and some lines are invisible to the observer. As in drawing, they are depicted with a dashed line. Designate a parallelepiped by its vertices:

ABCDA1 B 1 C 1 D 1 .

Two faces of a parallelepiped that have a common edge are called adjacent, and not having a common edge, - opposite. Two vertices that do not belong to the same face are called opposite. A line segment connecting opposite vertices is called parallel diagonal

The image of pyramids, in particular tetrahedra, was considered in § 8 in connection with their construction from segments.

! Consideration of images of flat and spatial

figures allows formulate requirements for images:

1) the image must be correct, that is, satisfy certain rules;

2) the image should be visual;

3) the image should be easy to follow.

The correctness of the image is ensured by observing the rules for constructing parallel projections. Clarity and simplicity are ensured by the choice of the design direction, that is, the "angle of view" on the figure and the location of the projection plane. Thus, the images of the SABC tetrahedron in Fig. 210, a), b) cannot be considered successful. In the first case, parallel projection onto the face plane ABC is used, and in the second case, the direction of projection is determined by the straight line AB. In both these cases, the volume of the figure is lost. As a rule, use the third image (Fig. 210, c). It is a flat quadrilateral ABCS with diagonals AC and SB. The invisible edge AC is shown by a dashed line.

An important means of ensuring the clarity of the image is the image of the elements of the figure (medians, bisectors, midlines, diagonals, etc.), as well as the simplest sections.

The construction of images of various figures is an integral part of solving stereometry problems.

Often, when solving problems, it is necessary to perform certain constructions on the image (draw a median, indicate the center of the inscribed circle, construct a section, etc.). These constructions are usually performed according to the properties of parallel design.

Example 1. On an arbitrary image of a right-angled isosceles triangle ABC (C \u003d 90 °), build an image: 1) center O of the circumscribed circle; 2) an inscribed square, two sides of which lie on the legs

triangle, and one of the vertices is on the hypotenuse BA.

 Let the image of a right-angled isosceles triangle ABC (Fig. 211, a) be a triangle A 1 B 1 C 1 (Fig. 211, b).

1) The center of a circle circumscribed about a right triangle is the midpoint of the hypotenuse. Therefore, its image is the middle of the image of the hypotenuse.

Construction. We divide the segment A 1 B 1 in half, the division point O 1 is the desired one (Fig. 211, c).

2) If from the middle O of the hypotenuse AB we draw perpendiculars to the legs (see Fig. 211, a), then we get a square that satisfies the condition of the assignment. The perpendiculars drawn are parallel to the legs. This is what we will use to build the desired image.

Construction. From point O 1 we draw segments O 1 E 1 and O 1 F 1, parallel to C 1 B 1 and C 1 A 1, respectively (Fig. 211, d). QuadrilateralC 1 E 1 O 1 F 1 is desired.

Example 2. On the image of a cube, construct its section by a plane passing through the midpoints of three parallel edges.

 In fig. 212 the midpoints of the ribs AA 1, BB 1, SS 1, DD 1 of the cube ABCDA 1 B 1 C 1 D 1 are denoted respectively by A 2, B 2, C 2, D 2. The images of these points lie at the midpoints of the images of the corresponding segments (why?). Let the cutting plane pass through the points A 2 , B 2 , D 2 . Since all the faces of the cube are squares, then the segment A 2 B 2 passing through the midpoints of the opposite

Then the line BE sets the right direction for the project

sides of the square AA 1 B 1 B, equal to side square AB (or the edge of a cube) and is parallel to this side.

Similarly, D 2 C 2 ||DC and D 2 C 2 =DC. Since AB ||DC , then, in accordance with the transitivity of the parallelism relation, A 2 B 2 ||D 2 C 2 . A single plane passes through parallel lines A 2 B 2 , D 2 C 2 . Points A 2, B 2, D 2 lie in this plane, therefore this plane is the desired secant. The cutting plane intersects the faces of the cube in equal segments A 2 B 2, B 2 C 2, C 2 D 2 and D 2 A 2. Therefore, the quadrilateral A 2 B 2 C 2 D 2, which is the desired section, has the shape of a rhombus. It is easy to see that the diagonals B 2 D 2 and A 2 C 2 of this rhombus are equal to each other. That is, quadrilateral A 2 B 2 C 2 D 2 - square. We not only built the section, but also established its shape.

Let us consider the substantiation of the above conclusions regarding the image of the main flat figures.

Theorem 1 (on the image of a triangle).

Any triangle can be an image of a given triangle.

Let triangle ABC be given. Take an arbitrary triangle KMN. It can be an image of the triangle ABC if there is a projection plane and a projection direction such that the parallel projection of the triangle ABC is similar to the triangle KMN.

We choose the projection plane α so that it intersects the plane of the triangle ABC along the straight line AC (Fig. 213). We need to choose the direction of projection so that the projection of the triangle ABC on the plane α is a triangle similar to the triangle KMN. To do this, we construct a triangle CAE in the plane α, similar to the triangle KMN with a coefficient sub-

biya MK AC

roving. Since the triangle CAE is a parallel projection of the triangle ABC, and the triangles CAE and KMN are similar, then the triangle KMN is an image of a triangular

ka ABC.

! This theorem opens wide opportunities to select images of this triangle, although, of course, you should not use images with properties that the original does not have. For example, it is inappropriate to depict an arbitrary triangle as a rectangular one.

Turning to images of other polygons, we note that, as a rule, theorems similar to Theorem 1 do not hold for them, although some of their properties are preserved in the image. First of all, we will talk about the parallelism of the sides (why?). In this connection, we present another important theorem.

Theorem 2 (on the image of a parallelogram).

Any parallelogram can be an image of a given parallelogram.

This theorem can be proved by dividing parallelograms by diagonals into triangles and using Theorem 1 (see Fig.

rice. 206, a, b)

We have already met situations where planimetric facts have analogs in space. And such cases will continue to occur. The simplest spatial figure - the tetrahedron - corresponds to a triangle on the plane. By Theorem 1, any triangle can be an image of a given triangle. On the other hand, the tetrahedron is projected into a quadrilateral, which, after drawing diagonals in it, becomes the image of the tetrahedron. The question arises: can an arbitrary quadrilateral be an image of a given tetrahedron? An affirmative answer to it is given by the theorem of the German mathematicians Polke K. (1810–1877) and Schwarz G. (1843–1921). Based on it, you can build an image of polyhedra. To do this, you need to select four vertices that do not lie in the same plane. They are the vertices of some tetrahedron. Then set the image of these points in an arbitrary way. And even then, complete the image of the entire figure, using the design properties.

Example 3. Construct an image of a regular hexagon.

 Consider a regular hexagon ABCDEF (Fig. 214, a). It has properties that should be preserved in its images. The sides of the hexagon are pairwise parallel (AB ||ED, BC ||EF, CD ||AF ). It has a center of symmetry O, and the segments connecting the point O with the vertices of the hexagon are equal to each other and equal to its side. Now it is easy to see that it is enough to construct an image of a parallelogram (even a rhombus) ABCO, in order to then complete the image of the entire hexagon to it.

Let the parallelogram A 1 B 1 C 1 O 1 be the image of the parallelogram ABCO (it can be an arbitrary parallelogram!). Extending 1 O 1 and C 1 O 1 beyond the point O 1 so that O 1 D 1 \u003d A 1 O 1, O 1 F 1 \u003d C 1 O 1, we construct a parallelogram F 1 O 1 D 1 E 1 (Fig. 214, b). In essence, a parallelogram has been built, centrally symmetrical to the parallelogram A 1 B 1 C 1 O 1 with respect to its top O 1 . By connecting the points A 1 and F 1, C 1 and D 1, we get the image of a regular hexagon (Fig. 214, c).

 Control questions

1. Which of the figures in Fig. 215, a)-d) is not an image of a square?

2. Which of the figures in Fig. 216, a)-d) is not an image of a cube?

3. Which of the Fig. 217, a) - d) the image of the cube is not correct?

4. Which of the Fig. 218, a) - d) the image of the tetrahedron is not correct?

5. Is the parallel projection of the figure its image?

6. Can a right triangle be considered an image of an isosceles triangle?

7. Is it true that the image of the midline of a triangle is middle line his pictures?

8. Can a parallelogram be an image of a trapezoid?

9. Can a triangle represent a tetrahedron?

10. Is it possible to draw a tetrahedron in such a way that exactly one of its faces is invisible?

11. What is the smallest number of cube edges that can be visible in the image? And the greatest?

12. What figure is the image: a) a segment; b) a triangle; c) trapezoid; d) parallelogram; e) n-gon?

Graphic exercises

1. Set which faces of the tetrahedron ABCD shown in fig. 219, belong to the points P , K, M ?

2. Which pairs of dots X ,Y, Z, T indicated on the image of the tetrahedron in fig. 220 do not lie on the same face?

3. What figure is a section of a cube by a plane passing through the points M, N, P, indicated in fig. 221, a)-d)?

174°. An image of an isosceles triangle in the form of a scalene triangle is given. Build an image on this image:

1) bisectors of the angle at the vertex;

2) a perpendicular to the base, drawn through the middle of the side; 3) a rhombus, two adjacent sides of which coincide with the side

the sides of the triangle.

175. On the image of an isosceles right-angled triangle, construct an image of a square lying in the plane of the triangle, if the side of the square is:

1°) leg of a given triangle; 2) the hypotenuse of the given triangle.

176. On an arbitrary image of an equilateral triangle ABC, build an image:

1°) points of intersection of the altitudes of the triangle; 2°) of a "circumscribed" rectangle, one of whose sides

coincides with some side of the triangle, and the other contains the opposite vertex; 3) bisectors outer corner triangle.

177. An image of a triangle and its two heights is given. Construct an image of the center of the circle circumscribed about this triangle.

178. In the image of a right triangle, one of sharp corners which is equal to 60 °, build an image: 1) the bisector of this angle; 2) the height drawn to the hypotenuse;

3) the center of the inscribed circle.

179°. Construct a picture of a rhombus and its height drawn from the vertex of an angle whose magnitude is 120°.

180. Build an image of a square, having an image of the point of intersection of its diagonals and two:

1°) neighboring vertices; 2*) opposite vertices. 181. On an arbitrary image of an isosceles trapezoid, the side of which is equal to the smaller base, build

image:

1°) axes of symmetry of the trapezium; 2) an inscribed rectangle, two vertices of which lie

reaps on a larger base and one of the sides coincides with a smaller base; 3) the center of the circle touching the sides and less

the base of the trapezoid.

182. An image of an isosceles trapezium is given, the angles at the base of which are equal to 45 °. Build the image:

1) the center of a circle circumscribed about a trapezoid;

2*) the center of the circle touching the smaller base and the sides.

183. An image of a circle and one of its diameters is given. Construct an image of the circle's radii perpendicular to this diameter.

184. An image of a cube ABCDA 1 B 1 C 1 D 1 is given.

1°) Construct a line of intersection of the planes DA 1 C 1 and B 1 D 1 D. 2) Find the length of the segment of this line contained in the cube if the edge of the cube is equal to a.

3) Construct a section of the cube by a plane passing through the centers of three pairwise adjacent faces.

185. An image of a tetrahedron ABCD is given, points K, M and P are the midpoints of DC, AD and BD, respectively.

1°) Construct the line of intersection of the ACP and BMK planes. 2) Find the length of the segment of this line contained in the tetrahedron if the lengths of all its edges are equal.

3) Construct a section of the tetrahedron by a plane passing through the intersection points of the medians of its three faces.

186. Construct a section of the tetrahedron SABC by a plane passing through:

1°) the midpoints of the ribs SA, SC and BC;

2) point M on AS (AM :AS = 1:2), point N on SC (CN :NS = 1:2)

and point P on BC (CP :PB = 1:2);

3) the midpoints of the edges AS, AB and the center of the face SBC ; 4*) face centers ASB, ABC and BSC.

187. Construct a section of the cube ABCDA 1 B 1 C 1 D 1 by a plane passing through:

1) edge CD and face center AA 1 B 1 B ;

2) the diagonal A 1 D and the center of the edge BCC 1 B 1 ;

3*) midpoints of edges AD , CD and point B ;

4*) the centers of the faces CDD 1 C 1 , CBB 1 C 1 and point A.

Exercises to repeat

188. Two parallel lines are intersected by a third line. One of the eight formed angles is equal to 50°. What is each of the other angles?

189. CubeABCDА 1 В 1 С 1 D 1 is given.

1) Select all edges parallel to edge AA 1 .

2) Prove that the edge DC is parallel to the intersection of planes ABC 1 and A 1 B 1 D .

4) Let a be an arbitrary segment in the face of a cube. Construct a segment parallel to segment a in a non-adjacent face of the cube.

Any parallelogram can be an image of a given parallelogram.

• To teach how to apply the acquired knowledge in practice, according to the model, algorithm, with a hint.
• To consolidate the skills of constructing sections using the axioms of stereometry.
• Develop students' spatial thinking.

During the classes.

I. Organizational part.

II. Analysis of homework.

Homework was divided into three levels of difficulty.

Task 1 and 2 - first level

Task 3 and 4 - second level

Task 5 and 6 - third level

Task 1. ABSA 1 C 1 – triangular prism, point F - the middle of the rib AB , dot O lies on the continuation of the rib Sun so FROM located between AT and O . Construct a section of a prism by a plane in 1 FO .

Task 2. Dot O - the middle of the rib DD 1 Cuba ABCDA 1 B 1 C 1 D 1. Plot the intersection points of the lines A 1 O and C1O with base plane ABCD and calculate the distance between them if the length of the edge of the cube is 2 cm.

Problem 3. Given a triangular pyramid SABC points R and R lie on the ribs SA and Sun, dot F lies on the continuation of the rib AU so that point FROM lies between the points BUT and F. Construct a section of the pyramid by a plane PRF

Task 4. SABCD- quadrangular pyramid. Dot R lies on the edge SCD, a dot F along the edge DC so that point D lies between F and FROM. PFB.

Task 5. DABC- a regular tetrahedron, the edge length of which is 4 cm. Point O - middle of the rib D.B.. Dot F lies on the continuation of the rib Sun so FROM - middle of the segment bf, dot T lies on the continuation of the rib AU so FROM - middle of the segment AT. Construct a section of a tetrahedron by a plane FTO and calculate its perimeter.

Task 6. DABC- triangular pyramid Point F lies on the edge D.B., dot T lies on the continuation of the rib AB so that point BUT located between the points T and AT, a dot R lies on the continuation of the rib CD so that point FROM lies between the points D and R. Construct a section of the pyramid by a plane TFR.

Each group is given tasks depending on the level of difficulty. Students complete these tasks, and then collectively discuss the solution.

Condition: whether the filled figures are sections of the depicted polyhedra by a plane PQR ? In cases where the section is shown incorrectly, find the correct solution.

The figures show regular parallelepipeds.

First level task:

Second level task:

Third level task:

Each group is given a main task and an additional one. In the additional task, the figures show triangular prisms (levels 1 and 2) and a triangular pyramid (level 3).

The work is evaluated by the teacher and then marked in the journal.

First level task:

• AT triangular pyramid DABC dot O - point of intersection of face medians DBC. Dot F lies on a straight line AB so AT lies between the points BUT and F, a dot E lies on a straight line AU so that point FROM lies between BUT and E. Construct a section of the pyramid by a plane OEF.

• PQR

Second level task:

• ABSA 1 IN 1 WITH 1 - triangular prism. Dot O lies on the edge A 1 C 1 ,. Dot F lies on the continuation of the rib AU so FROM lies between BUT and F. Dot To lies on the continuation of the rib AB so AT located between BUT and To. Construct a section of a prism by a plane OKF.

• Additional task: are the filled shapes the sections of the depicted polyhedra a plane PQR ? In cases where the section is shown incorrectly, find the correct solution.

Third level task:

• Base of a cuboid A BCDA l B 1 C 1 D 1 - a square whose side length is 2 cm. Point O - middle lateral rib DD 1 and points To and F lie on the continuation of the ribs Sun and AB respectively so that Sun = 2SK, AB = 2FA . Calculate Sectional Area of ​​Parallelepiped by Plane OFК , if DD 1 = 4 cm.

• Additional task: are the filled shapes the sections of the depicted polyhedra a plane PQR ? In cases where the section is shown incorrectly, find the correct solution.

Students choose the appropriate level of difficulty.

Task for the first level of difficulty:

Task for the second level of difficulty:

Task for the third level of difficulty: