How to make a section of a cube using three points. Section of a cube

Tasks on Constructing sections of a cubeD1
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Test work.

A cutting plane of a cube is any plane on both sides of which there are points of a given cube.

three given points that are the midpoints of the edges. Find the perimeter of the section if the edge

Construct a section of the cube with a plane passing through three given points, which are its vertices. Find the perimeter of the section if the edge of the cube

Construct a section of the cube with a plane passing through three given points. Find the perimeter of the section if the edge of the cube is equal to a.

Construct a section of the cube with a plane passing through three given points, which are the midpoints of its edges.

B1. V. Cube. Level B. Help. Construct a section of a cube with a plane passing through points A, K and E. Find the line of intersection of this plane a) with edge BB1; b) plane (CC1D). E. C1. K. A1. D1. C. D. A. Menu.

Geometry 10th grade

“Determination of dihedral angles” - The point on the edge can be arbitrary. Let's build BK. Task. Problem solving. Plane M. Rhombus. Definition and properties. Where can you see theorem of three perpendiculars. Ends of the segment. Let's cast a beam. Properties. Dihedral angles in the pyramids. Points M and K lie in different faces. Segments AC and BC. Property of a trihedral angle. Definition. Dihedral angles. Find the angle. Draw a perpendicular. Degree measure corner.

“Examples of central symmetry” - Plane. Axioms of planimetry. Dots. Central symmetry. One center of symmetry. Hotel "Pribaltiyskaya". Train capsule. Length of the segment. Examples of symmetry in plants. Central symmetry in architecture. Chamomile. A segment has a certain length. Segment. Axioms of stereometry and planimetry. Axioms of stereometry. Central symmetry in squares. Central symmetry in transport. Various straight lines.

"Equilateral polygons" - Octahedron The octahedron is made up of eight equilateral triangles. "Edra" - face of a "tetra" - 4 "hexa" - 6 "okta" - 8 "icos" - 20 "dedeka" - 12. A tetrahedron has 4 faces, 4 vertices and 6 edges. The dodecahedron has 12 faces, 20 vertices and 30 edges. The octahedron has 8 faces, 6 vertices and 12 edges. There are 5 types of regular polyhedra. Dodecahedron The dodecahedron is made up of twelve equilateral pentagons.

“Application of regular polyhedra” - Polyhedra in nature. Euler's theorem. Project objectives. Use in life. The world of regular polyhedra. Polyhedra in architecture. Polyhedra in art. Polyhedra in mathematics. Archimedes. Kepler. Theory of polyhedra. Golden ratio in the dodecahedron and icosahedron. Conclusion. Plato. Group "Historians". Euclid. The history of the emergence of regular polyhedra. The relationship between the “golden section” and the origin of polyhedra.

"Platonic solids" - Octahedron. Plato's solids. Hexahedron. Regular polyhedra. Plato. Dodecahedron. Duality. Icosahedron. Regular polyhedra or Platonic solids. Tetrahedron.

“Methods for constructing sections of polyhedra” - Rules for self-control. Construct a cross section of the prism. Ship. Polygons. The simplest tasks. The relative position of the plane and the polyhedron. Intersection points. Do the lines intersect? The cuts formed a pentagon. We make cuts. Laws of geometry. Axiomatic method. Trace of the cutting plane. Task. Cutting plane. Construction of sections of polyhedra. Section. Survey. Any plane. Sections of a parallelepiped.

" Mystery three points» Information and research project

Project goals: constructing sections in a cube passing through three points; composing problems on the topic “Section of a cube by a plane”; presentation design; preparing a speech.

Not in geometry royal road Euclid

Axioms of stereometry Through any three points in space that do not lie on the same straight line, there is a single plane.

To solve many geometric problems related to a cube, it is useful to be able to draw their sections in a drawing different planes. By section we mean any plane (let's call it a cutting plane), on both sides of which there are points of a given figure. A cutting plane intersects a polyhedron along segments. The polygon that will be formed by these segments is the cross section of the figure.

Rules for constructing sections of polyhedra: 1) draw straight lines through points lying in the same plane; 2) we are looking for direct intersections of the cutting plane with the faces of the polyhedron, for this: a) we are looking for the points of intersection of a straight line belonging to the cutting plane with a straight line belonging to one of the faces (lying in the same plane); b) the cutting plane intersects parallel faces along parallel straight lines.

The cube has six sides. Its cross-section can be: triangles, quadrangles, pentagons, hexagons.

Let's consider the construction of these sections.

Triangle

The resulting triangle EFG will be the desired section. Construct a section of the cube with a plane passing through points E, F, G lying on the edges of the cube.

Construct a section of the cube with a plane passing through points A, C and M.

To construct a section of a cube passing through points lying on the edges of the cube emerging from one vertex, it is enough to simply connect these points with segments. The cross section will form a triangle.

Quadrangle

Construct a section of the cube with a plane passing through points E, F, G lying on the edges of the cube.

The resulting rectangle BCFE will be the desired section. Construct a section of the cube with a plane passing through points E, F, G lying on the edges of the cube, for which AE = DF. Solution. To construct a section of a cube passing through points E, F, G, connect points E and F. Line EF will be parallel to AD and therefore BC. Let's connect points E and B, F and C.

Construct a section of the cube with a plane passing through points E, F lying on the edges of the cube and vertex B. Solution. To construct a section of a cube passing through points E, F and vertex B, connect points E and B, F and B with segments. Through points E and F we draw lines parallel to BF and BE, respectively.

The resulting parallelogram BFGE will be the desired section. Construct a section of the cube with a plane passing through points E, F lying on the edges of the cube and vertex B. Solution. To construct a section of a cube passing through points E, F and vertex B, connect points E and B, F and B with segments. Through points E and F we draw lines parallel to BF and BE, respectively.

The cutting plane is parallel to one of the edges of the cube or passes through the edge (rectangle) The cutting plane intersects four parallel edges of the cube (parallelogram)

Pentagon

The resulting pentagon EFSGQ will be the required section Construct a section of the cube with a plane passing through points E, F, G lying on the edges of the cube. Solution. To construct a section of a cube passing through points E, F, G, draw a straight line EF and designate P its point of intersection with AD. Let us denote by Q, R the points of intersection of straight line PG with AB and DC. Let us denote by S the point of intersection of FR with CC 1. Let us connect the points E and Q, G and S.

Through point P we draw a line parallel to MN. It intersects edge BB1 ​​at point S. PS is the trace of the cutting plane in the face (BCC1). We draw a straight line through points M and S lying in the same plane (ABB1). We received a trace of MS (visible). Planes (ABB1) and (CDD1) are parallel. There is already a straight line MS in the plane (ABB1), so through point N in the plane (CDD1) we draw a straight line parallel to MS. This line intersects edge D1C1 at point L. Its trace is NL (invisible). Points P and L lie in the same plane (A1B1C1), so we draw a straight line through them. Pentagon MNLPS is the required section.

When a cube is cut by a plane, the only pentagon that can be formed is one that has two pairs of parallel sides.

Hexagon

Construct a section of the cube with a plane passing through points E, F, G lying on the edges of the cube. Solution. To construct a section of a cube passing through points E, F, G, we find the point P of intersection of the straight line EF and the plane of the face ABCD. Let us denote by Q, R the points of intersection of straight line PG with AB and CD. Let's draw a line RF and denote S, T its points of intersection with CC 1 and DD 1. Let's draw a line TE and denote U its point of intersection with A 1 D 1. Connect points E and Q, G and S, F and U. The resulting hexagon EUFSGQ will be the desired section.

When a cube is cut by a plane, the only hexagon that can be formed is one that has three pairs of parallel sides.

Given: M€AA1 , N€B1C1,L€AD Build: (MNL)

Lesson type: Combined lesson.

Goals and objectives:

• educational – formation and development of spatial concepts in students; developing skills in solving problems involving constructing sections of the simplest polyhedra;
• educational - cultivate the will and perseverance to achieve final results when constructing sections of the simplest polyhedra; Foster a love and interest in learning mathematics.
• developing – student development logical thinking, spatial representations, development of self-control skills.

Equipment: computers with a specially developed program, handouts in the form of ready-made drawings with tasks, bodies of polyhedra, individual cards with homework.

Lesson structure:

1. State the topic and purpose of the lesson (2 min).
2. Instructions on how to complete tasks on a computer (2 min).
3. Update background knowledge and student skills (4 min).
4. Self-test (3 min).
5. Solving problems with an explanation of the solution by the teacher (15 min).
6. Independent work with self-test (10 min).
7. Staging homework(2 min).
8. Summing up (2 min).

Lesson progress

1. Communicating the topic and purpose of the lesson

After checking the class’s readiness for the lesson, the teacher reports that today there is a lesson on the topic “Constructing sections of polyhedra”; problems will be considered on constructing sections of some simple polyhedra with planes passing through three points belonging to the edges of the polyhedra. The lesson will be taught using a computer presentation made in Power Point.

2. Safety instructions when working in a computer lab

Teacher. I draw your attention to the fact that you are starting to work in a computer class, and you must follow the rules of conduct and work at the computer. Secure retractable tabletops and ensure proper fit.

3. Updating the basic knowledge and skills of students

Teacher. To solve many geometric problems related to polyhedra, it is useful to be able to construct their sections in a drawing using different planes, find the point of intersection of a given line with a given plane, and find the line of intersection of two given planes. In previous lessons, we looked at sections of polyhedra by planes parallel to the edges and faces of the polyhedra. In this lesson we will look at problems involving constructing sections with a plane passing through three points located on the edges of polyhedra. To do this, consider the simplest polyhedra. What are these polyhedra? (Models of a cube, tetrahedron, regular quadrangular pyramid, straight triangular prism).

Students must determine the type of polyhedron.

Teacher. Let's see how they look on the monitor screen. We move from image to image by pressing the left mouse button.

Images of the named polyhedra appear on the screen one after another.

Teacher. Let us remember what is called a section of a polyhedron.

Student. A polygon whose sides are segments belonging to the faces of the polyhedron, with ends on the edges of the polyhedron, obtained by intersecting the polyhedron with an arbitrary cutting plane.

Teacher. What polygons can be sections of these polyhedra.

Student. Sections of a cube: three - hexagons. Sections of a tetrahedron: triangles, quadrangles. Sections of a quadrangular pyramid and a triangular prism: three - pentagons.

4. Self-testing

Teacher. In accordance with the concept of sections of polyhedra, knowledge of the axioms of stereometry and the relative position of lines and planes in space, you are asked to answer the test questions. The computer will appreciate you. Maximum score 3 points – for 3 correct answers. On each slide you must click the button with the number of the correct answer. You work in pairs, so each of you will receive the same computer-specified number of points. Click the next slide indicator. You have 3 minutes to complete the task.

I. Which figure shows a section of a cube by a plane ABC?

II. Which figure shows a cross section of a pyramid with a plane passing through the diagonal of the base? BD parallel to the edge S.A.?

III. Which figure shows a cross section of a tetrahedron passing through a point M parallel to the plane ABS?

5. Solving problems with an explanation of the solution by the teacher

Teacher. Let's move on directly to solving problems. Click the next slide indicator.

Problem 1 This task Let's look at it orally with a step-by-step demonstration of the construction on the monitor screen. The transition is carried out by clicking the mouse.

Given a cube ABCDAA 1 B 1 C 1 D 1. On his edge BB 1 given point M. Find the point of intersection of a line C 1 M with the plane of the cube face ABCD.

Consider the image of a cube ABCDAA 1 B 1 C 1 D 1 with a dot M on the edge BB 1 Points M And WITH 1 belong to the plane BB 1 WITH 1 What can be said about the straight line C 1 M ?

Student. Straight C 1 M belongs to the plane BB 1 WITH 1

Teacher. Searched point X belongs to the line C 1 M, and therefore planes BB 1 WITH 1. What's it like relative position planes BB 1 WITH 1 and ABC?

Student. These planes intersect in a straight line B.C..

Teacher. That means everything common points planes BB 1 WITH 1 and ABC belong to the line B.C.. Searched point X must simultaneously belong to the planes of two faces: ABCD And BB 1 C 1 C; from this it follows that point X must lie on the line of their intersection, i.e. on the straight line Sun. This means that point X must lie on two straight lines simultaneously: WITH 1 M And Sun and, therefore, is their point of intersection. Let's look at the construction of the desired point on the monitor screen. You will see the construction sequence by pressing the left mouse button: continue WITH 1 M And Sun to the intersection at the point X, which is the desired intersection point of the line WITH 1 M with face plane ABCD.

Teacher. Use the next slide indicator to move to the next task. Let's consider this problem with a brief description of the construction.

A) Construct a section of a cube with a plane passing through the points A 1 , MD 1 C 1 and NDD 1 and b) Find the line of intersection of the cutting plane with the plane of the lower base of the cube.

Solution. I. The cutting plane has a face A 1 B 1 C 1 D 1 two common points A 1 and M and, therefore, intersects with it along a straight line passing through these points. Connecting the dots A 1 and M straight line segment, we find the line of intersection of the plane of the future section and the plane top edge. We will write this fact as follows: A 1 M. Press the left mouse button, pressing again will construct this straight line.

Similarly, we find the lines of intersection of the cutting plane with the faces AA 1 D 1 D And DD 1 WITH 1 WITH. By clicking the mouse button, you will see a brief recording and construction progress.

Thus, A 1 NM? the desired section.

Let's move on to the second part of the problem. Let's find the line of intersection of the cutting plane with the plane of the lower base of the cube.

II. The cutting plane intersects with the plane of the base of the cube in a straight line. To depict this line, it is enough to find two points belonging to this line, i.e. common points of the cutting plane and the face plane ABCD. Based on previous task such points will be: point X=. Press the key, you will see a short recording and construction. And period Y, what do you guys think, how to get it?

Student. Y =

Teacher. Let's look at its construction on the screen. Click the mouse button. Connecting the dots X And Y(Record X-Y), we obtain the desired straight line - the line of intersection of the cutting plane with the plane of the lower base of the cube. Click the left mouse button – short note and construction.

Problem 3 Construct a section of the cube with a plane passing through the points:

Also, by pressing the mouse button, you will see the construction progress and a short recording on the monitor screen. Based on the concept of a section, it is enough for us to find two points in the plane of each face to construct the line of intersection of the cutting plane and the plane of each face of the cube. Points M And N belong to the plane A 1 IN 1 WITH 1. By connecting them, we get the line of intersection of the cutting plane and the plane of the upper face of the cube (press the mouse button). Let's continue the straight lines MN And D 1 C 1 before the intersection. Let's get a point X, belonging to both the plane A 1 IN 1 WITH 1 and plane DD 1 C 1 (mouse click). Points N And TO belong to the plane BB 1 WITH 1. By connecting them, we get the line of intersection of the cutting plane and the face BB 1 WITH 1 WITH. (Mouse click). Connecting the dots X And TO, and continue straight HC to the intersection with the line DC. Let's get a point R and segment KR – line of intersection of the cutting plane and the face DD 1 C 1 C. (Mouse click). Continuing straight KR And DD 1 before intersection, we get a point Y, belonging to the plane AA 1 D 1. (Mouse click). In the plane of this face we need one more point, which we obtain as a result of the intersection of lines MN And A 1 D 1. This is the point . (Mouse click). Connecting the dots Y And Z, we get And . (Mouse click). Connecting Q And R, R And M, will we get it? the desired section.

Brief description of the construction:

2) ;

6) ;

7) ;

13) ? the desired section.

Problems involving constructing sections of a cube using a plane are, as a rule, simpler than, for example, problems involving sections of a pyramid.

We can draw a straight line through two points if they lie in the same plane. When constructing sections of a cube, another option is possible for constructing a trace of a cutting plane. Since the third plane intersects two parallel planes along parallel straight lines, then if a straight line has already been constructed in one of the faces, and in the other there is a point through which the section passes, then we can draw a line parallel to this point through this point.

Let's look at specific examples how to construct sections of a cube using a plane.

1) Construct a section of the cube with a plane passing through points A, C and M.

Problems of this type are the simplest of all problems for constructing sections of a cube. Since points A and C lie in the same plane (ABC), we can draw a straight line through them. Its trace is segment AC. It is invisible, so we depict AC with a stroke. Similarly, we connect points M and C, which lie in the same plane (CDD1), and points A and M, which lie in the same plane (ADD1). Triangle ACM is the required section.

2) Construct a section of the cube with a plane passing through points M, N, P.

Here only points M and N lie in the same plane (ADD1), so we draw a straight line through them and get a trace MN (invisible). Because opposite faces cubes lie in parallel planes, then the cutting plane intersects parallel planes (ADD1) and (BCC1) along parallel lines. We have already constructed one of the parallel lines - this is MN.

Through point P we draw a line parallel to MN. It intersects edge BB1 ​​at point S. PS is the trace of the cutting plane in the face (BCC1).

We draw a straight line through points M and S lying in the same plane (ABB1). We received a trace of MS (visible).

Planes (ABB1) and (CDD1) are parallel. There is already a straight line MS in the plane (ABB1), so through point N in the plane (CDD1) we draw a straight line parallel to MS. This line intersects edge D1C1 at point L. Its trace is NL (invisible). Points P and L lie in the same plane (A1B1C1), so we draw a straight line through them.

Pentagon MNLPS is the required section.

3) Construct a section of the cube with a plane passing through points M, N, P.

Points M and N lie in the same plane (ВСС1), so a straight line can be drawn through them. We get the trace MN (visible). The plane (BCC1) is parallel to the plane (ADD1), therefore, through the point P lying in (ADD1), we draw a line parallel to MN. It intersects edge AD at point E. We have obtained a trace PE (invisible).

There are no longer points lying in the same plane, or a straight line and points in parallel planes. Therefore, we need to continue one of the existing lines to get an additional point.

If we continue the line MN, then, since it lies in the plane (BCC1), we need to look for the point of intersection of MN with one of the lines of this plane. There are already intersection points with CC1 and B1C1 - these are M and N. What remains are straight lines BC and BB1. Let's continue BC and MN until they intersect at point K. Point K lies on line BC, which means it belongs to the plane (ABC), so we can draw a straight line through it and point E, which lies in this plane. It intersects edge CD at point H. EH is its trace (invisible). Since H and N lie in the same plane (CDD1), a straight line can be drawn through them. We get an HN (invisible) trace.

Planes (ABC) and (A1B1C1) are parallel. In one of them there is a line EH, in the other there is a point M. We can draw a line parallel to EH through M. We get the MF trace (visible). Draw a straight line through points M and F.

The hexagon MNHEPF is the required section.

If we were to continue straight line MN until it intersects with another straight plane (BCC1), BB1, we would get point G belonging to the plane (ABB1). This means that through G and P we can draw a straight line whose trace is PF. Next, we draw straight lines through points lying in parallel planes and arrive at the same result.

Working with straight PE gives the same section MNHEPF.

4) Construct a section of the cube with a plane passing through the point M, N, P.

Here we can draw a straight line through points M and N lying in the same plane (A1B1C1). Her footprint is MN (visible). There are no more points lying in the same plane or in parallel planes.

Let's continue the straight line MN. It lies in the plane (A1B1C1), so it can intersect only with one of the straight lines of this plane. There are already intersection points with A1D1 and C1D1 - N and M. Two more straight lines of this plane - A1B1 and B1C1. The point of intersection of A1B1 and MN is S. Since it lies on the line A1B1, it belongs to the plane (ABB1), which means that a straight line can be drawn through it and point P, which lies in the same plane. Line PS intersects edge AA1 at point E. PE is its trace (visible). Through points N and E, lying in the same plane (ADD1), you can draw a straight line, the trace of which is NE (invisible). In the plane (ADD1) there is a line NE, in the plane parallel to it (BCC1) there is a point P. Through point P we can draw a line PL parallel to NE. It intersects edge CC1 at point L. PL is the trace of this line (visible). Points M and L lie in the same plane (CDD1), which means that a straight line can be drawn through them. Her trail is ML (invisible). Pentagon MLPEN is the required section.

It was possible to continue the straight line NM in both directions and look for its intersection points not only with the straight line A1B1, but also with the straight line B1C1, which also lies in the plane (A1B1C1). In this case, through point P we draw two lines at once: one in the plane (ABB1) through points P and S, and the second in the plane (BCC1), through points P and R. After which it remains to connect the points lying in the same plane: M c L, E - with N.