Lesson objectives: consider solving problems on constructing sections if two points of the section belong to the same face.
During the classes
Learning new concepts
Definition 1.
The cutting plane of a polyhedron is any plane on both sides of which there are points of the given polyhedron.
Definition 2.
A section of a polyhedron is a polygon whose sides are the segments along which the cutting plane intersects the faces of the polyhedron.
Exercise. Name the segments along which the cutting plane intersects the faces of the parallelepiped (Fig. 1). Name the section of the parallelepiped.
Basic actions when constructing sections
Theoretical basis |
Answer |
|
| 1. How to check whether the section has been constructed or not | Section Definition | It must be a polygon whose sides belong to the faces of the polyhedron |
| 2. Before starting work, determine whether it is possible to construct a section based on the task data | Methods for defining a plane | It is possible if these elements uniquely define a plane, that is, three points are given that do not lie on the same line, a point and a line, etc. |
| 3. In the plane of some face there are two points of the cutting plane |
If two points belong to a plane, then the entire line belongs to the plane | Draw a straight line through these points |
| 4. In one of the parallel faces there is a section side, and in the other there is a section point | Property of parallel planes | Through this point draw a line parallel to this point |
| 5. There is a section point in one face and it is known that the cutting plane passes through a line parallel to this face | A sign of parallelism between a line and a plane. Property of parallel planes | Construct a line of intersection of planes parallel to a given line |
| 6. Two points of the section belong to one face, and the third point lies in the adjacent one | Axioms of stereometry | The cutting plane intersects the faces along the segments OC and AB, which are called the trace of the cutting plane on the faces |
|
||
Problem solving
Task 1. Which of the quadrangles, EFKM or EFKL, can be a section of this polyhedron (Fig. 2)? Why?
Task 2. The student drew a cross-section of a tetrahedron (Fig. 3). Is such a section possible?
Solution. It is necessary to prove that N, M and H, L lie in the same plane. Let points N and M belong to the back face, H and L to the bottom face, that is, the intersection point of NM and HL must lie on a line belonging to both faces, that is, AC. Let's extend the lines NM and HL and find the point of their intersection. This point will not belong to line AC. This means that the points N, M, L, H do not form a flat polygon. Impossible.
Task 3. Construct a section of the ABCS tetrahedron with a plane passing through points K, L, N, where K and N are the midpoints of the edges SA and SB, respectively (Fig. 4).
1. In which face can the sides of the section be constructed?
2. Select one of the points at which the section breaks.
Solution. Method I. Select point L.
We determine the face in which the selected point lies and in which it is necessary to construct a section.
We determine the face in which the straight line KN lies, not passing through the selected point L.
Find the line of intersection of faces ABC and ASB.
What is the relative position of the lines KN and AB (Fig. 5)?
[Parallel.]
What needs to be constructed if the cutting plane passes through a straight line parallel to the line of intersection of the planes?
[Draw a line parallel to AB through point L. This line intersects edge CB at point P.]
We connect points belonging to the same face. KLPN - the required section.
Method II. Select point N (Fig. 6).
We determine the faces in which point N and straight line KL lie.
The line of intersection of these planes will be straight line SC. Find the intersection point of lines KL and SC. Let's denote it Y.
Connect points N and Y. Line NY intersects edge CB at point P.
We connect points belonging to the same face.
KLNP - the required section.
Explain this decision.
One student works at the board, the rest in notebooks.
Problem 4. Construct a section of a parallelepiped passing through points M, P and H, H ` (A1B1C1) (Fig. 7).
Solution. 1. Connect the points belonging to the same face.
2. Which line and point do we choose to construct the section?
3. What do we determine next?
4. What is the relative position of the selected straight line and the line of intersection of the faces (Fig. 8)?
5. How to construct a trace of a cutting plane on the face B1C1D1A1 passing through point H?
6. Connect the points belonging to the same face.
7. Which line and point should be chosen to construct the trace of the cutting plane on the face AA1D1D?
8. What is the relative position of the faces BB1C1C and AA1D1D?
9. What property must be used to construct the trace of the cutting plane on the face AA1D1D?
10. Name the required section.
Task 5. Construct a section of the SABCD pyramid passing through points M, P and H,
H` (ABC) (Fig. 9).
Answer: See Figure 10.
Homework assignment
Task. How will the constructions change if exactly
How will H change its position? Construct sections using different options (Fig. 11).
Today we’ll look again at how construct a section of a tetrahedron with a plane.
Let's consider the simplest case (mandatory level), when 2 points of the section plane belong to one face, and the third point belongs to another face.
Let us remind you algorithm for constructing sections of this type (case: 2 points belong to the same face).
1. We are looking for a face that contains 2 points of the section plane. Draw a straight line through two points lying on the same face. We find the points of its intersection with the edges of the tetrahedron. The part of the straight line that ends up in the face is the side of the section.
2. If the polygon can be closed, the section has been constructed. If it is impossible to close, then we find the intersection point of the constructed line and the plane containing the third point.
1. We see that points E and F lie on the same face (BCD), draw a straight line EF in the plane (BCD). 
2. Find the point of intersection of the straight line EF with the edge of the tetrahedron BD, this is point H.
3. Now you need to find the point of intersection of the straight line EF and the plane containing the third point G, i.e. plane (ADC).
The straight line CD lies in the planes (ADC) and (BDC), which means it intersects the straight line EF, and point K is the point of intersection of the straight line EF and the plane (ADC).
4. Next, we find two more points lying in the same plane. These are points G and K, both lie in the plane of the left side face. We draw a line GK and mark the points at which this line intersects the edges of the tetrahedron. These are points M and L.
4. It remains to “close” the section, i.e. connect the points lying on the same face. These are points M and H, and also L and F. Both of these segments are invisible, we draw them with a dotted line.
The cross-section turned out to be a quadrangle MHFL. All its vertices lie on the edges of the tetrahedron. Let's select the resulting section.
Now let's formulate "properties" of a correctly constructed section:
1. All vertices of a polygon, which is a section, lie on the edges of a tetrahedron (parallelepiped, polygon).
2. All sides of the section lie on the faces of the polyhedron.
3. Each face of a polygon can contain no more than one (one or none!) side of the section
In this lesson we will look at the tetrahedron and its elements (tetrahedron edge, surface, faces, vertices). And we will solve several problems on constructing sections in a tetrahedron, using the general method for constructing sections.
Topic: Parallelism of lines and planes
Lesson: Tetrahedron. Problems on constructing sections in a tetrahedron
How to build a tetrahedron? Let's take an arbitrary triangle ABC. Any point D, not lying in the plane of this triangle. We get 4 triangles. The surface formed by these 4 triangles is called a tetrahedron (Fig. 1.). The internal points bounded by this surface are also part of the tetrahedron.
Rice. 1. Tetrahedron ABCD
Elements of a tetrahedron
A,B,
C,
D - vertices of a tetrahedron.
AB,
A.C.,
AD,
B.C.,
BD,
CD - tetrahedron edges.
ABC,
ABD,
BDC,
ADC - tetrahedron faces.
Comment: can be taken flat ABC behind tetrahedron base, and then point D is vertex of a tetrahedron. Each edge of the tetrahedron is the intersection of two planes. For example, rib AB- this is the intersection of planes ABD And ABC. Each vertex of a tetrahedron is the intersection of three planes. Vertex A lies in planes ABC, ABD, ADWITH. Dot A is the intersection of the three designated planes. This fact is written as follows: A= ABC ∩ ABD ∩ ACD.
Tetrahedron definition
So, tetrahedron is a surface formed by four triangles.
Tetrahedron edge- the line of intersection of two planes of the tetrahedron.
Make 4 equal triangles from 6 matches. It is impossible to solve the problem on a plane. And in space this is easy to do. Let's take a tetrahedron. 6 matches are its edges, four faces of the tetrahedron and will be four equal triangles. The problem is solved.
Given a tetrahedron ABCD. Dot M belongs to an edge of the tetrahedron AB, dot N belongs to an edge of the tetrahedron IND and period R belongs to the edge DWITH(Fig. 2.). Construct a section of a tetrahedron with a plane MNP.
Rice. 2. Drawing for problem 2 - Construct a section of a tetrahedron with a plane
Solution:
Consider the face of a tetrahedron DSun. On this face of the point N And P belong to the faces DSun, and therefore the tetrahedron. But according to the condition of the point N, P belong to the cutting plane. Means, N.P.- this is the line of intersection of two planes: the plane of the face DSun and cutting plane. Let's assume that straight lines N.P. And Sun not parallel. They lie in the same plane DSun. Let's find the point of intersection of the lines N.P. And Sun. Let's denote it E(Fig. 3.).
Rice. 3. Drawing for problem 2. Finding point E
Dot E belongs to the section plane MNP, since it lies on the straight line NP, and the straight line NP lies entirely in the section plane MNP.
Also point E lies in a plane ABC, because it lies on a straight line Sun out of plane ABC.
We get that EAT- line of intersection of planes ABC And MNP, since points E And M lie simultaneously in two planes - ABC And MNP. Let's connect the dots M And E, and continue straight EAT to the intersection with the line AC. Point of intersection of lines EAT And AC let's denote Q.
So in this case NPQМ- the required section.
Rice. 4. Drawing for problem 2. Solution of problem 2
Let us now consider the case when N.P. parallel B.C.. If straight N.P. parallel to some line, for example, a straight line Sun out of plane ABC, then straight N.P. parallel to the entire plane ABC.
The desired section plane passes through the straight line N.P., parallel to the plane ABC, and intersects the plane in a straight line MQ. So the line of intersection MQ parallel to the line N.P.. We get, NPQМ- the required section.
Dot M lies on the side edge ADIN tetrahedron ABCD. Construct a section of the tetrahedron with a plane that passes through the point M parallel to the base ABC.
Rice. 5. Drawing for problem 3 Construct a section of a tetrahedron with a plane
Solution:
Cutting plane φ
parallel to the plane ABC according to the condition, this means that this plane φ
parallel to lines AB, AC, Sun.
In plane ABD through the point M let's make a direct PQ parallel AB(Fig. 5). Straight PQ lies in a plane ABD. Similarly in the plane ACD through the point R let's make a direct PR parallel AC. Got a point R. Two intersecting lines PQ And PR plane PQR respectively parallel to two intersecting lines AB And AC plane ABC, which means planes ABC And PQR parallel. PQR- the required section. The problem is solved.
Given a tetrahedron ABCD. Dot M- internal point, point on the face of the tetrahedron ABD. N- internal point of the segment DWITH(Fig. 6.). Construct the intersection point of a line N.M. and planes ABC.
Rice. 6. Drawing for problem 4
Solution:
To solve this, we will construct an auxiliary plane DMN. Let it be straight DM intersects line AB at point TO(Fig. 7.). Then, SKD- this is a section of the plane DMN and tetrahedron. In plane DMN lies and straight N.M., and the resulting straight line SK. So if N.M. not parallel SK, then they will intersect at some point R. Dot R and there will be the desired intersection point of the line N.M. and planes ABC.
Rice. 7. Drawing for problem 4. Solution of problem 4
Given a tetrahedron ABCD. M- internal point of the face ABD. R- internal point of the face ABC. N- internal point of the edge DWITH(Fig. 8.). Construct a section of a tetrahedron with a plane passing through the points M, N And R.
Rice. 8. Drawing for problem 5 Construct a section of a tetrahedron with a plane
Solution:
Let us consider the first case, when the straight line MN not parallel to the plane ABC. In the previous problem we found the point of intersection of the line MN and planes ABC. This is the point TO, it is obtained using the auxiliary plane DMN, i.e. we do DM and we get a point F. We carry out CF and at the intersection MN we get a point TO.
Rice. 9. Drawing for problem 5. Finding point K
Let's make a direct KR. Straight KR lies both in the section plane and in the plane ABC. Getting the points P 1 And R 2. Connecting P 1 And M and as a continuation we get the point M 1. Connecting the dot R 2 And N. As a result, we obtain the desired section Р 1 Р 2 NM 1. The problem in the first case is solved.
Let's consider the second case, when the straight line MN parallel to the plane ABC. Plane MNP passes through a straight line MN parallel to the plane ABC and intersects the plane ABC along some straight line R 1 R 2, then straight R 1 R 2 parallel to the given line MN(Fig. 10.).
Rice. 10. Drawing for problem 5. The required section
Now let's draw a straight line R 1 M and we get a point M 1.Р 1 Р 2 NM 1- the required section.
So, we looked at the tetrahedron and solved some typical tetrahedron problems. In the next lesson we will look at a parallelepiped.
1. I. M. Smirnova, V. A. Smirnov. - 5th edition, corrected and expanded - M.: Mnemosyne, 2008. - 288 p. : ill. Geometry. Grades 10-11: textbook for students of general education institutions (basic and specialized levels)
2. Sharygin I.F. - M.: Bustard, 1999. - 208 p.: ill. Geometry. Grades 10-11: Textbook for general education institutions
3. E. V. Potoskuev, L. I. Zvalich. - 6th edition, stereotype. - M.: Bustard, 008. - 233 p. :il. Geometry. Grade 10: Textbook for general education institutions with in-depth and specialized study of mathematics
Additional web resources
2. How to construct a cross section of a tetrahedron. Mathematics ().
3. Festival of pedagogical ideas ().
Do problems at home on the topic “Tetrahedron”, how to find the edge of a tetrahedron, faces of a tetrahedron, vertices and surface of a tetrahedron
1. Geometry. Grades 10-11: textbook for students of general education institutions (basic and specialized levels) I. M. Smirnova, V. A. Smirnov. - 5th edition, corrected and expanded - M.: Mnemosyne, 2008. - 288 pp.: ill. Tasks 18, 19, 20 p. 50
2. Point E mid-rib MA tetrahedron MAVS. Construct a section of the tetrahedron with a plane passing through the points B, C And E.
3. In the tetrahedron MABC, point M belongs to the face AMB, point P belongs to the face BMC, point K belongs to the edge AC. Construct a section of the tetrahedron with a plane passing through the points M, R, K.
4. What shapes can be obtained as a result of the intersection of a tetrahedron with a plane?
Axioms of planimetry:
In different textbooks, the properties of lines and planes can be presented in different ways, in the form of an axiom, a corollary from it, a theorem, lemma, etc. Consider the textbook by Pogorelov A.V.
A straight line divides a plane into two half-planes.
0
From any half-line an angle with a given degree measure less than 180 can be plotted into a given half-plane. 0 , and only one.
Whatever a triangle is, there is an equal triangle in a given location relative to a given half-line.
Through a point not lying on a given line, it is possible to draw on the plane at most one straight line parallel to the given one.
Axioms of stereometry:
Whatever the plane, there are points that belong to this plane, and points that do not belong to this plane, and points that do not belong to it.
If two different planes have a common point, then they intersect along a straight line passing through this point.
If two different lines have a common point, then a plane can be drawn through them, and only one.
Whatever the line, there are points that belong to this line and points that do not belong to it.
Through any two points you can draw a straight line, and only one.
Of the three points on a line, one and only one lies between the other two.
Each segment has a certain length greater than zero. The length of a segment is equal to the sum of the lengths of the parts into which it is divided by any of its points.
A straight line belonging to a plane divides this plane into two half-planes.
Each angle has a certain degree measure greater than zero. The straight angle is 180 0 . The degree measure of an angle is equal to the sum of the degree measures of the angles into which it is divided by any ray passing between its sides.
On any half-line from its starting point, you can plot a segment of a given length, and only one.
From a half-line on the plane containing it, an angle with a given degree measure less than 180 can be plotted into a given half-plane 0 , and only one.
Whatever the triangle, there is an equal triangle in a given plane in a given location relative to a given half-line in that plane.
On a plane, through a given point that does not lie on a given line, it is possible to draw at most one straight line parallel to the given one.
Section
In space, two figures, for our case a plane and a polyhedron, can have the following relative positions: do not intersect, intersect at a point, intersect in a straight line and the plane intersects the polyhedron along its interior (Fig. 1), and at the same time form the following figures:
a) empty figure (do not intersect)
b) point
c) segment
d) polygon
If there is a polygon at the intersection of a polyhedron and a plane, then this polygoncalled a section of a polyhedron with a plane .
Fig.1
Definition. Section spatial body (for example, a polyhedron) is the figure resulting from the intersection of the body with a plane.
Cutting plane polyhedron let's call any plane on both sides of which there are points of a given polyhedron.
We will consider only the case when the plane intersects the polyhedron along its interior. In this case, the intersection of this plane with each face of the polyhedron will be a certain segment.
If the planes intersect in a straight line, then the straight line is calledfollowing one of these planes onto the other.
In general, the cutting plane of a polyhedron intersects the plane of each of its faces (as well as any other cutting plane of this polyhedron). It also intersects each of the lines on which the edges of the polyhedron lie.
The straight line along which the cutting plane intersects the plane of any face of the polyhedron is calledfollowing the cutting plane on the plane of this face, and the point at which the cutting plane intersects the line containing any edge of the polyhedron is calledfollowing the cutting plane onthis straight line. This point is also the trace of a line on the cutting plane. If the cutting plane directly intersects the face of the polyhedron, then we can talk about the trace of the cutting plane on the face, and, similarly, abouttrace of the cutting plane on the edge of the polyhedron, that is, about the trace of an edge on a cutting plane.
Since a straight line is uniquely determined by two points, to find the trace of a cutting plane on any other plane and, in particular, on the plane of any face of a polyhedron, it is sufficient to construct two common points of the planes
To construct the trace of a cutting plane, as well as to construct a section of a polyhedron with this plane, not only the polyhedron, but also the cutting plane must be specified. And the construction of the section plane depends on the specification of this plane. The main ways to define a plane, and in particular a cutting plane, are as follows:
three points not lying on the same line;
a straight line and a point not lying on it;
two parallel lines;
two intersecting lines;
a point and two intersecting lines;
Other ways of specifying a cutting plane are also possible.
Therefore, all methods for constructing sections of polyhedra can be divided into methods.
Methods for constructing sections of polyhedra
The method of sections of polyhedra in stereometry is used in construction problems. It is based on the ability to construct a section of a polyhedron and determine the type of section.
There are three main methods for constructing sections of polyhedra:
Axiomatic method:
Trace method.
Combined method.
Coordinate method.
Note that the trace method and the auxiliary section method are varietiesAxiomatic method for constructing sections.
We can also distinguish the following methods for constructing sections of polyhedra:
constructing a section of a polyhedron with a plane passing through a given point parallel to a given plane;
constructing a section passing through a given line parallel to another given line;
constructing a section passing through a given point parallel to two given intersecting lines;
constructing a section of a polyhedron with a plane passing through a given line perpendicular to a given plane;
constructing a section of a polyhedron with a plane passing through a given point perpendicular to a given straight line.
The main actions that make up the methods for constructing sections are finding the point of intersection of a line with a plane, constructing the line of intersection of two planes, constructing a straight line parallel to the plane, perpendicular to the plane. To construct a line of intersection of two planes, two of its points are usually found and a line is drawn through them. To construct the intersection point of a line and a plane, find a line in the plane that intersects the given one. Then the desired point is obtained at the intersection of the found line with the given one.
Let us consider separately the ones we have listedmethods for constructing sections of polyhedra:
Trace method.
Trace method is based (based on) the axioms of stereometry, the essence of the method is to construct an auxiliary line, which is an image of the line of intersection of the cutting plane with the plane of any face of the figure. It is most convenient to construct an image of the line of intersection of the cutting plane with the plane of the lower base. This linecalled the main trace of the cutting plane . Using a trace, it is easy to construct images of points of the cutting plane located on the lateral edges or faces of the figure. Consistently connecting the images of these points, we obtain an image of the desired section.
Note that that when constructing the main trace of a cutting plane, the following statement is used.
If the points belong to the cutting plane and do not lie on the same straight line, and their projection (central or parallel) onto the plane chosen as the main one, the points are respectively then the points of intersection of the corresponding lines, that is, the points and lie on the same line (Fig. 1, a, b).
Fig.1.a Fig.1.b
This straight line is the main trace of the cutting plane. Since the points lie on the main trace, to construct it it is enough to find two points out of these three.
Method of auxiliary sections.
This method of constructing sections of polyhedra is quite universal. In cases where the desired trace (or traces) of the cutting plane is outside the drawing, this method even has certain advantages. At the same time, it should be borne in mind that constructions performed using this method often turn out to be “crowded.” However, in some cases the method of auxiliary sections turns out to be the most rational.
Combined method
The essence of the combined method for constructing sections of polyhedra is the application of theorems on the parallelism of lines and planes in space in combination with the axiomatic method.
Coordinate method for constructing sections.
The essence of the coordinate method is to calculate the coordinates of the intersection points of the edges or polyhedron with the cutting plane, which is specified by the equation of the plane. The cutting plane equation is calculated based on the problem conditions.
Note , that this method of constructing a section of a polyhedron is acceptable for a computer, since it is associated with a large amount of calculations and therefore this method is advisable to implement using a computer.
Our main task will be to construct a section of a polyhedron with a plane, i.e. in constructing the intersection of these two sets.
Construction of sections of polyhedra
First of all, we note that a section of a convex polyhedron is a convex flat polygon, the vertices of which, in the general case, are the points of intersection of the cutting plane with the edges of the polyhedron, and the sides with its faces.
Examples of constructing sections:
The methods for defining a section are very diverse. The most common of them is the method of defining a cutting plane by three points that do not lie on the same straight line.
Example 1.
For parallelepiped ABCDA 1
B 1
C 1
D 1
. Construct a section passing through points M, N, L.
Solution:
Connect points M and L lying in plane AA 1 D 1 D.
Let us intersect the line ML (belonging to the section) with the edge A 1 D 1 1 D 1 D. Get point X 1 .
Point X1 lies on edge A 1 D 1 , and hence the plane A 1 B 1 C 1 D 1 , we connect it with a stitch N lying in the same plane.
X 1 N intersects edge A 1 B 1 at point K.
Connect points K and M lying in the same plane AA 1 B 1 B.
Let's find the straight line of intersection of the section plane with the DD plane 1
C 1
C:
Let us intersect the line ML (belonging to the section) with the edge DD 1 , they lie in the same plane AA 1 D 1 D, we get point X 2 .
Let us intersect the line KN (belonging to the section) with the edge D 1
C 1
, they lie in the same plane A 1
B 1
C 1
D 1
, we get point X3;
Points X2 and X3 lie in the DD plane 1 C 1 C. Draw a straight line X 2 X 3 , which intersects edge C 1 C at point T, and edge DC at point P. And connect points L and P lying in the plane ABCD.
Thus, the problem is considered solved if all the segments along which the plane intersects the faces of the polyhedron are found, which is what we did. MKNTPL - the required section.
Note. This same problem of constructing a section can be solved using the property of parallel planes.
From the above, you can create an algorithm (rule) for solving problems of this type.
Rules for constructing sections of polyhedra:
draw straight lines through points lying in the same plane;
We are looking for direct intersections of the section plane with the faces of the polyhedron, for this:
Example 2. DL, M
Let's solve using the axiomatic method:
Let's draw an auxiliary planeDKM, which intersects edges AB and BC at points E andF(progress of the solution in Fig. 2.). Let's construct a “trace” of the CM of the section plane on this auxiliary plane, find the point of intersection of the CM and EF– point P. Point P, likeL, lies in the ABC plane, and it is possible to draw a straight line along which the section plane intersects the ABC plane (“trace” of the section in the ABC plane).
Example 3. On the edges AB and AD of the MABCD pyramid, we define points P and Q, respectively, the midpoints of these edges, and on the edge MC we define a point R. Let us construct a section of the pyramid with a plane passing through points P, Q and R.
We will carry out the solution using a combined method:
1). It is clear that the main trace of the plane PQR is the straight line PQ.
2). Let us find the point K at which the MAC plane intersects the straight line PQ. Points K and R belong to both the PQR plane and the MAC plane. Therefore, by drawing the straight line KR, we get the line of intersection of these planes.
3). Let's find the point N=AC BD, draw a straight line MN and find the point F=KR MN.
4). Point F is the common point of the planes PQR and MDB, that is, these planes intersect along a straight line passing through point F. At the same time, since PQ is the midline of the triangle ABD, then PQ is parallel to BD, that is, the line PQ is parallel to the plane MDB. Then the plane PQR passing through the straight line PQ intersects the plane MDB along a straight line parallel to the straight line PQ, that is, parallel and straight BD. Therefore, in the plane MDB through point F we draw a line parallel to line BD.
5). Further constructions are clear from the figure. As a result, we obtain the polygon PQD"RB" - the desired section
Let's consider the cross sections of the prism
for simplicity, that is, convenience of logical thinking, let’s consider the sections of the cube (Fig. 3.a):
Rice. 3.a
Sections of a prism with planes parallel to the side edges are parallelograms. In particular, diagonal sections are parallelograms (Fig. 4).
Def. Diagonal section A prism is cut by a plane passing through two lateral edges that do not belong to the same face.
The polygon resulting from a diagonal section of a prism is a parallelogram. Question about the number of diagonal sectionsn-angle prism is more difficult than the question of the number of diagonals. There will be as many sections as there are diagonals at the base. We know that a convex prism has convex polygons at its bases, and a convex prismn-gon of diagonals. And so we can say that there are half as many diagonal sections as diagonals.
Note: When constructing sections of a parallelepiped in the figure, one should take into account the fact that if a cutting plane intersects two opposite faces along some segments, then these segments are parallel “by the property of a parallelepiped, i.e. The opposite faces of the parallelepiped are parallel and equal.”
We will give answers to frequently asked questions:
What polygons are obtained when a cube is cut by a plane?
"triangle, quadrangle, pentagon, hexagon."
Can a cube be cut by a plane into a heptagon? What about the octagon?
"can not".
3) The question arises: what is the largest number of sides of a polygon obtained by cutting a polyhedron with a plane?
The largest number of sides of a polygon obtained by cutting a polyhedron by a plane is equal to the number of faces of the polyhedron .
Example 3. Construct a cross section of prism A 1 B 1 C 1 D 1 ABCD by a plane passing through three points M, N, K.
Let us consider the case of the location of points M, N, K on the surface of the prism (Fig. 5).
Consider the case: In this case, it is obvious that M1 = B1.
Construction:
Example 4. Construct a section of the parallelepiped ABCDA 1 B 1 C 1 D 1 a plane passing through points M, N, P (the points are indicated in the drawing (Fig. 6)).
Solution:
Rice. 6
Points N and P lie in the section plane and in the plane of the lower base of the parallelepiped. Let's construct a straight line passing through these points. This straight line is the trace of the cutting plane onto the plane of the base of the parallelepiped.
Let us continue the straight line on which side AB of the parallelepiped lies. Lines AB and NP intersect at some point S. This point belongs to the section plane.
Since point M also belongs to the section plane and intersects line AA 1 at some point X.
Points X and N lie in the same plane of face AA 1 D 1 D, connect them and get straight line XN.
Since the planes of the faces of the parallelepiped are parallel, then through the point M we can draw a straight line to the face A 1 B 1 C 1 D 1 , parallel to the line NP. This line will intersect side B 1 WITH 1 at point Y.
Similarly, we draw straight line YZ, parallel to straight line XN. We connect Z with P and get the desired section - MYZPNX.
Sections of a pyramid by planes passing through its apex are triangles. In particular, triangles are diagonal sections. These are sections by planes passing through two non-adjacent lateral edges of the pyramid.
Example 4.
Construct a cross section of the pyramid ABCDplane passing through points K,L,
M.
Solution:
Let's draw another auxiliary planeDCKand construct the intersection point BLAndDK – point E. This point belongs to both auxiliary planes (Fig. 7, b);
Let's find the point of intersection of the segmentsL.M.and EC (these segments lie in the planeBLC, Fig. 7, c) – pointF. DotFlies in the section plane and in the planeDCK;
Let's make a directKFand find the point of intersection of this line withDC– pointN(dotNbelongs to the section). QuadrangleKLNM– the required section.
Let's solve this same example differently .
Let us assume that at points K,L, and M constructed sectionKLNM(Fig. 7). Let us denote byFpoint of intersection of the diagonals of a quadrilateralKLNM. Let's make a directDFand denote byF 1
its point of intersection with edge ABC. DotF 1
coincides with the point of intersection of straight lines AM and SC (F 1
simultaneously belongs to the planes AMDAndDSK). Full stopF 1
easy to build. Next we build a pointFas a point of intersectionDF 1
AndL.M.. Next we find the pointN.
The technique considered is calledinternal design method . (For our case we are talking about central design. QuadrangleKMSA is the projection of a quadrilateralKMNLfrom pointD. In this case, the point of intersection of the diagonalsKMNL- dotF– goes to the point of intersection of the diagonals of the quadrilateralKMSA - pointF 1 .
Sectional area of a polyhedron.
The problem of calculating the cross-sectional area of a polyhedron is usually solved in several stages. If the problem states that a section has been constructed (or that a cutting plane has been drawn, etc.), then at the first stage of the solution the type of figure obtained in the section is determined.
This must be done to select the appropriate formula for calculating the cross-sectional area. After the type of figure obtained in the section has been clarified and a formula has been selected for calculating the area of this figure, we proceed directly to the computational work.
In some cases, it may be easier if, without figuring out the type of figure obtained in the section, you go straight to calculating its area using the formula that follows from the theorem.
Theorem on the area of the orthogonal projection of a polygon: The area of the orthogonal projection of a polygon onto a plane is equal to the product of its area and the cosine of the angle between the plane of the polygon and the projection plane: .
The correct formula for calculating the sectional area is: where is the area of the orthogonal projection of the figure obtained in the section, and this is the angle between the cutting plane and the plane onto which the figure is projected. With this solution, it is necessary to construct an orthogonal projection of the figure obtained in the section and calculate
If the problem statement states that a section needs to be constructed and the area of the resulting section must be found, then at the first stage one should justifiably construct the given section, and then, naturally, determine the type of figure obtained in the section, etc.
Let us note the following fact: since sections of convex polyhedra are constructed, the section polygon will also be convex, so its area can be found by dividing it into triangles, that is, the section area is equal to the sum of the areas of the triangles from which it is composed.
Task 1.
– a regular triangular pyramid with a side of the base equal and a height equal to Construct a section of the pyramid with a plane passing through the points where is the middle of the side, and find its area (Fig. 8).
Solution.
The cross section of a pyramid is a triangle. Let's find its area.
Since the base of the pyramid is an equilateral triangle and the point is the midpoint of the side, it is the height and then, .
The area of a triangle can be found:
Task 2.
The lateral edge of a regular prism is equal to the side of the base. Construct sections of a prism with planes passing through a pointA, perpendicular to the straight line If we find the area of the resulting cross-section of the prism.
Solution.
Let's construct the given section. Let's do this from purely geometric considerations, for example, as follows.
In a plane passing through a given line and a given point, draw a line perpendicular to the line through this point (Fig. 9). For this purpose, let us use the fact that in the triangle that is, its median is also the height of this triangle. So it's straight.
Through the point we draw another line perpendicular to the line. Let us draw it, for example, in a plane passing through a straight line. It is clear that this line is the straight line
So, two intersecting lines are constructed, perpendicular to the line. These lines define a plane passing through a point perpendicular to the line, that is, a secant plane is specified.
Let's construct a section of the prism with this plane. Note that since, the line is parallel to the plane. Then the plane passing through the line intersects the plane along a line parallel to the line, that is, the line. Let's draw a straight line through the point and connect the resulting point with a dot.
Quadrilateral given section. Let's determine its area.
It is clear that a quadrilateral is a rectangle, that is, its area is
rice. 9
Do you know what is called the section of polyhedra by a plane? If you still doubt the correctness of your answer to this question, you can check yourself quite simply. We suggest you take a short test below.
Question. What is the number of the figure that shows the section of a parallelepiped by a plane?
So, the correct answer is in Figure 3.
If you answer correctly, it confirms that you understand what you are dealing with. But, unfortunately, even the correct answer to a test question does not guarantee you the highest grades in lessons on the topic “Sections of polyhedra.” After all, the most difficult thing is not recognizing sections in finished drawings, although this is also very important, but their construction.
To begin with, let us formulate the definition of a section of a polyhedron. So, a section of a polyhedron is a polygon whose vertices lie on the edges of the polyhedron, and whose sides lie on its faces.
Now let’s practice quickly and accurately constructing intersection points 
a given straight line with a given plane. To do this, let's solve the following problem.
Construct the intersection points of straight line MN with the planes of the lower and upper bases of the triangular prism ABCA 1 B 1 C 1, provided that point M belongs to the side edge CC 1, and point N belongs to edge BB 1.
Let's start by extending straight line MN in both directions in the drawing (Fig. 1). Then, in order to obtain the intersection points required by the problem, we extend the lines lying in the upper and lower bases. And now comes the most difficult moment in solving the problem: which lines in both bases need to be extended, since each of them has three lines.
In order to correctly complete the final step of construction, it is necessary to determine which of the direct bases are in the same plane as the straight line MN of interest to us. In our case, this is straight CB in the lower and C 1 B 1 in the upper bases. And it is precisely them that we extend until they intersect with the straight line NM (Fig. 2).
The resulting points P and P 1 are the points of intersection of the straight line MN with the planes of the upper and lower bases of the triangular prism ABCA 1 B 1 C 1 .
After analyzing the presented problem, you can proceed directly to constructing sections of polyhedra. The key point here will be reasoning that will help you arrive at the desired result. As a result, we will eventually try to create a template that will reflect the sequence of actions when solving problems of this type.
So, let's consider the following problem. Construct a section of a triangular prism ABCA 1 B 1 C 1 by a plane passing through points X, Y, Z belonging to edges AA 1, AC and BB 1, respectively.
Solution: Let's draw a drawing and determine which pairs of points lie in the same plane.
Pairs of points X and Y, X and Z can be connected, because they lie in the same plane.
Let's construct an additional point that will lie on the same face as point Z. To do this, we will extend the lines XY and CC 1, because they lie in the plane of the face AA 1 C 1 C. Let's call the resulting point P.
Points P and Z lie in the same plane - in the plane of the face CC 1 B 1 B. Therefore, we can connect them. The straight line PZ intersects the edge CB at a certain point, let's call it T. Points Y and T lie in the lower plane of the prism, connect them. Thus, the quadrilateral YXZT was formed, and this is the desired section. 
Summarize. To construct a section of a polyhedron with a plane, you must:
1) draw straight lines through pairs of points lying in the same plane.
2) find the lines along which the section planes and faces of the polyhedron intersect. To do this, you need to find the intersection points of a straight line belonging to the section plane with a straight line lying in one of the faces.
The process of constructing sections of polyhedra is complicated because it is different in each specific case. And no theory describes it from beginning to end. In fact, there is only one sure way to learn how to quickly and accurately construct sections of any polyhedra - this is constant practice. The more sections you build, the easier it will be for you to do this in the future.
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