What figure is the section of a cube by the kmo plane? “Section of a cube by a plane and their practical application in problems”

Select a polyhedron and difficulty level

Parallelepiped.

Tetrahedron.

Cube Level A.

Level A.

Parallelepiped.

Cube Level B.

Level B.

Tetrahedron.

Level B.

Parallelepiped.

Tetrahedron.

Cube Level C.

Level C.

Cube Level A.

points M, H and K, where KЄ(DCC 1 D 1 ).

V 1

WITH 1

D 1

Help

plane a) with edge BB 1 ; b) plane (SS 1 D).

Cube Level B.

V 1

WITH 1

D 1

Help

Cube Level C.

Construct a section of the cube with a plane passing through points K, E and M (M Є AB). Then find the intersection point of straight line BB 1 with this plane.

V 1

WITH 1

D 1

Cube Level A.

Construct a cross section of a tetrahedron passing through

points M, H and K, where KЄ(DCC 1 D 1 ).

V 1

WITH 1

EP ll MN

D 1

Cube Level B.

V 1

WITH 1

D 1

AN ll KE

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 BB 1 ; b) plane (SS 1 D).

Cube Level C.

Construct a section of the cube with a plane passing through points K, E and M (M Є AB). Then find the point of intersection of straight line BB1 with this plane.

V 1

WITH 1

D 1

PHKERF– required section

Level A. On ribs AA 1 and A 1 D 1 1 1 = 6, A 1 D 1 = 8, AB = 4 cm.

Help

Level B.

Help

LevelC. Three points S, R and L are given on the edges of the parallelepiped. Construct a section of the parallelepiped using the SRL plane.

Help

Level A. On ribs AA 1 and A 1 D 1 of the parallelepiped, the midpoints S and R are taken respectively. Construct a section of a parallelepiped using the plane SRВ 1 and find the cross-sectional area if AA 1 = 6, A 1 D 1 = 8, AB = 4 cm.

Note

Apply Heron's formula.

Level B

SRELZX– required section

Level C.

Tetrahedron.

Level A.

Help

Tetrahedron.

Construct a section of the tetrahedron with a plane passing

Level B.

Help

In the tetrahedron, at the heights of the faces (STA) and (ATV), points K and M are taken,

and point E lies in the plane (ABC). Draw a cross section of a tetrahedron

passing through these points.

Tetrahedron.

Level C.

Help

Construct a section of the tetrahedron with a plane passing

through the middles of the ribs ST, SA and point KЄTV. Define the view

quadrangle obtained in section.

Tetrahedron.

Level A.

Tetrahedron.

Construct a section of the tetrahedron with a plane passing

Level B.

through points M and H and point KЄ(ABC).

MNRE– required section

Tasks on Constructing sections of a cube D1
C1
E
A1
B1
D
A
F
B
WITH

Verification work.

1 option
Option 2
1. tetrahedron
1. parallelepiped
2. Properties of a parallelepiped

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

Secant
the plane intersects the faces of the cube along
segments.
A polygon whose sides are
These segments are called a section of the cube.
The sections of a cube can be triangles,
quadrilaterals, pentagons and
hexagons.
When constructing sections, one should take into account that
fact that if a cutting plane intersects two
opposite faces along some segments, then
these segments are parallel. (Explain why).

B1
C1
D1
A1
M
K
IMPORTANT!
B
WITH
D
If the cutting plane intersects
opposite edges, then it
K DCC1
intersects them in parallel
M BCC1
segments.

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 that are the midpoints of the edges.
Find the perimeter of the section if the edge of the cube is equal to a.
D1
N
K
A1
D
A
C1
B1
M
WITH
B

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 that are its vertices. Find
the perimeter of the section if the edge of the cube is equal to a.
D1
C1
A1
B1
D
A
WITH
B

D1
C1
A1
M
B1
D
A
WITH
B

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.

D1
C1
A1
B1
N
D
A
WITH
B

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

C1
D1
B1
A1
K
D
WITH
N
E
A
M
B

Tasks on Constructing sections of a cube D1
C1
E
A1
B1
D
A
F
B
WITH

Verification work.

1 option
Option 2
1. tetrahedron
1. parallelepiped
2. Properties of a parallelepiped

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

B1
C1
D1
A1
M
K
IMPORTANT!
B
WITH
D
If the cutting plane intersects
opposite edges, then it
K DCC1
intersects them in parallel
M BCC1
segments.

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 that are its vertices. Find
the perimeter of the section if the edge of the cube is equal to a.
D1
C1
A1
B1
D
A
WITH
B

D1
C1
A1
M
B1
D
A
WITH
B

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.

D1
C1
A1
B1
N
D
A
WITH
B

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

C1
D1
B1
A1
K
D
WITH
N
E
A
M
B

General education school of І-ІІІ levels No. 2

Department of Education of the Kirovskoye City Administration

“Cube section by plane

and their practical application in problems.”

Prepared by a math teacher

teacher-methodologist

Chumakova G.V.

2015

Introduction:

Problems on constructing sections of polyhedra occupy a significant place both in high school geometry courses and in exams at various levels. Solving this type of problem contributes to the assimilation of the axioms of stereometry, the systematization of knowledge and skills, the development of spatial understanding and constructive skills. The difficulties that arise when solving problems involving the construction of sections are well known.

The main actions that make up the method of constructing sections are finding the point of intersection of a straight line with a plane, constructing lines of intersection of two planes, constructing a straight line parallel to a plane, and constructing a straight line perpendicular to a plane.

I will illustrate the construction of a section using one problem from a school mathematics course:

№1. Construct at least two sections of the cubeABCDA 1 B 1 C 1 D 1 plane AM 1 C, if point M 1 moves along the segment BB 1 from B to B 1 . Find the boundaries of measuring the height of the section drawn from point M 1 .

Solution: Let's construct two required sections, taking point M 1 closer to point B, and point M 2 closer to B 1 . Both sections are shown in the figure. At the beginning of the movement when point M 1 just moved away from point B 1 , the section is a triangle with base AC and height M 1 O, which is slightly larger than the segment BO, i.e.
If point M 1 will take position M 2 located very close to point B 1 , That AM 2 C will almost coincide with AB 1 C, and its height is M 1 O – with segment B 1 O, whose length is
(OB 1 =
=
).

From here, for reasons of continuity, we conclude:

You should especially look at what happens if point M 1 takes the position of vertex B.

№2. Construct a section of the cube with a plane passing through three points A 1, E and L lying on the edges of the cube.

The planes of the faces A 1 ADD 1 and DD 1 C 1 C intersect along the straight line DD 1 , and the planes of the faces A 1 B 1 C 1 D 1 u DD 1 C 1 C intersect along the straight line D 1 C 1 . By connecting points A and E, we obtain a straight line of intersection of the cutting plane with the plane of the face AA 1 D 1 D, and continuing it, we find the point N, belonging to three planes: the cutting plane and the planes of the faces AA 1 D 1 D u DD 1 C 1 C.

Similarly, we find the point M common to three planes: the section plane and the planes of the faces A 1 B 1 C 1 D 1 u DD 1 C 1 C . Thus, points N u M belong to the cutting plane and the plane DD 1 C 1 C; straight line MN is the line of intersection of the section plane with the plane of the face DD 1 C 1 C, and F and K are the points of its intersection with the edges of the cube CD u CC 1. Consistently connecting points A 1 , E , F , K u L with straight lines, we obtain the pentagon A ! EFKL, which will give us the desired section.

When constructing a section of a cube using a plane X with an arbitrary arrangement of points in the section, the result is: triangle, trapezoid, rectangle, pentagon or hexagon. Naturally, the question arose of how the type of section depends on the type of location of the points defining this section

I decided to conduct a study to find out.

Construct sections of a cube by a plane when three points belonging to edges with one vertex are given.

Three points A 1 , D , C 1 are taken, which belong to the vertex D 1, and themselves are the vertices of the cube.

The cross-section results in an equilateral triangle, since A 1 C 1 , A 1 D u DC 1 are the diagonals of the faces of this cube.

Three dots: A 1 u C 1 are the vertices of the cube, and point F belongs to the edge of the cube DD 1. The points belong to the straight lines emerging from the vertex D 1 .

The cross-section results in an isosceles triangle, since F is equidistant from the points A 1 u C 1 .

Three dots: A 1 u C 1 are the vertices of the cube, and point F belongs to the straight line of the cube edge DD 1. The points belong to straight lines emerging from one vertex D 1 .

The cross-section results in an isosceles trapezoid, since F is equidistant from the points A 1 u C 1 , that is, LA 1 = KC 1 .

Three points belonging to edges with one vertex D 1. The points F u M belong to the continuations of the edges D 1 D u D 1 C, respectively, and the point A 1 is the vertex of the cube.

The cross-section results in a pentagon A 1 KLNG.

Three points F, M and Q are taken so that they lie on the continuation of the edges D 1 D, D 1 C 1, and D 1 A 1, respectively.

The cross-section results in a hexagon KLNGJH.

Three points lie on edges with one vertex D 1.

The cross-section results in an arbitrary triangle, but if the points are arranged so that D 1 Q =D 1 M =D 1 F , that is, if they were equidistant from the vertex D 1, then the cross-section would result in an equilateral triangle.

The cutting plane is defined by points H, Q and M. The cross-section produces a parallelogram, since KC ││ MP and MK ││ PC by the theorem on the intersection of two parallel planes with a third.

If points H, Q and M, define the cutting plane, distant from D, at a distance of 2a, where a is for the edge of the cube, then in the section a regular triangle ACB 1 is obtained.

Conclusion: the three points defining the section belong to three edges of the cube with a common vertex or are their continuation, then the section results in: triangle, pentagon, hexagon, trapezoid, parallelogram.

Constructing a section of a cube by a plane when three points are given, two of which lie on adjacent edges, and the third point lies on an edge not adjacent to them.

Three dots M, K u F, are taken so that M u F belong to edges with one vertex A 1, and point K lies on an edge not adjacent to them.

The cross-section results in a rectangle, since A 1 M = D 1 K and using the theorem of three perpendiculars it can be proven that MKLF is a rectangle, and if A 1 M D 1 K, then you can get a trapezoid or a pentagon.

Three points are taken so that K u L belong to edges emerging from one vertex A 1, and point N belongs to edge CC 1, not adjacent to them. K, L u N of the midpoints of the edges A 1 A, A 1 B 1 u CC 1 – respectively.

The cross-section results in a regular hexagon KLGNHM

Three points are taken so that K u L belong to edges emerging from one vertex A 1, and point T belongs to edge DC.

The cross-section results in a hexagon KLFRTZ.

Three points are taken so that K u L belong to the edges of the cube from one vertex A 1, and point M belongs to the edge DD 1.

The cross section results in a trapezoid LKQM.

Three dots K u L which belong to edges with one vertex A 1 and a point R which lies on edge BC.

The cross section results in a pentagon KLFRT.

Conclusion: If the cutting plane is defined by three points, two of which lie on adjacent edges, and the third on an edge not adjacent to them, then the section can result in a rectangle, pentagon, hexagon, trapezoid.

There is a parallelogram in the cross section of a cube and its special cases.

Points T, H, J defining the section are located so that T.H. AD, H.J. AD. The cross section results in a square HTKJ.

The section is specified by points C, F, L, with DF = FD 1, BL = LB 1. The cross-section produces a rhombus AFCL.

The section is defined by points C, G, H. B 1 H =DG. In cross section there is a parallelogram A 1 GCH.

The points defining the section are the vertices of the cube A, D, C 1. The cross section results in a rectangle

Regular polygons in a cross section of a cube

Triangle ABC 1 is equilateral, since its sides are the diagonals of the faces of the cube.

Triangle KMT is equilateral, since KV = MV = TV.

KMTE is a square, since the section is defined by points M, K, E and MK AD, E.K. AD.

The section has a regular hexagon KMTNEO, since the points H, E, K defining the section are the midpoints of the edges CC 1, DC, AA 1, respectively.

Cube and several problems on stereometry from the Unified State Exam.

In the manual “Unified State Exam 2005. Mathematics. Typical test problems” (Kornikova T. A. et al.) Contains 10 problems (C4) in stereometry, united by a common idea: a triangular prism ABCA is given 1 IN 1 WITH 1 the sides of the base AB and BC are mutually perpendicular and perpendicular to the edge BB 1 , AB=BC=BB 1 , vertex A is the top of the cone (or the center of one of the bases of the cylinder, or the center of the sphere), the base of the cone (the sphere or the second base of the cylinder) passes through the middle of one edge of the prism, its length is known. We need to find the volume or surface of a cone (sphere, cylinder).

General example solution:

Add this prism to a cube. Hexagon DEFKLM - a section of a cube by the plane of the base of a cone, the circle of which passes through the middle A 1 B 1, A is the vertex of the cone, or

DEFKLM is a section of a cube by the plane of the base of a cylinder, the circle of which passes through the middle of A 1 B 1, A is the center of the second base of the cylinder, or it is a section of a cube by the plane of a great circle of a sphere with center A, the sphere of which passes through the middle of A 1 B 1.

HexagonDEFKLM– section of a cube by a plane passing through the middle of edges A 1 IN 1 , BB 1 , VSZh when constructing points are obtainedK, L, M, which are the midpoints of the corresponding edges. The sides of this hexagon are the hypotenuses of the trianglesD.B. 1 E, EBF, FCK, KQL, LRM, M.A. 1 D, the legs of which are equal to half the edge of the cube. Then the center of this hexagon is the center of the circle circumscribed around it, which intersects the edges of the cube at the pointsD, E, F, K, Land M, the radius of this circle
, where A 1 IN 1 = A .

A.O. E.L. T. To. EAL – isosceles:AL = A.E. .

( ABE u EAL– rectangular,AB= AQ= A, BE = L.Q. = )

EO =OL as the midpoint of the diagonal EL of the hexagon DEFKLM, i.e. AO is the median, and according to the properties of an isosceles triangle, the height. AO is proved in a similar way DK. Since AO is perpendicular to two intersecting straight lines of the hexagon plane, then AO is perpendicular to the entire plane.

If A is the vertex of the cone, then AO is its height, if A is the center of the second base of the cylinder, then AO is the height of the cylinder.

ABC: AC=
, P – intersection points of the diagonals of the base of the cube, AP=
, RR 1 =AA 1 = A . OR=RR 1 = , then from the rectangular ROA JSC=
. And so AO=
.

Then, if we are talking about a cone:

(from
).

Answer:

If we are talking about a cylinder:

Answer:

If we are talking about the sphere:

Answer:

Kornikova T. A. and other typical test tasks. Unified State Examination - 2005

Option 6.

Task. Given is a prism ABCA 1 B 1 C 1 and a cylinder. Sides AB and BC of the base of the prism are perpendicular to edge BB 1 and mutually perpendicular. The center of the base of the cylinder is point A 1; the circle of the second base passes through the middle of edge A 1 B 1.

Find the total surface area of the cylinder if BB 1 =AB=BC=10. Find its volume.

Solution:

.
.

Lesson topic: Tasks on constructing sections.

The purpose of the lesson:

Develop skills in solving problems involving constructing sections of a tetrahedron and parallelogram.

During the classes

I. Organizational moment.

II. Checking homework

Answers to questions 14, 15.

14. Is there a tetrahedron with five right angles on its faces?

(Answer: no, because there are only 4 faces, they are triangles, and a triangle with two right angles does not exist.)

15. Is there a parallelepiped that has: a) only one face - a rectangle;

b) only two adjacent rhombus faces; c) all corners of the faces are sharp; d) all angles of the faces are right; e) the number of all sharp edges is not equal to the number of all obtuse angles of the faces?

(Answer: a) no (opposite sides are equal); b) no (for the same reason); c) no (such parallelograms do not exist); d) yes (rectangular parallelepiped); e) no (each face has two acute and two obtuse angles, or all straight lines).

III. Learning new material

Theoretical part. Practical part. Theoretical part.

To solve many geometric problems related to the tetrahedron and parallelepiped, it is useful to be able to draw their sections in 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 (that is, a tetrahedron or parallelepiped). The cutting plane intersects the tetrahedron (parallelepiped) along segments. The polygon that will be formed by these segments is the cross section of the figure. Since a tetrahedron has four faces, its cross-section can be triangles and quadrangles. The parallelepiped has six faces. Its cross-section can be triangles, quadrangles, pentagons, hexagons.

When constructing a section of a parallelepiped, we take into account the fact that if a cutting plane intersects two opposite faces along some segments, then these segments are parallel (property 1, paragraph 11: If two parallel planes are intersected by a third, then the lines of their intersection are parallel).

To construct a section, it is enough to construct the points of intersection of the cutting plane with the edges of the tetrahedron (parallelepiped), and then draw segments connecting each two constructed points lying on the same face.

Can a tetrahedron be cut by a plane into the quadrilateral shown in the figure?

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

E, F, G,

let's make a direct E.F. and denote P its point of intersection with AD.

Let's denote Q point of intersection of lines PG And AB.

Let's connect the dots E And Q, F And G.

The resulting trapezoid EFGQ will be the desired section.

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2.4. Construct a section of a cube with a plane passing through the points E, F, lying on the edges of the cube and the vertex B.

Solution. To construct a section of a cube passing through points E, F and the top B,

Let's connect the points with segments E And B, F And B.

Through dots E And F let's draw parallel lines B.F. And BE, respectively.

The resulting parallelogram BFGE will be the desired section.

2.5. Construct a section of a cube with a plane passing through the 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,

let's make a direct E.F. and denote P its point of intersection with AD.

Let's denote Q,R line intersection points PG With AB And DC.

Let's denote S intersection point FR c SS 1.

Let's connect the dots E And Q, G And S.

The resulting pentagon EFSGQ will be the desired section.

2.6. Construct a section of a cube with a plane passing through the 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,

let's find a point P intersection of a straight line E.F. and face plane ABCD.

Let's denote Q, R line intersection points PG With AB And CD.

Let's make a direct RF and denote S, T its points of intersection with CC 1 and DD 1.

Let's make a direct T.E. and denote U its point of intersection with A 1D 1.

Let's connect the dots E And Q, G And S, F and U.

The resulting hexagon EUFSGQ will be the desired section.

2.7. Construct a cross section of a tetrahedron ABCD AD and passing through the points E, F.

Solution. Let's connect the dots E And F. Through the pointF let's draw a straight lineFG, parallelA.D.

Let's connect the dots G And E.

The resulting triangle EFG will be the desired section.

2.8. Construct a cross section of a tetrahedron ABCD plane parallel to the edge CD and passing through the points E, F .

Solution. Through dots E And F let's draw straight lines E.G. And FH, parallel CD.

Let's connect the dots G And F, E And H.

The resulting triangle EFG will be the desired section.

2.9. Construct a cross section of a tetrahedron ABCD plane passing through the points E, F, G.

Solution. To construct a section of a tetrahedron passing through points E, F, G,

let's make a direct E.F. and denote P its point of intersection with BD.

Let's denote Q point of intersection of lines PG And CD.

Let's connect the dots F And Q, E And G.

The resulting quadrilateral EFQG will be the desired section.

IV. Lesson summary.

V. Homework p.14, p.27 No. 000 – option 1, 2.