A branch of mathematics that studies the properties of various shapes (points, lines, angles, two-dimensional and three-dimensional objects), their sizes and relative position. For the convenience of teaching, geometry is divided into planimetry and solid geometry. V… … Collier Encyclopedia

Geometry of spaces of dimension greater than three; the term is applied to those spaces whose geometry was originally developed for the case of three dimensions and only then generalized to the number of dimensions n> 3, primarily Euclidean space, ... ... Mathematical Encyclopedia

N dimensional Euclidean geometry is a generalization of Euclidean geometry to a space of more dimensions. Although the physical space is three-dimensional, and the human senses are designed to perceive three dimensions, N is dimensional ... ... Wikipedia

This term has other meanings, see Pyramidatsu (meanings). The reliability of this section of the article has been questioned. It is necessary to verify the accuracy of the facts stated in this section. There may be explanations on the talk page ... Wikipedia

- (Constructive Solid Geometry, CSG) technology used in modeling solids. Structural block geometry is often, but not always, a modeling technique in 3D graphics and CAD. It allows you to create a complex scene or ... Wikipedia

Constructive Solid Geometry (CSG) is a technology used in modeling solids. Structural block geometry is often, but not always, a modeling technique in 3D graphics and CAD. She ... ... Wikipedia

This term has other meanings, see Scope (meanings). Volume is an additive function of a set (measure) that characterizes the capacity of a region of space that it occupies. Initially, it arose and was applied without strict ... ... Wikipedia

Cube Type Regular polyhedron Face square Vertices Edges Faces ... Wikipedia

Volume is an additive function of a set (measure) that characterizes the capacity of a region of space that it occupies. Initially, it arose and was applied without a strict definition in relation to three-dimensional bodies of three-dimensional Euclidean space. ... ... Wikipedia

A portion of space bounded by a collection of a finite number of planar polygons (see GEOMETRY) connected in such a way that each side of any polygon is a side of exactly one other polygon (called ... ... Collier Encyclopedia

Prismatic polyhedron is a generalization of the prism in spaces of dimension 4 and higher. n-dimensional prismatic polyhedron is constructed from two ( n− 1 )-dimensional polytopes carried over to the next dimension.

Prismatic elements n-dimensional polyhedron are doubled from the elements ( n− 1 )-dimensional polyhedron, then new elements of the next level are created.

Let's take n-dimensional polyhedron with elements f i (\displaystyle f_(i)) (i-dimensional edge, i = 0, ..., n). Prismatic ( n + 1 (\displaystyle n+1))-dimensional polyhedron will have 2 f i + f − 1 (\displaystyle 2f_(i)+f_(-1)) dimension elements i(at f − 1 = 0 (\displaystyle f_(-1)=0), f n = 1 (\displaystyle f_(n)=1)).

By dimensions:

• Take a polygon n peaks and n parties. Get a prism with 2 n pinnacles, 3 n ribs and 2+n (\displaystyle 2+n) faces.
• We take a polyhedron with v peaks, e ribs and f faces. We get a (4-dimensional) prism with 2 v vertices, edges, faces and 2+f (\displaystyle 2+f) cells.
• We take a 4-dimensional polyhedron with v peaks, e ribs f faces and c cells. We get a (5-dimensional) prism with 2 v peaks, 2e+v (\displaystyle 2e+v) ribs 2 f + e (\displaystyle 2f+e)(2-dimensional) faces, 2 c + f (\displaystyle 2c+f) cells and 2+c (\displaystyle 2+c) hypercells.

Uniform prismatic polyhedra

Right n-polytope represented by the Schläfli symbol ( p, q, ..., t), can form a uniform prismatic polyhedron of dimension ( n+ 1 ) represented by the direct product of two Schläfli symbols: ( p, q, ..., t}×{}.

By dimensions:

• A prism from a 0-dimensional polyhedron is a line segment, represented by the empty Schläfli symbol ().
• A prism from a 1-dimensional polyhedron is a rectangle obtained from two line segments. This prism is represented as a product of the Schläfli symbols ()×(). If the prism is a square, the notation can be abbreviated: ()×() = (4).
• a polygonal prism is a 3-dimensional prism made from two polygons (one obtained by translating the other in parallel) that are connected by rectangles. From a regular polygon ( p) you can get a homogeneous n-coal prism represented by the product ( p)×(). If p= 4 , the prism becomes a cube: (4)×() = (4, 3).
• A 4-dimensional prism obtained from two polyhedra (one obtained by parallel translation of the other), with connecting 3-dimensional prismatic cells. From regular polyhedron {p, q) one can obtain a homogeneous 4-dimensional prism represented by the product ( p, q)×(). If the polyhedron is a cube and the sides of the prism are also cubes, the prism becomes a tesseract: (4, 3)×() = (4, 3, 3).

Higher-dimensional prismatic polyhedra also exist as direct products of any two polyhedra. The dimension of a prismatic polyhedron is equal to the product of the dimensions of the elements of the product. The first example of such a product exists in 4-dimensional space and is called duoprisms, which are obtained by multiplying two polygons. Regular duoprisms are represented by the symbol ( p}×{ q}.

Your privacy is important to us. For this reason, we have developed a Privacy Policy that describes how we use and store your information. Please read our privacy policy and let us know if you have any questions.

Collection and use of personal information

Personal information refers to data that can be used to identify a specific person or contact him.

You may be asked to provide your personal information at any time when you contact us.

The following are some examples of the types of personal information we may collect and how we may use such information.

What personal information we collect:

• When you submit an application on the site, we may collect various information, including your name, phone number, address Email etc.

How we use your personal information:

• Collected by us personal information allows us to contact you and inform you about unique offers, promotions and other events and upcoming events.
• From time to time, we may use your personal information to send you important notices and communications.
• We may also use personal information for internal purposes, such as conducting audits, data analysis and various research in order to improve the services we provide and provide you with recommendations regarding our services.
• If you enter a prize draw, contest or similar incentive, we may use the information you provide to administer such programs.

Disclosure to third parties

We do not disclose information received from you to third parties.

Exceptions:

• In the event that it is necessary - in accordance with the law, judicial order, in legal proceedings, and / or based on public requests or requests from state bodies in the territory of the Russian Federation - disclose your personal information. We may also disclose information about you if we determine that such disclosure is necessary or appropriate for security, law enforcement, or other public interest reasons.
• In the event of a reorganization, merger or sale, we may transfer the personal information we collect to the relevant third party successor.

Protection of personal information

We take precautions - including administrative, technical and physical - to protect your personal information from loss, theft, and misuse, as well as from unauthorized access, disclosure, alteration and destruction.

Maintaining your privacy at the company level

To ensure that your personal information is secure, we communicate privacy and security practices to our employees and strictly enforce privacy practices.

In the school curriculum for the course of solid geometry, the study of three-dimensional figures usually begins with a simple geometric body - a prism polyhedron. The role of its bases is performed by 2 equal polygons lying in parallel planes. A special case is a regular quadrangular prism. Its bases are 2 identical regular quadrangles, to which the sides are perpendicular, having the shape of parallelograms (or rectangles if the prism is not inclined).

What does a prism look like

A regular quadrangular prism is a hexagon, at the bases of which there are 2 squares, and the side faces are represented by rectangles. Another name for this geometric figure is a straight parallelepiped.

The figure, which depicts a quadrangular prism, is shown below.

You can also see in the picture essential elements, which make up the geometric body. They are commonly referred to as:

Sometimes in problems in geometry you can find the concept of a section. The definition will sound like this: a section is all points of a volumetric body that belong to the cutting plane. The section is perpendicular (crosses the edges of the figure at an angle of 90 degrees). For a rectangular prism, a diagonal section is also considered (the maximum number of sections that can be built is 2), passing through 2 edges and the diagonals of the base.

If the section is drawn in such a way that the cutting plane is not parallel to either the bases or the side faces, the result is a truncated prism.

Various ratios and formulas are used to find the reduced prismatic elements. Some of them are known from the course of planimetry (for example, to find the area of ​​the base of a prism, it is enough to recall the formula for the area of ​​a square).

Surface area and volume

To determine the volume of a prism using the formula, you need to know the area of ​​\u200b\u200bits base and height:

V = Sprim h

Since the base of a regular tetrahedral prism is a square with side a, You can write the formula in a more detailed form:

V = a² h

If we are talking about a cube - a regular prism with equal length, width and height, the volume is calculated as follows:

To understand how to find the lateral surface area of ​​a prism, you need to imagine its sweep.

It can be seen from the drawing that the side surface is made up of 4 equal rectangles. Its area is calculated as the product of the perimeter of the base and the height of the figure:

Sside = Pos h

Since the perimeter of a square is P = 4a, the formula takes the form:

Sside = 4a h

For cube:

Sside = 4a²

To calculate the area full surface prisms, you need to add 2 base areas to the lateral area:

Sfull = Sside + 2Sbase

As applied to a quadrangular regular prism, the formula has the form:

Sfull = 4a h + 2a²

For the surface area of ​​a cube:

Sfull = 6a²

Knowing the volume or surface area, you can calculate the individual elements of a geometric body.

Finding prism elements

Often there are problems in which the volume is given or the value of the lateral surface area is known, where it is necessary to determine the length of the side of the base or the height. In such cases, formulas can be derived:

• base side length: a = Sside / 4h = √(V / h);
• height or side rib length: h = Sside / 4a = V / a²;
• base area: Sprim = V / h;
• side face area: Side gr = Sside / 4.

To determine how much area a diagonal section has, you need to know the length of the diagonal and the height of the figure. For a square d = a√2. Therefore:

Sdiag = ah√2

To calculate the diagonal of the prism, the formula is used:

dprize = √(2a² + h²)

To understand how to apply the above ratios, you can practice and solve a few simple tasks.

Examples of problems with solutions

Here are some of the tasks that appear in the state final exams in mathematics.

Exercise 1.

Sand is poured into a box shaped like a regular quadrangular prism. The height of its level is 10 cm. What will the level of sand be if you move it into a container of the same shape, but with a base length 2 times longer?

It should be argued as follows. The amount of sand in the first and second containers did not change, i.e., its volume in them is the same. You can define the length of the base as a. In this case, for the first box, the volume of the substance will be:

V₁ = ha² = 10a²

For the second box, the length of the base is 2a, but the height of the sand level is unknown:

V₂ = h(2a)² = 4ha²

Insofar as V₁ = V₂, the expressions can be equated:

10a² = 4ha²

After reducing both sides of the equation by a², we get:

As a result, the new sand level will be h = 10 / 4 = 2.5 cm.

Task 2.

ABCDA₁B₁C₁D₁ is a regular prism. It is known that BD = AB₁ = 6√2. Find the total surface area of ​​the body.

To make it easier to understand which elements are known, you can draw a figure.

Since we are talking about a regular prism, we can conclude that the base is a square with a diagonal of 6√2. The diagonal of the side face has the same value, therefore, the side face also has the shape of a square equal to the base. It turns out that all three dimensions - length, width and height - are equal. We can conclude that ABCDA₁B₁C₁D₁ is a cube.

The length of any edge is determined through the known diagonal:

a = d / √2 = 6√2 / √2 = 6

The total surface area is found by the formula for the cube:

Sfull = 6a² = 6 6² = 216

Task 3.

The room is being renovated. It is known that its floor has the shape of a square with an area of ​​9 m². The height of the room is 2.5 m. What is the lowest cost of wallpapering a room if 1 m² costs 50 rubles?

Since the floor and ceiling are squares, that is, regular quadrilaterals, and its walls are perpendicular to horizontal surfaces, we can conclude that it is a regular prism. It is necessary to determine the area of ​​its lateral surface.

The length of the room is a = √9 = 3 m.

The square will be covered with wallpaper Sside = 4 3 2.5 = 30 m².

The lowest cost of wallpaper for this room will be 50 30 = 1500 rubles.

Thus, to solve problems for a rectangular prism, it is enough to be able to calculate the area and perimeter of a square and a rectangle, as well as to know the formulas for finding the volume and surface area.

How to find the area of ​​a cube

Diagonal sections The section of a prism by a plane passing through the diagonal of the base and two side edges adjacent to it is called the diagonal section of the prism. The section of the pyramid by a plane passing through the diagonal of the base and the top is called the diagonal section of the pyramid. Let the plane intersect the pyramid and be parallel to its base. The part of the pyramid enclosed between this plane and the base is called the truncated pyramid. The section of the pyramid is also called the base of the truncated pyramid.

Construction of sections When constructing sections of polyhedrons, the basic ones are the construction of the point of intersection of a straight line and a plane, as well as the line of intersection of two planes. If two points A and B of the line are given and their projections A' and B' on the plane are known, then the point C of the intersection of the given line and the plane will be the point of intersection of the lines AB and A'B' If three points A, B, C of the plane are given and known their projections A', B', C' onto another plane, then to find the line of intersection of these planes, find the points P and Q of intersection of lines AB and AC with the second plane. The line PQ will be the desired line of intersection of the planes.

Exercise 1 Construct a section of a cube by 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, Let's connect points E and B, F and B with segments. Draw lines through points E and F parallel to BF and BE, respectively. The resulting parallelogram BFGE will be the required section.

Exercise 2 Construct a section of a cube by 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 draw a line EF and denote by P its intersection point with AD. Let Q denote the point of intersection of lines PG and AB. Connect the points E and Q, F and G. The resulting trapezoid EFGQ will be the required section.

Exercise 3 Construct a section of a cube by 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 draw a line EF and denote by P its intersection point with AD. Let Q, R denote the points of intersection of the line PG with AB and DC. Denote by S the point of intersection of FR with СС 1. Connect the points E and Q, G and S. The resulting pentagon EFSGQ will be the required section.

Exercise 4 Construct a section of a cube by 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 the intersection of the line EF and the plane of the face ABCD. Denote by Q, R the points of intersection of the line PG with AB and CD. Draw line RF and denote by S, T its points of intersection with CC 1 and DD 1. Draw line TE and denote by U its point of intersection with A 1 D 1. Connect the points E and Q, G and S, U and F. The resulting hexagon EUFSGQ will be the required section.

Exercise 5 Construct a section of a cube by a plane passing through the points E, F, G belonging to the faces BB 1 C 1 C, CC 1 D 1 D, AA 1 B 1 B, respectively. Solution. From these points we drop the perpendiculars EE', FF', GG' onto the plane of the face ABCD, and find the points I and H of the intersection of the lines FE and FG with this plane. IH will be the line of intersection of the desired plane and the plane of the face ABCD. Denote by Q, R the points of intersection of the line IH with AB and BC. Draw lines PG and QE and denote by R, S their points of intersection with AA 1 and CC 1. Draw lines SU, UV and RV parallel to PR, PQ and QS. The resulting hexagon RPQSUV will be the required section.

Exercise 6 Construct a section of a cube by a plane passing through points E, F, lying on the edges of the cube, parallel to the diagonal BD. Solution. Let's draw lines FG and EH parallel to BD. Draw a line FP parallel to EG and connect the points P and G. Connect the points E and G, F and H. The resulting pentagon EGPFH will be the required section.

Construct a section of the prism ABCA 1 B 1 C 1 by a plane passing through the points E, F, G. Exercise 8 Solution. Connect the points E and F. Draw the line FG and denote its intersection point with CC 1 as H. Draw the line EH and denote its intersection point with A 1 C 1 as I. Connect the points I and G. The resulting quadrilateral EFGI will be the required section.

Construct a section of the prism ABCA 1 B 1 C 1 by a plane passing through points E, F, G. Exercise 9 Solution. Draw a line EG and denote by H and I its points of intersection with CC 1 and AC. Draw the line IF and denote its intersection point with AB as K. Draw the line FH and denote its intersection point with B 1 C 1 as L. Connect the points E and K, G and L. The resulting pentagon EKFLG will be the required section.

Construct a section of the prism ABCA 1 B 1 C 1 by a plane parallel to AC 1 passing through the points D 1. Exercise 10 Solution. Through the point D draw a line parallel to AC 1 and denote by E its point of intersection with the line BC 1. This point will belong to the plane of the face ADD 1 A 1. Draw the line DE and denote by F its point of intersection with the edge BC. Connect the points F and D by a segment. Through the point D draw a line parallel to the line FD and denote by G the point of its intersection with the edge A 1 C 1, H is the point of its intersection with the line A 1 B 1. Draw the line DH and denote by P its point of intersection with edge AA 1. Connect the points P and G with a segment. The resulting quadrilateral EFIK will be the required section.

Construct a section of the prism ABCA 1 B 1 C 1 by a plane passing through the points E on the edge BC, F on the face ABB 1 A 1 and G on the face ACC 1 A 1. Exercise 11 Solution. Draw the line GF and find the point H of its intersection with the plane ABC. Draw a line EH and denote by P and I its points of intersection with AC and AB. Draw lines PG and IF, and denote by S, R and Q their points of intersection with A 1 C 1, A 1 B 1 and BB 1. Connect the points E and Q, S and R. The resulting pentagon EQRSP will be the required section.

Construct a section of a regular hexagonal prism by a plane passing through points A, B, D 1. Exercise 12 Solution. Note that the section will pass through the point E 1. Draw the line AB and find its intersection points K and L with the lines CD and FE. Draw lines KD 1, LE 1 and find their intersection points P, Q with lines CC 1 and FF 1. Hexagon ABPD 1 E 1 Q will be the required section.

Construct a section of a regular hexagonal prism by a plane passing through points A, B', F'. Exercise 13 Solution. Let's draw segments AB' and AF'. Draw a line through the point B' parallel to AF', and denote its intersection point with EE 1 as E'. Through the point F' we draw a line parallel to AB', and its point of intersection with CC 1 is denoted by C'. Through the points E’ and C’ we draw lines parallel to AB’ and AF’, and their points of intersection with D 1 E 1 and C 1 D 1 will be denoted by D’, D”. Let's connect the points B', C'; D', D"; F', E'. The resulting heptagon AB’C’D”D’E’F’ will be the desired section.

Construct a section of a regular hexagonal prism by a plane passing through the points F', B', D'. Exercise 14 Solution. Draw the lines F'B' and F'D' and find their points of intersection P and Q with the plane ABC. Let's draw a line PQ. Denote by R the intersection point of PQ and FC. The intersection point of F'R and CC 1 will be denoted by C'. Connect points B', C' and C', D'. Through the point F' we draw lines parallel to C'D' and B'C', and their points of intersection with AA 1 and EE 1 will be denoted by A' and E'. Let's connect points A', B' and E', D'. The resulting hexagon A'B'C'D'E'F' will be the desired section.