A branch of mathematics that deals with the study of the properties of various figures (points, lines, angles, two-dimensional and three-dimensional objects), their sizes and relative positions. For ease of teaching, geometry is divided into planimetry and stereometry. IN… … Collier's Encyclopedia

Geometry of spaces of dimensions 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 physical space is three-dimensional, and 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. You must 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 solid bodies. Constructive block geometry is often, but not always, the way to model in 3D graphics and CAD. It allows you to create a complex scene or... Wikipedia

Constructive Solid Geometry (CSG) is a technology used in solid modeling. Constructive block geometry is often, but not always, the way to model in 3D graphics and CAD. She... ... Wikipedia

This term has other meanings, see Volume (meanings). Volume is an additive function of a set (a measure) characterizing the capacity of the area of ​​space that it occupies. Initially 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 (a measure) characterizing the capacity of the area 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's Encyclopedia

Diagonal sections The section of a prism by a plane passing through the diagonal of the base and the two side edges adjacent to it is called the diagonal section of the prism. A section of a pyramid with a plane passing through the diagonal of the base and the top is called a 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 a truncated pyramid. The cross section of a pyramid is also called the base of a truncated pyramid.

Construction of sections When constructing sections of polyhedra, 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 a line are given and their projections A' and B' onto the plane are known, then the point C of the intersection of the data of the 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 are known their projections A', B', C' onto another plane, then to find the line of intersection of these planes, the points P and Q of intersection of the lines AB and AC with the second plane are found. The straight line PQ will be the desired intersection line of the planes.

Exercise 1 Construct a section of a 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, we 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.

Exercise 2 Construct a section of a 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 denote P its point of intersection with AD. Let Q denote the point of intersection of lines PG and AB. Let's connect points E and Q, F and G. The resulting trapezoid EFGQ will be the desired section.

Exercise 3 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 denote 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. The resulting pentagon EFSGQ will be the desired section.

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

Exercise 5 Construct a section of the cube by a plane passing through 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 lower the perpendiculars EE’, FF’, GG’ to 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. Let us denote by Q, R the points of intersection of the straight line IH with AB and BC. Let's draw lines PG and QE and denote R, S their points of intersection with AA 1 and CC 1. Let's draw lines SU, UV and RV, parallel to PR, PQ and QS. The resulting hexagon RPQSUV will be the desired section.

Exercise 6 Construct a section of the cube with a plane passing through points E, F lying on the edges of the cube, parallel to the diagonal BD. Solution. Let us draw lines FG and EH parallel to BD. Let's draw a straight line FP parallel to EG and connect points P and G. Connect 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 with a plane passing through points E, F, G. Exercise 8 Solution. Let's connect points E and F. Let's draw a line FG and its point of intersection with CC 1, denote H. Let's draw a line EH and its point of intersection with A 1 C 1, denote I. Let's connect points I and G. The resulting quadrilateral EFGI will be the desired section.

Construct a section of the prism ABCA 1 B 1 C 1 with a plane passing through points E, F, G. Exercise 9 Solution. Let's draw a straight line EG and denote H and I its points of intersection with CC 1 and AC. Let's draw a straight line IF and its point of intersection with AB we'll denote K. We'll draw a line FH and its point of intersection with B 1 C 1 we'll denote L. Let's connect the points E and K, G and L. The resulting pentagon EKFLG will be the desired section.

Construct a section of the prism ABCA 1 B 1 C 1 with a plane parallel to AC 1 passing through points D 1. Exercise 10 Solution. Through point D we draw a line parallel to AC 1 and denote E its point of intersection with line BC 1. This point will belong to the plane of the face ADD 1 A 1. Draw a line DE and denote F its point of intersection with edge BC. Let us connect points F and D with a segment. Through point D we draw a line parallel to straight line FD and denote by G the point of its intersection with the edge A 1 C 1, H – the point of its intersection with line A 1 B 1. Let us draw a line DH and denote by P its point of intersection with edge AA 1. Connect 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 points E on edge BC, F on face ABB 1 A 1 and G on face ACC 1 A 1. Exercise 11 Solution. Let's draw a line GF and find the point H of its intersection with the plane ABC. Let us draw a straight line EH, and denote by P and I its points of intersection with AC and AB. Let's draw straight lines PG and IF, and denote S, R and Q their points of intersection with A 1 C 1, A 1 B 1 and BB 1. Let's connect points E and Q, S and R. The resulting pentagon EQRSP will be the desired section.

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

Construct a section of a regular hexagonal prism with a plane passing through points A, B’, F’. Exercise 13 Solution. Let’s draw segments AB’ and AF’. Through point B' we draw a line parallel to AF', and its point of intersection with EE 1 we denote E'. Through point F' we draw a line parallel to AB', and its point of intersection with CC 1 we denote as 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 we denote as D’, D”. Let’s connect 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 with a plane passing through points F’, B’, D’. Exercise 14 Solution. Let’s draw straight lines F’B’ and F’D’ and find their intersection points P and Q with plane ABC. Let's do a direct PQ. Let R denote the intersection point of PQ and FC. Let us denote the intersection point of F’R and CC 1 as C’. Let’s connect points B’, C’ and C’, D’. Through point F' we draw lines parallel to C'D' and B'C', and their points of intersection with AA 1 and EE 1 we denote as 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.

The answer to this question “what is a prism?”, as is the case with any geometric term, becomes clear if you study the properties of a given object. Of course, you can memorize a complex scientific term, according to which a prism is one of the types of polyhedra, the bases of which are parallel and the side faces are parallelograms, but it is easier to remember the properties of the object and then you can even independently formulate the concept of a prism.

Prism elements

The rather simple properties of a prism are difficult to understand without first studying a number of terms that are used to designate certain elements of a given geometric body. The following prism elements are distinguished:

• Each prism has two bases, they are polygons and are located in parallel planes.
• Side faces - all faces of the prism (except for the bases).
• Lateral surface - a set of lateral faces.
• A complete surface is a set of side faces and bases.
• Lateral edges are common to the side faces.
• Height is a segment drawn from one base to another perpendicular to the planes in which they are located.
• Diagonal - a segment drawn from one vertex of a prism to another.
• Diagonal plane - a plane that passes through one of the side edges of the prism and the diagonal of one of the bases.
• Diagonal section - a section formed by the intersection of a prism and a diagonal plane.
• Orthogonal section - a section formed by the intersection of a prism and a plane that is perpendicular to the side edge.
• Prism development - representation of all faces of a prism on one plane without distorting the sizes of the faces.

Prism properties

Now that you are familiar with the elements of a prism, you can consider its basic properties, as well as formulas that allow you to find the volume and area of ​​a figure:

• The bases of the prism are equal polygons.
• The lateral faces of the prism are parallelograms.
• All lateral edges of the prism are equal and parallel to each other.
• The orthogonal section is perpendicular to all lateral ribs.

Formulas for calculating area and volume

To find the volume of a prism, there is a very simple formula: V = S*h, where S is the area of ​​the prism, h is the height.

To find the total surface area of ​​a prism, you need to find the area of ​​its lateral surface and multiply the resulting value by twice the base area. In turn, to find the area of ​​the side surface, you can use the formula: S = P*l, where P is the perpendicular section perimeter, l is the length of the side rib.

Special types of prism

Some prisms have special distinctive properties, and special names have been invented for them:

• parallelepiped (sign - parallelograms at the base);
• straight prism (sign - the side ribs are perpendicular to the bases);
• regular prism (feature - a polygon with equal sides and angles at the base, rectangles at the bases);
• semi-regular prism (sign - squares at the bases).

Prism in optics

In optics, a prism is an object in the shape of a geometric body (prism) made of transparent material. The properties of prisms are widely used in optics, in particular in binoculars. Prismatic binoculars use a double Porro prism and an Abbe prism, named after their inventors. These prisms, due to their special structure and arrangement, create one or another optical effect.

A Porro prism is a prism whose base is an isosceles triangle. A double Porro prism is created due to the special arrangement in space of two Porro prisms. The double Porro prism allows you to flip the image, increase the optical distance between the lens and the eyepiece, while maintaining external dimensions.

An Abbe prism is a prism whose base is a triangle with angles of 30°, 60°, 90°. An Abbe prism is used when it is necessary to invert an image without deviating the line of sight to the object.

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Definition 1. A polyhedron, two of whose faces are polygons of the same name lying in parallel planes, and any two edges not lying in these planes are parallel, is called a prism. The term “prism” is of Greek origin and literally means “sawed off” (body). Polygons lying in parallel planes are called prism bases, and the remaining faces are called lateral faces. The surface of the prism thus consists of two equal polygons (bases) and parallelograms (side faces). There are triangular, quadrangular, pentagonal, etc. prisms. depending on the number of vertices of the base.

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All prisms are divided into straight and inclined. (Fig. 2) If the lateral edge of a prism is perpendicular to the plane of its base, then such a prism is called a straight one; If the lateral edge of a prism is perpendicular to the plane of its base, then such a prism is called inclined. A straight prism has rectangular side faces. A perpendicular to the planes of the bases, the ends of which belong to these planes, is called the height of the prism.

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Properties of a prism. 1. The bases of the prism are equal polygons. 2. The lateral faces of the prism are parallelograms. 3. The lateral edges of the prism are equal.

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The surface area of ​​the prism and the lateral surface area of ​​the prism. The surface of a polyhedron consists of a finite number of polygons (faces). The surface area of ​​a polyhedron is the sum of the areas of all its faces. The surface area of ​​the prisms (Spr) is equal to the sum of the areas of its lateral faces (side surface area Sside) and the areas of two bases (2Sbas) - equal polygons: Spop = Sside + 2Sbas. Theorem. The area of ​​the lateral surface of the prism is equal to the product of the perimeter of its perpendicular section and the length of the lateral edge.

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Proof. The lateral faces of a straight prism are rectangles, the bases of which are the sides of the base of the prism, and the heights are equal to the height h of the prism. Sside of the prism surface is equal to the sum S of the indicated triangles, i.e. equal to the sum of the products of the sides of the base and the height h. Taking the factor h out of brackets, we obtain in brackets the sum of the sides of the base of the prism, i.e. perimeter P. So, Sside = Ph. The theorem has been proven. Consequence. The lateral surface area of ​​a straight prism is equal to the product of the perimeter of its base and its height. Indeed, in a straight prism, the base can be considered as a perpendicular section, and the lateral edge is the height.

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Section of a prism 1. Section of a prism by a plane parallel to the base. The section forms a polygon equal to the polygon lying at the base. 2. Section of a prism by a plane passing through two non-adjacent lateral edges. A parallelogram is formed in the cross section. This section is called the diagonal section of the prism. In some cases, the result may be a diamond, rectangle or square.

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Definition 2. A right prism, the base of which is a regular polygon, is called a regular prism. Properties of a regular prism 1. The bases of a regular prism are regular polygons. 2. The lateral faces of a regular prism are equal rectangles. 3. The lateral edges of a regular prism are equal.

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Section of a regular prism. 1. Section of a regular prism with a plane parallel to the base. The section forms a regular polygon equal to the polygon lying at the base. 2. Section of a regular prism by a plane passing through two non-adjacent lateral edges. A rectangle is formed in the cross-section. In some cases, a square may form.

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Symmetry of a regular prism 1. The center of symmetry with an even number of sides of the base is the point of intersection of the diagonals of a regular prism (Fig. 6)

A triangular prism is a three-dimensional solid formed by connecting rectangles and triangles. In this lesson you will learn how to find the size of the inside (volume) and outside (surface area) of a triangular prism.

Triangular prism is a pentahedron formed by two parallel planes in which two triangles are located, forming two faces of a prism, and the remaining three faces are parallelograms formed from the sides of the triangles.

Elements of a triangular prism

Triangles ABC and A 1 B 1 C 1 are prism bases .

The quadrilaterals A 1 B 1 BA, B 1 BCC 1 and A 1 C 1 CA are lateral faces of the prism .

The sides of the faces are prism ribs(A 1 B 1, A 1 C 1, C 1 B 1, AA 1, CC 1, BB 1, AB, BC, AC), a triangular prism has 9 faces in total.

The height of a prism is the perpendicular segment that connects the two faces of the prism (in the figure it is h).

The diagonal of a prism is a segment that has ends at two vertices of the prism that do not belong to the same face. For a triangular prism such a diagonal cannot be drawn.

Base area is the area of ​​the triangular face of the prism.

is the sum of the areas of the quadrangular faces of the prism.

Types of triangular prisms

There are two types of triangular prism: straight and inclined.

A straight prism has rectangular side faces, and an inclined prism has parallelogram side faces (see figure)

A prism whose side edges are perpendicular to the planes of the bases is called a straight line.

A prism whose side edges are inclined to the planes of the bases is called inclined.

Basic formulas for calculating a triangular prism

Volume of a triangular prism

To find the volume of a triangular prism, you need to multiply the area of ​​its base by the height of the prism.

Prism volume = base area x height

V=S basic h

Prism lateral surface area

To find the lateral surface area of ​​a triangular prism, you need to multiply the perimeter of its base by its height.

Lateral surface area of ​​a triangular prism = base perimeter x height

S side = P main h

Total surface area of ​​the prism

To find the total surface area of ​​a prism, you need to add its base area and lateral surface area.

since S side = P main. h, then we get:

S full turn =P basic h+2S basic

Correct prism - a straight prism whose base is a regular polygon.

Prism properties:

The upper and lower bases of the prism are equal polygons.
The side faces of the prism have the shape of a parallelogram.
The lateral edges of the prism are parallel and equal.

Tip: When calculating a triangular prism, you must pay attention to the units used. For example, if the base area is indicated in cm 2, then the height should be expressed in centimeters and the volume in cm 3. If the base area is in mm 2, then the height should be expressed in mm, and the volume in mm 3, etc.

Prism example

In this example:
— ABC and DEF make up the triangular bases of the prism
- ABED, BCFE and ACFD are rectangular side faces
— The side edges DA, EB and FC correspond to the height of the prism.
— Points A, B, C, D, E, F are the vertices of the prism.

Problems for calculating a triangular prism

Problem 1. The base of a right triangular prism is a right triangle with legs 6 and 8, the side edge is 5. Find the volume of the prism.
Solution: The volume of a straight prism is equal to V = Sh, where S is the area of ​​the base and h is the side edge. The area of ​​the base in this case is the area of ​​a right triangle (its area is equal to half the area of ​​a rectangle with sides 6 and 8). Thus, the volume is equal to:

V = 1/2 6 8 5 = 120.

Task 2.

A plane parallel to the side edge is drawn through the middle line of the base of the triangular prism. The volume of the cut-off triangular prism is 5. Find the volume of the original prism.

Solution:

The volume of the prism is equal to the product of the area of ​​the base and the height: V = S base h.

The triangle lying at the base of the original prism is similar to the triangle lying at the base of the cut-off prism. The similarity coefficient is 2, since the section is drawn through the middle line (the linear dimensions of the larger triangle are twice as large as the linear dimensions of the smaller one). It is known that the areas of similar figures are related as the square of the similarity coefficient, that is, S 2 = S 1 k 2 = S 1 2 2 = 4S 1 .

The base area of ​​the entire prism is 4 times greater than the base area of ​​the cut-off prism. The heights of both prisms are the same, so the volume of the entire prism is 4 times the volume of the cut-off prism.

Thus, the required volume is 20.