How can you build a pyramid? On surface R Let's construct a polygon, for example the pentagon ABCDE. Out of plane R Let's take point S. By connecting point S with segments to all points of the polygon, we get the SABCDE pyramid (Fig.).

Point S is called top, and the polygon ABCDE is basis this pyramid. Thus, a pyramid with top S and base ABCDE is the union of all segments where M ∈ ABCDE.

Triangles SAB, SBC, SCD, SDE, SEA are called side faces pyramids, common sides of the lateral faces SA, SB, SC, SD, SE - lateral ribs.

The pyramids are called triangular, quadrangular, p-angular depending on the number of sides of the base. In Fig. Images of triangular, quadrangular and hexagonal pyramids are given.

The plane passing through the top of the pyramid and the diagonal of the base is called diagonal, and the resulting section is diagonal. In Fig. 186 one of the diagonal sections of the hexagonal pyramid is shaded.

The perpendicular segment drawn through the top of the pyramid to the plane of its base is called the height of the pyramid (the ends of this segment are the top of the pyramid and the base of the perpendicular).

The pyramid is called correct, if the base of the pyramid is a regular polygon and the vertex of the pyramid is projected at its center.

All lateral faces of a regular pyramid are congruent isosceles triangles. In a regular pyramid, all lateral edges are congruent.

The height of the lateral face of a regular pyramid drawn from its vertex is called apothem pyramids. All apothems of a regular pyramid are congruent.

If we designate the side of the base as A, and the apothem through h, then the area of one side face of the pyramid is 1/2 ah.

The sum of the areas of all the lateral faces of the pyramid is called lateral surface area pyramid and is designated by S side.

Since the lateral surface of a regular pyramid consists of n congruent faces, then

S side = 1/2 ahn=P h / 2 ,

where P is the perimeter of the base of the pyramid. Hence,

S side =P h / 2

i.e. The area of the lateral surface of a regular pyramid is equal to half the product of the perimeter of the base and the apothem.

The total surface area of the pyramid is calculated by the formula

S = S ocn. + S side. .

The volume of the pyramid is equal to one third of the product of the area of its base S ocn. to height H:

V = 1 / 3 S main. N.

The derivation of this and some other formulas will be given in one of the subsequent chapters.

Let's now build a pyramid in a different way. Let a polyhedral angle be given, for example, pentahedral, with vertex S (Fig.).

Let's draw a plane R so that it intersects all the edges of a given polyhedral angle at different points A, B, C, D, E (Fig.). Then the SABCDE pyramid can be considered as the intersection of a polyhedral angle and a half-space with the boundary R, in which the vertex S lies.

Obviously, the number of all faces of the pyramid can be arbitrary, but not less than four. When a trihedral angle intersects with a plane, a triangular pyramid is obtained, which has four sides. Any triangular pyramid is sometimes called tetrahedron, which means tetrahedron.

Truncated pyramid can be obtained if the pyramid is intersected by a plane parallel to the plane of the base.

In Fig. An image of a quadrangular truncated pyramid is given.

Truncated pyramids are also called triangular, quadrangular, n-gonal depending on the number of sides of the base. From the construction of a truncated pyramid it follows that it has two bases: upper and lower. The bases of a truncated pyramid are two polygons, the sides of which are parallel in pairs. The lateral faces of the truncated pyramid are trapezoids.

Height a truncated pyramid is a perpendicular segment drawn from any point of the upper base to the plane of the lower one.

Regular truncated pyramid called the part of a regular pyramid enclosed between the base and a section plane parallel to the base. The height of the side face of a regular truncated pyramid (trapezoid) is called apothem.

It can be proven that a regular truncated pyramid has congruent lateral edges, all lateral faces are congruent, and all apothems are congruent.

If in the correct truncated n-coal pyramid through A And b n indicate the lengths of the sides of the upper and lower bases, and through h is the length of the apothem, then the area of each side face of the pyramid is equal to

1 / 2 (A + b n) h

The sum of the areas of all the lateral faces of the pyramid is called the area of its lateral surface and is designated S side. . Obviously, for a correct truncated n-coal pyramid

S side = n 1 / 2 (A + b n) h.

Because pa= P and nb n= P 1 - the perimeters of the bases of the truncated pyramid, then

S side = 1 / 2 (P + P 1) h,

that is, the area of the lateral surface of a regular truncated pyramid is equal to half the product of the sum of the perimeters of its bases and the apothem.

Section parallel to the base of the pyramid

Theorem. If the pyramid is intersected by a plane parallel to the base, then:

1) the side ribs and height will be divided into proportional parts;

2) in cross-section you will get a polygon similar to the base;

3) the cross-sectional areas and bases are related as the squares of their distances from the top.

It is enough to prove the theorem for a triangular pyramid.

Since parallel planes are intersected by a third plane along parallel lines, then (AB) || (A 1 B 1), (BC) ||(B 1 C 1), (AC) || (A 1 C 1) (fig.).

Parallel lines cut the sides of an angle into proportional parts, and therefore

Therefore, ΔSAB ~ ΔSA 1 B 1 and

ΔSBC ~ ΔSB 1 C 1 and

Thus,

The corresponding angles of triangles ABC and A 1 B 1 C 1 are congruent, like angles with parallel and identical sides. That's why

ΔABC ~ ΔA 1 B 1 C 1

Squares similar triangles are related as the squares of the corresponding sides:

$$ \frac(S_(ABC))(S_(A_1 B_1 C_1))=\frac(\left|(AB)\right|^2)(\left|(A_(1)B_1)\right|^2 ) $$

Hence,

$$ \frac(S_(ABC))(S_(A_1 B_1 C_1))=\frac(\left|(SH)\right|^2)(\left|(SH_1)\right|^2) $$

Theorem. If two pyramids with equal heights are cut at the same distance from the top by planes parallel to the bases, then the areas of the sections are proportional to the areas of the bases.

Let (Fig. 84) B and B 1 be the areas of the bases of two pyramids, H be the height of each of them, b And b 1 - sectional areas by planes parallel to the bases and removed from the vertices at the same distance h.

According to the previous theorem we will have:

$$ \frac(b)(B)=\frac(h^2)(H^2)\: and \: \frac(b_1)(B_1)=\frac(h^2)(H^2) $ $
where
$$ \frac(b)(B)=\frac(b_1)(B_1)\: or \: \frac(b)(b_1)=\frac(B)(B_1) $$

Consequence. If B = B 1, then b = b 1, i.e. If two pyramids with equal heights have equal bases, then the sections equally spaced from the top are also equal.

Other materials

CHAPTER THREE

POLYhedra

1. PARALLELEPIPED AND PYRAMID

Properties of parallel sections in a pyramid

74. Theorem. If the pyramid (drawing 83) intersected by a plane parallel to the base, then:

1) the side ribs and height are divided by this plane into proportional parts;

2) in cross section it turns out to be a polygon (abcde ), similar to the base;

3) The cross-sectional areas and bases are related as the squares of their distances from the vertex.

1) Straight ab and AB can be considered as the line of intersection of two parallel planes (base and secant) with the third plane ASB; That's why ab||AB (§ 16). For the same reason bc||BC, CD||CD, ... and at||AM; Consequently

S a / a A=S b / b B=S c / c C=...=S m / m M

2) From the similarity of triangles ASB and a S b, then BSC and b S c etc. we output:

AB / ab= BS / bs; B.S. / bs= BC / bc ,

AB / ab= BC / bc

B.C. / bc=CS / cs; C.S. / cs= CD / CD from BC / bc= CD / CD .

We will also prove the proportionality of the remaining sides of the polygons ABCDE and abcde. Since, moreover, these polygons have equal corresponding angles (as formed by parallel and identically directed sides), then they are similar.

3) The areas of similar polygons are related as the squares of similar sides; That's why

75. Consequence. A regular truncated pyramid has an upper base that is a regular polygon similar to the lower base, and the side faces are equal and isosceles trapezoids(drawing 83).

The height of any of these trapezoids is called apothem regular truncated pyramid.

76. Theorem. If two pyramids with equal heights are cut at the same distance from the top by planes parallel to the bases, then the areas of the sections are proportional to the areas of the bases.

According to the previous theorem we will have:

77. Consequence. If B = B 1, then b = b 1, i.e. If two pyramids with equal heights have equal bases, then the sections equally spaced from the top are also equal.

Question:

The pyramid is intersected by a plane parallel to the base. The base area is 1690 dm2, and the cross-sectional area is 10 dm2. In what ratio, counting from the apex, does the cutting plane divide the height of the pyramid?

Answers:

a parallel plane cuts a pyramid similar to this one (h1/h)²=s1/s (h1/h)²=10/1690=1/169 h1/h=√1/169= 1/13 jndtn 1/13

1.3 Properties of parallel sections in a pyramid

1.3.1 Theorems on cuts in a pyramid

If the pyramid (1.11) is intersected by a plane parallel to the base, then:

1) the side ribs and height are divided by this plane into proportional parts;

2) in cross-section a polygon (abcde) similar to the base is obtained;

3) The cross-sectional areas and bases are related as the squares of their distances from the vertex.

1) Straight lines ab and AB can be considered as the lines of intersection of two parallel planes (base and secant) with the third plane ASB; therefore abkAB. For the same reason bckBC, cdkCD.... and amkAM; Consequently

aA Sa = bB Sb = cC Sc = ::: = mM Sm :

2) From the similarity of triangles ASB and aSb, then BSC and bSc, etc., we derive:

AB ab = BS bS ; BS bS = BC bc ;

AB ab = BC bc :

BC bc = CS cS ; CS cS = CD cd ;

BC bc = CD cd

We will also prove the proportionality of the remaining sides of the polygons ABCDE and abcde. Since, in addition, these polygons have equal corresponding angles (as formed by parallel and identically directed sides), then they are similar. The areas of similar polygons are related as the squares of similar sides; That's why

AB ab = AS as = M msS ;






set2D(1; 9; 1; 14);

;0 dash0 );

			Rice. 1.11: Pyramid
p5 = pointsPlot(
p5 = pointsPlot(

[ 0A 0; 0 B 0; 0 C 0; 0 D 0; 0 E 0; 0 a 0; 0 b 0; 0 c 0; 0 d 0; 0 M 0; 0 m 0; 0 S 0];

); showPlots(;0 noAxes0 );

1.3.2 Consequence

A regular truncated pyramid has an upper base that is a regular polygon similar to the lower base, and the side faces are equal and isosceles trapezoids (1.11).

The height of any of these trapezoids is called the apothem of a regular truncated pyramid.

1.3.3 Parallel section theorem in a pyramid

If two pyramids with equal heights are cut at the same distance from the top by planes parallel to the bases, then the areas of the sections are proportional to the areas of the bases.

Let (1.12) B and B1 be the areas of the bases of two pyramids, H the height of each of them, b and b1 the areas of sections by planes parallel to the bases and removed from the vertices at the same distance h.

According to the previous theorem we will have:

H2 B1

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