Pyramid. Truncated pyramid

Pyramid. Truncated pyramid

Pyramid is a polyhedron, one of whose faces is a polygon ( base ), and all other faces are triangles with a common vertex ( side faces ) (Fig. 15). The pyramid is called correct , if its basis is regular polygon and the top of the pyramid is projected into the center of the base (Fig. 16). A triangular pyramid with all edges equal is called tetrahedron .



Lateral rib of a pyramid is the side of the side face that does not belong to the base Height pyramid is the distance from its top to the plane of the base. All lateral edges of a regular pyramid are equal to each other, all lateral faces are equal isosceles triangles. The height of the side face of a regular pyramid drawn from the vertex is called apothem . Diagonal section is called a section of a pyramid by a plane passing through two lateral edges that do not belong to the same face.

Lateral surface area pyramid is the sum of the areas of all lateral faces. Area full surface is called the sum of the areas of all the side faces and the base.

Theorems

1. If in a pyramid all the lateral edges are equally inclined to the plane of the base, then the top of the pyramid is projected into the center of the circle circumscribed near the base.

2. If in a pyramid all lateral edges have equal lengths, then the top of the pyramid is projected into the center of the circle circumscribed near the base.

3. If all the faces in a pyramid are equally inclined to the plane of the base, then the top of the pyramid is projected into the center of a circle inscribed in the base.

To calculate the volume of an arbitrary pyramid, the correct formula is:

Where V- volume;

S base– base area;

H– height of the pyramid.

For a regular pyramid, the following formulas are correct:

Where p– base perimeter;

h a– apothem;

H- height;

S full

S side

S base– base area;

V– volume of a regular pyramid.

Truncated pyramid called the part of the pyramid enclosed between the base and a cutting plane parallel to the base of the pyramid (Fig. 17). Regular truncated pyramid called the part of a regular pyramid enclosed between the base and a cutting plane parallel to the base of the pyramid.

Grounds truncated pyramid - similar polygons. Side faces – trapezoids. Height of a truncated pyramid is the distance between its bases. Diagonal a truncated pyramid is a segment connecting its vertices that do not lie on the same face. Diagonal section is a section of a truncated pyramid by a plane passing through two lateral edges that do not belong to the same face.


For a truncated pyramid the following formulas are valid:

(4)

Where S 1 , S 2 – areas of the upper and lower bases;

S full– total surface area;

S side– lateral surface area;

H- height;

V– volume of a truncated pyramid.

For a regular truncated pyramid the formula is correct:

Where p 1 , p 2 – perimeters of the bases;

h a– apothem of a regular truncated pyramid.

Example 1. In a regular triangular pyramid, the dihedral angle at the base is 60º. Find the tangent of the angle of inclination of the side edge to the plane of the base.

Solution. Let's make a drawing (Fig. 18).


The pyramid is regular, which means that at the base there is an equilateral triangle and all the side faces are equal isosceles triangles. The dihedral angle at the base is the angle of inclination of the side face of the pyramid to the plane of the base. The linear angle is the angle a between two perpendiculars: etc. The top of the pyramid is projected at the center of the triangle (the center of the circumcircle and inscribed circle of the triangle ABC). The angle of inclination of the side edge (for example S.B.) is the angle between the edge itself and its projection onto the plane of the base. For the rib S.B. this angle will be the angle SBD. To find the tangent you need to know the legs SO And O.B.. Let the length of the segment BD equals 3 A. Dot ABOUT line segment BD is divided into parts: and From we find SO: From we find:

Answer:

Example 2. Find the volume of a regular truncated quadrangular pyramid if the diagonals of its bases are equal to cm and cm, and its height is 4 cm.

Solution. To find the volume of a truncated pyramid, we use formula (4). To find the area of ​​the bases, you need to find the sides of the base squares, knowing their diagonals. The sides of the bases are equal to 2 cm and 8 cm, respectively. This means the areas of the bases and Substituting all the data into the formula, we calculate the volume of the truncated pyramid:

Answer: 112 cm 3.

Example 3. Find the area of ​​the lateral face of a regular triangular truncated pyramid, the sides of the bases of which are 10 cm and 4 cm, and the height of the pyramid is 2 cm.

Solution. Let's make a drawing (Fig. 19).


The side face of this pyramid is an isosceles trapezoid. To calculate the area of ​​a trapezoid, you need to know the base and height. The bases are given according to the condition, only the height remains unknown. We'll find her from where A 1 E perpendicular from a point A 1 on the plane of the lower base, A 1 D– perpendicular from A 1 per AC. A 1 E= 2 cm, since this is the height of the pyramid. To find DE Let's make an additional drawing showing the top view (Fig. 20). Dot ABOUT– projection of the centers of the upper and lower bases. since (see Fig. 20) and On the other hand OK– radius inscribed in the circle and OM– radius inscribed in a circle:

MK = DE.

According to the Pythagorean theorem from

Side face area:


Answer:

Example 4. At the base of the pyramid lies an isosceles trapezoid, the bases of which A And b (a> b). Each side face forms an angle equal to the plane of the base of the pyramid j. Find the total surface area of ​​the pyramid.

Solution. Let's make a drawing (Fig. 21). Total surface area of ​​the pyramid SABCD equal to the sum of the areas and the area of ​​the trapezoid ABCD.

Let us use the statement that if all the faces of the pyramid are equally inclined to the plane of the base, then the vertex is projected into the center of the circle inscribed in the base. Dot ABOUT– vertex projection S at the base of the pyramid. Triangle SOD is the orthogonal projection of the triangle CSD to the plane of the base. By the theorem on the area of ​​orthogonal projection flat figure we get:


Likewise it means Thus, the problem was reduced to finding the area of ​​the trapezoid ABCD. Let's draw a trapezoid ABCD separately (Fig. 22). Dot ABOUT– the center of a circle inscribed in a trapezoid.


Since a circle can be inscribed in a trapezoid, then or From the Pythagorean theorem we have

is a polyhedron that is formed by the base of the pyramid and a section parallel to it. We can say that a truncated pyramid is a pyramid with the top cut off. This figure has many unique properties:

  • The lateral faces of the pyramid are trapezoids;
  • The lateral edges of a regular truncated pyramid are of the same length and inclined to the base at the same angle;
  • The bases are similar polygons;
  • In a regular truncated pyramid, the faces are identical isosceles trapezoids, whose area is equal. They are also inclined to the base at one angle.

The formula for the lateral surface area of ​​a truncated pyramid is the sum of the areas of its sides:

Since the sides of a truncated pyramid are trapezoids, to calculate the parameters you will have to use the formula trapezoid area. For a regular truncated pyramid, you can apply a different formula for calculating the area. Since all its sides, faces, and angles at the base are equal, it is possible to apply the perimeters of the base and the apothem, and also derive the area through the angle at the base.

If, according to the conditions in a regular truncated pyramid, the apothem (height of the side) and the lengths of the sides of the base are given, then the area can be calculated through the half-product of the sum of the perimeters of the bases and the apothem:

Let's look at an example of calculating the lateral surface area of ​​a truncated pyramid.
Given a regular pentagonal pyramid. Apothem l= 5 cm, the length of the edge in the large base is a= 6 cm, and the edge is at the smaller base b= 4 cm. Calculate the area of ​​the truncated pyramid.

First, let's find the perimeters of the bases. Since we are given a pentagonal pyramid, we understand that the bases are pentagons. This means that the bases contain a figure with five identical sides. Let's find the perimeter of the larger base:

In the same way we find the perimeter of the smaller base:

Now we can calculate the area of ​​a regular truncated pyramid. Substitute the data into the formula:

Thus, we calculated the area of ​​​​a regular truncated pyramid through the perimeters and apothem.

Another way to calculate the lateral surface area of ​​a regular pyramid is the formula through the angles at the base and the area of ​​these very bases.

Let's look at an example calculation. We remember that this formula applies only to a regular truncated pyramid.

Let a regular quadrangular pyramid be given. The edge of the lower base is a = 6 cm, and the edge of the upper base is b = 4 cm. The dihedral angle at the base is β = 60°. Find the lateral surface area of ​​a regular truncated pyramid.

First, let's calculate the area of ​​the bases. Since the pyramid is regular, all the edges of the bases are equal to each other. Considering that the base is a quadrilateral, we understand that it will be necessary to calculate area of ​​the square. It is the product of width and length, but when squared these values ​​are the same. Let's find the area of ​​the larger base:


Now we use the found values ​​to calculate the lateral surface area.

Knowing a few simple formulas, we easily calculated the area of ​​the lateral trapezoid of a truncated pyramid using various values.

This lesson will help you get an idea of ​​the topic “Pyramid. Regular and truncated pyramid." In this lesson we will get acquainted with the concept of a regular pyramid and give it a definition. Then we prove the theorem on the lateral surface of a regular pyramid and the theorem on the lateral surface of a regular truncated pyramid.

Theme: Pyramid

Lesson: Regular and truncated pyramids

Definition: A regular n-gonal pyramid is a pyramid that has a regular n-gon at its base, and the height is projected to the center of this n-gon (Fig. 1).

Rice. 1

Regular triangular pyramid

First, let's consider ∆ABC (Fig. 2), in which AB=BC=CA (that is, a regular triangle lies at the base of the pyramid). In a regular triangle, the centers of the inscribed and circumscribed circles coincide and are the center of the triangle itself. In this case, the center is found as follows: find the middle AB - C 1, draw a segment CC 1, which is the median, bisector and height; similarly, we find the middle of AC - B 1 and draw the segment BB 1. The intersection of BB 1 and CC 1 will be point O, which is the center of ∆ABC.

If we connect the center of the triangle O with the vertex of the pyramid S, we obtain the height of the pyramid SO ⊥ ABC, SO = h.

By connecting point S with points A, B and C we get the side edges of the pyramid.

We have obtained a regular triangular SABC pyramid (Fig. 2).

A pyramid is a polyhedron with a polygon at its base. All faces, in turn, form triangles that converge at one vertex. Pyramids are triangular, quadrangular, and so on. In order to determine which pyramid is in front of you, it is enough to count the number of angles at its base. The definition of “height of a pyramid” is very often found in geometry problems in school curriculum. In this article we will try to consider different ways her location.

Parts of the pyramid

Each pyramid consists of the following elements:

  • side faces, which have three corners and converge at the apex;
  • the apothem represents the height that descends from its apex;
  • the top of the pyramid is a point that connects the side ribs, but does not lie in the plane of the base;
  • the base is a polygon on which the vertex does not lie;
  • the height of a pyramid is a segment that intersects the top of the pyramid and forms a right angle with its base.

How to find the height of a pyramid if its volume is known

Through the formula V = (S*h)/3 (in the formula V is the volume, S is the area of ​​the base, h is the height of the pyramid) we find that h = (3*V)/S. To consolidate the material, let's immediately solve the problem. The triangular base is 50 cm 2 , whereas its volume is 125 cm 3 . The height of the triangular pyramid is unknown, which is what we need to find. Everything is simple here: we insert the data into our formula. We get h = (3*125)/50 = 7.5 cm.

How to find the height of a pyramid if the length of the diagonal and its edges are known

As we remember, the height of the pyramid forms a right angle with its base. This means that the height, edge and half of the diagonal together form Many, of course, remember the Pythagorean theorem. Knowing two dimensions, it will not be difficult to find the third quantity. Let's remember well-known theorem a² = b² + c², where a is the hypotenuse, and in our case the edge of the pyramid; b - the first leg or half of the diagonal and c - respectively, the second leg, or the height of the pyramid. From this formula c² = a² - b².

Now the problem: in a regular pyramid the diagonal is 20 cm, when the length of the edge is 30 cm. You need to find the height. We solve: c² = 30² - 20² = 900-400 = 500. Hence c = √ 500 = about 22.4.

How to find the height of a truncated pyramid

It is a polygon with a cross section parallel to its base. The height of a truncated pyramid is the segment that connects its two bases. The height can be found for a regular pyramid if the lengths of the diagonals of both bases, as well as the edge of the pyramid, are known. Let the diagonal of the larger base be d1, while the diagonal of the smaller base is d2, and the edge has length l. To find the height, you can lower the heights from the two upper opposite points of the diagram to its base. We see that we have two right triangle, it remains to find the lengths of their legs. To do this, subtract the smaller one from the larger diagonal and divide by 2. So we will find one leg: a = (d1-d2)/2. After which, according to the Pythagorean theorem, all we have to do is find the second leg, which is the height of the pyramid.

Now let's look at this whole thing in practice. We have a task ahead of us. A truncated pyramid has a square at the base, the diagonal length of the larger base is 10 cm, while the smaller one is 6 cm, and the edge is 4 cm. You need to find the height. First, we find one leg: a = (10-6)/2 = 2 cm. One leg is equal to 2 cm, and the hypotenuse is 4 cm. It turns out that the second leg or height will be equal to 16-4 = 12, that is, h = √12 = about 3.5 cm.

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 aspects side 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.

Because side surface 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

$$ \frac(\left|(SA)\right|)(\left|(SA_1)\right|)=\frac(\left|(SB)\right|)(\left|(SB_1)\right| )=\frac(\left|(SC)\right|)(\left|(SC_1)\right|) $$

Therefore, ΔSAB ~ ΔSA 1 B 1 and

$$ \frac(\left|(AB)\right|)(\left|(A_(1)B_1)\right|)=\frac(\left|(SB)\right|)(\left|(SB_1 )\right|) $$

ΔSBC ~ ΔSB 1 C 1 and

$$ \frac(\left|(BC)\right|)(\left|(B_(1)C_1)\right|)=\frac(\left|(SB)\right|)(\left|(SB_1 )\right|)=\frac(\left|(SC)\right|)(\left|(SC_1)\right|) $$

Thus,

$$ \frac(\left|(AB)\right|)(\left|(A_(1)B_1)\right|)=\frac(\left|(BC)\right|)(\left|(B_ (1)C_1)\right|)=\frac(\left|(AC)\right|)(\left|(A_(1)C_1)\right|) $$

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

The areas of similar triangles are related as the squares of the corresponding sides:

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

$$ \frac(\left|(AB)\right|)(\left|(A_(1)B_1)\right|)=\frac(\left|(SH)\right|)(\left|(SH_1 )\right|) $$

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.

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