- This is a polyhedron, which 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 a cut off top. This figure has many unique properties:

• The side faces of the pyramid are trapezoids;
• The lateral ribs 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, the area of ​​which is equal. They are also inclined to the base at one angle.

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

Since the sides of the truncated pyramid are trapezoids, you will have to use the formula to calculate the parameters trapezoid area. For a regular truncated pyramid, another formula for calculating the area can be applied. 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\u003d 5 cm, the length of the face in the large base is a\u003d 6 cm, and the face is at the smaller base b\u003d 4 cm. Calculate the area of ​​\u200b\u200bthe 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 are a figure with five identical sides. 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. We substitute the data in 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 corners at the base and the area of ​​\u200b\u200bthese very bases.

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

Let a regular quadrangular pyramid be given. The face of the lower base is a = 6 cm, and the face of the upper 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 faces of the bases are equal to each other. Given that the base is a quadrilateral, we understand that it will be necessary to calculate square area. It is the product of width and length, but squared, these values ​​​​are the same. 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 through various values.

Pyramid. Truncated pyramid

Pyramid is called 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 base is a regular polygon and the top of the pyramid is projected into the center of the base (Fig. 16). A triangular pyramid in which all edges are equal is called tetrahedron .

Side rib pyramid is called 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 side edges of a regular pyramid are equal to each other, all side faces are equal isosceles triangles. The height of the side face of a regular pyramid drawn from the vertex is called apothema . diagonal section A section of a pyramid is called a plane passing through two side edges that do not belong to the same face.

Side surface area pyramid is called the sum of the areas of all side faces. Full surface area is the sum of the areas of all the side faces and the base.

Theorems

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

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

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

where V- volume;

S main- base area;

H is the height of the pyramid.

For a regular pyramid, the following formulas are true:

where p- the perimeter of the base;

h a- apothem;

H- height;

S full

S side

S main- base area;

V is the volume of a regular pyramid.

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

Foundations truncated pyramid - similar polygons. Side faces - trapezoid. Height truncated pyramid is called 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 A section of a truncated pyramid is called a plane passing through two side edges that do not belong to the same face.

For a truncated pyramid, the formulas are valid:

(4)

where S 1 , S 2 - areas of the upper and lower bases;

S full is the total surface area;

S side is the lateral surface area;

H- height;

V is the volume of the truncated pyramid.

For a regular truncated pyramid, the following formula is true:

where p 1 , p 2 - base perimeters;

h a- the 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.

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

The pyramid is regular, which means that the base 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 will be the angle a between two perpendiculars: i.e. The top of the pyramid is projected at the center of the triangle (the center of the circumscribed circle and the inscribed circle in the triangle ABC). The angle of inclination of the side rib (for example SB) is the angle between the edge itself and its projection onto the base plane. For rib SB this angle will be the angle SBD. To find the tangent you need to know the legs SO and OB. Let the length of the segment BD is 3 a. dot O 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 cm and cm and the height is 4 cm.

Decision. To find the volume of a truncated pyramid, we use formula (4). To find the areas of the bases, you need to find the sides of the base squares, knowing their diagonals. The sides of the bases are 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 cm3.

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

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

The side face of this pyramid is an isosceles trapezium. To calculate the area of ​​a trapezoid, you need to know the bases and the height. The bases are given by condition, only the height remains unknown. Find it from where BUT 1 E perpendicular from a point BUT 1 on the plane of the lower base, A 1 D- perpendicular from BUT 1 on AC. BUT 1 E\u003d 2 cm, since this is the height of the pyramid. For finding DE we will make an additional drawing, in which we will depict a top view (Fig. 20). Dot O- projection of the centers of the upper and lower bases. since (see Fig. 20) and On the other hand OK is the radius of the inscribed circle and OM is the radius of the inscribed 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.

Decision. Let's make a drawing (Fig. 21). Total surface area of ​​the pyramid SABCD is 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 O- vertex projection S at the base of the pyramid. Triangle SOD is the orthogonal projection of the triangle CSD to the base plane. According to the theorem on the area of ​​the orthogonal projection of a flat figure, we get:

Similarly, it means Thus, the problem was reduced to finding the area of ​​the trapezoid ABCD. Draw a trapezoid ABCD separately (Fig. 22). Dot O is the center of a circle inscribed in a trapezoid.

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

A polyhedron in which one of the faces is a polygon, and all other faces are triangles with a common vertex, is called a pyramid.

These triangles that make up the pyramid are called side faces, and the remaining polygon is basis pyramids.

At the base of the pyramid lies a geometric figure - an n-gon. In this case, the pyramid is also called n-coal.

A triangular pyramid all edges of which are equal is called tetrahedron.

The edges of a pyramid that do not belong to the base are called lateral, and their common point is vertex pyramids. The other edges of the pyramid are commonly referred to as foundation parties.

The pyramid is called correct, if it has a regular polygon at its base, and all side edges are equal to each other.

The distance from the top of the pyramid to the plane of the base is called height pyramids. We can say that the height of the pyramid is a segment perpendicular to the base, the ends of which are at the top of the pyramid and on the plane of the base.

For any pyramid, the following formulas hold:

1) S full \u003d S side + S main, where

S full - the area of ​​the full surface of the pyramid;

S side - side surface area, i.e. the sum of the areas of all the side faces of the pyramid;

S base - the area of ​​the base of the pyramid.

2) V = 1/3 S main N, where

V is the volume of the pyramid;

H is the height of the pyramid.

For correct pyramid takes place:

S side = 1/2 P main h, where

P main - the perimeter of the base of the pyramid;

h is the length of the apothem, that is, the length of the height of the side face lowered from the top of the pyramid.

The part of the pyramid enclosed between two planes - the plane of the base and the secant plane, drawn parallel to the base, is called truncated pyramid.

The base of the pyramid and the section of the pyramid by a parallel plane are called grounds truncated pyramid. The rest of the faces are called lateral. The distance between the planes of the bases is called height truncated pyramid. Edges that do not belong to bases are called lateral.

In addition, the bases of the truncated pyramid similar n-gons. If the bases of a truncated pyramid are regular polygons, and all side edges are equal to each other, then such a truncated pyramid is called correct.

For arbitrary truncated pyramid the following formulas hold:

1) S full \u003d S side + S 1 + S 2, where

S full - total surface area;

S side - side surface area, i.e. the sum of the areas of all the side faces of the truncated pyramid, which are trapezoids;

S 1, S 2 - base areas;

2) V = 1/3(S 1 + S 2 + √(S 1 S 2))H, where

V is the volume of the truncated pyramid;

H is the height of the truncated pyramid.

For regular truncated pyramid we also have:

S side \u003d 1/2 (P 1 + P 2) h, where

P 1, P 2 - perimeters of the bases;

h - apothem (the height of the side face, which is a trapezoid).

Consider several problems on a truncated pyramid.

Task 1.

In a triangular truncated pyramid with a height of 10, the sides of one of the bases are 27, 29, and 52. Determine the volume of the truncated pyramid if the perimeter of the other base is 72.

Decision.

Consider the truncated pyramid ABCA 1 B 1 C 1 shown in Figure 1.

1. The volume of a truncated pyramid can be found by the formula

V = 1/3H (S 1 + S 2 + √(S 1 S 2)), where S 1 is the area of ​​one of the bases, can be found using the Heron formula

S = √(p(p – a)(p – b)(p – c)),

because The problem is given the lengths of three sides of a triangle.

We have: p 1 \u003d (27 + 29 + 52) / 2 \u003d 54.

S 1 \u003d √ (54 (54 - 27) (54 - 29) (54 - 52)) \u003d √ (54 27 25 2) \u003d 270.

2. The pyramid is truncated, which means that similar polygons lie at the bases. In our case, the triangle ABC is similar to the triangle A 1 B 1 C 1. In addition, the similarity coefficient can be found as the ratio of the perimeters of the considered triangles, and the ratio of their areas will be equal to the square of the similarity coefficient. Thus, we have:

S 1 /S 2 \u003d (P 1) 2 / (P 2) 2 \u003d 108 2 / 72 2 \u003d 9/4. Hence S 2 \u003d 4S 1 / 9 \u003d 4 270/9 \u003d 120.

So V = 1/3 10(270 + 120 + √(270 120)) = 1900.

Answer: 1900.

Task 2.

In a triangular truncated pyramid, a plane is drawn through the side of the upper base parallel to the opposite side edge. In what ratio is the volume of the truncated pyramid divided if the corresponding sides of the bases are related as 1:2?

Decision.

Consider ABCA 1 B 1 C 1 - a truncated pyramid depicted in rice. 2.

Since at the bases the sides are related as 1: 2, then the areas of the bases are related as 1: 4 (the triangle ABC is similar to the triangle A 1 B 1 C 1).

Then the volume of the truncated pyramid is:

V = 1/3h (S 1 + S 2 + √(S 1 S 2)) = 1/3h (4S 2 + S 2 + 2S 2) = 7/3 h S 2, where S 2 is the area of ​​the upper base, h is the height.

But the volume of the ADEA 1 B 1 C 1 prism is V 1 = S 2 h and, therefore,

V 2 \u003d V - V 1 \u003d 7/3 h S 2 - h S 2 \u003d 4/3 h S 2.

So, V 2: V 1 \u003d 3: 4.

Answer: 3:4.

Task 3.

The sides of the bases of a regular quadrangular truncated pyramid are 2 and 1, and the height is 3. A plane is drawn through the point of intersection of the diagonals of the pyramid parallel to the bases of the pyramid, dividing the pyramid into two parts. Find the volume of each of them.

Decision.

Consider the truncated pyramid ABCD 1 B 1 C 1 D 1 shown in rice. 3.

Let's denote O 1 O 2 \u003d x, then OO₂ \u003d O 1 O - O 1 O 2 \u003d 3 - x.

Consider the triangle B 1 O 2 D 1 and the triangle BO 2 D:

angle B 1 O 2 D 1 is equal to the angle BO 2 D as vertical;

the angle ВDO 2 is equal to the angle D 1 B 1 O 2 and the angle O 2 ВD is equal to the angle B 1 D 1 O 2 as lying crosswise at B 1 D 1 || BD and secants B₁D and BD₁, respectively.

Therefore, the triangle B 1 O 2 D 1 is similar to the triangle BO 2 D and the ratio of the sides takes place:

B1D 1 / BD \u003d O 1 O 2 / OO 2 or 1/2 \u003d x / (x - 3), whence x \u003d 1.

Consider triangle В 1 D 1 В and triangle LO 2 B: angle В is common, and there is also a pair of one-sided angles at B 1 D 1 || LM, then the triangle B 1 D 1 B is similar to the triangle LO 2 B, whence B 1 D: LO 2 \u003d OO 1: OO 2 \u003d 3: 2, i.e.

LO 2 \u003d 2/3 B 1 D 1, LN \u003d 4/3 B 1 D 1.

Then S KLMN = 16/9 S A 1 B 1 C 1 D 1 = 16/9.

So, V 1 \u003d 1/3 2 (4 + 16/9 + 8/3) \u003d 152/27.

V 2 \u003d 1/3 1 (16/9 + 1 + 4/3) \u003d 37/27.

Answer: 152/27; 37/27.

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