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 lateral 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.

Solution. 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.

Solution. 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.

Solution. 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.

Solution. 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.

We 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. By the area theorem orthogonal projection plane 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

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- 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. So, in the bases lies a figure with five the same 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 given 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.

On the this lesson we will consider a truncated pyramid, get acquainted with a regular truncated pyramid, and study their properties.

Let us recall the concept of an n-gonal pyramid using the example of a triangular pyramid. Triangle ABC is given. Outside the plane of the triangle, a point P is taken, connected to the vertices of the triangle. The resulting polyhedral surface is called a pyramid (Fig. 1).

Rice. 1. Triangular pyramid

Let us cut the pyramid with a plane parallel to the plane of the base of the pyramid. The figure obtained between these planes is called a truncated pyramid (Fig. 2).

Rice. 2. Truncated pyramid

Main elements:

Top base ;

Lower base ABC;

Side face ;

If PH is the height of the original pyramid, then is the height of the truncated pyramid.

The properties of a truncated pyramid follow from the method of its construction, namely from the parallelism of the planes of the bases:

All side faces of a truncated pyramid are trapezoids. Consider, for example, a face. It has the property of parallel planes (since the planes are parallel, they cut the side face of the original ABP pyramid along parallel lines), at the same time they are not parallel. Obviously, the quadrilateral is a trapezoid, like all the side faces of a truncated pyramid.

The ratio of the bases is the same for all trapezoids:

We have several pairs of similar triangles with the same similarity coefficient. For example, triangles and RAB are similar due to the parallelism of the planes and , the similarity coefficient:

At the same time, triangles and RCS are similar with similarity coefficient:

Obviously, the similarity coefficients for all three pairs of similar triangles are equal, so the ratio of the bases is the same for all trapezoids.

A regular truncated pyramid is a truncated pyramid obtained by cutting a regular pyramid with a plane parallel to the base (Fig. 3).

Rice. 3. Correct truncated pyramid

Definition.

A regular pyramid is called a pyramid, at the base of which lies a regular n-gon, and the vertex is projected into the center of this n-gon (the center of the inscribed and circumscribed circle).

AT this case at the base of the pyramid lies a square, and the vertex is projected to the point of intersection of its diagonals. The resulting regular quadrangular truncated pyramid ABCD - bottom base, - upper base. The height of the original pyramid - RO, truncated pyramid - (Fig. 4).

Rice. 4. Regular quadrangular truncated pyramid

Definition.

The height of a truncated pyramid is a perpendicular drawn from any point of one base to the plane of the second base.

The apothem of the original pyramid is RM (M is the middle of AB), the apothem of the truncated pyramid is (Fig. 4).

Definition.

The apothem of a truncated pyramid is the height of any side face.

It is clear that all the side edges of the truncated pyramid are equal to each other, that is, the side faces are equal isosceles trapezoids.

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

Proof (for a regular quadrangular truncated pyramid - Fig. 4):

So, we need to prove:

The lateral surface area here will consist of the sum of the areas of the lateral faces - trapezoids. Since the trapezoids are the same, we have:

The area of ​​an isosceles trapezoid is the product of half the sum of the bases and the height, the apothem is the height of the trapezoid. We have:

Q.E.D.

For an n-gonal pyramid:

Where n is the number of side faces of the pyramid, a and b are the bases of the trapezoid, is the apothem.

Sides of the base of a regular truncated quadrangular pyramid are equal to 3 cm and 9 cm, height - 4 cm. Find the area of ​​the lateral surface.

Rice. 5. Illustration for problem 1

Solution. Let's illustrate the condition:

Given: , ,

Draw a straight line MN through the point O parallel to the two sides of the lower base, similarly draw a straight line through the point (Fig. 6). Since the squares and constructions are parallel at the bases of the truncated pyramid, we get a trapezoid equal to the side faces. Moreover, its lateral side will pass through the middle of the upper and lower edges of the side faces and will be the epitome of a truncated pyramid.

Rice. 6. Additional constructions

Consider the resulting trapezoid (Fig. 6). In this trapezoid, the upper base, lower base and height are known. It is required to find the lateral side, which is the apothem of the given truncated pyramid. Draw perpendicular to MN. Let us drop the perpendicular NQ from the point. We get that the larger base is divided into segments of three centimeters (). Consider a right triangle, the legs in it are known, this is egyptian triangle, according to the Pythagorean theorem, we determine the length of the hypotenuse: 5 cm.

Now there are all the elements for determining the area of ​​the lateral surface of the pyramid:

The pyramid is crossed by a plane parallel to the base. Using the example of a triangular pyramid, prove that the side edges and the height of the pyramid are divided by this plane into proportional parts.

Proof. Let's illustrate:

Rice. 7. Illustration for problem 2

The pyramid RABC is given. RO is the height of the pyramid. The pyramid is dissected by a plane, a truncated pyramid is obtained, moreover. Point - the point of intersection of the height of the RO with the plane of the base of the truncated pyramid. It is necessary to prove:

The key to the solution is the property of parallel planes. Two parallel planes cut through any third plane so that the lines of intersection are parallel. From here: . The parallelism of the corresponding lines implies the presence of four pairs of similar triangles:

From the similarity of triangles follows the proportionality of the corresponding sides. Important feature is that the similarity coefficients for these triangles are the same:

Q.E.D.

A regular triangular pyramid RABC with a height and side of the base is dissected by a plane passing through the midpoint of the height PH parallel to the base ABC. Find the area of ​​the lateral surface of the resulting truncated pyramid.

Solution. Let's illustrate:

Rice. 8. Illustration for problem 3

DIA is a regular triangle, H is the center of this triangle (the center of the inscribed and circumscribed circles). RM is the apothem of the given pyramid. - the apothem of the truncated pyramid. According to the property of parallel planes (two parallel planes cut any third plane so that the intersection lines are parallel), we have several pairs of similar triangles with an equal similarity coefficient. In particular, we are interested in the relation:

Let's find NM. This is the radius of a circle inscribed in the base, we know the corresponding formula:

Now out right triangleРНМ according to the Pythagorean theorem, we find РМ - the apotheme of the original pyramid:

From the initial ratio:

Now we know all the elements for finding the lateral surface area of ​​a truncated pyramid:

So, we got acquainted with the concepts of a truncated pyramid and a regular truncated pyramid, gave basic definitions, considered properties, and proved the theorem on the area of ​​the lateral surface. The next lesson will focus on problem solving.

Bibliography

1. I. M. Smirnova, V. A. Smirnov. Geometry. Grade 10-11: a textbook for students of educational institutions (basic and profile levels) / I. M. Smirnova, V. A. Smirnov. - 5th ed., Rev. and additional - M.: Mnemosyne, 2008. - 288 p.: ill.
2. Sharygin I. F. Geometry. Grade 10-11: Textbook for general education educational institutions/ Sharygin I.F. - M.: Bustard, 1999. - 208 p.: ill.
3. E. V. Potoskuev, L. I. Zvalich. Geometry. Grade 10: Textbook for general educational institutions with in-depth and profile study of mathematics / E. V. Potoskuev, L. I. Zvalich. - 6th ed., stereotype. - M.: Bustard, 2008. - 233 p.: ill.

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