Lesson material.

In the previous lessons, we worked with pyramids. Let's remember what polyhedron is called a pyramid, what a regular pyramid is, recall the properties of a regular pyramid.

A polyhedron made up of an -gon and triangles is called pyramid .

The pyramid is called correct if its base is a regular polygon.

The lateral surface area of \u200b\u200ba regular pyramid is half the product of the base perimeter times the apothem.

All side edges of a regular pyramid are equal and the side edges are equal isosceles triangles.

Let us be given a pyramid PA 1 A 2 ... A n. Let us draw a cutting plane β parallel to the plane of the base of the pyramid and let this plane intersect the lateral edges at points B 1, B 2, ..., B n.

The plane β splits the pyramid into two figures: the pyramid PB 1 B 2… B n and the polyhedron. A polyhedron whose faces are n-gons A 1 A 2… A n and B 1 B 2… B n located in parallel planes and n quadrangles A 1 A 2 B 2 B 1, A 2 A 3 B 3 B 2,… , A n A 1 B 1 B n is called truncated pyramid.

There are many examples of truncated pyramids around us. The hood above the stove is in the shape of a truncated pyramid, keyboard keys and other items.

N-gons A 1 A 2 ... A n and B 1 B 2 ... B n are called respectively top and bottom base... Quadrangles A 1 A 2 B 2 B 1, A 2 A 3 B 3 B 2, ..., A n A 1 B 1 B n are called lateral faces.

The segments A 1 B 1, ..., A n B n are called lateral edges of the truncated pyramid.

The truncated pyramid is designated as A 1 A 2… A n B 1 B 2… B n. Take an arbitrary point C on the upper base and from this point we lower the perpendicular to the lower base. This perpendicular is called the height of the truncated pyramid.

Now let's prove that the side faces of the truncated pyramid are trapeziums.

For the proof, consider the face A 1 A 2 B 2 B 1. It is clear that the proof will be similar for other lateral faces.

Since the cutting plane was parallel to the plane of the base, it can be written that A 1 A 2 is parallel to B 1 B 2. Obviously, the other two sides of the quadrilateral A 1 A 2 B 2 B 1 are not parallel (they intersect at point P). We get that this quadrilateral is a trapezoid. Obviously, all other side faces will also be trapeziums.

As with the pyramid, the truncated pyramid can also be correct.

The truncated pyramid is called correct, if it is obtained by sectioning a regular pyramid with a plane parallel to the base.

The bases of the truncated pyramid are regular polygons, and the side faces are isosceles trapezoids.

The heights of these trapezoids are called apothems.

The union of side faces is called the lateral surface of the truncated pyramid, and the union of all faces is called the full surface of the truncated pyramid. Then the area of \u200b\u200bthe lateral surface of the pyramid is the sum of the areas of its lateral faces.

And the total surface area of \u200b\u200ba pyramid is the sum of the areas of all its faces.

Now let's state and prove the theorem on the lateral surface area of \u200b\u200ba regular truncated pyramid.

The lateral surface area of \u200b\u200ba regular truncated pyramid is equal to the product of the half-sum of the base perimeters and the apothem.

Evidence.

Let's write down the formula for finding the lateral surface area of \u200b\u200bthe truncated pyramid.

Since the truncated pyramid is correct, it means that its faces will be isosceles trapezoids.

The area of \u200b\u200ban isosceles trapezoid is equal to the product of the half-sum of the bases and the height. The height of the side face is nothing more than the apothem of the truncated pyramid.

Let's substitute everything in the original formula, put half of the apothem outside the brackets, and group the sides by bases in brackets. Then we get that the area of \u200b\u200bthe lateral surface will be equal to the product of the half-sum of the perimeters of the bases of the truncated pyramid by the apothem.

Q.E.D.

Let's solve several problems.

A task.Sides of the bases of a regular truncated quadrangular pyramid are equal to and. The height of the pyramid is. Find the lateral surface area.

Decision.

This lesson will help you get an idea of \u200b\u200bthe topic “Pyramid. Correct and Truncated Pyramid ". In this lesson we will get acquainted with the concept of a regular pyramid, we will 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-gon pyramid is a pyramid in which a regular n-gon lies at the base, and the height is projected to the center of this n-gon (Fig. 1).

Fig. 1

Regular triangular pyramid

To begin with, consider ∆ABC (Fig. 2), in which AB \u003d BC \u003d CA (that is, a regular triangle lies at the base of the pyramid). In a regular triangle, the center 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 find the middle of AC - B 1 and draw a segment BB 1. The intersection of BB 1 and CC 1 will be point O, which is the center of ∆ABS.

If we connect the center of triangle O to the top of the pyramid S, then we get the height of the pyramid SO ⊥ ABC, SO \u003d h.

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

We got a regular triangular SABC pyramid (Fig. 2).

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 to the center of the base (Fig. 16). A triangular pyramid in which all edges are equal is called tetrahedron .

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

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

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

3. If in the pyramid all the 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 - the height of the pyramid.

For the correct pyramid, the formulas are correct:

where p - base perimeter;

h a - apothem;

H - height;

S full

S side

S main - base area;

V - the volume of the correct pyramid.

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

Foundations truncated pyramids - similar polygons. Side faces - trapezoid. Height a truncated pyramid is the distance between its bases. Diagonal a truncated pyramid is called a segment connecting its vertices that do not lie on the same face. Diagonal section the section of a truncated pyramid is called a plane passing through two lateral edges that do not belong to one 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 - the volume of the truncated pyramid.

For a correct truncated pyramid, the formula is correct:

where p 1 , p 2 - base perimeters;

h a - the apothem of the 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, so 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. Linear angle is the angle a between two perpendiculars: and i.e. The top of the pyramid is projected in the center of the triangle (the center of the circumcircle and the inscribed circle in the triangle ABC). The angle of inclination of the lateral 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 equals 3 and... Dot ABOUT section 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 the truncated pyramid, we will use formula (4). To find the area of \u200b\u200bthe 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. So the areas of the bases and Having substituted all the data in the formula, we calculate the volume of the truncated pyramid:

Answer: 112 cm 3.

Example 3. Find the area of \u200b\u200bthe side 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.

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

The side face of this pyramid is an isosceles trapezoid. To calculate the area of \u200b\u200ba trapezoid, you need to know the base and height. The bases are given by condition, only the height remains unknown. We will find it from where AND 1 E perpendicular from point AND 1 on the plane of the lower base, A 1 D - perpendicular from AND 1 on AS. AND 1 E \u003d 2 cm, since this is the height of the pyramid. To find DE we will make an additional drawing, in which we will depict a top view (fig. 20). Point ABOUT - 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 - radius of the inscribed circle:

MK \u003d DE.

By the Pythagorean theorem from

Side face area:

Answer:

Example 4. At the base of the pyramid lies an isosceles trapezoid, the bases of which andand b (a> b). Each side face forms an angle with the base plane of the pyramid equal to j... Find the total surface area of \u200b\u200bthe pyramid.

Decision. Let's make a drawing (fig. 21). Total surface area of \u200b\u200bthe pyramid SABCD equal to the sum of the areas and area of \u200b\u200bthe 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 apex is projected to the center of the circle inscribed in the base. Point ABOUT - vertex projection S at the base of the pyramid. Triangle SOD is the orthogonal projection of the triangle CSD on the plane of the base. By the theorem on the area of \u200b\u200bthe orthogonal projection of a plane figure, we get:

Similarly, it means Thus, the task was reduced to finding the area of \u200b\u200bthe trapezoid ABCD... Draw a trapezoid ABCDseparately (fig. 22). Point ABOUT - the center of the circle inscribed in the trapezoid.

Since a circle can be inscribed into a trapezoid, either From, by the Pythagorean theorem, we have

In this lesson we will look at the truncated pyramid, get acquainted with the correct truncated pyramid, and study their properties.

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

Fig. 1. Triangular pyramid

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

Fig. 2. Truncated pyramid

Main elements:

Upper base;

Lower base ABC;

Side edge;

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 base planes:

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

The base ratio is the same for all trapezoids:

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

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

Obviously, the coefficients of similarity 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 by a plane parallel to the base (Fig. 3).

Fig. 3. Correct truncated pyramid

Definition.

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

In this case, a square lies at the base of the pyramid, and the top is projected to the intersection of its diagonals. The obtained regular quadrangular truncated pyramid ABCD has a lower base and an upper base. The height of the original pyramid - RO, the truncated pyramid - (Fig. 4).

Fig. 4. Regular quadrangular truncated pyramid

Definition.

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

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

Definition.

Apothem of the truncated pyramid - the height of any side face.

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

The lateral surface area of \u200b\u200ba regular truncated pyramid is equal to the product of the half-sum of the base perimeters and the apothem.

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

So, it is necessary to prove:

The area of \u200b\u200bthe lateral surface here will consist of the sum of the areas of the lateral faces - trapeziums. Since the trapezoids are the same, we have:

The area of \u200b\u200ban isosceles trapezoid is the product of the half-sum of the bases and the height, the apothem is the height of the trapezoid. We have:

Q.E.D.

For an n-sided pyramid:

Where n is the number of side faces of the pyramid, a and b are the base 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 lateral surface area.

Fig. 5. Illustration for problem 1

Decision. Let's illustrate the condition:

Given:,,

Through point O we draw a straight line MN parallel to the two sides of the lower base, similarly through the point we draw a straight line (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 be the apothem of the truncated pyramid.

Fig. 6. Additional constructions

Consider the resulting trapezoid (Fig. 6). In this trapezoid, the upper base, the lower base and the height are known. It is required to find the side that is the apothem of the given truncated pyramid. Let's 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-angled triangle, the legs in it are known, this is the 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 \u200b\u200bthe side surface of the pyramid:

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

Evidence. Let's illustrate:

Fig. 7. Illustration for problem 2

The RAVS pyramid is set. RO is the height of the pyramid. The pyramid is cut by a plane, a truncated pyramid is obtained, and. Point - the point of intersection of the RO height with the base plane of the truncated pyramid. It is necessary to prove:

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

The proportionality of the corresponding sides follows from the similarity of the triangles. An important feature is that the similarity coefficients for these triangles are the same:

Q.E.D.

A regular triangular pyramid RAVS with a height and a side of the base is dissected by a plane passing through the middle of the RN height parallel to the base ABC. Find the lateral surface area of \u200b\u200bthe resulting truncated pyramid.

Decision. Let's illustrate:

Fig. 8. Illustration for problem 3

ACB is a right triangle, H is the center of this triangle (the center of the inscribed and circumscribed circles). RM is the apothem of the given pyramid. - 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 coefficient of similarity. In particular, we are interested in the relation:

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

Now, from the right-angled triangle РНМ, according to the Pythagorean theorem, we find РМ - the apothem of the original pyramid:

From the initial ratio:

Now we know all the elements for finding the lateral surface area of \u200b\u200bthe 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 lateral surface area. The next lesson will be about problem solving.

List of references

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

1. Uztest.ru ().
2. Fmclass.ru ().
3. Webmath.exponenta.ru ().

Homework

- 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 truncated top. This shape has many unique properties:

• The side faces of the pyramid are trapeziums;
• Lateral ribs of a regular truncated pyramid are of equal length and inclined to the base at the same angle;
• The bases are like polygons;
• In a regular truncated pyramid, the faces are identical isosceles trapezoids, whose area is equal. They are also inclined to the base at the same angle.

The formula for the lateral surface area of \u200b\u200ba 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 the correct truncated pyramid, you can apply a different area formula. 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 deduce the area through the angle at the base.

If, according to the conditions in a regular truncated pyramid, the apothem (the height of the lateral 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 \u200b\u200ba truncated pyramid.
A regular pentagonal pyramid is given. Apothem l \u003d 5 cm, the length of the face in the large base is a \u003d 6 cm, and the edge in 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 a figure with five identical sides lies at the bases. 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 \u200b\u200bthe correct truncated pyramid. We substitute the data into the formula:

Thus, we calculated the area of \u200b\u200ba regular truncated pyramid through the perimeters and apothem.

Another way to calculate the lateral surface area of \u200b\u200ba regular pyramid is the formula through the corners at the base and the area of \u200b\u200bthese very bases.

Let's take a look at a calculation example. Remember that this formula applies only to the correct truncated pyramid.

Let a regular quadrangular pyramid be given. The edge of the lower base is a \u003d 6 cm, and the edge of the upper base is b \u003d 4 cm. The dihedral angle at the base is β \u003d 60 °. Find the lateral surface area of \u200b\u200ba regular truncated pyramid.

First, let's calculate the area of \u200b\u200bthe bases. Since the pyramid is correct, all the faces of the bases are equal to each other. Considering that there is a quadrangle at the base, we understand that it will be necessary to calculate square area... It is the product of width and length, but these values \u200b\u200bare the same squared. Find the area of \u200b\u200bthe larger base:


Now we use the found values \u200b\u200bto calculate the lateral surface area.

Knowing a few simple formulas, we easily calculated the area of \u200b\u200bthe lateral trapezium of the truncated pyramid through various values.