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 geometric figure– 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- this 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 occurs:

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.

Solution.

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?

Solution.

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 intersection point 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.

Solution.

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 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.

blog.site, with full or partial copying of the material, a link to the source is required.

- 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.

• 29.05.2016

  An oscillatory circuit is an electrical circuit containing an inductor, a capacitor and a source of electrical energy. When the circuit elements are connected in series, the oscillatory circuit is called serial, when parallel - parallel. An oscillatory circuit is the simplest system in which free electromagnetic oscillations. The resonant frequency of the circuit is determined by the so-called Thomson formula: ƒ = 1/(2π√(LC)) For …

• 20.09.2014

  The receiver is designed to receive signals in the LW range (150 kHz ... 300 kHz). main feature receiver in an antenna that has more inductance than a conventional magnetic antenna. That allows you to use the capacitance of the trimmer capacitor in the range of 4 ... 20pF, as well as such a receiver has acceptable sensitivity and a small gain in the RF path. The receiver for headphones (headphones) works, it is powered by ...

• 24.09.2014

  This device is designed to control the liquid level in tanks, as soon as the liquid rises to the set level, the device will start to give a continuous sound signal, when the liquid level reaches a critical level, the device will start to give an intermittent signal. The indicator consists of 2 generators, they are controlled by the sensor element E. It is placed in the tank at a level up to ...

• 22.09.2014

  KR1016VI1 is a digital multi-program timer designed to work with the ILTs3-5\7 indicator. It provides counting and displaying the current time in hours and minutes, the day of the week and the number of the control channel (9 alarm clocks). The scheme of the alarm clock is shown in the figure. The microcircuit is clocked. resonator Q1 at 32768 Hz. power is negative, the common plus goes to ...

The ability to calculate the volume of spatial figures is important in solving a number of practical problems in geometry. One of the most common shapes is the pyramid. In this article, we will consider the pyramids, both full and truncated.

Pyramid as a three-dimensional figure

Everyone knows about the Egyptian pyramids, so they have a good idea of ​​what figure will be discussed. Nevertheless, Egyptian stone structures are only a special case of a huge class of pyramids.

The geometric object under consideration in the general case is a polygonal base, each vertex of which is connected to some point in space that does not belong to the base plane. This definition leads to a figure consisting of one n-gon and n triangles.

Any pyramid consists of n+1 faces, 2*n edges and n+1 vertices. Since the figure under consideration is a perfect polyhedron, the numbers of marked elements obey the Euler equation:

2*n = (n+1) + (n+1) - 2.

The polygon located at the base gives the name of the pyramid, for example, triangular, pentagonal, and so on. A set of pyramids with different bases is shown in the photo below.

The point at which n triangles of the figure are connected is called the top of the pyramid. If a perpendicular is lowered from it to the base and it intersects it in the geometric center, then such a figure will be called a straight line. If this condition is not met, then there is an inclined pyramid.

A straight figure, the base of which is formed by an equilateral (equiangular) n-gon, is called regular.

Pyramid volume formula

To calculate the volume of the pyramid, we use the integral calculus. To do this, we break the figure parallel to the base cutting planes into an infinite number of thin layers. The figure below shows a quadrangular pyramid with height h and side length L, in which a thin sectional layer is marked with a quadrilateral.

The area of ​​each such layer can be calculated by the formula:

A(z) = A 0 *(h-z) 2 /h 2 .

Here A 0 is the area of ​​the base, z is the value of the vertical coordinate. It can be seen that if z = 0, then the formula gives the value A 0 .

To get the formula for the volume of the pyramid, you should calculate the integral over the entire height of the figure, that is:

V = ∫ h 0 (A(z)*dz).

Substituting the dependence A(z) and calculating the antiderivative, we arrive at the expression:

V = -A 0 *(h-z) 3 /(3*h 2)| h 0 \u003d 1/3 * A 0 * h.

We have obtained the formula for the volume of a pyramid. To find the value of V, it is enough to multiply the height of the figure by the area of ​​\u200b\u200bthe base, and then divide the result by three.

Note that the resulting expression is valid for calculating the volume of a pyramid of an arbitrary type. That is, it can be inclined, and its base can be an arbitrary n-gon.

and its volume

Received in paragraph above general formula for volume can be refined in the case of a pyramid with a regular base. The area of ​​such a base is calculated by the following formula:

A 0 = n/4*L 2 *ctg(pi/n).

Here L is the side length of a regular polygon with n vertices. The symbol pi is the number pi.

Substituting the expression for A 0 into the general formula, we obtain the volume of a regular pyramid:

V n = 1/3*n/4*L 2 *h*ctg(pi/n) = n/12*L 2 *h*ctg(pi/n).

For example, for a triangular pyramid, this formula leads to the following expression:

V 3 \u003d 3/12 * L 2 * h * ctg (60 o) \u003d √3 / 12 * L 2 * h.

For a regular quadrangular pyramid, the volume formula takes the form:

V 4 \u003d 4/12 * L 2 * h * ctg (45 o) \u003d 1/3 * L 2 * h.

Definition of volumes regular pyramids requires knowledge of the side of their base and the height of the figure.

Pyramid truncated

Suppose we have taken an arbitrary pyramid and cut off a part of its lateral surface containing the vertex. The remaining figure is called a truncated pyramid. It already consists of two n-gonal bases and n trapezoids that connect them. If the cutting plane was parallel to the base of the figure, then a truncated pyramid is formed with parallel similar bases. That is, the lengths of the sides of one of them can be obtained by multiplying the lengths of the other by some coefficient k.

The figure above shows a truncated regular one. It can be seen that its upper base, like the lower one, is formed by a regular hexagon.

The formula that can be derived using an integral calculus similar to the above is:

V = 1/3*h*(A 0 + A 1 + √(A 0 *A 1)).

Where A 0 and A 1 are the areas of the lower (large) and upper (small) bases, respectively. The variable h denotes the height of the truncated pyramid.

The volume of the pyramid of Cheops

It is curious to solve the problem of determining the volume that the largest Egyptian pyramid contains.

In 1984, British Egyptologists Mark Lehner and Jon Goodman established the exact dimensions of the Cheops pyramid. Its original height was 146.50 meters (currently about 137 meters). The average length of each of the four sides of the structure was 230.363 meters. The base of the pyramid is square with high accuracy.

Let's use the given figures to determine the volume of this stone giant. Since the pyramid is a regular quadrangular, then the formula is valid for it:

Plugging in the numbers, we get:

V 4 \u003d 1/3 * (230.363) 2 * 146.5 ≈ 2591444 m 3.

The volume of the pyramid of Cheops is almost 2.6 million m 3. For comparison, we note that the Olympic pool has a volume of 2.5 thousand m 3. That is, to fill the entire Cheops pyramid, more than 1000 such pools will be needed!