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 Egyptian pyramids, therefore, it is well represented 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. Set of pyramids with different grounds 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 divide the figure by secant planes parallel to the base into an infinite number of thin layers. The figure below shows a quadrangular pyramid of height h and side length L, in which the quadrangle marks thin layer sections.

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 specified in the case of a pyramid with right foundation. 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 the correct 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.

Determining the volumes of regular pyramids requires knowing 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 exact dimensions the Pyramid of Cheops. Its original height was 146.50 meters (currently about 137 meters). Average length 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!

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. area full surface 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. Dihedral angle at the base - this 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 but. 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.

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 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 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 but 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.

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 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 ABOUT 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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