Volume of a truncated pyramid formula. Pyramid

Pyramid. Truncated pyramid

Pyramid is 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 with all edges equal is called tetrahedron .

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

Lateral surface area pyramid is the sum of the areas of all lateral faces. Total surface area is called the sum of the areas of all the side faces and the base.

Theorems

1. If in a pyramid all the lateral 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 near the base.

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

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

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

Where V- volume;

S base– base area;

H– height of the pyramid.

For a regular pyramid, the following formulas are correct:

Where p– base perimeter;

h a– apothem;

H- height;

S full

S side

S base– base area;

V– volume of a regular pyramid.

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

Grounds truncated pyramid - similar polygons. Side faces – trapezoids. Height of a truncated pyramid is 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 is a section of a truncated pyramid by a plane passing through two lateral edges that do not belong to the same 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– volume of a truncated pyramid.

For a regular truncated pyramid the formula is correct:

Where p 1 , p 2 – perimeters of the bases;

h a– 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 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. The linear angle is the angle a between two perpendiculars: etc. The top of the pyramid is projected at the center of the triangle (the center of the circumcircle and inscribed circle of the triangle ABC). The angle of inclination of the side edge (for example S.B.) is the angle between the edge itself and its projection onto the plane of the base. For the rib S.B. this angle will be the angle SBD. To find the tangent you need to know the legs SO And O.B.. Let the length of the segment BD equals 3 A. Dot ABOUT 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 equal to cm and cm, and its height is 4 cm.

Solution. To find the volume of a truncated pyramid, we use formula (4). To find the area of ​​the bases, you need to find the sides of the base squares, knowing their diagonals. The sides of the bases are equal to 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 cm 3.

Example 3. Find the area of ​​the lateral 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.

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

The side face of this pyramid is an isosceles trapezoid. To calculate the area of ​​a trapezoid, you need to know the base and height. The bases are given according to the condition, only the height remains unknown. We'll find her from where A 1 E perpendicular from a point A 1 on the plane of the lower base, A 1 D– perpendicular from A 1 per AC. A 1 E= 2 cm, since this is the height of the pyramid. To find DE Let's make an additional drawing showing the 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– radius inscribed in the circle and OM– radius inscribed in a 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 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 plane of the base. Using the theorem on the area of ​​the orthogonal projection of a plane figure, we obtain:

Likewise it means Thus, the problem was reduced to finding the area of ​​the trapezoid ABCD. Let's draw a trapezoid ABCD separately (Fig. 22). Dot ABOUT– the center of a circle inscribed in a trapezoid.

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

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

Pyramid as a three-dimensional figure

Everyone knows about the Egyptian pyramids, so they have a good idea of ​​what kind of figure we will be talking about. However, 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 a certain point in space that does not belong to the plane of the base. 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 in question is a perfect polyhedron, the numbers of marked elements obey Euler’s equality:

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 a figure meet is called the vertex of the pyramid. If a perpendicular is lowered from it onto the base and it intersects it at the geometric center, then such a figure will be called a straight line. If this condition is not met, then an inclined pyramid occurs.

A right figure whose base is formed by an equilateral (equiangular) n-gon is called regular.

Formula for the volume of a pyramid

To calculate the volume of the pyramid, we will use integral calculus. To do this, we divide the figure by cutting 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 quadrilateral marks the thin layer of the section.

The area of ​​each such layer can be calculated using 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 obtain the formula for the volume of a 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 = 1/3*A 0 *h.

We have obtained the formula for the volume of a pyramid. To find the value of V, just multiply the height of the figure by the area of ​​the base, and then divide the result by three.

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

and its volume

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

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

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

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

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

Truncated pyramid

Let's assume that we took an arbitrary pyramid and cut off 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 similar parallel bases. That is, the lengths of the sides of one of them can be obtained by multiplying the lengths of the other by a certain 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 integral calculus similar to the one 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.

Volume of the Cheops pyramid

It is interesting to solve the problem of determining the volume that the largest Egyptian pyramid contains inside itself.

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 precision.

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

Substituting the numbers, we get:

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

The volume of the Cheops pyramid is almost 2.6 million m3. For comparison, we note that the Olympic swimming pool has a volume of 2.5 thousand m 3. That is, to fill the entire Cheops pyramid you will need more than 1000 such pools!

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