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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 in a pyramid all the lateral edges 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 is 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).
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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 the 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
is a polyhedron that 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 the top cut off. This figure has many unique properties:
- The lateral faces of the pyramid are trapezoids;
- The lateral edges 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 lateral surface area of a truncated pyramid is the sum of the areas of its sides: 
Since the sides of a truncated pyramid are trapezoids, to calculate the parameters you will have to use the formula trapezoid area. For a regular truncated pyramid, you can apply a different formula for calculating the area. 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= 5 cm, the length of the edge in the large base is a= 6 cm, and the edge is at the smaller base b= 4 cm. Calculate the area of the 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 contain a figure with five identical sides. Let's 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. Substitute the data into 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 angles at the base and the area of these very bases.
Let's look at an example calculation. We remember that this formula applies only to a regular truncated pyramid.
Let a regular quadrangular pyramid be given. The edge of the lower base is a = 6 cm, and the edge of the upper base is 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 edges of the bases are equal to each other. Considering that the base is a quadrilateral, we understand that it will be necessary to calculate area of the square. It is the product of width and length, but when squared these values are the same. Let's 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 using various values.
A polyhedron in which one of its 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 a geometric figure - an n-gon. In this case, the pyramid is also called n-carbon.
A triangular pyramid whose edges are all equal is called tetrahedron.
The edges of the pyramid that do not belong to the base are called lateral, and their common point is vertex pyramids. The other edges of the pyramid are usually called parties to the basis.
The pyramid is called correct, if it has a regular polygon at its base and all lateral 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 apply:
1) S full = S side + S main, Where
S total – area of the total surface of the pyramid;
S side – area of the lateral surface, i.e. the sum of the areas of all lateral faces of the pyramid;
S main – area of the base of the pyramid.
2) V = 1/3 S base N, Where
V is the volume of the pyramid;
H – height of the pyramid.
For regular pyramid occurs:
S side = 1/2 P main h, Where
P main – 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 cutting plane 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 reasons truncated pyramid. The remaining faces are called lateral. The distance between the planes of the bases is called height truncated pyramid. Edges that do not belong to the bases are called lateral.
In addition, the base of the truncated pyramid similar n-gons. If the bases of a truncated pyramid are regular polygons, and all lateral edges are equal to each other, then such a truncated pyramid is called correct.
For arbitrary truncated pyramid the following formulas apply:
1) S full = S side + S 1 + S 2, Where
S total – total surface area;
S side – area of the lateral surface, i.e. the sum of the areas of all lateral faces of a 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 – volume of the truncated pyramid;
H – height of the truncated pyramid.
For regular truncated pyramid we also have:
S side = 1/2(P 1 + P 2) h, Where
P 1, P 2 – perimeters of the bases;
h – apothem (height of the side face, which is a trapezoid).
Let's consider several problems involving a truncated pyramid.
Task 1.
In a triangular truncated pyramid with a height equal to 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 using 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 Heron’s formula
S = √(p(p – a)(p – b)(p – c)),
because The problem gives the lengths of the three sides of a triangle.
We have: p 1 = (27 + 29 + 52)/2 = 54.
S 1 = √(54(54 – 27)(54 – 29)(54 – 52)) = √(54 27 25 2) = 270.
2. The pyramid is truncated, which means that similar polygons lie at the bases. In our case, triangle ABC is similar to triangle A 1 B 1 C 1. In addition, the similarity coefficient can be found as the ratio of the perimeters of the triangles under consideration, and the ratio of their areas will be equal to the square of the similarity coefficient. Thus we have:
S 1 /S 2 = (P 1) 2 /(P 2) 2 = 108 2 /72 2 = 9/4. Hence S 2 = 4S 1 /9 = 4 270/9 = 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 a truncated pyramid divided if the corresponding sides of the bases are in the ratio 1:2?
Solution.
Consider ABCA 1 B 1 C 1 - a truncated pyramid shown in rice. 2.
Since the sides in the bases are in the ratio 1:2, the areas of the bases are in the ratio 1:4 (triangle ABC is similar to 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 – area of the upper base, h – height.
But the volume of the prism ADEA 1 B 1 C 1 is V 1 = S 2 h and, therefore,
V 2 = V – V 1 = 7/3 · h · S 2 - h · S 2 = 4/3 · h · S 2.
So, V 2: V 1 = 3: 4.
Answer: 3:4.
Task 3.
The sides of the bases of a regular quadrangular truncated pyramid are equal to 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 ABCDA 1 B 1 C 1 D 1 shown in rice. 3.
Let us denote O 1 O 2 = x, then OO₂ = O 1 O – O 1 O 2 = 3 – x.
Consider the triangle B 1 O 2 D 1 and the triangle BO 2 D:
angle B 1 O 2 D 1 is equal to angle BO 2 D as vertical;
angle BDO 2 is equal to angle D 1 B 1 O 2 and angle O 2 ВD is equal to angle B 1 D 1 O 2 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 side ratio is:
В1D 1 /ВD = О 1 О 2 /ОО 2 or 1/2 = x/(x – 3), whence x = 1.
Consider the triangle B 1 D 1 B and the triangle LO 2 B: angle B is common, and there is also a pair of one-sided angles at B 1 D 1 || LM, which means that triangle B 1 D 1 B is similar to triangle LO 2 B, from which B 1 D: LO 2 = OO 1: OO 2 = 3: 2, i.e.
LO 2 = 2/3 · B 1 D 1 , LN = 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 = 1/3 · 2(4 + 16/9 + 8/3) = 152/27.
V 2 = 1/3 · 1 · (16/9 + 1 + 4/3) = 37/27.
Answer: 152/27; 37/27.
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