How to find the area of a truncated pyramid. Truncated 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 side ribs regular pyramid 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 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.
Reasons 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 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 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. By the area theorem orthogonal projection flat figure we get:
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 at the base there is 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.
On this lesson we will look at a truncated pyramid, get acquainted with a regular truncated pyramid, and study their properties.
Let us recall the concept of an n-gonal pyramid using the example of a triangular pyramid. Triangle ABC is given. Outside the plane of the triangle, a point P is taken, connected to the vertices of the triangle. The resulting polyhedral surface is called a pyramid (Fig. 1).
Rice. 1. Triangular pyramid
Let's cut the pyramid with a plane parallel to the plane of the base of the pyramid. The figure obtained between these planes is called a truncated pyramid (Fig. 2).
Rice. 2. Truncated pyramid
Main elements:
Upper base;
ABC lower base;
Side face;
If PH is the height of the original pyramid, then it is the height of the truncated pyramid.
The properties of a truncated pyramid arise from the method of its construction, namely from the parallelism of the planes of the bases:
All lateral faces of a truncated pyramid are trapezoids. Consider, for example, the edge. It has the property of parallel planes (since the planes are parallel, they cut the side face of the original AVR pyramid along parallel straight lines), but at the same time they are not parallel. Obviously, the quadrilateral is a trapezoid, like all the lateral faces of the truncated pyramid.
The ratio of the bases is the same for all trapezoids:
We have several pairs of similar triangles with the same similarity coefficient. For example, triangles and RAB are similar due to the parallelism of the planes and , similarity coefficient:
At the same time, triangles and RVS are similar with the similarity coefficient:
Obviously, the similarity coefficients 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 with a plane parallel to the base (Fig. 3).
Rice. 3. Regular truncated pyramid
Definition.
A pyramid is called regular if its base is a regular n-gon, and its vertex is projected into the center of this n-gon (the center of the inscribed and circumscribed circle).
IN in this case At the base of the pyramid lies a square, and the top is projected at the intersection point of its diagonals. The resulting regular quadrangular truncated pyramid has ABCD - bottom base, - upper base. The height of the original pyramid is RO, the truncated pyramid is (Fig. 4).
Rice. 4. Regular quadrangular truncated pyramid
Definition.
The height of a truncated pyramid is a perpendicular drawn from any point of one base to the plane of the second base.
The apothem of the original pyramid is RM (M is the middle of AB), the apothem of the truncated pyramid is (Fig. 4).
Definition.
The apothem of a truncated pyramid is the height of any side face.
It is clear that all the side edges of the truncated pyramid are equal to each other, that is, the side faces are equal isosceles trapezoids.
The lateral surface area of a regular truncated pyramid is equal to the product of half the sum of the perimeters of the bases and the apothem.
Proof (for a regular quadrangular truncated pyramid - Fig. 4):
So, we need to prove:
The area of the side surface here will consist of the sum of the areas of the side faces - trapezoids. Since the trapezoids are the same, we have:
The area of an isosceles trapezoid is the product of half the sum of the bases and the height; the apothem is the height of the trapezoid. We have:
Q.E.D.
For an n-gonal pyramid:
Where n is the number of side faces of the pyramid, a and b are the bases of the trapezoid, and is the apothem.
Sides of the base of a regular truncated quadrangular pyramid 
equal 3 cm and 9 cm, height - 4 cm. Find the area of the lateral surface.
Rice. 5. Illustration for problem 1
Solution. Let us illustrate the condition:
Asked by: , ,
Through point O we draw a straight line MN parallel to the two sides of the lower base, and similarly through the point we draw a straight line (Fig. 6). Since the squares and constructions at the bases of the truncated pyramid are parallel, we obtain a trapezoid equal to the side faces. Moreover, its side side will pass through the middles of the upper and lower edges of the side faces and will be the apothem of the truncated pyramid.
Rice. 6. Additional constructions
Let's consider the resulting trapezoid (Fig. 6). In this trapezoid, the upper base, lower base and height are known. You need to find the side that is the apothem of a given truncated pyramid. Let's draw perpendicular to MN. From the point we lower the perpendicular NQ. We find that the larger base is divided into segments of three centimeters (). Consider a right triangle, the legs in it are known, this Egyptian triangle, using the Pythagorean theorem we determine the length of the hypotenuse: 5 cm.
Now there are all the elements to determine the area of the lateral surface of the pyramid:
The pyramid is intersected by a plane parallel to the base. Prove, using the example of a triangular pyramid, that the lateral edges and height of the pyramid are divided by this plane into proportional parts.
Proof. Let's illustrate:
Rice. 7. Illustration for problem 2
The RABC pyramid is given. PO - height of the pyramid. The pyramid is cut by a plane, a truncated pyramid is obtained, and. Point - the point of intersection of the height of the RO with the plane of the base of the truncated pyramid. It is necessary to prove:
The key to the solution is the property of parallel planes. Two parallel planes intersect any third plane so that the lines of intersection are parallel. From here: . The parallelism of the corresponding lines implies the presence of four pairs of similar triangles:
From the similarity of triangles follows the proportionality of the corresponding sides. Important Feature is that the similarity coefficients of these triangles are the same:
Q.E.D.
A regular triangular pyramid RABC with a height and side of the base is dissected by a plane passing through the middle of the height PH parallel to the base ABC. Find the lateral surface area of the resulting truncated pyramid.
Solution. Let's illustrate:
Rice. 8. Illustration for problem 3
ACB is a regular triangle, H is the center of this triangle (the center of the inscribed and circumscribed circles). RM is the apothem of a given pyramid. - apothem of a 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 similarity coefficient. In particular, we are interested in the relationship:
Let's find NM. This is the radius of a circle inscribed in the base; we know the corresponding formula:
Now from right triangle RNM using the Pythagorean theorem we find RM - apothem of the original pyramid:
From the initial ratio:
Now we know all the elements for finding the area of the lateral surface of a truncated pyramid:
So, we got acquainted with the concepts of a truncated pyramid and a regular truncated pyramid, gave basic definitions, examined the properties, and proved the theorem on the area of the lateral surface. The next lesson will focus on problem solving.
References
- I. M. Smirnova, V. A. Smirnov. Geometry. Grades 10-11: textbook for students of general education institutions (basic and profile levels) / I. M. Smirnova, V. A. Smirnov. - 5th ed., rev. and additional - M.: Mnemosyne, 2008. - 288 p.: ill.
- Sharygin I.F. Geometry. 10-11 grade: Textbook for general education educational institutions/ Sharygin I.F. - M.: Bustard, 1999. - 208 p.: ill.
- E. V. Potoskuev, L. I. Zvalich. Geometry. Grade 10: Textbook for general education institutions with in-depth and specialized study of mathematics /E. V. Potoskuev, L. I. Zvalich. - 6th ed., stereotype. - M.: Bustard, 2008. - 233 p.: ill.
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