Rectangular truncated pyramid corners at the base. Formulas and properties of a regular triangular pyramid. Truncated triangular pyramid. Correct truncated pyramid
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.
IN base of the pyramid lies geometric figure- n-gon. In this case, the pyramid is also called n-sided.
A triangular pyramid, all edges of which are 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 commonly referred to as parties of the basis.
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 = S side + S main, where
S full - the total surface area of the pyramid;
S side - lateral surface area, i.e. the sum of the areas of all side faces of the pyramid;
S main - the area of the base of the pyramid.
2) V = 1/3 S basic N, where
V is the volume of the pyramid;
H is the height of the pyramid.
For correct pyramid takes place:
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 dropped 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. Ribs that do not belong to the bases are called lateral.
Also, the base 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 an arbitrary truncated pyramid the following formulas hold:
1) S full = S side + S 1 + S 2, where
S full - total surface area;
S side - lateral 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 - the area of the bases;
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 correct truncated pyramid we also have:
S side = 1/2 (P 1 + P 2) h, where
P 1, P 2 - base perimeters;
h - apothem (the height of the side face, which is a trapezoid).
Let's consider several tasks for a truncated pyramid.
Objective 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 a truncated pyramid ABCA 1 B 1 C 1 shown in Figure 1.
1. The volume of the 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 by Heron's formula
S = √ (p (p - a) (p - b) (p - c)),
since in the problem, the lengths of the three sides of the triangle are given.
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, 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 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.
Objective 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 was the volume of the truncated pyramid divided if the respective sides of the bases are 1: 2?
Solution.
Consider ABCA 1 B 1 C 1 - a truncated pyramid shown in rice. 2.
Since the sides in the bases are related as 1: 2, the areas of the bases are related as 1: 4 (the ABC triangle is similar to the A1 B 1 C 1 triangle).
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 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.
Objective 3.
The sides of the bases of a regular quadrangular truncated pyramid are equal to 2 and 1, and the height is 3. Through the intersection of the pyramid's diagonals parallel to the pyramid's bases, a plane is drawn dividing the pyramid into two parts. Find the volume of each of them.
Solution.
Consider a truncated pyramid ABCDA 1 B 1 C 1 D 1 shown in rice. 3.
We denote O 1 O 2 = x, then OO₂ = O 1 O - O 1 O 2 = 3 - x.
Consider a triangle B 1 O 2 D 1 and a triangle B 2 D:
angle В 1 О 2 D 1 equal to the angle VO 2 D as vertical;
the angle BDO 2 is equal to the angle D 1 B 1 O 2 and the angle O 2 BD is equal to the angle B 1 D 1 O 2 as criss-crossing 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 = O 1 O 2 / OO 2 or 1/2 = x / (x - 3), whence x = 1.
Consider a triangle B 1 D 1 B and a triangle LO 2 B: angle B is common, and there is also a pair of one-sided angles for B 1 D 1 || LM, so the triangle B 1 D 1 B is similar to the triangle LO 2 B, whence 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.
site, with full or partial copying of the material, a link to the source is required.
Truncated pyramid is called a polyhedron whose vertices are the vertices of the base and the vertices of its section by a plane parallel to the base.
Truncated pyramid properties:
- The truncated pyramid bases are similar polygons.
- The side faces of the truncated pyramid are trapeziums.
- The side edges of a regular truncated pyramid are equal and equally inclined towards the base of the pyramid.
- The side faces of a regular truncated pyramid are equal isosceles trapezoids and are equally inclined towards the base of the pyramid.
- The dihedral angles at the lateral edges of a regular truncated pyramid are equal.
Surface area and volume of the truncated pyramid
Let - the height of the truncated pyramid, and - the perimeters of the bases of the truncated pyramid, and - the area of the bases of the truncated pyramid, - the area of the lateral surface of the truncated pyramid, - the total surface area of the truncated pyramid, - the volume of the truncated pyramid. Then the following relations hold:
.
If all dihedral angles at the base of the truncated pyramid are equal, and the heights of all side faces of the pyramid are equal, then
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 to the center of the base (Fig. 16). A triangular pyramid in which all edges are equal is called tetrahedron .
Side rib pyramid is the side of the side face that does not belong to the base Height pyramid is called 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 edges are equal isosceles triangles. The height of the side face of a regular pyramid drawn from the top is called apothem . Diagonal section the section of the pyramid is called a plane passing through two lateral edges that do not belong to one face.
Side surface area pyramid is called the sum of the areas of all side faces. Full surface area called the sum of the areas of all side faces and the base.
Theorems
1. If in a pyramid all side 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 about the base.
2. If in the pyramid all side edges have equal lengths, then the top of the pyramid is projected into the center of the circle circumscribed about the base.
3. If in the pyramid all the 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 following formula is correct:
where V- volume;
S main- base area;
H- the height of the pyramid.
For the correct pyramid, the formulas are correct:
where p- base perimeter;
h a- apothem;
H- height;
S full
S side
S main- base area;
V- the volume of the correct pyramid.
Truncated pyramid called the part of the pyramid, enclosed between the base and the secant plane parallel to the base of the pyramid (Fig. 17). Regular truncated pyramid is called the part of a regular pyramid, enclosed between the base and the secant plane parallel to the base of the pyramid.
Foundations truncated pyramids - similar polygons. Side faces - trapezoid. Height a truncated pyramid is the distance between its bases. Diagonal a truncated pyramid is called a segment connecting its vertices that do not lie on the same face. Diagonal section the section of a truncated pyramid is called a plane passing through two lateral edges that do not belong to one 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- the volume of the truncated pyramid.
For a correct truncated pyramid, the formula is correct:
where p 1 , p 2 - perimeters of the bases;
h a- the apothem of the 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, so at the base there is an equilateral triangle and all 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: and i.e. The top of the pyramid is projected in the center of the triangle (the center of the circumcircle and the inscribed circle in the triangle ABC). The angle of inclination of the lateral rib (for example SB) Is the angle between the edge itself and its projection onto the plane of the base. 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 equal to 3 but... Dot O line segment BD is divided into parts: and From we find SO: 
From we find: 
Answer:
Example 2. Find the volume of the correct truncated quadrangular pyramid, if the diagonals of its bases are equal to cm and cm, and the height is 4 cm.
Solution. To find the volume of the 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 2 cm and 8 cm, respectively. So the areas of the bases and Having substituted all the data in the formula, we calculate the volume of the truncated pyramid:
Answer: 112 cm 3.
Example 3. Find the area of the side 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 by condition, only the height remains unknown. We will find it from where BUT 1 E perpendicular from point BUT 1 on the plane of the lower base, A 1 D- perpendicular from BUT 1 on AS. BUT 1 E= 2 cm, since this is the height of the pyramid. To find DE we will make an additional drawing, in which we will depict a top view (fig. 20). Dot O- 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- radius of the inscribed circle:
MK = DE.
By 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 with the base plane of the pyramid equal to 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 apex is projected to the center of the circle inscribed in the base. Dot O- vertex projection S at the base of the pyramid. Triangle SOD is the orthogonal projection of the triangle CSD on the plane of the base. By the theorem on the area of the orthogonal projection of a plane figure, we get:
Similarly, it means 
Thus, the task was reduced to finding the area of the trapezoid ABCD... Draw a trapezoid ABCD separately (fig. 22). Dot O- the center of a circle inscribed in a trapezoid.
Since a circle can be inscribed into a trapezoid, either From, by the Pythagorean theorem, we have
Tasks on the topic: "Pyramid, Truncated Pyramid".
The height of a regular quadrangular pyramid is 6, and the apothem is 6.5. Find the perimeter of the base of this pyramid. Answer: 20.
The side surface of a regular pyramid is 24, and the base area is 12. At what angle are the side faces inclined to the base? Answer: 60
The volume of a regular quadrangular pyramid is 48, the height is 4. Find the area of the side surface of the pyramid. Answer: 60.
The height of the pyramid is 16. The area of the base is 512. At what distance from the base is the section parallel to it, if the section area is 50. Answer: 11
At the base of the pyramid lies a square with a diagonal equal to 6. One of the side edges is perpendicular to the base. The larger lateral rib is inclined towards the base at 45. What is the volume of the pyramid? Answer: 36.
In a triangular pyramid, the two side faces are mutually perpendicular. The areas of these faces are equal to P and Q, and the length of their common edge is equal to a. Determine the volume of the pyramid. Answer:
The base of the pyramid is a rectangle with sides 4 and 6. Each of the side edges is 7. Find the volume of the pyramid. Answer: 48.
In the pyramid, a plane of section parallel to the base divides the height in a ratio of 1: 1. Find the cross-sectional area if the base area is 60. Answer: 15
The side edges of the triangular pyramid are mutually perpendicular, each edge is equal to 3. Find the volume of the pyramid. Answer: 4.5
The volume of a regular quadrangular pyramid is 20, and its height is 1. Find the length of the apothem of the pyramid. Answer: 4
The height of a regular triangular pyramid is half the side of the base. Find the angle between the side face of the pyramid and the plane of the base. Answer: 60
Find the volume of a regular triangular pyramid if all lateral edges are inclined to the plane of the base at an angle of 45, and the median of the base is 6. Answer: 144
The height of the base of a regular triangular pyramid is 3, the lateral edge makes an angle of 30 with the height of the pyramid. Find the volume of the pyramid. Answer: 6
Find the area of the base of a regular triangular pyramid, whose height is 10, and the dihedral angle at the side of the base is 45. Answer: 900.
All side faces of the triangular pyramid make an angle of 45 with the base plane. Find the height of the pyramid if the sides of its base are 20.21 and 29. Answer: 6
At the base of the pyramid there is a triangle with sides 7,10 and 13. Height of the pyramid 4. Find the value of the dihedral angle at the base of the pyramid if all side faces are equally inclined to the plane of the base. Answer: 60
At the base of the pyramid lies an isosceles trapezoid, the lengths of the bases of which are 16 and 4. Find the height of the pyramid if each of its side faces makes an angle of 60 with the base. Answer: 4
The section of the pyramid by a plane parallel to the base divides the height of the pyramid in a ratio of 2: 3, counting from the top. The base area of the pyramid is 360. Find its cross-sectional area. Answer: 57.6
The base of the pyramid is a triangle with sides 5.5 and 6, the height of the pyramid passes through the center of the circle inscribed in this triangle and is equal to 2. Find the area of the side surface of the pyramid. Answer: 20.
The flat angles at the apex of the triangular pyramid are straight, the side edges of the pyramid are 5.6 and 7. Find the volume of the pyramid. Answer: 35
The sides of the bases of a regular truncated quadrangular pyramid are 4 and 6. Find the area of the diagonal section if the side edge forms an angle of 45 with a large base. Answer: 10
Find the height of a regular truncated quadrangular pyramid whose base sides are 14 and 10, and the diagonal is 18. Answer: 6.
The bases of the truncated pyramid have regular triangles with sides 2 and 6. Determine the height of this pyramid if its volume is 52. Answer: 12. В
The base of the pyramid is a rhombus with a side of 14 and an acute angle of 60. The dihedral angles at the base of the pyramid are 45 each. Calculate the volume of the pyramid. Answer: 343.
The area of the base of a regular quadrangular pyramid is 36, and its lateral surface is 60. Find the volume of the pyramid. Answer: 48
At the base of the pyramid is a triangle with sides 13, 14 and 15. Find the height of the pyramid if all the heights of the side faces are 14. Answer: 6
In what ratio does the plane parallel to the base divide the volume of the pyramid, if it divides the height in the ratio 3: 2? Answer: 27: 98
The base of the pyramid is a rhombus on side 6 and an acute angle of 30. Find the total surface area of the pyramid if each dihedral angle at the base is 60. Answer: 54.
At the base of the triangular pyramid FABC lies a regular triangle ABC with a side equal to FA =. The side faces of the pyramid have equal areas. Find the volume of the pyramid. Answer:
In a regular triangular pyramid, the side edge equal to 6 is inclined to the base at an angle of 30. Find the volume of the pyramid. Answer:
The height of a regular triangular pyramid is 2, and the side face makes an angle of 60 with the base plane. Find the volume of the pyramid. Answer: 24
Find the volume regular tetrahedron with an edge equal to a. Answer:, a = 5
The plane angle at the apex of a regular triangular pyramid is 90 *. The area of the side surface of the pyramid is 192. Find the radius of the circle circumscribed about the side face of the pyramid. Answer: 8
The angle between the side face and the plane of the base of a regular triangular pyramid is 45. The volume of the pyramid is equal. Find the side of the base of the pyramid. Answer: 2
The base of the pyramid is a rhombus with diagonals 6 and 8, the height of the pyramid passes through the intersection of the diagonals of the rhombus and is equal to 1. Find the side surface of the pyramid. Answer: 26
In a quadrangular pyramid, all lateral ribs are inclined to the plane of the base at an angle of 60. At its base lies an isosceles trapezium, the greater angle of which is 120. The diagonal of the trapezoid is the bisector of its acute angle. The height of the pyramid is 4. Find the larger base of the trapezoid. Answer: 8
Determine the volume of a regular quadrangular pyramid, knowing the angle = 30, made up by its lateral edge with the base plane, and the area of its diagonal section S =. Answer: 2.
The base of the pyramid is a regular triangle with a side. One of the side edges is perpendicular to the base, and the other two are inclined to the plane of the base at angles of 60. Find the area of the larger side face of the pyramid. Answer: 3.75
The base of the pyramid is a rectangle with an area of 81. Two side faces are perpendicular to the base plane, and the other two form angles 30 and 60 with it. Find the volume of the pyramid. Answer: 243
Find the volume of the pyramid, the base of which is an isosceles trapezoid with bases 10 and 20, and the side faces form dihedral angles equal to 60 with the base plane. Answer: 500
At the base of the pyramid lies a rectangular isosceles triangle with a hypotenuse with. Each edge of the pyramid is inclined to the base plane at an angle of 45. Find the total surface area of the pyramid. Answer:
The base side of a regular triangular pyramid is equal to a. The value of the angle formed by the height of the pyramid with the side face is 30. Find the total surface area of the pyramid. Answer:
The angle between the height of a regular quadrangular pyramid and its side edge is 60. Find the total surface area of the pyramid if its height is 10. Answer: 200 (3+)
The base of the pyramid is a rhombus with a larger diagonal of 12 and an acute angle of 60. All dihedral angles at the base of the pyramid are 45. Find the volume of the pyramid. Answer: 24
The bases of a regular truncated pyramid are squares with sides a and b (a> b). The lateral ribs are inclined to the plane of the base at an angle a. Determine the size of the dihedral angles at the sides of the bases. Answer : arctg (tga)
In a triangular truncated pyramid, the height is 10. The sides of one base are 27.29 and 52, and the perimeter of the other base is 72. Determine the volume of the truncated pyramid. Answer: 1900
At the bases of the truncated pyramid lie right-angled triangles with an acute angle of 60. The hypotenuses of these triangles are 6 and 4. The height of this pyramid. Find the volume of the learned pyramid. Answer: 9.5.
The sides of the bases of a regular quadrangular truncated pyramid are 4 and 4; the side face is inclined to the plane of the base at an angle of 60. Find the full surface of the pyramid. Answer: 128
The sides of the base of a regular quadrangular truncated pyramid are 3: 2. The height of the pyramid is 3. The lateral edge makes an angle of 60 with the base plane. Find the volume of the pyramid. Answer: 114
The lateral edge of a regular quadrangular truncated pyramid̊ is equal and inclined to the base plane at an angle of 60. The diagonal of the pyramid is perpendicular to the lateral edge. Find the area of the smaller base of the pyramid. Answer: 1.5
MUNICIPAL EDUCATIONAL INSTITUTION
"SCHOOL №2" OF THE CITY OF ALUSHTA
LESSON OUTLINE
Solving problems.
Pyramid. Truncated pyramid
Mathematic teacher
Pikhidchuk Irina Anatolyevna
2016 G.
LESSON
Geometry. Grade 11.
The lesson is designed for 3 hours. It is recommended to carry out with generalizing repetition.
TOPIC: Pyramid. Truncated pyramid. Solving problems.
THE MAIN TASK: Preparing for test work(identify problems; systematize and correct knowledge on the topic).
GOALS: 1) Check knowledge of the definitions: the angle between a straight line and a plane; linear angle of a dihedral angle (construction); correct pyramid.
Repeat the formulas: the volume of the pyramid; the radii of a circle inscribed and circumscribed about a polygon;
check drawing skills; the ability to justify the angles between the side edge and the plane of the base, between the side edge and the plane of the base.
reinforce computational skills.
DURING THE CLASSES:
Organizing time. Communication of the goals and objectives of the lesson.
Repetition.
Drawings on the flip-up board:
Assignment to the drawings: to formulate the definition of the angle between a straight line and a plane. Show the angle in the pictures and justify.
Main board
Show the angle between the side edge and the base plane of a regular triangular pyramid. Calculate the volume of the pyramid if the side of the base is a, the angle between the side edge and the plane of the base is a.
Find the volume of each of the given regular pyramids
CONCLUSION: 1) The angle between the lateral edge and the plane of the base is the angle between the lateral edge and the radius of the circle described near the base;
2) The angle between the lateral face and the plane of the base of the pyramid is the angle between the apothem and the radius of the circle inscribed in the base.
Homework on cards (assignment attached).
Geometry grade 11, (continued)
SOLUTION OF PROBLEMS: Pyramid. Truncated pyramid.
Problem number 1. At the base of the pyramid lies right triangle... The two faces containing the legs are perpendicular to the plane of the base. Show the angles between the side ribs and the plane of the base. Will they be equal if the triangle is isosceles.
Problem number 2. At the base of the pyramid is an isosceles triangle. The side ribs are inclined to the plane of the base at one angle. Plot the height of the pyramid and the angles between the side edges and the base plane (justify the construction)
Problem number 4. At the base of the pyramid lies a right-angled triangle. Each side rib forms the same angle with the base. Draw and justify the construction. Find the volume if the height of the pyramid is 7 cm and the angle between the side edge and the plane of the base is 60 0 .
CONCLUSION: The height of the pyramid is projected to the center of the circumscribed circle if: the side edges are equal; lateral ribs are inclined to the plane of the base at one angle; the pyramid is correct.
Homework. In a regular pyramid (triangular, quadrangular, hexagonal), build the angle between the side face and the base plane. Justify the construction.
