Stereometry developed from observations and solutions to issues that arose in the process of human practical activity. There is no doubt that even primitive man, having changed from a nomadic life to a settled life, having taken up agriculture, made attempts to estimate, at least in the roughest terms, the size of the harvest he had gathered by the masses of bread stacked in heaps, shocks or stacks. The builder of even the most ancient primitive buildings had to somehow take into account the material that he had at his disposal, and be able to calculate how much material would be needed to build a particular building. Stonecutting among the ancient Egyptians and Chaldeans required familiarity with the metric properties of at least the simplest geometric bodies: a cube, a parallelepiped, a prism, a cylinder, etc. The needs of agriculture, navigation, orientation in time pushed people to astronomical observations, and the latter to the study of the properties of the sphere and its parts, and, consequently, the laws of the relative position of planes and lines in space.

During the economic and cultural flourishing of ancient Greece and its colonies, geometry reached a high theoretical development. Of the outstanding geometers of Greece, Anaxagoras, Democritus, Hippocrates (5th century BC) were interested in stereometry. Hippocrates is among the first to solve the famous problem of antiquity - the Delhi problem of doubling the cube. In Plato's school, the problems of stereometry advanced considerably. One of the representatives of the school of Plato, Teetetus, considered the octahedron and the twenty-sided one and for the first time gave a theory of some properties of five regular polyhedra. Plato's student Menechme was the first to give some theory of conic sections. The greatest merit of Euclid is that he collected, processed and brought into a coherent system the material that had come down to him. Of the 13 books of his "Beginnings" of stereometry, XI-XIII books are assigned. The information about stereometry collected by Euclid was supplemented, deepened and expanded by the greatest mathematician of antiquity Archimedes. He gave thirteen semi-regular solids, each of which is limited by regular polygons, but not of the same kind, and calculated the volumes of solids of revolution. Thanks to the work of Archimedes, stereometry reached its culminating point, and elementary geometry in its modern sense was finally established.

After the fall of Greece, there is a long stagnation in the development of mathematics and stereometry in particular, which lasted for a thousand years. Much has been done by Kepler for the development of stereometry in modern times. In his "New Stereometry" - "stereometry of barrels" - he first used an infinitesimal quantity in geometry. The discovery of integral calculus by Newton and Leibniz finally solved the problem of quadrature and cubature.

Cylinder- a body that consists of two circles that do not lie in the same plane and are combined by parallel translation, and all segments connecting the corresponding points of these circles.

r is the radius of the cylinder;
d is the diameter of the cylinder;
l is the generatrix of the cylinder;
h is the height of the cylinder.

Note: in a right circular cylinder, the length of the generatrix is ​​equal to the length of the height.

Volume of a circular cylinder calculated by the formula:

V = π r 2 h, where

π – constant value (≈3.1415 );
r is the radius of the base of the cylinder;
h is the height of the cylinder.

Cube is a regular polyhedron, each face of which is a square. All edges of a cube are equal.

ABCDA 1 B 1 C 1 D 1 - cube;

A, B, C, D, A 1 , B 1 , C 1 , D 1- cube vertices;

a - the length of the edge of the cube.

Cube volume calculated by the formula:

V cube \u003d a 3, where

a is the length of the cube edge.

Tetrahedron is a regular polyhedron whose faces are four triangles.

ABCD - tetrahedron;

A, B, C, D - tetrahedron vertices;

AD, BD, CD, AB, BC, AC - edges of the tetrahedron;

ABD, BCD, ACD - faces of a tetrahedron.

Volume of a tetrahedron calculated by the formula:

a is the length of any edge of the tetrahedron.

Guidelines

To successfully complete tasks in this category, you must:

know the definitions of geometric bodies and their properties;

be able to perform actions with geometric shapes, coordinates and vectors;

be able to solve stereometric problems for finding geometric quantities (lengths, angles, areas, volumes);

know the formulas for calculating the areas and volumes of geometric bodies.

Tyes No. 8 Cylinder volume Option 1.

1. Find the volume of a cylinder with a height of 3 cm and a base diameter of 6 cm. a) 27π cm 3; b) 9π cm 3; c) 36π cm 3; d) 18π cm 3; e) 54π cm 3.

2. The volume of the cylinder is 27π. Find the diameter of the base of the cylinder if its total surface area is twice the lateral surface area.

a) 3; b) cannot be determined at 6; d) 2; e) 9.

3. The diagonal of the axial section of the cylinder makes an angle of 60˚ with the plane of the base of the cylinder. Find the volume of the cylinder if the area of ​​the axial section is 16√3 cm2.

a) 16π ​​cm 3; b) 16√3 cm 3; c) 32π√3 cm 3; d) 8π√3 cm 3; e) 16π√3 cm3.

4. A sphere of radius 1 cm is inscribed in a cylinder. Find the volume of the cylinder.

a) 4π cm 3; b) 2π cm 3; c) 8π cm 3; d) π cm 3; d) cannot be determined.

5. The volume of the cylinder is 120. Find the height of the cylinder with an accuracy of 0.01 if the radius of the base is 3 times greater than it.

a) 1.62; b) 1.63; c) 1.61; d) 1.6; e) 1.60.

6. The area of ​​​​the axial section of the cylinder is 21 cm 2, the area of ​​\u200b\u200bthe base is 18π cm 2. Find the volume of the cylinder.

a) 9π cm 3; b) 31.5π√2 cm 3; c) 21π cm 3; d) 63π cm 3; e) 31.5π√3 cm3.

7. Choose the correct statement.

a) The volume of a cylinder is half the product of the area of ​​the base and the height.

b) The volume of the cylinder is calculated by the formula V = πS/2, where S is the area of ​​the axial section of the cylinder;

c) the volume of an equilateral cylinder is V = 2πR 3 , where R is the radius of the base of the cylinder;

d) the volume of the cylinder is calculated by the formula V = Mh/2, where M is the area of ​​the lateral surface of the cylinder, and h is its height;

8. A section parallel to the axis of the cylinder cuts off an arc of 120˚ from the circumference of the base. The radius of the base of the cylinder is R, the angle between the diagonal of the section and the axis of the cylinder is 30˚. Find the volume of the cylinder a) 3πR 2 ; b) πR 3 √3; c) 3πR 3 ; d) πR 3 ; e) 3πR 3 √3.

9. Two planes are drawn through the generatrix of the cylinder. The angle between them is 120˚. The areas of the resulting sections are 1. The radius of the base of the cylinder is 1. Find the volume of the cylinder. a) π√3/3; b) 2π; c) π/2; d) pi; d) cannot be determined.

10. An aluminum wire with a diameter of 2 mm has a mass of 3.4 kg. Find the length of the wire to the nearest 1 cm if the density of aluminum is 2.6 g/cm3.

a) 41646; b) 43590; c) 41656; d) 41635; e) 41625.

Tyes No. 8 Cylinder volume Option 2.

1. Find the volume of a cylinder with a height of 6 cm and a base diameter of 3 cm. a) 13.5π cm 3; b) 9π cm 3; c) 27π cm 3; d) 18π cm 3; e) 54π cm 3.

2. The volume of the cylinder is 32π. Find the height of the cylinder if its total surface area is three times the lateral surface area.

a) 3; b) cannot be determined at 4; d) 8; D 2.

3. The diagonal of the axial section of the cylinder makes an angle of 60˚ with the plane of the base of the cylinder. Find the area of ​​the axial section if the volume of the cylinder is 16 π √3 cm 2.

a) 16 cm 2; b) 16√3 cm 2; c) 32√3 cm 2; d) 8√3 cm 2; e) 16π√3 cm2.

4. A sphere of radius 1 cm is described near the cylinder. Find the volume of the cylinder.

a) 4π√2 cm 3; b) 0.5π√2 cm 3; c) cannot be determined d) π cm 3; e) π√2 cm 3.

5. The volume of the cylinder is 120. Find the height of the cylinder with an accuracy of 0.01 if the radius of the base is 3 times less than it.

a) 2.3; b) 2.33; c) 2.35; d) 2.335; e) 2.34.

6. The area of ​​​​the axial section of the cylinder is 30 cm 2, the area of ​​\u200b\u200bthe base is 9π cm 2. Find the volume of the cylinder.

a) 45π cm 3; b) 22.5π cm 3; c) 23π cm 3; d) 9π cm 3; e) 30π cm 3.

7. Choose the wrong statement.

a) The volume of a cylinder is the product of the area of ​​the base and the height.

b) The volume of the cylinder is calculated by the formula V = 1/2πrS, where S is the area of ​​the axial section of the cylinder, and r is the radius of the cylinder;

c) the volume of an equilateral cylinder is calculated by the formula V = 1/4πh 3, where h is the height of the cylinder;

d) the volume of the cylinder is calculated by the formula V = 1/2Mr, where M is the area of ​​the lateral surface of the cylinder, and r is its radius;

e) the volume of an equilateral cylinder is calculated by the formula V = πh 3 /2, where h is the height of the cylinder.

8. A section parallel to the axis of the cylinder cuts off an arc of 120 0 from the circumference of the base. This section is removed from the axis of the cylinder by a distance equal to a. The diagonal of the section is 4a. Find the volume of the cylinder. a) 8pa 2 ; b) 4pa 3 ; c) 2πa 3 ; d) 16pa 3 ; e) 8πa 3 .

9. Two planes are drawn through the generatrix of the cylinder. The angle between them is 120˚. The areas of the resulting sections are 1. The height of the cylinder is 1. Find the volume of the cylinder. a) π/4; b) π/2; c) π; d) π/3; d) cannot be determined.

10. An aluminum wire with a diameter of 2 mm has a mass of 3.4 m. Find the mass of the wire with an accuracy of 1 g if the density of aluminum is 2.6 g / cm 3.

a) 278; b) 277; c) 29; d) 27; e) 28.

Job type: 8
Theme: Cylinder

Condition

In a cylindrical vessel, the liquid level reaches 20 cm. At what height will the liquid level be if it is poured into a second cylindrical vessel, the diameter of which is twice the diameter of the first? Express your answer in centimeters.

Show Solution

Decision

Let R be the radius of the base of the first vessel, then 2 R is the radius of the base of the second vessel. By condition, the volume of liquid V in the first and second vessel is the same. Denote by H - the level to which the liquid has risen in the second vessel. Then

V=\pi R^2 \cdot 20, and V=\pi (2R)^2H = 4\pi R^2H. From here \pi R^2 \cdot 20 = 4\pi R^2H, 20=4H H=5

Answer

Job type: 8
Theme: Cylinder

Condition

2000 cm 3 of water was poured into a cylindrical vessel. The liquid level turned out to be 15 cm. The part was completely immersed in water. At the same time, the liquid level in the vessel rose by 9 cm. What is the volume of the part? Express your answer in cm3.

Show Solution

Decision

Let R be the radius of the base of the cylinder, and h be the level of the water poured into the vessel. Then the volume of poured water is equal to the volume of a cylinder with base radius R and height h. V water \u003d S main. · h = \pi R^2\cdot h. According to the condition, the equality 2000=\pi R^2\cdot15 is fulfilled. From here, \pi R^2=\frac(2000)(15)=\frac(400)(3).

Let H be the water level in the vessel after the item is immersed in it. Then the total volume of water and the part is equal to the volume of a cylinder with a base radius R and a height H. By condition H=h+9=15+9=24. So V water + details = \pi R^2\cdot H=\frac(400)(3)\cdot24=3200. Therefore, V parts = V water + parts − V water = 3200-2000=1200.

Answer

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 8
Theme: Cylinder

Condition

Find the height of a cylinder if its base radius is 8 and its side surface area is 96\pi.

Show Solution

Decision

S=2\pi rh,

96\pi=2\pi\cdot8h,

h=\frac(96\pi)(16\pi)=6.

Answer

Source: "Mathematics. Preparation for the exam-2016. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 8
Theme: Cylinder

Condition

500 cubic meters were poured into a cylindrical vessel. see water. Determine the volume of the part completely submerged in water if, after immersion, the liquid level increased by 1.2 times. Express your answer in cube. cm.

Show Solution

Decision

Let V 1 denote the initial volume of liquid in the cylinder. After the part was immersed, the volume of liquid increased by 1.2 times, which means that the final volume of liquid is V 2 = 1.2 V 1. The volume of the part is equal to the difference between the volumes before and after immersion, which means V = V_2-V_1=1.2\cdot 500-500=100 cube cm.

Answer

When a liquid is overflowed, its initial volume does not change, i.e.: V 1 \u003d V 2, which means that the equality is true: \pi\left(\frac(d_1)(2)\right)^2h_1=\pi\left(\frac(3d_1)(2)\right)^2h_2

Substitute the values ​​from the condition, simplify the expression and find the desired height of the liquid of the second vessel h 2:

\pi \enspace\frac(d_1^(2))(4)\enspace 63=\pi \enspace\frac(9d_1^(2))(4)\enspace h_2

\frac(63)(4)=\frac(9)(4)h_2

h_2=\frac(63)(9)=7