The video course "Get an A" includes all the topics necessary for the successful passing of the exam in mathematics by 60-65 points. Completely all tasks 1-13 of the Profile USE in mathematics. Also suitable for passing the Basic USE in mathematics. If you want to pass the exam with 90-100 points, you need to solve part 1 in 30 minutes and without mistakes!

Preparation course for the exam for grades 10-11, as well as for teachers. Everything you need to solve part 1 of the exam in mathematics (the first 12 problems) and problem 13 (trigonometry). And this is more than 70 points on the Unified State Examination, and neither a hundred-point student nor a humanist can do without them.

All the necessary theory. Quick solutions, traps and secrets of the exam. All relevant tasks of part 1 from the Bank of FIPI tasks have been analyzed. The course fully complies with the requirements of the USE-2018.

The course contains 5 large topics, 2.5 hours each. Each topic is given from scratch, simply and clearly.

Hundreds of exam tasks. Text problems and probability theory. Simple and easy to remember problem solving algorithms. Geometry. Theory, reference material, analysis of all types of USE tasks. Stereometry. Cunning tricks for solving, useful cheat sheets, development of spatial imagination. Trigonometry from scratch - to task 13. Understanding instead of cramming. Visual explanation of complex concepts. Algebra. Roots, powers and logarithms, function and derivative. Base for solving complex problems of the 2nd part of the exam.

The volume of the prism. Problem solving

Geometry is the most powerful tool for the refinement of our mental faculties and enables us to think and reason correctly.

G. Galileo

The purpose of the lesson:

• to teach solving problems for calculating the volume of prisms, to summarize and systematize the information that students have about the prism and its elements, to form the ability to solve problems of increased complexity;
• develop logical thinking, the ability to work independently, the skills of mutual control and self-control, the ability to speak and listen;
• develop the habit of constant employment, some useful deed, education of responsiveness, diligence, accuracy.

Type of lesson: a lesson in the application of knowledge, skills and abilities.

Equipment: control cards, media projector, presentation “Lesson. Prism volume”, computers.

During the classes

• Lateral ribs of the prism (Fig. 2).
• The side surface of the prism (Figure 2, Figure 5).
• The height of the prism (Figure 3, Figure 4).
• Direct prism (Fig. 2,3,4).
• Inclined prism (Figure 5).
• Correct prism (Fig. 2, Fig. 3).
• Diagonal section of a prism (Fig. 2).
• Prism diagonal (Figure 2).
• Perpendicular section of the prism (pi3, fig4).
• The area of ​​the lateral surface of the prism.
• The total surface area of ​​the prism.
• The volume of the prism.

1. CHECK HOMEWORK (8 min)

Exchange notebooks, check the solution on the slides and mark the mark (mark 10 if the task is composed)

Draw a problem and solve it. The student defends the problem he has compiled at the blackboard. Figure 6 and Figure 7.





Chapter 2, §3
Task.2. The lengths of all edges of a regular triangular prism are equal to each other. Calculate the volume of the prism if its surface area is cm 2 (Fig. 8)



Chapter 2, §3
Task 5. The base of the straight prism ABCA 1B 1C1 is a right triangle ABC (angle ABC=90°), AB=4cm. Calculate the volume of the prism if the radius of the circumscribed triangle ABC is 2.5cm and the height of the prism is 10cm. (Figure 9).



Chapter 2, § 3
Problem 29. The length of the side of the base of a regular quadrangular prism is 3cm. The diagonal of the prism forms an angle of 30° with the plane of the side face. Calculate the volume of the prism (Figure 10).



2. Joint work of the teacher with the class (2-3 min.).

Purpose: summing up the results of the theoretical warm-up (students put down marks to each other), learning how to solve problems on the topic.

3. PHYSICAL MINUTE (3 min)
4. PROBLEM SOLVING (10 min)

At this stage, the teacher organizes frontal work on the repetition of methods for solving planimetric problems, planimetry formulas. The class is divided into two groups, some solve problems, others work at the computer. Then they change. Students are invited to solve all No. 8 (orally), No. 9 (orally). After they are divided into groups and transgress to solve problems No. 14, No. 30, No. 32.

Chapter 2, §3, page 66-67

Problem 8. All edges of a regular triangular prism are equal to each other. Find the volume of the prism if the cross-sectional area of ​​​​the plane passing through the edge of the lower base and the middle of the side of the upper base is cm (Fig. 11).



Chapter 2, §3, page 66-67
Problem 9. The base of a straight prism is a square, and its side edges are twice the side of the base. Calculate the volume of the prism if the radius of the circle circumscribed near the section of the prism by a plane passing through the side of the base and the middle of the opposite side edge is equal to (Fig. 12)



Chapter 2, §3, page 66-67
Task 14.The base of a straight prism is a rhombus, one of the diagonals of which is equal to its side. Calculate the perimeter of the section by a plane passing through the large diagonal of the lower base, if the volume of the prism is equal and all side faces are square (Fig. 13).



Chapter 2, §3, page 66-67
Problem 30.ABCA 1 B 1 C 1 is a regular triangular prism, all the edges of which are equal to each other, the point about the middle of the edge BB 1. Calculate the radius of the circle inscribed in the section of the prism by the AOS plane, if the volume of the prism is equal (Fig. 14).



Chapter 2, §3, page 66-67
Problem 32.In a regular quadrangular prism, the sum of the areas of the bases is equal to the area of ​​the lateral surface. Calculate the volume of the prism if the diameter of the circle circumscribed near the section of the prism by a plane passing through two vertices of the lower base and the opposite vertex of the upper base is 6 cm (Fig. 15).



While solving problems, students compare their answers with those shown by the teacher. This is an example of solving a problem with detailed comments ... Individual work of a teacher with “strong” students (10 min.).

5. Independent work of students on the test at the computer

1. The side of the base of a regular triangular prism is , and the height is 5. Find the volume of the prism.

1) 152) 45 3) 104) 125) 18

2. Choose the correct statement.

1) The volume of a right prism, the base of which is a right triangle, is equal to the product of the base area and the height.

2) The volume of a regular triangular prism is calculated by the formula V \u003d 0.25a 2 h - where a is the side of the base, h is the height of the prism.

3) The volume of a straight prism is equal to half the product of the area of ​​\u200b\u200bthe base and the height.

4) The volume of a regular quadrangular prism is calculated by the formula V \u003d a 2 h-where a is the side of the base, h is the height of the prism.

5) The volume of a regular hexagonal prism is calculated by the formula V \u003d 1.5a 2 h, where a is the side of the base, h is the height of the prism.

3. The side of the base of a regular triangular prism is equal to. A plane is drawn through the side of the lower base and the opposite top of the upper base, which passes at an angle of 45° to the base. Find the volume of the prism.

1) 92) 9 3) 4,54) 2,255) 1,125

4. The base of a straight prism is a rhombus, the side of which is 13, and one of the diagonals is 24. Find the volume of the prism if the diagonal of the side face is 14.

In the school curriculum for the course of solid geometry, the study of three-dimensional figures usually begins with a simple geometric body - a prism polyhedron. The role of its bases is performed by 2 equal polygons lying in parallel planes. A special case is a regular quadrangular prism. Its bases are 2 identical regular quadrangles, to which the sides are perpendicular, having the shape of parallelograms (or rectangles if the prism is not inclined).

What does a prism look like

A regular quadrangular prism is a hexahedron, at the bases of which there are 2 squares, and the side faces are represented by rectangles. Another name for this geometric figure is a straight parallelepiped.

The figure, which depicts a quadrangular prism, is shown below.

You can also see in the picture the most important elements that make up a geometric body. They are commonly referred to as:

Sometimes in problems in geometry you can find the concept of a section. The definition will sound like this: a section is all points of a volumetric body that belong to the cutting plane. The section is perpendicular (crosses the edges of the figure at an angle of 90 degrees). For a rectangular prism, a diagonal section is also considered (the maximum number of sections that can be built is 2), passing through 2 edges and the diagonals of the base.

If the section is drawn in such a way that the cutting plane is not parallel to either the bases or the side faces, the result is a truncated prism.

Various ratios and formulas are used to find the reduced prismatic elements. Some of them are known from the course of planimetry (for example, to find the area of ​​the base of a prism, it is enough to recall the formula for the area of ​​a square).

Surface area and volume

To determine the volume of a prism using the formula, you need to know the area of ​​\u200b\u200bits base and height:

V = Sprim h

Since the base of a regular tetrahedral prism is a square with side a, You can write the formula in a more detailed form:

V = a² h

If we are talking about a cube - a regular prism with equal length, width and height, the volume is calculated as follows:

To understand how to find the lateral surface area of ​​a prism, you need to imagine its sweep.

It can be seen from the drawing that the side surface is made up of 4 equal rectangles. Its area is calculated as the product of the perimeter of the base and the height of the figure:

Sside = Pos h

Since the perimeter of a square is P = 4a, the formula takes the form:

Sside = 4a h

For cube:

Sside = 4a²

To calculate the total surface area of ​​a prism, add 2 base areas to the side area:

Sfull = Sside + 2Sbase

As applied to a quadrangular regular prism, the formula has the form:

Sfull = 4a h + 2a²

For the surface area of ​​a cube:

Sfull = 6a²

Knowing the volume or surface area, you can calculate the individual elements of a geometric body.

Finding prism elements

Often there are problems in which the volume is given or the value of the lateral surface area is known, where it is necessary to determine the length of the side of the base or the height. In such cases, formulas can be derived:

• base side length: a = Sside / 4h = √(V / h);
• height or side rib length: h = Sside / 4a = V / a²;
• base area: Sprim = V / h;
• side face area: Side gr = Sside / 4.

To determine how much area a diagonal section has, you need to know the length of the diagonal and the height of the figure. For a square d = a√2. Therefore:

Sdiag = ah√2

To calculate the diagonal of the prism, the formula is used:

dprize = √(2a² + h²)

To understand how to apply the above ratios, you can practice and solve a few simple tasks.

Examples of problems with solutions

Here are some of the tasks that appear in the state final exams in mathematics.

Exercise 1.

Sand is poured into a box shaped like a regular quadrangular prism. The height of its level is 10 cm. What will the level of sand be if you move it into a container of the same shape, but with a base length 2 times longer?

It should be argued as follows. The amount of sand in the first and second containers did not change, i.e., its volume in them is the same. You can define the length of the base as a. In this case, for the first box, the volume of the substance will be:

V₁ = ha² = 10a²

For the second box, the length of the base is 2a, but the height of the sand level is unknown:

V₂ = h(2a)² = 4ha²

Insofar as V₁ = V₂, the expressions can be equated:

10a² = 4ha²

After reducing both sides of the equation by a², we get:

As a result, the new sand level will be h = 10 / 4 = 2.5 cm.

Task 2.

ABCDA₁B₁C₁D₁ is a regular prism. It is known that BD = AB₁ = 6√2. Find the total surface area of ​​the body.

To make it easier to understand which elements are known, you can draw a figure.

Since we are talking about a regular prism, we can conclude that the base is a square with a diagonal of 6√2. The diagonal of the side face has the same value, therefore, the side face also has the shape of a square equal to the base. It turns out that all three dimensions - length, width and height - are equal. We can conclude that ABCDA₁B₁C₁D₁ is a cube.

The length of any edge is determined through the known diagonal:

a = d / √2 = 6√2 / √2 = 6

The total surface area is found by the formula for the cube:

Sfull = 6a² = 6 6² = 216

Task 3.

The room is being renovated. It is known that its floor has the shape of a square with an area of ​​9 m². The height of the room is 2.5 m. What is the lowest cost of wallpapering a room if 1 m² costs 50 rubles?

Since the floor and ceiling are squares, that is, regular quadrilaterals, and its walls are perpendicular to horizontal surfaces, we can conclude that it is a regular prism. It is necessary to determine the area of ​​its lateral surface.

The length of the room is a = √9 = 3 m.

The square will be covered with wallpaper Sside = 4 3 2.5 = 30 m².

The lowest cost of wallpaper for this room will be 50 30 = 1500 rubles.

Thus, to solve problems on a rectangular prism, it is enough to be able to calculate the area and perimeter of a square and a rectangle, as well as to know the formulas for finding the volume and surface area.

How to find the area of ​​a cube

What is the volume of a prism and how to find it

The volume of a prism is the product of the area of ​​its base times its height.

However, we know that the base of a prism can have a triangle, a square, or some other polyhedron.

Therefore, to find the volume of a prism, you just need to calculate the area of ​​​​the base of the prism, and then multiply this area by its height.

That is, if there is a triangle at the base of the prism, then first you need to find the area of ​​\u200b\u200bthe triangle. If the base of the prism is a square or another polygon, then first you need to find the area of ​​the square or another polygon.

It should be remembered that the height of the prism is a perpendicular drawn to the bases of the prism.

What is a prism

Now let's remember the definition of a prism.

A prism is a polygon whose two faces (bases) are in parallel planes, and all edges outside these faces are parallel.

To put it simply, then:

A prism is any geometric figure that has two equal bases and flat faces.

The name of a prism depends on the shape of its base. When the base of a prism is a triangle, then such a prism is called triangular. A polyhedral prism is a geometric figure whose base is a polyhedron. A prism is also a kind of cylinder.

What are the types of prisms

If we look at the figure above, we can see that prisms are straight, regular and oblique.

Exercise

1. What is the correct prism?
2. Why is it called that?
3. What is the name of a prism whose bases are regular polygons?
4. What is the height of this figure?
5. What is the name of a prism whose edges are not perpendicular?
6. Define a triangular prism.
7. Can a prism be a parallelepiped?
8. What geometric figure is called a semi-regular polygon?

What elements does a prism consist of?

A prism consists of elements such as the bottom and top base, side faces, edges, and vertices.

Both bases of the prism lie in planes and are parallel to each other.
The side faces of the pyramid are parallelograms.
The lateral surface of the pyramid is the sum of the lateral faces.
The common sides of the side faces are nothing more than the side edges of this figure.
The height of the pyramid is the segment connecting the planes of the bases and is perpendicular to them.

Prism Properties

A geometric figure, like a prism, has a number of properties. Let's take a closer look at these properties:

First, the bases of a prism are called equal polygons;
Secondly, the side faces of the prism are presented in the form of a parallelogram;
Thirdly, this geometric figure has parallel and equal edges;
Fourth, the total surface area of ​​the prism is:

And now consider the theorem, which provides a formula by which to calculate the lateral surface area and proof.

Have you ever thought about such an interesting fact that a prism can be not only a geometric body, but also other objects around us. Even an ordinary snowflake, depending on the temperature regime, can turn into an ice prism, taking the form of a hexagonal figure.

But calcite crystals have such a unique phenomenon as to break up into fragments and take the shape of a parallelepiped. And what is most surprising, no matter how small the calcite crystals are crushed, the result is always the same, they turn into tiny parallelepipeds.

It turns out that the prism has gained popularity not only in mathematics, demonstrating its geometric body, but also in the field of art, since it is the basis of paintings created by such great artists as P. Picasso, Braque, Griss and others.

Job type: 8
Theme: Prism

Condition

In a regular triangular prism ABCA_1B_1C_1, the sides of the base are 4 , and the side edges are 10 . Find the sectional area of ​​the prism by the plane passing through the midpoints of edges AB, AC, A_1B_1 and A_1C_1.

Show Solution

Decision

Consider the following figure.

Segment MN is the midline of triangle A_1B_1C_1, so MN = \frac12 B_1C_1=2. Likewise, KL=\frac12BC=2. In addition, MK = NL = 10. This implies that the quadrilateral MNLK is a parallelogram. Since MK\parallel AA_1, then MK\perp ABC and MK\perp KL. Therefore, quadrilateral MNLK is a rectangle. S_(MNLK) = MK\cdot KL= 10\cdot 2 = 20.

Answer

Job type: 8
Theme: Prism

Condition

The volume of a regular quadrangular prism ABCDA_1B_1C_1D_1 is 24 . Point K is the middle of edge CC_1 . Find the volume of the pyramid KBCD.

Show Solution

Decision

According to the condition, KC is the height of the pyramid KBCD . CC_1 is the height of the prism ABCDA_1B_1C_1D_1 .

Since K is the midpoint of CC_1 , then KC=\frac12CC_1. Let CC_1=H , then KC=\frac12H. Note also that S_(BCD)=\frac12S_(ABCD). Then, V_(KBCD)= \frac13S_(BCD)\cdot\frac(H)(2)= \frac13\cdot\frac12S_(ABCD)\cdot\frac(H)(2)= \frac(1)(12)\cdot S_(ABCD)\cdot H= \frac(1)(12)V_(ABCDA_1B_1C_1D_1). Hence, V_(KBCD)=\frac(1)(12)\cdot24=2.

Answer

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

Job type: 8
Theme: Prism

Condition

Find the lateral surface area of ​​a regular hexagonal prism whose base side is 6 and its height is 8.

Show Solution

Decision

The area of ​​the lateral surface of the prism is found by the formula S side. = P main. · h = 6a\cdot h, where P main. and h are, respectively, the perimeter of the base and the height of the prism, equal to 8 , and a is the side of a regular hexagon, equal to 6 . Therefore, S side. = 6\cdot 6\cdot 8 = 288.

Answer

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

Job type: 8
Theme: Prism

Condition

Water is poured into a vessel shaped like a regular triangular prism. The water level reaches 40 cm. At what height will the water level be if it is poured into another vessel of the same shape, whose base side is twice that of the first? Express your answer in centimeters.

Show Solution

Decision

Let a be the side of the base of the first vessel, then 2 a is the side 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= \frac12\cdot a^2\cdot\sin60^(\circ)\cdot40= \frac(a^2\sqrt3)(4)\cdot40, and, V=\frac((2a)^2\sqrt3)(4)\cdot H. From here \frac(a^2\sqrt3)(4)\cdot40=\frac((2a)^2\sqrt3)(4)\cdot H, 40=4H, H=10.

Answer

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

Job type: 8
Theme: Prism

Condition

In a regular hexagonal prism ABCDEFA_1B_1C_1D_1E_1F_1 all edges are 2 . Find the distance between points A and E_1 .

Show Solution

Decision

Triangle AEE_1 is right-angled, since edge EE_1 is perpendicular to the plane of the base of the prism, angle AEE_1 will be a right angle.

Then by the Pythagorean theorem AE_1^2 = AE^2 + EE_1^2. Find AE from the triangle AFE using the cosine theorem. Each interior angle of a regular hexagon is 120^(\circ). Then AE^2= AF^2+FE^2-2\cdot AF\cdot FE\cdot\cos120^(\circ)= 2^2+2^2-2\cdot2\cdot2\cdot\left (-\frac12 \right).

Hence, AE^2=4+4+4=12,

AE_1^2=12+4=16,

AE_1=4.

Answer

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

Job type: 8
Theme: Prism

Condition

Find the area of ​​the lateral surface of a straight prism whose base is a rhombus with diagonals equal to 4\sqrt5 and 8 , and a side edge equal to 5 .

Show Solution

Decision

The area of ​​the lateral surface of a straight prism is found by the formula S side. = P main. · h = 4a\cdot h, where P main. and h, respectively, the perimeter of the base and the height of the prism, equal to 5, and a is the side of the rhombus. Let's find the side of the rhombus, using the fact that the diagonals of the rhombus ABCD are mutually perpendicular and the intersection point is divided in half.