The video course “Get an A” includes all the topics necessary for successful passing the Unified State Exam in mathematics for 60-65 points. Completely all tasks 1-13 of the Profile Unified State Exam in mathematics. Also suitable for passing the Basic Unified State Examination in mathematics. If you want to pass the Unified State Exam with 90-100 points, you need to solve part 1 in 30 minutes and without mistakes!
Preparation course for the Unified State Exam for grades 10-11, as well as for teachers. Everything you need to solve Part 1 of the Unified State Exam in mathematics (the first 12 problems) and Problem 13 (trigonometry). And this is more than 70 points on the Unified State Exam, and neither a 100-point student nor a humanities student can do without them.
All the necessary theory. Quick ways solutions, pitfalls and secrets of the Unified State Exam. All current tasks of part 1 from the FIPI Task Bank have been analyzed. The course fully complies with the requirements of the Unified State Exam 2018.
The course contains 5 large topics, 2.5 hours each. Each topic is given from scratch, simply and clearly.
Hundreds of Unified State Exam tasks. Word problems and probability theory. Simple and easy to remember algorithms for solving problems. Geometry. Theory, reference material, analysis of all types of Unified State Examination tasks. Stereometry. Tricky solutions, useful cheat sheets, development of spatial imagination. Trigonometry from scratch to problem 13. Understanding instead of cramming. Clear explanations of complex concepts. Algebra. Roots, powers and logarithms, function and derivative. Basis for solution complex tasks 2 parts of the Unified State Exam.
To solve geometry problems, you need to know formulas - such as the area of a triangle or the area of a parallelogram - as well as simple techniques that we will cover.
First, let's learn the formulas for the areas of figures. We have specially collected them in a convenient table. Print, learn and apply!
Of course, not all geometry formulas are in our table. For example, to solve problems in geometry and stereometry in the second part of the profile Unified State Exam in mathematics, other formulas for the area of a triangle are used. We will definitely tell you about them.
But what if you need to find not the area of a trapezoid or triangle, but the area of some complex figure? There are universal ways! We will show them using examples from the FIPI task bank.
1. How to find the area of a non-standard figure? For example, an arbitrary quadrilateral? A simple technique - let's divide this figure into those that we know everything about, and find its area - as the sum of the areas of these figures.
Divide this quadrilateral with a horizontal line into two triangles with a common base equal to . The heights of these triangles are equal to and . Then the area of the quadrilateral is equal to the sum of the areas of the two triangles: .
Answer: .
2. In some cases, the area of a figure can be represented as the difference of some areas.
It is not so easy to calculate what the base and height of this triangle are equal to! But we can say that its area is equal to the difference between the areas of a square with a side and three right triangles. Do you see them in the picture? We get: .
Answer: .
3. Sometimes in a task you need to find the area of not the entire figure, but part of it. Usually we are talking about the area of a sector - part of a circle. Find the area of a sector of a circle of radius whose arc length is equal to .
In this picture we see part of a circle. The area of the entire circle is equal to . It remains to find out which part of the circle is depicted. Since the length of the entire circle is equal (since), and the length of the arc of a given sector is equal, therefore, the length of the arc is one time less than the length of the entire circle. The angle at which this arc rests is also a factor of less than a full circle (that is, degrees). This means that the area of the sector will be several times smaller than the area of the entire circle.
And the ancient Egyptians used methods for calculating the areas of various figures, similar to our methods.
In my books "Beginnings" the famous ancient Greek mathematician Euclid described quite big number methods for calculating the areas of many geometric shapes. The first manuscripts in Rus' containing geometric information were written in the 16th century. They describe the rules for finding the areas of figures of various shapes.
Today with the help modern methods you can find the area of any figure with great accuracy.
Let's consider one of the simplest figures - a rectangle - and the formula for finding its area.
Rectangle area formula
Let's consider a figure (Fig. 1), which consists of $8$ squares with sides of $1$ cm. The area of one square with a side of $1$ cm is called a square centimeter and is written $1\ cm^2$.
The area of this figure (Fig. 1) will be equal to $8\cm^2$.
The area of a figure that can be divided into several squares with a side of $1\ cm$ (for example, $p$) will be equal to $p\ cm^2$.
In other words, the area of the figure will be equal to so many $cm^2$, into how many squares with side $1\ cm$ this figure can be divided.
Let's consider a rectangle (Fig. 2), which consists of $3$ stripes, each of which is divided into $5$ squares with a side of $1\ cm$. the entire rectangle consists of $5\cdot 3=15$ such squares, and its area is $15\cm^2$.
Picture 1.
Figure 2.
The area of figures is usually denoted by the letter $S$.
To find the area of a rectangle, you need to multiply its length by its width.
If we denote its length by the letter $a$, and its width by the letter $b$, then the formula for the area of a rectangle will look like:
Definition 1
The figures are called equal if, when superimposed on one another, the figures coincide. Equal figures have equal areas and equal perimeters.
The area of a figure can be found as the sum of the areas of its parts.
Example 1
For example, in Figure $3$, rectangle $ABCD$ is divided into two parts by line $KLMN$. The area of one part is $12\ cm^2$, and the other is $9\ cm^2$. Then the area of the rectangle $ABCD$ will be equal to $12\ cm^2+9\ cm^2=21\ cm^2$. Find the area of the rectangle using the formula:
As you can see, the areas found by both methods are equal.
Figure 3.
Figure 4.
The segment $AC$ divides the rectangle into two equal triangle: $ABC$ and $ADC$. This means that the area of each triangle is equal to half the area of the entire rectangle.
Definition 2
A rectangle with equal sides is called square.
If we denote the side of a square with the letter $a$, then the area of the square will be found by the formula:
Hence the name square of the number $a$.
Example 2
For example, if the side of a square is $5$ cm, then its area is:
Volumes
With the development of trade and construction back in the days of ancient civilizations, the need arose to find volumes. In mathematics, there is a branch of geometry that deals with the study of spatial figures, called stereometry. Mentions of this separate branch of mathematics were found already in the $IV$ century BC.
Ancient mathematicians developed a method for calculating the volume of simple figures - a cube and a parallelepiped. All buildings of those times were of this shape. But later methods were found to calculate the volume of figures of more complex shapes.
Volume of a rectangular parallelepiped
If you fill the mold with wet sand and then turn it over, you will get a three-dimensional figure that is characterized by volume. If you make several such figures using the same mold, you will get figures that have the same volume. If you fill the mold with water, then the volume of water and the volume of the sand figure will also be equal.
Figure 5.
You can compare the volumes of two vessels by filling one with water and pouring it into the second vessel. If the second vessel is completely filled, then the vessels have equal volumes. If water remains in the first, then the volume of the first vessel is greater than the volume of the second. If, when pouring water from the first vessel, it is not possible to completely fill the second vessel, then the volume of the first vessel is less than the volume of the second.
Volume is measured using the following units:
$mm^3$ -- cubic millimeter,
$cm^3$ -- cubic centimeter,
$dm^3$ -- cubic decimeter,
$m^3$ -- cubic meter,
$km^3$ -- cubic kilometer.
And the ancient Egyptians used methods for calculating the areas of various figures, similar to our methods.
In my books "Beginnings" The famous ancient Greek mathematician Euclid described a fairly large number of ways to calculate the areas of many geometric figures. The first manuscripts in Rus' containing geometric information were written in the 16th century. They describe the rules for finding the areas of figures of various shapes.
Today, using modern methods, you can find the area of any figure with great accuracy.
Let's consider one of the simplest figures - a rectangle - and the formula for finding its area.
Rectangle area formula
Let's consider a figure (Fig. 1), which consists of $8$ squares with sides of $1$ cm. The area of one square with a side of $1$ cm is called a square centimeter and is written $1\ cm^2$.
The area of this figure (Fig. 1) will be equal to $8\cm^2$.
The area of a figure that can be divided into several squares with a side of $1\ cm$ (for example, $p$) will be equal to $p\ cm^2$.
In other words, the area of the figure will be equal to so many $cm^2$, into how many squares with side $1\ cm$ this figure can be divided.
Let's consider a rectangle (Fig. 2), which consists of $3$ stripes, each of which is divided into $5$ squares with a side of $1\ cm$. the entire rectangle consists of $5\cdot 3=15$ such squares, and its area is $15\cm^2$.
Picture 1.
Figure 2.
The area of figures is usually denoted by the letter $S$.
To find the area of a rectangle, you need to multiply its length by its width.
If we denote its length by the letter $a$, and its width by the letter $b$, then the formula for the area of a rectangle will look like:
Definition 1
The figures are called equal if, when superimposed on one another, the figures coincide. Equal figures have equal areas and equal perimeters.
The area of a figure can be found as the sum of the areas of its parts.
Example 1
For example, in Figure $3$, rectangle $ABCD$ is divided into two parts by line $KLMN$. The area of one part is $12\ cm^2$, and the other is $9\ cm^2$. Then the area of the rectangle $ABCD$ will be equal to $12\ cm^2+9\ cm^2=21\ cm^2$. Find the area of the rectangle using the formula:
As you can see, the areas found by both methods are equal.
Figure 3.
Figure 4.
The line segment $AC$ divides the rectangle into two equal triangles: $ABC$ and $ADC$. This means that the area of each triangle is equal to half the area of the entire rectangle.
Definition 2
A rectangle with equal sides is called square.
If we denote the side of a square with the letter $a$, then the area of the square will be found by the formula:
Hence the name square of the number $a$.
Example 2
For example, if the side of a square is $5$ cm, then its area is:
Volumes
With the development of trade and construction back in the days of ancient civilizations, the need arose to find volumes. In mathematics, there is a branch of geometry that deals with the study of spatial figures, called stereometry. Mentions of this separate branch of mathematics were found already in the $IV$ century BC.
Ancient mathematicians developed a method for calculating the volume of simple figures - a cube and a parallelepiped. All buildings of those times were of this shape. But later methods were found to calculate the volume of figures of more complex shapes.
Volume of a rectangular parallelepiped
If you fill the mold with wet sand and then turn it over, you will get a three-dimensional figure that is characterized by volume. If you make several such figures using the same mold, you will get figures that have the same volume. If you fill the mold with water, then the volume of water and the volume of the sand figure will also be equal.
Figure 5.
You can compare the volumes of two vessels by filling one with water and pouring it into the second vessel. If the second vessel is completely filled, then the vessels have equal volumes. If water remains in the first, then the volume of the first vessel is greater than the volume of the second. If, when pouring water from the first vessel, it is not possible to completely fill the second vessel, then the volume of the first vessel is less than the volume of the second.
Volume is measured using the following units:
$mm^3$ -- cubic millimeter,
$cm^3$ -- cubic centimeter,
$dm^3$ -- cubic decimeter,
$m^3$ -- cubic meter,
$km^3$ -- cubic kilometer.
The video course “Get an A” includes all the topics necessary to successfully pass the Unified State Exam in mathematics with 60-65 points. Completely all tasks 1-13 of the Profile Unified State Exam in mathematics. Also suitable for passing the Basic Unified State Examination in mathematics. If you want to pass the Unified State Exam with 90-100 points, you need to solve part 1 in 30 minutes and without mistakes!
Preparation course for the Unified State Exam for grades 10-11, as well as for teachers. Everything you need to solve Part 1 of the Unified State Exam in mathematics (the first 12 problems) and Problem 13 (trigonometry). And this is more than 70 points on the Unified State Exam, and neither a 100-point student nor a humanities student can do without them.
All the necessary theory. Quick solutions, pitfalls and secrets of the Unified State Exam. All current tasks of part 1 from the FIPI Task Bank have been analyzed. The course fully complies with the requirements of the Unified State Exam 2018.
The course contains 5 large topics, 2.5 hours each. Each topic is given from scratch, simply and clearly.
Hundreds of Unified State Exam tasks. Word problems and probability theory. Simple and easy to remember algorithms for solving problems. Geometry. Theory, reference material, analysis of all types of Unified State Examination tasks. Stereometry. Tricky solutions, useful cheat sheets, development of spatial imagination. Trigonometry from scratch to problem 13. Understanding instead of cramming. Clear explanations of complex concepts. Algebra. Roots, powers and logarithms, function and derivative. A basis for solving complex problems of Part 2 of the Unified State Exam.