Formulas for finding areas and volumes of all figures. Volume of figures
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.
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 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 of modern methods, it is possible to find the area of any figure with great accuracy.
Consider one of the simplest shapes - a rectangle - and the formula for finding its area.
Rectangle area formula
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 as $1\cm^2$.
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 sides $1\ cm$ (for example, $p$) will be equal to $p\ cm^2$.
In other words, the area of the figure will be equal to as many $cm^2$ as the number of squares with side $1\ cm$ can be divided into this figure.
Consider a rectangle (Fig. 2) that consists of $3$ strips, each of which is divided into $5$ squares with sides $1\cm$. the whole rectangle consists of $5\cdot 3=15$ such squares, and its area is $15\cm^2$.
Picture 1.
Figure 2.
The area of the figures is usually denoted by the letter $S$.
To find the area of a rectangle, multiply its length by its width.
If we denote its length with the letter $a$, and the width with 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 $3$ the rectangle $ABCD$ is divided into two parts by the 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 a 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 triangles: $ABC$ and $ADC$. So the area of each of the triangles 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 the square by 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, there was a need to find volumes. In mathematics, there is a section of geometry that deals with the study of spatial figures, called stereometry. Mentions of this separate direction of mathematics were found already in the $4th 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 form. But in the future, ways were found to calculate the volume of figures of more complex shapes.
Volume of a cuboid
If you fill the mold with wet sand and then turn it over, you will get a three-dimensional figure, which 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 at the same time 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.
General review. Formulas of stereometry!
Hello dear friends! In this article, I decided to make a general overview of the problems in stereometry, which will be USE in mathematics e. It must be said that the tasks from this group are quite diverse, but not difficult. These are tasks for finding geometric quantities: lengths, angles, areas, volumes.
Considered: a cube, a rectangular parallelepiped, a prism, a pyramid, a compound polyhedron, a cylinder, a cone, a ball. It is sad that some graduates do not even take on such tasks at the exam itself, although more than 50% of them are solved elementarily, almost verbally.
The rest require little effort, knowledge and special techniques. In future articles, we will consider these tasks, do not miss it, subscribe to the blog update.
To solve, you need to know surface area and volume formulas parallelepiped, pyramid, prism, cylinder, cone and sphere. There are no complex tasks, they are all solved in 2-3 steps, it is important to "see" what formula needs to be applied.
All necessary formulas are presented below:
Ball or sphere. A spherical or spherical surface (sometimes simply a sphere) is the locus of points in space that are equidistant from one point - the center of the ball.
Ball volume equal to the volume of the pyramid, the base of which has the same area as the surface of the ball, and the height is the radius of the ball
The volume of a sphere is one and a half times less than the volume of a cylinder circumscribed around it.
A round cone can be obtained by rotating a right triangle around one of its legs, so a round cone is also called a cone of revolution. See also Surface area of a circular cone
Volume of a round cone is equal to one third of the product of the base area S and the height H:
(H - cube edge height)
A parallelepiped is a prism whose base is a parallelogram. The parallelepiped has six faces, and all of them are parallelograms. A parallelepiped whose four lateral faces are rectangles is called a right parallelepiped. A right box in which all six faces are rectangles is called a rectangular box.
Volume of a cuboid is equal to the product of the area of the base and the height:
(S is the area of the base of the pyramid, h is the height of the pyramid)
A pyramid is a polyhedron with one face - the base of the pyramid - an arbitrary polygon, and the rest - side faces - triangles with a common vertex, called the top of the pyramid.
A section parallel to the base of the pyramid divides the pyramid into two parts. The part of the pyramid between its base and this section is a truncated pyramid.
Volume of a truncated pyramid is equal to one third of the product of the height h (OS) by the sum of the areas of the upper base S1 (abcde), the lower base of the truncated pyramid S2 (ABCD) and the average proportional between them.
| 1. | V= |
n - the number of sides of a regular polygon - the bases of a regular pyramid
a - side of regular polygon - bases of regular pyramid
h - the height of the regular pyramid
A regular triangular pyramid is a polyhedron with one face - the base of the pyramid - a regular triangle, and the rest - side faces - equal triangles with a common vertex. The height descends to the center of the base from the top.
Volume of a regular triangular pyramid is equal to one third of the product of the area of an equilateral triangle, which is the base S (ABC) to the height h (OS)
a - side of a regular triangle - bases of a regular triangular pyramid
h - the height of a regular triangular pyramid
Derivation of the formula for the volume of a tetrahedron
The volume of a tetrahedron is calculated using the classical formula for the volume of a pyramid. It is necessary to substitute the height of the tetrahedron and the area of a regular (equilateral) triangle into it.
Volume of a tetrahedron- is equal to the fraction in the numerator of which the square root of two in the denominator is twelve, multiplied by the cube of the length of the edge of the tetrahedron
(h is the length of the side of the rhombus)
Circumference p is about three whole and one seventh the length of the diameter of a circle. The exact ratio of the circumference of a circle to its diameter is denoted by the Greek letter π
As a result, the perimeter of a circle or the circumference of a circle is calculated by the formula
| π rn |
(r is the radius of the arc, n is the central angle of the arc in degrees.)
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.
To solve problems in geometry, you need to know formulas - such as the area of a triangle or the area of \u200b\u200ba parallelogram - as well as simple tricks, which we will talk about.
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 exam in mathematics, other formulas for the area of a triangle are also 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 break this figure into those that we all know about, and find its area - as the sum of the areas of these figures.
Divide this quadrilateral by a horizontal line into two triangles with a common base equal to . The heights of these triangles are 
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 \u200b\u200bthe figure can be represented as the difference of any areas.
It is not so easy to calculate what the base and height in 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-angled triangles. See them in the picture? We get: .
Answer: .
3. Sometimes in a task it is necessary to find the area not of the whole figure, but of its part. Usually we are talking about the area of \u200b\u200bthe sector - part of the circle. Find the area of \u200b\u200bthe sector of the circle of radius , whose arc length is equal to .
In this picture we see part of a circle. The area of the whole circle is equal to , since . It remains to find out what part of the circle is depicted. Since the length of the entire circle is (since), and the length of the arc of this sector is , therefore, the length of the arc is several times less than the length of the entire circle. The angle on which this arc rests is also times less than a full circle (that is, degrees). This means that the area of the sector will be several times less than the area of the entire circle.
