How to find the area of ​​any figure. Area theorems for figures

Class: 5

In my opinion, the teacher’s task is not only to teach, but to develop cognitive interest in the student. Therefore, whenever possible, I connect lesson topics with practical tasks.

During the lesson, students, under the guidance of the teacher, draw up a plan for solving problems to find the area of ​​a “complex figure” (for calculating repair estimates), consolidate skills in solving problems to find the area; development of attention, ability to research activities, education of activity, independence.

Working in pairs creates a situation of communication between those who have knowledge and those who acquire it; This work is based on improving the quality of training in the subject. Promotes the development of interest in the learning process and deeper assimilation of educational material.

The lesson not only systematizes students’ knowledge, but also contributes to the development of creative and analytical abilities. The use of problems with practical content in the classroom allows us to show the relevance of mathematical knowledge in everyday life.

Lesson objectives:

Educational:

• consolidation of knowledge of formulas for the area of ​​a rectangle, right triangle;
• analysis of tasks for calculating the area of ​​a “complex” figure and methods for performing them;
• independent completion of tasks to test knowledge, skills and abilities.

Educational:

• development of methods of mental and research activity;
• developing the ability to listen and explain the course of a decision.

Educational:

• develop students' academic skills;
• cultivate a culture of oral and written mathematical speech;
• develop a friendly attitude in the classroom and the ability to work in groups.

Lesson type: combined.

Equipment:

• Mathematics: textbook for 5th grade. general education institutions/ N.Ya. Vilenkin, V.I. Zhokhov et al., M.: “Mnemosyne”, 2010.
• Cards for groups of students with shapes to calculate the area of ​​a complex shape.
• Drawing tools.

Lesson plan:

1. Organizing time.
2. Updating knowledge.
   a) Theoretical questions (test).
   b) Statement of the problem.
3. Learned new material.
   a) finding a solution to the problem;
   b) solution to the problem.
4. Fixing the material.
   a) collective problem solving;
   Physical education minute.
   b) independent work.
5. Homework.
6. Lesson summary. Reflection.

During the classes

I. Organizational moment.

We will begin the lesson with these parting words:

Math, friends,
Absolutely everyone needs it.
Work diligently in class
And success is sure to await you!

II. Updating knowledge.

A) Frontal work with signal cards (each student has cards with the numbers 1, 2, 3, 4; when answering a test question, the student raises a card with the number of the correct answer).

1. A square centimeter is:

1. area of ​​a square with a side of 1 cm;
2. square with side 1 cm;
3. square with a perimeter of 1 cm.

2. The area of ​​the figure shown in the figure is equal to:

1. 8 dm;
2. 8 dm 2;
3. 15 dm 2.

3. Is it true that equal figures have equal perimeters and equal areas?

4. The area of ​​a rectangle is determined by the formula:

1. S = a 2 ;
2. S = 2 (a + b);
3. S = a b.

5. The area of ​​the figure shown in the figure is equal to:

1. 12 cm;
2. 8 cm;
3. 16 cm.

b) (Formulation of the problem). Task. How much paint is needed to paint a floor that has the following shape (see figure), if 200 g of paint is consumed per 1 m2?

III. Learning new material.

What do we need to know to solve the last problem? (Find the area of ​​the floor that looks like a “complex figure.”)

Students formulate the topic and goals of the lesson (if necessary, the teacher helps).

Consider a rectangle ABCD. Let's draw a line in it KPMN, breaking the rectangle ABCD into two parts: ABNMPK And KPMNCD.

What is the area? ABCD? (15 cm 2)

What is the area of ​​the figure? ABMNPK? (7 cm 2)

What is the area of ​​the figure? KPMNCD? (8 cm 2)

Analyze your results. (15= = 7 + 8)

Conclusion? (The area of ​​the entire figure is equal to the sum of the areas of its parts.)

S = S 1 + S 2

How can we apply this property to solve our problem? (Let's break a complex figure into parts, find the areas of the parts, then the area of ​​the entire figure.)

S 1 = 7 2 = 14 (m 2)
S 2 = (7 – 4) (8 – 2 – 3) = 3 3 = 9 (m 2)
S 3 = 7 3 = 21 (m 2)
S = S 1 + S 2 + S 3 = 14 + 9 + 21 = 44 (m2)

Let's make up plan for solving problems to find the area of ​​a “complex figure”:

1. We break the figure into simple figures.
2. Finding the areas of simple figures.

a) Task 1. How many tiles will be needed to lay out a site of the following dimensions:

S = S 1 + S 2
S 1 = (60 – 30) 20 = 600 (dm 2)
S 2 = 30 50 = 1500 (dm 2)
S = 600 + 1500 = 2100 (dm 2)

Is there another way to solve? (We are considering the proposed options.)

Answer: 2100 dm 2.

Task 2. (collective decision on the board and in notebooks.) How many m2 of linoleum is required to renovate a room with the following shape:

S = S 1 + S 2
S 1 = 3 2 = 6 (m 2)
S 2 = ((5 – 3) 2) : 2 = 2 (m 2)
S = 6 + 2 = 8 (m2)

Answer: 8 m2.

Physical education minute.

And now, guys, stand up.
They quickly raised their hands up.
To the sides, forward, backward.
Turned right, left.
They sat down quietly and got back to work.

b) Independent work (educational) .

Students are divided into groups (No. 5–8 are stronger). Each group is a repair team.

Task for the teams: determine how much paint is needed to paint a floor that has the shape of the figure shown on the card, if 200 g of paint is required per 1 m2.

You build this figure in your notebook and write down all the data and begin the task. You can discuss the solution (but only in your group!). If some group copes with the task quickly, then they are given an additional task (after checking independent work).

Tasks for groups:

V. Homework.

paragraph 18, No. 718, No. 749.

Additional task. Plan diagram of the Summer Garden (St. Petersburg). Calculate its area.

VI. Lesson summary.

Reflection. Continue the sentence:

• Today I found out...
• It was interesting…
• It was difficult…
• Now I can…
• Gave me a lesson for life...

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 profile Unified State Examination 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 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 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 several times 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.

In the previous section devoted to the analysis of the geometric meaning definite integral, we received a number of formulas for calculating the area of ​​a curvilinear trapezoid:

S (G) = ∫ a b f (x) d x for a continuous and non-negative function y = f (x) on the interval [ a ; b ] ,

S (G) = - ∫ a b f (x) d x for a continuous and non-positive function y = f (x) on the interval [ a ; b ] .

These formulas are applicable to solving relatively simple problems. In reality, we will often have to work with more complex figures. In this regard, we will devote this section to an analysis of algorithms for calculating the area of ​​figures that are limited by functions in explicit form, i.e. like y = f(x) or x = g(y).

Theorem

Let the functions y = f 1 (x) and y = f 2 (x) be defined and continuous on the interval [ a ; b ] , and f 1 (x) ≤ f 2 (x) for any value x from [ a ; b ] . Then the formula for calculating the area of ​​the figure G, bounded by the lines x = a, x = b, y = f 1 (x) and y = f 2 (x) will look like S (G) = ∫ a b f 2 (x) - f 1 (x) d x .

A similar formula will be applicable for the area of ​​a figure bounded by the lines y = c, y = d, x = g 1 (y) and x = g 2 (y): S (G) = ∫ c d (g 2 (y) - g 1 (y) d y .

Proof

Let's look at three cases for which the formula will be valid.

In the first case, taking into account the property of additivity of area, the sum of the areas of the original figure G and the curvilinear trapezoid G 1 is equal to the area of ​​the figure G 2. It means that

Therefore, S (G) = S (G 2) - S (G 1) = ∫ a b f 2 (x) d x - ∫ a b f 1 (x) d x = ∫ a b (f 2 (x) - f 1 (x)) dx.

We can perform the last transition using the third property of the definite integral.

In the second case, the equality is true: S (G) = S (G 2) + S (G 1) = ∫ a b f 2 (x) d x + - ∫ a b f 1 (x) d x = ∫ a b (f 2 (x) - f 1 (x)) d x

The graphic illustration will look like:

If both functions are non-positive, we get: S (G) = S (G 2) - S (G 1) = - ∫ a b f 2 (x) d x - - ∫ a b f 1 (x) d x = ∫ a b (f 2 (x) - f 1 (x)) d x . The graphic illustration will look like:

Let's move on to consider the general case when y = f 1 (x) and y = f 2 (x) intersect the O x axis.

We denote the intersection points as x i, i = 1, 2, . . . , n - 1 . These points split the segment [a; b ] into n parts x i - 1 ; x i, i = 1, 2, . . . , n, where α = x 0< x 1 < x 2 < . . . < x n - 1 < x n = b . Фигуру G можно представить объединением фигур G i , i = 1 , 2 , . . . , n . Очевидно, что на своем интервале G i попадает под один из трех рассмотренных ранее случаев, поэтому их площади находятся как S (G i) = ∫ x i - 1 x i (f 2 (x) - f 1 (x)) d x , i = 1 , 2 , . . . , n

Hence,

S (G) = ∑ i = 1 n S (G i) = ∑ i = 1 n ∫ x i x i f 2 (x) - f 1 (x)) d x = = ∫ x 0 x n (f 2 (x) - f ( x)) d x = ∫ a b f 2 (x) - f 1 (x) d x

We can make the last transition using the fifth property of the definite integral.

Let us illustrate the general case on the graph.

The formula S (G) = ∫ a b f 2 (x) - f 1 (x) d x can be considered proven.

Now let's move on to analyzing examples of calculating the area of ​​figures that are limited by the lines y = f (x) and x = g (y).

We will begin our consideration of any of the examples by constructing a graph. The image will allow us to represent complex shapes as unions of simpler shapes. If constructing graphs and figures on them is difficult for you, you can study the section on basic elementary functions, geometric transformation of graphs of functions, as well as constructing graphs while studying a function.

Example 1

It is necessary to determine the area of ​​the figure, which is limited by the parabola y = - x 2 + 6 x - 5 and straight lines y = - 1 3 x - 1 2, x = 1, x = 4.

Solution

Let's draw the lines on the graph in the Cartesian coordinate system.

On the segment [ 1 ; 4 ] the graph of the parabola y = - x 2 + 6 x - 5 is located above the straight line y = - 1 3 x - 1 2. In this regard, to obtain the answer we use the formula obtained earlier, as well as the method of calculating the definite integral using the Newton-Leibniz formula:

S (G) = ∫ 1 4 - x 2 + 6 x - 5 - - 1 3 x - 1 2 d x = = ∫ 1 4 - x 2 + 19 3 x - 9 2 d x = - 1 3 x 3 + 19 6 x 2 - 9 2 x 1 4 = = - 1 3 4 3 + 19 6 4 2 - 9 2 4 - - 1 3 1 3 + 19 6 1 2 - 9 2 1 = = - 64 3 + 152 3 - 18 + 1 3 - 19 6 + 9 2 = 13

Answer: S(G) = 13

Let's look at a more complex example.

Example 2

It is necessary to calculate the area of ​​the figure, which is limited by the lines y = x + 2, y = x, x = 7.

Solution

In this case, we have only one straight line located parallel to the x-axis. This is x = 7. This requires us to find the second limit of integration ourselves.

Let's build a graph and plot on it the lines given in the problem statement.

Having the graph in front of our eyes, we can easily determine that the lower limit of integration will be the abscissa of the point of intersection of the graph of the straight line y = x and the semi-parabola y = x + 2. To find the abscissa we use the equalities:

y = x + 2 O DZ: x ≥ - 2 x 2 = x + 2 2 x 2 - x - 2 = 0 D = (- 1) 2 - 4 1 (- 2) = 9 x 1 = 1 + 9 2 = 2 ∈ O DZ x 2 = 1 - 9 2 = - 1 ∉ O DZ

It turns out that the abscissa of the intersection point is x = 2.

We draw your attention to the fact that in the general example in the drawing, the lines y = x + 2, y = x intersect at the point (2; 2), so such detailed calculations may seem unnecessary. We have provided such a detailed solution here only because in more complex cases the solution may not be so obvious. This means that it is always better to calculate the coordinates of the intersection of lines analytically.

On the interval [ 2 ; 7] the graph of the function y = x is located above the graph of the function y = x + 2. Let's apply the formula to calculate the area:

S (G) = ∫ 2 7 (x - x + 2) d x = x 2 2 - 2 3 · (x + 2) 3 2 2 7 = = 7 2 2 - 2 3 · (7 + 2) 3 2 - 2 2 2 - 2 3 2 + 2 3 2 = = 49 2 - 18 - 2 + 16 3 = 59 6

Answer: S (G) = 59 6

Example 3

It is necessary to calculate the area of ​​the figure, which is limited by the graphs of the functions y = 1 x and y = - x 2 + 4 x - 2.

Solution

Let's plot the lines on the graph.

Let's define the limits of integration. To do this, we determine the coordinates of the points of intersection of the lines by equating the expressions 1 x and - x 2 + 4 x - 2. Provided that x is not zero, the equality 1 x = - x 2 + 4 x - 2 becomes equivalent to the third degree equation - x 3 + 4 x 2 - 2 x - 1 = 0 with integer coefficients. To refresh your memory of the algorithm for solving such equations, we can refer to the section “Solving cubic equations.”

The root of this equation is x = 1: - 1 3 + 4 1 2 - 2 1 - 1 = 0.

Dividing the expression - x 3 + 4 x 2 - 2 x - 1 by the binomial x - 1, we get: - x 3 + 4 x 2 - 2 x - 1 ⇔ - (x - 1) (x 2 - 3 x - 1) = 0

We can find the remaining roots from the equation x 2 - 3 x - 1 = 0:

x 2 - 3 x - 1 = 0 D = (- 3) 2 - 4 · 1 · (- 1) = 13 x 1 = 3 + 13 2 ≈ 3 . 3; x 2 = 3 - 13 2 ≈ - 0 . 3

We found the interval x ∈ 1; 3 + 13 2, in which the figure G is contained above the blue and below the red line. This helps us determine the area of ​​the figure:

S (G) = ∫ 1 3 + 13 2 - x 2 + 4 x - 2 - 1 x d x = - x 3 3 + 2 x 2 - 2 x - ln x 1 3 + 13 2 = = - 3 + 13 2 3 3 + 2 3 + 13 2 2 - 2 3 + 13 2 - ln 3 + 13 2 - - - 1 3 3 + 2 1 2 - 2 1 - ln 1 = 7 + 13 3 - ln 3 + 13 2

Answer: S (G) = 7 + 13 3 - ln 3 + 13 2

Example 4

It is necessary to calculate the area of ​​the figure, which is limited by the curves y = x 3, y = - log 2 x + 1 and the abscissa axis.

Solution

Let's plot all the lines on the graph. We can get the graph of the function y = - log 2 x + 1 from the graph y = log 2 x if we position it symmetrically about the x-axis and move it up one unit. The equation of the x-axis is y = 0.

Let us mark the points of intersection of the lines.

As can be seen from the figure, the graphs of the functions y = x 3 and y = 0 intersect at the point (0; 0). This happens because x = 0 is the only real root of the equation x 3 = 0.

x = 2 is the only root of the equation - log 2 x + 1 = 0, so the graphs of the functions y = - log 2 x + 1 and y = 0 intersect at the point (2; 0).

x = 1 is the only root of the equation x 3 = - log 2 x + 1 . In this regard, the graphs of the functions y = x 3 and y = - log 2 x + 1 intersect at the point (1; 1). The last statement may not be obvious, but the equation x 3 = - log 2 x + 1 cannot have more than one root, since the function y = x 3 is strictly increasing, and the function y = - log 2 x + 1 is strictly decreasing.

The further solution involves several options.

Option #1

We can imagine the figure G as the sum of two curvilinear trapezoids located above the x-axis, the first of which is located below the midline on the segment x ∈ 0; 1, and the second is below the red line on the segment x ∈ 1; 2. This means that the area will be equal to S (G) = ∫ 0 1 x 3 d x + ∫ 1 2 (- log 2 x + 1) d x .

Option No. 2

Figure G can be represented as the difference of two figures, the first of which is located above the x-axis and below the blue line on the segment x ∈ 0; 2, and the second between the red and blue lines on the segment x ∈ 1; 2. This allows us to find the area as follows:

S (G) = ∫ 0 2 x 3 d x - ∫ 1 2 x 3 - (- log 2 x + 1) d x

In this case, to find the area you will have to use a formula of the form S (G) = ∫ c d (g 2 (y) - g 1 (y)) d y. In fact, the lines that bound the figure can be represented as functions of the argument y.

Let's solve the equations y = x 3 and - log 2 x + 1 with respect to x:

y = x 3 ⇒ x = y 3 y = - log 2 x + 1 ⇒ log 2 x = 1 - y ⇒ x = 2 1 - y

We get the required area:

S (G) = ∫ 0 1 (2 1 - y - y 3) d y = - 2 1 - y ln 2 - y 4 4 0 1 = = - 2 1 - 1 ln 2 - 1 4 4 - - 2 1 - 0 ln 2 - 0 4 4 = - 1 ln 2 - 1 4 + 2 ln 2 = 1 ln 2 - 1 4

Answer: S (G) = 1 ln 2 - 1 4

Example 5

It is necessary to calculate the area of ​​the figure, which is limited by the lines y = x, y = 2 3 x - 3, y = - 1 2 x + 4.

Solution

With a red line we plot the line defined by the function y = x. We draw the line y = - 1 2 x + 4 in blue, and the line y = 2 3 x - 3 in black.

Let's mark the intersection points.

Let's find the intersection points of the graphs of the functions y = x and y = - 1 2 x + 4:

x = - 1 2 x + 4 O DZ: x ≥ 0 x = - 1 2 x + 4 2 ⇒ x = 1 4 x 2 - 4 x + 16 ⇔ x 2 - 20 x + 64 = 0 D = (- 20) 2 - 4 1 64 = 144 x 1 = 20 + 144 2 = 16 ; x 2 = 20 - 144 2 = 4 Check: x 1 = 16 = 4, - 1 2 x 1 + 4 = - 1 2 16 + 4 = - 4 ⇒ x 1 = 16 not Is the solution to the equation x 2 = 4 = 2, - 1 2 x 2 + 4 = - 1 2 4 + 4 = 2 ⇒ x 2 = 4 is the solution to the equation ⇒ (4; 2) point of intersection i y = x and y = - 1 2 x + 4

Let's find the intersection point of the graphs of the functions y = x and y = 2 3 x - 3:

x = 2 3 x - 3 O DZ: x ≥ 0 x = 2 3 x - 3 2 ⇔ x = 4 9 x 2 - 4 x + 9 ⇔ 4 x 2 - 45 x + 81 = 0 D = (- 45 ) 2 - 4 4 81 = 729 x 1 = 45 + 729 8 = 9, x 2 45 - 729 8 = 9 4 Check: x 1 = 9 = 3, 2 3 x 1 - 3 = 2 3 9 - 3 = 3 ⇒ x 1 = 9 is the solution to the equation ⇒ (9 ; 3) point a s y = x and y = 2 3 x - 3 x 2 = 9 4 = 3 2, 2 3 x 1 - 3 = 2 3 9 4 - 3 = - 3 2 ⇒ x 2 = 9 4 There is no solution to the equation

Let's find the point of intersection of the lines y = - 1 2 x + 4 and y = 2 3 x - 3:

1 2 x + 4 = 2 3 x - 3 ⇔ - 3 x + 24 = 4 x - 18 ⇔ 7 x = 42 ⇔ x = 6 - 1 2 6 + 4 = 2 3 6 - 3 = 1 ⇒ (6 ; 1) point of intersection y = - 1 2 x + 4 and y = 2 3 x - 3

Method No. 1

Let us imagine the area of ​​the desired figure as the sum of the areas of individual figures.

Then the area of ​​the figure is:

S (G) = ∫ 4 6 x - - 1 2 x + 4 d x + ∫ 6 9 x - 2 3 x - 3 d x = = 2 3 x 3 2 + x 2 4 - 4 x 4 6 + 2 3 x 3 2 - x 2 3 + 3 x 6 9 = = 2 3 6 3 2 + 6 2 4 - 4 6 - 2 3 4 3 2 + 4 2 4 - 4 4 + + 2 3 9 3 2 - 9 2 3 + 3 9 - 2 3 6 3 2 - 6 2 3 + 3 6 = = - 25 3 + 4 6 + - 4 6 + 12 = 11 3

Method No. 2

The area of ​​the original figure can be represented as the sum of two other figures.

Then we solve the equation of the line relative to x, and only after that we apply the formula for calculating the area of ​​the figure.

y = x ⇒ x = y 2 red line y = 2 3 x - 3 ⇒ x = 3 2 y + 9 2 black line y = - 1 2 x + 4 ⇒ x = - 2 y + 8 s i n i a l i n e

So the area is:

S (G) = ∫ 1 2 3 2 y + 9 2 - - 2 y + 8 d y + ∫ 2 3 3 2 y + 9 2 - y 2 d y = = ∫ 1 2 7 2 y - 7 2 d y + ∫ 2 3 3 2 y + 9 2 - y 2 d y = = 7 4 y 2 - 7 4 y 1 2 + - y 3 3 + 3 y 2 4 + 9 2 y 2 3 = 7 4 2 2 - 7 4 2 - 7 4 1 2 - 7 4 1 + + - 3 3 3 + 3 3 2 4 + 9 2 3 - - 2 3 3 + 3 2 2 4 + 9 2 2 = = 7 4 + 23 12 = 11 3

As you can see, the values ​​are the same.

Answer: S (G) = 11 3

Results

To find the area of ​​a figure that is limited given lines we need to construct lines on the plane, find their intersection points, and apply the formula to find the area. In this section, we examined the most common variants of tasks.

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Area (in geometry)- Area, one of the main quantities associated with geometric shapes. In the simplest cases, it is measured by the number of unit squares filling a flat figure, that is, squares with a side equal to one unit of length. Calculation of P. was already in ancient times... ...

SQUARE- one of the quantitative characteristics of flat geometric shapes and surfaces. The area of ​​a rectangle is equal to the product of the lengths of two adjacent sides. The area of ​​a stepped figure (i.e. one that can be divided into several adjacent... ... Big Encyclopedic Dictionary

AREA (in geometry)- AREA, one of the quantitative characteristics of flat geometric shapes and surfaces. The area of ​​a rectangle is equal to the product of the lengths of two adjacent sides. The area of ​​a stepped figure (i.e. one that can be divided into several... ... encyclopedic Dictionary

SQUARE- AREA, squares, prev. about area and (obsolete) on area, plural. and areas, women. (book). 1. Part of a plane bounded by a broken or curved line (geom.). Area of ​​a rectangle. Area of ​​a curved figure. 2. only units. Space,… … Dictionary Ushakova

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Books

• Figures in mathematics, physics and nature. Squares, Triangles and Circles, Catherine Sheldrick-Ross. About the book Features of the book More than 75 unusual master classes will help turn the study of geometry into an exciting game The book describes the main figures in as much detail as possible: squares, circles and... Buy for 1206 rubles
• Figures in mathematics, physics and nature Squares, triangles and circles, Sheldrick-Ross K.. More than 75 unusual master classes will help turn the study of geometry into an exciting game. The book describes the main figures in as much detail as possible: squares, circles, triangles. The book will teach...

Knowledge of how to measure the Earth appeared in ancient times and gradually took shape in the science of geometry. This word is translated from Greek as “land surveying”.

The measure of the extent of a flat section of the Earth in length and width is area. In mathematics, it is usually denoted by the Latin letter S (from the English “square” - “area”, “square”) or the Greek letter σ (sigma). S denotes the area of ​​a figure on a plane or the surface area of ​​a body, and σ is the cross-sectional area of ​​a wire in physics. These are the main symbols, although there may be others, for example, in the field of strength of materials, A is the cross-sectional area of ​​the profile.

In contact with

Calculation formulas

Knowing the areas of simple figures, you can find the parameters of more complex ones.. Ancient mathematicians developed formulas that can be used to easily calculate them. Such figures are triangle, quadrangle, polygon, circle.

To find the area of ​​a complex plane figure, it is broken down into many simple figures such as triangles, trapezoids or rectangles. Then, using mathematical methods, a formula is derived for the area of ​​this figure. A similar method is used not only in geometry, but also in mathematical analysis to calculate the areas of figures bounded by curves.

Triangle

Let's start with the simplest figure - a triangle. They are rectangular, isosceles and equilateral. Take any triangle ABC with sides AB=a, BC=b and AC=c (∆ ABC). To find its area, let us recall the sine and cosine theorems known from the school mathematics course. Letting go of all calculations, we arrive at the following formulas:

• S=√ - Heron’s formula, known to everyone, where p=(a+b+c)/2 is the semi-perimeter of the triangle;
• S=a h/2, where h is the height lowered to side a;
• S=a b (sin γ)/2, where γ is the angle between sides a and b;
• S=a b/2, if ∆ ABC is rectangular (here a and b are legs);
• S=b² (sin (2 β))/2, if ∆ ABC is isosceles (here b is one of the “hips”, β is the angle between the “hips” of the triangle);
• S=a² √¾, if ∆ ABC is equilateral (here a is a side of the triangle).

Quadrangle

Let there be a quadrilateral ABCD with AB=a, BC=b, CD=c, AD=d. To find the area S of an arbitrary 4-gon, you need to divide it by the diagonal into two triangles, the areas of which S1 and S2 are not equal in the general case.

Then use the formulas to calculate them and add them, i.e. S=S1+S2. However, if a 4-gon belongs to a certain class, then its area can be found using previously known formulas:

• S=(a+c) h/2=e h, if the tetragon is a trapezoid (here a and c are the bases, e is the midline of the trapezoid, h is the height lowered to one of the bases of the trapezoid;
• S=a h=a b sin φ=d1 d2 (sin φ)/2, if ABCD is a parallelogram (here φ is the angle between sides a and b, h is the height dropped to side a, d1 and d2 are diagonals);
• S=a b=d²/2, if ABCD is a rectangle (d is a diagonal);
• S=a² sin φ=P² (sin φ)/16=d1 d2/2, if ABCD is a rhombus (a is the side of the rhombus, φ is one of its angles, P is the perimeter);
• S=a²=P²/16=d²/2, if ABCD is a square.

Polygon

To find the area of ​​an n-gon, mathematicians break it down into the simplest equal figures - triangles, find the area of ​​each of them and then add them. But if the polygon belongs to the class of regular, then use the formula:

S=a n h/2=a² n/=P²/, where n is the number of vertices (or sides) of the polygon, a is the side of the n-gon, P is its perimeter, h is the apothem, i.e. a segment drawn from the center of the polygon to one of its sides at an angle of 90°.

Circle

A circle is a perfect polygon with an infinite number of sides. We need to calculate the limit of the expression on the right in the formula for the area of ​​a polygon with the number of sides n tending to infinity. In this case, the perimeter of the polygon will turn into the length of a circle of radius R, which will be the boundary of our circle, and will become equal to P=2 π R. Substitute this expression into the above formula. We will get:

S=(π² R² cos (180°/n))/(n sin (180°/n)).

Let's find the limit of this expression as n→∞. To do this, we take into account that lim (cos (180°/n)) for n→∞ is equal to cos 0°=1 (lim is the sign of the limit), and lim = lim for n→∞ is equal to 1/π (we converted the degree measure into a radian, using the relation π rad=180°, and applied the first remarkable limit lim(sin x)/x=1 at x→∞). Substituting the obtained values ​​into the last expression for S, we arrive at the well-known formula:

S=π² R² 1 (1/π)=π R².

Units

Systemic and non-systemic units of measurement are used. System units belong to the SI (System International). This is a square meter (sq. meter, m²) and units derived from it: mm², cm², km².

In square millimeters (mm²), for example, they measure the cross-sectional area of ​​wires in electrical engineering, in square centimeters (cm²) - the cross-section of a beam in structural mechanics, in square meters (m²) - in an apartment or house, in square kilometers (km²) - in geography .

However, sometimes non-systemic units of measurement are used, such as: weave, ar (a), hectare (ha) and acre (ac). Let us present the following relations:

• 1 hundred square meters=1 a=100 m²=0.01 hectares;
• 1 ha=100 a=100 acres=10000 m²=0.01 km²=2.471 ac;
• 1 ac = 4046.856 m² = 40.47 a = 40.47 acres = 0.405 hectares.