Calculating the area of ​​a figure This is perhaps one of the most difficult problems in area theory. In school geometry, they are taught to find the areas of the main geometric shapes such as, for example, a triangle, a rhombus, a rectangle, a trapezoid, a circle, etc. However, one often has to deal with the calculation of the areas of more complex figures. It is in solving such problems that it is very convenient to use integral calculus.

Definition.

Curvilinear trapezoid some figure G is called, bounded by the lines y = f(x), y = 0, x = a and x = b, and the function f(x) is continuous on the segment [a; b] and does not change its sign on it (Fig. 1). The area of ​​a curvilinear trapezoid can be denoted by S(G).

The definite integral ʃ a b f(x)dx for the function f(x), which is continuous and non-negative on the segment [a; b], and is the area of ​​the corresponding curvilinear trapezoid.

That is, to find the area of ​​\u200b\u200bthe figure G, bounded by the lines y \u003d f (x), y \u003d 0, x \u003d a and x \u003d b, you need to calculate definite integralʃ a b f(x)dx.

In this way, S(G) = ʃ a b f(x)dx.

If the function y = f(x) is not positive on [a; b], then the area of ​​the curvilinear trapezoid can be found by the formula S(G) = -ʃ a b f(x)dx.

Example 1

Calculate the area of ​​\u200b\u200bthe figure bounded by the lines y \u003d x 3; y = 1; x = 2.

Solution.

The given lines form the figure ABC, which is shown by hatching on rice. 2.

The desired area is equal to the difference between the areas of the curvilinear trapezoid DACE and the square DABE.

Using the formula S = ʃ a b f(x)dx = S(b) – S(a), we find the limits of integration. To do this, we solve a system of two equations:

(y \u003d x 3,
(y = 1.

Thus, we have x 1 \u003d 1 - the lower limit and x \u003d 2 - the upper limit.

So, S = S DACE - S DABE = ʃ 1 2 x 3 dx - 1 = x 4 /4| 1 2 - 1 \u003d (16 - 1) / 4 - 1 \u003d 11/4 (square units).

Answer: 11/4 sq. units

Example 2

Calculate the area of ​​\u200b\u200bthe figure bounded by lines y \u003d √x; y = 2; x = 9.

Solution.

The given lines form the figure ABC, which is bounded from above by the graph of the function

y \u003d √x, and from below the graph of the function y \u003d 2. The resulting figure is shown by hatching on rice. 3.

The desired area is equal to S = ʃ a b (√x - 2). Let's find the limits of integration: b = 9, to find a, we solve the system of two equations:

(y = √x,
(y = 2.

Thus, we have that x = 4 = a is the lower limit.

So, S = ∫ 4 9 (√x – 2)dx = ∫ 4 9 √x dx –∫ 4 9 2dx = 2/3 x√x| 4 9 - 2x| 4 9 \u003d (18 - 16/3) - (18 - 8) \u003d 2 2/3 (square units).

Answer: S = 2 2/3 sq. units

Example 3

Calculate the area of ​​\u200b\u200bthe figure bounded by the lines y \u003d x 3 - 4x; y = 0; x ≥ 0.

Solution.

Let's plot the function y \u003d x 3 - 4x for x ≥ 0. To do this, we find the derivative y ':

y’ = 3x 2 – 4, y’ = 0 at х = ±2/√3 ≈ 1.1 are critical points.

If we draw the critical points on the real axis and place the signs of the derivative, we get that the function decreases from zero to 2/√3 and increases from 2/√3 to plus infinity. Then x = 2/√3 is the minimum point, the minimum value of the function y is min = -16/(3√3) ≈ -3.

Let's determine the intersection points of the graph with the coordinate axes:

if x \u003d 0, then y \u003d 0, which means that A (0; 0) is the point of intersection with the Oy axis;

if y \u003d 0, then x 3 - 4x \u003d 0 or x (x 2 - 4) \u003d 0, or x (x - 2) (x + 2) \u003d 0, from where x 1 \u003d 0, x 2 \u003d 2, x 3 \u003d -2 (not suitable, because x ≥ 0).

Points A(0; 0) and B(2; 0) are the intersection points of the graph with the Ox axis.

The given lines form the OAB figure, which is shown by hatching on rice. 4.

Since the function y \u003d x 3 - 4x takes on (0; 2) a negative value, then

S = |ʃ 0 2 (x 3 – 4x)dx|.

We have: ʃ 0 2 (x 3 - 4x)dx = (x 4 /4 - 4x 2 /2)| 0 2 \u003d -4, from where S \u003d 4 square meters. units

Answer: S = 4 sq. units

Example 4

Find the area of ​​the figure bounded by the parabola y \u003d 2x 2 - 2x + 1, the straight lines x \u003d 0, y \u003d 0 and the tangent to this parabola at the point with the abscissa x 0 \u003d 2.

Solution.

First, we compose the equation of the tangent to the parabola y \u003d 2x 2 - 2x + 1 at the point with the abscissa x₀ \u003d 2.

Since the derivative y' = 4x - 2, then for x 0 = 2 we get k = y'(2) = 6.

Find the ordinate of the touch point: y 0 = 2 2 2 – 2 2 + 1 = 5.

Therefore, the tangent equation has the form: y - 5 \u003d 6 (x - 2) or y \u003d 6x - 7.

Let's build a figure bounded by lines:

y \u003d 2x 2 - 2x + 1, y \u003d 0, x \u003d 0, y \u003d 6x - 7.

Г y \u003d 2x 2 - 2x + 1 - parabola. Points of intersection with the coordinate axes: A(0; 1) - with the Oy axis; with the Ox axis - there are no intersection points, because the equation 2x 2 - 2x + 1 = 0 has no solutions (D< 0). Найдем вершину параболы:

x b \u003d 2/4 \u003d 1/2;

y b \u003d 1/2, that is, the vertex of the parabola point B has coordinates B (1/2; 1/2).

So, the figure whose area is to be determined is shown by hatching on rice. five.

We have: S O A B D \u003d S OABC - S ADBC.

Find the coordinates of point D from the condition:

6x - 7 = 0, i.e. x \u003d 7/6, then DC \u003d 2 - 7/6 \u003d 5/6.

We find the area of ​​triangle DBC using the formula S ADBC ​​= 1/2 · DC · BC. In this way,

S ADBC ​​= 1/2 5/6 5 = 25/12 sq. units

S OABC = ʃ 0 2 (2x 2 - 2x + 1)dx = (2x 3 /3 - 2x 2 /2 + x)| 0 2 \u003d 10/3 (square units).

Finally we get: S O A B D \u003d S OABC - S ADBC ​​\u003d 10/3 - 25/12 \u003d 5/4 \u003d 1 1/4 (sq. units).

Answer: S = 1 1/4 sq. units

We have reviewed examples finding the areas of figures limited given lines . To successfully solve such problems, you need to be able to build lines and graphs of functions on a plane, find the points of intersection of lines, apply the formula for finding the area, which implies the ability and skills to calculate certain integrals.

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Theorem 1.

The area of ​​a square is equal to the square of its side.

Let us prove that the area S of a square with side a is equal to a 2 . Let's take a square with side 1 and divide it into n equal squares as shown in Figure 1. geometry area figure theorem

Picture 1.

Since the side of a square is 1, the area of ​​each small square equal. The side of each small square is equal, i.e. equal to a. It follows that. The theorem has been proven.

Theorem 2.

The area of ​​a parallelogram is equal to the product of its side by the height drawn to this side (Fig. 2.):

S = a * h.

Let ABCD be a given parallelogram. If it is not a rectangle, then one of its corners A or B is acute. Let, for definiteness, the angle A be acute (Fig. 2.).

Figure 2.

Let us drop the perpendicular AE from the vertex A to the line CB. The area of ​​the trapezoid AECD is equal to the sum of the areas of the parallelogram ABCD and the triangle AEB. Let us drop the perpendicular DF ​​from the vertex D to the line CD. Then the area of ​​the trapezoid AECD is equal to the sum of the areas of the rectangle AEFD and the triangle DFC. Right triangles AEB and DFC are congruent and therefore have equal areas. It follows that the area of ​​the parallelogram ABCD is equal to the area of ​​the rectangle AEFD, i.e. equals AE*AD. Segment AE is the height of the parallelogram lowered to the side AD, and, therefore, S = a * h. The theorem has been proven.

Theorem 3

The area of ​​a triangle is half the product of its side and the height drawn to it.(fig.3.):

Figure 3

Proof.

Let ABC be the given triangle. Let's add it to the parallelogram ABCD, as shown in the figure (Fig. 3.1.).

Figure 3.1.

The area of ​​a parallelogram is equal to the sum of the areas of triangles ABC and CDA. Since these triangles are congruent, the area of ​​the parallelogram is twice the area of ​​triangle ABC. The height of the parallelogram corresponding to side CB is equal to the height of the triangle drawn to side CB. This implies the assertion of the theorem. The theorem is proved.

Theorem 3.1.

The area of ​​a triangle is half the product of its two sides and the sine of the angle between them.(Figure 3.2.).

Figure 3.2.

Proof.

Let's introduce a coordinate system with the origin at point C so that B lies on the positive semi-axis C x , and point A has a positive ordinate. The area of ​​a given triangle can be calculated using the formula where h is the height of the triangle. But h is equal to the ordinate of point A, i.e. h=b sin C. Therefore, . The theorem has been proven.

Theorem 4.

The area of ​​a trapezoid is half the sum of its bases multiplied by its height(Fig.4.).

Figure 4

Proof.

Let ABCD be a given trapezoid (Fig. 4.1.).

Figure 4.1.

Diagonal AC of a trapezoid divides it into two triangles: ABC and CDA.

Therefore, the area of ​​a trapezoid is equal to the sum of the areas of these triangles.

The area of ​​triangle ACD is equal to the area of ​​triangle ABC. The altitudes AF and CE of these triangles are equal to the distance h between the parallel lines BC and AD, i.e. trapezium height. Consequently, . The theorem has been proven.

The areas of figures are of great importance in geometry, as in science. After all, area is one of the most important quantities in geometry. Without knowing the areas, it is impossible to solve many geometric problems, prove theorems, and substantiate axioms. The areas of figures were of great importance many centuries ago, but have not lost their significance in modern world. Area concepts are used in many professions. They are used in construction, design and in many other human activities. From this we can conclude that without the development of geometry, in particular the concepts of areas, humanity would not have been able to make such a big breakthrough in the field of science and technology.

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 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.

Class: 5

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

In the lesson, students, under the guidance of a teacher, draw up a plan for solving problems for finding the area of ​​\u200b\u200ba "complex figure" (for calculating repair estimates), consolidate the skills for solving problems for finding 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; the basis of such work is to improve the quality of training in the subject. Promotes the development of interest in the learning process and a deeper assimilation of educational material.

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

Lesson Objectives:

Educational:

• consolidation of knowledge of the formulas for the area of ​​\u200b\u200ba rectangle, right triangle;
• analysis of tasks for calculating the area of ​​\u200b\u200ba "complex" figure and methods for their implementation;
• independent performance of tasks to test knowledge, skills, abilities.

Developing:

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

Educational:

• to educate students in the skills of educational work;
• to cultivate a culture of oral and written mathematical speech;
• to cultivate friendship in the classroom and the ability to work in groups.

Lesson type: combined.

Equipment:

• Mathematics: textbook for 5 cells. general education institutions / N.Ya. Vilenkin, V.I. Zhokhov et al., M.: Mnemozina, 2010.
• Cards for groups of students with figures to calculate the area of ​​a complex figure.
• Drawing tools.

Lesson plan:

1. Organizing time.
2. Knowledge update.
   but) Theoretical questions(test).
   b) Statement of the problem.
3. Learned new material.
   a) finding a solution to the problem;
   b) solving the problem.
4. Fixing the material.
   a) collective problem solving;
   Fizkultminutka.
   b) independent work.
5. Homework.
6. Summary of the lesson. Reflection.

During the classes

I. Organizational moment.

Let's start the lesson with these words of encouragement:

Mathematics, friends,
Absolutely everyone needs it.
Work hard in class
And success is waiting for you!

II. Knowledge update.

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

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

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:

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

b) (Formulation of the problem). A task. How much paint is needed to paint a floor that has the following shape (see fig.), if 200 g of paint is consumed per 1 m 2?

III. Learning new material.

What do we need to know in order to solve the last problem? (Find the area of ​​the floor, which looks like a "complex figure.")

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

Consider a rectangle ABCD. Let's draw a line in it KPMN by 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 the results. (15==7+8)

Output? (The area of ​​the whole figure is equal to the sum of the areas of its parts.

S = S 1 + S 2

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

S 1 \u003d 7 2 \u003d 14 (m 2)
S 2 \u003d (7 - 4) (8 - 2 - 3) \u003d 3 3 \u003d 9 (m 2)
S 3 \u003d 7 3 \u003d 21 (m 2)
S \u003d S 1 + S 2 + S 3 \u003d 14 + 9 + 21 \u003d 44 (m 2)

Let's make up plan for solving problems for finding the area of ​​\u200b\u200ba "complex figure":

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

a) Task 1. How many tiles will be required to lay out a platform of the following sizes:

S = S 1 + S 2
S 1 \u003d (60 - 30) 20 \u003d 600 (dm 2)
S 2 \u003d 30 50 \u003d 1500 (dm 2)
S \u003d 600 + 1500 \u003d 2100 (dm 2)

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

Answer: 2100 dm 2.

Task 2. (collective decision on the board and in notebooks.) How much m 2 of linoleum is required to repair a room having the following shape:

S = S 1 + S 2
S 1 \u003d 3 2 \u003d 6 (m 2)
S 2 \u003d ((5 - 3) 2): 2 \u003d 2 (m 2)
S \u003d 6 + 2 \u003d 8 (m 2)

Answer: 8 m 2.

Fizkultminutka.

Now, guys, get up.
They quickly raised their hands.
Sideways, forward, backward.
Turned right, left.
We sat down quietly, back to business.

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 the floor that has the shape of the figure shown on the card, if 200 g of paint is required per 1 m 2.

You build this figure in your notebook and, writing down all the data, proceed to the task. You can discuss the solution (but only in your group!). If a group copes with the task quickly, then it - additional task (after verification of independent work).

Tasks for groups:

V. Homework.

item 18, no. 718, no. 749.

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

VI. Lesson results.

Reflection. Continue the phrase:

• Today I found out...
• It was interesting…
• It was difficult…
• Now I can…
• Lesson taught me for life...

If you plan to make repairs yourself, then you will need to make an estimate for building and finishing materials. To do this, you will need to calculate the area of ​​\u200b\u200bthe room in which you plan to carry out repairs. The main assistant in this is a specially designed formula. The area of ​​\u200b\u200bthe room, namely its calculation, will allow you to save a lot of money on building materials and direct the released financial resources in a more appropriate direction.

The geometric shape of the room

The formula for calculating the area of ​​​​a room directly depends on its shape. The most typical for domestic structures are rectangular and square rooms. However, during redevelopment, the standard form may be distorted. The rooms are:

• Rectangular.
• Square.
• Complex configuration (for example, round).
• With niches and ledges.

Each of them has its own calculation features, but, as a rule, the same formula is used. The area of ​​​​a room of any shape and size, one way or another, can be calculated.

Rectangular or square room

To calculate the area of ​​a rectangular or square room, just remember school lessons geometry. Therefore, it should not be difficult for you to determine the area of ​​\u200b\u200bthe room. The calculation formula looks like:

S rooms=A*B, where

A is the length of the room.

B is the width of the room.

To measure these values, you will need a regular tape measure. To get the most accurate calculations, it is worth measuring the wall on both sides. If the values ​​do not converge, take the average of the resulting data as a basis. But remember that any calculations have their own errors, so the material should be purchased with a margin.

A room with a complex configuration

If your room does not fall under the definition of "typical", i.e. has the shape of a circle, triangle, polygon, then you may need a different formula for calculations. You can try to conditionally divide the area of ​​\u200b\u200bthe room with such a characteristic into rectangular elements and make calculations in the standard way. If this is not possible for you, then use the following methods:

• The formula for finding the area of ​​a circle:

S room \u003d π * R 2, where

R is the radius of the room.

• The formula for finding the area of ​​a triangle is:

S room = √ (P (P - A) x (P - B) x (P - C)), where

P is the half-perimeter of the triangle.

A, B, C are the lengths of its sides.

Hence P \u003d A + B + C / 2

If in the process of calculating you have any difficulties, then it is better not to torture yourself and turn to professionals.

Room area with ledges and niches

Often the walls are decorated with decorative elements in the form of various niches or ledges. Also, their presence may be due to the need to hide some unaesthetic elements of your room. The presence of ledges or niches on your wall means that the calculation should be carried out in stages. Those. first, the area of ​​\u200b\u200ba flat section of the wall is found, and then the area of ​​\u200b\u200ba niche or ledge is added to it.

The area of ​​the wall is found by the formula:

S walls \u003d P x C, where

P - perimeter

C - height

You also need to consider the presence of windows and doors. Their area must be subtracted from the resulting value.

Room with multi-level ceiling

A multi-level ceiling does not complicate the calculations as much as it seems at first glance. If it has a simple design, then calculations can be made on the principle of finding the area of ​​\u200b\u200bwalls complicated by niches and ledges.

However, if the design of your ceiling has arcuate and wavy elements, then it is more appropriate to determine its area using the floor area. For this you need:

1. Find the dimensions of all straight sections of the walls.
2. Find the floor area.
3. Multiply the length and height of the vertical sections.
4. Sum the resulting value with the floor area.

Step-by-step instructions for determining the total

floor space

1. Free the room from unnecessary things. In the process of measuring, you will need free access to all areas of your room, so you need to get rid of everything that can interfere with this.
2. Visually divide the room into sections of regular and irregular shapes. If your room has a strictly square or rectangular shape, then this step can be skipped.
3. Make an arbitrary layout of the room. This drawing is needed so that all the data is always at your fingertips. Also, it will not give you the opportunity to get confused in numerous measurements.
4. Measurements must be taken several times. This is an important rule to avoid errors in calculations. Also if you are using make sure the beam lies flat on the wall surface.
5. Find the total area of ​​the room. The formula for the total area of ​​a room is to find the sum of all the areas of the individual sections of the room. Those. S total = S walls + S floors + S ceilings