The area of different figures. What is the area of a figure? Protection of personal information

Geometric area- a numerical characteristic of a geometric figure showing the size of this figure (part of the surface bounded by a closed contour of this figure). The size of the area is expressed by the number of square units contained in it.

Triangle area formulas

Triangle area formula for side and height
Area of a triangle equal to half the product of the length of a side of a triangle and the length of the altitude drawn to this side
The formula for the area of a triangle given three sides and the radius of the circumscribed circle
The formula for the area of a triangle given three sides and the radius of an inscribed circle
Area of a triangle is equal to the product of the half-perimeter of the triangle and the radius of the inscribed circle.

Square area formulas

The formula for the area of a square given the length of a side
square area is equal to the square of its side length.
The formula for the area of a square given the length of the diagonal
square area equal to half the square of the length of its diagonal.
S= 1 2
2

Rectangle area formula

Rectangle area

Formulas for the area of a parallelogram

Parallelogram area formula for side length and height
Parallelogram area
The formula for the area of a parallelogram given two sides and the angle between them
Parallelogram area is equal to the product of the lengths of its sides multiplied by the sine of the angle between them.
a b sinα

Formulas for the area of a rhombus

Rhombus area formula given side length and height
Rhombus area is equal to the product of the length of its side and the length of the height lowered to this side.
The formula for the area of a rhombus given the length of the side and the angle
Rhombus area is equal to the product of the square of the length of its side and the sine of the angle between the sides of the rhombus.
The formula for the area of a rhombus from the lengths of its diagonals
Rhombus area is equal to half the product of the lengths of its diagonals.

Trapezium area formulas

Heron's formula for a trapezoid
Where S is the area of the trapezoid,
- the length of the bases of the trapezoid,
- the length of the sides of the trapezoid,

How to find the area of a figure?

Knowing and being able to calculate the areas of various figures is necessary not only for solving simple geometric problems. You can not do without this knowledge when drawing up or checking estimates for the repair of premises, calculating the amount of necessary consumables. Therefore, let's figure out how to find the areas of different shapes.

The part of the plane enclosed within a closed contour is called the area of this plane. The area is expressed by the number of square units enclosed in it.

To calculate the area of basic geometric shapes, you must use the correct formula.

Area of a triangle

Designations:

If h, a are known, then the area of the desired triangle is determined as the product of the lengths of the side and the height of the triangle lowered to this side, divided in half: S=(a h)/2
If a, b, c are known, then the required area is calculated using the Heron formula: the square root taken from the product of half the perimeter of the triangle and three differences of half the perimeter and each side of the triangle: S = √ (p (p - a) (p - b) (p - c)).
If a, b, γ are known, then the area of the triangle is determined as half the product of 2 sides, multiplied by the value of the sine of the angle between these sides: S=(a b sin γ)/2
If a, b, c, R are known, then the required area is defined as dividing the product of the lengths of all sides of the triangle by the four radii of the circumscribed circle: S=(a b c)/4R
If p, r are known, then the desired area of the triangle is determined by multiplying half the perimeter by the radius of the circle inscribed in it: S = p r

square area

Designations:

If the side is known, then the area of this figure is determined as the square of the length of its side: S=a 2
If d is known, then the square area is defined as half the square of the length of its diagonal: S=d 2 /2

Rectangle area

Designations:

S - determined area,
a, b are the lengths of the sides of the rectangle.

If a, b are known, then the area of a given rectangle is determined by the product of the lengths of its two sides: S=a b
If the lengths of the sides are unknown, then the area of the rectangle must be divided into triangles. In this case, the area of a rectangle is defined as the sum of the areas of its constituent triangles.

Parallelogram area

Designations:

S - desired area,
a, b - side lengths,
h is the length of the height of the given parallelogram,
d1, d2 - lengths of two diagonals,
α - the angle between the sides,
γ is the angle between the diagonals.

If a, h are known, then the desired area is determined by multiplying the lengths of the side and the height lowered to this side: S = a h
If a, b, α are known, then the area of the parallelogram is determined by multiplying the lengths of the sides of the parallelogram and the value of the sine of the angle between these sides: S=a b sin α
If d 1 , d 2 , γ are known, then the area of the parallelogram is defined as half the product of the lengths of the diagonals and the value of the sine of the angle between these diagonals: S=(d 1 d 2 sinγ)/2

Rhombus area

Designations:

S - desired area,
a - side length,
h - height length,
α is the smaller angle between the two sides,
d1, d2 are the lengths of the two diagonals.

If a, h are known, then the area of the rhombus is determined by multiplying the length of the side by the length of the height that is lowered to this side: S = a h
If a, α are known, then the area of the rhombus is determined by multiplying the square of the side length by the sine of the angle between the sides: S=a 2 sin α
If d 1 and d 2 are known, then the desired area is determined as half the product of the lengths of the diagonals of the rhombus: S \u003d (d 1 d 2) / 2

Trapezium area

Designations:

If a, b, c, d are known, then the required area is determined by the formula: S= (a+b) /2 *√ .
With known a, b, h, the desired area is determined as the product of half the sum of the bases and the height of the trapezoid: S=(a+b)/2 h

Area of a convex quadrilateral

Designations:

If d 1 , d 2 , α are known, then the area of a convex quadrilateral is defined as half the product of the diagonals of the quadrilateral multiplied by the sine of the angle between these diagonals: S=(d 1 d 2 sin α)/2
With known p, r, the area of a convex quadrangle is defined as the product of the semiperimeter of the quadrangle and the radius of the circle inscribed in this quadrangle: S=p r
If a, b, c, d, θ are known, then the area of a convex quadrilateral is determined as the square root of the products of the difference of the semiperimeter and the length of each side minus the product of the lengths of all sides and the square of the cosine of half the sum of two opposite angles: S 2 = (p - a )(p - b)(p - c)(p - d) - abcd cos 2 ((α+β)/2)

Area of a circle

Designations:

If r is known, then the desired area is determined as the product of the number π and the radius squared: S=π r 2

If d is known, then the area of the circle is determined as the product of the number π times the square of the diameter, divided by four: S=(π d 2)/4

The area of a complex figure

The complex can be broken down into simple geometric shapes. The area of a complex figure is defined as the sum or difference of the component areas. Consider, for example, a ring.

Designation:

S is the area of the ring,
R, r are the radii of the outer circle and the inner one, respectively,
D, d are the diameters of the outer circle and the inner one, respectively.

To find the area of the ring, subtract the area from the area of the larger circle. smaller circle. S \u003d S1-S2 \u003d πR 2 -πr 2 \u003d π (R 2 -r 2).

Thus, if R and r are known, then the area of the ring is determined as the difference between the squares of the radii of the outer and inner circles, multiplied by the number pi: S=π(R 2 -r 2).

If D and d are known, then the area of the ring is determined as a quarter of the difference in the squares of the diameters of the outer and inner circles, multiplied by the number pi: S = (1/4) (D 2 -d 2) π.

Patch area

Suppose that inside one square (A) there is another (B) (smaller), and we need to find a filled cavity between the figures "A" and "B". Let's just say, a "frame" of a small square. For this:

Find the area of \u200b\u200bthe figure "A" (calculated by the formula for finding the area of a square).
Similarly, we find the area of \u200b\u200bthe figure "B".
Subtract from area "A" area "B". And thus we get the area of the shaded figure.

Now you know how to find the areas of different shapes.

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; there is a development of attention, the ability to research activities, the 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 a rectangle, a right-angled 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:

Organizing time.
Knowledge update.
a) Theoretical questions (test).
b) Statement of the problem.
Learned new material.
a) finding a solution to the problem;
b) solving the problem.
Fixing the material.
a) collective problem solving;
Fizkultminutka.
b) independent work.
Homework.
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.

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:

the area of a square with a side of 1 cm;
a square with a side of 1 cm;
square with a perimeter of 1 cm.

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

8 dm;
8 dm 2;
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:

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

5. The area of the figure shown in the figure is:

12 cm;
8 cm;
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)

Conclusion? (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":

We break the figure into simple figures.
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 will receive an 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...

In the previous section, devoted to the analysis of the geometric meaning of a definite integral, we obtained 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 segment [ a ; b] ,

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

These formulas are applicable for solving relatively simple problems. In fact, we often have to work with more complex shapes. In this regard, we will devote this section to the analysis of algorithms for calculating the area of figures, which are limited by functions in an 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 segment [ a ; b ] , and f 1 (x) ≤ f 2 (x) for any value x from [ a ; b] . Then the formula for calculating the area of \u200b\u200ba figure Gbounded by lines x \u003d a, x \u003d b, y \u003d f 1 (x) and y \u003d f 2 (x) will look like S (G) \u003d ∫ a b f 2 (x) - f 1 (x) d x .

A similar formula will be applicable for the area of \u200b\u200bthe figure bounded by the lines y \u003d c, y \u003d d, x \u003d g 1 (y) and x \u003d g 2 (y): S (G) \u003d ∫ c d (g 2 (y) - g 1 (y) d y .

Proof

We will analyze three cases for which the formula will be valid.

In the first case, taking into account the additivity property of the 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)) d x .

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 the consideration of the general case when y = f 1 (x) and y = f 2 (x) intersect the axis O x .

We will denote the intersection points as x i , i = 1 , 2 , . . . , n - 1 . These points break 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

Consequently,

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.

And now let's move on to the analysis of examples of calculating the area of \u200b\u200bfigures that are limited by the lines y \u003d f (x) and x \u003d g (y) .

Considering any of the examples, we will begin with the construction of a graph. The image will allow us to represent complex shapes as combinations of simpler shapes. If plotting graphs and shapes on them is difficult for you, you can study the section on basic elementary functions, geometric transformation of graphs of functions, as well as plotting during the study of a function.

Example 1

It is necessary to determine the area of \u200b\u200bthe figure, which is limited by the parabola y \u003d - x 2 + 6 x - 5 and straight lines y \u003d - 1 3 x - 1 2, x \u003d 1, x \u003d 4.

Solution

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

On the interval [ 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 an answer, we use the formula obtained earlier, as well as the method for calculating a 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 parallel to the x-axis. This is x = 7 . This requires us to find the second integration limit ourselves.

Let's build a graph and put on it the lines given in the condition of the problem.

Having a graph in front of our eyes, we can easily determine that the lower limit of integration will be the abscissa of the intersection point of the graph with a straight line y \u003d x and a semi-parabola y \u003d 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 D G x 2 = 1 - 9 2 = - 1 ∉ O D G

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 redundant. 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 better to always 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 . 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 \u200b\u200bthe figure, which is limited by the graphs of the functions y \u003d 1 x and y \u003d - x 2 + 4 x - 2.

Solution

Let's draw 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 equal to zero, the equality 1 x \u003d - x 2 + 4 x - 2 becomes equivalent to the equation of the third degree - x 3 + 4 x 2 - 2 x - 1 \u003d 0 with integer coefficients. You can refresh the memory of the algorithm for solving such equations by referring to the section “Solution of 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 \u003d 3 - 13 2 ≈ - 0. 3

We have found an interval x ∈ 1; 3 + 13 2 , where G is enclosed above the blue line 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) \u003d 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 \u003d x 3, y \u003d - log 2 x + 1 and the x-axis.

Solution

Let's put 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 place it symmetrically about the x-axis and move it up one unit. The equation of the x-axis y \u003d 0.

Let's denote the points of intersection of the lines.

As can be seen from the figure, the graphs of the functions y \u003d x 3 and y \u003d 0 intersect at the point (0; 0) . This is because x \u003d 0 is the only real root of the equation x 3 \u003d 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 \u003d x 3 and y \u003d - log 2 x + 1 intersect at the point (1; 1) . The last statement may not be obvious, but the equation x 3 \u003d - log 2 x + 1 cannot have more than one root, since the function y \u003d x 3 is strictly increasing, and the function y \u003d - log 2 x + 1 is strictly decreasing.

The next step involves several options.

Option number 1

We can represent the figure G as the sum of two curvilinear trapezoids located above the abscissa axis, the first of which is located below the midline on the segment x ∈ 0; 1 , and the second one 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 number 2

The 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 one is between the red and blue lines on the segment x ∈ 1 ; 2. This allows us to find the area like this:

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) \u003d ∫ c d (g 2 (y) - g 1 (y)) d y. In fact, the lines that bound the shape can be represented as functions of the y argument.

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 \u003d x, y \u003d 2 3 x - 3, y \u003d - 1 2 x + 4.

Solution

Draw a line on the chart with a red line, given by the function y = x . Draw the line y = - 1 2 x + 4 in blue, and mark the line y = 2 3 x - 3 in black.

Note the intersection points.

Find the intersection points of the graphs of 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 \u003d 144 x 1 \u003d 20 + 144 2 \u003d 16; x 2 = 20 - 144 2 = 4 i 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

Find the intersection point of the graphs of 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 \u003d 2 3 9 - 3 \u003d 3 ⇒ x 1 \u003d 9 is the solution to the equation ⇒ (9; 3) point and intersection 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 is not a solution to the equation

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 number 1

We represent 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 number 2

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

Then we solve the line equation for x, and only after that we apply the formula for calculating the area of \u200b\u200bthe 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 i l i n i i

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 match.

Answer: S (G) = 11 3

Results

To find the area of a figure that is bounded by given lines, we need to draw lines on a plane, find their intersection points, and apply the formula for finding the area. In this section, we have reviewed the most common options for tasks.

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

There is an infinite number of flat figures of various shapes, both regular and irregular. A common property of all figures is that any of them has an area. The areas of the figures are the dimensions of the part of the plane occupied by these figures, expressed in certain units. This value is always expressed as a positive number. The unit of measurement is the area of a square whose side is equal to a unit of length (for example, one meter or one centimeter). The approximate value of the area of any figure can be calculated by multiplying the number of unit squares into which it is divided by the area of one square.

Other definitions of this concept are as follows:

1. The areas of simple figures are scalar positive quantities that satisfy the conditions:

Equal figures have equal areas;

If a figure is divided into parts (simple figures), then its area is the sum of the areas of these figures;

A square with a side unit of measurement serves as a unit of area.

2. The areas of figures of complex shape (polygons) are positive quantities with the following properties:

Equal polygons have the same areas;

If a polygon consists of several other polygons, its area is equal to the sum of the areas of the latter. This rule is true for non-overlapping polygons.

As an axiom, the statement is accepted that the areas of figures (polygons) are positive values.

The definition of the area of a circle is given separately as the value to which the area of a given circle inscribed in a circle tends - despite the fact that the number of its sides tends to infinity.

The areas of irregularly shaped figures (arbitrary figures) do not have a definition, only methods for calculating them are determined.

Calculation of areas already in antiquity was an important practical task in determining the size of land plots. The rules for calculating areas for several hundred years were formulated by Greek scientists and set forth in Euclid's Elements as theorems. Interestingly, the rules for determining the areas of simple figures in them are the same as at present. The areas with a curvilinear contour were calculated using the limit transition.

Calculating the areas of simple rectangles, squares), familiar to everyone from school, is quite simple. It is not even necessary to memorize the formulas containing the letter designations for the areas of the figures. Just remember a few simple rules:

2. The area of a rectangle is calculated by multiplying its length by its width. In this case, it is necessary that the length and width be expressed in the same units of measurement.

3. We calculate the area of a complex figure by dividing it into several simple ones and adding the resulting areas.

4. The diagonal of a rectangle divides it into two triangles whose areas are equal and equal to half of its area.

5. The area of a triangle is calculated as half the product of its height and base.

6. The area of a circle is equal to the product of the square of the radius and the well-known number "π".

7. The area of a parallelogram is calculated as the product of adjacent sides and the sine of the angle lying between them.

8. The area of a rhombus is ½ the result of multiplying the diagonals by the sine of the internal angle.

9. The area of the trapezoid is found by multiplying its height by the length of the midline, which is equal to the arithmetic mean of the bases. Another option for determining the area of a trapezoid is to multiply its diagonals and the sine of the angle lying between them.

For clarity, children in elementary school are often given tasks: to find the area of a figure drawn on paper using a palette or a sheet of transparent paper, divided into cells. Such a sheet of paper is superimposed on the measured figure, the number of full cells (area units) that fit in its contour is counted, then the number of incomplete ones, which is divided in half.

The area of ​​different figures. What is the area of ​​a figure? Protection of personal information

Triangle area formulas

Square area formulas

Rectangle area formula

Formulas for the area of ​​a parallelogram

Formulas for the area of ​​a rhombus

Trapezium area formulas

Area of ​​a triangle

Designations:

square area

Designations:

Rectangle area

Designations:

Parallelogram area

Designations:

Rhombus area

Designations:

Trapezium area

Designations:

Area of ​​a convex quadrilateral

Designations:

Area of ​​a circle

Designations:

The area of ​​a complex figure

Designation:

Patch area

During the classes

I. Organizational moment.

II. Knowledge update.

III. Learning new material.

V. Homework.

VI. Lesson results.

Results

The area of different figures. What is the area of a figure? Protection of personal information

Formulas for the area of a parallelogram

Formulas for the area of a rhombus

Area of a triangle

Area of a convex quadrilateral

Area of a circle

The area of a complex figure