\[(\Large(\text(Basic facts about area)))\]

We can say that the area of ​​a polygon is a value that indicates the part of the plane that a given polygon occupies. The area unit is taken as the area of ​​a square with a side of \(1\) cm, \(1\) mm, etc. (single square). Then the area will be measured in cm\(^2\) , mm\(^2\) respectively.

In other words, we can say that the area of ​​​​a figure is a value whose numerical value shows how many times a unit square fits in a given figure.

Area Properties

1. The area of ​​any polygon is a positive value.

2. Equal polygons have equal areas.

3. If a polygon is composed of several polygons, then its area is equal to the sum of the areas of these polygons.

4. The area of ​​a square with side \(a\) is \(a^2\) .

\[(\Large(\text(Area of ​​rectangle and parallelogram)))\]

Theorem: area of ​​a rectangle

The area of ​​a rectangle with sides \(a\) and \(b\) is \(S=ab\) .

Proof

Let's build the rectangle \(ABCD\) to a square with side \(a+b\) , as shown in the figure:

This square consists of a rectangle \(ABCD\) , another rectangle equal to it, and two squares with sides \(a\) and \(b\) . In this way,

\(\begin(multline*) S_(a+b)=2S_(\text(pr-k))+S_a+S_b \Leftrightarrow (a+b)^2=2S_(\text(pr-k))+ a^2+b^2 \Leftrightarrow\\ a^2+2ab+b^2=2S_(\text(pr-k))+a^2+b^2 \Rightarrow S_(\text(pr-k) )=ab \end(multline*)\)

Definition

The height of a parallelogram is the perpendicular drawn from the vertex of the parallelogram to the side (or extension of the side) that does not contain that vertex.
For example, the height \(BK\) falls on the side \(AD\) , and the height \(BH\) falls on the extension of the side \(CD\) :

Theorem: area of ​​a parallelogram

The area of ​​a parallelogram is equal to the product of the height and the side to which this height is drawn.

Proof

Draw perpendiculars \(AB"\) and \(DC"\) as shown in the figure. Note that these perpendiculars are equal to the height of the parallelogram \(ABCD\) .

Then \(AB"C"D\) is a rectangle, hence \(S_(AB"C"D)=AB"\cdot AD\) .

Note that right triangles \(ABB"\) and \(DCC"\) are equal. In this way,

\(S_(ABCD)=S_(ABC"D)+S_(DCC")=S_(ABC"D)+S_(ABB")=S_(AB"C"D)=AB"\cdot AD.\)

\[(\Large(\text(Area of ​​triangle)))\]

Definition

We will call the side to which the altitude is drawn in the triangle the base of the triangle.

Theorem

The area of ​​a triangle is half the product of its base and the height drawn to that base.

Proof

Let \(S\) be the area of ​​the triangle \(ABC\) . Let's take the side \(AB\) as the base of the triangle and draw the height \(CH\) . Let's prove that \ We complete the triangle \(ABC\) to the parallelogram \(ABDC\) as shown in the figure:

Triangles \(ABC\) and \(DCB\) are equal in three sides (\(BC\) is their common side, \(AB = CD\) and \(AC = BD\) as opposite sides of the parallelogram \(ABDC\ ) ), so their areas are equal. Therefore, the area \(S\) of the triangle \(ABC\) is equal to half the area of ​​the parallelogram \(ABDC\) , i.e. \(S = \dfrac(1)(2)AB\cdotCH\).

Theorem

If two triangles \(\triangle ABC\) and \(\triangle A_1B_1C_1\) have equal heights, then their areas are related as the bases to which these heights are drawn.

Consequence

The median of a triangle divides it into two triangles of equal area.

Theorem

If two triangles \(\triangle ABC\) and \(\triangle A_2B_2C_2\) each have the same angle, then their areas are related as the products of the sides forming this angle.

Proof

Let \(\angle A=\angle A_2\) . Let's combine these corners as shown in the figure (the point \(A\) is aligned with the point \(A_2\) ):

Draw heights \(BH\) and \(C_2K\) .

The triangles \(AB_2C_2\) and \(ABC_2\) have the same height \(C_2K\) , therefore: \[\dfrac(S_(AB_2C_2))(S_(ABC_2))=\dfrac(AB_2)(AB)\]

The triangles \(ABC_2\) and \(ABC\) have the same height \(BH\) , therefore: \[\dfrac(S_(ABC_2))(S_(ABC))=\dfrac(AC_2)(AC)\]

Multiplying the last two equalities, we get: \[\dfrac(S_(AB_2C_2))(S_(ABC))=\dfrac(AB_2\cdot AC_2)(AB\cdot AC) \qquad \text( or ) \qquad \dfrac(S_(A_2B_2C_2))(S_ (ABC))=\dfrac(A_2B_2\cdot A_2C_2)(AB\cdot AC)\]

Pythagorean theorem

In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs:

The converse is also true: if in a triangle the square of the length of one side is equal to the sum of the squares of the lengths of the other two sides, then such a triangle is right-angled.

Theorem

The area of ​​a right triangle is half the product of the legs.

Theorem: Heron's formula

Let \(p\) be the semiperimeter of a triangle, \(a\) , \(b\) , \(c\) be the lengths of its sides, then its area is equal to \

\[(\Large(\text(Area of ​​a rhombus and a trapezoid)))\]

Comment

Because rhombus is a parallelogram, then the same formula is true for it, i.e. The area of ​​a rhombus is equal to the product of the height and the side to which this height is drawn.

Theorem

The area of ​​a convex quadrilateral whose diagonals are perpendicular is half the product of the diagonals.

Proof

Consider the quadrilateral \(ABCD\) . Denote \(AO=a, CO=b, BO=x, DO=y\) :

Note that this quadrilateral is made up of four right triangles, therefore, its area is equal to the sum of the areas of these triangles:

\(\begin(multline*) S_(ABCD)=\frac12ax+\frac12xb+\frac12by+\frac12ay=\frac12(ax+xb+by+ay)=\\ \frac12((a+b)x+(a+b) y)=\frac12(a+b)(x+y)\end(multline*)\)

Corollary: area of ​​a rhombus

The area of ​​a rhombus is half the product of its diagonals: \

Definition

The height of a trapezoid is a perpendicular drawn from the top of one base to the other base.

Theorem: area of ​​a trapezoid

The area of ​​a trapezoid is half the sum of the bases times the height.

Proof

Consider a trapezoid \(ABCD\) with bases \(BC\) and \(AD\) . Draw \(CD"\parallel AB\) as shown in the figure:

Then \(ABCD"\) is a parallelogram.

We also draw \(BH"\perp AD, CH\perp AD\) (\(BH"=CH\) are the heights of the trapezoid).

Then \(S_(ABCD")=BH"\cdot AD"=BH"\cdot BC, \quad S_(CDD")=\dfrac12CH\cdot D"D\)

Because a trapezoid consists of a parallelogram \(ABCD"\) and a triangle \(CDD"\) , then its area is equal to the sum of the areas of the parallelogram and the triangle, that is:

\ \[=\dfrac12 CH\left(BC+AD"+D"D\right)=\dfrac12 CH\left(BC+AD\right)\]

Everyone who studied mathematics and geometry at school knows these sciences at least superficially. But over time, if they are not practiced, knowledge is forgotten. Many even believe that they just wasted their time studying geometric calculations. However, they are wrong. Technical workers perform daily work related to geometric calculations. As for the calculation of the area of ​​a polygon, this knowledge also finds its application in life. They will be needed at least in order to calculate the area of ​​\u200b\u200bthe land. So let's learn how to find the area of ​​a polygon.

Polygon definition

First, let's define what a polygon is. This is a flat geometric figure, which was formed as a result of the intersection of three or more lines. Another simple definition: a polygon is a closed polyline. Naturally, at the intersection of lines, intersection points are formed, their number is equal to the number of lines that form a polygon. The points of intersection are called the vertices, and the segments formed from the straight lines are called the sides of the polygon. Adjacent segments of a polygon are not on the same straight line. Line segments that are nonadjacent are those that do not pass through common points.

The sum of the areas of triangles

How to find the area of ​​a polygon? The area of ​​a polygon is the inner part of the plane, which was formed at the intersection of the segments or sides of the polygon. Since a polygon is a combination of shapes such as a triangle, rhombus, square, trapezoid, there is simply no universal formula for calculating its area. In practice, the most universal method is the division of a polygon into simpler figures, the area of ​​which is not difficult to find. By adding the sums of the areas of these simple figures, we get the area of ​​the polygon.

Through the area of ​​the circle

In most cases, the polygon has a regular shape and forms a figure with equal sides and angles between them. Calculating the area in this case is very simple using the inscribed or circumscribed circle. If the area of ​​the circle is known, then it must be multiplied by the perimeter of the polygon, and then the resulting product divided by 2. As a result, the formula for calculating the area of ​​such a polygon is obtained: S = ½∙P∙r., where P is the area of ​​the circle, and r is the perimeter of the polygon .

The method of splitting a polygon into “convenient” shapes is the most popular in geometry, it allows you to quickly and correctly find the area of ​​a polygon. The 4th grade of high school usually learns such methods.

In this article, we will talk about how to express the area of ​​a polygon in which a circle can be inscribed in terms of the radius of this circle. It is immediately worth noting that not every polygon can be inscribed in a circle. However, if this is possible, then the formula by which the area of ​​such a polygon is calculated becomes very simple. Read this article to the end or watch the attached video tutorial and you will learn how to express the area of ​​a polygon in terms of the radius of a circle inscribed in it.

The formula for the area of ​​a polygon in terms of the radius of the inscribed circle

Let's draw a polygon A 1 A 2 A 3 A 4 A 5 , not necessarily correct, but one in which a circle can be inscribed. Let me remind you that an inscribed circle is a circle that touches all sides of the polygon. In the figure, this is a green circle centered at a point O:

We have taken a 5-gon here as an example. But in fact this is of no essential importance, since the further proof is valid for both the 6-gon and the 8-gon, and in general for any "gon" arbitrarily.

If you connect the center of the inscribed circle with all the vertices of the polygon, then it will be divided into as many triangles as there are vertices in the given polygon. In our case: 5 triangles. If we connect the dot O with all points of tangency of the inscribed circle with the sides of the polygon, you get 5 segments (in the figure below, these are the segments Oh 1 , Oh 2 , Oh 3 , Oh 4 and Oh 5), which are equal to the radius of the circle and are perpendicular to the sides of the polygon to which they are drawn. The latter is true, since the radius drawn to the point of contact is perpendicular to the tangent:

How to find the area of ​​our circumscribed polygon? The answer is simple. It is necessary to add up the areas of all the triangles obtained as a result of splitting:

Consider what is the area of ​​a triangle. In the picture below, it is highlighted in yellow:

It is equal to half the product of the base A 1 A 2 to the height Oh 1 drawn to this base. But, as we have already found out, this height is equal to the radius of the inscribed circle. That is, the formula for the area of ​​a triangle takes the form: , where r is the radius of the inscribed circle. Similarly, the areas of all the remaining triangles are found. As a result, the desired area of ​​the polygon is equal to:

It can be seen that in all terms of this sum there is a common factor , which can be taken out of brackets. The result is the following expression:

That is, in brackets there was simply the sum of all sides of the polygon, that is, its perimeter P. Most often, in this formula, the expression is simply replaced by p and call this letter "half-perimeter". As a result, the final formula becomes:

That is, the area of ​​a polygon in which a circle of known radius is inscribed is equal to the product of this radius and the semiperimeter of the polygon. This is the result we were aiming for.

Finally, he notes that a circle can always be inscribed in a triangle, which is a special case of a polygon. Therefore, for a triangle, this formula can always be applied. For other polygons with more than 3 sides, you first need to make sure that a circle can be inscribed in them. If so, you can safely use this simple formula and find the area of ​​\u200b\u200bthis polygon from it.

Prepared by Sergey Valerievich

Distance and Length Units Converter Area Units Converter Join © 2011-2017 Mikhail Dovzhik Copying of materials is prohibited. In the online calculator, you can use values ​​in the same units of measurement! If you have trouble converting units of measure, use the Distance and Length Unit Converter and the Area Unit Converter. Additional features of the quadrilateral area calculator

• You can move between input fields by pressing the right and left keys on the keyboard.

Theory. Area of ​​a quadrilateral A quadrilateral is a geometric figure consisting of four points (vertices), no three of which lie on the same straight line, and four segments (sides) connecting these points in pairs. A quadrilateral is called convex if the segment connecting any two points of this quadrilateral will be inside it.

How to find the area of ​​a polygon?

The formula for determining the area is determined by taking each edge of the polygon AB, and calculating the area of ​​the triangle ABO with a vertex at the origin O, through the coordinates of the vertices. When walking around a polygon, triangles are formed, including the inside of the polygon and located outside it. The difference between the sum of these areas is the area of ​​the polygon itself.

Therefore, the formula is called the surveyor's formula, since the "cartographer" is at the origin; if it walks the area counterclockwise, the area is added if it's on the left and subtracted if it's on the right in terms of the origin. The area formula is valid for any non-intersecting (simple) polygon, which may be convex or concave. Content

• 1 Definition
• 2 Examples
• 3 More complex example
• 4 Name explanation
• 5 See

Polygon area

Attention

It could be:

• triangle;
• quadrilateral;
• five- or hexagon and so on.

Such a figure will certainly be characterized by two positions:

1. Adjacent sides do not belong to the same line.
2. Non-adjacent ones have no common points, that is, they do not intersect.

To understand which vertices are adjacent, you need to see if they belong to the same side. If yes, then neighboring. Otherwise, they can be connected by a segment, which must be called a diagonal. They can only be drawn in polygons that have more than three vertices.

What kinds of them exist? A polygon with more than four corners can be convex or concave. The difference of the latter is that some of its vertices may lie on different sides of a straight line drawn through an arbitrary side of the polygon.

How to find the area of ​​a regular and irregular hexagon?

• Knowing the length of the side, multiply it by 6 and get the perimeter of the hexagon: 10 cm x 6 \u003d 60 cm
• Substitute the results in our formula:
Area \u003d 1/2 * perimeter * apothema Area \u003d ½ * 60cm * 5√3 Solve: Now it remains to simplify the answer to get rid of square roots, and indicate the result in square centimeters: ½ * 60 cm * 5√3 cm \u003d 30 * 5√3 cm =150 √3 cm =259.8 cm² Video on how to find the area of ​​a regular hexagon There are several options for determining the area of ​​an irregular hexagon:

• trapezoid method.
• A method for calculating the area of ​​irregular polygons using the coordinate axis.
• A method for splitting a hexagon into other shapes.

Depending on the initial data that you will know, the appropriate method is selected.

Important

Some irregular hexagons consist of two parallelograms. To determine the area of ​​a parallelogram, multiply its length by its width and then add the two already known areas. Video on how to find the area of ​​a polygon An equilateral hexagon has six equal sides and is a regular hexagon.

The area of ​​an equilateral hexagon is equal to 6 areas of the triangles into which a regular hexagonal figure is divided. All triangles in a regular hexagon are equal, so to find the area of ​​such a hexagon, it will be enough to know the area of ​​at least one triangle. To find the area of ​​an equilateral hexagon, of course, the formula for the area of ​​a regular hexagon, described above, is used.

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Decorating a home, clothing, drawing pictures contributed to the process of formation and accumulation of information in the field of geometry, which people of those times obtained empirically, bit by bit and passed on from generation to generation. Today, knowledge of geometry is necessary for a cutter, a builder, an architect, and every ordinary person in everyday life. Therefore, you need to learn how to calculate the area of ​​\u200b\u200bdifferent figures, and remember that each of the formulas can be useful later in practice, including the formula for a regular hexagon.
A hexagon is such a polygonal figure, the total number of angles of which is six. A regular hexagon is a hexagonal figure that has equal sides. The angles of a regular hexagon are also equal to each other.
In everyday life, we can often find objects that have the shape of a regular hexagon.

Irregular polygon area calculator by sides

You will need

• - roulette;
• — electronic rangefinder;
• - a sheet of paper and a pencil;
• - calculator.

Instruction 1 If you need the total area of ​​an apartment or a separate room, just read the technical passport for the apartment or house, it shows the footage of each room and the total footage of the apartment. 2 To measure the area of ​​a rectangular or square room, take a tape measure or an electronic rangefinder and measure the length of the walls. When measuring distances with a rangefinder, be sure to keep the beam direction perpendicular, otherwise the measurement results may be distorted. 3 Then multiply the resulting length (in meters) of the room by the width (in meters). The resulting value will be the floor area, it is measured in square meters.

Gauss area formula

If you need to calculate the floor area of ​​a more complex structure, such as a pentagonal room or a room with a round arch, sketch a schematic sketch on a piece of paper. Then divide the complex shape into several simple ones, such as a square and a triangle, or a rectangle and a semicircle. Use a tape measure or rangefinder to measure the size of all sides of the resulting figures (for a circle, you need to know the diameter) and enter the results on your drawing.

5 Now calculate the area of ​​each shape separately. The area of ​​rectangles and squares is calculated by multiplying the sides. To calculate the area of ​​a circle, divide the diameter in half and square (multiply it by itself), then multiply the result by 3.14.
If you only want half of the circle, divide the resulting area in half. To calculate the area of ​​a triangle, find P by dividing the sum of all sides by 2.

Formula for calculating the area of ​​an irregular polygon

If the points are numbered sequentially in a counterclockwise direction, then the determinants in the formula above are positive and the modulus in it can be omitted; if they are numbered in a clockwise direction, the determinants will be negative. This is because the formula can be viewed as a special case of Green's theorem. To apply the formula, you need to know the coordinates of the polygon vertices in the Cartesian plane.

For example, let's take a triangle with coordinates ((2, 1), (4, 5), (7, 8)). Take the first x-coordinate of the first vertex and multiply it by the y-coordinate of the second vertex, and then multiply the x-coordinate of the second vertex by the y-coordinate of the third. We repeat this procedure for all vertices. The result can be determined by the following formula: A tri.

The formula for calculating the area of ​​an irregular quadrilateral

A) _(\text(tri.))=(1 \over 2)|x_(1)y_(2)+x_(2)y_(3)+x_(3)y_(1)-x_(2) y_(1)-x_(3)y_(2)-x_(1)y_(3)|) where xi and yi denote the corresponding coordinate. This formula can be obtained by opening the brackets in the general formula for the case n = 3. Using this formula, you can find that the area of ​​a triangle is equal to half the sum of 10 + 32 + 7 - 4 - 35 - 16, which gives 3. The number of variables in the formula depends on the number of sides of the polygon. For example, the formula for the area of ​​a pentagon will use variables up to x5 and y5: A pent. = 1 2 | x 1 y 2 + x 2 y 3 + x 3 y 4 + x 4 y 5 + x 5 y 1 − x 2 y 1 − x 3 y 2 − x 4 y 3 − x 5 y 4 − x 1 y 5 | (\displaystyle \mathbf (A) _(\text(pent.))=(1 \over 2)|x_(1)y_(2)+x_(2)y_(3)+x_(3)y_(4 )+x_(4)y_(5)+x_(5)y_(1)-x_(2)y_(1)-x_(3)y_(2)-x_(4)y_(3)-x_(5 )y_(4)-x_(1)y_(5)|) A for a quad - variables up to x4 and y4: A quad.

1.1 Calculation of areas in antiquity

1.2 Different approaches to the study of the concepts of "area", "polygon", "area of ​​a polygon"

1.2.1 The concept of area. Area Properties

1.2.2 The concept of a polygon

1.2.3 The concept of the area of ​​a polygon. Descriptive definition

1.3 Various formulas for the areas of polygons

1.4 Derivation of polygon area formulas

1.4.1 Area of ​​a triangle. Heron's formula

1.4.2 Area of ​​a rectangle

1.4.3 Area of ​​a trapezoid

1.4.4 Area of ​​a quadrilateral

1.4.5 Universal formula

1.4.6 Area of ​​an n-gon

1.4.7 Calculating the area of ​​a polygon from the coordinates of its vertices

1.4.8 Pick Formula

1.5 The Pythagorean theorem on the sum of the areas of squares built on the legs of a right triangle

1.6 Equivalence of triangles. Bogliai-Gervin theorem

1.7 Ratio of areas of similar triangles

1.8 Figures with the largest area

1.8.1 Trapezoid or rectangle

1.8.2 A remarkable property of a square

1.8.3 Plots of different shape

1.8.4 Triangle with largest area

Chapter 2. Methodological features of studying the areas of polygons in mathematical classes

2.1 Thematic planning and features of teaching in classes with in-depth study of mathematics

2.2 Lesson methodology

2.3 Results of experimental work

Conclusion

Literature

Introduction

The topic "Area of ​​polygons" is an integral part of the school mathematics course, which is quite natural. Indeed, historically, the very emergence of geometry is associated with the need to compare land plots of one form or another. At the same time, it should be noted that the educational opportunities for the disclosure of this topic in secondary school are far from being fully used.

The main task of teaching mathematics at school is to ensure a strong and conscious mastery of the system of mathematical knowledge and skills necessary for every member of modern society in everyday life and work, sufficient to study related disciplines and continue education.

Along with the solution of the main task, an in-depth study of mathematics provides for the formation of a steady interest in the subject in students, the identification and development of their mathematical abilities, an orientation towards professions that are significantly related to mathematics, and preparation for studying at a university.

Qualification work includes the content of the mathematics course of a general education school and a number of additional questions that are directly adjacent to this course and deepen it along the main ideological lines.

The inclusion of additional questions serves two interrelated purposes. On the one hand, this is the creation, in conjunction with the main sections of the course, of a base to meet the interests and develop the abilities of students with a penchant for mathematics, on the other hand, the fulfillment of meaningful gaps in the main course, giving the content of in-depth study the necessary integrity.

The qualifying work consists of an introduction, two chapters, a conclusion and cited literature. The first chapter discusses the theoretical foundations of the study of the areas of polygons, and the second chapter deals directly with the methodological features of the study of areas.

Chapter 1

1.1Calculation of areas in antiquity

The rudiments of geometric knowledge related to the measurement of areas are lost in the depths of millennia.

Back in 4 - 5 thousand years ago, the Babylonians were able to determine the area of ​​a rectangle and a trapezoid in square units. The square has long served as a standard for measuring areas due to many of its remarkable properties: equal sides, equal and right angles, symmetry and general perfection of form. Squares are easy to build, or you can fill a plane without gaps.

In ancient China, the measure of area was a rectangle. When masons determined the area of ​​a rectangular house wall, they multiplied the height and width of the wall. This is the accepted definition in geometry: the area of ​​a rectangle is equal to the product of its adjacent sides. Both of these sides must be expressed in the same linear units. Their product will be the area of ​​the rectangle, expressed in the corresponding square units. Say, if the height and width of the wall are measured in decimeters, then the product of both measurements will be expressed in square decimeters. And if the area of ​​each facing Plot is a square decimeter, then the resulting product will indicate the number of tiles needed for facing. This follows from the statement underlying the measurement of areas: the area of ​​a figure made up of non-intersecting figures is equal to the sum of their areas.

The ancient Egyptians 4,000 years ago used almost the same techniques as we do to measure the area of ​​a rectangle, triangle, and trapezoid: the base of the triangle was divided in half, and multiplied by the height; for a trapezoid, the sum of the parallel sides was divided in half and multiplied by the height, and so on. To calculate the area

quadrilateral with sides (Fig. 1.1), the formula (1.1) was applied

those. half-sums of opposite sides were multiplied.

This formula is obviously incorrect for any quadrilateral; it follows from it, in particular, that the areas of all rhombuses are the same. Meanwhile, it is obvious that the areas of such rhombuses depend on the magnitude of the angles at the vertices. This formula is only valid for a rectangle. With its help, you can approximately calculate the area of ​​quadrilaterals, in which the angles are close to right.

To determine the area

an isosceles triangle (Fig. 1.2), in which the Egyptians used the approximate formula:
(1.2) Fig. 1.2 The error made in this case is the smaller, the smaller the difference between the side and the height of the triangle, in other words, the closer the top (and) to the base of the height from. That is why the approximate formula (1.2) is applicable only for triangles with a relatively small vertex angle.

But already the ancient Greeks knew how to correctly find the areas of polygons. In his Elements, Euclid does not use the word "area", since by the very word "figure" he understands a part of a plane bounded by one or another closed line. Euclid does not express the result of measuring the area as a number, but compares the areas of different figures with each other.

Like other scientists of antiquity, Euclid deals with the transformation of some figures into others, they are equal in size. The area of ​​a compound figure will not change if its parts are arranged differently, but without crossing. Therefore, for example, it is possible, based on the formulas for the area of ​​a rectangle, to find the formulas for the areas of other figures. So, the triangle is divided into such parts, from which you can then make a rectangle of equal area to it. From this construction it follows that the area of ​​a triangle is equal to half the product of its base and height. Resorting to such a redrawing, they find that the area of ​​the parallelogram is equal to the product of the base and the height, the area of ​​the trapezoid is the product of half the sum of the bases and the height.

When masons have to tile a wall of complex configuration, they can determine the area of ​​the wall by counting the number of tiles that went into tiling. Some tiles, of course, will have to be chipped so that the edges of the cladding coincide with the edge of the wall. The number of all tiles that went into work evaluates the wall area with an excess, the number of unbroken tiles - with a disadvantage. As the size of the cells decreases, the amount of waste decreases, and the area of ​​the wall, determined by the number of tiles, is calculated more and more accurately.

One of the late Greek mathematicians - encyclopedists, whose works were mainly applied in nature, was Heron of Alexandria, who lived in the 1st century. n. e. Being an outstanding engineer, he was also called "Heron the Mechanic". In his work Dioptrics, Heron describes various machines and practical measuring instruments.

One of Heron's books was named by him "Geometrics" and is a kind of collection of formulas and corresponding problems. It contains examples for calculating the areas of squares, rectangles and triangles. Heron writes about finding the area of ​​a triangle along its sides: “Let, for example, one side of a triangle have a length of 13 measured cords, the second 14 and the third 15. To find the area, do the following. Add 13, 14 and 15; you get 42. Half of that is 21. Subtract from this three sides one by one; first subtract 13 - it will remain 8, then 14 - it will remain 7, and finally 15 - it will remain 6. Now multiply them: 21 times 8 will give 168, take this 7 times - you get 1176, and this 6 more times - you get 7056. From here the square root will be 84. That's how many measuring cords will be in the area of ​​\u200b\u200bthe triangle.