Two figures having equal volumes are called equal in size. Equal and equal figures

VIII class: Topic 3. Areas of figures. Pythagorean theorem.

1. The concept of area. Equal-sized figures.

If length is a numerical characteristic of a line, then area is a numerical characteristic of a closed figure. Despite the fact that we are well familiar with the concept of area from everyday life, it is not easy to give a strict definition to this concept. It turns out that the area of ​​a closed figure can be called any non-negative quantity having the following properties of measuring the areas of figures:

Equal figures have equal areas. If a given closed figure is divided into several closed figures, then the area of ​​the figure is equal to the sum of the areas of its constituent figures (the figure in Figure 1 is divided into n figures; in this case, the area of ​​the figure, where Si- square i-th figure).

In principle, it would be possible to come up with a set of quantities that have the formulated properties, and therefore characterize the area of ​​the figure. But the most familiar and convenient value is the one that characterizes the area of ​​a square as the square of its side. Let's call this “agreement” the third property of measuring the areas of figures:

The area of ​​a square is equal to the square of its side (Figure 2).

With this definition, the area of ​​the figures is measured in square units ( cm 2, km 2, ha=100m 2).

Figures having equal areas are called equal in size .

Comment: Equal figures have equal areas, that is, equal figures are equal in size. But equal-sized figures are not always equal (for example, Figure 3 shows a square and an isosceles triangle made up of equal right-angled triangles (by the way, such figures called equally composed ); it is clear that the square and the triangle are equal in size, but not equal, since they do not overlap).

Next, we will derive formulas for calculating the areas of all main types of polygons (including the well-known formula for finding the area of ​​a rectangle), based on the formulated properties of measuring the areas of figures.

2. Area of ​​a rectangle. Area of ​​a parallelogram.

Formula for calculating the area of ​​a rectangle: The area of ​​a rectangle is equal to the product of its two adjacent sides (Figure 4).

Given:

ABCD- rectangle;

AD=a, AB=b.

Prove: SABCD=a× b.

Proof:

1. Extend the side AB for a segment B.P.=a, and the side AD- for a segment D.V.=b. Let's build a parallelogram APRV(Figure 4). Since Ð A=90°, APRV- rectangle. At the same time AP=a+b=AV, Þ APRV– a square with side ( a+b).

2. Let us denote B.C.Ç RV=T, CDÇ PR=Q. Then BCQP– a square with a side a, CDVT– a square with a side b, CQRT- rectangle with sides a And b.

Formula for calculating the area of ​​a parallelogram: The area of ​​a parallelogram is equal to the product of its height and its base (Figure 5).

Comment: The base of a parallelogram is usually called the side to which the height is drawn; It is clear that any side of a parallelogram can serve as a base.

Given:

ABCD– p/g;

B.H.^AD, HÎ AD.

Prove: SABCD=AD× B.H..

Proof:

1. Let's take it to the base AD height CF(Figure 5).

2. B.C.ïê HF, B.H.ïê CF, Þ BCFH- p/g by definition. Ð H=90°, Þ BCFH- rectangle.

3. BCFH– p/g, Þ according to the p/g property B.H.=CF, Þ D BAH=D CDF along the hypotenuse and leg ( AB=CD according to St. p/g, B.H.=CF).

4. SABCD=SABCF+S D CDF=SABCF+S D BAH=SBCFH=B.H.× B.C.=B.H.× AD. #

3. Area of ​​a triangle.

Formula for calculating the area of ​​a triangle: The area of ​​a triangle is equal to half the product of its height and its base (Figure 6).

Comment: In this case, the base of the triangle is the side to which the altitude is drawn. Any of the three sides of a triangle can serve as its base.

Given:

BD^A.C., DÎ A.C..

Prove: .

Proof:

1. Let's complete D ABC to p/y ABKC by passing through the vertex B direct B.K.ïê A.C., and through the top C– straight CKïê AB(Figure 6).

2. D ABC=D KCB on three sides ( B.C.– general, AB=KC And A.C.=K.B. according to St. p/g), Þ https://pandia.ru/text/78/214/images/image014_34.gif" width="107" height="36">).

Corollary 2: If we consider p/u D ABC with height A.H., drawn to the hypotenuse B.C., That . Thus, in p/u D-ke height drawn to the hypotenuse is equal to the ratio of the product of its legs to the hypotenuse . This relation is quite often used when solving problems.

4. Corollaries from the formula for finding the area of ​​a triangle: the ratio of the areas of triangles with equal heights or bases; equal triangles in figures; property of the areas of triangles formed by the diagonals of a convex quadrilateral.

From the formula for calculating the area of ​​a triangle, two consequences follow in an elementary way:

1. Ratio of areas of triangles with equal heights equal to the ratio of their bases (in Figure 8 ).

2. Ratio of areas of triangles with equal bases equal to the ratio of their heights (in Figure 9 ).

Comment: When solving problems, triangles with a common height are very often encountered. In this case, as a rule, their bases lie on the same straight line, and the vertex opposite the bases is common (for example, in Figure 10 S 1:S 2:S 3=a:b:c). You should learn to see the total height of such triangles.

Also, the formula for calculating the area of ​​a triangle yields useful facts that allow you to find equal triangles in figures:

1. The median of an arbitrary triangle divides it into two equal triangles (in Figure 11 at D A.B.M. and D ACM height A.H.– general, and the grounds B.M. And C.M. equal by definition of median; it follows that D A.B.M. and D ACM equal in size).

2. The diagonals of a parallelogram divide it into four equal triangles (in Figure 12 A.O.– median of the triangle ABD by the property of diagonals p/g, Þ due to the previous properties of triangles ABO And ADO equal in size; because B.O.– median of the triangle ABC, triangles ABO And BCO equal in size; because CO– median of the triangle BCD, triangles BCO And DCO equal in size; Thus, S D ADO=S D ABO=S D BCO=S D DCO).

3. The diagonals of a trapezoid divide it into four triangles; two of them, adjacent to the lateral sides, are equal in size (Figure 13).

Given:

ABCD– trapezoid;

B.C.ïê AD; A.C.Ç BD=O.

Prove: S D ABO=S D DCO.

Proof:

1. Let's draw the heights B.F. And CH(Figure 13). Then D ABD and D ACD base AD– general, and heights B.F. And CH equal; Þ S D ABD=S D ACD.

2. S D ABO=S D ABD ‑ S D AOD=S D ACD ‑ S D AOD=S D DCO. #

If you draw the diagonals of a convex quadrilateral (Figure 14), four triangles are formed, the areas of which are related by a very easy-to-remember ratio. The derivation of this relationship relies solely on the formula for calculating the area of ​​a triangle; however, it is found quite rarely in the literature. Being useful in solving problems, the relation that will be formulated and proven below deserves close attention:

Property of the areas of triangles formed by the diagonals of a convex quadrilateral: If the diagonals of a convex quadrilateral ABCD intersect at a point O, then (Figure 14).

ABCD– convex quadrangle;

https://pandia.ru/text/78/214/images/image025_28.gif" width="149" height="20">.

Proof:

1. B.F.– overall height D AOB and D BOC; Þ S D AOB:S D BOC=A.O.:CO.

2. D.H.– overall height D AOD and D C.O.D.; Þ S D AOD:S D C.O.D.=A.O.:CO.

5. Ratio of areas of triangles having equal angles.

Theorem on the ratio of the areas of triangles having equal angles: The areas of triangles having equal angles are related as the products of the sides enclosing these angles (Figure 15).

Given:

D ABC,D A 1B 1C 1;

Ð BAC=Ð B 1A 1C 1.

Prove:

.

Proof:

1. Lay it down on the ray AB segment AB 2=A 1B 1, and on the beam A.C.– segment A.C. 2=A 1C 1 (Figure 15). Then D AB 2C 2=D A 1B 1C 1 on two sides and the angle between them ( AB 2=A 1B 1 and A.C. 2=A 1C 1 by construction, and Р B 2A.C. 2=р B 1A 1C 1 by condition). Means, .

2. Connect the dots C And B 2.

3. CH– overall height D AB 2C and D ABC, Þ https://pandia.ru/text/78/214/images/image033_22.gif" width="81" height="43 src=">.

6. Property of the bisector of a triangle.

Using the theorems on the ratio of the areas of triangles having equal angles, and on the ratio of the areas of triangles with equal heights, we simply prove a fact that is extremely useful in solving problems and is not directly related to the areas of figures:

Triangle bisector property: The bisector of a triangle divides the side to which it is drawn into segments proportional to the sides adjacent to them.

Given:

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Proof:

1..gif" width="72 height=40" height="40">.

3. From points 1 and 2 we get: , Þ https://pandia.ru/text/78/214/images/image041_19.gif" width="61" height="37">. #

Comment: Since the extreme members or middle members can be swapped in the correct proportion, it is more convenient to remember the property of the bisector of a triangle in the following form (Figure 16): .

7. Area of ​​a trapezoid.

Formula for calculating the area of ​​a trapezoid: The area of ​​a trapezoid is equal to the product of its height and half the sum of its bases.

Given:

ABCD– trapezoid;

B.C.ïê AD;

B.H.- height.

https://pandia.ru/text/78/214/images/image044_21.gif" width="127" height="36">.

Proof:

1. Let's draw a diagonal BD and height DF(Figure 17). BHDF– rectangle, Þ B.H. = DF.

Consequence: The ratio of the areas of trapezoids with equal heights is equal to the ratio of their midlines (or the ratio of the sums of the bases).

8. Area of ​​a quadrilateral with mutually perpendicular diagonals.

Formula for calculating the area of ​​a quadrilateral with mutually perpendicular diagonals: The area of ​​a quadrilateral with mutually perpendicular diagonals is equal to half the product of its diagonals.

ABCD– quadrangle;

A.C.^BD.

https://pandia.ru/text/78/214/images/image049_20.gif" width="104" height="36">.

Proof:

1. Let us denote A.C.Ç BD=O. Since A.C.^BD, A.O.– height D ABD, A CO– height D CBD(Figures 18a and 18b for the cases of convex and non-convex quadrilaterals, respectively).

2.
(the signs “+” or “-” correspond to the cases of convex and non-convex quadrilaterals, respectively). #

The Pythagorean theorem plays an extremely important role in solving a wide variety of problems; it allows you to find the unknown side of a right triangle from its two known sides. There are many known proofs of the Pythagorean theorem. Let us present the simplest of them, based on formulas for calculating the areas of a square and a triangle:

Pythagorean theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs.

Given:

D ABC– p/u;

Ð A=90°.

Prove:

B.C. 2=AB 2+A.C. 2.

Proof:

1. Let us denote A.C.=a, AB=b. Let's put it on the ray AB segment B.P.=a, and on the beam A.C.– segment CV=b(Figure 19). Let's draw through the point P direct PRïê AV, and through the point V– straight VRïê AP. Then APRV- p/g by definition. Moreover, since Р A=90°, APRV- rectangle. And because AV=a+b=AP, APRV– a square with a side a+b, And SAPRV=(a+b)2. Next we will divide the side PR dot Q into segments PQ=b And QR=a, and the side RV– dot T into segments RT=b And TV=a.

2. D ABC=D PQB=D RTQ=D VCT on two sides, Þ Ð ACB=Ð PBQ=Ð RQT=Ð VTC, B.C.=QB=T.Q.=C.T., and https://pandia.ru/text/78/214/images/image055_17.gif" width="115" height="36">.

3. Because B.C.=QB=T.Q.=C.T., CBQT- rhombus At the same time QBC=180°-(р ABC+Ð PBQ)=180°-(Р ABC+Ð ACB)=Ð BAC=90°; Þ CBQT- square, and SCBQT=B.C. 2.

4. . So, B.C. 2=AB 2+A.C. 2. #

The inverse Pythagorean theorem is a sign of a right triangle, i.e., it allows you to check whether the triangle is right-angled using three known sides.

Converse Pythagorean theorem: If the square of a side of a triangle is equal to the sum of the squares of its other two sides, then the triangle is right-angled and its longest side is the hypotenuse.

Given:

B.C. 2=AB 2+A.C. 2.

Prove: D ABC– p/u;

Ð A=90°.

Proof:

1. Construct a right angle A 1 and put the segments on its sides A 1B 1=AB And A 1C 1=A.C.(Figure 20). In the received p/u D A 1B 1C 1 by Pythagorean theorem B 1C 12=A 1B 12+A 1C 12=AB 2+A.C. 2; but according to the condition AB 2+A.C. 2=B.C. 2; Þ B 1C 12=B.C. 2, Þ B 1C 1=B.C..

2. D ABC=D A 1B 1C 1 on three sides ( A 1B 1=AB And A 1C 1=A.C. by construction, B 1C 1=B.C. from point 1), Þ Ð A=Ð A 1=90°, Þ D ABC- p/u. #

Right triangles whose side lengths are expressed in natural numbers are called Pythagorean triangles , and the triplets of the corresponding natural numbers are Pythagorean triplets . Pythagorean triplets are useful to remember (the larger of these numbers is equal to the sum of the squares of the other two). Here are some Pythagorean triples:

3, 4, 5;

5, 12, 13;

8, 15, 17;

7, 24, 25;

20, 21, 29;

12, 35, 37;

9, 40, 41.

A right triangle with sides 3, 4, 5 was used in Egypt to construct right angles, and therefore such triangle called Egyptian .

10. Heron's formula.

Heron's formula allows you to find the area of ​​an arbitrary triangle from its three known sides and is indispensable in solving many problems.

Heron's formula: Area of ​​a triangle with sides a, b And c is calculated using the following formula: , where is the semi-perimeter of the triangle.

Given:

B.C.=a; A.C.=b; AB=c.). Then .

4. Substitute the resulting expression for height into the formula for calculating the area of ​​the triangle: . #

In everyday life, we are surrounded by many different objects. Some of them have the same size and the same shape. For example, two identical sheets or two identical bars of soap, two identical coins, etc.

In geometry, figures that have the same size and shape are called equal figures. The figure below shows two figures A1 and A2. To establish the equality of these figures, we need to copy one of them onto tracing paper. And then move the tracing paper and combine a copy of one figure with another figure. If they match, it means that these figures are the same figures. In this case, write A1 = A2 using the usual equal sign.

Determining the equality of two geometric figures

We can imagine that the first figure was superimposed on the second figure, and not a copy of it on tracing paper. Therefore, in the future we will talk about superimposing the figure itself, and not its copy, on another figure. Based on the foregoing, we can formulate a definition equality of two geometric figures.

Two geometric figures are called equal if they can be combined by superimposing one figure on the other. In geometry, for some geometric figures (for example, triangles), special characteristics are formulated, when fulfilled, we can say that the figures are equal.

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Figures are called equal if their shape and size are the same. From this definition it follows, for example, that if a given rectangle and square have equal areas, then they still do not become equal figures, since these are different figures in shape. Or, two circles definitely have the same shape, but if their radii are different, then these are also not equal figures, since their sizes do not match. Equal figures are, for example, two segments of the same length, two circles with the same radius, two rectangles with pairwise equal sides (the short side of one rectangle is equal to the short side of the other, the long side of one rectangle is equal to the long side of the other).

It can be difficult to determine by eye whether figures of the same shape are equal. Therefore, to determine the equality of simple figures, they are measured (using a ruler or compass). Segments have length, circles have radius, rectangles have length and width, squares have only one side. It should be noted here that not all figures can be compared. It is impossible, for example, to determine the equality of lines, since any line is infinite and, therefore, all lines can be said to be equal to each other. The same goes for rays. Although they have a beginning, they have no end.

If we are dealing with complex (arbitrary) figures, then it can even be difficult to determine whether they have the same shape. After all, figures can be inverted in space. Look at the picture below. It is difficult to say whether these figures are the same in shape or not.

Thus, it is necessary to have a reliable principle for comparing figures. It is like this: equal figures coincide when superimposed on each other.

To compare two depicted figures by superposition, apply tracing paper (transparent paper) to one of them and copy (draw) the shape of the figure onto it. They try to put a copy on tracing paper on the second figure so that the figures coincide. If this succeeds, then the given figures are equal. If not, then the figures are not equal. When applying, the tracing paper can be rotated as desired, as well as turned over.

If you can cut out the shapes themselves (or they are separate flat objects and not drawn), then tracing paper is not needed.

When studying geometric figures, you can notice many of their features related to the equality of their parts. So, if you fold a circle along the diameter, then its two halves will turn out to be equal (they will coincide by overlap). If you cut a rectangle diagonally, you get two right triangles. If one of them is rotated 180 degrees clockwise or counterclockwise, it will coincide with the second. That is, the diagonal divides the rectangle into two equal parts.

What figures are called equal?

Figures are called equal, which coincide when superimposed.

A common mistake when answering this question is to answer by mentioning equal sides and angles of a geometric figure. However, this does not take into account that the sides of a geometric figure are not necessarily straight. Therefore, only the coincidence of geometric figures when superimposed can be a sign of their equality.

In practice, this is easy to check using an overlay; they should match.

Everything is very simple and accessible, usually equal figures are immediately visible.

Equal figures are those whose geometry parameters coincide. These parameters are: length of sides, size of angles, thickness.

The easiest way to understand that the figures are equal is to use overlay. If the sizes of the figures are the same, they are called equal.

Equal Only those geometric figures that have exactly the same parameters are named:

1) perimeter;

2) area;

4) dimensions.

That is, if one figure is superimposed on another, they will coincide.

It is a mistake to assume that if figures have the same perimeter or area, then they are equal. In fact, geometric figures that have equal area are called equal in area.

Figures are called equal if they coincide when superimposed on each other. Equal figures have the same size, shape, area and perimeter. But figures that are equal in area may not be equal to each other.

In geometry, according to the rules, equal figures must have the same area and perimeter, that is, they must have absolutely the same shapes and sizes. And they must completely match when superimposed on each other. If there are any discrepancies, then these figures can no longer be called equal.

Figures can be called equal provided that they completely coincide when superimposed on each other, i.e. they have the same size, shape and therefore area and perimeter, as well as other characteristics. Otherwise, we cannot talk about the equality of the figures.

The very word equal contains the essence.

These are figures that are completely identical to each other. That is, they completely coincide. If a figure is placed one on top of another, then the figures will overlap themselves on all sides.

They are the same, that is, equal.

Unlike equal triangles (to determine which it is enough to fulfill one of the conditions - signs of equality), equal figures are those that have the same not only shape, but also dimensions.

You can determine whether one figure is equal to another using the superposition method. In this case, the figures must match both sides and corners. These will be equal figures.

Only such figures can be equal if, when superimposed, their sides and angles completely coincide. In fact, for all the simplest polygons, the equality of their areas also indicates the equality of the figures themselves. Example: a square with side a will always be equal to another square with the same side a. The same applies to rectangles and rhombuses - if their sides are equal to the sides of another rectangle, they are equal. A more complex example: triangles will be congruent if they have equal sides and corresponding angles. But these are only special cases. In more general cases, the equality of figures is still proved by superposition, and this superposition in planimetry is pompously called motion.

One of the basic concepts in geometry is figure. This term refers to a set of points on a plane limited by a finite number of lines. Some figures can be considered equal, which is closely related to the concept of movement. Geometric figures can be considered not in isolation, but in one way or another in relation to each other - their mutual arrangement, contact and adjacency, the position “between”, “inside”, the relationship expressed in the concepts of “more”, “less”, “equal” .Geometry studies the invariant properties of figures, i.e. those that remain unchanged under certain geometric transformations. Such a transformation of space, in which the distance between the points that make up a particular figure remains unchanged, is called movement. Movement can appear in different versions: parallel translation, identical transformation, rotation around an axis, symmetry relative to a straight line or plane, central, rotational, portable symmetry .

Movement and equal figures

Equal and equal figures