Different ways of proving the Pythagorean theorem

student of grade 9 "A"

MOU SOSH №8

Scientific adviser:

mathematic teacher,

MOU SOSH №8

Art. Novorozhdestvenskaya

Krasnodar Territory.

Art. Novorozhdestvenskaya

ANNOTATION.

The Pythagorean theorem is rightfully considered the most important in the course of geometry and deserves close attention. It is the basis for solving many geometric problems, the basis for studying the theoretical and practical course of geometry in the future. The theorem is surrounded by the richest historical material related to its appearance and methods of proof. Studying the history of the development of geometry instills a love for this subject, contributes to the development of cognitive interest, general culture and creativity, and also develops research skills.

As a result of the search activity, the goal of the work was achieved, which is to replenish and generalize knowledge on the proof of the Pythagorean theorem. I managed to find and consider different ways proof and deepen knowledge on the topic, going beyond the pages of a school textbook.

The collected material even more convinces that the Pythagorean theorem is the great theorem of geometry, has great theoretical and practical significance.

Introduction. History reference 5 Main part 8

3. Conclusion 19

4. Used literature 20
1. INTRODUCTION. HISTORY REFERENCE.

The essence of truth is that it is for us forever,

When we see the light at least once in her insight,

And the Pythagorean theorem after so many years

For us, as for him, it is indisputable, flawless.

To delight the gods, Pythagoras made a vow:

For having touched the endless wisdom,

He killed a hundred bulls, thanks to the eternal;

He offered prayers and praises to the victim after him.

Since then, the bulls, when they smell, pushing,

That the trail again leads people to the new truth,

They roar furiously, so there is no urine to listen,

Such a Pythagoras instilled terror in them forever.

Bulls, powerless to resist the new truth,

What remains? - Just closing your eyes, roar, tremble.

It is not known how Pythagoras proved his theorem. What is certain is that he discovered it under the strong influence of Egyptian science. A special case Pythagoras' theorems - the properties of a triangle with sides 3, 4 and 5 - were known to the builders of the pyramids long before the birth of Pythagoras, but he himself studied for more than 20 years with the Egyptian priests. A legend has survived, which says that, having proved his famous theorem, Pythagoras sacrificed a bull to the gods, and according to other sources, even 100 bulls. This, however, contradicts information about the moral and religious views of Pythagoras. In literary sources, you can read that he "forbade even to kill animals, and even more so to feed them, because animals have a soul, like us." Pythagoras ate only honey, bread, vegetables and occasionally fish. In connection with all this, the following entry can be considered more plausible: "... and even when he discovered that in a right-angled triangle the hypotenuse has a correspondence with the legs, he sacrificed a bull made from wheat dough."

The popularity of the Pythagorean theorem is so great that its proofs are found even in fiction, for example, in the story of the famous English writer Huxley "Young Archimedes". The same Proof, but for the particular case of an isosceles right-angled triangle, is given in Plato's Menon dialogue.

Fairy tale "House".

“Far, far away, where even airplanes do not fly, is the country of Geometry. In this unusual country there was one amazing city - the city of Theorem. Once I came to this city beautiful girl named Hypotenuse. She tried to rent a room, but wherever she turned, she was refused everywhere. Finally she went to the rickety house and knocked. She was opened by a man who called himself the Right Angle, and he invited Hypotenuse to live with him. The hypotenuse remained in the house where Right Angle and his two young sons named Cathety lived. Since then, life in the House of Right Angle has changed in a new way. The hypotenuse planted flowers on the window, and red roses in the front garden. The house took the shape of a right-angled triangle. Both legs really liked Hypotenuse and asked her to stay forever in their house. Lo in the evenings, this friendly family gathers at the family table. Sometimes Right Angle plays hide and seek with his kids. Most often he has to look, and Hypotenuse hides so skillfully that it can be very difficult to find her. One day while playing, Right Angle noticed interesting property: if he manages to find the legs, then it is not difficult to find the Hypotenuse. So Right Angle uses this pattern, I must say, very successfully. The Pythagorean theorem is based on the property of this right-angled triangle. "

(From the book by A. Okunev "Thank you for the lesson, children").

A playful formulation of the theorem:

If we are given a triangle

And, moreover, with a right angle,

Then the square of the hypotenuse

We will always easily find:

We erect the legs in a square,

We find the sum of the degrees -

And in such a simple way

We will come to the result.

Studying algebra and the beginnings of analysis and geometry in grade 10, I became convinced that in addition to the method of proving the Pythagorean theorem considered in grade 8, there are other ways of proving. I present them for your review.
2. MAIN PART.

Theorem. In a right triangle, a square

hypotenuse is equal to the sum squares of legs.

1 METHOD.

Using the properties of the areas of polygons, we will establish a remarkable relationship between the hypotenuse and the legs of a right-angled triangle.

Proof.

a, in and hypotenuse With(Fig. 1, a).

Let us prove that c² = a² + b².

Proof.

Let's build a triangle to a square with a side a + b as shown in fig. 1, b. The area S of this square is equal to (a + b) ². On the other hand, this square is made up of four equal right-angled triangles, each of which is ½ aw, and a square with side With, therefore S = 4 * ½ av + s² = 2av + s².

In this way,

(a + b) ² = 2 av + s²,

c² = a² + b².

The theorem is proved.
2 METHOD.

After studying the topic "Similar triangles" I found out that it is possible to apply the similarity of triangles to the proof of the Pythagorean theorem. Namely, I used the statement that the leg of a right-angled triangle is the proportional average for the hypotenuse and the segment of the hypotenuse enclosed between the leg and the height drawn from the top of the right angle.

Consider a right-angled triangle with a right angle С, СD– height (Fig. 2). Let us prove that AS² + CB² = AB² .

Proof.

Based on the statement about the leg of a right-angled triangle:

AC =, SV =.

Let's square and add the resulting equalities:

AC² = AB * AD, CB² = AB * DB;

AC² + CB² = AB * (AD + DB), where AD + DB = AB, then

AC² + SV² = AB * AB,

AC² + CB² = AB².

The proof is complete.
3 METHOD.

The definition of the cosine can be applied to the proof of the Pythagorean theorem acute angle right triangle. Consider fig. 3.

Proof:

Let ABC be a given right-angled triangle with right angle C. Draw the height CD from the vertex of right angle C.

By definition of the cosine of an angle:

cos A = AD / AC = AC / AB. Hence AB * AD = AC²

Likewise,

cos B = BD / BC = BC / AB.

Hence AB * BD = BC².

Adding the obtained equalities term by term and noting that AD + DB = AB, we get:

AS² + sun² = AB (AD + DB) = AB²

The proof is complete.
4 METHOD.

Having studied the topic "Relations between the sides and angles of a right-angled triangle", I think that the Pythagorean theorem can be proved in another way.

Consider a right-angled triangle with legs a, in and hypotenuse With... (fig. 4).

Let us prove that c² = a² + b².

Proof.

sin B = a / c ; cos B = a / s , then, squaring the obtained equalities, we get:

sin² B =в² / с²; cos² V= a² / c².

Adding them together, we get:

sin² V+ cos² B = b² / s² + a² / c², where sin² V+ cos² B = 1,

1 = (b² + a²) / c², therefore

c² = a² + b².

The proof is complete.

5 METHOD.

This proof is based on cutting the squares built on the legs (Fig. 5), and stacking the resulting parts on the square built on the hypotenuse.

6 METHOD.

For proof on the leg Sun build BCD ABC(fig. 6). We know that the areas of such figures are related as squares of their similar linear dimensions:

Subtracting the second equality from the first, we get

c2 = a2 + b2.

The proof is complete.

7 METHOD.

Given(fig. 7):

ABC,= 90 ° , Sun= a, AC =b, AB = c.

Prove:c2 = a2 +b2.

Proof.

Let the leg b a. Let's continue the segment SV per point V and build a triangle BMD so that the points M and A lay on one side of a straight line CD and besides, BD =b, BDM= 90 °, DM= a, then BMD= ABC on both sides and the corner between them. Points A and M connect by segments AM. We have MD CD and AC CD, means straight AS parallel to the straight line MD. Because MD< АС, then straight CD and AM not parallel. Consequently, AMDC - rectangular trapezoid.

In right triangles ABC and BMD 1 + 2 = 90 ° and 3 + 4 = 90 °, but since = =, then 3 + 2 = 90 °; then AVM= 180 ° - 90 ° = 90 °. It turned out that the trapezoid AMDC is divided into three non-overlapping right-angled triangles, then according to the axioms of areas

(a + b) (a + b)

Dividing all terms of the inequality by, we get

ab + c2 + ab = (a +b) , 2 ab+ c2 = a2+ 2ab+ b2,

c2 = a2 + b2.

The proof is complete.

8 METHOD.

This method is based on the hypotenuse and legs of a right-angled triangle. ABC. He constructs the corresponding squares and proves that the square built on the hypotenuse is equal to the sum of the squares built on the legs (Fig. 8).

Proof.

1) DBC= FBA= 90 °;

DBC + ABC= FBA + ABC, means, FBC = DBA.

In this way, FBC=ABD(on both sides and the corner between them).

2) , where AL DE, since BD is a common base, DL - overall height.

3) , since FB is a foundation, AB- overall height.

4)

5) Similarly, one can prove that

6) Adding term by term, we get:

, BC2 = AB2 + AC2 . The proof is complete.

9 METHOD.

Proof.

1) Let ABDE- a square (Fig. 9), the side of which is equal to the hypotenuse of a right-angled triangle ABC (AB= s, BC = a, AC =b).

2) Let DK BC and DK = BC, since 1 + 2 = 90 ° (like the acute corners of a right triangle), 3 + 2 = 90 ° (like the corner of a square), AB= BD(sides of the square).

Means, ABC= BDK(by hypotenuse and acute angle).

3) Let EL DK, AM EL. You can easily prove that ABC = BDK = DEL = EAM (with legs a and b). Then KS= CM= ML= LK= a -b.

4) SKB = 4S + SKLMC= 2ab+ (a - b),With2 = 2ab + a2 - 2ab + b2,c2 = a2 + b2.

The proof is complete.

10 METHOD.

The proof can be drawn on a figure jokingly called "Pythagorean pants" (Fig. 10). Its idea is to transform the squares built on the legs into equal triangles that together make up the square of the hypotenuse.

ABC we move, as shown by the arrow, and it takes the position KDN. The rest of the figure AKDCB equal area of ​​a square AKDC - this is a parallelogram AKNB.

Parallelogram model made AKNB... We shift the parallelogram as sketched in the content of the work. To show the transformation of a parallelogram into an equal-area triangle, in front of the students' eyes, cut off a triangle on the model and shift it down. Thus, the area of ​​the square AKDC turned out equal to the area of ​​the rectangle. Similarly, convert the area of ​​the square to the area of ​​the rectangle.

Let's make a transformation for a square built on a leg a(Fig. 11, a):

a) the square is transformed into an equal-area parallelogram (Fig. 11.6):

b) the parallelogram is rotated by a quarter of a turn (Fig. 12):

c) the parallelogram is transformed into an equal-sized rectangle (Fig. 13): 11 METHOD.

Proof:

PCL - straight line (Fig. 14);

KLOA= ACPF= ACED= a2;

LGBO= SVMR =CBNQ= b 2;

AKGB= AKLO +LGBO= c2;

c2 = a2 + b2.

The proof is over .

12 METHOD.

Rice. 15 illustrates another original proof of the Pythagorean theorem.

Here: triangle ABC with right angle C; section Bf perpendicular SV and is equal to it, the segment BE perpendicular AB and is equal to it, the segment AD perpendicular AS and equal to him; points F, C,D belong to one straight line; quadrangles ADFB and ACBE are equal, since ABF = ECB; triangles ADF and ACE equal areas; subtract from both equal-sized quadrangles the common triangle for them ABC, get

, c2 = a2 + b2.

The proof is complete.

13 METHOD.

The area of ​​this right-angled triangle, on the one hand, is equal to , with another, ,

3. CONCLUSION.

As a result of the search activity, the goal of the work was achieved, which is to replenish and generalize knowledge on the proof of the Pythagorean theorem. I managed to find and consider various ways to prove it and deepen knowledge on the topic, going beyond the pages of a school textbook.

The material I have collected even more convinces that the Pythagorean theorem is the great theorem of geometry, has enormous theoretical and practical significance. In conclusion, I would like to say: the reason for the popularity of the Pythagorean triplet theorem is beauty, simplicity and significance!

4. USED LITERATURE.

1. Entertaining algebra. ... Moscow "Science", 1978.

2. Weekly educational and methodological supplement to the newspaper "September 1", 24/2001.

3. Geometry 7-9. and etc.

4. Geometry 7-9. and etc.

Pythagorean theorem: The sum of the areas of the squares resting on the legs ( a and b) is equal to the area of ​​the square built on the hypotenuse ( c).

Geometric formulation:

Initially, the theorem was formulated as follows:

Algebraic formulation:

That is, denoting the length of the hypotenuse of a triangle by c, and the lengths of the legs through a and b :

a 2 + b 2 = c 2

Both statements of the theorem are equivalent, but the second statement is more elementary, it does not require the concept of area. That is, the second statement can be checked without knowing anything about the area and measuring only the lengths of the sides of a right-angled triangle.

The reverse Pythagorean theorem:

Proof

On the this moment in the scientific literature, 367 proofs of this theorem have been recorded. Probably the Pythagorean theorem is the only theorem with such an impressive number of proofs. This variety can be explained only by the fundamental meaning of the theorem for geometry.

Of course, conceptually all of them can be divided into a small number of classes. The most famous of them are: proofs by the area method, axiomatic and exotic proofs (for example, using differential equations).

Through similar triangles

The following proof of the algebraic formulation is the simplest of the proofs built directly from the axioms. In particular, it does not use the concept of the area of ​​a figure.

Let ABC there is a right-angled triangle with a right angle C... Let's draw the height from C and denote its base by H... Triangle ACH like a triangle ABC in two corners. Similarly, triangle CBH is similar ABC... Introducing the notation

we get

What is the equivalent

Adding, we get

Areas proof

The proofs below, despite their seeming simplicity, are not at all so simple. All of them use the properties of area, the proof of which is more difficult than the proof of the Pythagorean theorem itself.

Equal complementarity proof

1. Place four equal right-angled triangles as shown in Figure 1.
2. Quadrilateral with sides c is a square, since the sum of two acute angles is 90 °, and the unfolded angle is 180 °.
3. The area of ​​the entire figure is, on the one hand, the area of ​​a square with sides (a + b), and on the other hand, the sum of the areas of four triangles and two inner squares.

Q.E.D.

Evidence through scattering

Elegant proof by permutation

An example of one of such proofs is shown in the drawing on the right, where a square built on the hypotenuse is transformed by permutation into two squares built on the legs.

Euclid's proof

Drawing for Euclid's proof

Illustration for Euclid's proof

The idea behind Euclid's proof is as follows: let's try to prove that half of the area of ​​the square built on the hypotenuse is equal to the sum of the halves of the areas of the squares built on the legs, and then the areas of the large and two small squares are equal.

Consider the drawing on the left. On it, we built squares on the sides of a right-angled triangle and drawn a ray s from the vertex of the right angle C perpendicular to the hypotenuse AB, it cuts the square ABIK, built on the hypotenuse, into two rectangles - BHJI and HAKJ, respectively. It turns out that the areas of these rectangles are exactly equal to the areas of the squares built on the corresponding legs.

Let's try to prove that the area of ​​the square DECA is equal to the area of ​​the rectangle AHJK For this we use an auxiliary observation: The area of ​​a triangle with the same height and base as this rectangle is equal to half the area of ​​the given rectangle. This is a consequence of the definition of the area of ​​a triangle as half of the product of the base and the height. From this observation it follows that the area of ​​the triangle ACK is equal to the area of ​​the triangle AHK (not shown in the figure), which, in turn, is equal to half the area of ​​the rectangle AHJK.

Let us now prove that the area of ​​the triangle ACK is also equal to half the area of ​​the square DECA. The only thing that needs to be done for this is to prove the equality of the triangles ACK and BDA (since the area of ​​the triangle BDA is equal to half the area of ​​the square according to the above property). Equality is obvious, the triangles are equal on two sides and the angle between them. Namely - AB = AK, AD = AC - the equality of the angles CAK and BAD is easy to prove by the method of motion: we rotate the triangle CAK by 90 ° counterclockwise, then it is obvious that the corresponding sides of the two triangles under consideration will coincide (since the angle at the apex of the square is 90 °).

The reasoning about the equality of the areas of the square BCFG and the rectangle BHJI is completely analogous.

Thus, we have proved that the area of ​​the square built on the hypotenuse is the sum of the areas of the squares built on the legs. The idea behind this proof is further illustrated with the animation above.

Proof of Leonardo da Vinci

Proof of Leonardo da Vinci

The main elements of the proof are symmetry and motion.

Consider the drawing, as seen from the symmetry, the segment CI cuts the square ABHJ into two identical parts (since the triangles ABC and JHI are equal by construction). Rotating 90 degrees counterclockwise, we see that the shaded shapes are equal CAJI and GDAB ... Now it is clear that the area of ​​the shaded figure is equal to the sum of the halves of the areas of the squares built on the legs and the area of ​​the original triangle. On the other hand, it is equal to half the area of ​​the square built on the hypotenuse plus the area of ​​the original triangle. The final step in the proof is left to the reader.

Proof by the method of infinitesimal

The following proof using differential equations is often attributed to the famous English mathematician Hardy, who lived in the first half of the 20th century.

Looking at the drawing shown in the figure and observing the change of the side a, we can write the following relation for infinitely small increments of the sides With and a(using the similarity of triangles):

Proof by the method of infinitesimal

Using the method of separating variables, we find

A more general expression for changing the hypotenuse in the case of increments of both legs

Integrating this equation and using initial conditions, we get

c 2 = a 2 + b 2 + constant.

Thus, we arrive at the desired answer

c 2 = a 2 + b 2 .

As it is easy to see, the quadratic dependence in the final formula appears due to the linear proportionality between the sides of the triangle and the increments, while the sum is associated with independent contributions from the increments of different legs.

A simpler proof can be obtained if we assume that one of the legs does not experience an increment (in this case, the leg b). Then for the constant of integration we obtain

Variations and generalizations

• If instead of squares we construct other similar figures on the legs, then the following generalization of the Pythagorean theorem is true: In a right-angled triangle, the sum of the areas of similar figures built on the legs is equal to the area of ​​the figure built on the hypotenuse. In particular:
  • The sum of the areas of regular triangles built on the legs is equal to the area of ​​a regular triangle built on the hypotenuse.
  • The sum of the areas of the semicircles built on the legs (as in the diameter) is equal to the area of ​​the semicircle built on the hypotenuse. This example is used to prove the properties of figures bounded by arcs of two circles and bearing the name of hippocratic lunes.

Story

Chu-pei 500-200 BC. Left inscription: the sum of the squares of the lengths of the height and base is the square of the length of the hypotenuse.

The ancient Chinese book Chu-pei speaks of Pythagorean triangle with sides 3, 4 and 5: In the same book, a drawing is proposed that coincides with one of the drawings of the Hindu geometry of Baskhara.

Cantor (the largest German historian of mathematics) believes that the equality 3 ² + 4 ² = 5 ² was already known to the Egyptians around 2300 BC. e., during the time of King Amenemhat I (according to papyrus 6619 of the Berlin Museum). According to Cantor, the harpedonapts, or "rope pulls", built right angles using right-angled triangles with sides 3, 4, and 5.

It is very easy to reproduce their way of building. Take a rope 12 m long and tie it to it along a colored strip at a distance of 3 m. from one end and 4 meters from the other. The right angle will be enclosed between the sides 3 and 4 meters long. The Harpedonapts might argue that their way of building becomes superfluous, if you use, for example, the wooden square used by all carpenters. Indeed, there are known Egyptian drawings in which such a tool is found, for example, drawings depicting a carpentry workshop.

Somewhat more is known about the Babylonian Pythagorean theorem. In one text dating back to the time of Hammurabi, that is, to 2000 BC. BC, an approximate calculation of the hypotenuse of a right-angled triangle is given. From this we can conclude that in Mesopotamia they knew how to perform calculations with right-angled triangles, at least in some cases. Based, on the one hand, on the current level of knowledge about Egyptian and Babylonian mathematics, and on the other, on a critical study of Greek sources, Van der Waerden (Dutch mathematician) made the following conclusion:

Literature

In Russian

• Skopets Z.A. Geometric miniatures. M., 1990
• Yelensky Sch. In the footsteps of Pythagoras. M., 1961
• Van der Waerden B.L. Awakening science. Mathematics Ancient egypt, Babylon and Greece. M., 1959
• Glazer G.I. History of mathematics at school. M., 1982
• V. Litzman, "The Pythagorean Theorem" M., 1960.
  • A site about the Pythagorean theorem with a large number of proofs, the material is taken from the book of V. Litzman, a large number of drawings are presented in the form of separate graphic files.
• The Pythagorean theorem and Pythagorean triplets a chapter from the book by DV Anosov "A Look at Mathematics and Something From It"
• On the Pythagorean theorem and methods of its proof G. Glazer, Academician of the Russian Academy of Education, Moscow

In English

• The Pythagorean Theorem at WolframMathWorld
• Cut-The-Knot, a section on the Pythagorean theorem, about 70 proofs and a wealth of additional information

Wikimedia Foundation. 2010.

Average level

Right triangle... The Complete Illustrated Guide (2019)

RIGHT TRIANGLE. FIRST LEVEL.

In tasks, a right angle is not at all necessary - the lower left, so you need to learn how to recognize a right-angled triangle in this form,

and in such,

and in such

What good is there in a right triangle? Well ... first, there are special beautiful names for his parties.

Attention to the drawing!

Remember and don't confuse: legs - two, and the hypotenuse - only one(the one and only and the longest)!

Well, the names have been discussed, now the most important thing: the Pythagorean theorem.

Pythagorean theorem.

This theorem is the key to solving many problems involving a right-angled triangle. It was proved by Pythagoras in completely immemorial times, and since then it has brought many benefits to those who know it. And the best thing about her is that she is simple.

So, Pythagorean theorem:

Do you remember the joke: "Pythagorean pants are equal on all sides!"?

Let's draw these same Pythagorean pants and look at them.

Doesn't it look like some kind of shorts? Well, on which sides and where are they equal? Why and where did the joke come from? And this joke is connected precisely with the Pythagorean theorem, more precisely, with the way Pythagoras himself formulated his theorem. And he formulated it as follows:

"Sum squares built on legs is equal to square area built on the hypotenuse ”.

Doesn't it sound a little different? And so, when Pythagoras drew the statement of his theorem, just such a picture turned out.

In this picture, the sum of the areas of the small squares is equal to the area of ​​the large square. And so that the children better remember that the sum of the squares of the legs is equal to the square of the hypotenuse, someone witty and invented this joke about Pythagorean pants.

Why are we now formulating the Pythagorean theorem

Did Pythagoras suffer and talk about squares?

You see, in ancient times there was no ... algebra! There were no designations and so on. There were no inscriptions. Can you imagine how awful it was for the poor ancient disciples to memorize everything with words ??! And we can be glad that we have a simple formulation of the Pythagorean theorem. Let's repeat it again to remember it better:

It should be easy now:

The square of the hypotenuse is equal to the sum of the squares of the legs.

Well, the most important theorem about a right-angled triangle has been discussed. If you are interested in how it is proved, read the next levels of the theory, and now let's go further ... in dark forest... trigonometry! To the terrible words sine, cosine, tangent and cotangent.

Sine, cosine, tangent, cotangent in a right triangle.

In fact, it's not that scary at all. Of course, the "real" definitions of sine, cosine, tangent and cotangent should be found in the article. But I really don't want to, right? We can rejoice: to solve problems about a right-angled triangle, you can simply fill in the following simple things:

Why is it all about the corner? Where is the corner? In order to understand this, you need to know how statements 1 - 4 are written in words. Look, understand and remember!

1.
In fact, it sounds like this:

And what about the corner? Is there a leg that is opposite the corner, that is, the opposite (for the corner) leg? Of course have! This is a leg!

But what about the angle? Look closely. Which leg is adjacent to the corner? Of course, the leg. Hence, for the angle, the leg is adjacent, and

Now, attention! Look what we got:

You see how great:

Now let's move on to tangent and cotangent.

How can I write it down in words now? What is the leg in relation to the corner? Opposite, of course - it "lies" opposite the corner. And the leg? Adjacent to the corner. So what did we do?

See the numerator and denominator are reversed?

And now again the corners and made the exchange:

Summary

Let's briefly write down everything we have learned.

Pythagorean theorem:

The main theorem about a right-angled triangle is the Pythagorean theorem.

Pythagorean theorem

By the way, do you remember well what legs and hypotenuse are? If not, then look at the picture - refresh your knowledge

It is possible that you have already used the Pythagorean theorem many times, but have you ever wondered why such a theorem is true? How can I prove it? Let's do like the ancient Greeks. Let's draw a square with a side.

You see how cleverly we divided its sides into lengths and!

Now let's connect the marked points

Here we, however, have noted something else, but you yourself look at the drawing and think about why this is so.

What is the area of ​​the larger square? Right, . A smaller area? Certainly, . The total area of ​​the four corners remains. Imagine that we took them two at a time and leaned them against each other with hypotenuses. What happened? Two rectangles. This means that the area of ​​the "scraps" is equal to.

Let's put it all together now.

Let's transform:

So we visited Pythagoras - we proved his theorem in an ancient way.

Right triangle and trigonometry

For a right-angled triangle, the following relationships hold:

The sine of an acute angle is equal to the ratio of the opposite leg to the hypotenuse

The cosine of an acute angle is equal to the ratio of the adjacent leg to the hypotenuse.

The tangent of an acute angle is equal to the ratio of the opposite leg to the adjacent leg.

The cotangent of an acute angle is equal to the ratio of the adjacent leg to the opposite leg.

And once again, all this is in the form of a plate:

It is very comfortable!

Equality tests for right-angled triangles

I. On two legs

II. On the leg and hypotenuse

III. By hypotenuse and acute angle

IV. On a leg and a sharp corner

a)

b)

Attention! It is very important here that the legs are "appropriate". For example, if it is like this:

THEN TRIANGLES ARE NOT EQUAL, despite the fact that they have one of the same acute angle.

Need to in both triangles, the leg was adjacent, or in both triangles, opposite.

Have you noticed how the signs of equality of right triangles differ from the usual signs of equality of triangles? Take a look at the topic “and pay attention to the fact that for equality of“ ordinary ”triangles you need the equality of their three elements: two sides and an angle between them, two angles and a side between them or three sides. But for the equality of right-angled triangles, only two corresponding elements are enough. Great, isn't it?

The situation is approximately the same with the signs of the similarity of right-angled triangles.

Signs of the similarity of right-angled triangles

I. On a sharp corner

II. On two legs

III. On the leg and hypotenuse

Median in a right triangle

Why is this so?

Consider a whole rectangle instead of a right-angled triangle.

Let's draw a diagonal and consider a point - the point of intersection of the diagonals. What is known about the diagonals of a rectangle?

And what follows from this?

So it turned out that

1. - median:

Remember this fact! Helps a lot!

What's even more surprising is that the converse is also true.

What good can you get from the fact that the median drawn to the hypotenuse is equal to half of the hypotenuse? Let's look at the picture

Look closely. We have:, that is, the distances from the point to all three vertices of the triangle turned out to be equal. But in a triangle there is only one point, the distances from which about all three vertices of the triangle are equal, and this is the CENTER of the DESCRIBED CIRCLE. So what happened?

Let's start with this "besides ..."

Let's look at and.

But in such triangles all angles are equal!

The same can be said about and

Now let's draw it together:

What benefit can be derived from this "triple" similarity.

Well, for example - two formulas for the height of a right triangle.

Let's write down the relationship of the respective parties:

To find the height, we solve the proportion and get the first formula "Height in a right triangle":

So, let's apply the similarity:.

What happens now?

Again we solve the proportion and get the second formula:

Both of these formulas must be very well remembered and whichever is more convenient to apply. Let's write them down again

Pythagorean theorem:

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

Signs of equality of right-angled triangles:

• on two legs:
• on the leg and hypotenuse: or
• along the leg and adjacent acute angle: or
• along the leg and the opposite acute angle: or
• by hypotenuse and acute angle: or.

Signs of the similarity of right-angled triangles:

• one sharp corner: or
• from the proportionality of the two legs:
• from the proportionality of the leg and the hypotenuse: or.

Sine, cosine, tangent, cotangent in a right triangle

• The sine of an acute angle of a right triangle is the ratio of the opposite leg to the hypotenuse:
• The cosine of an acute angle of a right triangle is the ratio of the adjacent leg to the hypotenuse:
• The tangent of an acute angle of a right-angled triangle is the ratio of the opposite leg to the adjacent one:
• The cotangent of an acute angle of a right-angled triangle is the ratio of the adjacent leg to the opposite one:.

Height of a right triangle: or.

In a right-angled triangle, the median drawn from the vertex of the right angle is half the hypotenuse:.

Area of ​​a right triangle:

• through the legs:

Pythagorean theorem Is one of the fundamental theorems of Euclidean geometry, establishing the relation

between the sides of a right-angled triangle.

It is believed to have been proven by the Greek mathematician Pythagoras, after whom it was named.

Geometric formulation of the Pythagorean theorem.

Initially, the theorem was formulated as follows:

In a right-angled triangle, the area of ​​the square built on the hypotenuse is equal to the sum of the areas of the squares,

built on legs.

Algebraic formulation of the Pythagorean theorem.

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

That is, denoting the length of the hypotenuse of a triangle by c, and the lengths of the legs through a and b:

Both formulations Pythagorean theorems are equivalent, but the second formulation is more elementary, it is not

requires the concept of area. That is, the second statement can be checked without knowing anything about the area and

by measuring only the lengths of the sides of a right-angled triangle.

The converse theorem of Pythagoras.

If the square of one side of the triangle is equal to the sum of the squares of the other two sides, then

rectangular triangle.

Or, in other words:

For every three positive numbers a, b and c such that

there is a right-angled triangle with legs a and b and hypotenuse c.

Pythagoras' theorem for an isosceles triangle.

Pythagoras' theorem for an equilateral triangle.

Proofs of the Pythagorean theorem.

At the moment, 367 proofs of this theorem have been recorded in the scientific literature. Probably the theorem

Pythagoras is the only theorem with such an impressive number of proofs. Such diversity

can only be explained by the fundamental meaning of the theorem for geometry.

Of course, conceptually all of them can be divided into a small number of classes. The most famous of them:

proof area method, axiomatic and exotic evidence(For example,

via differential equations).

1. Proof of the Pythagorean theorem through similar triangles.

The following proof of the algebraic formulation is the simplest of the proofs under construction

directly from the axioms. In particular, it does not use the concept of the area of ​​a figure.

Let ABC there is a right-angled triangle with a right angle C... Let's draw the height from C and denote

its foundation through H.

Triangle ACH like a triangle AB C in two corners. Similarly, triangle CBH is similar ABC.

Introducing the notation:

we get:

,

which corresponds to -

By adding a 2 and b 2, we get:

or, as required to prove.

2. Proof of the Pythagorean theorem by the area method.

The proofs below, despite their seeming simplicity, are not at all so simple. All of them

use the properties of the area, the proof of which is more difficult than the proof of the Pythagorean theorem itself.

• Proof through equal complementarity.

Place four equal rectangular

triangle as shown in the figure

on right.

Quadrilateral with sides c- square,

since the sum of two acute angles is 90 °, and

expanded angle - 180 °.

The area of ​​the entire figure is, on the one hand,

area of ​​a square with side ( a + b), and on the other hand, the sum of the areas of the four triangles and

Q.E.D.

3. Proof of the Pythagorean theorem by the method of infinitesimal.

​

Considering the drawing shown in the figure, and

watching the side changea, we can

write the following relation for infinitely

small side incrementsWith and a(using the similarity

triangles):

Using the variable separation method, we find:

A more general expression for changing the hypotenuse in the case of increments of both legs:

Integrating this equation and using the initial conditions, we get:

Thus, we arrive at the desired answer:

As it is easy to see, the quadratic dependence in the final formula appears due to the linear

proportionality between the sides of the triangle and the increments, while the sum is related to independent

contributions from the increment of different legs.

A simpler proof can be obtained if we assume that one of the legs does not experience an increment

(in this case, the leg b). Then for the constant of integration we get:

Instructions

If you need to calculate according to the Pythagorean theorem, use the following algorithm: - Determine in the triangle which sides are the legs and the hypotenuse. The two sides forming an angle of ninety degrees are the legs, the remaining third is the hypotenuse. (cm) - Raise each leg of this triangle to the second power, that is, multiply by yourself. Example 1. Let it be necessary to calculate the hypotenuse, if one leg in a triangle is 12 cm, and the other is 5 cm. First, the squares of the legs are equal: 12 * 12 = 144 cm and 5 * 5 = 25 cm - Next, determine the sum of the squares legs. A certain number is hypotenuse, you need to get rid of the second power of the number to find the length this side of the triangle. To do this, remove from under square root the value of the sum of the squares of the legs. Example 1.14 + 25 = 169. The square root of 169 will be 13. Therefore, the length of the given hypotenuse is equal to 13 cm.

Another way to calculate length hypotenuse consists in the terminology of sine and angles in a triangle. By definition: sine of the angle alpha - the opposite leg to the hypotenuse. That is, looking at the figure, sin a = CB / AB. Hence, the hypotenuse AB = CB / sin a. Example 2. Let the angle be 30 degrees, and the opposite leg is 4 cm. You need to find the hypotenuse. Solution: AB = 4 cm / sin 30 = 4 cm / 0.5 = 8 cm. Answer: length hypotenuse is equal to 8 cm.

A similar way of finding hypotenuse from the definition of the cosine of an angle. The cosine of the angle is the ratio of the adjacent leg and hypotenuse... That is, cos a = AC / AB, hence AB = AC / cos a. Example 3. In the triangle ABC, AB is the hypotenuse, the angle BAC is 60 degrees, the leg AC is 2 cm. Find AB.
Solution: AB = AC / cos 60 = 2 / 0.5 = 4 cm. Answer: the hypotenuse is 4 cm in length.

Useful advice

When finding the value of the sine or cosine of an angle, use either the sine and cosine table or the Bradis table.

Tip 2: How to find the length of the hypotenuse in a right triangle

The hypotenuse is the longest side in a right-angled triangle, so it is not surprising that with Greek this word is translated as "stretched". This side always lies opposite an angle of 90 °, and the sides forming this angle are called legs. Knowing the lengths of these sides and the magnitude of the acute angles in different combinations these values ​​can be calculated and the length of the hypotenuse.

Instructions

If the lengths of both triangles (A and B) are known, then use the lengths of the hypotenuse (C), perhaps the most well-known mathematical postulate - the Pythagorean theorem. It says that the square of the length of the hypotenuse is the sum of the squares of the lengths of the legs, from which it follows that you should calculate the root of the sum of the squared lengths of the two sides: C = √ (A² + B²). For example, if the length of one leg is 15, a - 10 centimeters, then the length of the hypotenuse will be approximately 18.0277564 centimeters, since √ (15² + 10²) = √ (225 + 100) = √325≈18.0277564.

If the length of only one of the legs (A) in a right-angled triangle is known, as well as the value of the angle lying opposite it (α), then the length of the hypotenuse (C) can be done using one of the trigonometric functions - sine. To do this, divide the length of the known side by the sine of the known angle: C = A / sin (α). For example, if the length of one of the legs is 15 centimeters, and the angle at the opposite vertex of the triangle is 30 °, then the length of the hypotenuse will be 30 centimeters, since 15 / sin (30 °) = 15 / 0.5 = 30.

If in a right-angled triangle the value of one of the acute angles (α) and the length of the adjacent leg (B) are known, then to calculate the length of the hypotenuse (C), you can use another trigonometric function - cosine. You should divide the length of the known leg by the cosine of the known angle: C = B / cos (α). For example, if the length of this leg is 15 centimeters, and the acute angle adjacent to it is 30 °, then the length of the hypotenuse will be approximately 17.3205081 centimeters, since 15 / cos (30 °) = 15 / (0.5 * √3) = 30 / √3≈17.3205081.

It is customary to denote by length the distance between two points of any segment. It can be a straight, broken or closed line. You can calculate the length in a fairly simple way if you know some other indicators of the segment.

Instructions

If you need to find the length of a side of a square, then this will not work if you know its area S. Due to the fact that all sides of a square have