Equations by the Pythagorean theorem. Methods for proving the Pythagorean theorem. Similar geometric shapes on three sides

Make sure the triangle you are given is right-angled, as the Pythagorean theorem only applies to right-angled triangles. In right-angled triangles, one of the three angles is always 90 degrees.

A right angle in a right triangle is indicated by a square icon, not a curve, which is an oblique angle.

Add guidelines for the sides of the triangle. Label the legs as "a" and "b" (legs - sides intersecting at right angles), and the hypotenuse as "c" (hypotenuse - the largest side right trianglelying opposite right angle).

Determine which side of the triangle you want to find. The Pythagorean theorem allows you to find any side of a right triangle (if the other two sides are known). Determine which side (a, b, c) you need to find.

For example, given a hypotenuse equal to 5, and given a leg equal to 3. In this case, find the second leg. We will come back to this example later.
If the other two sides are unknown, it is necessary to find the length of one of the unknown sides in order to be able to apply the Pythagorean theorem. To do this, use the basic trigonometric functions (if you are given the value of one of the oblique angles).

Substitute in the formula a 2 + b 2 \u003d c 2 the values \u200b\u200byou give (or the values \u200b\u200byou found). Remember that a and b are legs and c is hypotenuse.

In our example, write: 3² + b² \u003d 5².

Square each side you know. Or leave the degrees - you can square the numbers later.

In our example, write: 9 + b² \u003d 25.

Isolate the unknown side on one side of the equation. To do this, transfer the known values \u200b\u200bto the other side of the equation. If you find the hypotenuse, then in the Pythagorean theorem it is already isolated on one side of the equation (so nothing needs to be done).

In our example, move 9 to the right side of the equation to isolate the unknown b². You will get b² \u003d 16.

Extract square root from both sides of the equation. At this stage, on one side of the equation there is an unknown (squared), and on the other side there is a free term (number).

In our example, b² \u003d 16. Take the square root of both sides of the equation and get b \u003d 4. So the second leg is 4 .

Use the Pythagorean theorem in your daily life, as it can be applied in a wide variety of practical situations. To do this, learn to recognize right-angled triangles in everyday life - in any situation in which two objects (or lines) intersect at right angles, and a third object (or line) connects (diagonally) the tops of the first two objects (or lines), you can use the Pythagorean theorem to find the unknown side (if the other two sides are known).

Example: given a staircase leaning against a building. The bottom of the stairs is 5 meters from the base of the wall. The top of the stairs is 20 meters from the ground (up the wall). How long are the stairs?
- "5 meters from the base of the wall" means that a \u003d 5; "Is 20 meters from the ground" means that b \u003d 20 (that is, you are given two legs of a right triangle, since the wall of the building and the surface of the Earth intersect at right angles). The length of the ladder is the length of the hypotenuse, which is unknown.
  - a² + b² \u003d c²
  - (5) ² + (20) ² \u003d c²
  - 25 + 400 \u003d c²
  - 425 \u003d c²
  - c \u003d √425
  - c \u003d 20.6. So the approximate length of the ladder is 20.6 meters.

The Pythagorean theorem states:

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

a 2 + b 2 \u003d c 2,

a and b - legs forming a right angle.
from - the hypotenuse of the triangle.

Pythagorean theorem formulas

a \u003d \\ sqrt (c ^ (2) - b ^ (2))
b \u003d \\ sqrt (c ^ (2) - a ^ (2))
c \u003d \\ sqrt (a ^ (2) + b ^ (2))

Proof of the Pythagorean theorem

The area of \u200b\u200ba right-angled triangle is calculated by the formula:

S \u003d \\ frac (1) (2) ab

To calculate the area of \u200b\u200ban arbitrary triangle, the area formula is:

p - semi-perimeter. p \u003d \\ frac (1) (2) (a + b + c),
r Is the radius of the inscribed circle. For rectangle r \u003d \\ frac (1) (2) (a + b-c).

Then we equate the right sides of both formulas for the area of \u200b\u200ba triangle:

\\ frac (1) (2) ab \u003d \\ frac (1) (2) (a + b + c) \\ frac (1) (2) (a + b-c)

2 ab \u003d (a + b + c) (a + b-c)

2 ab \u003d \\ left ((a + b) ^ (2) -c ^ (2) \\ right)

2 ab \u003d a ^ (2) + 2ab + b ^ (2) -c ^ (2)

0 \u003d a ^ (2) + b ^ (2) -c ^ (2)

c ^ (2) \u003d a ^ (2) + b ^ (2)

The reverse Pythagorean theorem:

If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is rectangular. That is, for any triple of positive numbers a, b and csuch that

a 2 + b 2 \u003d c 2,

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

Pythagorean theorem - one of the fundamental theorems of Euclidean geometry, establishing the relationship between the sides of a right-angled triangle. It was proved by the scientist mathematician and philosopher Pythagoras.

The meaning of the theorem in that it can be used to prove other theorems and solve problems.

Additional material:

Pythagorean theorem: The sum of the areas of the squares resting on the legs ( a and b), is equal to the area of \u200b\u200bthe 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:

Evidence

Of course, conceptually all of them can be divided into a small number of classes. The most famous of them: 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 \u200b\u200ba figure.

Let be ABC there is a right-angled triangle 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 equivalent

Adding, we get

Areas proof

The proofs given below, despite their apparent 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

Arrange four equal right-angled triangles as shown in Figure 1.
Quadrilateral with sides c is a square, since the sum of two acute angles is 90 °, and the unfolded angle is 180 °.
The area of \u200b\u200bthe entire figure is, on the one hand, the area of \u200b\u200ba 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 scaling

Elegant proof by permutation

An example of one of such proofs is shown in the drawing on the right, where the 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

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 \u200b\u200bthe square DECA is equal to the area of \u200b\u200bthe rectangle AHJK To do this, let's use an auxiliary observation: The area of \u200b\u200ba triangle with the same height and base as this rectangle is equal to half the area of \u200b\u200bthe given rectangle. This is a consequence of defining the area of \u200b\u200ba triangle as half the product of the base and the height. It follows from this observation that the area of \u200b\u200bthe triangle ACK is equal to the area of \u200b\u200bthe triangle AHK (not shown in the figure), which, in turn, is equal to half the area of \u200b\u200bthe rectangle AHJK.

Let us now prove that the area of \u200b\u200bthe triangle ACK is also equal to half the area of \u200b\u200bthe square DECA. The only thing that needs to be done for this is to prove that the triangles ACK and BDA are equal (since the area of \u200b\u200bthe triangle BDA is equal to half the area of \u200b\u200bthe square according to the above property). Equality is obvious, the triangles are equal on two sides and the angle between them. Namely - AB \u003d AK, AD \u003d 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 vertex 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 \u200b\u200bthe 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

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). By rotating 90 degrees counterclockwise, we see that the shaded shapes are equal CAJI and GDAB ... Now it is clear that the area of \u200b\u200bthe shaded figure is equal to the sum of the halves of the areas of the squares built on the legs and the area of \u200b\u200bthe original triangle. On the other hand, it is equal to half the area of \u200b\u200bthe square built on the hypotenuse, plus the area of \u200b\u200bthe original triangle. The last 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 in side a, we can write the following ratio for infinitely small increments of the sides from and a (using similarity to 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

By 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 related to 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 \u200b\u200bthe 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 \u200b\u200ba 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 \u200b\u200bthe 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.

History

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 a 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 Bashara.

Cantor (the largest German historian of mathematics) believes that the equality 3 ² + 4 ² \u003d 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 would become superfluous, if you use, for example, the wooden square used by all carpenters. Indeed, Egyptian drawings are known 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 of Ancient Egypt, Babylon and Greece. M., 1959
Glazer G.I. History of mathematics at school. M., 1982
V. Litzman, "The Pythagorean Theorem" M., 1960.
- The site about the Pythagorean theorem with a large number of proofs is taken from the book by 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 (eng.)
Cut-The-Knot, a section on the Pythagorean theorem, about 70 proofs and a wealth of additional information

Wikimedia Foundation. 2010.

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 \u200b\u200bthe 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 theoremsare 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 any triple of positive numbers a, b and csuch that

there is a right-angled triangle with legs a and band 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:

evidence area method, axiomatic and exotic evidence (eg,

through 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 area of \u200b\u200ba figure.

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

its foundation through H.

Triangle ACH like a triangle ABC 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 given below, despite their apparent 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.

Arrange 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 \u200b\u200bthe entire figure is, on the one hand,

area of \u200b\u200ba 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 incrementsfrom 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:

The fate of other theorems and problems is peculiar ... How can one explain, for example, such exceptional attention from mathematicians and amateurs of mathematics to the Pythagorean theorem? Why were many of them not content with the already known proofs, but found their own, bringing the number of proofs to several hundred in twenty-five comparatively foreseeable centuries?
When it comes to the Pythagorean theorem, the unusual begins with its name. It is believed that Pythagoras was not the first to formulate it. It is also considered doubtful that he gave her proof. If Pythagoras is a real person (some even doubt this!), Then he lived, most likely, in the VI-V centuries. BC e. He himself did not write anything, called himself a philosopher, which meant, in his understanding, "striving for wisdom", founded the Pythagorean Union, whose members were engaged in music, gymnastics, mathematics, physics and astronomy. Apparently, he was also an excellent orator, as evidenced by the following legend related to his stay in the city of Crotone: “The first appearance of Pythagoras before the people in Crotone began with a speech to the young men, in which he was so strict, but at the same time so fascinating outlined the responsibilities of the young men, that the elders in the city asked not to leave them without instruction. In this second speech, he pointed to lawfulness and purity of morals as the foundations of the family; in the next two he turned to children and women. The consequence last speech, in which he especially condemned luxury, was that thousands of precious dresses were delivered to the temple of Hera, for no woman dared to show herself in them on the street anymore ... ”Nevertheless, even in the second century AD, that is, After 700 years, quite real people lived and worked, outstanding scientists who were clearly under the influence of the Pythagorean union and who had great respect for what, according to legend, Pythagoras created.
Undoubtedly, the interest in the theorem is also caused by the fact that it occupies one of the central places in mathematics, and by the satisfaction of the authors of the proofs who overcame the difficulties, about which the Roman poet Quintus Horace Flaccus, who lived before our era, spoke well: "It is difficult to express well-known facts." ...
Initially, the theorem established the relationship between the areas of squares built on the hypotenuse and legs of a right triangle:
.
Algebraic formulation:
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 the triangle through c, and the lengths of the legs through a and b: a 2 + b 2 \u003d 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 converse theorem of Pythagoras. For any triple of positive numbers a, b, and c such that
a 2 + b 2 \u003d c 2, there is a right-angled triangle with legs a and b and hypotenuse c.

Evidence

At the moment, 367 proofs of this theorem have been recorded in the scientific literature. The Pythagorean theorem is probably 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: 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 area of \u200b\u200ba figure.
Let ABC be a right-angled triangle with right angle C. Draw the height from C and denote its base by H. Triangle ACH is similar to triangle ABC in two angles.
Likewise, triangle CBH is similar to ABC. Introducing the notation

we get

What is equivalent

Adding, we get

or

Areas proof

Equal complementarity proof

1. Place four equal right-angled triangles as shown in the figure.
2. A quadrilateral with sides c is a square, since the sum of two acute angles is 90 °, and the expanded angle is 180 °.
3. The area of \u200b\u200bthe whole figure is, on the one hand, the area of \u200b\u200ba square with sides (a + b), and on the other hand, the sum of the areas of four triangles and an inner square.

Q.E.D.

Evidence through scaling

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

Euclid's proof

The idea behind Euclid's proof is as follows: let's try to prove that half of the area of \u200b\u200bthe square built on the hypotenuse is equal to the sum of half 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 \u200b\u200bthe square DECA is equal to the area of \u200b\u200bthe rectangle AHJK For this we use an auxiliary observation: The area of \u200b\u200ba triangle with the same height and base as this rectangle is equal to half the area of \u200b\u200bthe given rectangle. This is a consequence of defining the area of \u200b\u200ba triangle as half the product of the base and the height. It follows from this observation that the area of \u200b\u200bthe triangle ACK is equal to the area of \u200b\u200bthe triangle AHK (not shown in the figure), which, in turn, is equal to half the area of \u200b\u200bthe rectangle AHJK. Let us now prove that the area of \u200b\u200bthe triangle ACK is also equal to half the area of \u200b\u200bthe 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 \u200b\u200bthe triangle BDA is equal to half the area of \u200b\u200bthe square according to the above property). Equality is obvious, the triangles are equal on two sides and the angle between them. Namely - AB \u003d AK, AD \u003d 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 vertex 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 \u200b\u200bthe square built on the hypotenuse is the sum of the areas of the squares built on the legs.

Proof of Leonardo da Vinci

The main elements of the proof are symmetry and motion.

Consider the drawing, as can be seen from the symmetry, the segment CI cuts the square ABHJ into two identical parts (since the triangles ABC and JHI are equal in construction). Using a 90 degree counterclockwise rotation, we see that the shaded figures CAJI and GDAB are equal. Now it is clear that the area of \u200b\u200bthe shaded figure is equal to the sum of the halves of the areas of the squares built on the legs and the area of \u200b\u200bthe original triangle. On the other hand, it is equal to half the area of \u200b\u200bthe square built on the hypotenuse, plus the area of \u200b\u200bthe original triangle. The last step in the proof is left to the reader.