When you first started learning square roots and how to solve irrational equations(equalities containing the unknown under the root sign), you probably got the first idea about them practical use. The ability to extract Square root of numbers is also necessary for solving problems on the application of the Pythagorean theorem. This theorem relates the lengths of the sides of any right triangle.

Let the lengths of the legs of a right triangle (those two sides that converge at a right angle) be denoted by the letters and , and the length of the hypotenuse (the longest side of the triangle opposite right angle) will be denoted by the letter . Then the corresponding lengths are related by the following relation:

This equation allows you to find the length of a side of a right triangle in the case when the length of its other two sides is known. In addition, it allows you to determine whether the considered triangle is a right triangle, provided that the lengths of all three sides known in advance.

Solving problems using the Pythagorean theorem

To consolidate the material, we will solve the following problems for the application of the Pythagorean theorem.

So given:

1. The length of one of the legs is 48, the hypotenuse is 80.
2. The length of the leg is 84, the hypotenuse is 91.

Let's get to the solution:

a) Substituting the data into the equation above gives the following results:

48 2 + b 2 = 80 2

2304 + b 2 = 6400

b 2 = 4096

b= 64 or b = -64

Since the length of a side of a triangle cannot be expressed negative number, the second option is automatically discarded.

Answer to the first picture: b = 64.

b) The length of the leg of the second triangle is found in the same way:

84 2 + b 2 = 91 2

7056 + b 2 = 8281

b 2 = 1225

b= 35 or b = -35

As in the previous case, the negative solution is discarded.

Answer to the second picture: b = 35

We are given:

1. The lengths of the smaller sides of the triangle are 45 and 55, respectively, and the larger ones are 75.
2. The lengths of the smaller sides of the triangle are 28 and 45, respectively, and the larger ones are 53.

We solve the problem:

a) It is necessary to check whether the sum of the squares of the lengths of the smaller sides of a given triangle is equal to the square of the length of the larger one:

45 2 + 55 2 = 2025 + 3025 = 5050

Therefore, the first triangle is not a right triangle.

b) The same operation is performed:

28 2 + 45 2 = 784 + 2025 = 2809

Therefore, the second triangle is a right triangle.

First we find the length longest segment, formed by points with coordinates (-2, -3) and (5, -2). For this we use known formula to find the distance between points in rectangular system coordinates:

Similarly, we find the length of the segment enclosed between the points with coordinates (-2, -3) and (2, 1):

Finally, we determine the length of the segment between points with coordinates (2, 1) and (5, -2):

Since there is an equality:

then the corresponding triangle is a right triangle.

Thus, we can formulate the answer to the problem: since the sum of the squares of the sides with the smallest length is equal to the square of the side with greatest length, the points are the vertices of a right triangle.

The base (located strictly horizontally), the jamb (located strictly vertically) and the cable (stretched diagonally) form a right triangle, respectively, the Pythagorean theorem can be used to find the length of the cable:

Thus, the length of the cable will be approximately 3.6 meters.

Given: the distance from point R to point P (the leg of the triangle) is 24, from point R to point Q (hypotenuse) - 26.

So, we help Vitya solve the problem. Since the sides of the triangle shown in the figure are supposed to form a right triangle, you can use the Pythagorean theorem to find the length of the third side:

So, the width of the pond is 10 meters.

Sergey Valerievich

Pythagorean theorem

The fate of other theorems and problems is peculiar... How can one explain, for example, such exceptional attention on the part of mathematicians and mathematicians to the Pythagorean theorem? Why many of them weren't content already known evidence, but found their own, bringing the amount of evidence to several hundred in twenty-five comparatively visible centuries?
When we are talking about the Pythagorean theorem, the unusual begins already with its name. It is believed that it was by no means Pythagoras who formulated it for the first time. It is also doubtful that he gave her proof. If Pythagoras - real face(some even doubt this!), then he most likely lived in the 6th-5th centuries. BC e. He himself did not write anything, he called himself a philosopher, which meant, in his understanding, “aspiring to wisdom”, founded the Pythagorean Union, whose members were engaged in music, gymnastics, mathematics, physics and astronomy. Apparently, he was also a great speaker, as evidenced by the following legend relating to his stay in the city of Croton: “The first appearance of Pythagoras before the people in Croton began with a speech to young men, in which he is so strict, but at the same time so fascinating outlined the duties of the young men, that the elders in the city asked not to leave them without teaching. In this second speech, he pointed to legality and purity of morals, as the foundations of the family; in the next two he addressed children and women. Consequence last speech in which he especially condemned luxury was that thousands of precious dresses were delivered to the temple of Hera, for not a single woman dared to show themselves in them on the street anymore ... ”Nevertheless, back in the second century of our era, i.e. . after 700 years, they lived and worked completely real people, outstanding scientists who were clearly influenced by the Pythagorean union and with great respect for what, according to legend, Pythagoras created.
It is also undoubted that the interest in the theorem is caused both 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, well said: “It is difficult to express well-known facts” .
Initially, the theorem established the relationship between the areas of squares built on the hypotenuse and the legs of a right triangle:
.
Algebraic formulation:
AT right triangle square of the length of the hypotenuse is equal to the sum squares of leg lengths.
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 formulations of the theorem are equivalent, but the second formulation is more elementary, it does not require the concept of area. That is, the second statement can be verified without knowing anything about the area and by measuring only the lengths of the sides of a right triangle.
Inverse theorem Pythagoras. For every trio positive numbers a, b and c such that
a 2 + b 2 = c 2 , there is a right triangle with legs a and b and hypotenuse c.

Proof of

At the moment in scientific literature 367 proofs of this theorem were recorded. Probably, the Pythagorean theorem is the only theorem with such an impressive number of proofs. Such a variety can only be explained by the fundamental significance 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: area proofs, 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 be a right triangle with right angle C. Draw a height from C and denote its base by H. Triangle ACH is similar to triangle ABC in two angles.
Similarly, triangle CBH is similar to ABC. Introducing the notation

we get

What is equivalent

Adding, we get

or

Area proofs

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

Proof via Equivalence

1. Arrange four equal right 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 straight angle is 180°.
3. The area of ​​the whole figure is equal, on the one hand, to the area of ​​a square with side (a + b), and on the other hand, to the sum four triangles and inner square.



Q.E.D.

Evidence through Equivalence

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

Euclid's proof

The idea of ​​Euclid's proof is as follows: let's try to prove that half the area of ​​the square built on the hypotenuse is equal to the sum of the half 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 triangle and drew a ray s from the vertex of 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 To do this, we use an auxiliary observation: The area of ​​a triangle with the same height and base as the given rectangle is equal to half the area of ​​the given rectangle. This is a consequence of defining the area of ​​a triangle as half the product of the base and the height. From this observation it follows that the area of ​​triangle ACK is equal to the area of ​​triangle AHK (not shown), which, in turn, is equal to half the area of ​​rectangle AHJK. Let us now prove that the area of ​​triangle ACK is also equal to half the area of ​​square DECA. The only thing that needs to be done for this is to prove the equality of triangles ACK and BDA (since the area of ​​triangle BDA is equal to half the area of ​​the square by the above property). This equality is obvious, the triangles are equal in 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 movement: let's rotate the triangle CAK 90 ° counterclockwise, then it is obvious that the corresponding sides of the two considered triangles will coincide (due to the fact that the angle at the vertex of the square is 90°). The argument 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.

Proof of Leonardo da Vinci

The main elements of the proof are symmetry and movement.

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 the equality of the shaded figures CAJI and GDAB. Now it is clear that the area of ​​the figure shaded by us is equal to the sum of half 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 last step in the proof is left to the reader.

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

between the sides of a right triangle.

It is believed that it was proved by the Greek mathematician Pythagoras, after whom it is named.

Geometric formulation of the Pythagorean theorem.

The theorem was originally formulated as follows:

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

built on catheters.

Algebraic formulation of the 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.

That is, denoting the length of the hypotenuse of the triangle through 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 does not

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

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

The inverse 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

triangle is rectangular.

Or, in other words:

For any triple of positive numbers a, b and c, such that

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

The Pythagorean theorem for an isosceles triangle.

Pythagorean 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 significance 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 of area method, axiomatic and exotic evidence(For example,

via differential equations).

1. Proof of the Pythagorean theorem in terms of similar triangles.

The following proof of the algebraic formulation is the simplest of the proofs constructed

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

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

its foundation through H.

Triangle ACH similar to a triangle AB C on two corners. Likewise, the triangle CBH similar ABC.

By introducing the notation:

we get:

,

which matches -

Having folded a 2 and b 2 , we get:

or , which was to be proved.

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

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

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

• Proof through equicomplementation.

Arrange four equal rectangular

triangle as shown in the picture

on right.

Quadrilateral with sides c- square,

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

the developed angle is 180°.

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

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

Q.E.D.

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

​

Considering the drawing shown in the figure, and

watching the side changea, we can

write the following relation for infinite

small side incrementswith and a(using similarity

triangles):

Using the method of separation of variables, we find:

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

Integrating given 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 the 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 leg b). Then for the integration constant we get:

(according to Papyrus 6619 of the Berlin Museum). According to Cantor, harpedonapts, or "string tensioners," built right angles using right triangles with sides 3, 4, and 5.

It is very easy to reproduce their method of construction. Let's 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. A right angle will be enclosed between sides 3 and 4 meters long. It might be objected to the Harpedonapts that their method of construction becomes redundant if, for example, the wooden square used by all carpenters is used. 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 Pythagorean theorem among the Babylonians. In one text dating back to the time of Hammurabi, that is, to 2000 BC. e. , an approximate calculation of the hypotenuse of a right triangle is given. From this we can conclude that in Mesopotamia they were able to perform calculations with right-angled triangles, at least in some cases. Based, on the one hand, on the current level of knowledge of Egyptian and Babylonian mathematics, and, on the other hand, on a critical study of Greek sources, Van der Waerden (a Dutch mathematician) concluded that there was a high probability that the hypotenuse square theorem was known in India already around the 18th century BC. e.

Around 400 BC. e., according to Proclus, Plato gave a method for finding Pythagorean triples, combining algebra and geometry. Around 300 BC. e. Euclid's Elements contains the oldest axiomatic proof of the Pythagorean theorem.

Wording

Geometric formulation:

The theorem was originally formulated as follows:

Algebraic formulation:

That is, denoting the length of the hypotenuse of the triangle through, and the lengths of the legs through and:

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

Inverse Pythagorean theorem:

Proof of

At the moment, 367 proofs of this theorem have been recorded in the scientific literature. Probably, the Pythagorean theorem is the only theorem with such an impressive number of proofs. Such a variety can only be explained by the fundamental significance 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 figure area.

Let be ABC there is a right angled triangle C. Let's draw a height from C and denote its base by H. Triangle ACH similar to a triangle ABC at two corners. Likewise, the triangle CBH similar ABC. Introducing the notation

we get

What is equivalent

Adding, we get

, which was to be proved

Area proofs

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

Proof via Equivalence

1. Arrange four equal right triangles as shown in Figure 1.
2. Quadrilateral with sides c is a square because the sum of two acute angles is 90° and the straight angle is 180°.
3. The area of ​​the whole figure is equal, on the one hand, to the area of ​​a square with a side (a + b), and on the other hand, the sum of the areas of four triangles and the area of ​​the inner square.

Q.E.D.

Euclid's proof

The idea of ​​Euclid's proof is as follows: let's try to prove that half the area of ​​the square built on the hypotenuse is equal to the sum of the half 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 triangle and drew a ray s from the vertex of 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 To do this, we use an auxiliary observation: The area of ​​a triangle with the same height and base as the given rectangle is equal to half the area of ​​the given rectangle. This is a consequence of defining the area of ​​a triangle as half the product of the base and the height. From this observation it follows that the area of ​​triangle ACK is equal to the area of ​​triangle AHK (not shown), which, in turn, is equal to half the area of ​​rectangle AHJK.

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

The argument 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. Idea this evidence further illustrated with the animation above.

Proof of Leonardo da Vinci

The main elements of the proof are symmetry and movement.

Consider the drawing, as can be seen from the symmetry, the segment cuts the square into two identical parts (since the triangles and are equal in construction).

Using a counterclockwise rotation of 90 degrees around the point , we see the equality of the shaded figures and .

Now it is clear that the area of ​​the figure we have shaded is equal to the sum of half the areas of small 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 large square (built on the hypotenuse) plus the area of ​​the original triangle. Thus, half the sum of the areas of the small squares is equal to half the area of ​​the large square, and therefore the sum of the areas of the squares built on the legs is equal to the area of ​​the square built on the hypotenuse.

Proof by the infinitesimal method

The following proof by means of differential equations is often attributed to the well-known English mathematician Hardy, who lived in the first half of the 20th century.

Considering the drawing shown in the figure and observing the change in side a, we can write the following relation for infinitesimal side increments with and a(using similar triangles):

Using the method of separation of variables, 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 obtain

Thus, we arrive at the desired answer

It is easy to see that 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 due to the 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). Then for the integration constant we get

Variations and Generalizations

Similar geometric shapes on three sides

Generalization for similar triangles, area of ​​green figures A + B = area of ​​blue C

Pythagorean theorem using similar right triangles

A generalization of the Pythagorean theorem was made by Euclid in his work Beginnings, expanding the areas of the squares on the sides to the areas of similar geometric shapes :

If we construct similar geometric figures (see Euclidean geometry) on the sides of a right triangle, then the sum of the two smaller figures will equal the area of ​​the larger figure.

The main idea of ​​this generalization is that the area of ​​such a geometric figure is proportional to the square of any of its linear dimensions and, in particular, to the square of the length of any side. Therefore, for similar figures with areas A, B and C built on sides with length a, b and c, we have:

But, according to the Pythagorean theorem, a 2 + b 2 = c 2 , then A + B = C.

Conversely, if we can prove that A + B = C for three similar geometric figures without using the Pythagorean theorem, then we can prove the theorem itself, moving to reverse direction. For example, the starting center triangle can be reused as a triangle C on the hypotenuse, and two similar right triangles ( A and B) built on the other two sides, which are formed as a result of dividing the central triangle by its height. The sum of the two smaller areas of the triangles is then obviously equal to the area of ​​the third, thus A + B = C and, following the previous proofs in reverse order, we get the Pythagorean theorem a 2 + b 2 = c 2 .

Cosine theorem

The Pythagorean theorem is special case more general theorem cosines, which relates the lengths of the sides in an arbitrary triangle:

where θ is the angle between the sides a and b.

If θ is 90 degrees then cos θ = 0 and the formula is simplified to the usual Pythagorean theorem.

Arbitrary triangle

To any chosen corner of an arbitrary triangle with sides a, b, c inscribe an isosceles triangle in such a way that equal angles at its base, θ equaled the chosen angle. Let us assume that the chosen angle θ is located opposite the side indicated c. As a result, we got a triangle ABD with angle θ, which is located opposite the side a and parties r. The second triangle is formed by the angle θ, which is opposite the side b and parties with length s, as it shown on the picture. Thabit Ibn Qurra stated that the sides in these three triangles are related as follows:

As the angle θ approaches π/2, the base isosceles triangle decreases, and the two sides r and s overlap each other less and less. When θ = π/2, ADB turns into a right triangle, r + s = c and we get the initial Pythagorean theorem.

Let's look at one of the arguments. Triangle ABC has the same angles as triangle ABD, but in reverse order. (The two triangles have a common angle at vertex B, both have angle θ, and also have the same third angle, by the sum of the angles of the triangle) Accordingly, ABC is similar to the reflection ABD of triangle DBA, as shown in the lower figure. Let us write the relation between opposite sides and adjacent to the angle θ,

So is the reflection of another triangle,

Multiply the fractions and add these two ratios:

Q.E.D.

Generalization for arbitrary triangles via parallelograms

Generalization for arbitrary triangles,
area of ​​green plot = area blue

Proof of the thesis that in the figure above

Let's make a further generalization for non-rectangular triangles, using parallelograms on three sides instead of squares. (squares are a special case.) The top figure shows that for acute triangle the area of ​​the parallelogram on the long side is equal to the sum of the parallelograms on the other two sides, provided that the parallelogram on the long side is built as shown in the figure (the dimensions marked with arrows are the same and determine the sides of the lower parallelogram). This replacement of squares by parallelograms bears a clear resemblance to the initial Pythagorean theorem and is believed to have been formulated by Pappus of Alexandria in 4 CE. e.

The bottom figure shows the progress of the proof. Let's look at the left side of the triangle. The left green parallelogram has the same area as the left side of the blue parallelogram because they have the same base b and height h. Also, the left green box has the same area as the left green box in the top picture because they have common ground(upper left-hand side triangle) and the total height perpendicular to that side of the triangle. Arguing similarly for the right side of the triangle, we prove that the lower parallelogram has the same area as the two green parallelograms.

Complex numbers

The Pythagorean theorem is used to find the distance between two points in a Cartesian coordinate system, and this theorem is true for all true coordinates: distance s between two points ( a, b) and ( c, d) equals

There are no problems with the formula if complex numbers are treated as vectors with real components x + i y = (x, y). . For example, the distance s between 0 + 1 i and 1 + 0 i calculate as modulus of vector (0, 1) − (1, 0) = (−1, 1), or

However, for operations with vectors with complex coordinates, it is necessary to make a certain improvement to the Pythagorean formula. Distance between points with complex numbers (a, b) and ( c, d); a, b, c, and d are all complex, we formulate using absolute values. Distance s based on vector difference (a − c, b − d) in following form: let the difference a − c = p+i q, where p is the real part of the difference, q is the imaginary part, and i = √(−1). Likewise, let b − d = r+ i s. Then:

where is the complex conjugate of . For example, the distance between points (a, b) = (0, 1) and (c, d) = (i, 0) , calculate the difference (a − c, b − d) = (−i, 1) and the result would be 0 if complex conjugates were not used. Therefore, using the improved formula, we get

The module is defined like this:

Stereometry

A significant generalization of the Pythagorean theorem for three-dimensional space is de Gua's theorem, named after J.-P. de Gua: if a tetrahedron has a right angle (as in a cube), then the square of the area of ​​the face opposite the right angle is equal to the sum of the squares of the areas of the other three faces. This conclusion can be summarized as " n-dimensional Pythagorean theorem":

Pythagoras's theorem three-dimensional space connects the diagonal AD with three sides.

Another generalization: The Pythagorean theorem can be applied to stereometry in the following form. Consider cuboid, as it shown on the picture. Find the length of the diagonal BD using the Pythagorean theorem:

where three sides form a right triangle. Use the horizontal diagonal BD and the vertical edge AB to find the length of the diagonal AD, again using the Pythagorean theorem:

or, if everything is written in one equation:

This result is a 3D expression for determining the magnitude of the vector v(diagonal AD) expressed in terms of its perpendicular components ( v k) (three mutually perpendicular sides):

This equation can be viewed as a generalization of the Pythagorean theorem for a multidimensional space. However, the result is actually nothing more than the repeated application of the Pythagorean theorem to a sequence of right triangles in successively perpendicular planes.

vector space

In the case of an orthogonal system of vectors, an equality takes place, which is also called the Pythagorean theorem:

If is the projection of the vector onto coordinate axes, then this formula coincides with the Euclidean distance - and means that the length of the vector is equal to the root square sum squares of its components.

An analogue of this equality in the case endless system vectors is called Parseval's equality.

Non-Euclidean geometry

The Pythagorean theorem is derived from the axioms of Euclidean geometry and, in fact, is not valid for non-Euclidean geometry, in the form in which it is written above. (That is, the Pythagorean theorem turns out to be a kind of equivalent to Euclid's postulate of parallelism) In other words, in non-Euclidean geometry, the ratio between the sides of the triangle will necessarily be in a form different from the Pythagorean theorem. For example, in spherical geometry, all three sides of a right triangle (say a, b and c), which bound an octant (an eighth part) of the unit sphere, have length π/2, which contradicts the Pythagorean theorem, because a 2 + b 2 ≠ c 2 .

Consider here two cases of non-Euclidean geometry - spherical and hyperbolic geometry; in both cases, as for the Euclidean space for right triangles, the result that replaces the Pythagorean theorem follows from the cosine theorem.

However, the Pythagorean theorem remains valid for hyperbolic and elliptic geometry if the requirement that the triangle is right-angled is replaced by the condition that the sum of two angles of the triangle must be equal to the third, say A+B = C. Then the ratio between the sides looks like this: the sum of the areas of circles with diameters a and b equal to the area of ​​a circle with a diameter c.

spherical geometry

For any right triangle on a sphere with radius R(for example, if the angle γ in the triangle is right) with sides a, b, c the relationship between the parties will look like this:

This equality can be derived as a special case spherical cosine theorem, which is valid for all spherical triangles:

where cosh is the hyperbolic cosine. This formula is a special case of the hyperbolic cosine theorem, which is valid for all triangles:

where γ is the angle whose vertex is opposite the side c.

where g ij is called the metric tensor. It can be a position function. Such curvilinear spaces include Riemannian geometry as general example. This formulation is also suitable for Euclidean space when using curvilinear coordinates. For example, for polar coordinates:

vector product

The Pythagorean theorem connects two expressions for the magnitude of a vector product. One approach to defining a cross product requires that it satisfy the equation:

this formula uses the dot product. The right side of the equation is called the Gram's determinant for a and b, which is equal to the area of ​​the parallelogram formed by these two vectors. Based on this requirement, as well as the requirement that the vector product be perpendicular to its components a and b it follows that, except for the trivial cases of 0- and 1-dimensional space, the vector product is only defined in three and seven dimensions. We use the definition of the angle in n-dimensional space:

this property of the vector product gives its value in the following form:

Through the fundamental trigonometric identity Pythagoras, we get a different form of writing its value:

An alternative approach to defining a cross product uses an expression for its magnitude. Then, arguing in reverse order, we obtain a connection with scalar product:

see also

Notes

1. History topic: Pythagoras's theorem in Babylonian mathematics
2. ( , p. 351) p. 351
3. ( , Vol I, p. 144)
4. Discussion historical facts given in (, p. 351) p. 351
5. Kurt Von Fritz (Apr., 1945). "The Discovery of Incommensurability by Hippasus of Metapontum". The Annals of Mathematics, Second Series(Annals of Mathematics) 46 (2): 242–264.
6. Lewis Carroll, "The story with knots", M., Mir, 1985, p. 7
7. Asger Aaboe Episodes from the early history of mathematics. - Mathematical Association of America, 1997. - P. 51. - ISBN 0883856131
8. Pythagorean Proposition by Elisha Scott Loomis
9. Euclid's Elements: Book VI, Proposition VI 31: "In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle."
10. Lawrence S. Leff cited work. - Barron's Educational Series. - P. 326. - ISBN 0764128922
11. Howard Whitley Eves§4.8:...generalization of Pythagorean theorem // Great moments in mathematics (before 1650) . - Mathematical Association of America, 1983. - P. 41. - ISBN 0883853108
12. Tâbit ibn Qorra (full name Thābit ibn Qurra ibn Marwan Al-Ṣābiʾ al-Ḥarrānī) (826-901 AD) was a physician living in Baghdad who wrote extensively on Euclid’s Elements and others mathematical subjects.
13. Aydin Sayili (Mar. 1960). "Thâbit ibn Qurra's Generalization of the Pythagorean Theorem". Isis 51 (1): 35–37. DOI:10.1086/348837.
14. Judith D. Sally, Paul Sally Exercise 2.10(ii) // Cited work . - P. 62. - ISBN 0821844032
15. For the details of such a construction, see George Jennings Figure 1.32: The generalized Pythagorean theorem // Modern geometry with applications: with 150 figures . - 3rd. - Springer, 1997. - P. 23. - ISBN 038794222X
16. Arlen Brown, Carl M. Pearcy item C: Norm for an arbitrary n-tuple ... // An introduction to analysis . - Springer, 1995. - P. 124. - ISBN 0387943692 See also pages 47-50.
17. Alfred Gray, Elsa Abbena, Simon Salamon Modern differential geometry of curves and surfaces with Mathematica . - 3rd. - CRC Press, 2006. - P. 194. - ISBN 1584884487
18. Rajendra Bhatia matrix analysis. - Springer, 1997. - P. 21. - ISBN 0387948465
19. Stephen W. Hawking cited work. - 2005. - P. 4. - ISBN 0762419229
20. Eric W. Weisstein CRC concise encyclopedia of mathematics. - 2nd. - 2003. - P. 2147. - ISBN 1584883472
21. Alexander R. Pruss

Middle level

Right triangle. Complete illustrated guide (2019)

RIGHT TRIANGLE. FIRST LEVEL.

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

and in such

and in such

What is good about a right triangle? Well... first of all, there are special beautiful names for his sides.

Attention to the drawing!

Remember and do not confuse: legs - two, and the hypotenuse - only one(the only, unique and longest)!

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

Pythagorean theorem.

This theorem is the key to solving many problems involving a right 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 very Pythagorean pants and look at them.

Does it really look like 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 like this:

"Sum area of ​​squares, built on the legs, is equal to square area built on the hypotenuse.

Doesn't it sound a little different, doesn't it? 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 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 signs and so on. There were no inscriptions. Can you imagine how terrible it was for the poor ancient students 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 better remember:

Now it should be easy:

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

Well, the most important theorem about a right triangle was discussed. If you're wondering how it's proven, read on. next levels theory, and now let's move on ... to dark forest... trigonometry! To the terrible words sine, cosine, tangent and cotangent.

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

In fact, everything is not so scary at all. Of course, the "real" definition of sine, cosine, tangent and cotangent should be looked at in the article. But you really don't want to, do you? We can rejoice: to solve problems about a right 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.
It actually sounds like this:

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

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

And now, attention! Look what we got:

See how great it is:

Now let's move on to tangent and cotangent.

How to put it into words now? What is the leg in relation to the corner? Opposite, of course - it "lies" opposite the corner. And the cathet? Adjacent to the corner. So what did we get?

See how the numerator and denominator are reversed?

And now again the corners and made the exchange:

Summary

Let's briefly write down what we have learned.

Pythagorean theorem:

The main right triangle theorem is the Pythagorean theorem.

Pythagorean theorem

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

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

You see how cunningly we divided its sides into segments of lengths and!

Now let's connect the marked points

Here we, however, noted something else, but you yourself look at the picture and think about why.

What is the area of ​​the larger square? Correctly, . What about the smaller area? Certainly, . The total area of ​​the four corners remains. Imagine that we took two of them and leaned against each other with hypotenuses. What happened? Two rectangles. So, the area of ​​"cuttings" is equal.

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 triangle, the following relations hold:

Sinus acute angle 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 in the form of a plate:

It is very comfortable!

Signs of equality of right triangles

I. On two legs

II. By leg and hypotenuse

III. By hypotenuse and acute angle

IV. Along the leg and acute angle

a)

b)

Attention! Here it is very important that the legs are "corresponding". For example, if it goes like this:

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

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

Have you noticed how the signs of equality of right triangles differ from the usual signs of equality of triangles? Look at the topic “and pay attention to the fact that for the 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. It's great, right?

Approximately the same situation with signs of similarity of right triangles.

Signs of similarity of right triangles

I. Acute corner

II. On two legs

III. By leg and hypotenuse

Median in a right triangle

Why is it so?

Consider a whole rectangle instead of a right triangle.

Let's draw a diagonal and consider a point - the point of intersection of the diagonals. What do you know about the diagonals of a rectangle?

And what follows from this?

So it happened that

1. - median:

Remember this fact! Helps a lot!

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

What good can be gained from the fact that the median drawn to the hypotenuse is equal to half 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 CIRCUM DEscribed. So what happened?

So let's start with this "besides...".

Let's look at i.

But in similar triangles all angles are equal!

The same can be said about and

Now let's draw it together:

What use can be drawn from this "triple" similarity.

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

We write the relations of the corresponding parties:

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

So, let's apply the similarity: .

What will happen now?

Again we solve the proportion and get the second formula:

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

Pythagorean theorem:

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

Signs of equality of right triangles:

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

Signs of similarity of right triangles:

• one sharp corner: or
• from the proportionality of the two legs:
• from the proportionality of the leg and 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 triangle is the ratio of the opposite leg to the adjacent one:
• The cotangent of an acute angle of a right triangle is the ratio of the adjacent leg to the opposite:.

Height of a right triangle: or.

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

Area of ​​a right triangle:

• through the catheters: