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 area method, axiomatic And exotic evidence(For example,
by using 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 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 of 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 this equation and using the initial conditions, we obtain:
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, the leg b). Then for the integration constant we get:
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 (Fig. 1):
$c^(2)=a^(2)+b^(2)$
Proof of the Pythagorean Theorem
Let triangle $A B C$ be a right triangle with right angle $C$ (Fig. 2).
Let's draw a height from the vertex $C$ to the hypotenuse $A B$, denote the base of the height as $H$ .
Right triangle $A C H$ is similar to triangle $A B C$ in two angles ($\angle A C B=\angle C H A=90^(\circ)$, $\angle A$ is common). Similarly, triangle $C B H$ is similar to $A B C$ .
Introducing the notation
$$B C=a, A C=b, A B=c$$
from the similarity of triangles we get that
$$\frac(a)(c)=\frac(H B)(a), \frac(b)(c)=\frac(A H)(b)$$
Hence we have that
$$a^(2)=c \cdot H B, b^(2)=c \cdot A H$$
Adding the obtained equalities, we obtain
$$a^(2)+b^(2)=c \cdot H B+c \cdot A H$$
$$a^(2)+b^(2)=c \cdot(H B+A H)$$
$$a^(2)+b^(2)=c \cdot A B$$
$$a^(2)+b^(2)=c \cdot c$$
$$a^(2)+b^(2)=c^(2)$$
Q.E.D.
Geometric formulation of the Pythagorean theorem
Theorem
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 the legs (Fig. 2):
Examples of problem solving
Example
Exercise. You are given a right triangle $A B C$ whose legs are 6 cm and 8 cm. Find the hypotenuse of this triangle.
Solution. According to the condition of the leg $a=6$ cm, $b=8$ cm. Then, according to the Pythagorean theorem, the square of the hypotenuse
$c^(2)=a^(2)+b^(2)=6^(2)+8^(2)=36+64=100$
Hence we get that the required hypotenuse
$c=\sqrt(100)=10$ (cm)
Answer. 10 cm
Example
Exercise. Find the area of a right triangle if it is known that one of its legs is 5 cm longer than the other, and the hypotenuse is 25 cm.
Solution. Let $x$ cm be the length of the smaller leg, then $(x+5)$ cm is the length of the larger one. Then, according to the Pythagorean theorem, we have:
$$x^(2)+(x+5)^(2)=25^(2)$$
We open the brackets, reduce similar ones and solve the resulting quadratic equation:
$x^(2)+5 x-300=0$
According to Vieta's theorem, we get that
$x_(1)=15$ (cm) , $x_(2)=-20$ (cm)
The value of $x_(2)$ does not satisfy the condition of the problem, which means that the smaller leg is 15 cm, and the larger one is 20 cm.
The area of a right triangle is half the product of the lengths of its legs, that is
$$S=\frac(15 \cdot 20)(2)=15 \cdot 10=150\left(\mathrm(cm)^(2)\right)$$
Answer.$S=150\left(\mathrm(cm)^(2)\right)$
Historical reference
Pythagorean theorem- one of the fundamental theorems of Euclidean geometry, establishing the relationship between the sides of a right triangle.
The ancient Chinese book "Zhou bi suan jing" speaks of a Pythagorean triangle with sides 3, 4 and 5. The largest German historian of mathematics Moritz Kantor (1829 - 1920) believes that the equality $3^(2)+4^(2)=5^ (2) $ was already known to the Egyptians around 2300 BC. According to the scientist, the builders then built right angles using right-angled triangles with sides 3, 4 and 5. Somewhat more is known about the Pythagorean theorem among the Babylonians. One text gives an approximate calculation of the hypotenuse of an isosceles right triangle.
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.
The potential for creativity is usually attributed to the humanities, leaving the natural scientific analysis, practical approach and dry language of formulas and numbers. Mathematics cannot be classified as a humanities subject. But without creativity in the "queen of all sciences" you will not go far - people have known about this for a long time. Since the time of Pythagoras, for example.
School textbooks, unfortunately, usually do not explain that in mathematics it is important not only to cram theorems, axioms and formulas. It is important to understand and feel its fundamental principles. And at the same time, try to free your mind from clichés and elementary truths - only in such conditions are all great discoveries born.
Such discoveries include the one that today we know as the Pythagorean theorem. With its help, we will try to show that mathematics not only can, but should be fun. And that this adventure is suitable not only for nerds in thick glasses, but for everyone who is strong in mind and strong in spirit.
From the history of the issue
Strictly speaking, although the theorem is called the "Pythagorean theorem", Pythagoras himself did not discover it. The right triangle and its special properties have been studied long before it. There are two polar points of view on this issue. According to one version, Pythagoras was the first to find a complete proof of the theorem. According to another, the proof does not belong to the authorship of Pythagoras.
Today you can no longer check who is right and who is wrong. It is only known that the proof of Pythagoras, if it ever existed, has not survived. However, there are suggestions that the famous proof from Euclid's Elements may belong to Pythagoras, and Euclid only recorded it.
It is also known today that problems about a right-angled triangle are found in Egyptian sources from the time of Pharaoh Amenemhet I, on Babylonian clay tablets from the reign of King Hammurabi, in the ancient Indian treatise Sulva Sutra and the ancient Chinese work Zhou-bi suan jin.
As you can see, the Pythagorean theorem has occupied the minds of mathematicians since ancient times. Approximately 367 various pieces of evidence that exist today serve as confirmation. No other theorem can compete with it in this respect. Notable evidence authors include Leonardo da Vinci and the 20th President of the United States, James Garfield. All this speaks of the extreme importance of this theorem for mathematics: most of the theorems of geometry are derived from it or, in one way or another, connected with it.
Proofs of the Pythagorean theorem
School textbooks mostly give algebraic proofs. But the essence of the theorem is in geometry, so let's first of all consider those proofs of the famous theorem that are based on this science.
Proof 1
For the simplest proof of the Pythagorean theorem for a right triangle, you need to set ideal conditions: let the triangle be not only right-angled, but also isosceles. There is reason to believe that it was such a triangle that was originally considered by ancient mathematicians.
Statement "a square built on the hypotenuse of a right triangle is equal to the sum of the squares built on its legs" can be illustrated with the following drawing:
Look at the isosceles right triangle ABC: On the hypotenuse AC, you can build a square consisting of four triangles equal to the original ABC. And on the legs AB and BC built on a square, each of which contains two similar triangles.
By the way, this drawing formed the basis of numerous anecdotes and cartoons dedicated to the Pythagorean theorem. Perhaps the most famous is "Pythagorean pants are equal in all directions":
Proof 2
This method combines algebra and geometry and can be seen as a variant of the ancient Indian proof of the mathematician Bhaskari.
Construct a right triangle with sides a, b and c(Fig. 1). Then build two squares with sides equal to the sum of the lengths of the two legs - (a+b). In each of the squares, make constructions, as in figures 2 and 3.
In the first square, build four of the same triangles as in Figure 1. As a result, two squares are obtained: one with side a, the second with side b.
In the second square, four similar triangles constructed form a square with a side equal to the hypotenuse c.
The sum of the areas of the constructed squares in Fig. 2 is equal to the area of the square we constructed with side c in Fig. 3. This can be easily verified by calculating the areas of the squares in Fig. 2 according to the formula. And the area of the inscribed square in Figure 3. by subtracting the areas of four equal right-angled triangles inscribed in the square from the area of \u200b\u200ba large square with a side (a+b).
Putting all this down, we have: a 2 + b 2 \u003d (a + b) 2 - 2ab. Expand the brackets, do all the necessary algebraic calculations and get that a 2 + b 2 = a 2 + b 2. At the same time, the area of the inscribed in Fig.3. square can also be calculated using the traditional formula S=c2. Those. a2+b2=c2 You have proved the Pythagorean theorem.
Proof 3
The very same ancient Indian proof is described in the 12th century in the treatise “The Crown of Knowledge” (“Siddhanta Shiromani”), and as the main argument the author uses an appeal addressed to the mathematical talents and powers of observation of students and followers: “Look!”.
But we will analyze this proof in more detail:
Inside the square, build four right-angled triangles as indicated in the drawing. The side of the large square, which is also the hypotenuse, is denoted With. Let's call the legs of the triangle A And b. According to the drawing, the side of the inner square is (a-b).
Use the square area formula S=c2 to calculate the area of the outer square. And at the same time calculate the same value by adding the area of the inner square and the area of \u200b\u200ball four right triangles: (a-b) 2 2+4*1\2*a*b.
You can use both options to calculate the area of a square to make sure they give the same result. And that gives you the right to write down that c 2 =(a-b) 2 +4*1\2*a*b. As a result of the solution, you will get the formula of the Pythagorean theorem c2=a2+b2. The theorem has been proven.
Proof 4
This curious ancient Chinese evidence was called the "Bride's Chair" - because of the chair-like figure that results from all the constructions:
It uses the drawing we have already seen in Figure 3 in the second proof. And the inner square with side c is constructed in the same way as in the ancient Indian proof given above.
If you mentally cut off two green right-angled triangles from the drawing in Fig. 1, move them to opposite sides of the square with side c and attach the hypotenuses to the hypotenuses of the lilac triangles, you get a figure called “bride’s chair” (Fig. 2). For clarity, you can do the same with paper squares and triangles. You will see that the "bride's chair" is formed by two squares: small ones with a side b and big with a side a.
These constructions allowed the ancient Chinese mathematicians and us following them to come to the conclusion that c2=a2+b2.
Proof 5
This is another way to find a solution to the Pythagorean theorem based on geometry. It's called the Garfield Method.
Construct a right triangle ABC. We need to prove that BC 2 \u003d AC 2 + AB 2.
To do this, continue the leg AC and build a segment CD, which is equal to the leg AB. Lower Perpendicular AD line segment ED. Segments ED And AC are equal. connect the dots E And IN, and E And WITH and get a drawing like the picture below:
To prove the tower, we again resort to the method we have already tested: we find the area of the resulting figure in two ways and equate the expressions to each other.
Find the area of a polygon ABED can be done by adding the areas of the three triangles that form it. And one of them ERU, is not only rectangular, but also isosceles. Let's also not forget that AB=CD, AC=ED And BC=CE- this will allow us to simplify the recording and not overload it. So, S ABED \u003d 2 * 1/2 (AB * AC) + 1 / 2BC 2.
At the same time, it is obvious that ABED is a trapezoid. Therefore, we calculate its area using the formula: SABED=(DE+AB)*1/2AD. For our calculations, it is more convenient and clearer to represent the segment AD as the sum of the segments AC And CD.
Let's write both ways to calculate the area of a figure by putting an equal sign between them: AB*AC+1/2BC 2 =(DE+AB)*1/2(AC+CD). We use the equality of segments already known to us and described above to simplify the right-hand side of the notation: AB*AC+1/2BC 2 =1/2(AB+AC) 2. And now we open the brackets and transform the equality: AB*AC+1/2BC 2 =1/2AC 2 +2*1/2(AB*AC)+1/2AB 2. Having finished all the transformations, we get exactly what we need: BC 2 \u003d AC 2 + AB 2. We have proved the theorem.
Of course, this list of evidence is far from complete. The Pythagorean theorem can also be proved using vectors, complex numbers, differential equations, stereometry, etc. And even physicists: if, for example, liquid is poured into square and triangular volumes similar to those shown in the drawings. By pouring liquid, it is possible to prove the equality of areas and the theorem itself as a result.
A few words about Pythagorean triplets
This issue is little or not studied in the school curriculum. Meanwhile, it is very interesting and is of great importance in geometry. Pythagorean triples are used to solve many mathematical problems. The idea of them can be useful to you in further education.
So what are Pythagorean triplets? So called natural numbers, collected in threes, the sum of the squares of two of which is equal to the third number squared.
Pythagorean triples can be:
- primitive (all three numbers are relatively prime);
- non-primitive (if each number of a triple is multiplied by the same number, you get a new triple that is not primitive).
Even before our era, the ancient Egyptians were fascinated by the mania for the numbers of Pythagorean triplets: in tasks they considered a right-angled triangle with sides of 3.4 and 5 units. By the way, any triangle whose sides are equal to the numbers from the Pythagorean triple is by default rectangular.
Examples of Pythagorean triples: (3, 4, 5), (6, 8, 10), (5, 12, 13), (9, 12, 15), (8, 15, 17), (12, 16, 20) ), (15, 20, 25), (7, 24, 25), (10, 24, 26), (20, 21, 29), (18, 24, 30), (10, 30, 34), (21, 28, 35), (12, 35, 37), (15, 36, 39), (24, 32, 40), (9, 40, 41), (27, 36, 45), (14 , 48, 50), (30, 40, 50) etc.
Practical application of the theorem
The Pythagorean theorem finds application not only in mathematics, but also in architecture and construction, astronomy, and even literature.
First, about construction: the Pythagorean theorem is widely used in it in problems of different levels of complexity. For example, look at the Romanesque window:
Let's denote the width of the window as b, then the radius of the great semicircle can be denoted as R and express through b: R=b/2. The radius of smaller semicircles can also be expressed in terms of b: r=b/4. In this problem, we are interested in the radius of the inner circle of the window (let's call it p).
The Pythagorean theorem just comes in handy to calculate R. To do this, we use a right-angled triangle, which is indicated by a dotted line in the figure. The hypotenuse of a triangle consists of two radii: b/4+p. One leg is a radius b/4, another b/2-p. Using the Pythagorean theorem, we write: (b/4+p) 2 =(b/4) 2 +(b/2-p) 2. Next, we open the brackets and get b 2 /16+ bp / 2 + p 2 \u003d b 2 / 16 + b 2 / 4-bp + p 2. Let's transform this expression into bp/2=b 2 /4-bp. And then we divide all the terms into b, we give similar ones to get 3/2*p=b/4. And in the end we find that p=b/6- which is what we needed.
Using the theorem, you can calculate the length of the rafters for a gable roof. Determine how high a mobile tower is needed for the signal to reach a certain settlement. And even steadily install a Christmas tree in the city square. As you can see, this theorem lives not only on the pages of textbooks, but is often useful in real life.
As far as literature is concerned, the Pythagorean theorem has inspired writers since antiquity and continues to do so today. For example, the nineteenth-century German writer Adelbert von Chamisso was inspired by her to write a sonnet:
The light of truth will not soon dissipate,
But, having shone, it is unlikely to dissipate
And, like thousands of years ago,
Will not cause doubts and disputes.
The wisest when it touches the eye
Light of truth, thank the gods;
And a hundred bulls, stabbed, lie -
The return gift of the lucky Pythagoras.
Since then, the bulls have been roaring desperately:
Forever aroused the bull tribe
event mentioned here.
They think it's about time
And again they will be sacrificed
Some great theorem.
(translated by Viktor Toporov)
And in the twentieth century, the Soviet writer Yevgeny Veltistov in his book "The Adventures of Electronics" devoted a whole chapter to the proofs of the Pythagorean theorem. And half a chapter of the story about the two-dimensional world that could exist if the Pythagorean theorem became the fundamental law and even religion for a single world. It would be much easier to live in it, but also much more boring: for example, no one there understands the meaning of the words “round” and “fluffy”.
And in the book “The Adventures of Electronics”, the author, through the mouth of the mathematics teacher Taratara, says: “The main thing in mathematics is the movement of thought, new ideas.” It is this creative flight of thought that generates the Pythagorean theorem - it is not for nothing that it has so many diverse proofs. It helps to go beyond the usual, and look at familiar things in a new way.
Conclusion
This article was created so that you can look beyond the school curriculum in mathematics and learn not only those proofs of the Pythagorean theorem that are given in the textbooks "Geometry 7-9" (L.S. Atanasyan, V.N. Rudenko) and "Geometry 7 -11” (A.V. Pogorelov), but also other curious ways to prove the famous theorem. And also see examples of how the Pythagorean theorem can be applied in everyday life.
Firstly, this information will allow you to claim higher scores in math classes - information on the subject from additional sources is always highly appreciated.
Secondly, we wanted to help you get a feel for how interesting mathematics is. To be convinced by specific examples that there is always a place for creativity in it. We hope that the Pythagorean theorem and this article will inspire you to do your own research and exciting discoveries in mathematics and other sciences.
Tell us in the comments if you found the evidence presented in the article interesting. Did you find this information helpful in your studies? Let us know what you think about the Pythagorean theorem and this article - we will be happy to discuss all this with you.
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According to van der Waerden, it is very likely that the ratio in general form was already known in Babylon around the 18th century BC. e.
Approximately 400 BC. e., according to Proclus, Plato gave a method for finding Pythagorean triples, combining algebra and geometry. Around 300 B.C. e. in the "Elements" of Euclid appeared the oldest axiomatic proof of the Pythagorean theorem.
Wording
The main formulation contains algebraic operations - in a right triangle, the lengths of the legs of which are equal a (\displaystyle a) And b (\displaystyle b), and the length of the hypotenuse is c (\displaystyle c), the relation is fulfilled:
.An equivalent geometric formulation is also possible, resorting to the concept of area figure: 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 the legs. In this form, the theorem is formulated in Euclid's Principia.
Inverse Pythagorean Theorem- the statement about the rectangularity of any triangle, the lengths of the sides of which are related by the relation a 2 + b 2 = c 2 (\displaystyle a^(2)+b^(2)=c^(2)). As a consequence, for any triple of positive numbers a (\displaystyle a), b (\displaystyle b) And c (\displaystyle c), such that a 2 + b 2 = c 2 (\displaystyle a^(2)+b^(2)=c^(2)), there is a right triangle with legs a (\displaystyle a) And b (\displaystyle b) and hypotenuse c (\displaystyle c).
Proof
At least 400 proofs of the Pythagorean theorem have been recorded in the scientific literature, which is explained both by the fundamental value for geometry and by the elementary nature of the result. The main directions of proofs are: algebraic use of the ratios of elements triangle (such, for example, is the popular similarity method), area method, there are also various exotic proofs (for example, using differential equations).
Through similar triangles
Euclid's classical proof aims to establish the equality of the areas between the rectangles formed by dissecting the square above the hypotenuse with the height from the right angle with the squares above the legs.
The construction used for the proof is as follows: for a right triangle with a right angle C (\displaystyle C), squares over the legs and and squares over the hypotenuse A B I K (\displaystyle ABIK) height is being built C H (\displaystyle CH) and the beam that continues it s (\displaystyle s), dividing the square above the hypotenuse into two rectangles and . The proof is aimed at establishing the equality of the areas of the rectangle A H J K (\displaystyle AHJK) with a square over the leg A C (\displaystyle AC); the equality of the areas of the second rectangle, which is a square above the hypotenuse, and the rectangle above the other leg is established in a similar way.
Equality of the areas of a rectangle A H J K (\displaystyle AHJK) And A C E D (\displaystyle ACED) established through the congruence of triangles △ A C K (\displaystyle \triangle ACK) And △ A B D (\displaystyle \triangle ABD), the area of each of which is equal to half the area of squares A H J K (\displaystyle AHJK) And A C E D (\displaystyle ACED) respectively, in connection with the following property: the area of a triangle is equal to half the area of a rectangle if the figures have a common side, and the height of the triangle to the common side is the other side of the rectangle. The congruence of triangles follows from the equality of two sides (sides of squares) and the angle between them (composed of a right angle and an angle at A (\displaystyle A).
Thus, the proof establishes that the area of the square above the hypotenuse, composed of rectangles A H J K (\displaystyle AHJK) And B H J I (\displaystyle BHJI), is equal to the sum of the areas of the squares above the legs.
Proof of Leonardo da Vinci
The area method also includes the proof found by Leonardo da Vinci. Let there be a right triangle △ A B C (\displaystyle \triangle ABC) right angle C (\displaystyle C) and squares A C E D (\displaystyle ACED), B C F G (\displaystyle BCFG) And A B H J (\displaystyle ABHJ)(see picture). In this proof on the side H J (\displaystyle HJ) the latter, a triangle is constructed to the outside, congruent △ A B C (\displaystyle \triangle ABC), moreover, reflected both relative to the hypotenuse and relative to the height to it (that is, J I = B C (\displaystyle JI=BC) And H I = A C (\displaystyle HI=AC)). Straight C I (\displaystyle CI) splits the square built on the hypotenuse into two equal parts, since triangles △ A B C (\displaystyle \triangle ABC) And △ J H I (\displaystyle \triangle JHI) are equal in construction. The proof establishes the congruence of quadrilaterals C A J I (\displaystyle CAJI) And D A B G (\displaystyle DABG), the area of each of which, on the one hand, is equal to the sum of half the areas of the squares on the legs and the area of the original triangle, on the other hand, to half the area of the square on the hypotenuse plus the area of the original triangle. In total, half the sum of the areas of the squares over the legs is equal to half the area of the square over the hypotenuse, which is equivalent to the geometric formulation of the Pythagorean theorem.
Proof by the infinitesimal method
There are several proofs using the technique of differential equations. In particular, Hardy is credited with a proof using infinitesimal leg increments a (\displaystyle a) And b (\displaystyle b) and hypotenuse c (\displaystyle c), and preserving the similarity with the original rectangle, that is, ensuring the fulfillment of the following differential relations:
d a d c = c a (\displaystyle (\frac (da)(dc))=(\frac (c)(a))), d b d c = c b (\displaystyle (\frac (db)(dc))=(\frac (c)(b))).By the method of separation of variables, a differential equation is derived from them c d c = a d a + b d b (\displaystyle c\ dc=a\,da+b\,db), whose integration gives the relation c 2 = a 2 + b 2 + C o n s t (\displaystyle c^(2)=a^(2)+b^(2)+\mathrm (Const) ). Application of initial conditions a = b = c = 0 (\displaystyle a=b=c=0) defines a constant as 0, which results in the assertion of the theorem.
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.
Variations and Generalizations
Similar geometric shapes on three sides
An important geometric generalization of the Pythagorean theorem was given by Euclid in the "Beginnings", moving from the areas of squares on the sides to the areas of arbitrary similar geometric figures: the sum of the areas of such figures built on the legs will be equal to the area of a figure similar to them, built on the hypotenuse.
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 (\displaystyle A), B (\displaystyle B) And C (\displaystyle C) built on legs with lengths a (\displaystyle a) And b (\displaystyle b) and hypotenuse c (\displaystyle c) accordingly, there is a relation:
A a 2 = B b 2 = C c 2 ⇒ A + B = a 2 c 2 C + b 2 c 2 C (\displaystyle (\frac (A)(a^(2)))=(\frac (B )(b^(2)))=(\frac (C)(c^(2)))\,\Rightarrow \,A+B=(\frac (a^(2))(c^(2) ))C+(\frac (b^(2))(c^(2)))C).Since according to the Pythagorean theorem a 2 + b 2 = c 2 (\displaystyle a^(2)+b^(2)=c^(2)), then it is done.
In addition, if it is possible to prove without using the Pythagorean theorem that for the areas of three similar geometric figures on the sides of a right triangle, the relation A + B = C (\displaystyle A+B=C), then using the reverse of the proof of Euclid's generalization, we can derive the proof of the Pythagorean theorem. For example, if on the hypotenuse we construct a right triangle congruent to the initial one with area C (\displaystyle C), and on the legs - two similar right-angled triangles with areas A (\displaystyle A) And B (\displaystyle B), then it turns out that the triangles on the legs are formed as a result of dividing the initial triangle by its height, that is, the sum of two smaller areas of the triangles is equal to the area of the third, thus A + B = C (\displaystyle A+B=C) and, applying the relation for similar figures, the Pythagorean theorem is derived.
Cosine theorem
The Pythagorean theorem is a special case of the more general cosine theorem which relates the lengths of the sides in an arbitrary triangle:
a 2 + b 2 − 2 a b cos θ = c 2 (\displaystyle a^(2)+b^(2)-2ab\cos (\theta )=c^(2)),where is the angle between the sides a (\displaystyle a) And b (\displaystyle b). If the angle is 90°, then cos θ = 0 (\displaystyle \cos \theta =0), and the formula simplifies to the usual Pythagorean theorem.
Arbitrary triangle
There is a generalization of the Pythagorean theorem to an arbitrary triangle, operating solely on the ratio of the lengths of the sides, it is believed that it was first established by the Sabian astronomer Sabit ibn Kurra. In it, for an arbitrary triangle with sides, an isosceles triangle with a base on the side c (\displaystyle c), the vertex coinciding with the vertex of the original triangle, opposite the side c (\displaystyle c) and angles at the base equal to the angle θ (\displaystyle \theta ) opposite side c (\displaystyle c). As a result, two triangles are formed, similar to the original one: the first one with sides a (\displaystyle a), the lateral side of the inscribed isosceles triangle far from it, and r (\displaystyle r)- side parts c (\displaystyle c); the second is symmetrical to it from the side b (\displaystyle b) with a party s (\displaystyle s)- the relevant part of the side c (\displaystyle c). As a result, the relation is fulfilled:
a 2 + b 2 = c (r + s) (\displaystyle a^(2)+b^(2)=c(r+s)),which degenerates into the Pythagorean theorem at θ = π / 2 (\displaystyle \theta =\pi /2). The ratio is a consequence of the similarity of the formed triangles:
c a = a r , c b = b s ⇒ c r + c s = a 2 + b 2 (\displaystyle (\frac (c)(a))=(\frac (a)(r)),\,(\frac (c) (b))=(\frac (b)(s))\,\Rightarrow \,cr+cs=a^(2)+b^(2)).Pappus area theorem
Non-Euclidean geometry
The Pythagorean theorem is derived from the axioms of Euclidean geometry and is invalid for non-Euclidean geometry - the fulfillment of the Pythagorean theorem is equivalent to the postulate of Euclidean parallelism.
In non-Euclidean geometry, the relationship between the sides of a right triangle will necessarily be in a form different from the Pythagorean theorem. For example, in spherical geometry, all three sides of a right triangle, which bound the octant of the unit sphere, have length π / 2 (\displaystyle \pi /2), which contradicts the Pythagorean theorem.
Moreover, the Pythagorean theorem is valid in hyperbolic and elliptic geometry, if the requirement that the triangle is rectangular is replaced by the condition that the sum of two angles of the triangle must be equal to the third.
spherical geometry
For any right triangle on a sphere with radius R (\displaystyle R)(for example, if the angle in the triangle is right) with sides a , b , c (\displaystyle a,b,c) the relationship between the sides is:
cos (c R) = cos (a R) ⋅ cos (b R) (\displaystyle \cos \left((\frac (c)(R))\right)=\cos \left((\frac (a)(R))\right)\cdot \cos \left((\frac (b)(R))\right)).This equality can be derived as a special case of the spherical cosine theorem, which is valid for all spherical triangles:
cos (c R) = cos (a R) ⋅ cos (b R) + sin (a R) ⋅ sin (b R) ⋅ cos γ (\displaystyle \cos \left((\frac ( c)(R))\right)=\cos \left((\frac (a)(R))\right)\cdot \cos \left((\frac (b)(R))\right)+\ sin \left((\frac (a)(R))\right)\cdot \sin \left((\frac (b)(R))\right)\cdot \cos \gamma ). ch c = ch a ⋅ ch b (\displaystyle \operatorname (ch) c=\operatorname (ch) a\cdot \operatorname (ch) b),Where ch (\displaystyle \operatorname (ch) )- hyperbolic cosine. This formula is a special case of the hyperbolic cosine theorem, which is valid for all triangles:
ch c = ch a ⋅ ch b − sh a ⋅ sh b ⋅ cos γ (\displaystyle \operatorname (ch) c=\operatorname (ch) a\cdot \operatorname (ch) b-\operatorname (sh) a\cdot \operatorname (sh) b\cdot \cos \gamma ),Where γ (\displaystyle \gamma )- an angle whose vertex is opposite to a side c (\displaystyle c).
Using the Taylor series for the hyperbolic cosine ( ch x ≈ 1 + x 2 / 2 (\displaystyle \operatorname (ch) x\approx 1+x^(2)/2)) it can be shown that if the hyperbolic triangle decreases (that is, when a (\displaystyle a), b (\displaystyle b) And c (\displaystyle c) tend to zero), then the hyperbolic relations in a right triangle approach the relation of the classical Pythagorean theorem.
Application
Distance in two-dimensional rectangular systems
The most important application of the Pythagorean theorem is to determine the distance between two points in a rectangular system coordinates: distance s (\displaystyle s) between points with coordinates (a , b) (\displaystyle (a,b)) And (c , d) (\displaystyle (c,d)) equals:
s = (a − c) 2 + (b − d) 2 (\displaystyle s=(\sqrt ((a-c)^(2)+(b-d)^(2)))).For complex numbers, the Pythagorean theorem gives a natural formula for finding the modulus complex number - for z = x + y i (\displaystyle z=x+yi) it is equal to the length
Pythagoras is a Greek scientist who lived about 2500 years ago (564-473 BC).
Let a right triangle be given whose sides A, b And With(Fig. 267).
Let's build squares on its sides. The areas of these squares are respectively A 2 , b 2 and With 2. Let's prove that With 2 = a 2 +b 2 .
Let us construct two squares MKOR and M'K'O'R' (Fig. 268, 269), taking for the side of each of them a segment equal to the sum of the legs of the right triangle ABC.
Having completed the constructions shown in Figures 268 and 269 in these squares, we will see that the MKOR square is divided into two squares with areas A 2 and b 2 and four equal right triangles, each of which is equal to right triangle ABC. The square M'K'O'R' is divided into a quadrilateral (it is shaded in Figure 269) and four right-angled triangles, each of which is also equal to the triangle ABC. The shaded quadrilateral is a square, since its sides are equal (each is equal to the hypotenuse of the triangle ABC, i.e. With), and the angles are straight lines ∠1 + ∠2 = 90°, whence ∠3 = 90°).
Thus, the sum of the areas of the squares built on the legs (in Figure 268 these squares are shaded) is equal to the area of the MKOR square without the sum of the areas of four equal triangles, and the area of the square built on the hypotenuse (in Figure 269 this square is also shaded) is equal to the area of the square M'K'O'R', equal to the square of MKOR, without the sum of the areas of four similar triangles. Therefore, the area of the square built on the hypotenuse of a right triangle is equal to the sum of the areas of the squares built on the legs.
We get the formula With 2 = a 2 +b 2 , where With- hypotenuse, A And b- legs of a right triangle.
The Pythagorean theorem can be summarized as follows:
The square of the hypotenuse of a right triangle is equal to the sum of the squares of the legs.
From the formula With 2 = a 2 +b 2 you can get the following formulas:
A 2 = With 2 - b 2 ;
b 2 = With 2 - A 2 .
These formulas can be used to find the unknown side of a right triangle given two of its sides.
For example:
a) if legs are given A= 4 cm, b\u003d 3 cm, then you can find the hypotenuse ( With):
With 2 = a 2 +b 2 , i.e. With 2 = 4 2 + 3 2 ; with 2 = 25, whence With= √25 = 5(cm);
b) if the hypotenuse is given With= 17 cm and leg A= 8 cm, then you can find another leg ( b):
b 2 = With 2 - A 2 , i.e. b 2 = 17 2 - 8 2 ; b 2 = 225, whence b= √225 = 15 (cm).
Corollary: If in two right triangles ABC and A 1 B 1 C 1 hypotenuse With And With 1 are equal, and the leg b triangle ABC is greater than the leg b 1 triangle A 1 B 1 C 1,
then the leg A triangle ABC is less than the leg A 1 triangle A 1 B 1 C 1 .
Indeed, based on the Pythagorean theorem, we get:
A 2 = With 2 - b 2 ,
A 1 2 = With 1 2 - b 1 2
In the written formulas, the minuends are equal, and the subtrahend in the first formula is greater than the subtrahend in the second formula, therefore, the first difference is less than the second,
i.e. A 2 a 1 2 . Where A a 1 .
