When you first started learning square roots and how to solve irrational equations (equalities containing an unknown under the root sign), you probably got the first idea about them. practical use... Ability to extract Square root of numbers is also necessary for solving problems on the application of the Pythagorean theorem. This theorem connects the lengths of the sides of any right triangle.
Let the lengths of the legs of a right-angled triangle (those two sides that converge at right angles) be denoted by the letters and, and the length of the hypotenuse (the longest side of the triangle opposite right angle) will be indicated by a letter. Then the corresponding lengths are related by the following relation:
This equation allows you to find the length of the side of a right-angled triangle in the case when the length of its other two sides is known. In addition, it allows you to determine whether the triangle in question is right-angled, provided that the lengths of all three sides are known in advance.
Solving problems using the Pythagorean theorem
To consolidate the material, we will solve the following problems on the application of the Pythagorean theorem.
So, given:
- The length of one of the legs is 48, the hypotenuse is 80.
- The length of the leg is 84, the hypotenuse is 91.
Let's start solving:
a) Substitution of data into the above equation gives the following results:
48 2 + b 2 = 80 2
2304 + b 2 = 6400
b 2 = 4096
b= 64 or b = -64
Since the side length of a triangle cannot be expressed negative number, the second option is automatically discarded.
Answer to the first figure: 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 decision is discarded.
Answer to the second figure: b = 35
We are given:
- The lengths of the smaller sides of the triangle are 45 and 55, respectively, and the larger ones are 75.
- 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 the 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 right-angled.
b) The same operation is performed:
28 2 + 45 2 = 784 + 2025 = 2809
Therefore, the second triangle is right-angled.
First, find the length of the largest segment formed by the points with coordinates (-2, -3) and (5, -2). To do this, we use the well-known formula for finding 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 the points with coordinates (2, 1) and (5, -2):
Since the equality holds:
then the corresponding triangle is right-angled.
Thus, we can formulate the answer to the problem: since the sum of the squares of the sides with the shortest length is equal to the square of the side with the greatest length, the points are the vertices of a right-angled triangle.
The base (located strictly horizontally), the jamb (located strictly vertically) and the cable (extended diagonally) form a right-angled 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 (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-angled triangle, the Pythagorean theorem can be used to find the length of the third side:
So, the width of the pond is 10 meters.
Sergey Valerievich
Different ways of proving the Pythagorean theorem
student of grade 9 "A"
MOU SOSH №8
Supervisor:
mathematic teacher,
MOU SOSH №8
Art. Novorozhdestvenskaya
Krasnodar Territory.
Art. Novorozhdestvenskaya
ANNOTATION.
The Pythagorean theorem is rightfully considered the most important in the course of geometry and deserves close attention. It is the basis for solving many geometric problems, the basis for studying the theoretical and practical course of geometry in the future. The theorem is surrounded by the richest historical material related to its appearance and methods of proof. Studying the history of the development of geometry instills a love for this subject, contributes to the development of cognitive interest, general culture and creativity, and also develops skills research work.
As a result of the search activity, the goal of the work was achieved, which is to replenish and generalize knowledge on the proof of the Pythagorean theorem. I managed to find and consider different ways proof and deepen knowledge on the topic, going beyond the pages of a school textbook.
The collected material even more convinces that the Pythagorean theorem is the great theorem of geometry, has great theoretical and practical significance.
Introduction. Historical reference 5 Main part 8
3. Conclusion 19
4. Used literature 20
1. INTRODUCTION. HISTORICAL REFERENCE.
The essence of truth is that it is for us forever,
When we see the light at least once in her insight,
And the Pythagorean theorem after so many years
For us, as for him, it is indisputable, flawless.
To delight the gods, Pythagoras made a vow:
For having touched the endless wisdom,
He killed a hundred bulls, thanks to the eternal;
He offered prayers and praises to the victim after him.
Since then, the bulls, when they smell, pushing,
That the trail again leads people to the new truth,
They roar furiously, so there is no urine to listen,
Such a Pythagoras instilled terror in them forever.
Bulls, powerless to resist the new truth,
What remains? - Just closing your eyes, roar, tremble.
It is not known how Pythagoras proved his theorem. What is certain is that he discovered it under the strong influence of Egyptian science. A special case Pythagoras' theorems - the properties of a triangle with sides 3, 4 and 5 - were known to the builders of the pyramids long before the birth of Pythagoras, but he himself studied for more than 20 years with the Egyptian priests. A legend has survived, which says that, having proved his famous theorem, Pythagoras sacrificed a bull to the gods, and according to other sources, even 100 bulls. This, however, contradicts information about the moral and religious views of Pythagoras. In literary sources, you can read that he "forbade even to kill animals, and even more so to feed them, because animals have a soul, like us." Pythagoras ate only honey, bread, vegetables and occasionally fish. In connection with all this, the following entry can be considered more plausible: "... and even when he discovered that in a right-angled triangle the hypotenuse has a correspondence with the legs, he sacrificed a bull made from wheat dough."
The popularity of the Pythagorean theorem is so great that its proofs are found even in fiction, for example, in the story of the famous English writer Huxley "Young Archimedes". The same Proof, but for the particular case of an isosceles right-angled triangle, is given in Plato's Menon dialogue.
Fairy tale "House".
“Far, far away, where even airplanes do not fly, is the country of Geometry. In this unusual country there was one amazing city - the city of Theorem. Once I came to this city beautiful girl named Hypotenuse. She tried to rent a room, but wherever she turned, she was refused everywhere. Finally she went to the rickety house and knocked. She was opened by a man who called himself the Right Angle, and he invited Hypotenuse to live with him. The hypotenuse remained in the house where Right Angle and his two young sons named Cathety lived. Since then, life in the House of Right Angle has changed in a new way. The hypotenuse planted flowers on the window, and red roses in the front garden. The house took the shape of a right-angled triangle. Both legs really liked Hypotenuse and asked her to stay forever in their house. Lo in the evenings, this friendly family gathers at the family table. Sometimes Right Angle plays hide and seek with his kids. Most often he has to look, and Hypotenuse hides so skillfully that it can be very difficult to find her. One day while playing, Right Angle noticed interesting property: if he manages to find the legs, then it is not difficult to find the Hypotenuse. So Right Angle uses this pattern, I must say, very successfully. The Pythagorean theorem is based on the property of this right-angled triangle. "
(From the book by A. Okunev "Thank you for the lesson, children").
A playful formulation of the theorem:
If we are given a triangle
And, moreover, with a right angle,
Then the square of the hypotenuse
We will always easily find:
We erect the legs in a square,
We find the sum of the degrees -
And in such a simple way
We will come to the result.
Studying algebra and the beginnings of analysis and geometry in 10th grade, I made sure that in addition to what was discussed in 8th grade There are other ways of proving the Pythagorean theorem. I present them for your review.
2. MAIN PART.
Theorem. In a right triangle, a square
hypotenuse is equal to the sum squares of legs.
1 METHOD.
Using the properties of the areas of polygons, we will establish a remarkable relationship between the hypotenuse and the legs of a right-angled triangle.
Proof.
a, in and hypotenuse with(Fig. 1, a).
Let us prove that c² = a² + b².
Proof.
Let's build a triangle to a square with a side a + b as shown in fig. 1, b. The area S of this square is equal to (a + b) ². On the other hand, this square is made up of four equal right-angled triangles, each of which is ½ aw, and a square with side with, therefore S = 4 * ½ av + s² = 2av + s².
Thus,
(a + b) ² = 2 av + s²,
c² = a² + b².
The theorem is proved.
2 METHOD.
After studying the topic "Similar triangles" I found out that it is possible to apply the similarity of triangles to the proof of the Pythagorean theorem. Namely, I used the statement that the leg of a right-angled triangle is the proportional average for the hypotenuse and the segment of the hypotenuse enclosed between the leg and the height drawn from the top of the right angle.
Consider a right-angled triangle with a right angle С, СD– height (Fig. 2). Let us prove that AS² + CB² = AB²
.
Proof.
Based on the statement about the leg of a right-angled triangle:
AC =, SV =.
Let's square and add the resulting equalities:
AC² = AB * AD, CB² = AB * DB;
AC² + CB² = AB * (AD + DB), where AD + DB = AB, then
AC² + SV² = AB * AB,
AC² + CB² = AB².
The proof is complete.
3 METHOD.
The definition of the cosine of an acute angle of a right triangle can be applied to the proof of the Pythagorean theorem. Consider fig. 3.
Proof:
Let ABC be a given right-angled triangle with right angle C. Draw the height CD from the vertex of right angle C.
By definition of the cosine of an angle:
cos A = AD / AC = AC / AB. Hence AB * AD = AC²
Likewise,
cos B = BD / BC = BC / AB.
Hence AB * BD = BC².
Adding the obtained equalities term by term and noting that AD + DB = AB, we get:
AS² + sun² = AB (AD + DB) = AB²
The proof is complete.
4 METHOD.
Having studied the topic "Relations between the sides and angles of a right-angled triangle", I think that the Pythagorean theorem can be proved in another way.
Consider a right-angled triangle with legs a, in and hypotenuse with... (fig. 4).
Let us prove that c² = a² + b².
Proof.
sin B = a / c ; cos B = a / s , then, squaring the obtained equalities, we get:
sin² B =в² / с²; cos² V= a² / c².
Adding them together, we get:
sin² V+ cos² B = b² / s² + a² / c², where sin² V+ cos² B = 1,
1 = (b² + a²) / c², therefore
c² = a² + b².
The proof is complete.
5 METHOD.
This proof is based on cutting the squares built on the legs (Fig. 5) and laying the resulting pieces on the square built on the hypotenuse.
6 METHOD.
For proof on the leg Sun build BCD ABC(fig. 6). We know that the areas of such figures are related as squares of their similar linear dimensions:
Subtracting the second equality from the first, we get
c2 = a2 + b2.
The proof is complete.
7 METHOD.
Given(fig. 7):
ABC,= 90 ° , Sun= a, AC =b, AB = c.
Prove:c2 = a2 +b2.
Proof.
Let the leg b a. Let's continue the segment SV per point V and build a triangle BMD so that the points M and A lay on one side of a straight line CD and besides, BD =b, BDM= 90 °, DM= a, then BMD= ABC on both sides and the corner between them. Points A and M connect by segments AM. We have MD CD and AC CD, means straight AS parallel to the straight line MD. Because MD< АС, then straight CD and AM not parallel. Consequently, AMDC - rectangular trapezoid.
In right triangles ABC and BMD 1 + 2 = 90 ° and 3 + 4 = 90 °, but since = =, then 3 + 2 = 90 °; then AVM= 180 ° - 90 ° = 90 °. It turned out that the trapezoid AMDC is divided into three non-overlapping right-angled triangles, then by axioms squares
(a + b) (a + b)
Dividing all terms of the inequality by, we get
ab + c2 + ab = (a +b) , 2 ab+ c2 = a2+ 2ab+ b2,
c2 = a2 + b2.
The proof is complete.
8 METHOD.
This method is based on the hypotenuse and legs of a right-angled triangle. ABC. He constructs the corresponding squares and proves that the square built on the hypotenuse is equal to the sum of the squares built on the legs (Fig. 8).
Proof.
1) DBC= FBA= 90 °;
DBC + ABC= FBA + ABC, means, FBC = DBA.
Thus, FBC=ABD(on both sides and the corner between them).
2) 
,
where AL DE, since BD is a common base, DL - overall height.
3) 
, since FB is a foundation, AB- overall height.
4) 
5) Similarly, one can prove that 
6) Adding term by term, we get:
, BC2
= AB2 + AC2
.
The proof is complete.
9 METHOD.
Proof.
1) Let ABDE- a square (Fig. 9), the side of which is equal to the hypotenuse of a right-angled triangle ABC (AB= s, BC = a, AC =b).
2) Let DK BC and DK = BC, since 1 + 2 = 90 ° (like the acute corners of a right triangle), 3 + 2 = 90 ° (like the corner of a square), AB= BD(sides of the square).
Means, ABC= BDK(by hypotenuse and acute angle).
3) Let EL DK, AM EL. You can easily prove that ABC = BDK = DEL = EAM (with legs a and b). Then KS= CM= ML= LK= a -b.
4) SKB = 4S + SKLMC= 2ab+ (a - b),with2 = 2ab + a2 - 2ab + b2,c2 = a2 + b2.
The proof is complete.
10 METHOD.
The proof can be drawn on a figure jokingly called "Pythagorean pants" (Fig. 10). Its idea is to transform the squares built on the legs into equal triangles that together make up the square of the hypotenuse.
ABC we move, as shown by the arrow, and it takes the position KDN. The rest of the figure AKDCB equal area of a square AKDC - this is a parallelogram AKNB.
Parallelogram model made AKNB... We shift the parallelogram as sketched in the content of the work. To show the transformation of a parallelogram into an equal-area triangle, in front of the students' eyes, cut off a triangle on the model and shift it down. Thus, the area of the square AKDC turned out equal to the area of the rectangle. Similarly, convert the area of the square to the area of the rectangle.
Let's make a transformation for a square built on a leg a(Fig. 11, a):
a) the square is transformed into an equal-area parallelogram (Fig. 11.6):
b) the parallelogram is rotated by a quarter of a turn (Fig. 12):
c) the parallelogram is transformed into an equal-sized rectangle (Fig. 13): 11 METHOD.
Proof:
PCL - straight line (Fig. 14);
KLOA= ACPF= ACED= a2;
LGBO= SVMR =CBNQ= b 2;
AKGB= AKLO +LGBO= c2;
c2 = a2 + b2.
The proof is over .
12 METHOD.
Rice. 15 illustrates another original proof of the Pythagorean theorem.
Here: triangle ABC with right angle C; section Bf perpendicular SV and is equal to it, the segment BE perpendicular AB and is equal to it, the segment AD perpendicular AS and equal to him; points F, C,D belong to one straight line; quadrangles ADFB and ACBE are equal, since ABF = ECB; triangles ADF and ACE equal areas; subtract from both equal-sized quadrangles the common triangle for them ABC, get
, c2 = a2 +
b2.
The proof is complete.
13 METHOD.
The area of this right-angled triangle, on the one hand, is equal to ,
with another, ,
3. CONCLUSION.
As a result of the search activity, the goal of the work was achieved, which is to replenish and generalize knowledge on the proof of the Pythagorean theorem. I managed to find and consider various ways to prove it and deepen knowledge on the topic, going beyond the pages of a school textbook.
The material I have collected even more convinces that the Pythagorean theorem is the great theorem of geometry, has enormous theoretical and practical significance. In conclusion, I would like to say: the reason for the popularity of the Pythagorean triplet theorem is beauty, simplicity and significance!
4. USED LITERATURE.
1. Entertaining algebra. ... Moscow "Science", 1978.
2. Weekly educational and methodological supplement to the newspaper "September 1", 24/2001.
3. Geometry 7-9. and etc.
4. Geometry 7-9. and etc.
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 are the sides intersecting at right angles), and the hypotenuse as "c" (the hypotenuse is the most big side a right-angled triangle opposite a 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, you need to find the second leg. We'll 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 = c 2 the values given to you (or the values you found). Remember that a and b are legs and c is hypotenuse.
- In our example, write: 3² + b² = 5².
Square each side you know. Or leave the degrees - you can square the numbers later.
- In our example, write: 9 + b² = 25.
Isolate the unknown side on one side of the equation. To do this, transfer known values to 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 = 16.
Extract the square root of both sides of the equation after there is an unknown (squared) on one side of the equation and an intercept (number) on the other side.
- In our example, b² = 16. Take the square root of both sides of the equation and get b = 4. So the second leg is 4.
Use the Pythagorean theorem in Everyday life since 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. Bottom part the staircase 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 = 5; "Is 20 meters from the ground" means that b = 20 (that is, you are given two legs of a right-angled 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 = c²
- (5) ² + (20) ² = c²
- 25 + 400 = c²
- 425 = c²
- c = √425
- s = 20.6. Thus, the approximate length of the stairs is 20.6 meters.
- "5 meters from the base of the wall" means that a = 5; "Is 20 meters from the ground" means that b = 20 (that is, you are given two legs of a right-angled 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.
Pythagorean theorem: The sum of the areas of the squares resting on the legs ( a and b) is equal to the area of the square built on the hypotenuse ( c).
Geometric formulation:
Initially, the theorem was formulated as follows:
Algebraic formulation:
That is, denoting the length of the hypotenuse of a triangle by c, and the lengths of the legs through a and b :
a 2 + b 2 = c 2Both statements of the theorem are equivalent, but the second statement is more elementary, it does not require the concept of area. That is, the second statement can be checked without knowing anything about the area and measuring only the lengths of the sides of a right-angled triangle.
The reverse Pythagorean theorem:
Proof
On this moment in the scientific literature, 367 proofs of this theorem have been recorded. Probably the Pythagorean theorem is the only theorem with such an impressive number of proofs. This variety can be explained only by the fundamental meaning of the theorem for geometry.
Of course, conceptually all of them can be divided into a small number of classes. The most famous of them are: proofs by the area method, axiomatic and exotic proofs (for example, using differential equations).
Through similar triangles
The following proof of the algebraic formulation is the simplest of the proofs built directly from the axioms. In particular, it does not use the concept of the area of a figure.
Let be ABC there is a right-angled triangle with a right angle C... Let's draw the height from C and denote its base by H... Triangle ACH like a triangle ABC in two corners. Similarly, triangle CBH is similar ABC... Introducing the notation
we get
What is the equivalent
Adding, we get
Areas proof
The proofs below, despite their seeming simplicity, are not at all so simple. All of them use the properties of area, the proof of which is more difficult than the proof of the Pythagorean theorem itself.
Equal complementarity proof
- Place 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 the entire figure is, on the one hand, the area of a square with sides (a + b), and on the other hand, the sum of the areas of four triangles and two inner squares.
Q.E.D.
Evidence through scattering
Elegant proof by permutation
An example of one of such proofs is shown in the drawing on the right, where a square built on the hypotenuse is transformed by permutation into two squares built on the legs.
Euclid's proof
Drawing for Euclid's proof
Illustration for Euclid's proof
The idea behind Euclid's proof is as follows: let's try to prove that half of the area of the square built on the hypotenuse is equal to the sum of the halves of the areas of the squares built on the legs, and then the areas of the large and two small squares are equal.
Consider the drawing on the left. On it, we built squares on the sides of a right-angled triangle and drawn a ray s from the vertex of the right angle C perpendicular to the hypotenuse AB, it cuts the square ABIK, built on the hypotenuse, into two rectangles - BHJI and HAKJ, respectively. It turns out that the areas of these rectangles are exactly equal to the areas of the squares built on the corresponding legs.
Let's try to prove that the area of the square DECA is equal to the area of the rectangle AHJK To do this, let's use an auxiliary observation: The area of a triangle with the same height and base as this rectangle is equal to half the area of the given rectangle. This is a consequence of 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 the triangle ACK is equal to the area of the triangle AHK (not shown in the figure), which, in turn, is equal to half the area of the rectangle AHJK.
Let us now prove that the area of the triangle ACK is also equal to half the area of the square DECA. The only thing that needs to be done for this is to prove the equality of the triangles ACK and BDA (since the area of the triangle BDA is equal to half the area of the square according to the above property). Equality is obvious, the triangles are equal on two sides and the angle between them. Namely - AB = AK, AD = AC - the equality of the angles CAK and BAD is easy to prove by the method of motion: we rotate the triangle CAK 90 ° counterclockwise, then it is obvious that the corresponding sides of the two triangles under consideration will coincide (since the angle at the apex of the square is 90 °).
The reasoning about the equality of the areas of the square BCFG and the rectangle BHJI is completely analogous.
Thus, we have proved that the area of the square built on the hypotenuse is the sum of the areas of the squares built on the legs. The idea behind this proof is further illustrated with the animation above.
Proof of Leonardo da Vinci
Proof of Leonardo da Vinci
The main elements of the proof are symmetry and motion.
Consider the drawing, as seen from the symmetry, the segment CI cuts the square ABHJ into two identical parts (since the triangles ABC and JHI are equal by construction). Rotating 90 degrees counterclockwise, we see that the shaded shapes are equal CAJI and GDAB ... Now it is clear that the area of the shaded figure is equal to the sum of the halves of the areas of the squares built on the legs and the area of the original triangle. On the other hand, it is equal to half the area of the square built on the hypotenuse plus the area of the original triangle. The final step in the proof is left to the reader.
Proof by the method of infinitesimal
The following proof using differential equations is often attributed to the famous English mathematician Hardy, who lived in the first half of the 20th century.
Looking at the drawing shown in the figure and observing the change of the side a, we can write the following relation for infinitely small increments of the sides with and a(using the similarity of triangles):
Proof by the method of infinitesimal
Using the method of separating variables, we find
A more general expression for changing the hypotenuse in the case of increments of both legs
Integrating this equation and using initial conditions, we get
c 2 = a 2 + b 2 + constant.Thus, we arrive at the desired answer
c 2 = a 2 + b 2 .As it is easy to see, the quadratic dependence in the final formula appears due to the linear proportionality between the sides of the triangle and the increments, while the sum is 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 the figure built on the hypotenuse. In particular:
- The sum of the areas of regular triangles built on the legs is equal to the area of a regular triangle built on the hypotenuse.
- The sum of the areas of the semicircles built on the legs (as in the diameter) is equal to the area of the semicircle built on the hypotenuse. This example is used to prove the properties of figures bounded by arcs of two circles and bearing the name of hippocratic lunes.
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 Pythagorean triangle with sides 3, 4 and 5: In the same book, a drawing is proposed that coincides with one of the drawings of the Hindu geometry of Baskhara.
Cantor (the largest German historian of mathematics) believes that the equality 3 ² + 4 ² = 5 ² was already known to the Egyptians around 2300 BC. e., during the time of King Amenemhat I (according to papyrus 6619 of the Berlin Museum). According to Cantor, the harpedonapts, or "rope pulls", built right angles using right-angled triangles with sides 3, 4, and 5.
It is very easy to reproduce their way of building. Take a rope 12 m long and tie it to it along a colored strip at a distance of 3 m. from one end and 4 meters from the other. The right angle will be enclosed between the sides 3 and 4 meters long. The Harpedonapts might argue that their way of building becomes superfluous, if you use, for example, the wooden square used by all carpenters. Indeed, there are known Egyptian drawings in which such a tool is found, for example, drawings depicting a carpentry workshop.
Somewhat more is known about the Babylonian Pythagorean theorem. In one text dating back to the time of Hammurabi, that is, to 2000 BC. BC, an approximate calculation of the hypotenuse of a right-angled triangle is given. From this we can conclude that in Mesopotamia they knew how to perform calculations with right-angled triangles, at least in some cases. Based, on the one hand, on the current level of knowledge about Egyptian and Babylonian mathematics, and on the other, on a critical study of Greek sources, Van der Waerden (Dutch mathematician) made the following conclusion:
Literature
In Russian
- Skopets Z.A. Geometric miniatures. M., 1990
- Yelensky Sch. In the footsteps of Pythagoras. M., 1961
- Van der Waerden B.L. Awakening science. Maths Ancient egypt, Babylon and Greece. M., 1959
- Glazer G.I. History of mathematics at school. M., 1982
- V. Litzman, "The Pythagorean Theorem" M., 1960.
- A site about the Pythagorean theorem with a large number of proofs, the material is taken from the book of V. Litzman, a large number of drawings are presented in the form of separate graphic files.
- The Pythagorean theorem and Pythagorean triplets a chapter from the book by DV Anosov "A Look at Mathematics and Something From It"
- On the Pythagorean theorem and methods of its proof G. Glazer, Academician of the Russian Academy of Education, Moscow
In English
- The Pythagorean Theorem at WolframMathWorld
- Cut-The-Knot, a section on the Pythagorean theorem, about 70 proofs and a wealth of additional information
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