What is the Pythagorean theorem. Pythagorean theorem: the square of the hypotenuse is the sum of the legs squared

When you first started learning about square roots and how to solve irrational equations (equalities containing an unknown under the root sign), you probably got the first idea of ​​\u200b\u200btheir practical use. The ability to extract the 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 located opposite the 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 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 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 as a 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, find the length of the largest segment formed by points with coordinates (-2, -3) and (5, -2). To do this, we use the well-known formula for finding the distance between points in a rectangular coordinate system:

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 shortest length is equal to the square of the side with the longest 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

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-angled 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 instance:

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 .

Story

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

In the ancient Chinese book Chu-pei ( English) (Chinese 周髀算經) speaks of a Pythagorean triangle with sides 3, 4 and 5. In the same book, a drawing is proposed that coincides with one of the drawings of the Hindu geometry of Bashara.

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

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 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

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. The idea behind this proof is further illustrated with the animation above.

Proof of Leonardo da Vinci

The main elements of the proof are symmetry and 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 using differential equations is often attributed to the famous 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 in the opposite 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, performing the previous proofs in reverse order, we obtain the Pythagorean theorem a 2 + b 2 = c 2 .

Cosine theorem

The Pythagorean theorem is a special case of the more general cosine theorem that 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 we inscribe an isosceles triangle in such a way that equal angles at its base θ are equal to 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 of the isosceles triangle decreases and the two sides r and s overlap 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 the opposite sides and those 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 upper figure shows that for an 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 constructed as shown in the figure (the dimensions marked with arrows are the same and determine 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 a common base (upper left side of the triangle) and a common 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 all complex, we formulate using absolute values. Distance s based on vector difference (a − c, b − d) in the 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":

The Pythagorean theorem in three dimensions relates the diagonal AD to three sides.

Another generalization: The Pythagorean theorem can be applied to stereometry in the following form. Consider a rectangular box, as shown in the figure. 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 - these are projections of the vector onto the coordinate axes, then this formula coincides with the Euclidean distance - and means that the length of the vector is equal to the square root of the sum of the squares of its components.

The analogue of this equality in the case of an infinite system of 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) that bound the octant (an eighth) 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 of the 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 a common 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 of Pythagoras, we obtain another 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 the 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. A discussion of historical facts is 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 other 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

The Pythagorean theorem says:

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

a 2 + b 2 = c 2,

• a and b- legs forming a right angle.
• With is the hypotenuse of the triangle.

Formulas of the Pythagorean theorem

• a = \sqrt(c^(2) - b^(2))
• b = \sqrt (c^(2) - a^(2))
• c = \sqrt (a^(2) + b^(2))

Proof of the Pythagorean Theorem

The area of ​​a right triangle is calculated by the formula:

S = \frac(1)(2)ab

To calculate the area of ​​an arbitrary triangle, the area formula is:

• p- semiperimeter. p=\frac(1)(2)(a+b+c) ,
• r is the radius of the inscribed circle. For a rectangle r=\frac(1)(2)(a+b-c).

Then we equate the right sides of both formulas for the area of ​​a triangle:

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

2 ab = (a+b+c) (a+b-c)

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

2ab = a^(2)+2ab+b^(2)-c^(2)

0=a^(2)+b^(2)-c^(2)

c^(2) = a^(2)+b^(2)

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 the triangle is a right triangle. That is, for any triple of 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.

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

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

Additional material:

The Pythagorean theorem says:

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

a 2 + b 2 = c 2,

• a and b- legs forming a right angle.
• With is the hypotenuse of the triangle.

Formulas of the Pythagorean theorem

• a = \sqrt(c^(2) - b^(2))
• b = \sqrt (c^(2) - a^(2))
• c = \sqrt (a^(2) + b^(2))

Proof of the Pythagorean Theorem

The area of ​​a right triangle is calculated by the formula:

S = \frac(1)(2)ab

To calculate the area of ​​an arbitrary triangle, the area formula is:

• p- semiperimeter. p=\frac(1)(2)(a+b+c) ,
• r is the radius of the inscribed circle. For a rectangle r=\frac(1)(2)(a+b-c).

Then we equate the right sides of both formulas for the area of ​​a triangle:

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

2 ab = (a+b+c) (a+b-c)

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

2ab = a^(2)+2ab+b^(2)-c^(2)

0=a^(2)+b^(2)-c^(2)

c^(2) = a^(2)+b^(2)

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 the triangle is a right triangle. That is, for any triple of 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.

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

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

Additional material: