However, the name was received in honor of the scientist only for the reason that he was the first and even the only person who was able to prove the theorem.
The German mathematical historian Cantor claimed that the theorem was known to the Egyptians around 2300 BC. e. He believed that right angles were previously constructed using right triangles with sides of 3, 4 and 5.
The famous scientist Kepler said that geometry has an irreplaceable treasure - this is the Pythagorean theorem, thanks to which most theorems in geometry can be deduced.
Previously, the Pythagorean theorem was called the “bride’s theorem” or “the nymph’s theorem.” And the whole point is that her drawing was very similar to a butterfly or nymph. The Arabs, when translating the text of the theorem, decided that nymph meant bride. And so it appeared interesting name at the theorem.
Pythagorean theorem, formula
Theorem
– in a right triangle, the sum of the squares of the legs () is equal to the square of the hypotenuse (). This is one of the fundamental theorems of Euclidean geometry.
Formula:
As already mentioned, there are many different proofs of the theorem with diverse mathematical approaches. However, area theorems are more often used.
Let's construct squares on the triangle ( blue, green, red)
That is, the sum of the areas of squares built on the legs is equal to the area of the square built on the hypotenuse. Accordingly, the areas of these squares are equal to – . This is the geometric explanation of Pythagoras.
Proof of the theorem using the area method: 1st way
Let's prove that .
Let's consider the same triangle with legs a, b and hypotenuse c.
- We complete the right triangle to a square. From leg “a” we continue the line upward to the distance of leg “b” (red line).
- Next, draw the line of the new leg “a” to the right (green line).
- We connect the two legs with the hypotenuse “c”.
It turns out the same triangle, only upside down.
We build similarly on the other side: from leg “a” we draw a line of leg “b” and down “a” and “b” And from the bottom of leg “b” we draw a line of leg “a”. A hypotenuse “c” was drawn in the center from each leg. Thus the hypotenuses formed a square in the center.
This square consists of 4 identical triangles. And the area of each right triangle is half the product of its legs. Respectively, . And the area of the square in the center = , since all 4 hypotenuses have side . The sides of a quadrilateral are equal and the angles are right. How can we prove that the angles are right? Very simple. Let's take the same square:
We know that these two angles shown in the figure are 90 degrees. Since the triangles are equal, it means that the next angle of leg “b” is equal to the previous leg “b”:
The sum of these two angles = 90 degrees. Accordingly, the previous angle is also 90 degrees. Of course, it’s similar on the other side. Accordingly, we really have a square with right angles.
Since the acute angles of a right triangle add up to 90 degrees, the angle of a quadrilateral will also equal 90 degrees, because 3 angles add up to 180 degrees.
Accordingly, the area of a square is the sum of four areas of identical right triangles and the area of the square formed by the hypotenuses.
Thus, we got a square with side . We know that the area of a square with a side is the square of its side. That is . This square consists of four identical triangles.
And this means that we have proven the Pythagorean theorem.
IMPORTANT!!! If we find the hypotenuse, then we add the two legs, and then we derive the answer from the root. When finding one of the legs: from the square of the length of the second leg, subtract the square of the length of the hypotenuse and find Square root.
Examples of problem solving
Example 1
Task
Given: a right triangle with legs 4 and 5.
Find the hypotenuse. For now we’ll denote it “c”
Solution
The sum of the squares of the legs is equal to the square of the hypotenuse. In our case - .
Let's use the Pythagorean theorem:
So, , and . The legs add up to 41.
Then . That is, the square of the hypotenuse is 41.
Square of 41 = 6.4.
We found the hypotenuse.
Answer
Hypotenuse = 6.4
Every schoolchild knows that the square of the hypotenuse is always equal to the sum of the legs, each of which is squared. This statement is called the Pythagorean theorem. It is one of the most famous theorems of trigonometry and mathematics in general. Let's take a closer look at it.
The concept of a right triangle
Before moving on to consider the Pythagorean theorem, in which the square of the hypotenuse is equal to the sum of the legs that are squared, we should consider the concept and properties of a right triangle for which the theorem is valid.
Triangle - flat figure having three angles and three sides. A right triangle, as its name suggests, has one right angle, that is, this angle is equal to 90 o.
From general properties for all triangles, it is known that the sum of all three angles of this figure is 180 o, which means that for a right triangle, the sum of two angles that are not right angles is 180 o - 90 o = 90 o. This last fact means that any angle in a right triangle that is not right will always be less than 90 o.
The side that lies against right angle, is usually called the hypotenuse. The other two sides are the legs of the triangle, they can be equal to each other, or they can be different. From trigonometry we know that the greater the angle against which a side of a triangle lies, the greater the length of that side. This means that in a right triangle the hypotenuse (lies opposite the 90 o angle) will always be greater than any of the legs (lie opposite the angles< 90 o).
Mathematical notation of Pythagorean theorem
This theorem states that the square of the hypotenuse is equal to the sum of the legs, each of which is previously squared. To write this formulation mathematically, consider a right triangle in which sides a, b and c are the two legs and the hypotenuse, respectively. In this case, the theorem, which is formulated as the square of the hypotenuse is equal to the sum of the squares of the legs, can be represented by the following formula: c 2 = a 2 + b 2. From here other formulas important for practice can be obtained: a = √(c 2 - b 2), b = √(c 2 - a 2) and c = √(a 2 + b 2).
Note that in the case of a right-angled equilateral triangle, that is, a = b, the formulation: the square of the hypotenuse is equal to the sum of the legs, each of which is squared, will be mathematically written as follows: c 2 = a 2 + b 2 = 2a 2, which implies the equality: c = a√2.
Historical reference
The Pythagorean theorem, which states that the square of the hypotenuse is equal to the sum of the legs, each of which is squared, was known long before the famous Greek philosopher paid attention to it. Many papyri Ancient Egypt, as well as clay tablets of the Babylonians confirm that these peoples used the noted property of the sides of a right triangle. For example, one of the first Egyptian pyramids, the pyramid of Khafre, the construction of which dates back to the 26th century BC (2000 years before the life of Pythagoras), was built based on knowledge of the aspect ratio in a right triangle 3x4x5.
Why then does the theorem now bear the name of the Greek? The answer is simple: Pythagoras is the first to mathematically prove this theorem. The surviving Babylonian and Egyptian written sources only speak of its use, but do not provide any mathematical proof.
It is believed that Pythagoras proved the theorem in question by using the properties of similar triangles, which he obtained by drawing the height in a right triangle from an angle of 90 o to the hypotenuse.
An example of using the Pythagorean theorem
Let's consider simple task: it is necessary to determine the length of the inclined staircase L, if it is known that it has a height H = 3 meters, and the distance from the wall against which the staircase rests to its foot is P = 2.5 meters.
In this case, H and P are the legs, and L is the hypotenuse. Since the length of the hypotenuse is equal to the sum of the squares of the legs, we get: L 2 = H 2 + P 2, whence L = √(H 2 + P 2) = √(3 2 + 2.5 2) = 3.905 meters or 3 m and 90, 5 cm.
Story
Chu-pei 500-200 BC. On the left is the inscription: the sum of the squares of the lengths of the height and base is the square of the length of the hypotenuse.
In the ancient Chinese book Chu-pei ( English) (Chinese 周髀算經) talks about a Pythagorean triangle with sides 3, 4 and 5. The same book offers a drawing that coincides with one of the drawings of the Hindu geometry of Bashara.
Around 400 BC. BC, according to Proclus, Plato gave a method for finding Pythagorean triplets, combining algebra and geometry. Around 300 BC. e. The oldest axiomatic proof of the Pythagorean theorem appeared in Euclid's Elements.
Formulations
Geometric formulation:
The theorem was originally formulated as follows:
Algebraic formulation:
That is, denoting the length of the hypotenuse of the triangle by , and the lengths of the legs by 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.
Converse Pythagorean theorem:
|
For every triple of positive numbers , and , such that , there exists a right triangle with legs and and hypotenuse . |
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 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: 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, constructed directly from the axioms. In particular, it does not use the concept of area of a figure.
Let ABC there is a right triangle with a right angle C. Let's draw the height from C and denote its base by H. Triangle ACH similar to a triangle ABC at two corners. Likewise, triangle CBH similar ABC. By introducing the notation
we get
What is equivalent
Adding it up, we get
, which is what needed to be provenProofs using the area method
The proofs below, despite their apparent simplicity, are not so simple at all. They all use properties of area, the proof of which is more complex than the proof of the Pythagorean theorem itself.
Proof via equicomplementation
- Let's arrange four equal right triangle as shown in Figure 1.
- Quadrangle with sides c is a square, since the sum of two acute angles is 90°, and the straight angle is 180°.
- The area of the entire figure is equal, on the one hand, to the area of a square with side (a + b), and on the other hand, to the sum of the areas of the 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.
Let's look at the drawing on the left. On it we constructed squares on the sides of a right triangle and drew 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, we will 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 in the figure), 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 according to the above property). This equality is obvious: the triangles are equal on both 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 in question will coincide (due to the fact that the angle at the vertex of the square is 90°).
The reasoning for the equality of the areas of the square BCFG and the rectangle BHJI is completely similar.
Thus, we proved that the area of a square built on the hypotenuse is composed of the areas of squares built on the legs. The idea behind this proof is further illustrated by the animation above.
Proof of Leonardo da Vinci
The main elements of the proof are symmetry and motion.
Let's consider the drawing, as can be seen from the symmetry, the segment cuts the square into two identical parts (since the triangles are equal in construction).
Using a 90-degree counterclockwise rotation 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 the 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 small squares is equal to half the area of the large square, and therefore the sum of the areas of 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.
Looking at 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 triangle similarity):
Using the method of separation of variables, we find
More general expression to change the hypotenuse in case of increments of both legs
Integrating this equation and using the initial conditions, we obtain
Thus we arrive at the desired answer
As 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 associated with independent contributions from the increment of different legs.
A simpler proof can be obtained if we assume that one of the legs does not experience an increment (in this case leg). Then for the integration constant we obtain
Variations and generalizations
Similar geometric shapes on three sides
Generalization for similar triangles, area of green shapes A + B = area of blue C
Pythagorean theorem using similar right triangles
Euclid generalized the Pythagorean theorem in his work Beginnings, expanding the areas of the squares on the sides to 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 be equal to 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 by division central triangle its height. The sum of the two smaller triangles' areas is then obviously equal to the area of the third, thus A + B = C and, fulfilling the previous proof in reverse order, we obtain the Pythagorean theorem a 2 + b 2 = c 2 .
Cosine theorem
The Pythagorean theorem is special case a more general theorem of cosines, which relates the lengths of the sides in an arbitrary triangle:
where θ is the angle between the sides a And b.
If θ is 90 degrees then cos θ = 0 and the formula simplifies to the usual Pythagorean theorem.
Free Triangle
To any selected corner of an arbitrary triangle with sides a, b, c inscribe an isosceles triangle in such a way that equal angles at its base θ was equal to the selected angle. Let us assume that the selected angle θ is located opposite the side designated c. As a result, we got 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 located opposite the side b and parties With length s, as it shown on the picture. Thabit Ibn Qurra argued that the sides in these three triangles are related as follows:
As the angle θ approaches π/2, the base of the isosceles triangle becomes smaller and the two sides r and s overlap each other less and less. When θ = π/2, ADB becomes a right triangle, r + s = c and we obtain the initial Pythagorean theorem.
Let's consider 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 an angle θ and also have the same third angle, based on 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 down the relationship between opposite sides and those adjacent to the angle θ,
Also a reflection of another triangle,
Let's multiply the fractions and add these two ratios:
Q.E.D.
Generalization for arbitrary triangles via parallelograms
Generalization for arbitrary triangles,
green area plot = area blue
Proof of the thesis that in the figure above
Let's make a further generalization for non-right triangles by using parallelograms on three sides instead of squares. (squares are a special case.) The top 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 indicated by the arrows are the same and determine sides of the lower parallelogram). This replacement of squares with parallelograms bears a clear resemblance to the initial theorem of Pythagoras, thought to have been formulated by Pappus of Alexandria in 4 AD. 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. Additionally, the left green parallelogram has the same area as the left green parallelogram in the top picture because they share a common base (the top left side of the triangle) and a common height perpendicular to that side of the triangle. Using similar reasoning for the right side of the triangle, we will 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 valid 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, distance s between 0 + 1 i and 1 + 0 i calculated as the modulus of the vector (0, 1) − (1, 0) = (−1, 1), or
However, for operations with vectors with complex coordinates, it is necessary to make some improvements to the Pythagorean formula. Distance between points with complex numbers (a, b) And ( c, d); a, b, c, And d all complex, let us 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- 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 number for . For example, the distance between points (a, b) = (0, 1) And (c, d) = (i, 0) , let's 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 as follows:
Stereometry
A significant generalization of the Pythagorean theorem for three-dimensional space is de Goy's theorem, named after J.-P. de Gois: 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-dimensional space relates the diagonal AD to three sides.
Another generalization: The Pythagorean theorem can be applied to stereometry in the following form. Consider a rectangular parallelepiped as shown in the figure. Let's find the length of the diagonal BD using the Pythagorean theorem:
where the three sides form a right triangle. We use the horizontal diagonal BD and the vertical edge AB to find the length of the diagonal AD, for this we again use the Pythagorean theorem:
or, if we write everything in one equation:
This result is a three-dimensional 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 considered as a generalization of the Pythagorean theorem for multidimensional space. However, the result is actually nothing more than 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, there is an equality, 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 relationship between the sides of a triangle will necessarily be in a form different from the Pythagorean theorem. For example, in spherical geometry, all three sides of a right triangle (say a, b And c), which limit the octant (eighth part) of the unit sphere, have a length of π/2, which contradicts the Pythagorean theorem, because a 2 + b 2 ≠ c 2 .
Let us consider here two cases of non-Euclidean geometry - spherical and hyperbolic geometry; in both cases, as for Euclidean space for right triangles, the result, which 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 rectangular 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 relationship 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 diameter c.
Spherical geometry
For any right triangle on a sphere with radius R(for example, if the angle γ in a triangle is right) with sides a, b, c The relationship between the parties will look like this:
This equality can be derived as a special case spherical cosine theorem, which is valid for all spherical triangles:
where cosh is the hyperbolic cosine. This formula is a special case of the hyperbolic cosine theorem, which is valid for all triangles:
where γ is the angle whose vertex is opposite to the side c.
Where g ij called a metric tensor. It may be a function of position. Such curvilinear spaces include Riemannian geometry as general example. This formulation is also suitable for Euclidean space when using curvilinear coordinates. For example, for polar coordinates:
Vector artwork
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 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 is perpendicular to its components a And b it follows that, except for trivial cases from 0- and 1-dimensional space, the cross product is defined only in three and seven dimensions. We use the definition of the angle in n-dimensional space:
This property of a cross product gives its magnitude as follows:
Through the fundamental trigonometric identity of Pythagoras we obtain another form of writing its value:
An alternative approach to defining a cross product is to use an expression for its magnitude. Then, reasoning in reverse order, we obtain a connection with the scalar product:
see also
Notes
- History topic: Pythagoras’s theorem in Babylonian mathematics
- ( , p. 351) p. 351
- ( , Vol I, p. 144)
- Discussion historical facts given in (, p. 351) p. 351
- 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.
- Lewis Carroll, “The Story with Knots”, M., Mir, 1985, p. 7
- Asger Aaboe Episodes from the early history of mathematics. - Mathematical Association of America, 1997. - P. 51. - ISBN 0883856131
- Python Proposition by Elisha Scott Loomis
- 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.”
- Lawrence S. Leff cited work. - Barron's Educational Series. - P. 326. - ISBN 0764128922
- Howard Whitley Eves§4.8:...generalization of Pythagorean theorem // Great moments in mathematics (before 1650). - Mathematical Association of America, 1983. - P. 41. - ISBN 0883853108
- 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.
- Aydin Sayili (Mar. 1960). "Thâbit ibn Qurra"s Generalization of the Pythagorean Theorem." Isis 51 (1): 35–37. DOI:10.1086/348837.
- Judith D. Sally, Paul Sally Exercise 2.10 (ii) // Cited work. - P. 62. - ISBN 0821844032
- 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
- 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.
- Alfred Gray, Elsa Abbena, Simon Salamon Modern differential geometry of curves and surfaces with Mathematica. - 3rd. - CRC Press, 2006. - P. 194. - ISBN 1584884487
- Rajendra Bhatia Matrix analysis. - Springer, 1997. - P. 21. - ISBN 0387948465
- Stephen W. Hawking cited work. - 2005. - P. 4. - ISBN 0762419229
When you first started learning about square roots and how to solve irrational equations (equalities involving an unknown under the root sign), you probably got your first taste of their practical uses. The ability to take the square root of numbers is also necessary to solve problems using 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 meet at right angles) be designated by the letters and, and the length of the hypotenuse (the longest side of the triangle located opposite the right angle) will be designated 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 when the length of its other two sides is known. In addition, it allows you to determine whether the triangle in question is a right triangle, 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 using 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 get to the solution:
a) Substituting the 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 length of a side of a triangle cannot be expressed negative number, the second option is automatically discarded.
Answer to the first picture: b = 64.
b) The length of the leg of the second triangle is found in the same way:
84 2 + b 2 = 91 2
7056 + b 2 = 8281
b 2 = 1225
b= 35 or b = -35
As in the previous case, a negative decision is discarded.
Answer to the second picture: b = 35
We are given:
- The lengths of the smaller sides of the triangle are 45 and 55, respectively, and the larger sides are 75.
- The lengths of the smaller sides of the triangle are 28 and 45, respectively, and the larger sides are 53.
Let's solve the problem:
a) It is necessary to check whether the sum of the squares of the lengths of the shorter sides is equal given triangle the square of the greater length:
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, let's 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 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 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 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, to find the length of the cable the Pythagorean theorem can be used:
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) is 26.
So, let’s help Vita 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
G. Glaser,
Academician of the Russian Academy of Education, Moscow
About the Pythagorean theorem and methods of proving it
The area of a square built on the hypotenuse of a right triangle is equal to the sum of the areas of the squares built on its legs...
This is one of the most famous geometric theorems of antiquity, called the Pythagorean theorem. Almost everyone who has ever studied planimetry knows it even now. It seems to me that if we want to let extraterrestrial civilizations know about the existence of intelligent life on Earth, then we should send an image of the Pythagorean figure into space. I think that if thinking beings can accept this information, then without complex signal decoding they will understand that there is a fairly developed civilization on Earth.
The famous Greek philosopher and mathematician Pythagoras of Samos, after whom the theorem is named, lived about 2.5 thousand years ago. The biographical information that has reached us about Pythagoras is fragmentary and far from reliable. Many legends are associated with his name. It is reliably known that Pythagoras traveled a lot in the countries of the East, visiting Egypt and Babylon. In one of the Greek colonies of Southern Italy, he founded the famous “Pythagorean School”, which played an important role in scientific and political life ancient Greece. It is Pythagoras who is credited with proving the famous geometric theorem. Based on legends spread by famous mathematicians (Proclus, Plutarch, etc.), for a long time it was believed that this theorem was not known before Pythagoras, hence the name - the Pythagorean theorem.
There is no doubt, however, that this theorem was known many years before Pythagoras. Thus, 1500 years before Pythagoras, the ancient Egyptians knew that a triangle with sides 3, 4 and 5 is right-angled, and used this property (i.e. the theorem converse of the theorem Pythagoras) for constructing right angles when planning land plots and building structures. Even today, rural builders and carpenters, when laying the foundation of a hut and making its parts, draw this triangle to obtain a right angle. The same thing was done thousands of years ago in the construction of magnificent temples in Egypt, Babylon, China, and probably in Mexico. The oldest Chinese mathematical and astronomical work that has come down to us, Zhou Bi, written about 600 years before Pythagoras, contains, among other proposals related to the right triangle, the Pythagorean theorem. Even earlier this theorem was known to the Hindus. Thus, Pythagoras did not discover this property of a right triangle; he was probably the first to generalize and prove it, thereby transferring it from the field of practice to the field of science. We don't know how he did it. Some historians of mathematics assume that Pythagoras’s proof was not fundamental, but only a confirmation, a test of this property on a number of particular types of triangles, starting with an isosceles right triangle, for which it obviously follows from Fig. 1.
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ancient times Mathematicians are finding more and more new proofs of the Pythagorean theorem, more and more new plans for its proof. More than one hundred and fifty such proofs - more or less strict, more or less visual - are known, but the desire to increase their number has remained. I think that independent “discovery” of proofs of the Pythagorean theorem will be useful for modern schoolchildren.
Let's look at some examples of evidence that can suggest the direction of such searches.
Pythagorean proof
";A square built on the hypotenuse of a right triangle is equal to the sum of the squares built on its legs."; The simplest proof of the theorem is obtained in the simplest case of an isosceles right triangle. This is probably where the theorem began. In fact, it is enough just to look at the mosaic of isosceles right triangles to be convinced of the validity of the theorem. For example, for DABC: a square built on the hypotenuse AC, contains 4 original triangles, and squares built on legs of two. The theorem has been proven.
Proofs based on the use of the concept of equal size of figures.
In this case, we can consider evidence in which a square built on the hypotenuse of a given right triangle is “composed” of the same figures as squares built on the sides. We can also consider proofs that use rearrangements of the summands of the figures and take into account a number of new ideas.
In Fig. 2 shows two equal squares. The length of the sides of each square is a + b. Each of the squares is divided into parts consisting of squares and right triangles. It is clear that if the quadruple area of a right triangle with legs a, b is subtracted from the area of the square, then equal areas will remain, i.e. c 2 = a 2 + b 2 . However, the ancient Hindus, to whom this reasoning belongs, usually did not write it down, but accompanied the drawing with only one word: “look!” It is quite possible that Pythagoras offered the same proof.
Additive evidence.
These proofs are based on the decomposition of squares built on the legs into figures from which one can add a square built on the hypotenuse.
Here: ABC is a right triangle with right angle C; CMN; CKMN; PO||MN; EF||MN.
Independently prove the pairwise equality of triangles obtained by partitioning squares built on the legs and hypotenuse.
Prove the theorem using this partition.
Based on the proof of al-Nayriziyah, another decomposition of squares into pairwise equal figures was carried out (Fig. 5, here ABC is a right triangle with right angle C).
Another proof by the method of decomposing squares into equal parts, called the “wheel with blades,” is shown in Fig. 6. Here: ABC is a right triangle with right angle C; O is the center of a square built on a large side; dotted lines passing through point O are perpendicular or parallel to the hypotenuse.
This decomposition of squares is interesting because its pairwise equal quadrilaterals can be mapped onto each other by parallel translation. Many other proofs of the Pythagorean theorem can be offered using the decomposition of squares into figures.
Evidence by the method of completion.
The essence of this method is that equal figures are added to the squares built on the legs and to the square built on the hypotenuse in such a way that equal figures are obtained.
The validity of the Pythagorean theorem follows from the equal size of the hexagons AEDFPB and ACBNMQ. Here CEP, line EP divides the hexagon AEDFPB into two equal quadrilaterals, line CM divides the hexagon ACBNMQ into two equal quadrilaterals; Rotating the plane 90° around center A maps the quadrilateral AEPB onto the quadrilateral ACMQ.
In Fig. 8 The Pythagorean figure is completed to a rectangle, the sides of which are parallel to the corresponding sides of the squares built on the sides. Let's divide this rectangle into triangles and rectangles. From the resulting rectangle, we first subtract all the polygons 1, 2, 3, 4, 5, 6, 7, 8, 9, leaving a square built on the hypotenuse. Then from the same rectangle we subtract rectangles 5, 6, 7 and the shaded rectangles, we get squares built on the legs. |
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Now let us prove that the figures subtracted in the first case are equal in size to the figures subtracted in the second case.
KLOA = ACPF = ACED = a 2 ;
LGBO = CBMP = CBNQ = b 2 ;
AKGB = AKLO + LGBO = c 2 ;
hence c 2 = a 2 + b 2 .
OCLP = ACLF = ACED = b 2 ; 
CBML = CBNQ = a 2 ;
OBMP = ABMF = c 2 ;
OBMP = OCLP + CBML;
c 2 = a 2 + b 2 .
Algebraic method of proof.
Rice. 12 illustrates the proof of the great Indian mathematician Bhaskari (famous author Lilavati, X |
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Let us present in a modern presentation one of these proofs, due to Pythagoras. |
II century). The drawing was accompanied by only one word: LOOK! Among the proofs of the Pythagorean theorem N 
and fig. 13 ABC – rectangular, C – right angle, CMAB, b 1 – projection of leg b onto the hypotenuse, a 1 – projection of leg a onto the hypotenuse, h – altitude of the triangle drawn to the hypotenuse.
From the fact that ABC is similar to ACM it follows
b 2 = cb 1 ; (1)
from the fact that ABC is similar to BCM it follows
a 2 = ca 1 . (2)
Adding equalities (1) and (2) term by term, we obtain a 2 + b 2 = cb 1 + ca 1 = c(b 1 + a 1) = c 2 .
If Pythagoras did offer such a proof, then he was also familiar with a number of important geometric theorems that modern historians of mathematics usually attribute to Euclid.
Moehlmann's proof (Fig. 14). |
The area of a given right triangle, on the one hand, is equal to the other, where p is the semi-perimeter of the triangle, r is the radius of the circle inscribed in it 
We have:whence it follows that c 2 =a 2 +b 2.
in the second 
Equating these expressions, we obtain the Pythagorean theorem.
Combined method
Equality of triangles
c 2 = a 2 + b 2 . (3)
Comparing relations (3) and (4), we obtain that
c 1 2 = c 2, or c 1 = c.
Thus, the triangles - the given one and the one constructed - are equal, since they have three respectively equal sides. Angle C 1 is right, so angle C of this triangle is also right.
Ancient Indian evidence.
Mathematicians Ancient India noticed that to prove the Pythagorean theorem it is enough to use the inside of an ancient Chinese drawing. In the treatise “Siddhanta Shiromani” (“Crown of Knowledge”), written on palm leaves, by the greatest Indian mathematician of the 18th century. Bha-skaras drawing is placed (Fig. 4)
characteristic of Indian evidence is the word “look!” As you can see, right triangles are laid here with the hypotenuse facing outwards and a square With 2 transferred to the “bride’s chair” With 2 -b 2 . Note that special cases of the Pythagorean theorem (for example, constructing a square whose area is twice as large Fig.4 area of a given square) are found in the ancient Indian treatise ";Sulva";
We solved a right triangle and squares built on its legs, or, in other words, figures made up of 16 identical isosceles right triangles and therefore fitting into a square. That's how lily is. a small fraction of the wealth hidden in the pearl of ancient mathematics - the Pythagorean theorem.
Ancient Chinese evidence.
Mathematical treatises Ancient China came to us in the edition of P.V. BC. The fact is that in 213 BC. Chinese Emperor Shi Huang Di, trying to eliminate previous traditions, ordered all ancient books to be burned. In P century BC. In China, paper was invented and at the same time the reconstruction of ancient books began. The most important of the surviving astronomical works is the book “Mathematics” containing a drawing (Fig. 2, a) proving the Pythagorean theorem. The key to this proof is not difficult to find. In fact, in the ancient Chinese drawing there are four equal right-angled triangles with legs a, b and a hypotenuse With stacked G) so that their outer contour forms Fig. 2 a square with side a+b, and the inner one is a square with side c, built on the hypotenuse (Fig. 2, b). If a square with side c is cut out and the remaining 4 shaded triangles are placed in two rectangles (Fig. 2, V), then it is clear that the resulting void, on the one hand, is equal to WITH 2 , and on the other - With 2 +b 2 , those. c 2= 2 +b 2 . The theorem has been proven. Note that with this proof, the constructions inside the square on the hypotenuse, which we see in the ancient Chinese drawing (Fig. 2, a), are not used. Apparently, ancient Chinese mathematicians had a different proof. Precisely if in a square with side With two shaded triangles (Fig. 2, b) cut off and attach the hypotenuses to the other two hypotenuses (Fig. 2, G), then it is easy to discover that
The resulting figure, sometimes called the "bride's chair", consists of two squares with sides A And b, those. c 2 == a 2 +b 2 .
N 
and Figure 3 reproduces a drawing from the treatise “Zhou-bi...”. Here the Pythagorean theorem is considered for the Egyptian triangle with legs 3, 4 and hypotenuse 5 units of measurement. The square on the hypotenuse contains 25 cells, and the square inscribed in it on the larger leg contains 16. It is clear that the remaining part contains 9 cells. This will be the square on the smaller side.