The solution of the problem:

252 = 242 + 72, which means the triangle is right-angled and its area is equal to half the product of its legs, i.e. S = hс * с: 2, where с is the hypotenuse, hс ​​is the height drawn to the hypotenuse, then hс = = = 6.72 (cm)

Answer: 6.72 cm.

Purpose of the stage:

Slide number 4

“4” - 1 incorrect answer

“3” - the answers are incorrect.

I suggest doing:

Slide number 5

Purpose of the stage:

At the end of the lesson:

The following phrases are written on the board:

The lesson is useful, everything is clear.

You still have to work hard.

Yes, it’s still difficult to study!

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"Mathematics lesson project "Theorem inverse to the Pythagorean theorem""

Lesson project “Theorem inverse to the Pythagorean theorem”

A lesson in “discovering” new knowledge

Lesson objectives:

activity: developing in students the ability to independently construct new methods of action based on the method of reflexive self-organization;

educational: expansion of the conceptual base by including new elements into it.

Stage of motivation for learning activities (5 min)

Mutual greeting of the teacher and students, checking readiness for the lesson, organizing attention and internal readiness, quickly integrating students into the business rhythm by solving problems using ready-made drawings:

Find BC if ABCD is a rhombus.

ABCD is a rectangle. AB:AD = 3:4. Find AD.

Find AD.

Find AB.

Find the sun.

Answers to problems based on ready-made drawings:

1.BC = 3; 2.BP = 4cm; 3.AB = 3√2cm.

Stage of “discovery” of new knowledge and methods of action (15 min)

Purpose of the stage: formulation of the topic and goals of the lesson using introductory dialogue (the “problem situation” technique).

Formulate statements inverse to the data and find out whether they are true:slide number 1

In the latter case, students can formulate a statement that is the opposite of the given one.

Instructions for working in pairs to study the proof of the theorem inverse to the Pythagorean theorem.

I instruct students about the method of activity, about the location of the material.

Assignment for couples: slide number 2

Independent work in pairs to study the proof of the theorem inverse to the Pythagorean theorem. Public protection of evidence.

One of the pairs begins their presentation by stating the theorem. There is an active discussion of the proof, during which one or another option is justified with the help of questions from the teacher and students.

Comparing the proof of the theorem with the teacher's proof

The teacher works at the blackboard, addressing the students who are working in their notebooks.

Given: ABC – triangle, AB 2 = AC 2 + BC 2

Find out whether ABC is rectangular. Proof:

Consider A 1 B 1 C 1 such that ˂C = 90 0, A 1 C 1 = AC, B 1 C 1 = BC. Then, by the Pythagorean theorem, A 1 B 1 2 = A 1 C 1 2 + B 1 C 1 2.

Since A 1 C 1 = AC, B 1 C 1 = BC, then: A 1 C 1 2 + B 1 C 1 2 = AC 2 + BC 2 = AB 2, therefore, AB 2 = A 1 B 1 2 and AB = A 1 B 1.

A 1 B 1 C 1 = ABC on three sides, from where ˂C = ˂C 1 = 90 0, that is, ABC is rectangular. So, 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 right-angled.

This statement is called a theorem converse to the Pythagorean theorem.

Public speech by one of the students about Pythagorean triangles (prepared information).

Slide number 3

After the information, I ask the students a few questions.

Are the following triangles Pythagorean triangles?

with hypotenuse 25 and leg 15;

with legs 5 and 4?

Stage of primary consolidation with pronunciation in external speech (10 min)

Purpose of the stage: demonstrate the application of the inverse theorem to the Pythagorean theorem in the process of solving problems.

I propose to solve problem No. 499 a) from the textbook. One of the students is invited to the board, solves the problem with the help of the teacher and students, pronouncing the solution in external speech. During the guest student's presentation, I ask several questions:

How to check if a triangle is right angled?

To which side will the shorter altitude of the triangle be drawn?

What method of calculating the height of a triangle is often used in geometry?

Using the formula for calculating the area of ​​a triangle, find the desired height.

The solution of the problem:

25 2 = 24 2 + 7 2, which means the triangle is right-angled and its area is equal to half the product of its legs, i.e. S = h с * с: 2, where с is the hypotenuse, h с is the height drawn to the hypotenuse, then h с = = = 6.72 (cm)

Answer: 6.72 cm.

Stage of independent work with self-test according to the standard (10 min)

Purpose of the stage: improve independent activity in the classroom by carrying out self-tests, learn to evaluate activities, analyze, and draw conclusions.

Independent work is proposed with a proposal to adequately evaluate your work and give an appropriate rating.

Slide number 4

Grading criteria: “5” - all answers are correct

“4” - 1 incorrect answer

“3” - the answers are incorrect.

The stage of informing students about homework, instructions on how to complete it (3 min).

I inform students about their homework, explain how to complete it, and check their understanding of the content of the work.

I suggest doing:

Slide number 5

Reflection stage of educational activities in the lesson (2 min)

Purpose of the stage: teach students to assess their readiness to detect ignorance, find the causes of difficulties, and determine the result of their activities.

At this stage, I invite each student to choose only one of the guys to whom I would like to say thank you for their cooperation and explain how exactly this cooperation manifested itself.

The teacher's word of thanks is final. At the same time, I choose those who received the least number of compliments.

At the end of the lesson:

The following phrases are written on the board:

The lesson is useful, everything is clear.

There's just one thing that's a little unclear.

You still have to work hard.

Yes, it’s still difficult to study!

Children come up and put a sign (tick) next to the words that suit them best at the end of the lesson.

According to Van der Waerden, it is very likely that the ratio in general form was known in Babylon around the 18th century BC. e.

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

The basic formulation contains algebraic operations - in a right triangle, the lengths of which are equal a (\displaystyle a) And b (\displaystyle b), and the length of the hypotenuse is c (\displaystyle c), the following relation is satisfied:

.

An equivalent geometric formulation is also possible, resorting to the concept of area of ​​a figure: in a right triangle, the area of ​​the square built on the hypotenuse is equal to the sum of the areas of the squares built on the legs. The theorem is formulated in this form in Euclid’s Elements.

Converse Pythagorean theorem- a statement about the rectangularity of any triangle, the lengths of the sides of which are related by the relation a 2 + b 2 = c 2 (\displaystyle a^(2)+b^(2)=c^(2)). As a consequence, for every triple of positive numbers a (\displaystyle a), b (\displaystyle b) And c (\displaystyle c), such that a 2 + b 2 = c 2 (\displaystyle a^(2)+b^(2)=c^(2)), there is a right triangle with legs a (\displaystyle a) And b (\displaystyle b) and hypotenuse c (\displaystyle c).

Proof

There are at least 400 proofs of the Pythagorean theorem recorded in the scientific literature, which is explained by both its fundamental significance for geometry and the elementary nature of the result. The main directions of proofs are: algebraic use of relations between the elements of a triangle (for example, the popular method of similarity), the method of areas, there are also various exotic proofs (for example, using differential equations).

Through similar triangles

The classical proof of Euclid is aimed at establishing the equality of areas between rectangles formed by dissecting the square above the hypotenuse by the height of the right angle with the squares above the legs.

The construction used for the proof is as follows: for a right triangle with a right angle C (\displaystyle C), squares over the legs and and squares over the hypotenuse A B I K (\displaystyle ABIK) height is being built CH and the ray that continues it s (\displaystyle s), dividing the square above the hypotenuse into two rectangles and . The proof aims to establish the equality of the areas of the rectangle A H J K (\displaystyle AHJK) with a square over the leg A C (\displaystyle AC); the equality of the areas of the second rectangle, constituting the square above the hypotenuse, and the rectangle above the other leg is established in a similar way.

Equality of areas of a rectangle A H J K (\displaystyle AHJK) And A C E D (\displaystyle ACED) is established through the congruence of triangles △ A C K ​​(\displaystyle \triangle ACK) And △ A B D (\displaystyle \triangle ABD), the area of ​​each of which is equal to half the area of ​​the squares A H J K (\displaystyle AHJK) And A C E D (\displaystyle ACED) accordingly, in connection with the following property: the area of ​​a triangle is equal to half the area of ​​a rectangle if the figures have a common side, and the height of the triangle to the common side is the other side of the rectangle. The congruence of triangles follows from the equality of two sides (sides of squares) and the angle between them (composed of a right angle and an angle at A (\displaystyle A).

Thus, the proof establishes that the area of ​​a square above the hypotenuse, composed of rectangles A H J K (\displaystyle AHJK) And B H J I (\displaystyle BHJI), is equal to the sum of the areas of the squares over the legs.

Proof of Leonardo da Vinci

The area method also includes a proof found by Leonardo da Vinci. Let a right triangle be given △ A B C (\displaystyle \triangle ABC) with right angle C (\displaystyle C) and squares A C E D (\displaystyle ACED), B C F G (\displaystyle BCFG) And A B H J (\displaystyle ABHJ)(see picture). In this proof on the side HJ (\displaystyle HJ) of the latter, a triangle is constructed on the outer side, congruent △ A B C (\displaystyle \triangle ABC), moreover, reflected both relative to the hypotenuse and relative to the height to it (that is, J I = B C (\displaystyle JI=BC) And H I = A C (\displaystyle HI=AC)). Straight C I (\displaystyle CI) splits the square built on the hypotenuse into two equal parts, since triangles △ A B C (\displaystyle \triangle ABC) And △ J H I (\displaystyle \triangle JHI) equal in construction. The proof establishes the congruence of quadrilaterals C A J I (\displaystyle CAJI) And D A B G (\displaystyle DABG), the area of ​​each of which turns out to be, on the one hand, equal to the sum of half the areas of the squares on the legs and the area of ​​the original triangle, on the other hand, half the area of ​​the square on the hypotenuse plus the area of ​​the original triangle. In total, half the sum of the areas of the squares over the legs is equal to half the area of ​​the square over the hypotenuse, which is equivalent to the geometric formulation of the Pythagorean theorem.

Proof by the infinitesimal method

There are several proofs using the technique of differential equations. In particular, Hardy is credited with a proof using infinitesimal increments of legs a (\displaystyle a) And b (\displaystyle b) and hypotenuse c (\displaystyle c), and preserving similarity with the original rectangle, that is, ensuring the fulfillment of the following differential relations:

d a d c = c a (\displaystyle (\frac (da)(dc))=(\frac (c)(a))), d b d c = c b (\displaystyle (\frac (db)(dc))=(\frac (c)(b))).

Using the method of separating variables, a differential equation is derived from them c d c = a d a + b d b (\displaystyle c\ dc=a\,da+b\,db), whose integration gives the relation c 2 = a 2 + b 2 + C o n s t (\displaystyle c^(2)=a^(2)+b^(2)+\mathrm (Const) ). Application of initial conditions a = b = c = 0 (\displaystyle a=b=c=0) defines the constant as 0, which results in the statement of the theorem.

The quadratic dependence in the final formula appears due to the linear proportionality between the sides of the triangle and the increments, while the sum is associated with independent contributions from the increment of different legs.

Variations and generalizations

Similar geometric shapes on three sides

An important geometric generalization of the Pythagorean theorem was given by Euclid in the Elements, moving from the areas of squares on the sides to the areas of arbitrary similar geometric figures: the sum of the areas of such figures built on the legs will be equal to the area of ​​a similar figure built on the hypotenuse.

The main idea of ​​this generalization is that the area of ​​such a geometric figure is proportional to the square of any of its linear dimensions and, in particular, to the square of the length of any side. Therefore, for similar figures with areas A (\displaystyle A), B (\displaystyle B) And C (\displaystyle C), built on legs with lengths a (\displaystyle a) And b (\displaystyle b) and hypotenuse c (\displaystyle c) Accordingly, the following relation holds:

A a 2 = B b 2 = C c 2 ⇒ A + B = a 2 c 2 C + b 2 c 2 C (\displaystyle (\frac (A)(a^(2)))=(\frac (B )(b^(2)))=(\frac (C)(c^(2)))\,\Rightarrow \,A+B=(\frac (a^(2))(c^(2) ))C+(\frac (b^(2))(c^(2)))C).

Since according to the Pythagorean theorem a 2 + b 2 = c 2 (\displaystyle a^(2)+b^(2)=c^(2)), then done.

In addition, if it is possible to prove without invoking the Pythagorean theorem that the areas of three similar geometric figures on the sides of a right triangle satisfy the relation A + B = C (\displaystyle A+B=C), then using the reverse of the proof of Euclid's generalization, one can derive a proof of the Pythagorean theorem. For example, if on the hypotenuse we construct a right triangle congruent with the initial one with an area C (\displaystyle C), and on the sides - two similar right-angled triangles with areas A (\displaystyle A) And B (\displaystyle B), then it turns out that triangles on the sides are formed as a result of dividing the initial triangle by its height, that is, the sum of the two smaller areas of the triangles is equal to the area of ​​the third, thus A + B = C (\displaystyle A+B=C) and, applying the relation for similar figures, the Pythagorean theorem is derived.

Cosine theorem

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

a 2 + b 2 − 2 a b cos ⁡ θ = c 2 (\displaystyle a^(2)+b^(2)-2ab\cos (\theta )=c^(2)),

where is the angle between the sides a (\displaystyle a) And b (\displaystyle b). If the angle is 90°, then cos ⁡ θ = 0 (\displaystyle \cos \theta =0), and the formula simplifies to the usual Pythagorean theorem.

Free Triangle

There is a generalization of the Pythagorean theorem to an arbitrary triangle, operating solely on the ratio of the lengths of the sides, it is believed that it was first established by the Sabian astronomer Thabit ibn Qurra. In it, for an arbitrary triangle with sides, an isosceles triangle with a base on the side fits into it c (\displaystyle c), the vertex coinciding with the vertex of the original triangle, opposite the side c (\displaystyle c) and angles at the base equal to the angle θ (\displaystyle \theta ), opposite side c (\displaystyle c). As a result, two triangles are formed, similar to the original one: the first - with sides a (\displaystyle a), the side farthest from it of the inscribed isosceles triangle, and r (\displaystyle r)- side parts c (\displaystyle c); the second - symmetrically to it from the side b (\displaystyle b) with the side s (\displaystyle s)- the corresponding part of the side c (\displaystyle c). As a result, the following relation is satisfied:

a 2 + b 2 = c (r + s) (\displaystyle a^(2)+b^(2)=c(r+s)),

degenerating into the Pythagorean theorem at θ = π / 2 (\displaystyle \theta =\pi /2). The relationship is a consequence of the similarity of the formed triangles:

c a = a r , c b = b s ⇒ c r + c s = a 2 + b 2 (\displaystyle (\frac (c)(a))=(\frac (a)(r)),\,(\frac (c) (b))=(\frac (b)(s))\,\Rightarrow \,cr+cs=a^(2)+b^(2)).

Pappus's theorem on areas

Non-Euclidean geometry

The Pythagorean theorem is derived from the axioms of Euclidean geometry and is not valid for non-Euclidean geometry - the fulfillment of the Pythagorean theorem is equivalent to the Euclidian parallelism postulate.

In non-Euclidean geometry, the relationship between the sides of a right triangle will necessarily be in a form different from the Pythagorean theorem. For example, in spherical geometry, all three sides of a right triangle, which bound the octant of the unit sphere, have a length π / 2 (\displaystyle \pi /2), which contradicts the Pythagorean theorem.

Moreover, the Pythagorean theorem is valid in hyperbolic and elliptic geometry if the requirement that the triangle is rectangular is replaced by the condition that the sum of two angles of the triangle must be equal to the third.

Spherical geometry

For any right triangle on a sphere with radius R (\displaystyle R)(for example, if the angle in a triangle is right) with sides a , b , c (\displaystyle a,b,c) the relationship between the sides is:

cos ⁡ (c R) = cos ⁡ (a R) ⋅ cos ⁡ (b R) (\displaystyle \cos \left((\frac (c)(R))\right)=\cos \left((\frac (a)(R))\right)\cdot \cos \left((\frac (b)(R))\right)).

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

cos ⁡ (c R) = cos ⁡ (a R) ⋅ cos ⁡ (b R) + sin ⁡ (a R) ⋅ sin ⁡ (b R) ⋅ cos ⁡ γ (\displaystyle \cos \left((\frac ( c)(R))\right)=\cos \left((\frac (a)(R))\right)\cdot \cos \left((\frac (b)(R))\right)+\ sin \left((\frac (a)(R))\right)\cdot \sin \left((\frac (b)(R))\right)\cdot \cos \gamma ). ch ⁡ c = ch ⁡ a ⋅ ch ⁡ b (\displaystyle \operatorname (ch) c=\operatorname (ch) a\cdot \operatorname (ch) b),

Where ch (\displaystyle \operatorname (ch) )- hyperbolic cosine. This formula is a special case of the hyperbolic cosine theorem, which is valid for all triangles:

ch ⁡ c = ch ⁡ a ⋅ ch ⁡ b − sh ⁡ a ⋅ sh ⁡ b ⋅ cos ⁡ γ (\displaystyle \operatorname (ch) c=\operatorname (ch) a\cdot \operatorname (ch) b-\operatorname (sh) a\cdot \operatorname (sh) b\cdot \cos \gamma ),

Where γ (\displaystyle \gamma )- an angle whose vertex is opposite to the side c (\displaystyle c).

Using the Taylor series for the hyperbolic cosine ( ch ⁡ x ≈ 1 + x 2 / 2 (\displaystyle \operatorname (ch) x\approx 1+x^(2)/2)) it can be shown that if a hyperbolic triangle decreases (that is, when a (\displaystyle a), b (\displaystyle b) And c (\displaystyle c) tend to zero), then the hyperbolic relations in a right triangle approach the relation of the classical Pythagorean theorem.

Application

Distance in two-dimensional rectangular systems

The most important application of the Pythagorean theorem is determining the distance between two points in a rectangular coordinate system: distance s (\displaystyle s) between points with coordinates (a , b) (\displaystyle (a,b)) And (c , d) (\displaystyle (c,d)) equals:

s = (a − c) 2 + (b − d) 2 (\displaystyle s=(\sqrt ((a-c)^(2)+(b-d)^(2)))).

For complex numbers, the Pythagorean theorem gives a natural formula for finding the modulus of a complex number - for z = x + y i (\displaystyle z=x+yi) it is equal to the length

It is remarkable that the property specified in the Pythagorean theorem is a characteristic property of a right triangle. This follows from the theorem converse to the Pythagorean theorem.

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 right-angled.

Heron's formula

Let us derive a formula expressing the plane of a triangle in terms of the lengths of its sides. This formula is associated with the name of Heron of Alexandria - an ancient Greek mathematician and mechanic who probably lived in the 1st century AD. Heron paid much attention to the practical applications of geometry.

Theorem. The area S of a triangle whose sides are equal to a, b, c is calculated by the formula S=, where p is the semi-perimeter of the triangle.

Proof.

Given: ?ABC, AB= c, BC= a, AC= b. Angles A and B are acute. CH - height.

Prove:

Proof:

Consider triangle ABC, in which AB=c, BC=a, AC=b. Every triangle has at least two acute angles. Let A and B be acute angles of triangle ABC. Then the base H of altitude CH of the triangle lies on side AB. Let us introduce the following notation: CH = h, AH=y, HB=x. by the Pythagorean theorem a 2 - x 2 = h 2 =b 2 -y 2, whence

Y 2 - x 2 = b 2 - a 2, or (y - x) (y + x) = b 2 - a 2, and since y + x = c, then y- x = (b2 - a2).

Adding the last two equalities, we get:

2y = +c, whence

y=, and, therefore, h 2 = b 2 -y 2 =(b - y)(b+y)=

Subject: The theorem converse to the Pythagorean theorem.

Lesson objectives: 1) consider the theorem converse to the Pythagorean theorem; its application in the process of problem solving; consolidate the Pythagorean theorem and improve problem solving skills for its application;

2) develop logical thinking, creative search, cognitive interest;

3) to cultivate in students a responsible attitude to learning and a culture of mathematical speech.

Lesson type. A lesson in learning new knowledge.

During the classes

І. Organizing time

ІІ. Update knowledge

Lesson for mewouldI wantedstart with a quatrain.

Yes, the path of knowledge is not smooth

But we know from our school years,

There are more mysteries than answers,

And there is no limit to the search!

So, in the last lesson you learned the Pythagorean theorem. Questions:

The Pythagorean theorem is true for which figure?

Which triangle is called a right triangle?

State the Pythagorean theorem.

How can the Pythagorean theorem be written for each triangle?

Which triangles are called equal?

Formulate the criteria for the equality of triangles?

Now let’s do a little independent work:

Solving problems using drawings.

№1

(1 b.) Find: AB.

№2

(1 b.) Find: VS.

№3

( 2 b.)Find: AC

№4

(1 point)Find: AC

№5 Given by: ABCDrhombus

(2 b.) AB = 13 cm

AC = 10 cm

Find inD

Self-test No. 1. 5

№2. 5

№3. 16

№4. 13

№5. 24

ІІІ. Studying new material.

The ancient Egyptians built right angles on the ground in this way: they divided the rope into 12 equal parts with knots, tied its ends, after which the rope was stretched on the ground so that a triangle was formed with sides of 3, 4 and 5 divisions. The angle of the triangle that lay opposite the side with 5 divisions was right.

Can you explain the correctness of this judgment?

As a result of searching for an answer to the question, students should understand that from a mathematical point of view the question is posed: will the triangle be right-angled?

We pose a problem: how to determine, without making measurements, whether a triangle with given sides will be rectangular. Solving this problem is the goal of the lesson.

Write down the topic of the lesson.

Theorem. If the sum of the squares of two sides of a triangle is equal to the square of the third side, then the triangle is right-angled.

Prove the theorem independently (make a proof plan using the textbook).

From this theorem it follows that a triangle with sides 3, 4, 5 is right-angled (Egyptian).

In general, numbers for which equality holds , are called Pythagorean triplets. And triangles whose side lengths are expressed by Pythagorean triplets (6, 8, 10) are Pythagorean triangles.

Consolidation.

Because , then a triangle with sides 12, 13, 5 is not right-angled.

Because , then a triangle with sides 1, 5, 6 is right-angled.

№ 430 (a, b, c)

( - is not)

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

between the sides of a right triangle.

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

Geometric formulation of the Pythagorean theorem.

The theorem was originally formulated as follows:

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

built on legs.

Algebraic formulation of the Pythagorean theorem.

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

That is, denoting the length of the hypotenuse of the triangle by c, and the lengths of the legs through a And b:

Both formulations Pythagorean theorem are equivalent, but the second formulation is more elementary, it does not

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

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

Converse 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

right triangle.

Or, in other words:

For every triple of positive numbers a, b And c, such that

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

Pythagorean theorem for an isosceles triangle.

Pythagorean theorem for an equilateral triangle.

Proofs of the Pythagorean theorem.

Currently, 367 proofs of this theorem have been recorded in the scientific literature. Probably the theorem

Pythagoras is the only theorem with such an impressive number of proofs. Such diversity

can only be explained by the fundamental significance of the theorem for geometry.

Of course, conceptually all of them can be divided into a small number of classes. The most famous of them:

proof area method, axiomatic And exotic evidence(For example,

by using differential equations).

1. Proof of the Pythagorean theorem using 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 foundation through H.

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

By introducing the notation:

we get:

,

which corresponds to -

Folded a 2 and b 2, we get:

or , which is what needed to be proven.

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

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

use properties of area, the proofs of which are more complex than the proof of the Pythagorean theorem itself.

• Proof through equicomplementarity.

Let's arrange four equal rectangular

triangle as shown in the figure

on right.

Quadrangle with sides c- square,

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

unfolded angle - 180°.

The area of ​​the entire figure is equal, on the one hand,

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

Q.E.D.

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

​

Looking at the drawing shown in the figure and

watching the side changea, we can

write the following relation for infinitely

small side incrementsWith And a(using similarity

triangles):

Using the variable separation method, we find:

A more general expression for the change in the hypotenuse in the case of increments on both sides:

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 related to the independent

contributions from the increment of different legs.

A simpler proof can be obtained if we assume that one of the legs does not experience an increase

(in this case the leg b). Then for the integration constant we obtain: