Pythagorean theorem direct and inverse with proof. Lesson "theorem - the inverse of the Pythagorean theorem"

Lesson Objectives:

general education:

  • check the theoretical knowledge of students (properties of a right triangle, the Pythagorean theorem), the ability to use them in solving problems;
  • creating problem situation, bring students to the “discovery” of the inverse Pythagorean theorem.

developing:

  • development of skills to apply theoretical knowledge in practice;
  • development of the ability to formulate conclusions during observations;
  • development of memory, attention, observation:
  • development of learning motivation through emotional satisfaction from discoveries, through the introduction of elements of the history of the development of mathematical concepts.

educational:

  • to cultivate a steady interest in the subject through the study of the life of Pythagoras;
  • fostering mutual assistance and objective assessment of classmates' knowledge through peer review.

Lesson form: class-lesson.

Lesson plan:

  • Organizing time.
  • Examination homework. Knowledge update.
  • Solving practical problems using the Pythagorean theorem.
  • New topic.
  • Primary consolidation of knowledge.
  • Homework.
  • Lesson results.
  • Independent work (according to individual cards with guessing the aphorisms of Pythagoras).

During the classes.

Organizing time.

Checking homework. Knowledge update.

Teacher: What task did you do at home?

Students: Given two sides of a right-angled triangle, find the third side, arrange the answers in the form of a table. Repeat the properties of a rhombus and a rectangle. Repeat what is called the condition and what is the conclusion of the theorem. Prepare reports on the life and work of Pythagoras. Bring a rope with 12 knots tied to it.

Teacher: Check answers to homework according to the table

(data are in black, responses are in red).

Teacher: Statements are written on the board. If you agree with them on the sheets of paper opposite the corresponding question number, put “+”, if you do not agree, then put “-”.

Statements are written on the board.

  1. The hypotenuse is larger than the leg.
  2. Sum sharp corners right triangle is 180 0 .
  3. Area of ​​a right triangle with legs A And V calculated by the formula S=ab/2.
  4. The Pythagorean theorem is true for all isosceles triangles.
  5. In a right triangle, the leg opposite the angle 30 0 is equal to half the hypotenuse.
  6. The sum of the squares of the legs is equal to the square of the hypotenuse.
  7. The square of the leg is equal to the difference of the squares of the hypotenuse and the second leg.
  8. The side of a triangle is equal to the sum of the other two sides.

Works are checked by peer review. Controversial statements are discussed.

Key to theoretical questions.

Students rate each other according to the following system:

8 correct answers “5”;
6-7 correct answers “4”;
4-5 correct answers “3”;
less than 4 correct answers “2”.

Teacher: What did we talk about in the last lesson?

Student: About Pythagoras and his theorem.

Teacher: Formulate the Pythagorean theorem. (Several students read the wording, at this time 2-3 students prove it at the blackboard, 6 students at the first desks on the sheets).

Mathematical formulas are written on the magnetic board on the cards. Choose those that reflect the meaning of the Pythagorean theorem, where A And V - catheters, With - hypotenuse.

1) c 2 \u003d a 2 + b 2 2) c \u003d a + b 3) a 2 \u003d from 2 - to 2
4) c 2 \u003d a 2 - in 2 5) in 2 \u003d c 2 - a 2 6) a 2 \u003d c 2 + in 2

While the students proving the theorem at the blackboard and in the field are not ready, the floor is given to those who prepared reports on the life and work of Pythagoras.

Schoolchildren working in the field hand over leaflets and listen to the evidence of those who worked at the blackboard.

Solving practical problems using the Pythagorean theorem.

Teacher: I offer you practical tasks using the studied theorem. We will visit the forest first, after the storm, then in the countryside.

Task 1. After the storm, the spruce broke. The height of the remaining part is 4.2 m. The distance from the base to the fallen top is 5.6 m. Find the height of the spruce before the storm.

Task 2. The height of the house is 4.4 m. The width of the lawn around the house is 1.4 m. How long should the ladder be made so that it does not step on the lawn and reaches the roof of the house?

New topic.

Teacher:(music plays) Close your eyes, for a few minutes we will plunge into history. We are with you in Ancient Egypt. Here in the shipyards the Egyptians build their famous ships. But land surveyors, they measure plots of land, the boundaries of which were washed away after the flood of the Nile. Builders build grandiose pyramids that still amaze us with their magnificence. In all these activities, the Egyptians needed to use right angles. They knew how to build them using a rope with 12 knots tied at the same distance from each other. Try and you, arguing like the ancient Egyptians, build right-angled triangles with the help of your ropes. (Solving this problem, the guys work in groups of 4 people. After a while, someone shows the construction of a triangle on the tablet at the blackboard).

The sides of the resulting triangle are 3, 4 and 5. If you tie one more knot between these knots, then its sides will become 6, 8 and 10. If two each - 9, 12 and 15. All these triangles are rectangular because.

5 2 \u003d 3 2 + 4 2, 10 2 \u003d 6 2 + 8 2, 15 2 \u003d 9 2 + 12 2, etc.

What property must a triangle have in order to be a right triangle? (Students try to formulate the inverse Pythagorean theorem themselves, finally, someone succeeds).

How is this theorem different from the Pythagorean theorem?

Student: The condition and the conclusion are reversed.

Teacher: At home, you repeated what such theorems are called. So what are we up to now?

Student: With the inverse Pythagorean theorem.

Teacher: Write down the topic of the lesson in your notebook. Open your textbooks on page 127, read this statement again, write it down in your notebook and analyze the proof.

(After several minutes of independent work with the textbook, if desired, one person at the blackboard gives a proof of the theorem).

  1. What is the name of a triangle with sides 3, 4 and 5? Why?
  2. What triangles are called Pythagorean triangles?
  3. What triangles did you work with in your homework? And in problems with a pine tree and a ladder?

Primary consolidation of knowledge

.

This theorem helps solve problems in which it is necessary to find out whether triangles are right triangles.

Tasks:

1) Find out if a triangle is right-angled if its sides are equal:

a) 12.37 and 35; b) 21, 29 and 24.

2) Calculate the heights of a triangle with sides 6, 8 and 10 cm.

Homework

.

Page 127: Inverse Pythagorean theorem. No. 498 (a, b, c) No. 497.

Lesson results.

What new did you learn in the lesson?
  • How did the Egyptians use the inverse Pythagorean theorem?
  • What tasks is it used for?
  • What triangles did you meet?
  • What do you remember and like the most?
  • Independent work (carried out on individual cards).

    Teacher: At home, you repeated the properties of a rhombus and a rectangle. List them (there is a conversation with the class). In the last lesson, we talked about the fact that Pythagoras was a versatile person. He was engaged in medicine, and music, and astronomy, and was also an athlete and participated in Olympic Games. Pythagoras was also a philosopher. Many of his aphorisms are still relevant to us today. Now you will perform independent work. For each task, several answers are given, next to which fragments of Pythagorean aphorisms are written. Your task is to solve all the tasks, make a statement from the received fragments and write it down.

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

    Approximately 400 BC. e., according to Proclus, Plato gave a method for finding Pythagorean triples, combining algebra and geometry. Around 300 B.C. e. in the "Elements" of Euclid appeared the oldest axiomatic proof of the Pythagorean theorem.

    Wording

    The main wording is algebraic actions- in a right triangle, the lengths of the legs of which are equal a (\displaystyle a) And b (\displaystyle b), and the length of the hypotenuse is c (\displaystyle c), the relation is fulfilled:

    .

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

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

    Proof

    IN scientific literature at least 400 proofs of the Pythagorean theorem have been recorded, which is explained both by its fundamental importance for geometry and by the elementarity of the result. The main directions of proofs are: algebraic use of the ratios of elements triangle (such, for example, is the popular similarity method), area method, there are also various exotic proofs (for example, using differential equations).

    Through similar triangles

    Euclid's classic proof aims to establish the equality of areas between rectangles formed by dissecting a square over a hypotenuse with a height of right angle with squares over the legs.

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

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

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

    Proof of Leonardo da Vinci

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

    Proof by the infinitesimal method

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

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

    By the method of separation of variables, one derives from them differential equation 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 initial conditions a = b = c = 0 (\displaystyle a=b=c=0) defines a constant as 0, which results in the assertion of the theorem.

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

    Variations and Generalizations

    Similar geometric shapes on three sides

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

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

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

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

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

    Cosine theorem

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

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

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

    Arbitrary triangle

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

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

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

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

    Pappus area theorem

    Non-Euclidean geometry

    The Pythagorean theorem is derived from the axioms of Euclidean geometry and is invalid for non-Euclidean geometry - the fulfillment of the Pythagorean theorem is equivalent to the postulate of Euclidean parallelism.

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

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

    spherical geometry

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

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

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

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

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

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

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

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

    Application

    Distance in two-dimensional rectangular systems

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

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

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

    Consideration of topics school curriculum with the help of video lessons is a convenient way to study and assimilate the material. Video helps to focus students' attention on the main theoretical points and not to miss important details. If necessary, students can always listen to the video lesson again or go back a few topics.

    This 8th grade video lesson will help students learn new theme by geometry.

    In the previous topic, we studied the Pythagorean theorem and analyzed its proof.

    There is also a theorem which is known as the inverse Pythagorean theorem. Let's consider it in more detail.

    Theorem. A triangle is right-angled if it satisfies the equality: the value of one side of the triangle squared is the same as the sum of the other two sides squared.

    Proof. Suppose we are given a triangle ABC, in which the equality AB 2 = CA 2 + CB 2 is true. We need to prove that angle C is 90 degrees. Consider a triangle A 1 B 1 C 1 in which angle C 1 is 90 degrees, side C 1 A 1 is equal to CA and side B 1 C 1 is equal to BC.

    Applying the Pythagorean theorem, we write the ratio of the sides in the triangle A 1 C 1 B 1: A 1 B 1 2 = C 1 A 1 2 + C 1 B 1 2 . By replacing the expression with equal sides, we get A 1 B 1 2 = CA 2 + CB 2 .

    We know from the conditions of the theorem that AB 2 = CA 2 + CB 2 . Then we can write A 1 B 1 2 = AB 2 , which implies that A 1 B 1 = AB.

    We found that in triangles ABC and A 1 B 1 C 1 three sides are equal: A 1 C 1 = AC, B 1 C 1 = BC, A 1 B 1 = AB. So these triangles are congruent. From the equality of triangles it follows that the angle C is equal to the angle C 1 and, accordingly, is equal to 90 degrees. We have determined that triangle ABC is a right triangle and its angle C is 90 degrees. We have proved this theorem.

    The author then gives an example. Suppose we are given an arbitrary triangle. The dimensions of its sides are known: 5, 4 and 3 units. Let's check the statement from the theorem converse to the Pythagorean theorem: 5 2 = 3 2 + 4 2 . If the statement is correct, then the given triangle is a right triangle.

    In the following examples, the triangles will also be right-angled if their sides are equal:

    5, 12, 13 units; the equality 13 2 = 5 2 + 12 2 is true;

    8, 15, 17 units; the equation 17 2 = 8 2 + 15 2 is true;

    7, 24, 25 units; the equation 25 2 = 7 2 + 24 2 is true.

    The concept of the Pythagorean triangle is known. It is a right triangle whose side values ​​are integers. If the legs of the Pythagorean triangle are denoted by a and c, and the hypotenuse b, then the values ​​of the sides of this triangle can be written using the following formulas:

    b \u003d k x (m 2 - n 2)

    c \u003d k x (m 2 + n 2)

    where m, n, k are any integers, and the value of m is greater than the value of n.

    An interesting fact: a triangle with sides 5, 4 and 3 is also called the Egyptian triangle, such a triangle was known in ancient Egypt.

    In this video tutorial, we got acquainted with the theorem, the converse of the Pythagorean theorem. Consider the proof in detail. Students also learned which triangles are called Pythagorean triangles.

    Students can easily get acquainted with the topic "Theorem, converse theorem Pythagoras" independently with the help of this video tutorial.

    Lesson Objectives:

    Educational: formulate and prove the Pythagorean theorem and the converse of the Pythagorean theorem. Show their historical and practical significance.

    Developing: develop attention, memory, logical thinking students, the ability to reason, compare, draw conclusions.

    Educational: to cultivate interest and love for the subject, accuracy, the ability to listen to comrades and teachers.

    Equipment: Portrait of Pythagoras, posters with tasks for consolidation, textbook "Geometry" grades 7-9 (I.F. Sharygin).

    Lesson plan:

    I. Organizational moment - 1 min.

    II. Checking homework - 7 min.

    III. introduction teachers, historical background - 4-5 min.

    IV. Formulation and proof of the Pythagorean theorem - 7 min.

    V. Formulation and proof of the theorem converse to the Pythagorean theorem - 5 min.

    Fixing new material:

    a) oral - 5-6 minutes.
    b) written - 7-10 min.

    VII. Homework - 1 min.

    VIII. Summing up the lesson - 3 min.

    During the classes

    I. Organizational moment.

    II. Checking homework.

    p.7.1, No. 3 (at the board according to the finished drawing).

    Condition: The height of a right triangle divides the hypotenuse into segments of length 1 and 2. Find the legs of this triangle.

    BC = a; CA=b; BA=c; BD = a 1 ; DA = b 1 ; CD = hC

    Additional question: write down the ratios in a right triangle.

    item 7.1, No. 5. Cut the right triangle into three triangles similar to each other.

    Explain.

    ASN ~ ABC ~ SVN

    (draw students' attention to the correct recording of the corresponding vertices of similar triangles)

    III. Introductory speech of the teacher, historical background.

    The truth will remain eternal, as soon as a weak person knows it!

    And now the Pythagorean theorem is true, as in his distant age.

    It is no coincidence that I began my lesson with the words of the German novelist Chamisso. Our lesson today is about the Pythagorean theorem. Let's write the topic of the lesson.

    Before you is a portrait of the great Pythagoras. Born in 576 BC. Having lived for 80 years, he died in 496 BC. Known as an ancient Greek philosopher and teacher. He was the son of the merchant Mnesarchus, who often took him on his trips, thanks to which the boy developed curiosity and a desire to learn new things. Pythagoras is a nickname given to him for his eloquence (“Pythagoras” means “persuasive speech”). He himself did not write anything. All his thoughts were recorded by his students. As a result of the first lecture he gave, Pythagoras acquired 2,000 students who, together with their wives and children, formed a huge school and created a state called “Great Greece”, which is based on the laws and rules of Pythagoras, revered as divine commandments. He was the first to call his reasoning about the meaning of life philosophy (philosophy). He was prone to mystification and demonstrative behavior. Once Pythagoras hid underground, and learned about everything that was happening from his mother. Then, withered like a skeleton, he declared in the public assembly that he had been in Hades, and showed amazing awareness of earthly events. For this, the touched inhabitants recognized him as God. Pythagoras never cried and was generally inaccessible to passions and excitement. He believed that he comes from a seed that is better compared to human. The whole life of Pythagoras is a legend that has come down to our time and told us about the most talented man of the ancient world.

    IV. Formulation and proof of the Pythagorean theorem.

    The formulation of the Pythagorean theorem is known to you from the course of algebra. Let's remember her.

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

    However, this theorem was known many years before Pythagoras. For 1500 years before Pythagoras, the ancient Egyptians knew that a triangle with sides 3, 4 and 5 is rectangular and used this property to build right angles when planning land and constructing buildings. In the oldest Chinese mathematical and astronomical work “Zhiu-bi” that has come down to us, written 600 years before Pythagoras, among other sentences related to a right triangle, the Pythagorean theorem is also contained. Even earlier, this theorem was known to the Hindus. Thus, Pythagoras did not discover this property of a right-angled triangle; he was probably the first to generalize and prove it, to transfer it from the field of practice to the field of science.

    Since ancient times, mathematicians have been finding more and more proofs of the Pythagorean theorem. There are over a hundred and fifty known. Let's recall the algebraic proof of the Pythagorean theorem, known to us from the course of algebra. (“Mathematics. Algebra. Functions. Data analysis” G.V. Dorofeev, M., “Bubblehead”, 2000).

    Invite students to remember the proof for the drawing and write it on the board.

    (a + b) 2 \u003d 4 1/2 a * b + c 2 b a

    a 2 + 2a * b + b 2 \u003d 2a * b + c 2

    a 2 + b 2 = c 2 a a b

    The ancient Hindus, to whom this reasoning belongs, usually did not write it down, but accompanied the drawing with only one word: “Look.”

    Let us consider in a modern presentation one of the proofs belonging to Pythagoras. At the beginning of the lesson, we remembered the theorem on ratios in a right triangle:

    h 2 \u003d a 1 * b 1 a 2 \u003d a 1 * c b 2 \u003d b 1 * c

    We add the last two equalities term by term:

    b 2 + a 2 \u003d b 1 * c + a 1 * c \u003d (b 1 + a 1) * c 1 \u003d c * c \u003d c 2; a 2 + b 2 = c 2

    Despite the apparent simplicity of this proof, it is far from being the simplest one. After all, for this it was necessary to draw a height in a right-angled triangle and consider similar triangles. Please write down this proof in your notebook.

    V. Statement and proof of the theorem converse to the Pythagorean theorem.

    What is the inverse of this theorem? (... if the condition and conclusion are reversed.)

    Let's now try to formulate the theorem, the reverse of the Pythagorean theorem.

    If in a triangle with sides a, b and c the equality with 2 \u003d a 2 + b 2 is true, then this triangle is right-angled, and the right angle is opposite to side c.

    (Proof of the inverse theorem on a poster)

    ABC, BC = a,

    AC = b, BA = c.

    a 2 + b 2 = c 2

    Prove:

    ABC - rectangular,

    Proof:

    Consider a right triangle A 1 B 1 C 1,

    where C 1 \u003d 90 °, A 1 C 1 \u003d a, A 1 C 1 \u003d b.

    Then, according to the Pythagorean theorem, B 1 A 1 2 \u003d a 2 + b 2 \u003d c 2.

    That is, B 1 A 1 \u003d c A 1 B 1 C 1 \u003d ABC on three sides of ABC - rectangular

    C = 90°, which was to be proved.

    VI. Consolidation of the studied material (orally).

    1. According to the poster with ready-made drawings.

    Fig.1: find AD if BD = 8, BDA = 30°.

    Fig. 2: find CD if BE = 5, BAE = 45°.

    Fig.3: find BD if BC = 17, AD = 16.

    2. Is a triangle right-angled if its sides are expressed by numbers:

    5 2 + 6 2 ? 7 2 (no)

    9 2 + 12 2 = 15 2 (yes)

    15 2 + 20 2 = 25 2 (yes)

    What are the triples of numbers in the last two cases called? (Pythagorean).

    VI. Problem solving (in writing).

    No. 9. The side of an equilateral triangle is equal to a. Find the height of this triangle, the radius of the circumscribed circle, the radius of the inscribed circle.

    № 14. Prove that in a right triangle the radius of the circumscribed circle is equal to the median drawn to the hypotenuse and equal to half of the hypotenuse.

    VII. Homework.

    Item 7.1, pp. 175-177, analyze Theorem 7.4 (generalized Pythagorean theorem), No. 1 (oral), No. 2, No. 4.

    VIII. Lesson results.

    What new did you learn at the lesson today? …………

    Pythagoras was first and foremost a philosopher. Now I want to read you a few of his sayings, which are relevant in our time for you and me.

    • Do not raise dust on the path of life.
    • Do only what in the future will not upset you and will not force you to repent.
    • Never do what you do not know, but learn everything you need to know, and then you will lead a quiet life.
    • Don't close your eyes when you want to sleep without understanding all your actions on the previous day.
    • Learn to live simply and without luxury.

    Subject: Theorem inverse to the Pythagorean theorem.

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

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

    3) to educate students in a responsible attitude to learning, a culture of mathematical speech.

    Lesson type. A lesson in learning new knowledge.

    During the classes

    І. Organizing time

    ІІ. Update knowledge

    Lesson to mewouldwantedstart with a quatrain.

    Yes, the path of knowledge is not smooth

    But we know from school years

    More mysteries than riddles

    And there is no limit to the search!

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

    The Pythagorean theorem is valid for which figure?

    Which triangle is called a right triangle?

    Formulate the Pythagorean theorem.

    How will the Pythagorean theorem be written for each triangle?

    What triangles are called equal?

    Formulate signs of equality of triangles?

    And now let's do a little independent work:

    Solving problems according to drawings.

    1

    (1 b.) Find: AB.

    2

    (1 b.) Find: BC.

    3

    ( 2 b.)Find: AC

    4

    (1 b.)Find: AC

    5 Given: ABCDrhombus

    (2 b.) AB \u003d 13 cm

    AC = 10 cm

    Find inD

    Self Check #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, which 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: will the triangle be right-angled.

    We pose the problem: how, without making measurements, to determine whether a triangle with given sides is right-angled. Solving this problem is the purpose 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 a right triangle.

    Independently prove the theorem (make up a proof plan according to the textbook).

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

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

    Consolidation.

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

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

      430 (a, b, c)

    ( - is not)