Inverse Pythagorean theorem formula and proof. Lesson "theorem inverse to the Pythagorean theorem"

Lesson objectives:

general education:

developing:

  • development of skills to apply theoretical knowledge in practice;
  • developing the ability to formulate conclusions from 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 sustainable interest in the subject through the study of the life activity of Pythagoras;
  • fostering mutual assistance and objective assessment of classmates’ knowledge through mutual testing.

Lesson format: class-lesson.

Lesson plan:

Progress of the lesson.

Organizational moment.

Checking homework. Updating knowledge.

Teacher: What task did you do at home?

Students: Using two given sides of a right triangle, find the third side and present the answers in table form. 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 on it.

Teacher: Check the answers to your homework using the table

(data are highlighted in black, answers are in red).

Teacher: Statements are written on the board. If you agree with them, put “+” on the pieces of paper next to the corresponding question number; if you don’t agree, then put “–”.

Statements are pre-written on the board.

  1. The hypotenuse is longer than the leg.
  2. The sum of the acute angles of a 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 30 0 angle 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 between the squares of the hypotenuse and the second leg.
  8. A side of a triangle is equal to the sum of the other two sides.

The work is checked using mutual verification. Statements that have caused controversy are discussed.

Key to theoretical questions.

Students grade each other using 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: State the Pythagorean theorem. (Several students read the formulation, at this time 2-3 students prove it at the blackboard, 6 students at the first desks on pieces of paper).

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

1) c 2 = a 2 + b 2 2) c = a + b 3) a 2 = from 2 – in 2
4) with 2 = a 2 – in 2 5) in 2 = c 2 – a 2 6) a 2 = c 2 + c 2

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

Schoolchildren working in the field hand in pieces of paper and listen to the evidence of those who worked at the board.

Solving practical problems using the Pythagorean theorem.

Teacher: I offer you practical problems using the theorem being studied. We will first visit the forest, after the storm, then in a suburban area.

Problem 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.

Problem 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 interfere with the lawn and reaches the roof of the house?

New topic.

Teacher:(music sounds) 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 surveyors measure areas of land whose boundaries were washed away after the Nile flood. Builders build grandiose pyramids that still amaze us with their magnificence. In all of these activities, the Egyptians needed to use right angles. They knew how to build them using a rope with 12 knots tied at equal distances from each other. Try, thinking like the ancient Egyptians, to build right triangles with your ropes. (To solve this problem, the guys work in groups of 4. After a while, someone shows the construction of a triangle on a tablet near the board).

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 there are two each – 9, 12 and 15. All these triangles are right-angled because

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

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

How does this theorem differ from the Pythagorean theorem?

Student: The condition and the conclusion have changed places.

Teacher: At home you repeated what such theorems are called. So what have we met now?

Student: With the inverse Pythagorean theorem.

Teacher: Let's write down the topic of the lesson in our notebook. Open your textbooks to page 127, read this statement again, write it down in your notebook and analyze the proof.

(After a few 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? What about problems with a pine tree and a ladder?

Primary consolidation of knowledge

.

This theorem helps solve problems in which you need to find out whether triangles are right-angled.

Quests:

1) Find out whether 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 summary.

 What new did you learn in the lesson?
  • How was the inverse Pythagorean theorem used in Egypt?
  • What problems is it used to solve?
  • What triangles did you meet?
  • What do you remember and like most?

    • Independent work (carried out using 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 how Pythagoras was a versatile personality. He studied medicine, music, and astronomy, and was also an athlete and participated in the Olympic Games. Pythagoras was also a philosopher. Many of his aphorisms are still relevant for us today. Now you will do independent work. For each task, several answer options are given, next to which fragments of Pythagoras’ aphorisms are written. Your task is to solve all the tasks, compose a statement from the received fragments and write it down.

    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 three 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

    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:

    The Pythagorean theorem states:

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

    a 2 + b 2 = c 2,

    • a And b– legs forming a right angle.
    • With– hypotenuse of the triangle.

    Formulas of the Pythagorean theorem

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

    Proof of the Pythagorean Theorem

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

    S = \frac(1)(2) ab

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

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

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

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

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

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

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

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

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

    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 the triangle is right-angled. That is, for any triple of positive numbers a, b And c, such that

    a 2 + b 2 = c 2,

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

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

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

    Additional material:

    Lesson objectives:

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

    Developmental: 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” for grades 7-9 (I.F. Sharygin).

    Lesson plan:

    I. Organizational moment – ​​1 min.

    II. Checking homework – 7 min.

    III. Introductory speech by the teacher, 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.

    Consolidating new material:

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

    VII. Homework – 1 min.

    VIII. Summing up the lesson – 3 min.

    Lesson progress

    I. Organizational moment.

    II. Checking homework.

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

    Condition: The altitude 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 = h C

    Additional question: write the ratios in a right triangle.

    Section 7.1, No. 5. Cut the right triangle into three similar triangles.

    Explain.

    ASN ~ ABC ~ SVN

    (draw students’ attention to the correctness of writing the corresponding vertices of similar triangles)

    III. Introductory speech by the teacher, historical background.

    The truth will remain eternal as soon as a weak person recognizes 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 down the topic of the lesson.

    Before you is a portrait of the great Pythagoras. Born in 576 BC. Having lived 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 by 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 2000 students, who, together with their wives and children, formed a huge school and created a state called “Greater Greece,” which was 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. One day Pythagoras hid underground, and learned about everything that was happening from his mother. Then, withered like a skeleton, he declared in a public meeting that he had been to Hades, and showed an amazing knowledge of earthly events. For this, the touched residents recognized him as God. Pythagoras never cried and was generally inaccessible to passions and excitement. He believed that he came from a seed that was better than a human one. 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.

    You know the formulation of the Pythagorean theorem from your algebra course. 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. 1500 years before Pythagoras, the ancient Egyptians knew that a triangle with sides 3, 4 and 5 is rectangular and used this property to construct right angles when planning land plots and constructing buildings. In the oldest Chinese mathematical and astronomical work that has come down to us, “Zhiu-bi,” written 600 years before Pythagoras, among other proposals relating to the right triangle, the Pythagorean theorem is contained. 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, to transfer it from the field of practice to the field of science.

    WITH ancient times mathematicians are finding more and more proofs of the Pythagorean theorem. More than one and a half hundred of them are known. Let's remember the algebraic proof of the Pythagorean theorem, known to us from the algebra course. (“Mathematics. Algebra. Functions. Data analysis” G.V. Dorofeev, M., “Drofa”, 2000).

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

    (a + b) 2 = 4 1/2 a * b + c 2 b a

    a 2 + 2a * b + b 2 = 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 about relations in a right triangle:

    h 2 = a 1* b 1 a 2 = a 1* c b 2 = b 1* c

    Let's add the last two equalities term by term:

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

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

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

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

    Let's now try to formulate the theorem converse to the Pythagorean theorem.

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

    (Proof of the converse theorem on the 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 = 90°, A 1 C 1 = a, A 1 C 1 = b.

    Then, by the Pythagorean theorem, B 1 A 1 2 = a 2 + b 2 = c 2.

    That is, B 1 A 1 = c A 1 B 1 C 1 = ABC on three sides ABC is rectangular

    C = 90°, which is what needed to be proven.

    VI. Consolidation of the studied material (orally).

    1. Based on a poster with ready-made drawings.

    Fig. 1: find AD if ВD = 8, ВDA = 30°.

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

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

    2. Is a triangle rectangular 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 names of triplets of numbers in the last two cases? (Pythagorean).

    VI. Solving problems (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, and the radius of the inscribed circle.

    No. 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 the hypotenuse.

    VII. Homework.

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

    VIII. Lesson summary.

    What new did you learn in class today? …………

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

    • Don't raise dust on life's path.
    • Do only what will not upset you later and will not force you to repent.
    • Never do what you don’t 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 having sorted out all your actions of the past day.
    • Learn to live simply and without luxury.

    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:

    Let's consider triangle ABC, in which AB=c, BC=a, AC=b. Every triangle has at least two acute angles. Let A and B be sharp corners 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)=

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