Drawing of a parallelogram with the diagonal of the sides of the parallelogram are equal. Property of the diagonals of a parallelogram. Complete lessons - Knowledge Hypermarket. Diagonals bisected by intersection point
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Lesson summary.
Algebra grade 8
Teacher Sysoy A.K.
School 1828
Lesson topic: "Parallelogram and its properties"
Lesson type: combined
Lesson objectives:
1) Ensure the assimilation of a new concept - a parallelogram and its properties
2) Continue the development of skills and abilities for solving geometric problems;
3) Development of a culture of mathematical speech
Lesson plan:
1. Organizational moment
(Slide 1)
The slide demonstrates a statement by Lewis Carroll. The students are informed about the purpose of the lesson. Pupils' readiness for the lesson is checked.
2. Updating knowledge
(Slide 2)
On the blackboard tasks for oral work. The teacher invites the students to think about these problems and raise their hand to those who have understood how to solve the problem. After solving two problems, a student is called to the blackboard to prove the theorem on the sum of angles, who independently makes additional constructions on the drawing and verbally proves the theorem.
Students use the formula for the sum of the angles of a polygon:
3. Main part
(Slide 3)
On the board is the definition of a parallelogram. The teacher talks about the new figure and formulates the definition, making the necessary explanations with the help of the drawing. Then, on the checkered part of the presentation, using a marker and a ruler, he shows how you can draw a parallelogram (several cases are possible)
(Slide 4)
The teacher formulates the first property of a parallelogram. Asks the students to say, from the drawing, what is given and what needs to be proved. After that, tasks are given on the board. Students guess (maybe with the help of a teacher) that the sought equalities must be proved through the equalities of the triangles, which can be obtained by drawing a diagonal (a diagonal appears on the board). Further, the students guess why the triangles are equal and call the sign of equality of triangles (the corresponding shape appears). Verbally communicate the facts that are necessary for the equality of the triangles (as they name them, the corresponding visualization appears). Further, the students formulate the property of equal triangles, it appears in the form of point 3 of the proof and then independently complete the proof of the theorem orally.
(Slide 5)
The teacher formulates the second property of the parallelogram. A parallelogram drawing appears on the board. The teacher offers to say from the drawing what is given, what needs to be proved. After the students correctly report what is given and what needs to be proved, the condition of the theorem appears. Students guess that the equality of the parts of the diagonals can be proved through the equality of trianglesAOB and COD... Using the previous parallelogram property, one guesses about the equality of the sidesAB and CD... Then they understand that it is necessary to find equal angles and, using the properties of parallel straight lines, prove the equality of the angles adjacent to equal sides. These stages are visualized on a slide. The truth of the theorem follows from the equality of the triangles - the students say on the slide, the corresponding visualization appears.
(Slide 6)
The teacher formulates the third property of the parallelogram. Depending on the time that remains until the end of the lesson, the teacher can give the students the opportunity to independently prove this property, or limit themselves to its formulation, and leave the proof itself to the students as homework. The proof can be based on the sum of the angles of the inscribed polygon, which was repeated at the beginning of the lesson, or on the sum of the inner one-sided angles for two parallel linesAD and BC, and a secant, for exampleAB.
4. Securing the material
At this stage, students use previously learned theorems to solve problems. Students select ideas for solving the problem independently. Because possible options There is a lot of design and they all depend on how the students will look for a solution to the problem, there is no visualization of the solution to the problems, and the students independently draw up each stage of the solution on a separate board with the solution written in a notebook.
(Slide 7)
The task condition appears. The teacher proposes to formulate "Given" according to the condition. After the disciples have correctly compiled short note conditions, "Given" appears on the board. The progress of solving the problem may look like this:
Let's draw the height BH (rendered)
Triangle AHB is rectangular. Angle A equal to the angle C and is equal to 30 0 (by the property of opposite angles in a parallelogram). 2BH = AB (according to the property of the leg, which lies opposite the angle of 30 0 in right triangle). So AB = 13 cm.
AB = CD, BC = AD (by the property of opposite sides in a parallelogram) So AB = CD = 13cm. Since the perimeter of the parallelogram is 50 cm, then BC = AD = (50 - 26): 2 = 12cm.
Answer: AB = CD = 13 cm, BC = AD = 12 cm.
(Slide 8)
The task condition appears. The teacher proposes to formulate "Given" according to the condition. Then "Given" appears on the screen. With the help of red lines, a quadrilateral is selected, about which you need to prove that it is a parallelogram. The progress of solving the problem may look like this:
Because BK and MD are perpendiculars to one straight line, then straight lines BK and MD are parallel.
Through adjacent angles, it can be shown that the sum of the inner one-sided angles for straight lines BM and KD and secant MD is equal to 180 0. Therefore, these lines are parallel.
Since the opposite sides of the BMDK quadrilateral are pairwise parallel, this quadrilateral is a parallelogram.
5. End of the lesson. Outcome behavior.
(Slide 8)
Questions on a new topic appear on the slide, to which students answer.
In which the opposite sides are parallel, that is, lie on parallel lines. Special cases of a parallelogram are rectangle, square and rhombus.
Properties
- The opposite sides of the parallelogram are equal.
- The opposite angles of the parallelogram are equal.
- The sum of the angles adjacent to one side is 180 ° (by the property of parallel lines).
- The parallelogram diagonals intersect, and the intersection point divides them in half: .
- The intersection point of the diagonals is the center of symmetry of the parallelogram.
- The parallelogram is divided by the diagonal into two equal triangles.
- The middle lines of a parallelogram intersect at the point of intersection of its diagonals. At this point, two of its diagonals and two of its middle lines are divided in half.
- Parallelogram identity: the sum of the squares of the diagonals of a parallelogram is equal to twice the sum of the squares of its two adjacent sides: let a be the length of side AB, b - the length of side BC, and - the length of the diagonals; then
- An affine transform always transforms a parallelogram into a parallelogram. For any parallelogram, there is an affine transformation that maps it to a square.
Parallelogram signs
Quadrilateral ABCD is a parallelogram if one of the following conditions is met (in this case, all the others are also met):
- A quadrilateral without self-intersections has two opposite sides simultaneously equal and parallel: .
- All opposite angles are equal in pairs: .
- For a quadrilateral without self-intersections, all opposite sides are pairwise equal: .
- All opposite sides are parallel in pairs: .
- The diagonals are halved at their intersection: .
- The sum of adjacent angles is 180 degrees: .
- The sum of the distances between the midpoints of opposite sides of a convex quadrilateral is equal to its semiperimeter.
- The sum of the squares of the diagonals is equal to the sum of the squares of the sides of a convex quadrilateral: .
Parallelogram area
Here are the formulas peculiar to the parallelogram. See also formulas for the area of arbitrary quadrangles.The area of a parallelogram is equal to the product of its base and height:
, where - side, is the height drawn to this side.
The area of a parallelogram is equal to the product of its sides by the sine of the angle between them:
, where and - sides, and - the angle between sides a and b.
see also
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Excerpt from Parallelogram
“The doctor says there’s no danger,” said the Countess, but as she spoke, she raised her eyes with a sigh, and there was an expression in this gesture that contradicted her words.- Where is he? Can I see him, can I? - asked the princess.
- Now, princess, now, my friend. Is this his son? - she said, referring to Nikolushka, who entered with Desal. - We can all fit, the house is big. Oh, what a lovely boy!
The countess ushered the princess into the drawing-room. Sonya talked to m lle Bourienne. The Countess caressed the boy. The old count entered the room, greeting the princess. The old count has changed tremendously since the last time the princess saw him. Then he was a lively, cheerful, self-confident old man, now he seemed a pitiful, lost person. As he spoke to the princess, he constantly looked around, as if asking everyone if he was doing what was needed. After the devastation of Moscow and his estate, knocked out of his usual rut, he apparently lost consciousness of his importance and felt that he had no place in life.
In spite of the excitement in which she was, in spite of one desire to see her brother as soon as possible and her annoyance that at that moment, when she only wanted to see him, she was occupied and pretended to praise her nephew, the princess noticed everything that was done around her, and felt the need for a while to submit to this new order into which she was entering. She knew that all this was necessary, and it was difficult for her, but she did not annoy them.
“This is my niece,” said the count, introducing Sonya. “You don’t know her, princess?”
The princess turned to her and, trying to extinguish the hostile feeling that had risen in her soul towards this girl, kissed her. But it became hard for her because the mood of everyone around her was so far from what was in her soul.
- Where is he? She asked again, addressing everyone.
"He's downstairs, Natasha is with him," Sonya answered, blushing. - Let's go find out. I think you are tired, princess?
Tears of annoyance came to the princess's eyes. She turned away and wanted to ask the countess again where to go to him, as light, swift, as if merry steps were heard in the doorway. The princess looked around and saw Natasha almost running in, that Natasha who had so disliked her on that long-standing meeting in Moscow.
But before the princess had time to look at the face of this Natasha, she realized that this was her sincere comrade in grief, and therefore her friend. She rushed to meet her and, embracing her, wept on her shoulder.
As soon as Natasha, who was sitting at the head of Prince Andrei, learned of the arrival of Princess Marya, she quietly left his room with those quick, as it seemed to Princess Marya, as if with merry steps and ran to her.
On her agitated face, when she ran into the room, there was only one expression - an expression of love, boundless love for him, for her, for everything that was close to a loved one, an expression of pity, suffering for others and a passionate desire to give all of herself for in order to help them. It was evident that at that moment not a single thought about herself, about her relationship to him was in Natasha's soul.
The sensitive Princess Marya understood all this from the first glance on Natasha's face and cried with sorrowful pleasure on her shoulder.
“Let's go, let's go to him, Marie,” Natasha said, leading her to another room.
Princess Marya raised her face, wiped her eyes and turned to Natasha. She felt that from her she would understand and learn everything.
Proof
The first step is to draw a diagonal AC. Two triangles are obtained: ABC and ADC.
Since ABCD is a parallelogram, the following is true:
AD || BC \ Rightarrow \ angle 1 = \ angle 2 as lying crosswise.
AB || CD \ Rightarrow \ angle3 = \ angle 4 as lying crosswise.
Therefore, \ triangle ABC = \ triangle ADC (according to the second criterion: and AC is common).
And, therefore, \ triangle ABC = \ triangle ADC, then AB = CD and AD = BC.
Proven!
2. Opposite angles are the same.
Proof
According to the evidence properties 1 We know that \ angle 1 = \ angle 2, \ angle 3 = \ angle 4... Thus, the sum of opposite angles is: \ angle 1 + \ angle 3 = \ angle 2 + \ angle 4... Considering that \ triangle ABC = \ triangle ADC we get \ angle A = \ angle C, \ angle B = \ angle D.
Proven!
3. The diagonals are bisected by the intersection point.
Proof
Let's draw one more diagonal.
By property 1 we know that the opposite sides are identical: AB = CD. Once again, mark the intersecting equal angles.
Thus, you can see that \ triangle AOB = \ triangle COD by the second sign of equality of triangles (two angles and a side between them). That is, BO = OD (opposite angles \ angle 2 and \ angle 1) and AO = OC (opposite angles \ angle 3 and \ angle 4, respectively).
Proven!
Parallelogram signs
If only one feature is present in your task, then the figure is a parallelogram and you can use all the properties of this figure.
For better memorization, we note that the parallelogram sign will answer the following question - "how to find out?"... That is, how do you know that a given figure is a parallelogram.
1. A parallelogram is a quadrilateral in which two sides are equal and parallel.
AB = CD; AB || CD \ Rightarrow ABCD - parallelogram.
Proof
Let's take a closer look. Why AD || BC?
\ triangle ABC = \ triangle ADC by property 1: AB = CD, AC - total and \ angle 1 = \ angle 2 as criss-cross at parallel AB and CD and secant AC.
But if \ triangle ABC = \ triangle ADC, then \ angle 3 = \ angle 4 (lie opposite AB and CD, respectively). And hence AD || BC (\ angle 3 and \ angle 4 are also equal).
The first sign is correct.
2. A parallelogram is a quadrilateral in which opposite sides are equal.
AB = CD, AD = BC \ Rightarrow ABCD - parallelogram.
Proof
Consider this feature. Draw the diagonal AC again.
By property 1\ triangle ABC = \ triangle ACD.
It follows that: \ angle 1 = \ angle 2 \ Rightarrow AD || BC and \ angle 3 = \ angle 4 \ Rightarrow AB || CD, that is, ABCD is a parallelogram.
The second sign is correct.
3. A parallelogram is a quadrilateral in which opposite angles are equal.
\ angle A = \ angle C, \ angle B = \ angle D \ Rightarrow ABCD- parallelogram.
Proof
2 \ alpha + 2 \ beta = 360 ^ (\ circ)(since ABCD is a quadrilateral and \ angle A = \ angle C, \ angle B = \ angle D by condition).
So, \ alpha + \ beta = 180 ^ (\ circ). But \ alpha and \ beta are internal one-sided with a secant AB.
And the fact that \ alpha + \ beta = 180 ^ (\ circ) also says that AD || BC.
In this case, \ alpha and \ beta are internal one-sided with a secant AD. And that means AB || CD.
The third sign is correct.
4. A parallelogram is a quadrilateral in which the diagonals are bisected by the intersection point.
AO = OC; BO = OD \ Rightarrow parallelogram.
Proof
BO = OD; AO = OC, \ angle 1 = \ angle 2 as vertical \ Rightarrow \ triangle AOB = \ triangle COD, \ Rightarrow \ angle 3 = \ angle 4, and \ Rightarrow AB || CD.
Similarly BO = OD; AO = OC, \ angle 5 = \ angle 6 \ Rightarrow \ triangle AOD = \ triangle BOC \ Rightarrow \ angle 7 = \ angle 8, and \ Rightarrow AD || BC.
The fourth sign is correct.
