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1. Parallelogram
Compound word "parallelogram"? And behind it is a very simple figure.
Well, that is, we took two parallel lines:
Crossed by two more:
And inside - a parallelogram!
What are the properties of a parallelogram?
Parallelogram properties.
That is, what can be used if a parallelogram is given in the problem?
This question is answered by the following theorem:
Let's draw everything in detail.
What does first point of the theorem? And the fact that if you HAVE a parallelogram, then by all means
The second paragraph means that if there is a parallelogram, then, again, by all means:
Well, and finally, the third point means that if you HAVE a parallelogram, then be sure:
See what a wealth of choice? What to use in the task? Try to focus on the question of the task, or just try everything in turn - some kind of “key” will do.
And now let's ask ourselves another question: how to recognize a parallelogram "in the face"? What must happen to a quadrilateral in order for us to have the right to give it the “title” of a parallelogram?
This question is answered by several signs of a parallelogram.
Features of a parallelogram.
Attention! Begin.
Parallelogram.
Pay attention: if you have found at least one sign in your problem, then you have exactly a parallelogram, and you can use all the properties of a parallelogram.
2. Rectangle
I don't think it will be news to you at all.
The first question is: is a rectangle a parallelogram?
Of course it is! After all, he has - remember, our sign 3?
And from here, of course, it follows that for a rectangle, like for any parallelogram, and, and the diagonals are divided by the intersection point in half.
But there is a rectangle and one distinctive property.
Rectangle property
Why is this property distinctive? Because no other parallelogram has equal diagonals. Let's formulate it more clearly.
Pay attention: in order to become a rectangle, a quadrilateral must first become a parallelogram, and then present the equality of the diagonals.
3. Diamond
And again the question is: is a rhombus a parallelogram or not?
With full right - a parallelogram, because it has and (remember our sign 2).
And again, since a rhombus is a parallelogram, then it must have all the properties of a parallelogram. This means that a rhombus has opposite angles equal, opposite sides are parallel, and the diagonals are bisected by the point of intersection.
Rhombus Properties
Look at the picture:
As in the case of a rectangle, these properties are distinctive, that is, for each of these properties, we can conclude that we have not just a parallelogram, but a rhombus.
Signs of a rhombus
And pay attention again: there should be not just a quadrilateral with perpendicular diagonals, but a parallelogram. Make sure:
No, of course not, although its diagonals and are perpendicular, and the diagonal is the bisector of angles u. But ... the diagonals do not divide, the intersection point in half, therefore - NOT a parallelogram, and therefore NOT a rhombus.
That is, a square is a rectangle and a rhombus at the same time. Let's see what comes out of this.
Is it clear why? - rhombus - the bisector of angle A, which is equal to. So it divides (and also) into two angles along.
Well, it's quite clear: the rectangle's diagonals are equal; rhombus diagonals are perpendicular, and in general - parallelogram diagonals are divided by the point of intersection in half.
MIDDLE LEVEL
Properties of quadrilaterals. Parallelogram
Parallelogram properties
Attention! The words " parallelogram properties» means that if you have a task there is parallelogram, then all of the following can be used.
Theorem on the properties of a parallelogram.
In any parallelogram:
Let's see why this is true, in other words WE WILL PROVE theorem.
So why is 1) true?
Since it is a parallelogram, then:
- like lying crosswise
- as lying across.
Hence, (on the II basis: and - general.)
Well, once, then - that's it! - proved.
But by the way! We also proved 2)!
Why? But after all (look at the picture), that is, namely, because.
Only 3 left).
To do this, you still have to draw a second diagonal.
And now we see that - according to the II sign (the angle and the side "between" them).
Properties proven! Let's move on to the signs.
Parallelogram features
Recall that the sign of a parallelogram answers the question "how to find out?" That the figure is a parallelogram.
In icons it's like this:
Why? It would be nice to understand why - that's enough. But look:
Well, we figured out why sign 1 is true.
Well, that's even easier! Let's draw a diagonal again.
Which means:
And is also easy. But… different!
Means, . Wow! But also - internal one-sided at a secant!
Therefore the fact that means that.
And if you look from the other side, then they are internal one-sided at a secant! And therefore.
See how great it is?!
And again simply:
Exactly the same, and.
Pay attention: if you found at least one sign of a parallelogram in your problem, then you have exactly parallelogram and you can use everyone properties of a parallelogram.
For complete clarity, look at the diagram:
Properties of quadrilaterals. Rectangle.
Rectangle properties:
Point 1) is quite obvious - after all, sign 3 () is simply fulfilled
And point 2) - very important. So let's prove that
So, on two legs (and - general).
Well, since the triangles are equal, then their hypotenuses are also equal.
Proved that!
And imagine, the equality of the diagonals is a distinctive property of a rectangle among all parallelograms. That is, the following statement is true
Let's see why?
So, (meaning the angles of the parallelogram). But once again, remember that - a parallelogram, and therefore.
Means, . And, of course, it follows from this that each of them After all, in the amount they should give!
Here we have proved that if parallelogram suddenly (!) will be equal diagonals, then this exactly a rectangle.
But! Pay attention! This is about parallelograms! Not any a quadrilateral with equal diagonals is a rectangle, and only parallelogram!
Properties of quadrilaterals. Rhombus
And again the question is: is a rhombus a parallelogram or not?
With full right - a parallelogram, because it has and (Remember our sign 2).
And again, since a rhombus is a parallelogram, it must have all the properties of a parallelogram. This means that a rhombus has opposite angles equal, opposite sides are parallel, and the diagonals are bisected by the point of intersection.
But there are also special properties. We formulate.
Rhombus Properties
Why? Well, since a rhombus is a parallelogram, then its diagonals are divided in half.
Why? Yes, that's why!
In other words, the diagonals and turned out to be the bisectors of the corners of the rhombus.
As in the case of a rectangle, these properties are distinctive, each of them is also a sign of a rhombus.
Rhombus signs.
Why is that? And look
Hence, and both these triangles are isosceles.
In order to be a rhombus, a quadrilateral must first "become" a parallelogram, and then already demonstrate feature 1 or feature 2.
Properties of quadrilaterals. Square
That is, a square is a rectangle and a rhombus at the same time. Let's see what comes out of this.
Is it clear why? Square - rhombus - the bisector of the angle, which is equal to. So it divides (and also) into two angles along.
Well, it's quite clear: the rectangle's diagonals are equal; rhombus diagonals are perpendicular, and in general - parallelogram diagonals are divided by the point of intersection in half.
Why? Well, just apply the Pythagorean Theorem to.
SUMMARY AND BASIC FORMULA
Parallelogram properties:
- Opposite sides are equal: , .
- Opposite angles are: , .
- The angles at one side add up to: , .
- The diagonals are divided by the intersection point in half: .
Rectangle properties:
- The diagonals of a rectangle are: .
- Rectangle is a parallelogram (all properties of a parallelogram are fulfilled for a rectangle).
Rhombus properties:
- The diagonals of the rhombus are perpendicular: .
- The diagonals of a rhombus are the bisectors of its angles: ; ; ; .
- A rhombus is a parallelogram (all properties of a parallelogram are fulfilled for a rhombus).
Square properties:
A square is a rhombus and a rectangle at the same time, therefore, for a square, all the properties of a rectangle and a rhombus are fulfilled. As well as:
Well, the topic is over. If you are reading these lines, then you are very cool.
Because only 5% of people are able to master something on their own. And if you have read to the end, then you are in the 5%!
Now the most important thing.
You've figured out the theory on this topic. And, I repeat, it's ... it's just super! You are already better than the vast majority of your peers.
The problem is that this may not be enough ...
For what?
For the successful passing of the exam, for admission to the institute on the budget and, MOST IMPORTANTLY, for life.
I will not convince you of anything, I will just say one thing ...
People who have received a good education earn much more than those who have not received it. This is statistics.
But this is not the main thing.
The main thing is that they are MORE HAPPY (there are such studies). Perhaps because much more opportunities open up before them and life becomes brighter? Don't know...
But think for yourself...
What does it take to be sure to be better than others on the exam and be ultimately ... happier?
FILL YOUR HAND, SOLVING PROBLEMS ON THIS TOPIC.
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You will need solve problems on time.
And, if you haven’t solved them (LOTS!), you will definitely make a stupid mistake somewhere or simply won’t make it in time.
It's like in sports - you need to repeat many times to win for sure.
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Lesson outline.
Algebra Grade 8
Teacher Sysoi 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 developing skills and abilities to solve geometric problems;
3) Development of a culture of mathematical speech
Lesson plan:
1. Organizational moment
(Slide 1)
The slide shows the statement of Lewis Carroll. Students are informed about the purpose of the lesson. The readiness of students for the lesson is checked.
2. Updating knowledge
(Slide 2)
On the board tasks for oral work. The teacher invites students to think about these problems and raise their hands to those who understand how to solve the problem. After solving two problems, a student is called to the board to prove the theorem on the sum of angles, who independently makes additional constructions on the drawing and proves the theorem orally.
Students use the formula for the sum of the angles of a polygon:
3. Main body
(Slide 3)
On the board is the definition of a parallelogram. The teacher talks about a new figure and formulates a definition, making the necessary explanations using the drawing. Then, on the checkered part of the presentation, using a marker and a ruler, shows how to draw a parallelogram (several cases are possible)
(Slide 4)
The teacher formulates the first property of a parallelogram. Invites students to say, according to the picture, what is given and what needs to be proved. After that, the given task appears on the board. Students guess (perhaps with the help of a teacher) that the required equalities must be proved through the equalities of triangles, which can be obtained by drawing a diagonal (a diagonal appears on the board). Next, the students guess why the triangles are equal and call the sign of the equality of triangles (the corresponding form appears). Verbally communicate the facts that are necessary for the equality of triangles (as they name them, the corresponding visualization appears). Next, 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 a parallelogram. A drawing of a parallelogram appears on the board. The teacher offers to say from the picture 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 parts of the diagonals can be proved through the equality of trianglesAOB and COD. Using the previous property of a parallelogram, guess 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 lines, they prove the equality of angles adjacent to equal sides. These stages are visualized on the slide. The truth of the theorem follows from the equality of triangles - the students say on the slide the corresponding visualization appears.
(Slide 6)
The teacher formulates the third property of a parallelogram. Depending on the time that remains until the end of the lesson, the teacher can give the students the opportunity to prove this property on their own, or limit it 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 interior one-sided angles for two parallel linesAD and BC, and a secant, for exampleAB.
4. Fixing the material
At this stage, students, using previously studied theorems, solve problems. Ideas for solving the problem are selected by students on their own. Since there are many possible design options 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 suggests formulating “Given” according to the condition. After the students correctly write down the condition, “Given” appears on the board. The problem solving process might look like this:
Draw height BH (rendered)
Triangle AHB is a right triangle. Angle A is equal to 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 opposite the angle of 30 0 in a right triangle). So AB = 13 cm.
AB \u003d CD, BC \u003d AD (by the property of opposite sides in a parallelogram) So AB \u003d CD \u003d 13cm. Since the perimeter of the parallelogram is 50 cm, then BC \u003d AD \u003d (50 - 26): 2 \u003d 12 cm.
Answer: AB=CD=13cm, BC=AD=12cm.
(Slide 8)
The task condition appears. The teacher suggests formulating “Given” according to the condition. Then “Dano” 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 problem solving process might look like this:
Because BK and MD are perpendicular to the same line, then lines BK and MD are parallel.
Through adjacent angles, it can be shown that the sum of internal one-sided angles for lines BM and KD and secant MD is 180 0 . Therefore, these lines are parallel.
Since the opposite sides of the quadrilateral BMDK 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.
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A parallelogram is a quadrilateral whose opposite sides are parallel in pairs. The area of a parallelogram is equal to the product of its base (a) and its height (h). You can also find its area through two sides and an angle and through the diagonals.
Parallelogram properties
1. Opposite sides are identical.
First of all, draw the diagonal \(AC \) . Two triangles are obtained: \(ABC \) and \(ADC \) .
Since \(ABCD \) is a parallelogram, the following is true:
\(AD || BC \Rightarrow \angle 1 = \angle 2 \) like lying across.
\(AB || CD \Rightarrow \angle3 = \angle 4 \) like lying across.
Therefore, (on the second basis: and \(AC\) is common).
And, therefore, \(\triangle ABC = \triangle ADC \), then \(AB = CD \) and \(AD = BC \) .
2. Opposite angles are identical.
According to the proof properties 1 We know that \(\angle 1 = \angle 2, \angle 3 = \angle 4 \). So the sum of the opposite angles is: \(\angle 1 + \angle 3 = \angle 2 + \angle 4 \). Given that \(\triangle ABC = \triangle ADC \) we get \(\angle A = \angle C \) , \(\angle B = \angle D \) .
3. The diagonals are bisected by the intersection point.
By property 1 we know that opposite sides are identical: \(AB = CD \) . Once again we note the equal angles lying crosswise.
Thus, it is seen that \(\triangle AOB = \triangle COD \) according to the second criterion for the equality of triangles (two angles and a side between them). That is, \(BO = OD \) (opposite the corners \(\angle 2 \) and \(\angle 1 \) ) and \(AO = OC \) (opposite the corners \(\angle 3 \) and \( \angle 4 \) respectively).
Parallelogram features
If only one sign is present in your problem, then the figure is a parallelogram and you can use all the properties of this figure.
For better memorization, note that the sign of a parallelogram will answer the following question - "how to find out?". That is, how to find out that a given figure is a parallelogram.
1. A parallelogram is a quadrilateral whose two sides are equal and parallel.
\(AB = CD \) ; \(AB || CD \Rightarrow ABCD \)- parallelogram.
Let's consider in more detail. Why \(AD || BC \) ?
\(\triangle ABC = \triangle ADC \) on property 1: \(AB = CD \) , \(\angle 1 = \angle 2 \) as crosswise with parallel \(AB \) and \(CD \) and secant \(AC \) .
But if \(\triangle ABC = \triangle ADC \), then \(\angle 3 = \angle 4 \) (they lie opposite \(AD || BC \) (\(\angle 3 \) and \(\angle 4 \) - lying opposite are also equal).
The first sign is correct.
2. A parallelogram is a quadrilateral whose opposite sides are equal.
\(AB = CD \) , \(AD = BC \Rightarrow ABCD \) is a parallelogram.
Let's 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 whose opposite angles are equal.
\(\angle A = \angle C \) , \(\angle B = \angle D \Rightarrow ABCD \)- parallelogram.
\(2 \alpha + 2 \beta = 360^(\circ) \)(because \(\angle A = \angle C \) , \(\angle B = \angle D \) by definition).
It turns out, \(\alpha + \beta = 180^(\circ) \). But \(\alpha \) and \(\beta \) are internal one-sided at the secant \(AB \) .
