In a rectangular parallelogram. Calculate the sum of angles and area of ​​a parallelogram: properties and characteristics

Compound word "parallelogram"? And behind it lies a very simple figure.

Well, that is, we took two parallel lines:

Crossed by two more:

And inside there is a parallelogram!

What properties does a parallelogram have?

Properties of a parallelogram.

That is, what can you use if the problem is given a parallelogram?

The following theorem answers this question:

Let's draw everything in detail.

What does it mean first point of the theorem? And the fact is that if you HAVE a parallelogram, then you will certainly

The second point means that if there IS a parallelogram, then, again, certainly:

Well, and finally, the third point means that if you HAVE a parallelogram, then be sure to:

Do you see what a wealth of choice there is? What to use in the problem? Try to focus on the question of the task, or just try everything one by one - some “key” will do.

Now let’s ask ourselves another question: how can we recognize a parallelogram “by sight”? What must happen to a quadrilateral so that we have the right to give it the “title” of a parallelogram?

Several signs of a parallelogram answer this question.

Signs of a parallelogram.

Attention! Begin.

Parallelogram.

Please note: if you found at least one sign in your problem, then you definitely have a parallelogram, and you can use all the properties of a parallelogram.

2. Rectangle

I think that it will not be news to you at all that

First question: 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 in a rectangle, like in any parallelogram, the diagonals are divided in half by the point of intersection.

But the rectangle also has one distinctive property.

Rectangle property

Why is this property distinctive? Because no other parallelogram has equal diagonals. Let's formulate it more clearly.

Please note: in order to become a rectangle, a quadrilateral must first become a parallelogram, and then demonstrate the equality of the diagonals.

3. Diamond

And again the question: is a rhombus a parallelogram or not?

With full right - a parallelogram, because it has and (remember our feature 2).

And again, since a rhombus is a parallelogram, then it must have all the properties of a parallelogram. This means that in a rhombus, opposite angles are equal, opposite sides are parallel, and the diagonals bisect at the point of intersection.

Properties of a rhombus

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 this is not just a parallelogram, but a rhombus.

Signs of a diamond

And again, pay attention: there must be not just a quadrilateral whose diagonals are perpendicular, but a parallelogram. Make sure:

No, of course, although its diagonals are perpendicular, and the diagonal is the bisector of the angles and. But... diagonals are not divided in half by the point of intersection, 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 happens.

Is it clear why? - rhombus is the bisector of angle A, which is equal to. This means it divides (and also) into two angles along.

Well, it's quite clear: the diagonals of a rectangle are equal; The diagonals of a rhombus are perpendicular, and in general, a parallelogram of diagonals is divided in half by the point of intersection.

AVERAGE LEVEL

Properties of quadrilaterals. Parallelogram

Properties of a parallelogram

Attention! Words " properties of a parallelogram"mean that if in your task There is parallelogram, then all of the following can be used.

Theorem on the properties of a parallelogram.

In any parallelogram:

Let's understand why this is all true, in other words WE'LL PROVE theorem.

So why is 1) true?

If it is a parallelogram, then:

• lying criss-cross
• lying like crosses.

This means (according to criterion II: and - general.)

Well, that’s it, that’s it! - proved.

But by the way! We also proved 2)!

Why? But (look at the picture), that is, precisely because.

Only 3 left).

To do this, you still have to draw a second diagonal.

And now we see that - according to the II characteristic (angles and the side “between” them).

Properties proven! Let's move on to the signs.

Signs of a parallelogram

Recall that the parallelogram sign answers the question “how do you know?” that a 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, it's even easier! Let's draw a diagonal again.

Which means:

AND It's also easy. But...different!

Means, . Wow! But also - internal one-sided with a secant!

Therefore the fact that means that.

And if you look from the other side, then - internal one-sided with a secant! And therefore.

Do you see how great it is?!

And again simple:

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

This means on two sides (and - general).

Well, since the triangles are equal, then their hypotenuses are also equal.

Proved that!

And imagine, equality of diagonals is a distinctive property of a rectangle among all parallelograms. That is, this statement is true^

Let's understand why?

This means (meaning the angles of a parallelogram). But let us remember once again that it is a parallelogram, and therefore.

Means, . Well, of course, it follows that each of them! After all, they have to give in total!

So they proved that if parallelogram suddenly (!) the diagonals turn out to be equal, then this exactly a rectangle.

But! Pay attention! This is about parallelograms! Not just anyone a quadrilateral with equal diagonals is a rectangle, and only parallelogram!

Properties of quadrilaterals. Rhombus

And again the question: is a rhombus a parallelogram or not?

With full right - a parallelogram, because it has (Remember our feature 2).

And again, since a rhombus is a parallelogram, it must have all the properties of a parallelogram. This means that in a rhombus, opposite angles are equal, opposite sides are parallel, and the diagonals bisect at the point of intersection.

But there are also special properties. Let's formulate it.

Properties of a rhombus

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 turned out to be 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.

Signs of a diamond.

Why is this? And look,

That means both These triangles are isosceles.

To be a rhombus, a quadrilateral must first “become” a parallelogram, and then exhibit 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 happens.

Is it clear why? A square - a rhombus - is the bisector of an angle that is equal to. This means it divides (and also) into two angles along.

Well, it's quite clear: the diagonals of a rectangle are equal; The diagonals of a rhombus are perpendicular, and in general, a parallelogram of diagonals is divided in half by the point of intersection.

Why? Well, let's just apply the Pythagorean theorem to...

SUMMARY AND BASIC FORMULAS

Properties of a parallelogram:

1. Opposite sides are equal: , .
2. Opposite angles are equal: , .
3. The angles on one side add up to: , .
4. The diagonals are divided in half by the point of intersection: .

Rectangle properties:

1. The diagonals of the rectangle are equal: .
2. A rectangle is a parallelogram (for a rectangle all the properties of a parallelogram are fulfilled).

Properties of a rhombus:

1. The diagonals of a rhombus are perpendicular: .
2. The diagonals of a rhombus are the bisectors of its angles: ; ; ; .
3. A rhombus is a parallelogram (for a rhombus all the properties of a parallelogram are fulfilled).

Properties of a square:

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. And:

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A parallelogram is a quadrilateral whose opposite sides are parallel in pairs (Fig. 233).

For an arbitrary parallelogram the following properties hold:

1. Opposite sides of a parallelogram are equal.

Proof. In the parallelogram ABCD we draw the diagonal AC. Triangles ACD and AC B are equal, as having a common side AC and two pairs of equal angles adjacent to it:

(like crosswise angles with parallel lines AD and BC). This means, and like the sides of equal triangles lying opposite equal angles, which is what needed to be proven.

2. Opposite angles of a parallelogram are equal:

3. Adjacent angles of a parallelogram, i.e., angles adjacent to one side, add up, etc.

The proof of properties 2 and 3 is immediately obtained from the properties of angles for parallel lines.

4. The diagonals of a parallelogram bisect each other at their intersection point. In other words,

Proof. Triangles AOD and BOC are congruent, since their sides AD and BC are equal (property 1) and the angles adjacent to them (like crosswise angles for parallel lines). From here it follows that the corresponding sides of these triangles are equal: AO, which is what needed to be proven.

Each of these four properties characterizes a parallelogram, or, as they say, is its characteristic property, i.e., every quadrilateral that has at least one of these properties is a parallelogram (and, therefore, has all the other three properties).

Let us carry out the proof for each property separately.

1". If the opposite sides of a quadrilateral are equal in pairs, then it is a parallelogram.

Proof. Let the quadrilateral ABCD have sides AD and BC, AB and CD respectively equal (Fig. 233). Let's draw the diagonal AC. Triangles ABC and CDA will be congruent as having three pairs of equal sides.

But then the angles BAC and DCA are equal and . The parallelism of sides BC and AD follows from the equality of angles CAD and ACB.

2. If a quadrilateral has two pairs of opposite angles equal, then it is a parallelogram.

Proof. Let . Since then both sides AD and BC are parallel (based on the parallelism of lines).

3. We leave the formulation and proof to the reader.

4. If the diagonals of a quadrilateral bisect each other at the point of intersection, then the quadrilateral is a parallelogram.

Proof. If AO = OS, BO = OD (Fig. 233), then the triangles AOD and BOC are equal, as having equal angles (vertical!) at the vertex O, enclosed between pairs of equal sides AO and CO, BO and DO. From the equality of triangles we conclude that sides AD and BC are equal. The sides AB and CD are also equal, and the quadrilateral turns out to be a parallelogram according to the characteristic property G.

Thus, in order to prove that a given quadrilateral is a parallelogram, it is enough to verify the validity of any of the four properties. The reader is invited to independently prove another characteristic property of a parallelogram.

5. If a quadrilateral has a pair of equal, parallel sides, then it is a parallelogram.

Sometimes any pair of parallel sides of a parallelogram is called its bases, then the other two are called lateral sides. A straight line segment perpendicular to two sides of a parallelogram, enclosed between them, is called the height of the parallelogram. Parallelogram in Fig. 234 has a height h drawn to the sides AD and BC, its second height is represented by the segment .

Lesson summary.

Algebra 8th grade

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 developing the 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 a statement by 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 are 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 part

(Slide 3)

Definition of a parallelogram on the board. The teacher talks about a new figure and formulates a definition, making the necessary explanations using a drawing. Then, on the checkered part of the presentation, using a marker and a ruler, he shows how to draw a parallelogram (several cases are possible)

(Slide 4)

The teacher formulates the first property of a parallelogram. Invites students to tell from the drawing what is given and what needs to be proven. After this, the given task appears on the board. Students guess (maybe with the help of the teacher) that the required equalities must be proven through the equalities of triangles, which can be obtained by drawing a diagonal (a diagonal appears on the board). Next, students guess why the triangles are equal and name the sign that triangles are equal (the corresponding shape appears). They verbally communicate the facts that are necessary to make the triangles equal (as they name them, a corresponding visualization appears). Next, students formulate the property of congruent triangles, it appears as 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 suggests using the picture to tell what is given and what needs to be proven. After students correctly report what is given and what needs to be proven, the condition of the theorem appears. Students guess that the equality of the parts of the diagonals can be proven through the equality of trianglesAOB And C.O.D.. Using the previous property of a parallelogram, one guesses that the sides are equalAB And CD. Then they understand that they need to find equal angles and, using the properties of parallel lines, 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 the triangles - the students say it and a corresponding visualization appears on the slide.

(Slide 6)

The teacher formulates the third property of a parallelogram. Depending on the time remaining until the end of the lesson, the teacher can give the students the opportunity to prove this property on their own, 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 an inscribed polygon, which was repeated at the beginning of the lesson, or on the sum of the internal one-sided angles of two parallel linesAD And B.C., and a secant, for exampleAB.

4. Fixing the material

At this stage, students use previously learned theorems to solve problems. Students select ideas for solving the problem independently. 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 recording the solution in a notebook.

(Slide 7)

The task condition appears. The teacher suggests formulating “Given” according to the condition. After the students correctly write down a short statement of the condition, “Given” appears on the board. The process for solving the problem might look like this:

Let's draw the height BH (visualized)

Triangle AHB is a right triangle. Angle A is equal to angle C and equals 30 0 (according to the property of opposite angles in a parallelogram). 2BH =AB (by the property of the leg lying opposite the 30 0 angle in a right triangle). So AB = 13 cm.

AB = CD, BC = AD (according to the property of opposite sides in a parallelogram) So AB = CD = 13 cm. Since the perimeter of the parallelogram is 50 cm, then BC = AD = (50 – 26): 2 = 12 cm.

Answer: AB = CD = 13 cm, BC = AD = 12 cm.

(Slide 8)

The task condition appears. The teacher suggests formulating “Given” according to the condition. Then “Given” appears on the screen. Using red lines, a quadrilateral is highlighted, about which you need to prove that it is a parallelogram. The process for solving the problem might look like this:

Because BK and MD are perpendicular to one line, then lines BK and MD are parallel.

Through adjacent angles it can be shown that the sum of the internal one-sided angles at straight lines BM and KD and the secant MD is equal to 180 0. Therefore, these lines are parallel.

Since the quadrilateral BMDK has opposite sides parallel in pairs, then this quadrilateral is a parallelogram.

5. End of the lesson. Behavior of the results.

(Slide 8)

Questions on the new topic appear on the slide, to which students answer.

Proof

First of all, let's draw the diagonal AC. We get two triangles: ABC and ADC.

Since ABCD is a parallelogram, the following is true:

AD || BC \Rightarrow \angle 1 = \angle 2 like lying crosswise.

AB || CD\Rightarrow\angle3 =\angle 4 like 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 identical.

Proof

According to the proof 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 divided in half by the intersection point.

Proof

Let's draw another diagonal.

By property 1 we know that opposite sides are identical: AB = CD. Once again, note the crosswise lying equal angles.

Thus, it is clear that \triangle AOB = \triangle COD according to the second criterion for the equality of triangles (two angles and the 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).

Proven!

Signs of a parallelogram

If only one feature 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 parallelogram sign 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 is a parallelogram.

Proof

Let's take a closer look. Why AD || BC?

\triangle ABC = \triangle ADC by property 1: AB = CD, AC - common and \angle 1 = \angle 2 lying crosswise with 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 therefore AD || BC (\angle 3 and \angle 4 - those lying crosswise 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.

Proof

Let's consider this sign. Let's draw the diagonal AC again.

By property 1\triangle ABC = \triangle ACD .

It follows that: \angle 1 = \angle 2 \Rightarrow AD || B.C. 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.

Proof

2 \alpha + 2 \beta = 360^(\circ)(since ABCD is a quadrilateral, and \angle A = \angle C , \angle B = \angle D by condition).

It turns out that \alpha + \beta = 180^(\circ) . But \alpha and \beta are internal one-sided at the secant AB.

And the fact that \alpha + \beta = 180^(\circ) also means that AD || B.C.

Moreover, \alpha and \beta are internal one-sided at the secant AD . And that means AB || CD.

The third sign is correct.

4. A parallelogram is a quadrilateral whose diagonals are divided in half by the point of intersection.

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 || B.C.

The fourth sign is correct.