Average level

Parallelogram, rectangle, rhombus, square (2019)

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.

AVERAGE 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:

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

Rectangle properties:

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

Rhombus properties:

1. The diagonals of the rhombus are perpendicular: .
2. The diagonals of a rhombus are the bisectors of its angles: ; ; ; .
3. 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.

A parallelogram is a quadrilateral whose opposite sides are parallel, that is, they lie on parallel lines (Fig. 1).

Theorem 1. On the properties of sides and angles of a parallelogram. In a parallelogram, opposite sides are equal, opposite angles are equal, and the sum of the angles adjacent to one side of the parallelogram is 180°.

Proof. In this parallelogram ABCD, draw a diagonal AC and get two triangles ABC and ADC (Fig. 2).

These triangles are equal, since ∠ 1 = ∠ 4, ∠ 2 = ∠ 3 (cross-lying angles at parallel lines), and side AC is common. From the equality Δ ABC = Δ ADC it follows that AB \u003d CD, BC \u003d AD, ∠ B \u003d ∠ D. The sum of the angles adjacent to one side, for example, angles A and D, is equal to 180 ° as one-sided with parallel lines. The theorem has been proven.

Comment. The equality of the opposite sides of a parallelogram means that the segments of the parallel ones cut off by the parallel ones are equal.

Corollary 1. If two lines are parallel, then all points of one line are at the same distance from the other line.

Proof. Indeed, let a || b (Fig. 3).

Let us draw from some two points B and C of the line b the perpendiculars BA and CD to the line a. Since AB || CD, then the figure ABCD is a parallelogram, and therefore AB = CD.

The distance between two parallel lines is the distance from an arbitrary point on one of the lines to the other line.

By what has been proved, it is equal to the length of the perpendicular drawn from some point of one of the parallel lines to the other line.

Example 1 The perimeter of the parallelogram is 122 cm. One of its sides is 25 cm longer than the other. Find the sides of the parallelogram.

Solution. By Theorem 1, opposite sides of a parallelogram are equal. Let's denote one side of the parallelogram as x, the other as y. Then by condition $$\left\(\begin(matrix) 2x + 2y = 122 \\x - y = 25 \end(matrix)\right.$$ Solving this system, we get x = 43, y = 18. Thus Thus, the sides of the parallelogram are 18, 43, 18 and 43 cm.

Example 2

Solution. Let figure 4 correspond to the condition of the problem.

Denote AB by x and BC by y. By condition, the perimeter of the parallelogram is 10 cm, i.e. 2(x + y) = 10, or x + y = 5. The perimeter of the triangle ABD is 8 cm. And since AB + AD = x + y = 5, then BD = 8 - 5 = 3 . So BD = 3 cm.

Example 3 Find the angles of the parallelogram, knowing that one of them is 50° greater than the other.

Solution. Let figure 5 correspond to the condition of the problem.

Let us denote the degree measure of angle A as x. Then the degree measure of the angle D is x + 50°.

Angles BAD and ADC are internal one-sided with parallel lines AB and DC and secant AD. Then the sum of these named angles will be 180°, i.e.
x + x + 50° = 180°, or x = 65°. Thus, ∠ A = ∠ C = 65°, a ∠ B = ∠ D = 115°.

Example 4 The sides of the parallelogram are 4.5 dm and 1.2 dm. A bisector is drawn from the vertex of an acute angle. What parts does it divide the long side of the parallelogram into?

Solution. Let figure 6 correspond to the condition of the problem.

AE is the bisector of the acute angle of the parallelogram. Therefore, ∠ 1 = ∠ 2.

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As in Euclidean geometry, the point and the line are the main elements of the theory of planes, so the parallelogram is one of the key figures of convex quadrilaterals. From it, like threads from a ball, flow the concepts of "rectangle", "square", "rhombus" and other geometric quantities.

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Definition of a parallelogram

convex quadrilateral, consisting of segments, each pair of which is parallel, is known in geometry as a parallelogram.

What a classic parallelogram looks like is a quadrilateral ABCD. The sides are called the bases (AB, BC, CD and AD), the perpendicular drawn from any vertex to the opposite side of this vertex is called the height (BE and BF), the lines AC and BD are the diagonals.

Attention! Square, rhombus and rectangle are special cases of parallelogram.

Sides and angles: ratio features

Key properties, by and large, predetermined by the designation itself, they are proved by the theorem. These characteristics are as follows:

1. Sides that are opposite are identical in pairs.
2. Angles that are opposite to each other are equal in pairs.

Proof: consider ∆ABC and ∆ADC, which are obtained by dividing quadrilateral ABCD by line AC. ∠BCA=∠CAD and ∠BAC=∠ACD, since AC is common to them (vertical angles for BC||AD and AB||CD, respectively). It follows from this: ∆ABC = ∆ADC (the second criterion for the equality of triangles).

Segments AB and BC in ∆ABC correspond in pairs to lines CD and AD in ∆ADC, which means that they are identical: AB = CD, BC = AD. Thus, ∠B corresponds to ∠D and they are equal. Since ∠A=∠BAC+∠CAD, ∠C=∠BCA+∠ACD, which are also identical in pairs, then ∠A = ∠C. The property has been proven.

Characteristics of the figure's diagonals

Main feature these parallelogram lines: the point of intersection bisects them.

Proof: let m. E be the intersection point of the diagonals AC and BD of the figure ABCD. They form two commensurate triangles - ∆ABE and ∆CDE.

AB=CD since they are opposite. According to lines and secants, ∠ABE = ∠CDE and ∠BAE = ∠DCE.

According to the second sign of equality, ∆ABE = ∆CDE. This means that the elements ∆ABE and ∆CDE are: AE = CE, BE = DE and, moreover, they are commensurate parts of AC and BD. The property has been proven.

Features of adjacent corners

At adjacent sides, the sum of the angles is 180°, since they lie on the same side of the parallel lines and the secant. For quadrilateral ABCD:

∠A+∠B=∠C+∠D=∠A+∠D=∠B+∠C=180º

Bisector properties:

1. , dropped to one side, are perpendicular;
2. opposite vertices have parallel bisectors;
3. the triangle obtained by drawing the bisector will be isosceles.

Determining the characteristic features of a parallelogram by the theorem

The features of this figure follow from its main theorem, which reads as follows: quadrilateral is considered a parallelogram in the event that its diagonals intersect, and this point divides them into equal segments.

Proof: Let lines AC and BD of quadrilateral ABCD intersect in t. E. Since ∠AED = ∠BEC, and AE+CE=AC BE+DE=BD, then ∆AED = ∆BEC (by the first sign of equality of triangles). That is, ∠EAD = ∠ECB. They are also the interior crossing angles of the secant AC for lines AD and BC. Thus, by definition of parallelism - AD || BC. A similar property of the lines BC and CD is also derived. The theorem has been proven.

Calculating the area of ​​a figure

The area of ​​this figure found in several ways one of the simplest: multiplying the height and the base to which it is drawn.

Proof: Draw perpendiculars BE and CF from vertices B and C. ∆ABE and ∆DCF are equal since AB = CD and BE = CF. ABCD is equal to the rectangle EBCF, since they also consist of proportionate figures: S ABE and S EBCD, as well as S DCF and S EBCD. It follows that the area of ​​this geometric figure is the same as that of a rectangle:

S ABCD = S EBCF = BE×BC=BE×AD.

To determine the general formula for the area of ​​a parallelogram, we denote the height as hb, and the side b. Respectively:

Other ways to find area

Area calculations through the sides of the parallelogram and the angle, which they form, is the second known method.

,

Spr-ma - area;

a and b are its sides

α - angle between segments a and b.

This method is practically based on the first, but in case it is unknown. always cuts off a right triangle whose parameters are found by trigonometric identities, i.e. . Transforming the ratio, we get . In the equation of the first method, we replace the height with this product and obtain a proof of the validity of this formula.

Through the diagonals of a parallelogram and an angle, which they create when they intersect, you can also find the area.

Proof: AC and BD intersecting form four triangles: ABE, BEC, CDE and AED. Their sum is equal to the area of ​​this quadrilateral.

The area of ​​each of these ∆ can be found from the expression , where a=BE, b=AE, ∠γ =∠AEB. Since , then a single value of the sine is used in the calculations. That is . Since AE+CE=AC= d 1 and BE+DE=BD= d 2 , the area formula reduces to:

.

Application in vector algebra

The features of the constituent parts of this quadrilateral have found application in vector algebra, namely: the addition of two vectors. The parallelogram rule states that if given vectorsandnotare collinear, then their sum will be equal to the diagonal of this figure, the bases of which correspond to these vectors.

Proof: from an arbitrarily chosen beginning - that is. - we build vectors and . Next, we build a parallelogram OASV, where the segments OA and OB are sides. Thus, the OS lies on the vector or sum.

Formulas for calculating the parameters of a parallelogram

The identities are given under the following conditions:

1. a and b, α - sides and the angle between them;
2. d 1 and d 2 , γ - diagonals and at the point of their intersection;
3. h a and h b - heights lowered to sides a and b;

Parameter	Formula
Finding sides
along the diagonals and the cosine of the angle between them
diagonally and sideways
through height and opposite vertex
Finding the length of the diagonals
on the sides and the size of the top between them

A parallelogram is a quadrilateral in which opposite sides are pairwise parallel.

A parallelogram has all the properties of quadrilaterals, but it also has its own distinctive features. Knowing them, we can easily find both sides and angles of a parallelogram.

Parallelogram properties

1. The sum of the angles in any parallelogram, as in any quadrilateral, is 360°.
2. The middle lines of a parallelogram and its diagonals intersect at one point and bisect it. This point is called the center of symmetry of the parallelogram.
3. Opposite sides of a parallelogram are always equal.
4. Also, this figure always has opposite angles equal.
5. The sum of the angles adjacent to either side of a parallelogram is always 180°.
6. 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. This is expressed by the formula:
   • d 1 2 + d 2 2 = 2 (a 2 + b 2), where d 1 and d 2 are diagonals, a and b are adjacent sides.
7. The cosine of an obtuse angle is always less than zero.

How to find the angles of a given parallelogram, applying these properties in practice? And what other formulas can help us with this? Consider specific tasks that require: find the angles of the parallelogram.

Finding the corners of a parallelogram

Case 1. The measure of an obtuse angle is known, it is required to find an acute angle.

Example: In parallelogram ABCD, angle A is 120°. Find the measure of the remaining angles.

Solution: Using property No. 5, we can find the measure of the angle B adjacent to the angle given in the task. It will be equal to:

• 180°-120°= 60°

And now, using property #4, we determine that the two remaining angles C and D are opposite to the angles we have already found. Angle C is opposite to angle A, angle D is opposite to angle B. Therefore, they are equal in pairs.

• Answer: B=60°, C=120°, D=60°

Case 2. The lengths of the sides and the diagonal are known

In this case, we need to use the cosine theorem.

We can first use the formula to calculate the cosine of the angle we need, and then use a special table to find what the angle itself is equal to.

For an acute angle, the formula is:

• cosa \u003d (A² + B² - d²) / (2 * A * B), where
• a is the desired acute angle,
• A and B are sides of a parallelogram
• d - smaller diagonal

For an obtuse angle, the formula changes slightly:

• cosß \u003d (A² + B² - D²) / (2 * A * B), where
• ß is an obtuse angle,
• A and B are sides
• D - large diagonal

Example: you need to find the acute angle of a parallelogram whose sides are 6 cm and 3 cm, and the smaller diagonal is 5.2 cm

We substitute the values ​​​​into the formula for finding an acute angle:

• cosa = (6 2 + 3 2 - 5.2 2) / (2 * 6 * 3) = (36 + 9 - 27.04) / (2 * 18) = 17.96/36 ~ 18/36 ~1/2
• cosa = 1/2. According to the table, we find out that the desired angle is 60 °.