Parallelogram and its area. We calculate the sum of angles and the area of ​​a parallelogram: properties and signs. Features of adjacent corners

Parallelogram area

Theorem 1

The area of ​​a parallelogram is defined as the product of the length of its side times the height drawn to it.

where $a$ is the side of the parallelogram, $h$ is the height drawn to this side.

Proof.

Let us be given a parallelogram $ABCD$ with $AD=BC=a$. Let's draw the heights $DF$ and $AE$ (Fig. 1).

Picture 1.

It is obvious that the figure $FDAE$ is a rectangle.

\[\angle BAE=(90)^0-\angle A,\ \] \[\angle CDF=\angle D-(90)^0=(180)^0-\angle A-(90)^0 =(90)^0-\angle A=\angle BAE\]

Therefore, since $CD=AB,\ DF=AE=h$, $\triangle BAE=\triangle CDF$, by $I$ the triangle equality test. Then

So according to the rectangle area theorem:

The theorem has been proven.

Theorem 2

The area of ​​a parallelogram is defined as the product of the length of its adjacent sides times the sine of the angle between those sides.

Mathematically, this can be written as follows

where $a,\ b$ are the sides of the parallelogram, $\alpha $ is the angle between them.

Proof.

Let us be given a parallelogram $ABCD$ with $BC=a,\ CD=b,\ \angle C=\alpha $. Draw the height $DF=h$ (Fig. 2).

Figure 2.

By definition of the sine, we get

Hence

Hence, by Theorem $1$:

The theorem has been proven.

Area of ​​a triangle

Theorem 3

The area of ​​a triangle is defined as half the product of the length of its side and the height drawn to it.

Mathematically, this can be written as follows

where $a$ is the side of the triangle, $h$ is the height drawn to this side.

Proof.

Figure 3

So by Theorem $1$:

The theorem has been proven.

Theorem 4

The area of ​​a triangle is defined as half the product of the length of its adjacent sides times the sine of the angle between those sides.

Mathematically, this can be written as follows

where $a,\ b$ are the sides of the triangle, $\alpha $ is the angle between them.

Proof.

Let us be given a triangle $ABC$ with $AB=a$. Draw the height $CH=h$. Let's build it up to the parallelogram $ABCD$ (Fig. 3).

Obviously, $\triangle ACB=\triangle CDB$ by $I$. Then

So by Theorem $1$:

The theorem has been proven.

Trapezium area

Theorem 5

The area of ​​a trapezoid is defined as half the product of the sum of the lengths of its bases times its height.

Mathematically, this can be written as follows

Proof.

Let us be given a trapezoid $ABCK$, where $AK=a,\ BC=b$. Let us draw the heights $BM=h$ and $KP=h$ in it, as well as the diagonal $BK$ (Fig. 4).

Figure 4

By Theorem $3$, we get

The theorem has been proven.

Task example

Example 1

Find the area of ​​an equilateral triangle if the length of its side is $a.$

Solution.

Since the triangle is equilateral, all its angles are equal to $(60)^0$.

Then, by Theorem $4$, we have

Answer:$\frac(a^2\sqrt(3))(4)$.

Note that the result of this problem can be used to find the area of ​​any equilateral triangle with a given side.

As in Euclidean geometry, the point and the straight 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, about. - 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
along the sides and one of the diagonals	 Conclusion The parallelogram, as one of the key figures of geometry, is used in life, for example, in construction when calculating the area of ​​​​a site or other measurements. Therefore, knowledge about the distinguishing features and methods for calculating its various parameters can be useful at any time in life.

When solving problems on this topic, in addition to basic properties parallelogram and the corresponding formulas, you can remember and apply the following:

1. The bisector of the interior angle of a parallelogram cuts off an isosceles triangle from it
2. Bisectors of internal angles adjacent to one of the sides of a parallelogram are mutually perpendicular
3. Bisectors coming from opposite internal angles of a parallelogram, parallel to each other or lie on one straight line
4. The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides
5. The area of ​​a parallelogram is half the product of the diagonals times the sine of the angle between them.

Let's consider the tasks in the solution of which these properties are used.

Task 1.

The bisector of angle C of parallelogram ABCD intersects side AD at point M and the continuation of side AB beyond point A at point E. Find the perimeter of the parallelogram if AE \u003d 4, DM \u003d 3.

Solution.

1. Triangle CMD isosceles. (Property 1). Therefore, CD = MD = 3 cm.

2. Triangle EAM is isosceles.
Therefore, AE = AM = 4 cm.

3. AD = AM + MD = 7 cm.

4. Perimeter ABCD = 20 cm.

Answer. 20 cm

Task 2.

Diagonals are drawn in a convex quadrilateral ABCD. It is known that the areas of triangles ABD, ACD, BCD are equal. Prove that the given quadrilateral is a parallelogram.

Solution.

1. Let BE be the height of triangle ABD, CF be the height of triangle ACD. Since, according to the condition of the problem, the areas of the triangles are equal and they have a common base AD, then the heights of these triangles are equal. BE = CF.

2. BE, CF are perpendicular to AD. Points B and C are located on the same side of the line AD. BE = CF. Therefore, the line BC || AD. (*)

3. Let AL be the altitude of triangle ACD, BK the altitude of triangle BCD. Since, according to the condition of the problem, the areas of the triangles are equal and they have a common base CD, then the heights of these triangles are equal. AL = BK.

4. AL and BK are perpendicular to CD. Points B and A are located on the same side of the straight line CD. AL = BK. Therefore, the line AB || CD (**)

5. Conditions (*), (**) imply that ABCD is a parallelogram.

Answer. Proven. ABCD is a parallelogram.

Task 3.

On the sides BC and CD of the parallelogram ABCD, the points M and H are marked, respectively, so that the segments BM and HD intersect at the point O;<ВМD = 95 о,