How to find the area of ​​a parallelogram? How is the area of ​​a parallelogram found?

Parallelogram is a quadrilateral whose sides are parallel in pairs.

In this figure, opposite sides and angles are equal to each other. The diagonals of a parallelogram intersect at one point and bisect it. Formulas for the area of ​​a parallelogram allow you to find the value using the sides, height and diagonals. A parallelogram can also be presented in special cases. They are considered a rectangle, square and rhombus.
First, let's look at an example of calculating the area of ​​a parallelogram by height and the side to which it is lowered.

This case is considered a classic and does not require additional investigation. It’s better to consider the formula for calculating the area through two sides and the angle between them. The same method is used in calculations. If the sides and the angle between them are given, then the area is calculated as follows:

Suppose we are given a parallelogram with sides a = 4 cm, b = 6 cm. The angle between them is α = 30°. Let's find the area:

Area of ​​a parallelogram through diagonals

The formula for the area of ​​a parallelogram using the diagonals allows you to quickly find the value.
For calculations, you will need the size of the angle located between the diagonals.

Let's consider an example of calculating the area of ​​a parallelogram using diagonals. Let a parallelogram be given with diagonals D = 7 cm, d = 5 cm. The angle between them is α = 30°. Let's substitute the data into the formula:

An example of calculating the area of ​​a parallelogram through the diagonal gave us an excellent result - 8.75.

Knowing the formula for the area of ​​a parallelogram through the diagonal, you can solve many interesting problems. Let's look at one of them.

Task: Given a parallelogram with an area of ​​92 square meters. see Point F is located in the middle of its side BC. Let's let's find the area trapezoid ADFB, which will lie in our parallelogram. First, let's draw everything we received according to the conditions.
Let's get to the solution:

According to our conditions, ah =92, and accordingly, the area of ​​our trapezoid will be equal to

What is a parallelogram? A parallelogram is a quadrilateral whose opposite sides are parallel in pairs.

1. The area of ​​a parallelogram is calculated by the formula:

\[ \LARGE S = a \cdot h_(a)\]

Where:
a is the side of the parallelogram,
h a – height drawn to this side.

2. If the lengths of two adjacent sides of a parallelogram and the angle between them are known, then the area of ​​the parallelogram is calculated by the formula:

\[ \LARGE S = a \cdot b \cdot sin(\alpha) \]

3. If the diagonals of a parallelogram are given and the angle between them is known, then the area of ​​the parallelogram is calculated by the formula:

\[ \LARGE S = \frac(1)(2) \cdot d_(1) \cdot d_(2) \cdot sin(\alpha) \]

Properties of a parallelogram

In a parallelogram, opposite sides are equal: \(AB = CD\), \(BC = AD\)

In a parallelogram, opposite angles are equal: \(\angle A = \angle C\), \(\angle B = \angle D\)

The diagonals of a parallelogram at the intersection point are divided in half \(AO = OC\) , \(BO = OD\)

The diagonal of a parallelogram divides it into two equal triangles.

The sum of the angles of a parallelogram adjacent to one side is 180 o:

\(\angle A + \angle B = 180^(o)\), \(\angle B + \angle C = 180^(o)\)

\(\angle C + \angle D = 180^(o)\), \(\angle D + \angle A = 180^(o)\)

The diagonals and sides of a parallelogram are related by the following relationship:

\(d_(1)^(2) + d_(2)^2 = 2a^(2) + 2b^(2) \)

In a parallelogram, the angle between the heights is equal to its acute angle: \(\angle K B H =\angle A\) .

The bisectors of angles adjacent to one side of a parallelogram are mutually perpendicular.

The bisectors of two opposite angles of a parallelogram are parallel.

Signs of a parallelogram

A quadrilateral will be a parallelogram if:

\(AB = CD\) and \(AB || CD\)

\(AB = CD\) and \(BC = AD\)

\(AO = OC\) and \(BO = OD\)

\(\angle A = \angle C\) and \(\angle B = \angle D\)

Area of ​​a parallelogram

Theorem 1

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

where $a$ is a 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 us draw the heights $DF$ and $AE$ (Fig. 1).

Picture 1.

Obviously, the $FDAE$ figure is a rectangle.

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

Consequently, since $CD=AB,\ DF=AE=h$, by the $I$ criterion for the equality of triangles $\triangle BAE=\triangle CDF$. Then

So, according to the theorem on the area of ​​a rectangle:

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 these 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 $. Let us draw the height $DF=h$ (Fig. 2).

Figure 2.

By definition of sine, we get

Hence

So, 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 altitude drawn to it.

Mathematically this can be written as follows

where $a$ is a 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 and the sine of the angle between these 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 $AB=a$. Let's find the height $CH=h$. Let's build it up to a parallelogram $ABCD$ (Fig. 3).

Obviously, by the $I$ criterion for the equality of triangles, $\triangle ACB=\triangle CDB$. Then

So, by Theorem $1$:

The theorem has been proven.

Area of ​​trapezoid

Theorem 5

The area of ​​a trapezoid is defined as half the product of the sum of the lengths of its bases and 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 in it the heights $BM=h$ and $KP=h$, as well as the diagonal $BK$ (Fig. 4).

Figure 4.

By Theorem $3$, we get

The theorem has been proven.

Sample task

Example 1

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

Formula for the area of ​​a parallelogram

The area of ​​a parallelogram is equal to the product of its side and the height of that side.

Proof

If the parallelogram is a rectangle, then the equality is satisfied by the theorem on the area of ​​a rectangle. Next, we assume that the angles of the parallelogram are not right.

Let $\angle BAD$ be an acute angle in parallelogram $ABCD$ and $AD > AB$. Otherwise, we will rename the vertices. Then the height $BH$ from the vertex $B$ to the line $AD$ falls on the side $AD$, since the leg $AH$ is shorter than the hypotenuse $AB$, and $AB< AD$. Основание $K$ высоты $CK$ из точки $C$ на прямую $AB$ лежит на продолжении отрезка $AD$ за точку $D$, так как угол $\angle BAD$ острый, а значит $\angle CDA$ тупой. Вследствие параллельности прямых $BA$ и $CD$ $\angle BAH = \angle CDK$. В параллелограмме противоположные стороны равны, следовательно, по стороне и двум углам, треугольники $\triangle ABH = \triangle DCK$ равны.

Let's compare the area of ​​the parallelogram $ABCD$ and the area of ​​the rectangle $HBCK$. The area of ​​a parallelogram is greater by area $\triangle ABH$, but less by area $\triangle DCK$. Since these triangles are equal, their areas are equal. This means that the area of ​​a parallelogram is equal to the area of ​​a rectangle with sides length to side and the height of the parallelogram.

Formula for the area of ​​a parallelogram using sides and sine

The area of ​​a parallelogram is equal to the product of adjacent sides and the sine of the angle between them.

Proof

The height of the parallelogram $ABCD$ dropped onto side $AB$ is equal to the product of the segment $BC$ and the sine of the angle $\angle ABC$. It remains to apply the previous statement.

Formula for the area of ​​a parallelogram using the diagonals

The area of ​​a parallelogram is equal to half the product of the diagonals and the sine of the angle between them.

Proof

Let the diagonals of the parallelogram $ABCD$ intersect at the point $O$ at an angle $\alpha$. Then $AO=OC$ and $BO=OD$ by the parallelogram property. The sines of the angles that add up to $180^\circ$ are equal, $\angle AOB = \angle COD = 180^\circ - \angle BOC = 180^\circ - \angle AOD$. This means that the sines of the angles at the intersection of the diagonals are equal to $\sin \alpha$.

$S_(ABCD)=S_(\triangle AOB) + S_(\triangle BOC) + S_(\triangle COD) + S_(\triangle AOD)$

according to the axiom of area measurement. We apply the triangle area formula $S_(ABC) = \dfrac(1)(2) \cdot AB \cdot BC \sin \angle ABC$ for these triangles and angles when the diagonals intersect. The sides of each are equal to half the diagonals, and the sines are also equal. Therefore, the areas of all four triangles are equal to $S = \dfrac(1)(2) \cdot \dfrac(AC)(2) \cdot \dfrac(BD)(2) \cdot \sin \alpha = \dfrac(AC \ cdot BD)(8) \sin \alpha$. Summing up all of the above, we get

$S_(ABCD) = 4S = 4 \cdot \dfrac(AC \cdot BD)(8) \sin \alpha = \dfrac(AC \cdot BD \cdot \sin \alpha)(2)$

Just as in Euclidean geometry, a point and a straight line are the main elements of the theory of planes, so a 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 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 depicted by a quadrilateral ABCD. The sides are called bases (AB, BC, CD and AD), the perpendicular drawn from any vertex to the side opposite to this vertex is called height (BE and BF), lines AC and BD are called diagonals.

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

Sides and angles: features of the relationship

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

1. The sides that are opposite are identical in pairs.
2. Angles opposite each other are equal in pairs.

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

The segments AB and BC in ∆ABC correspond in pairs to the 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 pairwise identical, then ∠A = ∠C. The property has been proven.

Characteristics of the diagonals of a figure

Main feature of these lines of a parallelogram: the point of intersection divides them in half.

Proof: Let i.e. be the intersection point of diagonals AC and BD of figure ABCD. They form two commensurate triangles - ∆ABE and ∆CDE.

AB=CD since they are opposites. According to the lines and the secant, ∠ABE = ∠CDE and ∠BAE = ∠DCE.

By the second criterion of equality, ∆ABE = ∆CDE. This means that the elements ∆ABE and ∆CDE: AE = CE, BE = DE and at the same time they are proportional parts of AC and BD. The property has been proven.

Features of adjacent corners

Adjacent sides have a sum of angles equal to 180°, since they lie on the same side of parallel lines and a transversal. For quadrilateral ABCD:

∠A+∠B=∠C+∠D=∠A+∠D=∠B+∠C=180º

Properties of the bisector:

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

Determination of the characteristic features of a parallelogram using the theorem

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

Proof: let the lines AC and BD of the quadrilateral ABCD intersect at i.e. Since ∠AED = ∠BEC, and AE+CE=AC BE+DE=BD, then ∆AED = ∆BEC (by the first criterion for the equality of triangles). That is, ∠EAD = ∠ECB. They are also the internal cross angles of the secant AC for lines AD and BC. Thus, by definition of parallelism - AD || B.C. A similar property of lines BC and CD is also derived. The theorem has been proven.

Calculating the area of ​​a figure

Area of ​​this figure found by several methods 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 in size to rectangle EBCF, since they consist of commensurate figures: S ABE and S EBCD, as well as S DCF and S EBCD. It follows from this that the area of ​​this geometric figure is located in the same way as a rectangle:

S ABCD = S EBCF = BE×BC=BE×AD.

For determining general formula The area of ​​the parallelogram is denoted by 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 is the angle between segments a and b.

This method is practically based on the first, but in case it is unknown. always cuts off right triangle, whose parameters are found by trigonometric identities, that is, . Transforming the relation, 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 the angle, which they create when they intersect, you can also find the area.

Proof: AC and BD intersect to 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 by the expression , where a=BE, b=AE, ∠γ =∠AEB. Since , the calculations use a single sine value. 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 - i.e. - construct vectors and . Next, we construct 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 the sides
along the diagonals and the cosine of the angle between them
along diagonals and sides
through the height and the opposite vertex
Finding the length of diagonals
on the sides and the size of the apex 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 distinctive features and ways to calculate its various parameters can be useful at any time in life.