What is the area of ​​any parallelogram. How to find the area of ​​a parallelogram

Task 4.

One of the diagonals of the parallelogram, 4√6 long, makes an angle of 60 ° with the base, and the second diagonal makes an angle of 45 ° with the same base. Find the second diagonal.

Solution.

1. AO = 2√6.

2. We apply the theorem of sines to the triangle AOD.

AO / sin D = OD / sin A.

2√6 / sin 45 о = OD / sin 60 о.

ОD = (2√6sin 60 о) / sin 45 о = (2√6 √3 / 2) / (√2 / 2) = 2√18 / √2 = 6.

Answer: 12.

Task 5.

A parallelogram with sides 5√2 and 7√2 has a smaller angle between the diagonals equal to the smaller angle of the parallelogram. Find the sum of the lengths of the diagonals.

Solution.

Let d 1, d 2 be the diagonals of the parallelogram, and the angle between the diagonals and the smaller angle of the parallelogram is equal to φ.

1. Let's count two different
ways of its area.

S ABCD = AB AD sin A = 5√2 7√2 sin φ,

S ABCD = 1/2 AС ВD sin AОВ = 1/2 d 1 d 2 sin ф.

We obtain the equality 5√2 7√2 sin ф = 1 / 2d 1 d 2 sin ф or

2 5√2 7√2 = d 1 d 2;

2. Using the ratio between the sides and the diagonals of the parallelogram, we write the equality

(AB 2 + AD 2) 2 = AC 2 + BD 2.

((5√2) 2 + (7√2) 2) 2 = d 1 2 + d 2 2.

d 1 2 + d 2 2 = 296.

3. Let's compose the system:

(d 1 2 + d 2 2 = 296,
(d 1 + d 2 = 140.

We multiply the second equation of the system by 2 and add it to the first.

We get (d 1 + d 2) 2 = 576. Hence Id 1 + d 2 I = 24.

Since d 1, d 2 are the lengths of the diagonals of the parallelogram, then d 1 + d 2 = 24.

Answer: 24.

Task 6.

The sides of the parallelogram are 4 and 6. The acute angle between the diagonals is 45 °. Find the area of ​​the parallelogram.

Solution.

1. From the triangle AOB, using the cosine theorem, we write the relationship between the side of the parallelogram and the diagonals.

AB 2 = AO 2 + BO 2 2 AO VO cos AOB.

4 2 = (d 1/2) 2 + (d 2/2) 2 - 2 (d 1/2) (d 2/2) cos 45 o;

d 1 2/4 + d 2 2/4 - 2 (d 1/2) (d 2/2) √2 / 2 = 16.

d 1 2 + d 2 2 - d 1 d 2 √2 = 64.

2. Similarly, we write down the relation for the triangle AOD.

Let's take into account that<АОD = 135 о и cos 135 о = -cos 45 о = -√2/2.

We get the equation d 1 2 + d 2 2 + d 1 d 2 √2 = 144.

3. We have a system
(d 1 2 + d 2 2 - d 1 d 2 √2 = 64,
(d 1 2 + d 2 2 + d 1 d 2 √2 = 144.

Subtracting the first from the second equation, we get 2d 1 d 2 √2 = 80 or

d 1 d 2 = 80 / (2√2) = 20√2

4.S ABCD = 1/2 AC · ВD · sin AОВ = 1/2 · d 1 d 2 sin α = 1/2 · 20√2 · √2 / 2 = 10.

Note: In this and in the previous problem, there is no need to completely solve the system, foreseeing that in this problem, to calculate the area, we need the product of diagonals.

Answer: 10.

Task 7.

The area of ​​the parallelogram is 96 and its sides are 8 and 15. Find the square of the smaller diagonal.

Solution.

1.S ABCD = AB · AD · sin BAD. Let's make a substitution in the formula.

We get 96 = 8 15 sin BAD. Hence sin ВAD = 4/5.

2. Find cos BAD. sin 2 BAD + cos 2 BAD = 1.

(4/5) 2 + cos 2 BAD = 1.cos 2 BAD = 9/25.

According to the problem statement, we find the length of the smaller diagonal. The BD diagonal will be smaller if the BAD angle is sharp. Then cos BAD = 3/5.

3. From the triangle ABD by the cosine theorem we find the square of the diagonal BD.

BD 2 = AB 2 + AD 2 - 2 · AB · BD · cos BAD.

ВD 2 = 8 2 + 15 2 - 2 8 15 3/5 = 145.

Answer: 145.

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Before we know how to find the area of ​​a parallelogram, we need to remember what a parallelogram is and what is called its height. Parallelogram - a quadrangle, the opposite sides of which are pairwise parallel (lie on parallel lines). The perpendicular drawn from an arbitrary point on the opposite side to the straight line containing this side is called the height of the parallelogram.

Square, rectangle and rhombus are special cases of the parallelogram.

The parallelogram area is indicated as (S).

Formulas for finding the area of ​​a parallelogram

S = a * h, where a is the base, h is the height that is drawn to the base.

S = a * b * sinα, where a and b are bases and α is the angle between bases a and b.

S = p * r, where p is a semi-perimeter, r is the radius of the circle that is inscribed in the parallelogram.

The area of ​​the parallelogram, which is formed by the vectors a and b, is equal to the modulus of the product of the given vectors, namely:

Consider example # 1: Given a parallelogram, the side of which is 7 cm, and the height is 3 cm. How to find the area of ​​a parallelogram, we need a formula for the solution.

So S = 7x3. S = 21. Answer: 21 cm 2.

Consider example # 2: Given bases are 6 and 7 cm, and the angle between the bases is 60 degrees. How to find the area of ​​a parallelogram? Formula used to solve:

Thus, we first find the sine of the angle. Sine 60 = 0.5, respectively S = 6 * 7 * 0.5 = 21 Answer: 21 cm 2.

I hope that these examples will help you in solving problems. And remember, the main thing is knowledge of formulas and attentiveness.

Parallelogram is a geometric figure that is often found in the tasks of the geometry course (section planimetry). The key features of this quadrilateral are the equality of opposite angles and the presence of two pairs of parallel opposite sides. Special cases of a parallelogram are rhombus, rectangle, square.

The calculation of the area of ​​this type of polygon can be done in several ways. Let's consider each of them.

Find the area of ​​a parallelogram if the side and height are known

To calculate the area of ​​a parallelogram, you can use the values ​​of its side, as well as the length of the height dropped on it. In this case, the data obtained will be reliable both for the case of the known side - the base of the figure, and if you have the side of the figure at your disposal. In this case, the desired value will be obtained by the formula:

S = a * h (a) = b * h (b),

• S is the area to be determined
• a, b - known (or calculated) side,
• h is the height dropped on it.

Example: the value of the base of a parallelogram is 7 cm, the length of the perpendicular dropped on it from the opposite vertex is 3 cm.

Solution: S = a * h (a) = 7 * 3 = 21.

Find the area of ​​a parallelogram if 2 sides and the angle between them are known

Consider the case when you know the values ​​of the two sides of the figure, as well as the degree measure of the angle that they form with each other. The provided data can also be used to find the area of ​​a parallelogram. In this case, the formula expression will look like this:

S = a * c * sinα = a * c * sinβ,

• a - lateral side,
• c - known (or calculated) base,
• α, β - angles between sides a and c.

Example: the base of a parallelogram is 10 cm, its side is 4 cm less. The obtuse angle of the figure is 135 °.

Solution: determine the value of the second side: 10 - 4 = 6 cm.

S = a * c * sinα = 10 * 6 * sin135 ° = 60 * sin (90 ° + 45 °) = 60 * cos45 ° = 60 * √2 / 2 = 30√2.

Find the area of ​​a parallelogram if the diagonals and the angle between them are known

The presence of the known values ​​of the diagonals of a given polygon, as well as the angle that they form as a result of their intersection, makes it possible to determine the size of the area of ​​the figure.

S = (d1 * d2) / 2 * sinγ,
S = (d1 * d2) / 2 * sinφ,

S is the area to be determined,
d1, d2 - known (or calculated) diagonals,
γ, φ are the angles between the diagonals d1 and d2.

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

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

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

where:
a - side of the parallelogram,
h a - the height drawn to this side.

2. If the lengths of two adjacent sides of the 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 the 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) \]

Parallelogram properties

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

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

The parallelogram diagonals at the intersection point are halved \ (AO = OC \), \ (BO = OD \)

The diagonal of the 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 the 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 the angles adjacent to one side of the parallelogram are mutually perpendicular.

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

Parallelogram signs

A quadrilateral is 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 \)

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

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Defining a parallelogram

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

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

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

Sides and corners: ratio features

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

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

Proof: Consider ∆ABC and ∆ADC, which are obtained by dividing the quadrilateral ABCD by the 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 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 their identity: AB = CD, BC = AD. So ∠B corresponds to ∠D and they are equal. Since ∠A = ∠BAC + ∠CAD, ∠C = ∠BCA + ∠ACD, which are also pairwise the same, then ∠A = ∠C. The property is proven.

Characteristics of the diagonals of the figure

The main feature these parallelogram lines: the point of intersection divides them in half.

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

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

According to 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 is proven.

Features of adjacent corners

Adjacent sides have 180 ° sum of angles as they lie on the same side of parallel lines and a 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.

Determination of the characteristic features of a parallelogram by the theorem

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

Proof: let in point E the lines AC and BD of the quadrangle ABCD intersect. 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 internal cross-sectional angles AC for lines AD and BC. Thus, by definition of parallelism - AD || BC. A similar property of lines BC and CD is also displayed. The theorem is proved.

Calculating the area of ​​a shape

The area of ​​this figure is 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 the rectangle EBCF, since they also 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 found in the same way as the 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 is b... Respectively:

Other ways to find the area

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

,

Sпр-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 a right-angled triangle, the parameters of which are found by trigonometric identities, that is. Transforming the relation, we obtain. In the equation of the first method, we replace the height with this product and we obtain a proof of the validity of this formula.

Through the parallelogram diagonals and the angle, which they create when crossing, 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 quadrangle.

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

.

Applications 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 the given vectorsandnotcollinear, 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. - we build vectors and. Next, we build a parallelogram OACB, where the segments OA and OB are sides. Thus, the OS lies on a vector or sum.

Formulas for calculating parallelogram parameters

Identities are given under the following conditions:

1. a and b, α - sides and 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 parties
along the diagonals and the cosine of the angle between them
diagonally and side
through the height and the opposite vertex
Finding the length of the diagonals
along the sides and the size of the tops between them
on the sides and one of the diagonals	 Conclusion The parallelogram, as one of the key figures of geometry, finds application in life, for example, in construction when calculating the area of ​​a plot or other measurements. Therefore, knowledge about the distinctive features and methods of calculating its various parameters can be useful at any time in life.