The area of ​​a parallelogram if the sides are known. Perimeter and area of ​​a parallelogram

Task 4.

One of the diagonals of a parallelogram of length 4√6 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. Apply the sine theorem to the triangle AOD.

AO/sin D = OD/sin A.

2√6/sin 45 o = OD/sin 60 o.

OD = (2√6sin 60 o) / sin 45 o = (2√6 √3/2) / (√2/2) = 2√18/√2 = 6.

Answer: 12.

Task 5.

For a parallelogram with sides 5√2 and 7√2, the smaller angle between the diagonals is 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 be φ.

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

S ABCD \u003d AB AD sin A \u003d 5√2 7√2 sin f,

S ABCD \u003d 1/2 AC BD sin AOB \u003d 1/2 d 1 d 2 sin f.

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

2 5√2 7√2 = d 1 d 2 ;

2. Using the ratio between the sides and 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 make a system:

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

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 o. 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 \u003d AO 2 + VO 2 2 AO VO cos AOB.

4 2 \u003d (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 the relation for the triangle AOD.

We 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 \u003d 1/2 AC BD sin AOB \u003d 1/2 d 1 d 2 sin α \u003d 1/2 20√2 √2/2 \u003d 10.

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

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 \u003d AB AD sin VAD. Let's do a substitution in the formula.

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

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

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

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

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

BD 2 \u003d AB 2 + AD 2 - 2 AB BD cos BAD.

ВD 2 \u003d 8 2 + 15 2 - 2 8 15 3 / 5 \u003d 145.

Answer: 145.

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Parallelogram It is called a quadrilateral whose opposite sides are parallel to each other. The main tasks at school on this topic are to calculate the area of ​​a parallelogram, its perimeter, height, diagonals. These quantities and formulas for their calculation will be given below.

Parallelogram properties

Opposite sides of a parallelogram and opposite angles are equal to each other:
AB=CD, BC=AD ,

The diagonals of a parallelogram at the point of intersection are divided into two equal parts:

AO=OC, OB=OD.

Angles adjacent to either side (adjacent angles) add up to 180 degrees.

Each of the diagonals of a parallelogram divides it into two triangles of equal area and geometric dimensions.

Another remarkable property that is often used in solving problems is that the sum of the squares of the diagonals in a parallelogram is equal to the sum of the squares of all sides:

AC^2+BD^2=2*(AB^2+BC^2) .

The main features of parallelograms:

1. A quadrilateral whose opposite sides are pairwise parallel is a parallelogram.
2. A quadrilateral with equal opposite sides is a parallelogram.
3. A quadrilateral with equal and parallel opposite sides is a parallelogram.
4. If the diagonals of the quadrilateral at the point of intersection are divided in half, then this is a parallelogram.
5. A quadrilateral whose opposite angles are equal in pairs is a parallelogram

Bisectors of a parallelogram

Bisectors of opposite angles in a parallelogram can be parallel or coincide.

Bisectors of adjacent angles (adjacent to the same side) intersect at right angles (perpendicular).

Parallelogram height

Parallelogram height- this is a segment that is drawn from an angle perpendicular to the base. It follows from this that two heights can be drawn from each angle.

Parallelogram area formula

Parallelogram area is equal to the product of a side and the height drawn to it. The area formula is as follows

The second formula is no less popular in calculations and is defined as follows: the area of ​​\u200b\u200ba parallelogram is equal to the product of adjacent sides by the sine of the angle between them

Based on the above formulas, you will know how to calculate the area of ​​a parallelogram.

Parallelogram perimeter

The formula for calculating the perimeter of a parallelogram is

that is, the perimeter is twice the sum of the sides. Tasks on a parallelogram will be considered in neighboring materials, but for now, study the formulas. Most of the tasks for calculating the sides, diagonals of a parallelogram are quite simple and come down to knowing the sine theorem and the Pythagorean theorem.

Note. This is part of the lesson with problems in geometry (parallelogram section). If you need to solve a problem in geometry, which is not here - write about it in the forum. To denote the action of extracting a square root in solving problems, the symbol √ or sqrt () is used, and the radical expression is indicated in brackets.

Theoretical material

Explanations to the formulas for finding the area of ​​a parallelogram:

1. The area of ​​a parallelogram is equal to the product of the length of one of its sides and the height on that side.
2. The area of ​​a parallelogram is equal to the product of its two adjacent sides and the sine of the angle between them
3. The area of ​​a parallelogram is equal to half the product of its diagonals and the sine of the angle between them

Problems for finding the area of ​​a parallelogram

Solution.
Let's denote the smaller height of the parallelogram ABCD, lowered from the point B to the larger base AD as BK.
Find the value of the leg of a right triangle ABK formed by a smaller height, a smaller side and a part of a larger base. According to the Pythagorean theorem:

AB 2 = BK 2 + AK 2
82 = 9 2 + AK 2
AK 2 = 82 - 81
AK=1

Let us extend the upper base of the parallelogram BC and drop the height AN on it from its lower base. AN = BK as sides of rectangle ANBK. In the resulting right triangle ANC we find the leg NC.
AN 2 + NC 2 = AC 2
9 2 + NC 2 = 15 2
NC 2 = 225 - 81
NC2 = √144
NC = 12

Now let's find the larger base BC of parallelogram ABCD.
BC=NC-NB
We take into account that NB = AK as the sides of the rectangle, then
BC=12 - 1=11

The area of ​​a parallelogram is equal to the product of the base and the height to this base.
S=ah
S=BC * BK
S=11*9=99

Answer: 99 cm2.

A task

Solution.
Let us drop one more perpendicular DK onto the diagonal AC.
Accordingly, triangles AOB and DKC, COB and AKD are pairwise congruent. One of the sides is the opposite side of the parallelogram, one of the angles is a right one, since it is perpendicular to the diagonal, and one of the remaining angles is an internal cross lying for the parallel sides of the parallelogram and the secant of the diagonal.

Thus, the area of ​​the parallelogram is equal to the area of ​​the indicated triangles. That is
Sparall = 2S AOB +2S BOC

The area of ​​a right triangle is half the product of the legs. Where
S \u003d 2 (1/2 8 * 4) + 2 (1/2 6 * 4) \u003d 56 cm 2
Answer: 56 cm2.

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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