Formulas for finding the area of ​​a parallelogram abcd. Parallelogram area

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.

Decision.

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.

Decision.

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.

Decision.

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.

Decision.

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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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. I.e . 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 quadrangle 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

Geometric area- a numerical characteristic of a geometric figure showing the size of this figure (part of the surface bounded by a closed contour of this figure). The size of the area is expressed by the number of square units contained in it.

Triangle area formulas

1. Triangle area formula for side and height
   Area of ​​a triangle equal to half the product of the length of a side of a triangle and the length of the altitude drawn to this side
   
2. The formula for the area of ​​a triangle given three sides and the radius of the circumscribed circle
   
3. The formula for the area of ​​a triangle given three sides and the radius of an inscribed circle
   Area of ​​a triangle is equal to the product of the half-perimeter of the triangle and the radius of the inscribed circle.
   
where S is the area of ​​the triangle,
- the lengths of the sides of the triangle,
- the height of the triangle,
- the angle between the sides and,
- radius of the inscribed circle,
R - radius of the circumscribed circle,

Square area formulas

1. The formula for the area of ​​a square given the length of a side
   square area is equal to the square of its side length.
2. The formula for the area of ​​a square given the length of the diagonal
   square area equal to half the square of the length of its diagonal.
   

   S=	1	2
   	2

where S is the area of ​​the square,
is the length of the side of the square,
is the length of the diagonal of the square.

Rectangle area formula

Rectangle area is equal to the product of the lengths of its two adjacent sides

where S is the area of ​​the rectangle,
are the lengths of the sides of the rectangle.

Formulas for the area of ​​a parallelogram

1. Parallelogram area formula for side length and height
   Parallelogram area
2. The formula for the area of ​​a parallelogram given two sides and the angle between them
   Parallelogram area is equal to the product of the lengths of its sides multiplied by the sine of the angle between them.
   

   a b sinα

where S is the area of ​​the parallelogram,
are the lengths of the sides of the parallelogram,
is the height of the parallelogram,
is the angle between the sides of the parallelogram.

Formulas for the area of ​​a rhombus

1. Rhombus area formula given side length and height
   Rhombus area is equal to the product of the length of its side and the length of the height lowered to this side.
2. The formula for the area of ​​a rhombus given the length of the side and the angle
   Rhombus area is equal to the product of the square of the length of its side and the sine of the angle between the sides of the rhombus.
3. The formula for the area of ​​a rhombus from the lengths of its diagonals
   Rhombus area is equal to half the product of the lengths of its diagonals.
where S is the area of ​​the rhombus,
- length of the side of the rhombus,
- the length of the height of the rhombus,
- the angle between the sides of the rhombus,
1, 2 - the lengths of the diagonals.

Trapezium area formulas

1. Heron's formula for a trapezoid

   Where S is the area of ​​the trapezoid,
   - the length of the bases of the trapezoid,
   - the length of the sides of the trapezoid,
   

A parallelogram is a quadrangular figure whose opposite sides are pairwise parallel and pairwise equal. Its opposite angles are also equal, and the intersection point of the diagonals of the parallelogram divides them in half, while being the center of symmetry of the figure. Special cases of a parallelogram are such geometric shapes as a square, a rectangle and a rhombus. The area of ​​a parallelogram can be found in various ways, depending on what initial data is accompanied by the formulation of the problem.

1. In the simplest case, the area of ​​a parallelogram is defined as the product of its base and its height.
   

   S = DC ∙ h

   
   

   where S is the area of ​​the parallelogram;
   a - base;
   h is the height drawn to the given base.

   This formula is very easy to understand and remember if you look at the following figure.

   

   As you can see from this image, if we cut off an imaginary triangle to the left of the parallelogram and attach it to the right, then as a result we get a rectangle. And as you know, the area of ​​a rectangle is found by multiplying its length by its height. Only in the case of a parallelogram, the length will be the base, and the height of the rectangle will be the height of the parallelogram lowered to this side.

2. The area of ​​a parallelogram can also be found by multiplying the lengths of two adjacent bases and the sine of the angle between them:
   

   S = AD∙AB∙sinα

   
   

   where AD, AB are adjacent bases that form the intersection point and the angle a between themselves;
   α is the angle between the bases AD and AB.

   

3. Also, the area of ​​a parallelogram can be found by dividing in half the product of the lengths of the diagonals of the parallelogram by the sine of the angle between them.
   

   S = ½∙AC∙BD∙sinβ

   
   

   where AC, BD are the diagonals of the parallelogram;
   β is the angle between the diagonals.

   

4. There is also a formula for finding the area of ​​a parallelogram in terms of the radius of a circle inscribed in it. It is written as follows:

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